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Shane, Well, I've cooled off a bit since I wrote that last and I'll have to apologize if I sounded miffed, but, well, darn it, I was. I hate seeing something I really enjoy "improved" so much it becomes unusable and it seems to me, at least, that Mint Fluxbox might be heading that way. Please - please for gosh sake - take into account when you create your distros that you're not just making them for the new kids, but an established and VERY APPRECIATIVE user base, too. Honestly, the Fluxbox Mint distro was basically similar to what I would have rolled out myself and was actually working toward when it came out, so I owe you a great deal of thanks for that if nothing else:). Recently I decided to ditch my dual Win/Linux and go mount my /home on my other drive with Mint as the main and that's why I thought I'd try the new version. Not to harp a point, but you must have realized when you started this that the whole reason for using Fluxbox as the WM was for it's speed, utility and elegance - so why detract from that by burying what it does best amidst glamorous effects? I'm not so sure that "looking good" is a feature the same way that speed, utility and configurability are. You might want to reconsider that one. But to each their own and I thank you again for at least allowing us "old fuddies" the option of turning it off and getting it out of the way. Sadly, I have no Python skills to speak of. But I'll hold out with Elyssa until Gloria comes out and I can give that a try. Again, thanks and I didn't mean to stomp on toes - just kind of stomp in general.

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\begin{document} \title{Verbal quandles with one parameter} \author{Elizaveta Markhinina, Timur Nasybullov} \maketitle \begin{abstract} We find all words $W(x,y,z)$ in the free group $F(x,y,z)$, such that for every group $G$ and an element $c\in G$ the algebraic system $(G,*_{W,c})$ with the binary operation $*_{W,c}$ given by $a*_{W,c}b=W(a,b,c)$ for $a,b\in G$ is a quandle. Such quandles are called verbal quandles with one parameter. ~\\ \emph{Keywords:} quandle, verbal quandle, group. \end{abstract} \section{Introduction and preliminaries} A quandle is an algebraic system with a binary operation whose three axioms encode the Reidemeister moves on knot diagrams. More formally, a quandle $Q$ is an algebraic system with one binary algebraic operation $(x,y)\mapsto x*y$ which satisfies the following three axioms: \begin{itemize} \item[(q1)] $x*x=x$ for all $x\in Q$, \item[(q2)] the map $S_x:y\mapsto y*x$ is a bijection of $Q$ for all $x\in Q$, \item[(q3)] $(x*{y})*z=(x*z)*({y*z})$ for all $x,y,z\in Q$. \end{itemize} First examples of quandles date back to 1940s when Takasaki defined keys, objects which were later known as involutory quandles \cite{Tak}. The concept of quandles was explicitly presented in the independent works of Joyce \cite{Joy} and Matveev \cite{Mat}. To each oriented diagram $D_K$ of an oriented knot $K$ in $\mathbb{R}^3$ they associate the quandle $Q(K)$ which does not change if we apply the Reidemeister moves to the diagram $D_K$. Joyce and Matveev proved that two knot quandles $Q(K_1)$ and $Q(K_2)$ are isomorphic if and only if there is a (probably, reversing orientation) homeomorphism of the ambient space $\mathbb{R}^3$ which maps $K_1$ to $K_2$. The knot quandle is a very strong invariant for knots in $\mathbb{R}^3$, however, usually it is very difficult to determine if two knot quandles are isomorphic. In \cite{brimi} it is shown that the isomorphism problem for quandles is, from the perspective of Borel reducibility, fundamentally difficult (Borel complete). Sometimes homomorphisms from knot quandles to simpler quandles provide useful information that helps determine whether two knot quandles are isomorphic. This potential utility leads to the necessity of constructing quandles which are convenient to work with. Another area where quandles find applications is the theory of the set-theoretical Yang-Baxter equation. The Yang-Baxter equation first appeared in theoretical physics and statistical mechanics in the works of Yang \cite{Yan} and Baxter \cite{Bax1,Bax2}. The notion of the set-theoretical Yang-Baxter equation, was introduced by V.~Drinfel'd in the context of quantum groups (see \cite{Dri}). Recall that a set-theoretic solution of the Yang-Baxter equation is a pair $(X, r)$, where $X$ is a set and $r:X\times X\to X\times X$ is a bijective map such that $$(r\times id)(id \times r)(r\times id)=(id\times r)(r \times id)(id\times r).$$ If $Q$ is a quandle with the operation $*$, then the pair $(Q,r)$, where $r:Q\times Q\to Q\times Q$ is the map given by \begin{equation}\label{ybeonquandle} r(x,y)=(y*x,x) \end{equation} for $x,y\in Q$ is a set-theoretic solution of the Yang-Baxter equation. In the recent papers \cite{BarNas1,BarNas3} there was constructed a general method how a given solution of the set-theoretical Yang-Baxter equation on an arbitrary algebraic system $X$ can be used for constructing a representation of the virtual braid group $VB_n$ by automorphisms of the algebraic system $X$. In the papers \cite{byaka2,byaka1} these this method was used for constructing new representations of vertual braid groups. In the papers \cite{BarNas2, BarNas3} there was introduced a method how a given solution of the set-theoretical Yang-Baxter equation on an algebraic system $X$ can be used for constructing an invariant for virtual knots and links which is an algebraic system from the same category as $X$. If $X$ is a quandle, and the solution of the set-theoretical Yang-Baxter equation on $X$ is given by (\ref{ybeonquandle}), then using procedures described in \cite{BarNas1,BarNas2,BarNas3} it is possible to construct a representation $VB_n\to {\rm Aut}(X)$, and a quandle invariant for virtual links. However, in order to apply procedures described in \cite{BarNas1,BarNas2,BarNas3} to $X$ and $r$, the quandle $X$ must be cinvenient to work with, hence it is necessary to construct such quandles. Quandles were also investigated from an algebraic point of view and relations to other algebraic structures such as Lie algebras \cite{CarCra2}, Frobenius algebras and Yang-Baxter equation \cite{CarCra2}, Hopf algebras \cite{AndGra}, quasigroups and Moufang loops \cite{Moh1}, representation theory \cite{Moh2} and ring theory \cite{BarPas}. In all of these areas it is always useful to have examples of quandles which are convenient to work with. A lot of examples of quandles which are convenient to work with come from groups. The most common example of a quandle is the conjugation quandle ${\rm Conj}(G)$ of a group $G$, i.~e. the quandle $(G,*)$ with $x*y=y^{-1}xy$ for $x,y\in G$. If we define another operation $*$ on the group $G$, namely $x*y=yx^{-1}y$, then the set $G$ with this operation also forms a quandle. This quandle is called the core quandle of a group $G$ and is denoted by ${\rm Core}(G)$. In particular, if $G$ is an abelian group, then the quandle ${\rm Core}(G)$ is called the Takasaki quandle of the abelian group $G$ and is denoted by $T(G)$. Such quandles were studied by Takasaki in \cite{Tak}. Other examples of quandles arising from groups can be found, for example, in \cite{BarDeySin, BarNasSin, BarNas6}. The so-called verbal quandlós were introduced in the paper \cite{BarNas4}. Let $w = w(x,y)$ be an element of the free group $F(x,y)$ with two generators. If $G$ is a group, then $w$ defines a map $w:G\times G\to G$ which maps a pair $(a,b)$ to $w(a,b)$. For $a,b\in G$ denote by $a *_w b = w(a, b)$. We can think about~$*_w$ as about new binary operation on the group $G$, so, $(G,*_w)$ is an algebraic system. If the algebraic system $(G,*_w)$ is a quandle, then we call this quandle a verbal quandle defined by the word~$w$. It is clear that ${\rm Conj}(G)$ and ${\rm Core}(G)$ are verbal quandles. The following theorem which classifies verbal quandles is proved in \cite{BarNas4}. \begin{ttt}\label{verbrack} Let $w\in F_2$ be an element from the free group. If $Q = (G, *_{w})$ is a quandle for every group $G$, then one of the following conditions holds \begin{enumerate} \item $w(x,y)=yx^{-1}y$, \item $w(x,y)=y^{-s}xy^{s}$ for $s\in\mathbb{Z}$. \end{enumerate} \end{ttt} In the present paper we introduce the notion of a verbal quandle with $n$ parameters which generalizes the notion of a verbal quandle. Let $W = W(x,y, z_1, z_2,\dots, z_{n})$ be a word from the free group with $(n+2)$ free generators. For a group $G$ and fixed elements $c_1,c_2,\dots,c_n\in G$ denote by $*_{W,c_1,c_2,\dots,c_n}$ the binary operation on $G$ defined by the rule $$a*_{W,c_1,c_2,\dots,c_n}b=W(a,b,c_1,c_2,\dots,c_n).$$ If $(G,*_{W,c_1,c_2,\dots,c_n})$ is a quandle, then this quandles is called a verbal quandle with $n$ parameters. If $n=0$, then this quandle is just a verbal quandle. The main result of the present paper is the following theorem which classifies verbal quandles with one parameter. \begin{ttt}\label{th22233} Let $W=W(x,y,z)$ be an element from the free group $F(x,y,z)$. If $Q = (G, *_{W,c})$ is a quandle for every group $G$ and an element $c\in G$, then one of the following conditions holds \begin{enumerate} \item $W(x,y,z)=yx^{-1}y$, \item $W(x,y,z)=y^{-s}xy^{s}$ for $s\in\mathbb{Z}$, \item $W(x,y,z)=yz^{-s}y^{-1}xz^{s}$ for $s\in \mathbb{Z}$, \item $W(x,y,z)=yz^{-s}xy^{-1}z^s$ for $s\in\mathbb{Z}$, \item $W(x,y,z)=z^{-s}y^{-1}xz^sy$ for $s\in\mathbb{Z}$, \item $W(x,y,z)=z^{-s}xy^{-1}z^sy$ for $s\in\mathbb{Z}$. \end{enumerate} \end{ttt} It is clear that Theorem~\ref{verbrack} follows from Theorem~\ref{th22233}. In the consequent paper we are going to apply solutions of the set-theoretic Yang-Baxter equation constructed by verbal quandles with one parameter using formula (\ref{ybeonquandle}) in order to construct new group invariants for virtual knots and links using constructions from \cite{BarNas1,BarNas2,BarNas3}. ~\\ \noindent\textbf{Acknowledgment.} The work is supported by Mathematical Center in Akademgorodok under agreement 075-15-2022-281 with the Ministry of Science and Higher Education of the Russian Federation. \section{Proof of the main theorem} Let $W= x^{\alpha_1} y^{\beta_1} z^{\gamma_1}x^{\alpha_2} y^{\beta_2} z^{\gamma_2}\ldots x^{\alpha_k} y^{\beta_k}z^{\gamma_k}$ for $\alpha_i, \beta_i,\gamma_i \in \mathbb{Z}$ be a reduced word in the free group $F(x,y,z)$. Since $(G,*_{W,c})$ is a quandle, from the axiom (q2) of a quandle follows that for every $a, b \in G$ there exists an element $d \in G$ such that \begin{equation}\label{start}d *_{W,c} a =W(d,a,c)=d^{\alpha_1} a^{\beta_1} c^{\gamma_1}d^{\alpha_2} a^{\beta_2} c^{\gamma_2}\ldots d^{\alpha_k} a^{\beta_k}c^{\gamma_k}= b. \end{equation} It meant that we can express $d$ from equation (\ref{start}) via $a,b,c$. It is possible if and only if $W = u(y,z) x^{\varepsilon} w(y,z)$ for $u(y,z),w(y,z)\in F(y,z)$, $\varepsilon \in \{\pm 1 \}$. From the axiom (q1) it follows that for every element $a\in G$ the equality $a*_{W,c}a=a$ holds. This equality can be rewritten in details in the following form $$a=a*_{W,c}a=W(a,a,c)=u(a,c)a^{\varepsilon}w(a,c),$$ and therefore $u(a,c)=a(w(a,c))^{-1}a^{-\varepsilon}$ for every group $G$, fixed element $c\in G$ and element $a\in G$. Hence we have $u(y,z)=y(w(y,z))^{-1}y^{-\varepsilon}$ and \begin{equation}\label{generalequation} W(x,y,z)=y(w(y,z))^{-1}y^{-\varepsilon}x^{\varepsilon}w(y,z). \end{equation} Let us understand what are the possible options for the word $w(y,z)\in F(y,z)$. In order to do this express the word $w(y,z)$ as a reduced word $$w = y^{q_1} z^{s_1}y^{q_2}z^{s_1} \ldots y^{q_n} y^{s_n},$$ where $q_i, s_i \in \mathbb{Z}$ are non-zero integers with possible exceptions for $q_1$ and $s_n$. Denote by $|w(y,z)|$ the syllabic length of the word $w(y,z)$. Depending on the word $w(y,z)$ and the parameter $\varepsilon\in\{\pm1\}$ we consider the cases collected in the following table. \begin{center} \begin{tabular}[t]{|l|l|l|l|} \hline $|w|$ & $w\in F(y,z)$ & $\varepsilon\in\{\pm1\}$& $q_i,s_i\in\mathbb{Z}$ \\ \hline 1: $|w|=0$ & $1$ & & \\\hline 2: $|w|=1$ & 2.1: $y^q$& & \\\cline{2-4} & 2.2: $z^s$& 2.2.1: $\varepsilon=1$& \\\cline{3-4} & & 2.2.2: $\varepsilon=-1$& \\\hline 3: $|w|=2$ & 3.1: $y^qz^s$ & 3.1.1: $\varepsilon=1$ &3.1.1.1: $q=-1$ \\\cline{4-4} & & &3.1.1.2: $q\neq-1$ \\\cline{3-4} & & 3.1.2: $\varepsilon=-1$ &3.1.2.1: $q=1$ \\\cline{4-4} & & &3.1.2.2: $q\neq1$ \\\cline{2-4} & 3.2: $z^sy^q$ & 3.2.1: $\varepsilon=1$ &3.2.1.1: $q=1$ \\\cline{4-4} & & &3.2.1.2: $q\neq1$ \\\cline{3-4} & & 3.2.2: $\varepsilon=-1$ & \\\hline 4: $|w|=3$ & 4.1: $y^{q_1}z^sy^{q_2}$ & 4.1.1: $\varepsilon=1$&4.1.1.1: $q_2=1$, $q_1=-1$ \\\cline{4-4} && &4.1.1.2: $q_2=1$, $q_1\neq-1$ \\\cline{4-4} && &4.1.1.3: $q_2\neq1$ \\\cline{3-4} && 4.1.2: $\varepsilon=-1$&4.1.2.1: $q_2=1$ \\\cline{4-4} && &4.1.2.2: $q_2\neq1$ \\\cline{2-4} & 4.2: $z^{s_1}y^qz^{s_2}$ & &4.2.1: $|q|\neq 1$ \\\cline{4-4} & & &4.2.2: $|q|=1$ \\\hline 5: $|w|\geq 4$ & & &\\\hline \end{tabular} \end{center} From this table it is clear that the cases we are going to consider are all possible cases (the empty fields mean that the corresponding parameters can be arbitrary). We are going to prove that for the word $W(x,y,z)$ given by formula (\ref{generalequation}) the algebraic system $Q = (G, *_{W,c})$ is a quandle for every group $G$ and an element $c\in G$ if and only if the word $w(y,z)$ in formula (\ref{generalequation}) is given by one of the cases 1, 2.1, 2.2.1, 3.1.1.1, 3.2.1.1, 4.1.1.1. These cases give six possibilities for the word $W(x,y,z)$ described in the formulation of the theorem. \textbf{Case 1:} $|w(y,z)|=0$. In this case $w(y,z)=1$, and the word $$W(x,y,z)=y^{1-\varepsilon}x^{\varepsilon}$$ doesn't depend on $z$, hence for every group $G$ we have $(G,*_{W,c})=(G,*_{W})$. From Theorem~\ref{verbrack} it follows that we have either $W(x,y,z)=yx^{-1}y$, or $W(x,y,z)=y^{-s}xy^{s}$ for $s\in \mathbb{Z}$. It means that in Case 1 we have a quandle. \textbf{Case 2:} $|w(y,z)|=1$. In this case we have either $w(y,z)=y^q$ for non-zero $q\in\mathbb{Z}$, or $w(y,z)=z^s$ for non-zero $s\in\mathbb{Z}$. If $w(y,z)=y^q$ for non-zero $q\in\mathbb{Z}$ (case 2.1), then the word $$W(x,y,z)=y^{1-q-\varepsilon}x^{\varepsilon}y^q$$ doesn't depend on $z$, hence for every group $G$ we have $(G,*_{W,c})=(G,*_{W})$. From Theorem~\ref{verbrack} it follows that we have either $W(x,y,z)=yx^{-1}y$, or $W(x,y,z)=y^{-s}xy^{s}$ for $s\in \mathbb{Z}$. It means that in Case 2.1 we have a quandle. If $w(y,z)=z^s$ for non-zero $s\in\mathbb{Z}$ (case 2.2), then $W(x,y,z)=yz^{-s}y^{-\varepsilon}x^{\varepsilon}z^s$. Let us understand when the operation $*_{W,c}$ satisfies the third quandle axiom (q3) for each group $G$ and an element $c\in G$ \begin{align} \notag(a*_{W,c}b)*_{W,c}d&=(W(a,b,c))*_{W,c}d\\ \notag&=(bc^{-s}b^{-\varepsilon}a^{\varepsilon}c^s)*_{W,c}d\\ \notag&=W(bc^{-s}b^{-\varepsilon}a^{\varepsilon}c^s,d,c)\\ \label{left1}&=dc^{-s}d^{-\varepsilon}(bc^{-s}b^{-\varepsilon}a^{\varepsilon}c^s)^{\varepsilon}c^s,\\ \notag (a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=(W(a,d,c))*_{W,c}(W(b,d,c))\\ \notag&=(dc^{-s}d^{-\varepsilon}a^{\varepsilon}c^s)*_{W,c}(dc^{-s}d^{-\varepsilon}b^{\varepsilon}c^s)\\ \notag&=W(dc^{-s}d^{-\varepsilon}a^{\varepsilon}c^s,dc^{-s}d^{-\varepsilon}b^{\varepsilon}c^s,c)\\ \label{right1}&=(dc^{-s}d^{-\varepsilon}b^{\varepsilon}c^s)c^{-s}(dc^{-s}d^{-\varepsilon}b^{\varepsilon}c^s)^{-\varepsilon}(dc^{-s}d^{-\varepsilon}a^{\varepsilon}c^s)^{\varepsilon}c^s. \end{align} If $\varepsilon=1$ (case 2.2.1), then from formulas (\ref{left1}), (\ref{right1}) it follows that \begin{align*} (a*_{W,c}b)*_{W,c}d&=dc^{-s}d^{-1}bc^{-s}b^{-1}ac^{2s},\\ (a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=dc^{-s}d^{-1}bc^sc^{-s}c^{-s}b^{-1}dc^sd^{-1}dc^{-s}d^{-1}ac^sc^s\\ &=dc^{-s}d^{-1}bc^{-s}b^{-1}ac^{2s}, \end{align*} i.~e. the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds independently on $s$. It means that in Case 2.2.1 we have a quandle. If $\varepsilon=-1$ (case 2.2.2), then from formulas (\ref{left1}), (\ref{right1}) it follows that \begin{align} \label{left2} (a*_{W,c}b)*_{W,c}d&=dc^{-s}dc^{-s}ab^{-1}c^sb^{-1}c^s,\\ \notag (a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=dc^{-s}db^{-1}c^sc^{-s}dc^{-s}db^{-1}c^sc^{-s}ad^{-1}c^sd^{-1}c^s\\ \label{right2} &=dc^{-s}db^{-1}dc^{-s}db^{-1}ad^{-1}c^sd^{-1}c^s. \end{align} Let us look to $a,b,c,d$ as to the generators of the free group $F(a,b,c,d)$ on four generators. In this situation from equalities (\ref{left2}), (\ref{right2}) it is clear that $$(a*_{W,c}b)*_{W,c}d\neq (a*_{W,c}d)*_{W,c}(b*_{W,c}d)$$ independently on $s$, i.~e. there exists a group $G=F(a,b,c,d)$ and an element $c\in G$ such that in $(G,*_{W,c})$ the third quandle axiom (q3) doen't hold, i.~e. $(G,*_{W,c})$ is not a quandle. It means that in Case 2.2.2 we don't have a quandle. \textbf{Case 3:} $|w(y,z)|=2$. In this case $w(y,z)$ is either equal to $w(y,z)=y^qz^s$ or to $w(y,z)=z^sy^q$ for non-zero integers $q,s\in\mathbb{Z}$. If $w(y,z)=y^qz^s$ (case 3.1), then $W(x,y,z)=yz^{-s}y^{-q-\varepsilon}x^{\varepsilon}y^qz^s$. Let us understand when the operation $*_{W,c}$ satisfies the third quandle axiom (q3) for each group $G$ and an element $c\in G$ \begin{align} \notag(a*_{W,c}b)*_{W,c}d&=(W(a,b,c))*_{W,c}d\\ \notag&=(bc^{-s}b^{-q-\varepsilon}a^{\varepsilon}b^qc^s)*_{W,c}d\\ \notag&=W(bc^{-s}b^{-q-\varepsilon}a^{\varepsilon}b^qc^s,d,c)\\ \label{left3}&=dc^{-s}d^{-q-\varepsilon}(bc^{-s}b^{-q-\varepsilon}a^{\varepsilon}b^qc^s)^{\varepsilon}d^qc^s,\\ \notag (a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=(W(a,d,c))*_{W,c}(W(b,d,c))\\ \notag&=(dc^{-s}d^{-q-\varepsilon}a^{\varepsilon}d^qc^s)*_{W,c}(dc^{-s}d^{-q-\varepsilon}b^{\varepsilon}d^qc^s)\\ \notag&=W(dc^{-s}d^{-q-\varepsilon}a^{\varepsilon}d^qc^s,dc^{-s}d^{-q-\varepsilon}b^{\varepsilon}d^qc^s,c)\\ \notag&=(dc^{-s}d^{-q-\varepsilon}b^{\varepsilon}d^qc^s)c^{-s}(dc^{-s}d^{-q-\varepsilon}b^{\varepsilon}d^qc^s)^{-q-\varepsilon}\\ \label{right3}&~~~~\cdot(dc^{-s}d^{-q-\varepsilon}a^{\varepsilon}d^qc^s)^{\varepsilon}(dc^{-s}d^{-q-\varepsilon}b^{\varepsilon}d^qc^s)^qc^s. \end{align} Let us consider the case when $\varepsilon=1$ (case 3.1.1). From equalities (\ref{left3}), (\ref{right3}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{equation}\label{equality2} c^{-s}b^{-q-1}ab^qc^sd^q=d^q(dc^{-s}d^{-q-1}bd^qc^s)^{-q-1}(dc^{-s}d^{-q-1}ad^qc^s)(dc^{-s}d^{-q-1}bd^qc^s)^q. \end{equation} If $q=-1$ (case 3.1.1.1), then equality (\ref{equality2}) clearly holds, i.~e. for $W(x,y,z)=yz^{-s}xy^{-1}z^s$ the algebraic system $(G,*_{W,c})$ is a quandle for every group $G$ and an element $c\in G$. It means that in Case 3.1.1.1 we have a quandle. If $q\neq 1$ (case 3.1.1.2), i.~e. $q<-1$ or $q>0$, then look to $a,b,c,d$ as to the generators of the free group $F(a,b,c,d)$. The reduced word on the left side of equality (\ref{equality2}) begins with $c^{-s}$, while the word on the right side of equality (\ref{equality2}) begins either with $d^{q+1}$ (if $q<-1$) or with $d^q$ (of $q>0$), hence equality (\ref{equality2}) doesn't hold. It means that in Case 3.1.1.2 we don't have a quandle. Let us consider the case when $\varepsilon=-1$ (case 3.1.2). From equalities (\ref{left3}), (\ref{right3}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{multline}\label{equality22} c^{-s}b^{-q}ab^{q-1}c^sb^{-1}d^q=\\ =b^{-1}d^q(dc^{-s}d^{-q+1}b^{-1}d^qc^s)^{-q+1}c^{-s}d^{-q}ad^{q-1}c^sd^{-1}(dc^{-s}d^{-q+1}b^{-1}d^qc^s)^q. \end{multline} If $q=1$ (case 3.1.2.1), then equality (\ref{equality22}) can be rewritten in the following form $$ c^{-s}b^{-1}ac^sb^{-1}d= b^{-1}dc^{-s}d^{-1}ab^{-1}dc^s. $$ Looking to $a,b,c,d$ as to the generators of the free group it is clear that the last equality doesn't hold for all non-zero integers $s$, hence, if $q=1$, then there exists a group $G$ and an element $c\in G$ such that $(G,*_{W,c})$ is not a quandle. It means that in Case 3.1.2.1 we don't have a quandle. If $q\neq 1$, i.~e. $q<0$ or $q>1$ then look to $a,b,c,d$ as to the generators of the free group. The reduced word on the left side of equality (\ref{equality22}) begins with $c^{-s}$, while the word on the right side of equality (\ref{equality22}) begins with $b^{-1}$, hence, equality (\ref{equality22}) doesn't hold. So, in this case there exists a group $G$ and an element $c\in G$ such that $(G,*_{W,c})$ is not a quandle. It means that in Case 3.1.2.2 we don't have a quandle. Let us now consider the case when $|w(y,z)|=2$ and $w(y,z)=z^sy^q$ for non-zero integers $q,s\in\mathbb{Z}$ (case 3.2). In this case the word $W(x,y,z)$ has the following form $$W(x,y,z)=y^{1-q}z^{-s}y^{-\varepsilon}x^{\varepsilon}z^sy^q.$$ Let us understand when the operation $*_{W,c}$ satisfies the third quandle axiom (q3) for each group $G$ and an element $c\in G$ \begin{align} \notag(a*_{W,c}b)*_{W,c}d&=(W(a,b,c))*_{W,c}d\\ \notag&=(b^{1-q}c^{-s}b^{-\varepsilon}a^{\varepsilon}c^sb^q)*_{W,c}d\\ \notag&=W(b^{1-q}c^{-s}b^{-\varepsilon}a^{\varepsilon}c^sb^q,d,c)\\ \label{left33}&=d^{1-q}c^{-s}d^{-\varepsilon}\left(b^{1-q}c^{-s}b^{-\varepsilon}a^{\varepsilon}c^sb^q\right)^{\varepsilon}c^sd^q,\\ \notag (a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=(W(a,d,c))*_{W,c}(W(b,d,c))\\ \notag&=(d^{1-q}c^{-s}d^{-\varepsilon}a^{\varepsilon}c^sd^q)*_{W,c}(d^{1-q}c^{-s}d^{-\varepsilon}b^{\varepsilon}c^sd^q)\\ \notag&=W(d^{1-q}c^{-s}d^{-\varepsilon}a^{\varepsilon}c^sd^q,d^{1-q}c^{-s}d^{-\varepsilon}b^{\varepsilon}c^sd^q,c)\\ \notag &=\left(d^{1-q}c^{-s}d^{-\varepsilon}b^{\varepsilon}c^sd^q\right)^{1-q}c^{-s}\left(d^{1-q}c^{-s}d^{-\varepsilon}b^{\varepsilon}c^sd^q\right)^{-\varepsilon}\\ \label{right33}&~~~~\cdot\left(d^{1-q}c^{-s}d^{-\varepsilon}a^{\varepsilon}c^sd^q\right)^{\varepsilon}c^s\left(d^{1-q}c^{-s}d^{-\varepsilon}b^{\varepsilon}c^sd^q\right)^q \end{align} Consider the case when $\varepsilon=1$ (case 3.2.1). From equalities (\ref{left33}), (\ref{right33}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{multline}\label{longerthing} d^{1-q}c^{-s}d^{-1}b^{1-q}c^{-s}b^{-1}ac^sb^qc^sd^q=\\ =\left(d^{1-q}c^{-s}d^{-1}bc^sd^q\right)^{1-q}c^{-s}d^{-q}c^{-s}b^{-1}ac^sd^qc^s\left(d^{1-q}c^{-s}d^{-}bc^sd^q\right)^q. \end{multline} If $q=1$ (case 3.2.1.1), then equality (\ref{longerthing}) clearly holds independently on $s$. It means that in Case 3.2.1.1 we don't have a quandle. If $q\neq 1$ (case 3.2.1.2), i.~e. $q<0$ or $q>1$, then look to $a,b,c,d$ as to the generators of the free group. If $q<0$, then the reduced word on the left side of equality (\ref{longerthing}) ends with $d^{q}$, while the word on the right side of equality (\ref{longerthing}) ends with $d^{q-1}$, hence, equality (\ref{longerthing}) doesn't hold. If $q>1$, then the reduced word on the left side of equality (\ref{longerthing}) begins with $d^{1-q}$, while the word on the right side of equality (\ref{longerthing}) begins with $d^{-q}$, hence, equality (\ref{longerthing}) doesn't hold. It means that in Case 3.2.1.2 we don't have a quandle. Let us consider the case when $\varepsilon=-1$ (case 3.2.2). From equalities (\ref{left33}), (\ref{right33}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{multline}\label{anotherlong} d^{1-q}c^{-s}db^{-q}c^{-s}ab^{-1}c^sb^{q-1}c^sd^q=\\ =\left(d^{1-q}c^{-s}db^{-1}c^sd^q\right)^{1-q}c^{-s}d^{1-q}c^{-s}db^{-1} ad^{-1}c^sd^{q-1}c^s\left(d^{1-q}c^{-s}db^{-1}c^sd^q\right)^q. \end{multline} Look to $a,b,c,d$ as to the generators of the free group. If $q<0$, then the reduced word on the left side of equality (\ref{anotherlong}) ends with $d^{q}$, while the word on the right side of equality (\ref{anotherlong}) ends with $d^{q-1}$, hence, equality (\ref{anotherlong}) doesn't hold. If $q=1$, then using direct calculations it is clear that equality (\ref{anotherlong}) doesn't hold. If $q>1$, then the reduced word on the left side of equality (\ref{anotherlong}) begins with $d^{1-q}$, while the word on the right side of equality (\ref{anotherlong}) begins with $d^{-q}$, hence, equality (\ref{anotherlong}) doesn't hold. It means that in Case 3.2.2 we don't have a quandle. \textbf{Case 4:} $|w(y,z)|=3$. In this case $w(y,z)$ is either $w(y,z)=y^{q_1}z^{s}y^{q_2}$ for non-zero integers $q_1,q_2,s$ or $w(y,z)=z^{s_1}y^{q}z^{s_2}$ for non-zero integers $q,s_1,s_2$. Consider the case when $w(y,z)=y^{q_1}z^{s}y^{q_2}$ for non-zero integers $q_1,q_2,s$ (case~4.1). In this case $W(x,y,z)=y\left(y^{q_1}z^{s}y^{q_2}\right)^{-1}y^{-\varepsilon}x^{\varepsilon}\left(y^{q_1}z^{s}y^{q_2}\right)=y^{1-q_2}z^{-s}y^{-\varepsilon-q_1}x^{\varepsilon}y^{q_1}z^sy^{q_2}$. Let us understand when the operation $*_{W,c}$ satisfies the third quandle axiom (q3) for each group $G$ and an element $c\in G$ \begin{align} \notag(a&*_{W,c}b)*_{W,c}d=\\ \notag&=(W(a,b,c))*_{W,c}d\\ \notag&=(b^{1-q_2}c^{-s}b^{-\varepsilon-q_1}a^{\varepsilon}b^{q_1}c^sb^{q_2})*_{W,c}d\\ \notag&=W\left(b^{1-q_2}c^{-s}b^{-\varepsilon-q_1}a^{\varepsilon}b^{q_1}c^sb^{q_2},d,c\right)\\ \label{left7}&=d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}\left(b^{1-q_2}c^{-s}b^{-\varepsilon-q_1}a^{\varepsilon}b^{q_1}c^sb^{q_2}\right)^{\varepsilon}d^{q_1}c^sd^{q_2}\\ \notag (a&*_{W,c}d)*_{W,c}(b*_{W,c}d)=\\ \notag&=(W(a,d,c))*_{W,c}(W(b,d,c))\\ \notag&=\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}a^{\varepsilon}d^{q_1}c^sd^{q_2}\right)*_{W,c}\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2}\right)\\ \notag&=W(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}a^{\varepsilon}d^{q_1}c^sd^{q_2}, d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2},c)\\ \notag&=\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2}\right)^{1-q_2}c^{-s}\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2}\right)^{-\varepsilon-q_1}\\ \notag&~~~~\cdot\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}a^{\varepsilon}d^{q_1}c^sd^{q_2}\right)^{\varepsilon}\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2}\right)^{q_1}\\ \label{right7}&~~~~\cdot c^s\left(d^{1-q_2}c^{-s}d^{-\varepsilon-q_1}b^{\varepsilon}d^{q_1}c^sd^{q_2}\right)^{q_2} \end{align} Consider the case when $\varepsilon=1$ (case 4.1.1). From equalities (\ref{left7}), (\ref{right7}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{align} \notag d^{1-q_2}c^{-s}&d^{-1-q_1}b^{1-q_2}c^{-s}b^{-1-q_1}ab^{q_1}c^sb^{q_2}d^{q_1}c^sd^{q_2}=\\ \notag&=\left(d^{1-q_2}c^{-s}d^{-1-q_1}bd^{q_1}c^sd^{q_2}\right)^{1-q_2}c^{-s}\left(d^{1-q_2}c^{-s}d^{-1-q_1}bd^{q_1}c^sd^{q_2}\right)^{-1-q_1}\\ \notag&~~~~\cdot d^{1-q_2}c^{-s}d^{-1-q_1}ad^{q_1}c^sd^{q_2}\left(d^{1-q_2}c^{-s}d^{-1-q_1}bd^{q_1}c^sd^{q_2}\right)^{q_1}\\ \label{q2=1}&~~~~\cdot c^s\left(d^{1-q_2}c^{-s}d^{-1-q_1}bd^{q_1}c^sd^{q_2}\right)^{q_2} \end{align} If $q_2=1$, then equality (\ref{q2=1}) can be rewritten in the following form \begin{multline}\label{q2=11} d^{-1-q_1}c^{-s}b^{-1-q_1}ab^{q_1}c^s=\\ =\left(c^{-s}d^{-1-q_1}bd^{q_1}c^sd\right)^{-1-q_1}c^{-s}d^{-1-q_1}ad^{q_1}c^sd\left(c^{-s}d^{-1-q_1}bd^{q_1}c^sd\right)^{q_1}d^{-1-q_1} \end{multline} If $q_1=-1$ (case 4.1.1.1), then equality (\ref{q2=11}) holds independently on $s$. It means that in Case 4.1.1.1 we have a quandle. If $q_1\neq 1$ (case 4.1.1.2), i.~e. $q_1<-1$ or $q_1>0$, then look to $a,b,c,d$ as to the generators of the free group. If $q_1<-1$, then the reduced word on the left side of equality (\ref{q2=11}) end with $c^{s}$, while the word on the right side of equality (\ref{q2=11}) end with $d^{-1-q_1}$, hence, equality (\ref{q2=11}) doesn't hold. If $q_1>0$, then the reduced word on the left side of equality (\ref{q2=11}) ends with $c^s$, while the word on the right side of equality (\ref{q2=11}) ends with $d^{-q_1}$, hence, equality (\ref{q2=11}) doesn't hold. It means that in Case 4.1.1.2 we don't have a quandle. If $q_2\neq1$ (case 4.1.1.3), i.~e. $q_2<0$ or $q_2>1$, then look to $a,b,c,d$ as to the generators of the free group. If $q_2<0$, then the word on the left side of equality (\ref{q2=1}) ends with $d^{q_2}$, while the word on the right side of equality (\ref{q2=1}) ends with $d^{q_2-1}$, hence, equality (\ref{q2=1}) doesn't hold. If $q_2>1$, then the word on the left side of equality (\ref{q2=1}) begins with $d^{1-q_2}$, while the word on the right side of equality (\ref{q2=1}) begins with $d^{-q_2}$, hence, equality (\ref{q2=1}) doesn't hold. It means that in Case 4.1.1.3 we don't have a quandle. Codsider the case when $\varepsilon=-1$ (case 4.1.2). From equalities (\ref{left7}), (\ref{right7}) we see that the equality $(a*_{W,c}b)*_{W,c}d=(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$ holds if an only if \begin{align} \notag d^{1-q_2}c^{-s}&d^{1-q_1}b^{-q_2}c^{-s}b^{-q_1}ab^{q_1-1}c^sb^{q_2-1}d^{q_1}c^sd^{q_2}=\\ \notag&=\left(d^{1-q_2}c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd^{q_2}\right)^{1-q_2}c^{-s}\left(d^{1-q_2}c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd^{q_2}\right)^{1-q_1}\\ \notag&~~~~\cdot d^{-q_2}c^{-s}d^{-q_1}ad^{q_1-1}c^sd^{q_2-1}\left(d^{1-q_2}c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd^{q_2}\right)^{q_1}\\ \label{e-1last} &~~~~\cdot c^s\left(d^{1-q_2}c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd^{q_2}\right)^{q_2} \end{align} If $q_2=1$ (case 4.1.2.1), then equality (\ref{e-1last}) can be rewritten in the following form \begin{align} \notag &d^{1-q_1}b^{-1}c^{-s}b^{-q_1}ab^{q_1-1}c^s=\\ \label{e-1last2} &=\left(c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd\right)^{1-q_1}d^{-1}c^{-s}d^{-q_1}ad^{q_1-1}c^s\left(c^{-s}d^{1-q_1}b^{-1}d^{q_1}c^sd\right)^{q_1}d^{1-q_1}b^{-1} \end{align} If we look to $a,b,c,d$ as to the generators of the free group, then it is clear that the word on the left side of equality (\ref{e-1last2}) ends with $c^{s}$, while the word on the right side of equality (\ref{e-1last2}) ends with $b^{-1}$, hence, equality (\ref{e-1last2}) doesn't hold. It means that in Case 4.1.2.1 we don't have a quandle. If $q_2\neq1$ (case 4.1.2.2), i.~e. $q_2<0$ or $q_2>1$, then look to $a,b,c,d$ as to the generators of the free group. If $q_2<0$, then the word on the left side of equality (\ref{e-1last}) ends with $d^{q_2}$, while the word on the right side of equality (\ref{e-1last}) ends with $d^{q_2-1}$, hence, equality (\ref{e-1last}) doesn't hold. If $q_2>1$, then the word on the left side of equality (\ref{e-1last}) begins with $d^{1-q_2}$, while the word on the right side of equality (\ref{e-1last}) begins with $d^{-q_2}$, hence, equality (\ref{e-1last}) doesn't hold. It means that in Case 4.1.2.2 we don't have a quandle. Let us consider the case when $w(y,z)=z^{s_1}y^{q}z^{s_2}$ for non-zero integers $q,s_1,s_2$ (case~4.2). In this case \begin{equation}\label{generalbigguy} W(x,y,z)=y\left(z^{s_1}y^{q}z^{s_2}\right)^{-1}y^{-\varepsilon}x^{\varepsilon}\left(z^{s_1}y^{q}z^{s_2}\right)=yz^{-s_2}y^{-q}z^{-s_1}y^{-\varepsilon}x^{\varepsilon}z^{s_1}y^{q}z^{s_2}. \end{equation} Let us understand when the operation $*_{W,c}$ satisfies the third quandle axiom (q3) for each group $G$ and an element $c\in G$ \begin{align} \notag(a*_{W,c}b)*_{W,c}d&=(W(a,b,c))*_{W,c}d\\ \label{leftpartalmostfinal}&=W(W(a,b,c),d,c)\\ \notag(a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=(W(a,d,c))*_{W,c}(W(b,d,c))\\ \label{righttpartalmostfinal}&=W(W(a,d,c),W(b,d,c),c) \end{align} Let us look to the elements $a,b,c,d$ as to the generators of the free group. For a word $A\in F(a,b,c,d)$ denote by $|A|_c$ the number of syllables which are the powers of the element $c$ in the reduced form of $A$. Let us evaluate the values $|(a*_{W,c}b)*_{W,c}d|_c$ and $|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$ given by formulas (\ref{leftpartalmostfinal}), (\ref{righttpartalmostfinal}). First, let us find the upper bound for the value $|(a*_{W,c}b)*_{W,c}d|_c$. Denote by $A_1=W(a,b,c)$. From equality (\ref{leftpartalmostfinal}) and equality (\ref{generalbigguy}) it follows that \begin{align} \label{leftpartevaluation}(a*_{W,c}b)*_{W,c}d&=W(A,d,c)=dc^{-s_2}d^{-q}c^{-s_1}d^{-\varepsilon}A_1^{\varepsilon}c^{s_1}d^{q}c^{s_2} \end{align} In order to calculate the upper bound for $|(a*_{W,c}b)*_{W,c}d|_c$ we have to calculate the number of syllables which are the powers of the element $c$ in the right part of equality (\ref{leftpartevaluation}) without thinking if some powers of $c$ can be reduced with each other (since we try to calculate the upper bound). We have the following upper bound \begin{align} \notag |(a*_{W,c}b)*_{W,c}d|_c\leq 1+1+|A_1|+1+1=|A_1|+4. \end{align} Let us now calculate the lower bound for the value $|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$. Denote by $A_2=W(a,d,c)$, $B=W(b,d,c)$. From equality (\ref{righttpartalmostfinal}) and equality (\ref{generalbigguy}) it follows that \begin{align} \label{rightpartevaluation}(a*_{W,c}d)*_{W,c}(b*_{W,c}d)&=W(A_2,B,c)=Bc^{-s_2}B^{-q}c^{-s_1}B^{-\varepsilon}A_2^{\varepsilon}c^{s_1}B^{q}c^{s_2}. \end{align} In order to calculate the lower bound for $|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$ we have to calculate the number of syllables which are powers of $c$ in the right part of equality (\ref{rightpartevaluation}) which are guaranteed not to be reduced or merged with other syllables which are powers of $c$. From the equality $B=W(b,d,c)$ and equality (\ref{generalbigguy}) it follows that if $|q|\neq1$ (case 4.2.1), then in the word $B^{\pm q}$ there are at least $4(q-1)$ syllables which are powers of $c$ which are guaranteed not to be reduced or merged with other syllables which are powers of $c$ in the right part of equality (\ref{rightpartevaluation}). This is due to the fact that between $2$ consequent syllables which are powers of $b$ in the word $B^{\pm q}=W(b,d,c)^{\pm q}$ there are exactly $4$ syllables which are powers of $c$ which are guaranteed not to be reduced or merged with other syllables which are powers of $c$, and in the word $B^{\pm q}=W(b,d,c)^{\pm q}$ there are exactly $q$ syllables which are powers of $b$, hence, there are at least $4(q-1)$ syllables which are powers of $c$. Due to this remark in order to calculate the lower bound for $|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$ let in the right part of equality (\ref{rightpartevaluation}) calculate only syllables which are powers of $c$ in the word $B^q$, $B^{-q}$ and $A_2^{\varepsilon}$ (these syllables which are powers of $c$ are guaranteed not to be reduced or merged with other syllables which are powers of $c$). We have the following lower bound \begin{align} \notag |(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c\geq 4(q-1)+|A_2|_c+4(q-1)=|A_2|_c+8(q-1). \end{align} Hence we have proved that the following inequalities hold \begin{align} \notag |(a*_{W,c}b)*_{W,c}d|_c&\leq |A_1|_c+4,\\ \notag |(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c&\geq |A_2|_c+8(q-1). \end{align} Noting that $|A_1|_c=|W(a,b,c)|_c=|W(a,d,c)|_c=|A_2|_c$ we conclude that $$|(a*_{W,c}b)*_{W,c}d|_c\neq |(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$$ and hence $(a*_{W,c}b)*_{W,c}d\neq (a*_{W,c}d)*_{W,c}(b*_{W,c}d)$, i.~e. in this case there exists a group $G$ and an element $c\in G$ such that $(G,*_{W,c})$ is not a quandle. It means that in Case 4.2.1 we don't have a quandle. Let us now consider the case when $|q|=1$ (case 4.2.2), i.~e. the case when either $q=1$, or $q=-1$. In this case we can directly calculate all syllabels which are powers of $c$ in both $(a*_{W,c}b)*_{W,c}d$ and $(a*_{W,c}d)*_{W,c}(b*_{W,c}d)$, and make sure that $|(a*_{W,c}b)*_{W,c}d|_c\neq |(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c$. Hence $$(a*_{W,c}b)*_{W,c}d\neq (a*_{W,c}d)*_{W,c}(b*_{W,c}d),$$ i.~e. in this case there exists a group $G$ and an element $c\in G$ such that $(G,*_{W,c})$ is not a quandle. It means that in Case 4.2.2 we don't have a quandle. \textbf{Case 5:} $|w(y,z)|\geq4$. In this case the word $w(y,z)$ has at least two syllables which are powers of $z$. Using the same technique of evaluating the values \begin{align*}|(a*_{W,c}b)*_{W,c}d|_c&& \text{and}&&|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c \end{align*} as we did in case 4.2.1, repeating the proof of case 4.2.1 almost word by word we conclude that \begin{align*}|(a*_{W,c}b)*_{W,c}d|_c<|(a*_{W,c}d)*_{W,c}(b*_{W,c}d)|_c, \end{align*} there exists a group $G$ and an element $c\in G$ such that $(G,*_{W,c})$ is not a quandle. It means that in Case 5 we don't have a quandle.\hfill$\square$ \section{Verbal quandles with $n$ parameters} In this short section, we note that there is no sense to classify verbal quands with more than one parameter. More precisely, using the technique used for proving Theorem~\ref{th22233}, we can show that the following statement about verbal quands with $n$ parameters holds. \begin{ttt} Let $W\in F(x,y,z_1,z_2,\dots,z_n)$ be such that $Q=(G,*_{W,c_1,\dots,c_n})$ is a quandle for every group $G$ and elements $c_1,c_2,\dots,c_n\in G$. Then $W$ has one of the following forms \linespread{0} \begin{enumerate} \item $W=yx^{-1}y$, \item $W=y^{-n}xy^n$ for $n\in\mathbb{N}$, \item $W=yu^{-1}xy^{-1}u$ for $u\in F(z_1,z_2,\dots,z_n)$, \item $W=u^{-1}xy^{-1}uy$ for $u\in F(z_1,z_2,\dots,z_n)$, \item $W=yu^{-1}y^{-1}xu$ for $u\in F(z_1,z_2,\dots,z_n)$, \item $W=u^{-1}y^{-1}xuy$ for $u\in F(z_1,z_2,\dots,z_n)$. \end{enumerate} \end{ttt} \footnotesize

Monday 27th July We woke to our first cooler day since we came to Italy with the temperature around 30 degrees. The papers here are saying 2015 is shaping to be southern Europe’s hottest summer for 30 years so we picked it! After a slow start as both of us had a restless night with noisy neighbours we mooched off to the southern end of Florence again to track down the Brancacci Chapel. Orsanmichele church in Florence, formerly a grain market! Donatello sculpture of St George, one of 14 sculptures around the exterior On the way we visited the Orsanmichele Church in the Via de Calzaioli which was formerly a grain market but in the C14th was converted to a Church. The arcades of the former market were bricked up but in 14 niches around the exterior sculptors were commissioned by guilds of the day to add some class to the external appearance. Donatello, Verrochio and Ghiberti were among those who contributed sculptures. The internal appearance is simple with some impressive stained glass and an elaborate high altar with works by C14th and C16th artists. The Brancacci Chapel is part of the Carmelite convent church of Santa Maria del Carmine in a working class and residential area of Florence south of the Pitti Palace. Entry into the large church is not permitted but the Capella Brancacci chapel which is open to visitors and is frescoed largely by Massachio, having been begun by Masolino and in the way of these things finished off by Filipino Lippi. The major frescoes tell a complicated story of the life of Peter following the resurrection but the sequence is not coherent although the frescoes have been well preserved . Brancacci Chapel Carmelite Monastery Florence; Massachio fresco of Adam and Eve before the “Fall” but thinking! Massachio fresco of Adam and Eve after “the Fall” in Brancacci Chapel, Carmelite monastery Florence. These genuine expressions of despair and grief influenced the work of Da Vinci and Michelangelo The major interest is two facing relatively small panels of Adam and Eve on the edge of the chapel, On the right hand side facing the altar is a healthy and happy pair of humans but a bit edgy about the tempting offering coming between them in the form of a serpent. The facing panel on the left has Adam and Eve leaving the garden in complete despair having earned God’s displeasure by their disobedience. Their powerfully graphic expression of sorrow and anguish was unique in its day and it is said that both Da Vinci and Michelangelo came to see these frescoes and to learn from Massachio’s technique. They still have great power to communicate after all this time but they are certainly overwhelmed by the surrounding Peter frescoes. In the afternoon we walked the streets and lanes of Florence along the Fiume River and into the high end shopping centre of the Via De’ Tourabuoni which would give 5th Avenue New York more than a run for its money. There were some seriously amazing shops including a porcelain shop with its own internal rooms and garden, more like a palace than a shop. It was siesta time so two vast churches were passed over…San Trinita and a huge church next to the equally huge Palazzo Strozzi. Each floor of this vast palace is about 3 times the height of a normal floor in a building. One church which was open was the even larger Church of St Maria Novella with its own equally large frescoed Chapter House and other now museum buildings. The original Romanesque church was doubled in size by Alberti with a Gothic addition and contains two frescoed Strozzi chapels and many other impressive chapels and paintings. Massachio’s fading but still impressive Trinity is here with its carefully hidden Holy Spirit dove looking a bit like the Father’s collar! This was a long walking day for us and we were happy to seek refuge in the air conditioned comfort of our room.

World Wide Sachars:Information about Louis (1) Sachar Louis (1) Sachar (b. 8/4/1887, d. 9/7/1953) More About Louis (1) Sachar: Education: 9/8/1953, NY Times Obit:"Louis Sachar, VP of M. Wolf's Sons, Inc. shoe manufacturers of. Fact 6: 281 State St., Brooklyn, died yesterday at his home, 25 Lefferts Ave., Brooklyn.. Fact 7: 1887, He was 66 years old.Mr. Sachar was president of the Guild of Better Shoe Manu. Fact 8: factures. He leaves his wife, Lena; two sone, Robert and Richard and a daughter,. Fact 9: Mrs. Joyce Michaels.. More About Louis (1) Sachar and Lena Youngerman: Marriage: 8/19/1917, New York, NY. Children of Louis (1) Sachar and Lena Youngerman are: - +Richard Sachar, b. 5/15/1918, Brooklyn, NY. - +Robert Sachar, b. 6/4/1921, d. 1972. - +Joyce Sachar, b. 11/20/1925, d. 4/20/1983, Brooklyn, NY.

\begin{document} \title [logarithmically completely monotonic functions]{ A Class of logarithmically completely monotonic functions relating the $q$-gamma function and applications\\} \author[K. Mehrez]{ Khaled Mehrez } \address{Khaled Mehrez. D\'epartement de Math\'ematiques ISSAT Kasserine, Tunisia.} \email{k.mehrez@yahoo.fr} \begin{abstract} In this paper, the logarithmically complete monotonicity property for a functions involving $q$-gamma function is investigated for $q\in(0,1).$ As applications of this results, some new inequalities for the $q$-gamma function are established. Furthermore, let the sequence $r_n$ be defined by $$n!=\sqrt{2\pi n}(n/e)^n e^{r_n}.$$ We establish new estimates for Stirling's formula remainder $r_n.$ \end{abstract} \maketitle \noindent{ Keywords:} Completely monotonic functions, Logarithmically completely monotonic functions, $q$-gamma function, Stirling's formula, Inequalities. \\ \noindent Mathematics Subject Classification (2010): 33D05, 26D07, 26A48\\ \section{\textbf{Introduction}} A real valued function $f$, defined on an interval $I,$ is called completely monotonic, if f has derivatives of all orders and satisfies \begin{equation} (-1)^n f^{(n)}(x)\geq 0,\;\;\;n\in\mathbb{N}_0,\;x\in I, \end{equation} where $\mathbb{N}$ the set of all positive integers. A positive function $f$ is said to be logarithmically completely monotonic on an interval $I$ if its logarithm $\log f$ satisfies $$(-1)^n \Big(\log f(x)\Big)^{(n)}(x)\geq 0,$$ for all $x\in I$ and $n\in\mathbb{N}.$ Completely monotonic functions have remarkable applications in different branches of mathematics. For instance, they play a role in potential theory, probability theory, physics, numerical and asymptotic analysis, and combinatorics (see \cite{C1} and the references given therein). The $q$-analogue of the gamma function is defined as \begin{equation} \Gamma_q(x)=(1-q)^{1-x}\prod_{j=0}^{\infty}\frac{1-q^{j+1}}{1-q^{j+x}},\:0<q<1, \end{equation} and \begin{equation} \Gamma_q(x)=(q-1)^{1-x}q^{\frac{x(x-1)}{2}}\prod_{j=0}^{\infty}\frac{1-q^{-(j+1)}}{1-q^{-(j+x)}},\:q>1. \end{equation} The $q-$gamma function $\Gamma_q(z)$ has the following basic properties: \begin{equation} \lim_{q\longrightarrow1^{-}}\Gamma_{q}(z)=\lim_{q\longrightarrow1^{+}}\Gamma_{q}(z)=\Gamma(z), \end{equation} and \begin{equation}\label{369} \Gamma_{q}(z)=q^{\frac{(x-1)(x-2)}{2}}\Gamma_{\frac{1}{q}}(z). \end{equation} The $q-$digamma function $\psi_q,$ the $q-$analogue of the psi or digamma function $\psi$ is defined for $0<q<1$ by \begin{equation}\label{ttt} \begin{split} \psi_q(x)&=\frac{\Gamma^{'}_q(x)}{\Gamma_q(z)}\\ &=-\log(1-q)+\log q \sum_{k=0}^{\infty}\frac{q^{k+x}}{1-q^{k+x}}\\ &=-\log(1-q)+\log q \sum_{k=1}^{\infty}\frac{q^{kx}}{1-q^{k}}. \end{split} \end{equation} For $q>1$ and $x>0$, the $q-$digamma function $\psi_q$ is defined by \begin{equation*} \begin{split} \psi_q(x)&=-\log(q-1)+\log q\left[x-\frac{1}{2}-\sum_{k=0}^{\infty}\frac{q^{-(k+x)}}{1-q^{-(k+x)}}\right]\\ &=-\log(q-1)+\log q\left[x-\frac{1}{2}-\sum_{k=1}^{\infty}\frac{q^{-kx}}{1-q^{-kx}}\right] \end{split} \end{equation*} From the previous definitions, for a positive $x$ and $q>1$, we get $$\psi_q(x)=\frac{2x-3}{2}\log q+\psi_{1/q}(x).$$ Using the Euler-Maclaurin formula, Moak \cite{M} obtained the following q-analogue of Stirling formula \begin{equation} \log \Gamma_q(x)\sim \left(x-\frac{1}{2}\right)\log\left(\frac{1-q^x}{1-q}\right)+\frac{\Li_2(1-q^x)}{\log q}+\frac{1}{2}H(q-1)\log q+C_{\hat{q}}+\sum_{k=1}^{\infty}\frac{B_{2k}}{(2k)!}\left(\frac{\log \hat{q}}{\hat{q}^{x} -1}\right)^{2k-1} \hat{q}^{x} P_{2k-3}(\hat{q}^{x}) \end{equation} as $x\longrightarrow\infty$ where $H(.)$ denotes the Heaviside step function, $B_k,\;k=1,2,...$ are the Bernoulli numbers, $$\hat{q}=\begin{cases} q & if\,\,0<q<1\\ 1/q & if\: q>1\end{cases}$$ $\Li_2(z)$ is the dilogarithm function defined for complex argument $z$ as \cite{abra} \begin{equation} \Li_{2}(z)=-\int_{0}^{z}\frac{\log(1-t)}{t}dt,\;z\notin (0,\infty) \end{equation} $P_k$ is a polynomial of degree $k$ satisfying \begin{equation} P_k(z)=(z-z^2)P^{'}_{k-1}(z)+(kz+1)P_{k-1}(z),\;P_0=P_{-1}=1,\; k=1,2,... \end{equation} and $$C_k=\frac{1}{2}\log (2\pi)+\frac{1}{2}\log\left(\frac{q-1}{\log q}\right)-\frac{1}{24}\log q+\log\left(\sum_{m=-\infty}^{\infty}r^{m(6m+1)}-r^{(2m+1)(3m+1)}\right),$$ where $r=\exp(4\pi^2/\log q).$ It is easy to see that $$\lim_{q\longrightarrow 1}C_q=\frac{1}{2}\log(2\pi),\;\;\textrm{and}\;\;\lim_{q\longrightarrow 1}\frac{\Li_2(1-q^x)}{\log q}=-x.$$ Stirling's formula \begin{equation} n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n, \;n\in\mathbb{N} \end{equation} has many applications in statistical physics, probability theory and number theory. Actually, it was first discovered in 1733 by the French mathematician Abraham de Moivre (1667-1754) in the form $$n!\sim constant \sqrt{n}\left(\frac{n}{e}\right)^n$$ when he was studying the Gaussian distribution and the central limit theorem. Afterwards, the Scottish mathematician James Stirling (1692-1770) found the missing constant $\sqrt{2\pi}$ when he was trying to give the normal approximation of the binomial distribution. In 1940 Hummel \cite{H} defined the sequence $r_n$ by \begin{equation} n!=\sqrt{2\pi n}(n/e)^n e^{r_n}, \end{equation} and established \begin{equation}\label{333} \frac{11}{2}<r_n+\log \sqrt{2\pi}<1 \end{equation} After the inequality (\ref{333}) was published, many improvements have been given. For example, Robbins \cite{R} established \begin{equation} \frac{1}{12n+1}<r_n<\frac{1}{12n} \end{equation} The main aim of this paper is to investigate the logarithmic complete monotonicity property of the function \begin{equation}\label{001} f_{\alpha,\beta}(x;q)=\frac{\Gamma_{q}(x+\beta)\exp\left(\frac{-\Li_{2}(1-q^{x})}{\log q}\right)}{\left(\frac{1-q^x}{1-q}\right)^{x+\beta-\alpha}},\;x>0, \end{equation} for all reals $\alpha, \beta$ and $q$ such that $q\in(0,1).$ As applications of these results, sharp bounds for the $q$-gamma function are derived. In addition, we present new estimate for for Stirling's formula remainder $r_n,$ (see Corollary \ref{c4}). Some results are shown to be a generalization of results which were obtained by Chen and Qi \cite{qi}. \section{\textbf{logarithmically completely monotonic function related the $q$-gamma function}} In order to study the function defined by (\ref{001}) we need the following lemma which is considered the main tool to arrive at our results. \begin{lemme}\label{salem2}\cite{salem} For every $x,q\in\mathbb{R}_+$, there exists at least one real number $a \in [0, 1]$ such that \begin{equation} \psi_q(x)=\log\left(\frac{1-q^{x+a}}{1-q}\right)+\frac{q^{x}\,\log q}{1-q^{x}}-\left(\frac{1}{2}-a\right)H(q-1)\log q \end{equation} where $H(.)$ is the Heaviside step function. \end{lemme} \begin{theorem}\label{t1} Let $\alpha$ be a real number. The function $f_{\alpha,1}(q;x)$ is logarithmically completely monotonic on $(0,\infty)$, if and only if $2\alpha\leq1.$ \end{theorem} \begin{proof} Taking logarithm of $f_{\alpha, 1} (x;q)$ leads to \begin{equation} \log f_{\alpha,1} (x;q)=\log\Gamma_{q}(x+1)-(x+1-\alpha)\log\left(\frac{1-q^{x}}{1-q}\right)-\frac{\Li_{2}(1-q^x)}{\log q}. \end{equation} Differentiation yields \begin{equation} \left(\log f_{\alpha,1}(x;q)\right)^{'}=\psi_{q}(x+1)-\log\left(\frac{1-q^{x}}{1-q}\right)+\left(1-\alpha\right)\frac{\log q\;q^{x}}{1-q^{x}}. \end{equation} From the series expansion $$\frac{1}{(1-x)^2}=\sum_{k=0}^{\infty}(k+1)x^k,$$ for $x\in(0,1)$ and (\ref{ttt}) we get \begin{equation} \begin{split} \left(\log f_{\alpha,1}(x;q)\right)^{''}&=\psi_{q}^{'}(x+1)+(1-\alpha)\frac{q^x}{(1-q^x)^2}+\log q\frac{q^x}{1-q^x}\\ &=\big(\log q\big)^2\sum_{k=1}^{\infty}\frac{kq^{k(x+1)}}{1-q^k}+(1-\alpha)\big(\log q\big)^2\sum_{k=1}^{\infty}kq^{kx}+\log q\sum_{k=1}^{\infty} q^{kx}\\ &=\sum_{k=1}^{\infty}\frac{\log q q^{kx}}{1-q^k}\Phi_{\alpha,1}(q^k), \end{split} \end{equation} where $$\Phi_{\alpha,1}(y)=y\log y+(1-\alpha)(1-y)\log y+(1-y),\; y=q^k,\;k=1,2,...$$ In order to determine the sign of the function $\Phi_{\alpha,1}(y)$, we have \begin{equation}\label{kh1} \begin{split} \Phi_{\alpha,1}(y)&=y\big(-\log(1/y)+(1-\alpha)\log(1/y)(1-1/y)+1/y-1 \big)\\ &=y\sum_{k=2}^{\infty}\frac{(\log(1/y))^k}{(k-1)!}\big[\alpha-1+\frac{1}{k}\big]. \end{split} \end{equation} Therefore, the function $\Phi_{\alpha,1}(y)$ is less than zero if $2\alpha\leq1.$ Thus implies that the function $\left(\log f_{\alpha,1}(x;q)\right)^{''}$ is completely monotonic on $(0,\infty).$ This can be rewritten as $$(-1)^n\left(\log f_{\alpha,1}(x;q)\right)^{(n)}\geq0,\;n\geq2.$$ In particular, $\left(\log f_{\alpha,1}(x;q)\right)^{''}\geq0,$ so $\left(\log f_{\alpha,1}(x;q)\right)^{'}$ is increasing on $(0,\infty),$ and consequently \begin{equation*} \begin{split} \left(\log f_{\alpha,1}(x;q)\right)^{(1)}&\leq \lim_{x\longrightarrow\infty}\left(\log f_{\alpha,1}(x;q)\right)^{(1)}\\ &=\lim_{x\longrightarrow\infty}\Bigg(\psi_{q}(x+1)-\log\left(\frac{1-q^{x}}{1-q}\right)+\left(1-\alpha\right)\frac{\log q\; q^{x}}{1-q^{x}}\Bigg)\\ &=0. \end{split} \end{equation*} So $f_{\alpha,1}$ is logarithmically completely monotonic on $(0,\infty)$ if $2\alpha\leq1.$ Conversely, If the function $f_{\alpha,1}(x;q)$ is logarithmically completely monotonic on $(0,\infty)$, then for all real $x>0$, \begin{equation}\label{mp} \left(\log f_{\alpha,1}(x;q)\right)^{'}=\psi_q(x+1)-\log\left(\frac{1-q^{x}}{1-q}\right)+\left(1-\alpha\right)\frac{\log q\;q^{x}}{1-q^{x}}\leq 0. \end{equation} From the equation (\ref{mp}) and along with the identity \begin{equation} \psi_q(x+1)=\psi_q(x)-\frac{q^x \log q}{1-q^x}, \end{equation} we have $$ \left(\log f_{\alpha,1}(x;q)\right)^{'}=\psi_q(x)-\log\left(\frac{1-q^{x}}{1-q}\right)-\alpha\log q\frac{q^{x}}{1-q^{x}}\leq0,$$ which is equivalent to \begin{equation}\label{rr1} \psi_q(x)-\log\left(\frac{1-q^{x}}{1-q}\right)\leq\alpha\frac{\log q\;q^{x}}{1-q^{x}}. \end{equation} It is worth mentioning that, Moak \cite{M} proved the following approximation for the $q$-digamma function $$\psi_q(x)=\log\left(\frac{1-q^x}{1-q}\right)+\frac{1}{2}\frac{\log q\;q^{x}}{1-q^{x}}+O\left(\frac{\log^2 q\;q^{x}}{(1-q^{x})^2}\right)$$ holds for all $q>0$ and $x>0$ and so $\psi_q(x)\sim I(x;q)$ on $(0,\infty)$ where \begin{equation}\label{0rr2} I(x;q)=\log\left(\frac{1-q^x}{1-q}\right)+\frac{1}{2}\frac{\log q\;q^{x}}{1-q^{x}} \end{equation} Combining (\ref{rr1}) and (\ref{0rr2}) we have $$\alpha\leq \frac{1}{2}.$$ The proof is complete. \end{proof} \begin{theorem} \label{t2}Let $\alpha$ be a real number. The function $[f_{\alpha,1}(q;x)]^{-1}$ is logarithmically completely monotonic on $(0,\infty)$, if and only if $\alpha\geq1.$ \end{theorem} \begin{proof} From (\ref{kh1}), we conclude that the function $\Phi_{\alpha,1}(y)\geq0$ if $\alpha\geq1$ we conclude that $$(-1)^n\Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(n)}\geq0$$ for all $x>0,\;\alpha\geq1,\;q\in(0,1).$ and $n\geq2.$ So, $$\Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(1)}=\log\left(\frac{1-q^{x}}{1-q}\right)-\psi_q(x)+\alpha\log q\frac{q^{x}}{1-q^{x}},$$ is increasing, thus \begin{equation*} \begin{split} \Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(1)} &<\lim_{x\longrightarrow\infty}\Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(1)}\\ &=\lim_{x\longrightarrow\infty} \Bigg(\log\left(\frac{1-q^{x}}{1-q}\right)-\psi_q(x)+\alpha\log q\frac{q^{x}}{1-q^{x}}\Bigg)\\ &=0. \end{split} \end{equation*} Hence, For $\alpha\geq1$ and $n\in\mathbb{N},$ \begin{equation*} (-1)^n \Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(n)}\geq0, \end{equation*} on $(0,\infty).$ Now, assume that $\frac{1}{f_{\alpha,1}(q;x)}$ is logarithmically completely monotonic on $(0,\infty)$, by definition, this give us that for all $q\in(0,1)$ and $x>0,$ $$\Bigg(\log\frac{1}{f_{\alpha,1}(q;x)}\Bigg)^{(1)}=\log\left(\frac{1-q^{x}}{1-q}\right)-\psi_q(x)+\alpha\log q\frac{q^{x}}{1-q^{x}}\leq0,$$ which implies that \begin{equation}\label{ll} \alpha \geq \frac{1-q^x}{\log q\;q^x}\Bigg(\psi_q(x)-\log\Bigg(\frac{1-q^x}{1-q}\Bigg)\Bigg). \end{equation} In view of Lemma \ref{salem2} and inequality (\ref{ll}), we see that for all $x>0$ and $q\in(0,1)$ there exists at least one real number $a\in[0,1]$ such that \begin{equation*} \alpha \geq \frac{1-q^x}{\log q\;q^x}\Bigg(\log\Bigg(\frac{1-q^{x+a}}{1-q^x}\Bigg)+\frac{\log q\;q^x}{1-q^x}\Bigg), \end{equation*} and consequently $$\alpha\geq1$$ as $x\longrightarrow\infty.$ This ends the proof. \end{proof} \begin{theorem} \label{t3} Let $\alpha$ be a real number and $\beta\geq0.$ Then, the function $f_{\alpha,\beta}(x;q)$ is logarithmically completely monotonic function on $(0,\infty)$ if $2\alpha\leq1\leq\beta.$ \end{theorem} \begin{proof} Standard calculations lead us to \begin{equation*} \begin{split} \left(\log f_{\alpha,\beta}(x;q)\right)^{''}&=\psi_{q}^{'}(x+\beta)+(\beta-\alpha)\frac{q^x}{(1-q^x)^2}+\log q\frac{q^x}{1-q^x}\\ &=\big(\log q\big)^2\sum_{k=1}^{\infty}\frac{kq^{k(x+\beta)}}{1-q^k}+(\beta-\alpha)\big(\log q\big)^2\sum_{k=1}^{\infty}kq^{kx}+\log q\sum_{k=1}^{\infty} q^{kx}\\ &=\sum_{k=1}^{\infty}\frac{\log q.q^{kx}}{1-q^k}\Phi_{\alpha,\beta}(q^k), \end{split} \end{equation*} where $$\Phi_{\alpha,\beta}(y)=y^\beta\log y+(\beta-\alpha)(1-y)\log y+(1-y),\; y=q^k,\;k=1,2,...$$ Thus $$\Phi_{\alpha,\beta}(y)=y^\beta\left(\sum_{k=2}^{\infty}\frac{\left(\log(1/y)\right)^k}{(k-1)!}\left[\frac{\beta^k-(\beta-1)^k}{k}+(\beta-\alpha)[(\beta-1)^{k-1}-\beta^{k-1}]\right]\right).$$ From the inequality \cite{qi} $$\beta^k-(\beta-1)^k<k(\beta-\alpha)(\beta^{k-1}-(\beta-1)^{k-1})$$ we conclude that $\Phi_{\alpha,\beta}(y)\leq0.$ This implies that for $n\geq2$ \begin{equation}\label{xx} (-1)^{n}\left(\log f_{\alpha,\beta}(x;q)\right)^{(n)}\geq0 \end{equation} on $(0,\infty)$ for $2\alpha\leq1\leq\beta.$ As $\left(\log f_{\alpha,\beta}(x;q)\right)^{(2)}\geq0$, it follows that $\left(\log f_{\alpha,\beta}(x;q)\right)^{(1)}$ is increasing on $(0,\infty)$, and consequently \begin{equation*} \begin{split} \left(\log f_{\alpha,\beta}(x;q)\right)^{(1)}&\leq \lim_{x\longrightarrow\infty}\left(\log f_{\alpha,\beta}(x;q)\right)^{(1)}\\ &=\lim_{x\longrightarrow\infty}\Bigg(\psi_{q}(x+\beta)-\log\left(\frac{1-q^{x}}{1-q}\right)+\left(\beta-\alpha\right)\frac{\log q\; q^{x}}{1-q^{x}}\Bigg)\\ &=0. \end{split} \end{equation*} In conclusion, (\ref{xx}) is true also $n=1$, and we conclude that the function $f_{\alpha,\beta}(x;q)$ is logarithmically completely monotonic on $(0,\infty)$ for $2\alpha\leq 1\leq \beta.$ The proof is now completed. \end{proof} \section{\textbf{Inequalities}} As applications of the logarithmic complete monotonicity properties of the function (\ref{001}) which are proved in Theorem \ref{t1}, Theorem \ref{t2} and Theorem \ref{t3}, we can provide the following inequalities for the $q$-gamma functions. \begin{coro} Let $q\in(0,1), \;n\in\mathbb{N}$ and $x_k>0\;\;(1\leq k\leq n).$ Suppose that $$\sum_{k=1}^n p_k=1\;\;(p_k\geq0).$$ If $2\alpha\leq1\leq\beta$, then \begin{equation}\label{111} \begin{split} \frac{\Gamma_{q}\Big(\sum_{k=1}^{n}p_{k}x_{k}+\beta\Big)}{\prod_{k=1}^{n}\Big[\Gamma_{q}(x_{k}+\beta)\Big]^{p_{k}}}&\leq \frac{\left(\frac{1-q^{\sum_{k=1}^{n}p_{k}x_{k}}}{1-q}\right)^{\sum_{k=1}^{n}p_{k}x_{k}+\beta-\alpha}}{\prod_{k=1}^{n}\left(\frac{1-q^{x_{k}}}{1-q}\right)^{p_{k}(x_{k}+\beta-\alpha)}}\exp\left(\frac{Li_{2}\left(1-q^{\sum_{k=1}^{n}p_{k}x_{k}}\right)-\sum_{k=1}^{n}p_{k}Li_{2}(1-q^{x_{k}})}{\log q}\right) \end{split} \end{equation} \end{coro} \begin{proof} From Theorem \ref{t3}, $f_{\alpha,\beta}(x;q)$ is logarithmically completely monotonic on the interval $(0,\infty),$ which also implies that the function $f_{\alpha,\beta}(x;q)$ is logarithmically convex. Combining this fact with Jensen's inequality for convex functions yields \begin{equation}\label{pp} \log f_{\alpha,\beta}\left(\sum_{k=1}^{n}p_kx_k;q\right)\leq \sum_{k=1}^{n} p_k\log f_{\alpha,\beta}(x_k;q). \end{equation} Rearranging (\ref{pp}) can lead to the inequality (\ref{111}). \end{proof} \begin{coro} Let $q\in(0,1), \;n\in\mathbb{N}$ and $x_k>0\;\;(1\leq k\leq n).$ Suppose that $$\sum_{k=1}^n p_k=1\;\;(p_k\geq0).$$ Then, the following inequalities holds \begin{equation}\label{222} \begin{split} \frac{\left(\frac{1-q^{\sum_{k=1}^{n}p_{k}x_{k}}}{1-q}\right)^{\sum_{k=1}^{n}p_{k}x_{k}}}{\prod_{k=1}^{n}\left(\frac{1-q^{x_{k}}}{1-q}\right)^{p_{k}x_{k}}}\exp\left(\frac{\Li_{2}\left(1-q^{\sum_{k=1}^{n}p_{k}x_{k}}\right)-\sum_{k=1}^{n}p_{k}\Li_{2}(1-q^{x_{k}})}{\log q}\right) \end{split} \end{equation} $$\leq \frac{\Gamma_{q}\Big(\sum_{k=1}^{n}p_{k}x_{k}+1\Big)}{\prod_{k=1}^{n}\Big[\Gamma_{q}(x_{k}+1)\Big]^{p_{k}}}$$ $$\leq \frac{\left(\frac{1-q^{\sum_{k=1}^{n}p_{k}x_{k}}}{1-q}\right)^{\sum_{k=1}^{n}p_{k}x_{k}+1/2}}{\prod_{k=1}^{n}\left(\frac{1-q^{x_{k}}}{1-q}\right)^{p_{k}(x_{k}+1/2)}}\exp\left(\frac{\Li_{2}\left(1-q^{\sum_{k=1}^{n}p_{k}x_{k}}\right)-\sum_{k=1}^{n}p_{k}\Li_{2}(1-q^{x_{k}})}{\log q}\right)$$ \end{coro} \begin{proof} The right side inequality of (\ref{222}) follows by inequality (\ref{111}). From Theorem \ref{t2}, the function $f_{1,1}(x;q)$ is logarithmically concave. Combining this fact with Jensen's inequality for convex functions we obtain the left side inequality of (\ref{222}). \end{proof} \begin{coro} Let $q\in(0,1)$ and $a,b$ be a reals numbers such that $0<a<b$. Then the following inequalities \begin{equation}\label{sss} \frac{\left(\frac{1-q^{b}}{1-q}\right)^{b-1}}{\left(\frac{1-q^{a}}{1-q}\right)^{a-1}}\exp\left(\frac{\Li_{2}(1-q^{b})-Li_{2}(1-q^{a})}{\log q}\right) \leq\frac{\Gamma_{q}(b)}{\Gamma_{q}(a)}\leq\frac{\left(\frac{1-q^{b}}{1-q}\right)^{b-\frac{1}{2}}}{\left(\frac{1-q^{a}}{1-q}\right)^{a-\frac{1}{2}}}\exp\left(\frac{\Li_{2}(1-q^{b})-\Li_{2}(1-q^{a})}{\log q}\right) \end{equation} holds. \end{coro} \begin{proof} From the monotonicity of the functions $f_{\frac{1}{2},1}(q;x)$ and $[f_{1,1}(q;x)]^{-1}$ and the recurrence formula \begin{equation} \Gamma_{q}(x+1)=\frac{1-q^x}{1-q}\Gamma_q(x), \end{equation} we obtain the inequalities (\ref{sss}). \end{proof} \begin{coro}\label{c4} Let $q\in(0,1)$ the following inequalities \begin{equation}\label{kh22} \left(\frac{1-q^{x}}{1-q}\right)^{x}\exp\left(-\frac{\Li_{2}(1-q^{})}{\log q}\right)\exp\left(\frac{\Li_{2}(1-q^{x})}{\log q}\right) \leq\Gamma_q(x+1) \end{equation} $$\leq \left(\frac{1-q^{x}}{1-q}\right)^{x+1/2}\exp\left(-\frac{\Li_{2}(1-q^{})}{\log q}\right)\exp\left(\frac{\Li_{2}(1-q^{x})}{\log q}\right)$$ holds for all $x\in[1,\infty).$ \end{coro} \begin{proof} As the function $f_{1/2,1}(q;x)$ is logarithmically completely monotonic, $f_{1/2,1}(q;x)$ is also decreasing. The following inequality hold true for every $x\geq1:$ \begin{equation}\label{kh97} f_{1/2,1}(q;x)\leq f_{1/2,1}(q;1). \end{equation} In addition, as $1/f_{1,1}(q;x)$ is logarithmically completely monotonic, we deduce that $f_{1,1}(q;x)$ is increasing. The following inequality hold true for all $x\geq1:$ \begin{equation}\label{kh98} f_{1,1}(q;1)\leq f_{1,1}(q;x). \end{equation} Combining inequalities(\ref{kh97}) and (\ref{kh98}) we obtain the inequalities (\ref{kh22}). \end{proof} In the next Corollary we present new estimates for Stirling's formula remainder $r_n$. \begin{coro}The following inequalities hold true for every integer $n\geq1:$ \begin{equation}\label{rr} e.\left(\frac{n}{e}\right)^n\leq n!\leq e.\sqrt{n}\left(\frac{n}{e}\right)^n, \end{equation} and \begin{equation}\label{rr4} 1-\frac{\log 2\pi n}{2}\leq r_n\leq 1-\frac{\log 2\pi }{2} \end{equation} In each of the above inequalities equality hold if and only if $n=1.$ \end{coro} \begin{proof} Replacing $x$ by $n\geq1$ and letting $q\longrightarrow1$ in (\ref{kh22}) we get (\ref{rr}), and after some standard computations we obtain inequalities (\ref{rr4}). \end{proof} \section{\textbf{Concluding Remarks}} \noindent1. It is worth mentioning that the inequality (\ref{sss}) when letting $q$ tends to $1$, returns to the inequalities \cite{qi} \begin{equation} \frac{b^{b-1}}{a^{a-1}}e^{a-b}<\frac{\Gamma(a)}{\Gamma(b)}<\frac{b^{b-1/2}}{a^{a-1/2}}e^{a-b}, \end{equation} for $b>a>0.$\\ 2. Let $n=2$ and $p_k=1/2,\;k=1,2$ in inequalities (\ref{222}) we obtain the lower bounds for the q-analogue for Gurland's ratio \cite{M7} as follows \begin{equation}\label{kh7} \begin{split} \frac{\left(\frac{1-q^{\frac{x+y}{2}}}{1-q}\right)^{x+y}}{\left(\frac{1-q^{x}}{1-q}\right)^{x}\left(\frac{1-q^{y}}{1-q}\right)^{y}}&\leq\frac{\Gamma_{q}^{2}\left(\frac{x+y+2}{2}\right)}{\Gamma_{q}(x+1)\Gamma_{q}(y+1)}\exp\left(\frac{\Li_{2}(1-q^{x})+\Li_{2}(1-q^{y})-2\Li_{2}(1-q^{x+y/2})}{\log q}\right)\\ &\leq \frac{\left(\frac{1-q^{\frac{x+y}{2}}}{1-q}\right)^{x+y+1}}{\left(\frac{1-q^{x}}{1-q}\right)^{x+1/2}\left(\frac{1-q^{y}}{1-q}\right)^{y+1/2}} \end{split} \end{equation} where $x,y\in(0,\infty).$ In particular, let $q$ tends to $1$ in (\ref{kh7}) we get \begin{equation}\label{sana} \frac{\left(\frac{x+y}{2}\right)^{x+y}}{x^x y^y}\leq\frac{\Gamma^2\left(\frac{x+y+2}{2}\right)}{\Gamma(x+1)\Gamma(y+1)}\leq\frac{\left(\frac{x+y}{2}\right)^{x+y+1}}{x^{x+1/2} y^{y+1/2}}. \end{equation} The left hand side inequalities of (\ref{sana}) has proved first by Mortici \cite{MO} and the right hand side of inequalities (\ref{sana}) is new.

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"I hump the wild, because there is no bag limit on happiness." Do you know who said that? That quotation is attributed to our very special guest, the Motor City Madman himself, Sweaty Teddy Nugent! Wango zee tango!!! Ted is what you call a Renaissance Man: he's a musician, a songwriter, an avid hunter, entrepreneur, carnivore, and vocal supporter of the Second Amendment. And, they spoke Latin during the Renaissance...or they tried to, anyway, until someone bastardized it into French! It's only logical that a Renaissance Man be here to further educate us in the ways of the Latin language. His piercing voice and guitar riffs have been featured on 31 fucking albums! Nugent averages about 130+ shows a year and just this past year, The Nuge played his 6,000th show--aptly enough--on the Fourth of July. Uncle Ted's come a long way from the Amboy Dukes, I tell you what. So, it is with great honor that I turn the reins over to Ted Nugent today for the Friday Morning Latin Lesson. I don't know how you're doing it, Ted, but you sure do it good; I'm glad you're doing it for free. Cum catapultae proscriptae erunt tum soli proscripti catapultas habebunt! Pronounced: "Coom cah-tah-pool-tie pro-screep-tie eh-roont toom soul-ee pro-screep-tee cah-tah-pool-tahss hobb-abe-oont!" 10 comments: Years ago, I was watching "Politically Incorrect" with Bill Mahr, and the guests were Ted, dressed like Ted dresses, and 3 suits. The subject was that the Make A Wish Foundation wanted to recind a wish they granted to a boy, because he wanted to go on a bear hunt with his dad. It's an eye opening moment, when you're sitting there, and the guy who's supposed to be the raving lunitic is destroying the Holier Than Thou crowd with LOGIC and REASON. And while it scared me, I sided with him. Okay, he may got too an extreme I'm not comfortable with personanally, but I certainly defend his rights to do and say it. And sorrow shall befall the person who tries to take away the Motor City Madman's rights. And the man is deputized. That I love. Great. Now I have CatScratch Fever. You said 'cum!' heheheheheheh @ Scope: While doing research for this post, I saw a quote by someone (from Snopes, I believe) that said, "You may not always agree with Ted Nugent, but you will always be entertained by his interviews." While I don't take it to the "extreme", I do side with many of his views. Gay marriage, not so much, but the right to shoot and eat something? Yep, I'm right there. @ moooooog35 Better than having a dog encrusted in red shit, I'll bet. @ Chemgeek: Excellent. Further proof that I'm dragging my readership down to my level. I'm still partial to the quote, "On the 8th day god said, Kill 'em Uncle Ted." Growing up in a big hunting family and having a wife who is a hunter, we love our Ted Nugent. I love me some Nuge! He's the best and most eloquent (shocking) defender of hunting. He just lays it out there. "Why is it better for you to buy meat in a supermarket than for me to kill it and eat my kill? Because then you don't have to get your hands dirty. That's it." I support Ted and I own a shotgun for home defense! Wango-tango & white buffalo dreams to the USA! @ Jidai: I grew up in a hunting family and even shot (and ate) a few critters of my own. Let me just say this: Mourning dove = damned tasty. @ Words^3: It makes sense. Plus, most hunters probably end their prey's lives more humanely than a lot of slaughterhouses. Not that I'm knocking the meatpacking industry, but it's sadly probably true. @ Alex G.: My shotguns are all up in Indiana, since my wife (who didn't grow up in a house with guns) is mortified of them. My home defense is a beating stick and my fists of fury. I love Uncle Ted. That man is a bowl full of brilliance complimented by a flava sava, long scraggly hair, a side of juicy beef and a big double-barreled shotgun. You can see why I adore him.

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TITLE: Characteristic polynomial of a generic n*n matric QUESTION [6 upvotes]: Let $K$ be a field, and $F_K$ be the fraction field of the polynomial ring $R_K$ in $n^2$ indeterminates $X_{11},X_{12},...,X_{nn}$ over $K$. Now set $A = (X_{ij})_{i,j} \in M_n (F_K)$, and let $\chi_A$ be the characteristic polynomial of $A$. Question : Is it always true that $\chi_A$ is irreducible over $F_K$ ? Some thoughts : Since $R_K$ is a UFD, it is sufficient to check irreducibility in $R_K[X]$. If $ |K| \geq n$, then $\chi_A$ is separable (by evaluation of $A$ to a diagonal matrix with distinct diagonal entries). If $K[X]$ contains some irreducible $f$ of degree $n$, then this is true (just evaluate $A$ to the companion matrix of $f$ and observe that the evaluation morphism cannot increase degrees). For example, this yields the result when $K = \mathbb{Q}$ or when $K$ is a finite field. More generally, a non trivial factor of $\chi_A$ is unlikely to exist, since it would give a generic factorization of the characteristic polynomial of a $n \times n$ matrix with coefficients in $K$. Even in the case $K=\mathbb{C}$, I cannot prove (or disprove) the irreducibility of $\chi_A$. REPLY [9 votes]: If it were reducible, all $n\times n$ matrices over any field extension of $K$ would have reducible characteristic polynomials. But consider the companion matrix of an irreducible polynomial over some not algebraically closed field extension of $K$.

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\section{$W$-cap preconditioner}\label{sec:Vcap} To circumvent the need for mappings involving fractional Sobolev spaces we shall next study a different preconditioner for \eqref{eq:weak_form}. As will be seen the new preconditioner $W$-cap preconditioner \eqref{eq:operator_preconditionersV} is still robust with respect to the material and discretization parameters. Consider operator $\op{A}$ from problem \eqref{eq:op_long} as a mapping $W\times Q \rightarrow W^*\times Q^*$, with spaces $W, Q$ defined as \begin{equation}\label{eq:Vcap_spaces} \begin{aligned} W&=\left(H^1_0\left(\Omega\right)\cap\epsilon\HG\right)\times H^1_0\left(\Gamma\right),\\ Q&= H^{-1}\left(\Gamma\right). \end{aligned} \end{equation} The spaces are equipped with norms \begin{equation}\label{eq:Vcap_norms} \norm{w}^2_W = \HnormO{u}^2 + \epsilon^2\HnormG{T_{\Gamma} u}^2 + \HnormG{v}^2 \quad \text{ and }\quad \norm{p}^2_Q = \norm{p}^2_{-1, \Gamma}. \end{equation} Note that the trace of functions from space $U$ is here controlled in the norm $\HnormG{\cdot}$ and not the fractional norm $\norm{\cdot}_{\half, \Gamma}$ as was the case in \S \ref{sec:Qcap}. Also note that the space $W$ now is dependent on $\epsilon$ while $Q$ is not. The following result establishes well posedness of \eqref{eq:weak_form} with the above spaces. \begin{theorem}\label{thm:stab_inv} Let $W$ and $Q$ be the spaces \eqref{eq:Vcap_spaces}. The operator $\op{A}:W\times Q\rightarrow W^*\times Q^*$, defined in \eqref{eq:op_long} is an isomorphism and the condition number of $\op{A}$ is bounded independently of $\epsilon >0$. \end{theorem} \begin{proof} The proof proceeds by verifying the Brezzi conditions \ref{thm:brezzi}. With $w=(u, v)$, $\omega=(\phi, \psi)$ application of the Cauchy-Schwarz inequality yields \[ \begin{split} \brack{A w, \omega }_{\Omega} &= \inner{\nabla u }{\nabla \phi}_{\Omega} + \inner{\nabla v}{\nabla \psi}_{\Gamma} \\ & \leq \HnormO{u}\HnormO{\phi} + \HnormG{v}\HnormG{\psi} \\ & \leq \HnormO{u}\HnormO{\phi} + \epsilon^2\HnormG{T_{\Gamma} u}\HnormG{\phi} + \HnormG{v}\HnormG{\psi} \\ & \leq \norm{w}_W\norm{\omega}_W. \end{split} \] Therefore $A$ is bounded with $\norm{A}=1$ and \eqref{eq:Brezzi_A_bounded} holds. The coercivity of $A$ on $\ker B$ for \eqref{eq:Brezzi_A_coercivity} is obtained from \[ \begin{split} \inf_{w\in \ker B}\frac{\brack{Aw, w}_{\Omega}}{\norm{w}^2_W} &= \inf_{w\in \ker B}\frac{\HnormO{u}^2 + \HnormG{v}^2} { \HnormO{u}^2+\epsilon^2\HnormG{T_\Gamma u}^2+\HnormG{v}^2 } \\ &= \inf_{w\in \ker B}\frac{\HnormO{u}^2+\HnormG{v}^2} {\HnormO{u}^2 + 2\HnormG{v}^2} \geq \frac{1}{2}, \end{split} \] where we used that $\epsilon T_\Gamma u = v$ a.e. on the kernel. Consequently $\alpha=\tfrac{1}{2}$. Boundedness of $B$ in \eqref{eq:Brezzi_B_bounded} with a constant $\norm{B}=\sqrt{2}$ follows from the Cauchy-Schwarz inequality \[ \begin{split} \brack{B w, q}_{\Gamma} &\leq \norm{q}_{-1, \Gamma}\epsilon\HnormG{T_\Gamma u}+\norm{q}_{-1, \Gamma}\HnormG{v} \\ & \leq \sqrt{2}\norm{q}_Q\sqrt{\epsilon^2\HnormG{T_\Gamma u}^2 + \HnormG{v}^2} \\ & \leq \sqrt{2}\norm{q}_Q\sqrt{\HnormO{u}^2 + \epsilon^2\HnormG{T_{\Gamma} u}^2 + \HnormG{v}^2} \\ & \leq \sqrt{2}\norm{q}_Q\norm{w}_W. \end{split} \] To show that the inf-sup condition holds compute \[ \begin{split} \sup_{w\in W}\,\frac{\brack{B w, q}_{\Gamma}}{\norm{w}_W} &= \sup_{w\in W}\,\frac{\brack{q, \epsilon T_\Gamma u - v}_{\Gamma}}{\sqrt{ \HnormO{u}^2 + \epsilon^2 \HnormG{T_\Gamma u}^2 + \HnormG{v}^2} } \\ &\stackrel{u=0}\geq \sup_{v\in V} \frac{\brack{q, v}_{\Gamma}}{\HnormG{v}}=\norm{q}_Q. \end{split} \] Thus $\beta=1$ in condition \eqref{eq:Brezzi_infsup}. \qquad\end{proof} Following Theorem \ref{thm:stab_inv} the operator $\mathcal{A}$ is a symmetric isomorphism between spaces $W\times Q$ and $W^{*}\times Q^{*}$. As a preconditioner we shall consider a symmetric positive-definite isomorphism $W^{*}\times Q^{*}\rightarrow W\times Q$ \begin{equation}\label{eq:Wcap} \BBW= \begin{bmatrix} \inv{\open{-\Delta_{\Omega} + T_{\Gamma}^{*}\open{-\epsilon^2 \Delta_{\Gamma}} T_{\Gamma}}} & &\\ & \inv{\open{-\Delta_{\Gamma}}} & \\ & & {-\Delta_{\Gamma}}\\ \end{bmatrix}. \end{equation} \subsection{Discrete preconditioner}\label{eq:Vcap_discr} Similar to \S \ref{sec:Qcap_discrete} we shall construct discretizations $W_h \times Q_h$ of space $W\times Q$ \eqref{eq:Vcap_spaces} such that the finite dimensional operator $\mathcal{A}_h$ defined by considering $\mathcal{A}$ from \eqref{eq:op_long} on the constructed spaces satisfies the Brezzi conditions \ref{thm:brezzi}. Let $W_h \subset W$ and $Q_h \subset Q$ the spaces \eqref{eq:hspaces} of continuous piecewise linear polynomials. Then $A_h$, $B_h$ are continuous with respect to norms \eqref{eq:Vcap_norms} and it remains to verify conditions \eqref{eq:Brezzi_A_bounded} and \eqref{eq:Brezzi_infsup}. First, coercivity of $A_h$ is considered. \begin{lemma}\label{Vcap_coerc} Let $W_h, Q_h$ the spaces \eqref{eq:hspaces} and $A_h, B_h$ such that\\ $\brack{A w, \omega_h}_{\Omega}=\brack{A_h w_h, \omega_h}_{\Omega}$, $\brack{B w, q_h}_{\Gamma}=\brack{B_h w_h, q_h}_{\Gamma}$, for $\omega_h, w_h\in W_h$, $w\in W$ and $q_h\in Q_h$. Then there exists a constant $\alpha > 0$ such that for all $z_h \in \ker B_h$ \[ \brack{A_h z_h, z_h} \geq \alpha \norm{z_h}_{W}, \] where $\norm{\cdot}_{W}$ is defined in \eqref{eq:Vcap_norms}. \end{lemma} \begin{proof} The claim follows from coercivity of $A$ over $\ker B$ (cf. Theorem \ref{thm:stab_inv}) and the property $\ker B_h \subset \ker B$. To see that the inclusion holds, let $z_h \in \ker B_h$. Since $z_h$ is continuous on $\Gamma$ we have from definition $\brack{z_h, q_h}_{\Gamma}=0$ for all $q_h \in Q_h$ that $z_h|_\Gamma=0$. But then $\brack{z_h, q}=0$ for all $q\in Q$ and therefore $z_h \in \ker B$.\qquad \end{proof} Finally, to show that the discretization $W_h \times Q_h$ is stable we show that the inf-sup condition for $B_h$ holds. \begin{lemma}\label{Vcap_infsup} Let spaces $W_h, Q_h$ and operator $B_h$ from Lemma \ref{Vcap_coerc}. Then there exists $\beta > 0$ such that \begin{equation}\label{eq:discrete_inf_sup0} \inf_{q_h\in Q_h}\,\sup_{w_h\in W_h}\, \frac{\brack{B_h w_h, q_h}_{\Gamma}}{\norm{w_h}_W\norm{q_h}_Q} \geq \beta, \end{equation} where $\norm{\cdot}_{Q}$ is defined in \eqref{eq:Vcap_norms}. \end{lemma} \begin{proof} We first proceed as in the proof of Theorem \ref{thm:stab_inv} and compute \begin{equation}\label{eq:estim0} \sup_{w_h \in W_h} \frac{\brack{q_h, \epsilon T_{\Gamma} u_h - v_h}_{\Gamma}} {\norm{w_h}_{W}} \stackrel{u_h=0}{\geq} \sup_{v_h \in V_h} \frac{\brack{v_h, q_h}_{\Gamma}} {\semi{v_h}_{1, \Gamma}}. \end{equation} Next, for each $p\in H^1_{0}\open{\Gamma}$ let $v_h=\Pi p$ the element of $V_h$ defined in the proof of Lemma \ref{Qcap_discrete_inf_sup}. In particular, it holds that \[ \brack{{p - v_h}, q_h}_{\Gamma} = 0,\quad q_h\in Q_h \] and $\semi{v_h}_{1, \Gamma}\leq C \semi{p}_{1, \Gamma}$ for some constant $C$ depending only on $\Omega$ and $\Gamma$. Then \[ \norm{q_h}_{-1, \Gamma} = \sup_{p\in H^1_0\open{\Gamma}} \frac{\brack{q_h, p}_{\Gamma}} {\semi{p}_{1, \Gamma}} \leq C \sup_{v_h\in V_h} \frac{\brack{q_h, v_h}_{\Gamma}} {\semi{v_h}_{1, \Gamma}}. \] The estimate together with \eqref{eq:estim0} proves the claim of the lemma. \qquad \end{proof} Let now $\mat{A}_{U}, \mat{A}_{V}$ and $\mat{B}_{U}, \mat{B}_{V}$ the matrices defined in \eqref{eq:mat_defs} as representations of the corresponding finite dimensional operators in the basis of the stable spaces $W_h$ and $Q_h$. We shall represent the preconditioner $\BBW$ by a matrix \begin{equation}\label{eq:Vcap_precond} \BBWh = \begin{bmatrix} \inv{\open{\Amat_{U} + \epsilon^{2}\transp{\Tmat}\Amat\Tmat}} & & \\ & \inv{\open{\Amat_{V}}} & \\ & & \inv{\Hmat{-1}}\\ \end{bmatrix}, \end{equation} where $\inv{\Hmat{-1}}=\inv{\mat{M}}\mat{A}\inv{\mat{M}}$, cf. \eqref{eq:H_def}, and $\mat{M}$, $\mat{A}$ the matrices inducing $L^2$ and $H^1_0$ inner products on $Q_h$. Let us point out that there is an obvious correspondence between the matrix preconditioner $\BBWh$ and the operator $\BBW$ defined in \eqref{eq:operator_preconditionersV}. On the other hand it is not entirely straight forward that the matrix $\BBWh$ represents the $W$-cap preconditioner defined here in \eqref{eq:Wcap}. In particular, since the isomorphism from $Q^{*}=H^1_0(\Gamma)$ to $Q=H^{-1}(\Gamma)$ is realized by the Laplacian a case could be made for using the stiffness matrix $\mat{A}$ as a suitable representation of the operator. Let us first argue for $\mat{A}$ not being a suitable representation for preconditioning. Note that the role of matrix $\mathbb{A}\in\reals^{m\times n}$ in a linear system $\mathbb{A}\vec{x}=\vec{b}$ is to transform vectors from the solution space $\reals^n$ to the residual space $\reals^m$. In case the matrix is invertible the spaces concide. However, to emphasize the conceptual difference between the spaces, let us write $\mathbb{A}:\reals^{n}\rightarrow\reals^{n*}$. Then a preconditioner matrix is a mapping $\mathbb{B}:\reals^{n*}\rightarrow\reals^{n}$. The stiffness matrix $\mat{A}$, however, is such that $\mat{A}:\reals^{n_Q}\rightarrow\reals^{n_Q*}$. It remains to show that $\inv{\mat{M}}\mat{A}\inv{\mat{M}}$ is the correct representation of $A=-\Delta_{\Gamma}$. Recall that $Q_h\subset Q^*$ and $\mat{A}$ is the matrix representation of operator $A_h: Q_h\rightarrow Q^{*}_h$. Further, mappings $\pi_h: Q_h\rightarrow \reals^{n_Q}$, $\mu_h: Q^{*}_h\rightarrow \reals^{n_Q*}$ \[ p_h = \sum_j(\pi_h p_h)_j \chi_j, \quad p_h \in Q_h \quad\mbox{ and }\quad (\mu_h f_h)_j = \brack{f_j, \chi_j}, \quad f_h\in Q^{*}_h \] define isomorphisms between\footnotemark spaces $Q_h$, $\reals^{n_Q}$ and $Q^*_h$, $\reals^{n_Q*}$ \footnotetext{Note that in \S \ref{sec:intro} the mapping $\mu_h$ was considered as $\mu_h:Q^{*}_h\rightarrow \reals^{n_Q}$. The definition used here reflects the conceptual distinction between spaces $\reals^{n_Q}$ and $\reals^{n_Q*}$. That is, $\mu_h$ is viewed as a map from the space of right-hand sides of the operator equation $A_h p_h = L_h$ to the space of right-hand sides of the corresponding matrix equation $\mat{A}\vec{p}=\vec{b}$. } respectively. We can uniquely associate each $p_h\in Q_h$ with a functional in $Q^{*}_h$ via the Riesz map $I_h:Q_h\rightarrow Q^{*}_h$ defined as $\brack{I_h p_h, q_h}_{\Gamma}=\inner{p_h}{q_h}_{\Gamma}$. Since \[ \open{\mu_h I_h p_h}_j = \open{I_h p_h, \chi_j}_{\Gamma} = \sum_i \open{\pi_h p_h}_i \inner{\chi_i}{\chi_j}_{\Gamma} \] the operator $I_h$ is represented as the mass matrix $\mat{M}$. The matrix then provides a natural isomorphism from $\reals^{n_Q}$ to $\reals^{n_Q*}$. In turn $\inv{\mat{M}}\mat{A}\inv{\mat{M}}:\reals^{n_Q*}\rightarrow\reals^{n_Q}$ has the desired mapping properties. In conclusion, the inverse of the mass matrix was used in \eqref{eq:Vcap_precond} as a natural adapter to obtain a matrix operating between spaces suitable for preconditioning. Finally, we make a few observations about the matrix preconditioner $\BBWh$. Recall that the $Q$-cap preconditioner $\BBQh$ could be related to the Schur complement based preconditioner \eqref{eq:Schur_mat_preconditioner} obtained by factorizing $\AAh$ in \eqref{eq:eigen_matrix}. The relation of $\AAh$ to the $W$-cap preconditioner matrix \eqref{eq:Vcap_precond} is revealed in the following calculation \begin{equation}\label{eq:aux} \mathbb{U} \mathbb{L} \AAh = \begin{bmatrix} \Amat_{V} + \epsilon^{2}\transp{\Tmat}\Amat_{}\Tmat & & \\ & \tau^{2}\Amat_{} & -\Mmat_{}\\ -\epsilon\Mmat_{}\Tmat& & \Mmat_{}\inv{\Amat_{}}\Mmat_{}\\ \end{bmatrix}, \end{equation} where \[ \mathbb{U} = \begin{bmatrix} \mat{I} & \phantom{\transp{\Tmat}\Amat_{}\inv{\Mmat_{}}} & -\transp{\Tmat}\epsilon\Amat_{}\inv{\Mmat_{}}\\ & \mat{I} & \\ & & \mat{I}\\ \end{bmatrix}\, \quad\text{ and }\quad \mathbb{L} = \begin{bmatrix} \mat{I} & &\\ & \mat{I} & \\ & -\Mmat_{}\inv{\mat{A}_{}} & -\mat{I}\\ \end{bmatrix}. \] Here the matrix $\mathbb{L}$ introduces a Schur complement of a submatrix of $\AAh$ corresponding to spaces $V_h, Q_h$. The matrix $\mathbb{U}$ then eliminates the constraint on the space $U_h$. Preconditioner $\BBWh$ could now be interpreted as coming from the diagonal of the resulting matrix in \eqref{eq:aux}. Futher, note that the action of the $Q_h$-block can be computed cheaply by Jacobi iterations with a diagonally preconditioned mass matrix (cf. \cite{wathen_M}). \begin{table}[ht!] \begin{center} \caption{Spectral condition numbers of matrices ${\BBWh\AAh}$ for the system assembled on geometry (a) in Figure \ref{fig:scheme}. } \footnotesize{ \begin{tabular}{l|ccccccc} \hline \multirow{2}{*}{size} & \multicolumn{7}{c}{$\log_{10}\epsilon$}\\ \cline{2-8} & $-3$ & $-2$ & $-1$ & $0$ & $1$ & $2$ & $3$\\ \hline 99 & 2.619 & 2.627 & 2.546 & 3.615 & 3.998 & 4.044 & 4.048\\ 323 & 2.623 & 2.653 & 2.780 & 3.813 & 4.023 & 4.046 & 4.049\\ 1155 & 2.631 & 2.692 & 3.194 & 3.925 & 4.036 & 4.048 & 4.049\\ 4355 & 2.644 & 2.740 & 3.533 & 3.986 & 4.042 & 4.048 & 4.049\\ 16899 & 2.668 & 2.788 & 3.761 & 4.017 & 4.046 & 4.049 & 4.049\\ 66563 & 2.703 & 3.066 & 3.896 & 4.033 & 4.047 & 4.049 & 4.049\\ \hline \end{tabular} \label{tab:eigs_Vcap} } \end{center} \end{table} \subsection{Numerical experiments} Parameter robust properties of the $W$-cap preconditioner are demonstrated by the two numerical experiments used to validate the $Q$-cap preconditioner in \S \ref{sec:Qcap_numerics}. Both the experiments use discretization of domain (a) from Figure \ref{fig:scheme}. First, using the \textit{exact} preconditioner we consider the spectral condition numbers of matrices ${\BBWh \AAh}$. Next, using an approximation of $\BBWh$ the linear system $\overbar{\mathbb{B}}_W \AAh \vec{x}=\overbar{\mathbb{B}}_W\vec{f}$ is solved with the minimal residual method. The operator $\overbar{\mathbb{B}}_W$ is defined as \begin{equation}\label{eq:B2h_approx} \overbar{\mathbb{B}}_W = \begin{bmatrix} \text{AMG}{\open{\Amat_U + \epsilon^{2}\transp{\Tmat}\Amat\Tmat}} & & \\ & \text{LU}{\open{\Amat}} & \\ & & \text{LU}\open{\Mmat}\,\Amat\,\text{LU}\open{\Mmat}\\ \end{bmatrix}. \end{equation} The spectral condition numbers of matrices ${\BBWh \AAh}$ for different values of material parameter $\epsilon$ are listed in Table \ref{tab:eigs_Vcap}. For all the considered discretizations the condition numbers are bounded with respect to $\epsilon$. We note that the mesh convergence of the condition numbers appears to be faster and the obtained values are in general smaller than in case of the $Q$-cap preconditioner (cf. Table \ref{tab:eigs_Qcap}). Table \ref{tab:Vcap_iters} reports the number of iterations required for convergence of the minimal residual method for the linear system $\overbar{\mathbb{B}}_W \AAh \vec{x}=\overbar{\mathbb{B}}_W\vec{f}$. Like for the $Q$-cap preconditioner the method is started from a random initial vector and the condition $\transp{\vec{r}_k}\overbar{\mathbb{B}}_W \vec{r}_k < 10^{-12}$ is used as a stopping criterion. We find that the iteration counts with the $W$-cap preconditioner are again bounded for all the values of the parameter $\epsilon$. Consistent with the observations about the spectral condition number, the iteration count is in general smaller than for the system preconditioned with the $Q$-cap preconditioner. \begin{table}[ht!] \begin{center} \caption{Iteration count for system $\overbar{\mathbb{B}}_W\AAh \vec{x}=\overbar{\mathbb{B}}_W\vec{f}$ solved with the minimal residual method. The problem is assembled on geometry (a) from Figure \ref{fig:scheme}. Comparison to the number of iterations with the $Q$-cap preconditioned system is shown in the brackets (cf. also Table \ref{tab:Qcap_iters}). } \footnotesize{ \begin{tabular}{l|ccccccc} \hline \multirow{2}{*}{size} & \multicolumn{7}{c}{$\log_{10}\epsilon$}\\ \cline{2-8} & $-3$ & $-2$ & $-1$ & $0$ & $1$ & $2$ & $3$\\ \hline 66563 & 17(-3) & 33(-1)& 40(3) & 30(-2) & 20(-8) & 14(-10) & 12(-9)\\ 264195 & 19(-3) & 35(1) & 39(5) & 28(-2) & 19(-7) & 14(-10) & 11(-9)\\ 1052675 & 22(-2) & 34(1) & 37(5) & 27(-1) & 19(-7) & 14(-8) & 11(-7)\\ 4202499 & 24(-2) & 34(2) & 34(4) & 25(-1) & 17(-7) & 12(-8) & 9(-8)\\ 8398403 & 25(-1) & 32(2) & 32(2) & 24(-2) & 16(-6) & 11(-8) & 8(-7)\\ 11075583 & 25(-1) & 32(2) & 32(2) & 25(-1) & 16(-6) & 13(-6) & 11(-4)\\ \hline \end{tabular} \label{tab:Vcap_iters} } \end{center} \end{table} We note that the observations from \S \ref{sec:Qcap_numerics} about the difference in $\epsilon$-dependence of condition numbers and iteration counts of the $Q$-cap preconditioner apply to the $W$-cap preconditioner as well. Before addressing the question of computational costs of the proposed preconditioners let us remark that the $Q$-cap preconditioner and the $W$-cap preconditioners are not spectrally equivalent. Further, both preconditioners yield numerical solutions with linearly(optimaly) converging error, see Appendix \ref{sec:appendix_eoc}.

This weekend I dragged out all my paraphernalia to set up my outdoor canning kitchen. I set it up on the sidewalk just outside my kitchen door, and it will be getting quite a bit of use over the next few weeks. It’s a very simple setup, and keeps my house from becoming hot and steamy during the dog days of August. The burner/stand is from an old turkey fryer. I use one big pot to boil water for scalding tomatoes, peaches, corn, etc. The other big pot is filled with cold water, which I use to quickly cool said scalded tomatoes, peaches, corn, etc. From my staging area outdoors, I move my scalded/blanched produce into the house to finish processing. This weekend I’ve been working on a wheelbarrow full of organic sweet corn that was GIVEN to me. Yep, that’s right, it was free! I also got some help this time around. Since we moved our youngest son to college last weekend, my husband has been especially attentive to me, as we’ve been adjusting to our empty nest. I also got a little bit of help from the family cat. He’s always got to be in the center of whatever is going on around here. I finally had to give him a small ear of corn to get him out of my hair. He’s a raw fed kitty, so who would have figured he was a sucker for sweet corn? My setup for cutting corn is very basic. I use a bowl turned upside down in a shallow pan and a very sharp, comfortable knife. I keep a knife sharpener handy, and run the knife over it every dozen or so ears of corn. I use a regular sharp edged teaspoon to scrape the cobs after I’ve cut the kernels off. Once I have all the corn off the ears, I package it up in freezer bags, lay the bags flat on cookie sheets, and then stack the cookie sheets in my big deep freeze until the corn is frozen. Now, I need to go finish the corn. I’ll let you know how much I end up “putting by” later in the week when I share some more of my canning adventures. I’m going to have close to 50 pounds of tomatoes to deal with on Tuesday.

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Co-morbidities in elderly patients with hip fracture: recommendations of the ISFR-IOF hip fracture outcomes working group Abstract Introduction Hip fractures are the second leading cause of hospitalization in the aged and by 2041, epidemiologists forecast an increase in economic cost to $2.4 billion. The hip patient population often presents with comorbidities causing these patients to receive less aggressive medical treatment and have a low quality of life. We believe that physical function is a patient-important outcome for many medical and surgical interventions. The functional co-morbidity index (FCI), unlike prior co-morbidity indices, was developed with physical function as an outcome instead of being designed for administrative purposes or to predict mortality. Our objective was to evaluate the perceptions of practitioners in hip fracture care about the impact of comorbidities on physical function as primary outcome. Methods We piloted and then distributed a self-administered survey to members of the International Society for Fracture Repair hip fracture outcomes working group. For each of the 18 diagnoses included in the FCI index, we asked in our survey whether the presence of the co-morbidity and whether the severity of the co-morbidity was perceived to impact physical function in patients following a hip fracture. Results Seventeen out of 20 respondents completed the questionnaire. The presence and severity of arthritis was ‘strongly’ believed to predict physical function in those with hip fracture (69 and 85.7 %, respectively). Respondents ‘agreed’ (range 53–73 %) that 10/18 diagnoses would predict changes in physical function following hip fracture treatment. Whereas, 63 % of the practitioners‘strongly disagreed’ that diabetes types I and II would change physical function scores. Furthermore, dementia was listed as an additional diagnosis that would affect physical function. Conclusion The FCI may provide a useful instrument to predict functional outcome after hip fracture; however, the index may need to be modified for this specific population.

None of the members marked this couple as a favourite Add to your favourites We have recorded: 114 of their results as Professional results | compare results Ballroom - 1973 see graph Based on 114 results. Last on Wednesday, 14 October 2009 See 553 photos in the photogallery Upload photos Supadance Ambassador Victor Fung & Anna Mikhed started dancing together in 2003 (December) as Professional They do not compete together anymore and were last registered as a couple in USA. They danced in Professional . Their first competition that is recorded in our database was in England - Bournemouth on 2004 (January). So far their best mark in our database is getting into 1 in the Professional Ballroom (Imperial 2009 in England - Frimley Green on Saturday, 10 October 2009). They joined together after Victor and Liene Apale split in 2003 (December) - see profile and Anna and Dariusz Michalski split in 2003 (August) - see profile currently 0 comments | Add comment USAProfessional01 Dec 2003 #{text}

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Another manic set of phonecalls (Thanks Joan) see's us confirm two weekends running Fund Raising at Pets At Home in Cramlington. Thank you to Joan, Carol and Nicola for agreeing to give up parts of their weekend at such short notice, you have made it possible for us to be there this Saturday 17th and Sunday 18th April. We will also be there next weekend too (24th and 25th) so it will be great to make some new friends, and of course to also see some old ones! We have had a busy few weeks Fund Raising, and of course we have the Race Night at the Border Terrier on the 30th. Latest update on that is the we have a signed Newcastle United shirt to raffle off. Please jot the date in your diary if you fancy your chances of winning it.

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Israeli air force jets have bombed the Islamic University in the Gaza Strip, a significant cultural symbol for Hamas.. Israel's behaviour is nothing short of crimes against humanity. Blatant violation of human rights. The UN Security Council has called an emergency session. 28 December 2008 – The.” The international community needs to condemn Israel for war crimes. Now. There are no words. 2 comments: The international community needs to condemn Israel for war crimes. Now. There are no words. And they need to condemn the US for greenlighting everything that israel does. And such condemnation must include diplomatic and financial consequences. I wish that the international community had the power to impose such consequences but the power disparity makes that impossible. This is one of the reasons that we need a powerful UN with intervention capabilities. But that requires cooperation and that's not a commodity in abundance.

Also See: Kenova, WV ZIP Codes & ZIP Code Maps | Local Area Photos The City of Kenova had a population of 3,142 as of July 1, 2015. Kenova ranks in the upper quartile for Population Density when compared to the other cities, towns and Census Designated Places (CDPs) in West Virginia. See peer rankings below. The primary coordinate point for Kenova is located at latitude 38.4027 and longitude -82.5819 in Wayne County. The formal boundaries for the City of Kenova (see map below) encompass a land area of 1.26 sq. miles and a water area of 0.34 sq. miles. Wayne County is in the Eastern time zone (GMT -5). The elevation is 554 feet. The City of Kenova (GNIS ID: 2390604) has a C1 Census Class Code which indicates an active incorporated place that does not serve as a county subdivision equivalent. It also has a Functional Status Code of "A" which identifies an active government providing primary general-purpose functions. Beneath the boundary map are tables with Kenova population, income and housing data, five-year growth projections and peer comparisons for key demographic data. The estimates are for July 1, 2015. Alternate Unofficial Names for Kenova: Morganza, Virginia Point. The table below compares Kenova.

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TITLE: $\mathbb Z_{p^2}$ is not a non trivial semidirect product. QUESTION [0 upvotes]: I am trying to prove that the group $\mathbb Z_{p^2}$ (p prime) is not a non trivial semidirect product. Since a group $G \cong K \rtimes H$ if and only if for all short exact sequences $$0 \rightarrow H \rightarrow G \rightarrow K \rightarrow 0$$splits on the right, then I thought that I could prove that the appropiate sequence doesn't split on the right. The only non trivial group of $\mathbb Z_{p^2}$ is $\mathbb Z_p$, so the only possible non trivial semidirect product is $\mathbb Z_p \rtimes \mathbb Z_p$. I define the sequence $$0\longrightarrow \mathbb Z_{p} \overset{f}{\longrightarrow} \mathbb Z_{p^2} \overset{g}{\longrightarrow} \mathbb Z_p \longrightarrow 0 $$ Where $f$ is the inclusion an $g$ is the "projection". I would like to show that there is no such $\beta: \mathbb Z_p \to \mathbb Z_{p^2}$ with $g \circ \beta =Id_{\mathbb Z_{p}}$. Any help with this part of the solution would be appreciated. REPLY [1 votes]: I think you meant $\;G:=\Bbb Z_{p^2}\;$ is the cyclic group of order $\;p^2\;$ , and thus any "semi" direct product would in fact be direct, so direct sequences, split or not, are way too heavy and unnecessary for this. Since the only non-trivial subgroup of $\;G\;$ is the cyclic one of order $\;p\;$, any non-trivial direct product will have exponent at most $\;p\;$ and this is impossible.

Ever feel like this: “HELP. ME. My baby won’t nap unless being held & I feel like I’m going to die from exhaustion.”? Inside: Does having a baby who doesn’t nap make you frustrated? Tired? Short-tempered and behind on all household chores? Learn how to get baby to nap & get your sanity back. This post contains affiliate links. Learn more here. Have you ever felt that pang of envy? “How does he sleep?” a mom asked me, pointing at the restless sack of potatoes trying to jump out of my arms. My heart sank. I don’t want to complain. I don’t mean to complain. But I’m tired. So tired, worn down, and burned out from having a baby who refuses to nap, and demands my attention all day long. “Oh, he sleeps alright,” I answer, struggling to keep my son from wriggling his way off my hip. What followed was a brief conversation (lecture?) about how wonderful it is to have a baby that sleeps well, and that I should feel so lucky to have a good sleeper because “not all moms have it that easy.” Envy gripped my throat and made the words hard to escape when she went on to tell me how well her son sleeps – in fact, he sleeps 15 hours every night and still has 3 solid daytime naps with no issues. “Isn’t it great to have babies that sleep so well?” Walking away I tried to hold the tears back. I was tired, exhausted, running on nothing – not even fumes – and I didn’t know what I was doing wrong. I tried, really tried, to implement healthy sleep habits into my son’s life, so why did we struggle so much with naps? Why couldn’t he be a good sleeper? What was I doing wrong? The Unfortunate Side Effect of a Baby That Won’t Nap Because my son hates napping, he’s chronically overtired. When he gets overtired, he gets cranky. When he’s cranky, he needs to be held 100% of the time or he loses it. When I have to hold and entertain him all day, I get nothing done. When mama gets nothing done, mama gets uptight. So the real issue with my son’s naptime problems isn’t just that he’s grumpier because he’s lacking on sleep, it’s that he demands my attention 100% of the time and if I DO dare try to put him down for just a moment to change into a shirt that isn’t covered in yesterday’s spit up, he thinks the world is ending. It’s a crazy cycle, and these days it feels like it’s never going to end. “My Baby Won’t Nap!” + A Quick Fix for Parents It took me just about a year of struggling with baby sleep to learn this, but once I learned it everything changed… A baby’s brain is 25% of its adult size at birth. By 3 months, it has grown to 55% of its adult size. By 3 years, baby’s brain will be 80% of its adult size. Birth to age 3 sees the fastest rate of brain development in the entire human life span.MentalFloss.com This may sound like nothing more than a bunch of boring statistics, but what it says to me is this: Babies’ brains are growing rapidly, and the way a baby’s brain develops and grows is through play, exposure, and experience. Their brain is always turned “on”. Everything they come in contact with sends new sensations through their body and triggers their brain to connect what they just learned to their memory. Which means, since they’re constantly surrounded by new, exciting things, going from “on” to “off” is about as easy as riding a roller coaster then taking a nap. It just doesn’t happen. After you get off that roller coaster ride you’re wound up. Your body is hopped up on endorphins and telling it to “turn off” and take a nap just isn’t going to happen. It’s the same for babies. When they’re playing, their body is releasing endorphins and dopamine – and taking them from an environment where they are wound up and placing them into their crib expecting them to soundly drift off to sleep is next to impossible. They Are Too Wound up to Sleep. On top of constantly learning new things, babies are also incredibly social beings. Putting your 6-month-old down for a nap while their 6-year-old sister gets to stay up and watch Paw Patrol or gets to make Play-Doh pizzas just doesn’t seem fair, and they’ll make sure you know that. Their young brain is telling them to keep learning… but what they don’t know is that they need sleep to be able to continue learning. Sleep is such a vital part of brain development that sometimes it’s hard to find a healthy balance: You want your baby to stay up and play so they continue learning new things, but you also want them to take long naps and sleep through the night so they have healthy brain growth. Because your baby’s brain is always on the go, learning new things every second, it is crucial to practice a simple bedtime routine not only before putting your baby down for the night but also before putting them down for a nap. The naptime routine will help your baby predict what comes next. If the routine starts out with a diaper change in their bedroom with the curtains closed and putting on their sleep sack, eventually they will learn as soon as you take them into the bedroom to start anticipating what comes next: naptime. Using a naptime routine will remove your child from their stimulating surroundings and cue to their body that it’s time to switch off the go-go-go mentality and to cue their body to start releasing melatonin to help them fall asleep. >> If you have a particularly difficult sleeper, using melatonin supplements can help your child sleep better. Always talk to your doctor before giving your child any new supplements. *These supplements are recommended only for kids ages 3+. To make the naptime routine you choose really work, the key is to be consistent with it. Doing it only some days will confuse your baby and won’t help them learn how to predict what comes next. The best way to create a naptime routine for your baby is just like when you create a bedtime routine for them, although a naptime routine will typically be shorter. You can learn how to create a bedtime and naptime routine for your baby here. How to Get Baby to Nap It took a while, but eventually, I learned that some babies are just better sleepers than others. Sometimes, despite doing all the “best baby sleep practices”, sleep can still be difficult to come by at times. But I was determined to lessen the struggle and find something that helped at least a bit with our baby sleep problems. That’s when I learned the following 5 steps that have helped with our baby nap problems. 1. Understand the Science Behind the Way Babies Sleep Understanding a few key factors behind the way babies sleep will give you a better understanding of how babies sleep so that you can help your baby sleep better and will hopefully help solve your infant nap problems. - Understanding baby sleep cycles is important. All babies — and adults — have sleep cycles. Young babies will have sleep cycles of about 40 minutes long with them lengthening with age. Each sleep cycle is made up of active sleep and deep sleep. During the active sleep stage, baby will move around more, grunt or make noises, and will be much more likely to wake up from noises or movement that they sense. During the deep sleep period, the baby will typically be quiet and won’t stir. - Babies are easily overstimulated. Overstimulation is one of the leading causes of a baby who struggles to sleep. This is because as parents, we often don’t recognize the signs of overstimulation. Trying to put an overstimulated baby down for a nap is about as easy and pain-free as trying to stuff a rabid racoon into a cardboard box. - Babies have a 24-hour clock, but sometimes it needs to be tuned. Quite often in new babies, you’ll see daytime and nighttime confusion, where they think the daytime is for sleeping, and nighttime is for being awake and playing (and that’s when you need one of these so that you can lay on the couch while your little babe works all his energy out). Fine-tune your baby’s internal clock by exposing them to activities during the day, and keeping the house quiet, calm, and dark at night. Doing this will help baby get used to the idea that naps take place during the day, and the longer sleeping is for nighttime. “Sleep comes easier to some babies and harder to others, and if your baby struggles with sleep, it doesn’t necessarily mean that you’re doing anything wrong.” – Alice Callahan, PhD 2. Focus on Wake Times, Not Nap Length The biggest mistake I made when it came to my son’s naps were focusing so much on the length of his naps and not enough on the amount of time he was spending awake between his naps. This resulted in a very cranky and overtired baby because I was keeping him up for way too long. And when he got to this point of overtiredness, naps were a thing of the past. I set his nap schedule based on how long I thought his naps should be. His nap schedule looked something like this: - 7:00am: wake up - 9:00am: nap - 10:00am: wake up - 12:00pm: nap - 2:00pm: wake up - 3:30pm: nap - 4:30pm: wake up - 7:30pm: bedtime The problem with a schedule like the one above is that most days, things didn’t go as planned. Most days, he wouldn’t wake up right at 7 o’clock. And the problem was that even if he woke up earlier or later, for some reason I had it ground into my mind that I could not change the times he went for his nap, or I was a bad mom. So whether he woke up at 6:00 or 8:30 in the morning, I still (tried to) put him down for his nap at 9. This resulted in either a very overtired baby who fought sleep because he was past the point of exhaustion, or an under-tired baby who fought sleep because he had just woken up and wasn’t ready to go back to bed. But he needed to. I had created his nap schedule and we were sticking to it. That’s what everyone told me I had to do to have a good sleeper – set a schedule and don’t change it no matter what. This didn’t result in my son being a good sleeper. What it DID result in was him fighting naptime and me crying over a bucket of chocolate ice cream while I watched him on our baby monitor wondering what I was doing wrong. Now I know. I was focusing way too much on having the perfect nap schedule, and not enough on just paying attention to the amount of time he was spending awake. After learning about wake times and how to use them correctly, I felt like we had cracked the naptime code. Finally, I was able to put my son down for his nap when he was ready, instead of when I thought he should be ready. The trick with wake times is that they change periodically for each baby and at each stage. While one 6-month-old may be able to stay awake for 2.5 hours just fine, the next 6-month-old may only be able to make it to 1.75 hours before getting grumpy. Once you can find the sweet spot with your baby’s wake times and get them to bed right as they start to get tired, but before they are overtired, you’ll be able to have fuss-free nap times and bedtimes and you’ll even be able to dig into your secret chocolate stash without being interrupted by a crying baby. Download your free Wake Time Chart by Age here. 3. Stop Harmful Sleep Associations Sleep associations are anything that helps your baby fall asleep, but they’re not just for babies. Children and adults have their own sleep associations, too. This means that if your baby needs to be rocked to sleep in your arms and wakes up in the middle of the night – or naptime – in their crib, they will likely cry because they need their sleep association (being rocked) to fall back asleep. However, not all sleep associations are harmful. Harmful – or negative – sleep associations are those that require you to help your baby fall asleep. Positive sleep associations are ones that baby can do by himself to help himself fall back asleep without the help of a parent. Here’s what Traci Gleeson of Dream Team Baby has to say about negative sleep associations: “They’re not negative because they don’t work,” Traci says. “They’re negative because the baby associates that parent’s particular action with going to sleep. Then they wake up in the middle of the night and can’t figure out how to get back to sleep without it.” – Nanit Negative sleep associations require a parent to do any of the following actions to help baby fall asleep, which means when baby stirs every 40-minutes as his sleep cycle resets, there’s a greater chance of him waking up and not being able to fall back asleep without the help of a parent again… - Nursing baby to sleep - Bottle feeding baby to sleep - Rocking baby to sleep - Singing baby to sleep - Holding baby’s hand/rubbing their back until they fall asleep - Yoga balling baby to sleep (bouncing on one of these until baby falls asleep) - Laying beside baby until they’re asleep - Driving with baby to help them sleep - Singing lullabies until baby is asleep - Walking with them until they’re asleep - Pushing them in the stroller to help them sleep All the actions mentioned above are negative sleep associations because they require a parent or caregiver to help the baby fall asleep, which means when baby wakes during their sleep cycle, they won’t know how to put themselves back to sleep. If you find yourself doing any of the negative sleep associations mentioned above, try to wean them out and replace them with any of these positive sleep associations: - Give your baby a lovey to snuggle with – as long as they’re 1 year+ - Turn on a sound machine in their room (one like this that connects to your phone so you don’t have to go into the room to adjust it) 4. Stress is Contagious When you’re stressed and anxious, sleep is just about impossible. The same goes for babies, but babies don’t only struggle to sleep from their own stress, they can also sense when you’re stressed and struggle to sleep because of it. Try to be as calm and de-stressed as possible while getting your baby ready for their naps. Take your time, turn the lights down, and talk in a soothing voice to cue that it’s almost time to go to sleep. By doing this, and by leaving stress out of the bedroom, you’ll be helping your baby have a much easier time settling down and falling asleep, which brings me to the next step… 5. Create a Good Sleep Environment Creating a good sleep environment will help your baby fall asleep faster and stay asleep longer. A good sleep environment is when your baby knows what to expect next. - When the noisy toys are taken away and replaced with books. - When the TV is shut off and replaced with quiet play. - When the blinds are shut and the white noise is turned on. All these situations will help your baby slow down, calm down, and lay down. It will help their body go from go-go-go to ready for sleep. A good sleep environment will help your baby anticipate sleep, not fight it. When asked what the ideal sleep environment is, these were the three most common answers from sleep experts: A good sleep environment is, - Cool - Dark - Quiet Creating a good sleep environment is especially important for babies that only nap in arms. Learning how to get baby to nap in the crib during the day will be much easier if you keep their bedroom a quiet, calm place where they go to sleep – where the “good sleep environment” is always set. How to Use S-L-E-E-P to Help Your Baby Sleep Better I struggled with consistency when it came to my son’s nap times. I had a solid bedtime routine down pat, but I failed to realize that a naptime routine was just as important. I came up with a short acronym for the word SLEEP to help me remember everything I needed to do to get my son ready for bed… Sound – keep the noise level down and talk in a lower, quieter voice. Also, turn on the sound machine if applicable. Lightweight sack – keep the room temperature cool (61 – 68 degrees Fahrenheit is the ideal room temperature and put baby in a lightweight sleep sack. Environment – keep the room calm, dark, and quiet to set a good sleeping environment. Energy – is he too stimulated to sleep? Does he need more time to wind down? Is he showing signs of overstimulation? Pattern – keep a consistent bedtime routine before every nap. The way your baby sleeps now will impact their sleeping habits later on in life. A study found that the way a mother approached their baby’s sleep at 12 months also impacted how their children slept at four years old. “Multiple regression analysis revealed that 12 months maternal cognitions reflecting difficulties with limiting parental nighttime involvement were a statistically significant predictor of fragmented child’s sleep and of parental bedtime involvement at four years. More objective infant night-wakings at 12 months predicted lower sleep efficiency at four years.” – NCBI Help! My Baby Won’t Nap + Answers for You Before we “cracked the code” to our naptime problems — although some days ARE still a fight, but not nearly as many as used to be — I was drowning in questions… - Where should my baby sleep during the day? - How do I get my baby to sleep without being held? - Should my baby nap in the dark, or should it be light when he naps? - Where should my baby nap during the day? Here we’ll go over all the common questions that come along with baby nap problems. If your question isn’t answered, feel free to leave it in a comment. Should I make daytime sleep different from nighttime sleep for my baby? If your baby has daytime/nighttime confusion (they are awake at night but sleep most of the day away), then yes, you should make naptime different from bedtime. However, I got so caught up in doing this when I really didn’t need to be doing it, that it was ruining my son’s sleep. My son is a light sleeper. The slightest noise will cause him to wake up (he definitely doesn’t get that from his dad), so by trying so hard to make his naptimes different from his nighttime sleep, I was causing him to: - Sleep horribly - Have troubles falling asleep - Have short naps I tried to make his naps different from night time by having him sleep in his swing (though now I know letting babies sleep in swings in not recommended) or by moving his bassinet into the living room so he was sleeping in different surroundings than when it was nighttime. But, this meant I needed to sneak around the house as quietly as possible, and if I did make a noise, he would wake up. So use your judgment – if your baby isn’t struggling with daytime/nighttime confusion, there’s no problem with keeping naptimes and bedtimes similar. Where should my baby nap during the day? As previously mentioned, if you want to differentiate naptime from nighttime, you can move your baby’s bassinet into a different room or simply open up the curtains in his room to let the light in. But, if your babe doesn’t struggle with daytime/nighttime confusion, allowing them to nap in their room as they would at bedtime is no problem. In fact, letting them fall asleep in a dark room like they do at bedtime may have an advantage if your baby is struggling with naps, because the atmosphere and everything will remind them of when they sleep at night, which will – hopefully – encourage them to sleep better and longer at naptime. How long should my baby be awake between naps and how long should baby’s naps be? Focusing on wake times is far more important than focusing on nap length, and finding the sweet spot between not tired and overtired is key in successful nap times. You can grab the free wake times by age cheat-sheet here. You can grab the free wake times by age cheat-sheet here. Should babies nap in the dark? This depends on a few things – whether or not your baby struggles to sleep when it’s light out, how light of a sleeper baby is, and whether or not they have daytime and nighttime confusion. If your baby struggles to sleep when it’s light out, don’t try and force it just because some sleep expert told you to make your baby nap in the light. If it doesn’t work for your baby, it doesn’t work. And that’s okay. Each baby is different, which is why you want to find what works for your baby and keep doing it. Baby sleeps at night but won’t nap – what should I do? If your baby is sleeping good at night but nap times are a struggle, try to duplicate the nighttime sleep as much as possible. Try to incorporate your baby’s regular bedtime routine into the day before baby goes down for a nap. Make sure you’ve created a good sleeping environment for your baby by turning off screens at least half an hour before naptime, taking away noisy toys, and if possible sitting and reading to them, cuddling with them, or doing other non-stimulating activities. Winding down in the middle of the day can be difficult – especially for babies who don’t want to miss a thing. Work on making the half-hour before naptime as calm, quiet, and relaxing as possible to set the tone and prepare them for naptime. How to nap train baby Just like with sleep training baby for nighttime sleep, there are dozens of different sleep training methods from the Becoming BabyWise method to the Ferber method. Before you can start nap training your baby, you need to choose the sleep training method you will use: - Cry it out (CIO) or extinction - No-cry method - Chair method - Ferber method - Wake-and-sleep method - Fading method You can find a breakdown of each sleep training method here. Studies show that babies who are able to self-soothe at 12 months of age are less likely to have sleep-related problems at 2 years old. If your baby won’t nap unless being held, your best bet to turn your child into a good sleeper is choosing one of the mentioned sleep training methods and implementing it right away. How long should baby naps be? The length of time your baby should be napping will vary greatly from baby to baby. Some babies will nap for 2 to 3 hours at a time and take fewer naps a day, while other babies will nap for only 30 minutes at a time but nap more frequently throughout the day. Here’s a breakdown of how much sleep babies need during a 24-hour period, based on age: - 0 – 6 months: 14 to 17 hours of sleep. - 6 – 12 months: 14 to 15 hours of sleep. Typically, babies will transition from 3 to 2 naps around 6 – 9 months old. Then, the transition from 2 naps down to 1 happens around 12 – 24 months old. It will vary greatly from one baby to the next. How long is a baby’s sleep cycle & what are the stages of baby sleep? Depending on age, a baby’s sleep cycle is typically 40 minutes long. Through those 40 minutes, your baby will go through 5 different stages of sleep. La Lune Consulting says it best: - Stage 1 – Very light sleep. Your baby’s eyes are heavy and start to close. They appear drowsy and start to dose-off. - Stage 2 – Light sleep. Your baby appears to be falling asleep yet can be startled awake very easily. This is why your baby wakes immediately upon laying them down in their crib after being “rocked to sleep” even though they seemingly appear fast asleep. - Stage 3 – Deep sleep. Your baby’s brain waves start to slow as they slip into a deep sleep. - Stage 4 – Deepest sleep. This is where restoration of muscles, tissue repair, immune system building, and all other development related benefits kick in. - Stage 5 – REM sleep. Your baby is now in a light but active state of sleep. REM sleep is where dreaming occurs, so their brains are active but their bodies are still in a paralyzed state. “The completion of all of the stages is considered one sleep cycle. While it is not always consistent, sometimes babies (and adults too) start with stages 1-4 then bounce back to 3, or 2, eventually they get to stage 5. At the completion of stage 5, either your baby wakes up or the cycle starts all over again at stage 1. As the night goes on, REM sleep becomes longer and longer.” – La Lune Consulting Does TV light affect baby sleep? Miranda Hitti says that a new study has been done that links watching TV to irregular bedtimes and naptimes in kids under 3 years old. It is recommended time and time again that children under 2 years old do not watch any TV as this is such a critical time for brain development, and watching TV or playing on a phone doesn’t teach the real-life skills kids need. .” – Source The blue light let off by screens, whether that be TV, cellphones, or computers, hinders the production of melatonin, the “sleepy hormone”. The hindrance of melatonin makes falling asleep – and staying asleep – much more difficult for babies, older children, and adults alike. My Baby Won’t Nap… Your Turn Have you struggled with baby sleep in the past? Do you have a baby that doesn’t nap? What tips to make your baby nap have you picked up along the way? Feel free to share them in a comment below. Download Your Free Cheat Sheet to Ditch the sleep-deprivation and finally feel like yourself again. - Download the free cheat sheet: Baby Sleep Cheat Sheet: Rock your baby’s sleep by optimizing their wake times, creating a fail-proof routine, and learning how to calm them before they go to bed.. >> Learn how to help my baby sleep today Helpful Articles: This is the Perfect Bedtime Routine to Help Any Kid Sleep Better How to be a Good Mom & Stop Being a Drag So very glad I found this article. My baby’s naps are…. sporadic. Some days she’ll take 2 2-hour naps, others she won’t nap at all and she’ll scream all day. When I told an acquaintance I was having trouble getting her to nap, she told me “She doesn’t need a nap. YOU need her to nap.” Ummm no. It’s been 40 years since you had a baby. Spend some time with my over exhausted 3 month old and come back to me. 🤣 I am currently struggling with my 9 month baby girl going through a major sleep regression and am so glad I found your article! It was very thorough and provided some very helpful tips I am going to try! Thank you!

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TITLE: First Course in Linear algebra books that start with basic algebra? QUESTION [5 upvotes]: I'm 30 years old, and the only math I can remember from college is basic algebra and some probabilities. Next month, I have a machine learning project I'd like to work on, but I'll need a solid footing in linear algebra first. Are there any books or tutorials that can take me from the spotty math knowledge I have now to linear algebra? Your help would be appreciated. REPLY [1 votes]: Use Kolman/Hill. The math is really easier than calculus. If you remember systems of 2 or 3 equations and unknowns and a little bit of vectors, that is sort of the mindset you need. Anyhow, it starts from the very basics and is gentle. Is perfect for what you need, will support the machine learning class.

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TITLE: Image of a W*-Algebra under an injective *-homomorphism QUESTION [2 upvotes]: Proposition 1.16.2 of Sakai's "$\mathcal{C}^\ast$-Algebras and $\mathcal{W}^\ast$-Algebras" states Let $\phi$ be a weakly-$\ast$ continuous $\ast$-homomorphism of a $\mathcal{W}^*$-algebra $\mathcal{M}$ (i.e. a $\mathcal{C}^*$-algebra that is isometrically isomorphic to the topological dual of a Banach space) into another $\mathcal{W}^*$-algebra $\mathcal{N}$. Then the image $\phi(\mathcal{M})$ is closed with respect to the weak-* topology on $\mathcal{N}$. However, consider the following: Let $\mathcal{M}$ be a $\mathcal{W}^*$-algebra, and let $(\pi,\mathcal{H})$ be its universal representation. Then $\pi: \mathcal{M} \to \mathcal{B}(\mathcal{H})$ is an injective $\ast$-homomorphism. In particular, $\pi(\mathcal{M})$ is also a $\mathcal{W}^*$-algebra. On the other hand, all $\ast$-isomorphisms between $\mathcal{W}^*$-algebras are normal (i.e. suprema-preserving) and therefore weakly-$\ast$ continuous by $\mathcal{W}^\ast$-theory. Using the above proposition, this would imply that $\pi(\mathcal{M})$ is already weakly-$\ast$ closed, i.e. it coincides with the enveloping von Neumann algebra. However, this clearly contradicts the Sherman–Takeda theorem, which states that the enveloping von Neumann algebra can identified with the double dual of $\mathcal{M}$, where the latter is strictly larger than $\mathcal{M}$ if $\mathcal{M}$ has infinite-dimension. Where is the mistake in my reasoning? REPLY [3 votes]: The mistake in the reasoning is that the normality occurs on the image and not in the codomain. So you cannot conclude that it is continuous with the ambient weak$^*$-topology. For example, consider a non-type I von Neumann algebra $R$, say the hyperfinite II$_1$ for instance. As any other C$^*$-algebra, $R$ has an irreducible representation $\rho$. Because $R$ is simple, $\rho$ is faithful. So $\rho(R)\subset B(H_\rho)$ is weak$^*$-dense, which implies that $\rho(R)$ is not closed. Precisely because of this is that people in the past used to make the distinction between a $W^*$-algebra (as you defined), which is abstract, and a von Neumann algebra. These days (almost?) no one cares because if you start with a $W^*$-algebra you take the normal universal representation (i.e., the sum of the GNS of the normal states) and you get a bona-fide von Neumann algebra.

Your organization is somewhere on this ladder, and you will benefit by moving up. Content curation is a powerful approach to making your association more valuable than ever to members whose needs have changed dramatically. Start by identifying where you are today, and then move gradually up the maturity ladder. Even if you haven’t been focused on content curation, you are still probably gathering and sharing industry news with your members—an effort likely led by your communications or publications team. That’s the first rung on the ladder. The next rung is to be more selective about what you share, formalizing the process for making those choices. The third level is to allow members to choose what they receive. This involves new skillsets on your association’s part, better technology used to deliver this content (whether through an e-newsletter or web pages), and members’ willingness to actually make those selections. At the next step, more powerful technology, likely driven by artificial intelligence, will deliver content based on members’ actual behavior as well as your decisions; this is an easier effort for the audience, so it will likely be received well and have high engagement. Level 5 on the ladder is adding a true editorial voice, providing context that only your association can. This is a time commitment for your staff, but will result in an endless stream of useful, relevant content. The next rung up from that is to refine your organization’s voice and ensure that the voice of your curated content is consistent with the organization’s overall messaging and tone. Bringing the curated content into the fold of your organization’s content means a much more powerful experience for members. And you may find that more departments embrace the approach of curating content. The very top spot is reserved for associations that add transparency to their curated content, showing members how they selected the content they share. To spot when and how to proceed: - Use the analytics from your existing communications - Conduct member survey about what kinds of information your members want Members’ information needs are changing quickly, so keep a close eye on your analytics, and revisit the approach every year. Your association’s goal should be to provide value to your audiences by helping them make sense of information and their environment. Rather than being only writers, the communications department has to become the organization’s team of managing editors – setting the schedules, developing and enforcing style, making sure that the right people are disseminating and receiving the right information. They need to be editors-in-chief, planning, assigning and overseeing rather than doing. For more depth on this topic, download the whitepaper on content curation that I recently co-authored withElizabeth Engel, CAE, of Spark Consulting.

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TITLE: When do you use binomial and poisson distribution? How to compute for big factorials? QUESTION [0 upvotes]: Player 1 scored 10 points in a hockey game. Player 2 has the most points of any hockey player in history, averaging 2.629 points/game in the 394 regular season games played in 5 seasons '81-'82 to '85-'86. Computer the probability that Player 2 had at least 10 points in at least one of the 394 games. Textbook Solution: Poisson Distribution Formula: $$ {\rm P}(Y \geq 10) = \sum\limits_{y = 10}^\infty {\frac{{e^{ - \mu } \mu^y }}{{y!}}} =1-\sum\limits_{y = 0}^9 {\frac{{e^{ - \mu } \mu^y }}{{y!}}} = 1-0.9996=0.0004. $$ Binomial Distribution Formula: $$ {\rm P}(X \geq 1) = 1-{\rm P}(X=0) = 1-C^{394}_0 (.0004)^0 (1-.0004)^{394} = 1-0.8542=0.1458. $$ The first step is to use poisson distribution formula since the mean is given to find at least 10 points in a game in one of the 394 games played (space). I don't understand why we need to further use the binomial distribution formula when the probability of one game having at least 10 points is 0.0004? Also...how do I calculate big factorials for the binomial formula? I seem to get an error. I looked it up and I think Stirling’s formula can be used. This isn't covered in the course content and unsure if there is an easier way to calculate big factorials... REPLY [0 votes]: The question is asking for the probability that Player 2 has at least one game in which he scored at least $10$ points, among the $394$ games played. If you knew the probability that he would score at least $10$ points for any single game, then it's clear that the desired probability would be found through a binomial distribution calculation, just like if I asked you, if you roll a die $394$ times, what is the probability you would obtain at least one six--you'd need to know the probability of obtaining a six for a single roll. But you don't know the single-game probability of Player 2 scoring at least $10$ goals; instead, you are given the average goals per game, which is $2.629$. That means you have to use another model, one that applies to a single game's performance. To this end, we treat this number as an average rate of goals per game, and the random number of goals is modeled by a Poisson distribution, thus letting us compute the single-game probability of scoring at least $10$ goals. Regarding your second question, you do not actually need to compute such a large factorial. Since $\Pr[X = 0] = \binom{394}{0} (0.0004)^0 (1 - 0.0004)^{394 - 0}$, we can see that $$\binom{394}{0} = \frac{394!}{0! (394-0)!} = \frac{394!}{394!} = 1.$$ It's only when $n$ is large and $x$ (and $n-x$) are not small that the binomial coefficient $\binom{n}{x}$ is difficult to calculate exactly; e.g., $\binom{394}{2} = \frac{(394)(393)}{2}$ is still easy, but $\binom{394}{157}$ is not. In such a case, Stirling's approximation is appropriate.

Global Standards for Social Work Education and Training The Global Minimum Qualifying Standards Committee was set up as a joint initiative of the International Association of Schools of Social Work (IASSW) and the International Federation of Social Workers (IFSW) at the joint IASSW/IFSW Conference in Montreal, Canada in July 2000. Please click here to view the latest Global Standards for Social Work Education and Training document The final version of the Global Standards for Social Work Education and Training was adopted by IASSW and IFSW at their General Assemblies in Adelaide, Australia in October 2004. Vishanthie Sewpaul who was Chair of the Global Standards Committee writes: We are indebted to our international colleagues for their responsiveness and engagement in making the Global Standards for Social Work Education and Training possible. We are particularly indebted to all those colleagues who translated the document into several languages. The idea of Global Standards was conceived well before my entry into it through the visionary leadership of persons such as Lena Dominelli who was Chair of what was then called the Global Qualifying Standards Committee from January 2000 until January 2001, when I was appointed as the chair. On first hearing about the possibility of formulating global standards for social work education and training I was appalled by what I thought to be a far too presumptuous and ambitious project. I immediately questioned its potential to reinforce Western imperialism and hegemonic discourses and expressed my reservation about getting engaged in such as process. I was told that as I was aware of the complexities of such an initiative I would be well suited to approach it with the kinds of sensitivities that was required. I entered the terrain by beginning dialogue with members of the Global Standards Committee and with as many colleagues across the world as possible. I initially asked colleagues what they thought about the idea of developing global standards, what might be its potential advantages and disadvantages, what should constitute the contents of such a document, should it materialize. To my surprise, I found that the majority of colleagues were in favour of developing global standards. Their recommendation, I thought, was ‘a tall order’ that such a document that details certain universals be sufficiently flexible to be applicable to any context and allow for interpretations of locally specific social work education and practice. Having obtained the mandate to continue with such an initiative, on the input of the Committee and colleagues, a review of available national and regional standards and a review of literature a first draft was produced in January 2002. Various consultative processes, all of which are detailed in the Global Standards document, and several reviews later culminated in the document that was adopted at the General Assemblies of IASSW and IFSW in Adelaide in October 2004, with the proviso that the concerns of social pedagogues be incorporated into in the document, with the social pedagogues providing the language to embrace their concerns. When the social pedagogues provided such a language they insisted that all reference to “social work” should read as the “social work profession.” Thus, the final document refers to: “Global Standards for the Education and Training of the Social Work Profession.” Despite the flaws inherent in the process of representation the document, which has been developed through an inclusive a process as possible, does represent, to the best of our ability, the views of IASSW and IFSW membership. While the vision of global standards was initially conceived by IASSW and IFSW leadership, its substance was determined by a broad constituency. The document is not intended to be a finite, static end product and in the interests of deepening our commitment to social justice, human rights, inclusivity, international dialogue and responsiveness to service users we have to consistently question the value of what we are doing and how we are doing it. Thus, there is a call for our colleagues across the globe to critically engage with the document, assess its relevance for their particular historical, socio-economic, political and cultural contexts and engage in cross national and cross regional dialogue about social work education and practice. The Global Standards have stimulated a great deal of debate as seen in the number of publications related to it. See for example, the special themed issue on Global Standards in Social Work Education, Volume 23, No. 5, October 2004 and the International Journal of Social Welfare, Volume14, No. 3, July 2005. Interesting debates continue. For example, I was recently asked to write a response to a paper written by two UK colleagues for the International Journal of Social Welfare regarding the applicability of the international definition of social work and the global standards to the Chinese context.

TITLE: ODE standard form. QUESTION [3 upvotes]: I noticed that whenever mathematicians talk about Legendre polynomials they bring the ODE to the form $(1-x^2)f''(x)-2xf'(x)+n(n+1)f(x)=0$. When solving Poisson's equation, this form is not the most intuitive one, because you only get this one after substituting $x=\cos(t)$. My question is: Is there a reason Mathematicians always tend to have an ODE in a particular form and if yes: What determines this standard form? If anything is unclear, please let me know. REPLY [0 votes]: whenever mathematicians talk about Legendre polynomials they bring the ODE to the form ... in which its solutions are polynomials. Quite a natural thing to do. Yes, the polynomials are of most interest on $[-1,1]$ and yes, the composition $P_n(\cos t)$ is what comes up from separating the 3D Laplacian in coordinates. But the algebraic polynomials are considered the simplest, most basic of all functions. At least I find $\frac12(3x^2-1)$ easier to think about than $\frac12(3\cos^2t-1)$. Legendre polynomials have important orthogonality relation $\int_{-1}^1 P_n(x)P_m(x)\,dx =0 $ ($m\ne n$) which becomes something a bit more complicated if you plug in $x=\cos t$. This orthogonality is useful for polynomial approximation in $L^2[-1,1]$. And the solutions of $P_n(x)=0$ are the sample points of Gauss-Legendre quadrature, a widely used numerical integration algorithm. These are reasons to look at Legendre polynomials as algebraic polynomials, not as trigonometric ones.

\begin{document} \title{Binary trees, coproducts, and integrable systems} \author{B Erbe$^1$ and H J Schmidt$^2$} \address{$^1$Department of Physics, University of Regensburg, Regensburg, Germany} \ead{bjoern.erbe@physik.uni-regensburg.de} \address{$^2$Department of Physics, University of Osnabrueck, Osnabrueck, Germany} \ead{hschmidt@uos.de} \begin{abstract} We provide a unified framework for the treatment of special integrable systems which we propose to call ``generalized mean field systems". Thereby previous results on integrable classical and quantum systems are generalized. Following Ballesteros and Ragnisco, the framework consists of a unital algebra with brackets, a Casimir element, and a coproduct which can be lifted to higher tensor products. The coupling scheme of the iterated tensor product is encoded in a binary tree. The theory is exemplified by the case of a spin octahedron. The relation to other generalizations of the coalgebra approach is discussed. \end{abstract} \pacs{02.30.Ik, 75.10.Jm 20.00, 75.10.Hk } \vspace{2pc} This is an author-created, un-copyedited version of an article accepted for publication in Journal of Physics A: Mathematical and Theoretical. IOP Publishing Ltd is not responsible for any errors or omissions in this version of the manuscript or any version derived from it. The definitive publisher authenticated version is available online at 10.1088/1751-8113/43/8/085215. \maketitle \section{Introduction}\label{sec:I} In classical mechanics ``complete integrability" can be precisely defined in terms of the Arnol'd-Liouville theorem \cite{A78}. The corresponding generalization of this concept to quantum theory has not yet been achieved. Nevertheless, there exists a rich literature on integrable quantum systems under various headlines such as Yang-Baxter equations \cite{Bax82}, algebraic Bethe ansatz \cite{Fad95} and quantum groups \cite{Dri87}. Aside from this mainstream of research there are different theories of integrable systems which could be characterized as ``generalized mean field systems" (GMFS) \cite{BR98}, \cite{SS09}. It is the aim of the present article to provide a general framework for the description of such systems.\\ The prototype of the systems in question is a spin system where the spins are (Heisenberg-) coupled to each other with equal strength. It turns out that each spin will move exactly as if it would be under the influence of a uniform magnetic field. This justifies the above characterization as (generalized) ``mean field systems". The first generalizing step would be to consider systems which consist of uniformly coupled integrable subsystems. This property can be recursively applied. The underlying sequence of partial uniform couplings is most conveniently encoded in a binary tree, the leaves of which correspond to the smallest constituents of the system, see \cite{SS09}. For example, the uniform coupling of three pairs of spins can be described by the tree of figure \ref{fig0} and gives rise to an integrable spin octahedron, see figure \ref{fig4}.\\ Another generalization into a different direction is based upon the insight that at the core a GMFS consists of a unital algebra $\A$ with a bracket $[,\}$ and a co-multiplication $\Delta$, which can be lifted to tensor products of $\A$ and applied to a Casimir (central element) $c\in\A$, see \cite{BR98}. One then considers representations of $\A$ generated by certain commutation relations where the bracket $[,\}$ will either be represented by a Poisson bracket (classical case) or by the commutator of matrices (quantum case).\\ In our paper we simplify and generalize the approaches of \cite{BR98} and \cite{SS09}. Thereby the restriction to Heisenberg spin systems in \cite{SS09} is abolished by encorporating the coalgebra ansatz of \cite{BR98}. Vice versa, the theory of \cite{BR98} will be reformulated by using the language of binary trees, and generalized from ``homogeneous trees" to general ones. We also found that the postulate in \cite{BR98} of $\Delta$ being ``co-associative" is superfluous, but see Appendix A. After the first publication of the coalgebra approach \cite{BR98} various generalizations have been proposed, see \cite{BMR02}, \cite{BHMR04}, \cite{BB08}, \cite{M09a} and \cite{M09b}. We will comment on the relation of our approach to these generalization at the appropriate places in the paper and in two appendices. The obvious generalization of assuming several Casimir elements instead of a single one will be neglected here.\\ The paper is organized as follows. In section \ref{sec:T} we collect some definitions concerning binary trees which are needed later. Section \ref{sec:C} is devoted to the algebraic prerequisites including the coproduct $\Delta$ and its lift $\Delta^T$ to higher tensor products given by a binary tree $T$. In section \ref{sec:M} we apply these tools to the theory of integrable systems and prove the main result, theorem $1$, which is analogous to prop.~$1$ of \cite{BR98} and provides a number of commuting observables which is in many examples sufficient to guarantee complete integrability. In section \ref{sec:E} we discuss the elementary example of a Heisenberg spin octahedron in order to illustrate the application of the abstract theory. Some remarks of the corresponding Gaudin spin system and on the connection to other approaches follow. Two appendices on the issues of superintegrability and the recent loop coproduct approach close the paper. \section{Trees\label{sec:T}} We consider finite, binary trees $T$, shortly called ``trees". Recall that these consist of a set of ``nodes" $\mathcal{N }(T)$, such that all nodes $n\in\mathcal{N }(T)$, except the ``leaves" $\ell\in\mathcal{L}(T)$, are connected to exactly two ``children" $c_1(n),c_2(n)$, mixing the metaphors of horticulture and genealogy. We have to distinguish between the ``left child" $c_1$ and the ``right child" $c_2$. Due to this distinction, the leaves of a binary tree can be arranged in a natural order from left to right and hence be labelled by ``$\ell_1$" to ``$\ell_L$". All nodes, except the ``root" $r(T)$, are children of other nodes. By definition, different nodes have different children, see figure \ref{fig0}. As a tree, $T$ is a connected graph without cycles.\\ \begin{figure} \includegraphics[width=8cm]{FIG0} \caption{\label{fig0}Example of a binary tree with root $r$, six leaves $\ell_1,\ldots,\ell_6$ and further nodes $\alpha,\beta,\gamma,\delta$.} \end{figure} The simplest tree $\bullet$ consists of only one root. The next simplest one ${\sf V}$ has three nodes, that is, one root and two leaves. If $T_1$ and $T_2$ are (disjoint) trees, then ${\sf V}(T_1,T_2)$ will be the tree obtained by identifying the leaves of ${\sf V}$ with the roots $r(T_1)$ and $r(T_2)$, see figure \ref{fig1}. \begin{figure} \includegraphics[width=8cm]{FIG1} \caption{\label{fig1}Union ${\sf V}(T_1,T_2)$ of two binary trees $T_1$ and $T_2$.} \end{figure} Obviously, each tree can be obtained from copies of $\bullet$ by recursively applying the operation ${\sf V}(T_1,T_2)$. This opens the possibility to provide recursive definitions and proofs in the theory of trees. A tree $T$ will be called ``homogeneous" if it is of the form \begin{eqnarray}\label{T1a} T&=&{\sf V}(\ldots({\sf V}({\sf V}(\bullet,\bullet),\bullet),\ldots,\bullet)\\ \label{T1b} \mbox{or } T&=& {\sf V}(\bullet,{\sf V}(\bullet,\ldots ,V(\bullet,\bullet)\ldots) \;, \end{eqnarray} see, for example, figure \ref{fig2}. The tree of figure \ref{fig0} is not homogeneous.\\ \begin{figure} \includegraphics[width=8cm]{FIG2} \caption{\label{fig2}Examples of ``homogeneous" binary trees.} \end{figure} Binary trees are used in various parts of physics, e.~g.~in chaos theory \cite{KK09}, computational physics \cite{WP02}, or in the theory of spin networks \cite{TG09}. Here we utilize these structures for encoding the coupling schemes of certain integrable spin systems, similarly as in \cite{FPJ94}, \cite{RJ09} and \cite{SS09}.\\ The following lemma can be easily proved. \begin{lemma}\label{L1} $N(T)\equiv|\mathcal{N}(T)|=2|\mathcal{L}(T)|-1\equiv 2L(T)-1\;.$ \end{lemma} A sub-tree $S\subset T$ is given by a subset of nodes of $T$, which, according to their connections inherited from $T$ again form a tree. For example, if $n\in \mathcal{N}(T)$ then $T(n)$ will denote the maximal sub-tree of $T$ with the root $n$. Let $\mathcal{L}(n)\equiv \mathcal{L}(T(n))$. If $n,m\in\mathcal{N}(T)$, then either $\mathcal{L}(n)\subset\mathcal{L}(m)$ or $\mathcal{L}(m)\subset\mathcal{L}(n)$ , or $\mathcal{L}(n)\cap\mathcal{L}(m)=\emptyset$ . In the former two cases $m$ and $n$ will be called ``connected", in the latter case ``disjoint".\\ If $T_1$ and $T_2$ are (disjoint) trees and $\ell\in\mathcal{L}(T_1)$, then $T=T_1\circ_\ell T_2$ will denote the tree obtained by ``grafting", i.~e.~by identifying the root $r(T_2)$ with the leave $\ell$ of $T_1$, see figure \ref{fig3}. \begin{figure} \includegraphics[width=8cm]{FIG3} \caption{\label{fig3}Binary tree $T=T_1\circ_\ell T_2$ obtained by grafting $T_2$ on $T_1$.} \end{figure} \section{Coproducts\label{sec:C}} In this paper we often will consider the classical and the quantum case simultaneously. In both cases, the physical observables are obtained by suitable representations of an abstract unital algebra $(\mathcal{A},e)$ and its tensor products. In the quantum case, $\mathcal{A}$ will be an associative, non-abelian algebra with commutator $[a,b]=ab-ba,\; a,b\in\mathcal{A}$, and its physical representation is given in terms of finite-dimensional matrices. Typical examples are cases where $\mathcal{A}$ is defined as the universal enveloping algebra of some semi-simple Lie algebra. In the classical case, $\mathcal{A}$ will be an abelian algebra together with an abstract Poisson bracket $\{,\}$, see \cite{BR98}. Representations of $\mathcal{A}$ are then given by the algebra of smooth functions of some phase space together with the usual Poisson bracket.\\ To cover both cases, the commutator/Poisson bracket will be denoted by $[a,b\},\; a,b\in\mathcal{A}$. It makes $(\mathcal{A},[,\})$ into a Lie algebra and will act as an derivation on the associative product on $\mathcal{A}$. We will always consider algebras endowed with a bracket of one of these two kinds and the corresponding homomorphisms, that is, linear algebra homomorphism w.~r.~t.~both multiplications.\\ If $\mathcal{A}_1, \mathcal{A}_2$ are two algebras as explained above, then $\mathcal{A}_1\otimes \mathcal{A}_2$ will denote the algebraic tensor product, physically describing a composite system. It will be again a unital algebra with brackets upon linearly extending the definitions \begin{equation}\label{C1} (a\otimes b)(c\otimes d)=(ac)\otimes(bd) \end{equation} and \begin{equation}\label{C2} [(a\otimes b),(c\otimes d)\}=[a,c\}\otimes\frac{bd+db}{2}+ \frac{ac+ca}{2}\otimes [b,d\} \;. \end{equation} If $A:\mathcal{A}_1\longrightarrow\mathcal{A}_1$ and $B:\mathcal{A}_2\longrightarrow\mathcal{A}_2$ are morphisms as explained above, then also $A\otimes B:\mathcal{A}_1\otimes \mathcal{A}_2\longrightarrow\mathcal{A}_1\otimes \mathcal{A}_2$ will be such a morphism. \\ The ``coproduct" $\Delta$ will be a morphism \begin{equation}\label{C3} \Delta:\mathcal{A}\longrightarrow\mathcal{A}\otimes \mathcal{A} \;, \end{equation} that is, a linear algebra morphism plus a Poisson bracket morphism in the classical case. Usually, a coproduct is additionally required to be ``co-associative", see \cite{BR98}, but this property will not be needed in the main part of the present paper, hence here we use the term ``coproduct" in a more general sense. In Appendix A, co-associativity and co-commutativity of $\Delta$ will be assumed to extend the set of integrals of motion (superintegrability).\\ The crucial construction for integrability, as considered here, is the lift of the coproduct to higher order tensor products given by a tree $T$. To this end we first define $\mathcal{A}^T$ recursively by \begin{eqnarray}\label{C4a} \mathcal{A}^\bullet &=& \mathcal{A}\\ \label{C4b} \mathcal{A}^{{\sf V}(T_1,T_2)} &=& \mathcal{A}^{T_1}\otimes \mathcal{A}^{T_2} \;. \end{eqnarray} Sometimes it will be convenient to use the identification \begin{eqnarray}\label{C5} \mathcal{A}^{T_1} &=& \mathcal{A}^{T_2} =\underbrace{\mathcal{A}\otimes\ldots\otimes\mathcal{A}}_{L\mbox{ times}} \;, \end{eqnarray} if $L(T_1)=L(T_2)\equiv L$. W.~r.~t.~this identification the canonical embedding \begin{equation}\label{C6} j_n:\mathcal{A}^{T(n)}\longrightarrow \mathcal{A}^T,\quad n\in\mathcal{N}(T) \end{equation} can be defined by \begin{equation}\label{C7} j_n(a)=e\otimes\ldots \otimes a\otimes\ldots \otimes e,\quad a\in\mathcal{A}^{T(n)}\;. \end{equation} In the next step we define the lift of the coproduct $\Delta^T:\mathcal{A}\longrightarrow\mathcal{A}^T$ recursively by \begin{eqnarray}\label{C8a} \Delta^\bullet&=&\mbox{id}_{\A} \\ \label{C8b} \Delta^{{\sf V}(T_1,T_2)}&=& (\Delta^{T_1}\otimes\Delta^{T_2})\circ \Delta \;, \end{eqnarray} and conclude: \begin{lemma}\label{L2} $\Delta^T:\mathcal{A}\longrightarrow\mathcal{A}^T$ is a (Poisson) algebra morphism. \end{lemma} \noindent{\em Proof.} By induction over $T$. The claim follows since the tensor product and the composition of (Poisson) algebra morphisms is again a (Poisson) algebra morphism. \hfill$\Box$\\ Before formulating the main result we still need another definition. Let $n\in\mathcal{N}(T)$, then \begin{equation}\label{C9} \Delta^n\equiv j_n\circ \Delta^{T(n)}:\mathcal{A}\longrightarrow\mathcal{A}^T \;. \end{equation} We note that the generalization of the coalgebra approach to comodule algebras \cite{BMR02} where $\Delta^n$ is replaced by a suitable map ${\mathcal A}\longrightarrow {\mathcal A}\otimes{\mathcal B}\otimes\ldots\otimes{\mathcal B}$ is only possible for homogeneous trees.\\ \section{Integrable systems\label{sec:M}} Remember the grafting of trees $T=T_1\circ_\ell T_2$ explained in section \ref{sec:T}. It gives rise to a corresponding composition of lifted coproducts in the following sense: \begin{lemma}\label{L3} Let $T=T_1\circ_\ell T_2$, $x\in\mathcal{A}$ and write $\Delta^{T_1}(x)=\sum_i x_{i,\ell_1}\otimes\ldots x_{i,\ell_\mu}\otimes\ldots\otimes x_{i,\ell_{L_1}}$, where $\ell_1,\ldots,\ell_{L_1}$ denote the leaves of $T_1$ and $\ell_\mu=\ell$. Then $\Delta^T(x)=\sum_i x_{i,\ell_1}\otimes\ldots \Delta^{T_2}(x_{i,\ell_\mu})\otimes\ldots\otimes x_{i,\ell_{L_1}}$. \end{lemma} \noindent{\em Proof.} By induction over $T_1$. \hfill$\Box$\\ Further one can easily prove the following lemma: \begin{lemma}\label{L4} Let $n,m\in\mathcal{N}(T)$ be disjoint, i.~e.~$\mathcal{L}(n)\cap\mathcal{L}(m)=\emptyset$ and $x,y\in\mathcal{A}$. Then $[\Delta^n(x),\Delta^m(y)\}=0$. \end{lemma} \noindent{\em Proof.} $\Delta^n(x)$ and $\Delta^m(y)$ live in disjoint factors of the tensor product $\mathcal{A}^T$, since they are of the form $\Delta^n(x)=\sum_i e\otimes\ldots\bigotimes_{\ell\in\mathcal{L}(n)} x_{i,\ell} \otimes\ldots\otimes e$ and $\Delta^m(y) =\sum_j e \otimes \ldots \bigotimes_{\ell\in\mathcal{L}(m)} y_{j,\ell} \otimes\ldots\otimes e$ . Hence they commute. \hfill$\Box$\\ Now we are ready to formulate the main result. \begin{theorem}\label{M0} Let $n,m\in\mathcal{N}(T)$ such that $\mathcal{L}(m)\subset\mathcal{L}(n)$, $x\in\mathcal{A}$ and $c\in\mathcal{A}$ be a central element, i.e. $[y,c\}=0$ for all $y \in \mathcal{A}$. Then $[\Delta^n(x),\Delta^m(c)\}=0$. \end{theorem} \noindent{\em Proof.} We introduce the canonical partial embedding $j_m^n:\mathcal{A}^{T(m)}\longrightarrow\mathcal{A}^{T(n)}$ such that $j_m=j_n\circ j_m^n$. It follows that \begin{eqnarray} [\Delta^n(x),\Delta^m(c)\} &=& [j_n\circ\Delta^{T(n)}(x),j_m\circ \Delta^{T(m)}(c)\}\\ \label{M1a} &=&j_n\left( [\Delta^{T(n)}(x),j_m^n\circ \Delta^{T(m)}(c)\} \right)\label{M1b} \end{eqnarray} and thus it suffices to show that \begin{equation}\label{M2} [\Delta^{T(n)}(x),j_m^n\circ \Delta^{T(m)}(c)\}=0 \;. \end{equation} By applying lemma \ref{L3} to the sub-tree $T(n)$ we write \begin{equation}\label{M3} T(n)=T_1 \circ_m T(m),\quad m=\ell_\mu\in\mathcal{L}(T_1) \end{equation} and conclude \begin{eqnarray} \Delta^{T(n)}(x) &=& \sum_i x_{i,1}\otimes\ldots \Delta^{T(m)}(x_{i,\mu})\otimes\ldots x_{i,L_1}\;.\label{M4} \end{eqnarray} Hence \begin{eqnarray} [\Delta^{T(n)}(x),j_m^n\circ \Delta^{T(m)}(c)\} \nonumber &=& \sum_i x_{i,1}\otimes\ldots [\Delta^{T(m)}(x_{i,\mu}),\Delta^{T(m)}(c)\}\otimes\ldots x_{i,L_1}\\ \nonumber &=& \sum_i x_{i,1}\otimes\ldots \Delta^{T(m)}\left([x_{i,\mu},c\}\right)\otimes\ldots x_{i,L_1}\\ \label{M5} &=&0\;, \end{eqnarray} since $[x_{i,\mu},c\}=0$. \hfill$\Box$\\ This theorem generalizes prop.~$1$ of \cite{BR98} to arbitrary, not necessarily homogeneous trees. In order to guarantee complete integrability in the sense of the Arnol'd-Liouville theorem for $2L$-dimensional phase spaces (which is satisfied for spin systems, see section \ref{sec:E}) we would need $L$ pairwise commuting observables (``integrals in involution"). These are provided by the $\Delta^n(c)$ for each node $n$ which is not a leave. By lemma 1, there are exactly $N(T)-L(T)=L(T)-1$ such nodes. In \cite{BR98} the remaining observable is chosen as the Hamiltonian $H$. In the context of quantum spin systems another choice would be more appropriate, namely $\Delta^r(x)$ with a suitable $x\in{\mathcal A}$. The Hamiltonian $H$ could then be chosen as any element of the algebra generated by the $\Delta^n(c)$ and $\Delta^r(x)$.\\ In the general case the dimension of the phase space depends on the symplectic realization of $({\mathcal A},\{,\})$ and the choice of the symplectic leaves. A thorough discussion of these questions, which also applies to the binary tree approach, including issues of superintegrability can be found in \cite{BB08}. \\ \section{Examples and Outlook\label{sec:E}} In order to explain the application of theorem $1$ to integrability of quantum systems we consider the elementary example of a spin octahedron, figure \ref{fig4}, with Heisenberg Hamiltonian, following \cite{SS09}. We chose as $\A$ the universal enveloping algebra of the Lie algebra $su(2)$. More concretely: We consider three generators $X_1,X_2,X_3$ satisfying the abstract commutations relations \begin{equation}\label{M6} [X_j,X_k]=i \sum_{\ell=1}^3 \epsilon_{jk\ell} \, X_\ell \;, \end{equation} where $\epsilon_{jk\ell}$ denotes the completely anti-symmetric Levi-Civita symbol. $\A$ is the set of all finite polynomials $X$ of the standard form \begin{equation}\label{M7} X=\sum_{klm}c_{klm}\,X_1^k\,X_2^l\,X_3^m \;. \end{equation} The product $XY$ of two such polynomials is brought into the standard form (\ref{M7}) by successively applying the commutation relations (\ref{M6}). The unit element in $\A$ is $e=X_1^0\,X_2^0\,X_3^0$. It follows that $c\equiv X_1^2+X_2^2+X_3^2$ commutes with all $X\in\A$.\\ The coproduct $\Delta$ is defined on the generators by \begin{equation}\label{M8} \Delta(X_i)=e\otimes X_i+X_i\otimes e \end{equation} and then extended to general elements of the form (\ref{M7}) by employing the property of $\Delta$ being an algebra homomorphism. Thus, for example, \begin{eqnarray} \Delta(c)&=& \Delta(X_1^2+X_2^2+X_3^2)\\ &=& \Delta(X_1)^2+\Delta(X_2)^2+\Delta(X_3)^2\\ &=& \sum_{i=1}^3 (e\otimes X_i+X_i\otimes e)^2\\ &=& e\otimes c + c\otimes e + 2\sum_{i=1}^3 X_i\otimes X_i \;. \end{eqnarray} \begin{figure} \includegraphics[width=8cm]{FIG4} \caption{\label{fig4}The octahedral spin graph corresponding to the integrable Heisenberg Hamiltonian (\ref{M9}). Its coupling scheme is encoded in the binary tree of figure \ref{fig0} as explained in the text.} \end{figure} We further choose $T$ as the binary tree of figure \ref{fig0} and obtain the corresponding various commutations relations of theorem $1$ being valid in the $6$-fold tensor product $\bigotimes_{i=1}^6 \A$ where one usually chooses $\Delta^r(x)=\Delta^r(X_3)$. Next we consider the well-known $(2s+1)$-dimensional irreducible matrix representation of (\ref{M6}) and denote the representations of the generators $X_i$ by $\mathbf{S}_i$ (``spin operator components"). In the $6$-fold tensor product we denote the single spin components by $\mathbf{S}_i^\mu,\,\mu=1,\ldots,6$. In this representation, $c=s(s+1)\mathbf{1}$ and all commutation relations of theorem $1$ remain valid. Note that $\Delta(c)$ becomes $\left(\mathbf{S}^{(1)}+ \mathbf{S}^{(2)}\right)^2$, which is no longer a constant, analogously for higher tensor products $\Delta^n(c)$. This shows, by the way, why it is advantageous to work in an abstract setting and to consider concrete representations only after the coproduct is defined. Note further, that we could slightly generalize the example by considering different $s$ for each factor of the tensor product.\\ Let $(\mathcal{V},\mathcal{E})$ be the the octahedral spin graph of figure \ref{fig4} with its set of $6$ vertices $\mathcal{V}$ and the set of $12$ edges $\mathcal{E}$. The corresponding Heisenberg Hamiltonian $H$ can be written in various ways: \begin{eqnarray}\label{M9} &&H= 2 J \sum_{(\mu,\nu)\in\mathcal{E}}\mathbf{S}^{(\mu)}\cdot\mathbf{S}^{(\nu)}\\ \nonumber &&= J\left( \left(\sum_{\mu\in\mathcal{V}} \mathbf{S}^{(\mu)}\right)^2 -\left(\mathbf{S}^{(1)}+\mathbf{S}^{(2)}\right)^2 -\left(\mathbf{S}^{(3)}+\mathbf{S}^{(4)}\right)^2 -\left(\mathbf{S}^{(5)}+\mathbf{S}^{(6)}\right)^2 \right)\\ &&\\ &&=J\left( \Delta^r(c)-\Delta^\alpha(c)-\Delta^\beta(c)-\Delta^\gamma(c) \right) \;, \end{eqnarray} where the root $r$ and the nodes $\alpha,\beta,\gamma$ refer to the binary tree of figure \ref{fig0}. $J$ is some appropriate coupling constant. It is crucial that $H$ can be written as a linear combination of commuting observables according to theorem $1$. In this respect the octahedral Heisenberg Hamiltonian (\ref{M9}) is only the simplest case; for example, a Zeeman term proportional to $\Delta^r(\mathbf{S}_3)$ could be added without loosing integrability. The eigenvalues and common eigenvectors of the system of commuting observables $\Delta^n(c),\,n\in\mathcal{N}(T),$ result from the well-known rules of coupling angular momenta involving Clebsch-Gordan coefficients. An explicit formula for the eigenvalues and eigenvectors of $H$ and arbitrary binary trees has been given in \cite{SS09}. The example of the spin octahedron clearly shows the physical meaning of the binary tree $T$ on which theorem $1$ depends: $T$ encodes the coupling scheme of systems which are completely integrable due to their structure of uniformly coupled subsystems.\\ With exactly the same algebraic considerations and the same tree as above, theorem \ref{M0} provides us with another very interesting integrable model: \begin{eqnarray} \label{Gaudin} \nonumber H&=&A\left(\mathbf{S}^{(1)}\cdot \sum_{i=3}^6 \mathbf{S}^{(i)} + \mathbf{S}^{(2)}\cdot \sum_{i=3}^6 \mathbf{S}^{(i)}\right) + 2 (A+J)\mathbf{S}^{(1)} \cdot \mathbf{S}^{(2)} \\ &=& A\left(\Delta^r (c)-\Delta^{\delta}(c)\right)+J \Delta^{\alpha}(c), \end{eqnarray} The (Gaudin) Hamiltonian $H$ describes a central spin system with two central spins of exchange $2 (A+J)$, coupled homogenously to a bath of four spins. Such a system can serve for example as a simplified model for the hyperfine interaction in a double quantum dot, see \cite{SKL03}.\\ Apart from this physical meaning, it is interesting from a formal point of view. Besides the approach presented in this paper, systems can be integrable in the sense of algebraic Bethe ansatz. According to the ground breaking work of V.~Drinfel´d \cite{Dri87}, this is based on quasicocommutative bialgebras, which essentially means that there is an element $R \in \A \otimes \A$ with \begin{equation} \left(\tau \circ \Delta\right)(x) \cdot R=R \cdot \Delta(x) \end{equation} for all $x\in \A$. $\tau$ denotes the switch operator defined by linearly extending $\tau(a\otimes b)=b\otimes a$ and $\Delta$ a usual coproduct. \newline As this algebraic structure is somewhat similar to the one presented in this paper, the question arises, whether there is a connection between systems integrable in either sense. The above system, in contrast to the central spin system with one central spin, is not integrable by means of algebraic Bethe ansatz. Hence adding a second central spin destroys the Bethe ansatz, whereas integrability in the sense of theorem \ref{M0} remains unaffected.\\ Recently, a framework for integrability using so-called ``loop coproducts" has been proposed \cite{M09a}, \cite{M09b} which contains different previous approaches to integrability as special cases. It is, however, confined to the classical case. Some remarks on the relation between this approach and the present paper are included in the appendix B. \section*{Acknowledgements} We are indebted to a referee for hints to the literature concerning generalizations of \cite{BR98} and to Roman Schnalle for references on coupling trees. \appendix \section{Superintegrability} The approach \cite{BR98} to integrability via coalgebras has subsequently be extended to ``superintegrability" \cite{BHMR04}, \cite{BB08}. This roughly means that one is seeking for additional integrals of motion which, however, do not longer commute with the old ones. ``Additional" means, in the classical case, that the new integrals of motion are functionally independent of the old ones. Typically, this functional independence cannot be shown in the general setting, but only in concrete examples, see \cite{BHMR04}, \cite{BB08}. Also in theses references the role of co-associativity of $\Delta$ in connection to superintegrability has been stressed.\\ The question arises, whether these ideas can be transferred to the more general situation where the binary trees are not necessarily of homogeneous type. To this end we slightly refine the binary tree construct, following, for example, \cite{FPJ94}. Recall that, due to the distinction between ``left child" and ``right child", the leaves of a binary tree can be arranged in a natural order from left to right and hence be labeled by ``$\ell_1$" to ``$\ell_L$". Now assume that this labeling can be arbitrarily permuted. We will call the resulting structure a \underline{coupling tree}. For example, the only binary tree with three nodes, {\sf V}, gives rise to two different coupling trees, denoted by ${\sf V}(\ell_1,\ell_2)$ and ${\sf V}(\ell_2,\ell_1)$. Generally, a coupling tree $\tilde{T}$ can be represented as a pair $\tilde{T}=(T,\pi)$, where $T$ is a binary tree with $L$ leaves and $\pi\in{\mathcal S}_L$, a permutation of $L$ elements. A coupling tree with $L$ leaves can alternatively be construed as a monomic expression in the abstract variables $\ell_1,\ldots,\ell_L$, such that each variable occurs exactly once. For example, the tree of figure \ref{fig0}, conceived as a coupling tree, corresponds to the expression $(\ell_1\,\ell_2)((\ell_3\,\ell_4)(\ell_5\,\ell_6))$.\\ Most definitions and propositions of the sections \ref{sec:T} and \ref{sec:C} can be taken over directly or with minor modifications. It will be appropriate to reserve the union and grafting operations, see figures \ref{fig1} and \ref{fig3}, to binary trees, and to obtain the corresponding coupling trees by adducing a suitable permutation of the leaves, as explained above. Note, that the definition of ${\mathcal A}^T$ remains unchanged. We will extend the definitions (\ref{C4a}) and (\ref{C4b}) to coupling trees by \begin{equation}\label{A1.1} \Delta^{\tilde{T}}=\tilde{\pi}\circ\Delta^{T} \;, \end{equation} where $\tilde{T}=(T,\pi)$ and $\tilde{\pi}:{\mathcal A}^T\longrightarrow {\mathcal A}^T$ denotes the natural representation of $\pi$ by a permutation of factors of the tensor product. Following \cite{FPJ94} we consider two operations on coupling trees, namely \begin{eqnarray}\label{A1.2a} \mbox{exchange:}&\quad& {\sf V}(\ell_1,\ell_2) \rightleftarrows {\sf V}(\ell_2,\ell_1)\\ \label{A1.2b} \mbox{flop:}&\quad& {\sf V}({\sf V}(\ell_1,\ell_2),\ell_3) \rightleftarrows {\sf V}(\ell_1,{\sf V}(\ell_2,\ell_3))) \;, \end{eqnarray} see figure \ref{fig5}. We have the following \begin{proposition}\label{P1} Let $T_1$ and $T_2$ be two coupling trees with ${\mathcal L}(T_1)={\mathcal L}(T_2)$. Then $T_1$ can be transformed into $T_2$ by a finite sequence of exchanges and flops operating on subtrees. \end{proposition} We will skip the proof which is lengthy but straight forward. Note that, in the language of monomials, the proposition says that any two monomials with the variables $\ell_1,\ldots,\ell_L$ occurring exactly once can be transformed into each other by applying the rules of commutativity and associativity of the multiplication.\\ \begin{figure} \includegraphics[width=10cm]{FIG5} \caption{\label{fig5}Illustration of the operations ``exchange" and ``flop" on coupling trees.} \end{figure} We will say that the coproduct $\Delta:{\mathcal A}\longrightarrow{\mathcal A}\otimes {\mathcal A}$ is \underline{co-commutative} iff $\Delta^{{\sf V}(\ell_1,\ell_2)}=\Delta^{{\sf V}(\ell_2,\ell_1)}$ and \underline{co-associative} iff $\Delta^{{\sf V}({\sf V}(\ell_1,\ell_2),\ell_3)}=\Delta^{{\sf V}(\ell_1,{\sf V}(\ell_2,\ell_3))}$. Note that the coproduct defined in (\ref{M8}) is co-commutative as well as co-associative. Obviously, $\Delta$ is co-commutative and co-associative iff $\Delta^T$ is invariant under exchanges and flops operating on sub-trees of $T$. Together with proposition \ref{P1} we obtain: \begin{proposition}\label{P2} If $\Delta $ is co-commutative and co-associative and ${\mathcal L}(T_1)={\mathcal L}(T_2)$ then $\Delta^{T_1}=\Delta^{T_2}$. Moreover, if $n_1\in{\mathcal N}(T_1)$ and $n_2\in{\mathcal N}(T_2)$ such that ${\mathcal L}(n_1)={\mathcal L}(n_2)$ then $\Delta^{n_1}=\Delta^{n_2}$. \end{proposition} Now let ${\mathcal L}(T_1)={\mathcal L}(T_2)$ and consider the involutive sub-algebra ${\mathcal C}_1\subset {\mathcal A}^{T_1}$ generated by the elements $\Delta^n(c),\; n\in {\mathcal N}(T_1)$ and $\Delta^{r_1}(x)$, analogously for ${\mathcal C}_2\subset {\mathcal A}^{T_2}$, see section \ref{sec:M}. Define \begin{equation}\label{A1.3} CN(T_1,T_2)=\{(n_1,n_2)|n_1\in {\mathcal N}(T_1),\; n_2\in {\mathcal N}(T_2),\;{\mathcal L}(n_1)={\mathcal L}(n_2)\} \;, \end{equation} and ${\mathcal C}_{12}$ as the sub-algebra of ${\mathcal C}_1\cap{\mathcal C}_2$ generated by the elements $\Delta^{n_1}(c),\;\Delta^{r_1}(x)$ or, equivalently, by the $\Delta^{n_2}(c),\;\Delta^{r_2}(x)$, where $(n_1,n_2)$ runs through $CN(T_1,T_2)$. Then we conclude the main result of this appendix: \begin{theorem}\label{T3} Let $\Delta$ be co-commutative and co-associative and $H\in{\mathcal C}_{12}$, then $[H,K\}=0$ for all $K\in{\mathcal C}_1\cup{\mathcal C}_2$. \end{theorem} The scenario for superintegrability considered in \cite{BHMR04}, \cite{BB08} results as a special case of theorem \ref{T3} in the following sense. Let $T_1$ be the ``left-homogeneous tree" and $T_2$ the ``right-homogeneous tree" represented in figure \ref{fig2}. Then $CN(T_1,T_2)=\{(r_1,r_2)\}$ and ${\mathcal C}_{12}$ is the algebra generated by $\Delta^{r_1}(x)=\Delta^{r_2}(x)$. Note that in this case $T_1$ can be transformed into $T_2$ using only flops operating on sub-trees, hence the assumption of $\Delta$ being co-commutative will be superfluous. \section{Loop coproducts} Recently, a framework for integrability using so-called ``loop coproducts" has been proposed by F.~Musso \cite{M09a}, \cite{M09b} which contains different previous approaches to integrability as special cases, namely the coalgebra approach \cite{BR98}, the linear r-matrix formulation and formulations using Sklyanin or reflection algebras. It is, however, confined to the classical case. Nevertheless, one may ask whether, in the classical case, the loop coproduct approach also includes the generalization of the coalgebra approach we have given in this paper.\\ At first glance, the answer seems to be ``no", since the corresponding derivation in \cite{M09a} of the coalgebra as a special case utilizes the co-associativity of $\Delta$, which is not needed in our theorem \ref{M0}. Here we neglect the differences due to the assumption in \cite{M09a} and \cite{M09b} that the algebra ${\mathcal A}$ has a finite number of generators. A closer inspection, however, reveals that co-associativity is not necessary.\\ The loop coproduct approach \cite{M09b} is based on a family of maps $\Delta^{(k)}:{\mathcal A}\longrightarrow {\mathcal B},\;k=1,\ldots,m$ and postulates different properties of these maps for the cases $i<k$ (or $k<i$) and $i=k$. For comparison we have to set ${\mathcal B}={\mathcal A}^T$. In our approach, the set of nodes is only partially ordered by the definition $i\prec k$ iff ${\mathcal L}(i)\subset{\mathcal L}(k)$. However, $\prec$ can be extended to a linear order $<$ such that $i<k$ implies ${\mathcal L}(i)\subset{\mathcal L}(k)$. For $i<k$ we have either ${\mathcal L}(i)\cap {\mathcal L}(k)=\emptyset$ or ${\mathcal L}(i)\subset{\mathcal L}(k)$. In the first case $[\Delta^i(x),\Delta^k(y)\}=0$ for all $x\in{\mathcal A}$ due to lemma \ref{L4}. In the second case lemma \ref{L3} implies \begin{equation}\label{A2.1} [\Delta^i(x),\Delta^k(y)\}=\sum_j f_j\;[\Delta^i(x_j),\Delta^i(y)\} \end{equation} for all $x,y\in{\mathcal A}$ and some suitable $f_j\in{\mathcal B},\;x_j\in{\mathcal A}$. Hence, in both cases condition ($4$) of \cite{M09b} is satisfied. If $i=k$, condition ($5$) of \cite{M09b} follows since $\Delta^i$ is a (Poisson) algebra homomorphism in our theory.\\ We conclude that, in the case of classical mechanics and up to minor differences in the formulations, the loop coproduct theory \cite{M09a,M09b} contains the binary tree approach as a special case. Nevertheless, the binary tree approach has, to our opinion, its virtues as a constructive method particularly adapted to quantum spin systems. \section*{References}

\begin{document} \maketitle \tableofcontents \begin{abstract} The character theory of symmetric groups, and the theory of symmetric functions, both make use of the combinatorics of Young tableaux, such as the Robinson-Schensted algorithm, Sch\"utzenberger's {\em ``jeu de taquin''}, and evacuation. In 1995 Poirier and the second author introduced some algebraic structures, different from the plactic monoid, which induce some products and coproducts of tableaux, with homomorphisms. Their starting point are the two dual Hopf algebras of permutations, introduced by the authors in 1995. In 2006 Aguiar and Sottile studied in more detail the Hopf algebra of permutations: among other things, they introduce a new basis, by M\"obius inversion in the poset of weak order, that allows them to describe the primitive elements of the Hopf algebra of permutations. In the present Note, by a similar method, we determine the primitive elements of the Poirier-Reutenauer algebra of tableaux, using a partial order on tableaux defined by Taskin. \end{abstract} \section{Introduction} In 1995 the authors of the present paper introduced two dual Hopf algebra structures on permutations \cite{MR}. The products and coproducts of permutations originated from the concatenation Hopf algebra and shuffle Hopf algebra on $\mathbb Z\langle \mathbb N^{>0} \rangle$, the module generated by words of positive integers, from Solomon's descent algebra \cite{Sol} and Gessel's (internal) coalgebra \cite{G} of quasi--symmetric functions. The two Hopf structures on $\mathbb Z S$, the module with $\mathbb Z$--basis all the permutations in $S=\cup_{n\geq 0} S_n$, for $S_n$ the symmetric group on $\{1,\ldots,n\}$, are autodual \cite{BS}. In the same year \cite{PR}, carrying on these themes, Poirier and the second author proved that the two dual Hopf structures on $\mathbb Z S$ are free associative algebras. By restriction on the plactic classes they obtained two dual structures of Hopf algebras on the $\mathbb Z$--module $\mathbb Z S$ with basis the set $T$ of all standard Young tableaux. The product and coproduct are described there in term of Sch\"utzenberger's {\em ``jeu de taquin''} \cite{LS}. They also provided different morphisms between these structures and the descent algebras, symmetric functions and quasi-symmetric functions. In particular, the map sending a permutation into its left tableau in the Schensted correspondence is a Hopf morphism. Loday and Ronco \cite{LR} characterized the product of two permutations by the use of the weak order of permutations: it is the sum of all permutations in some interval for this order. In 2005 \cite{AS}, Aguiar and Sottile studied in further and thorough details the structure of the Hopf algebras of permutations, giving explicit formulas for its antipode, proving that it is a cofree algebra and determining its primitive elements. For the latter task, they introduced a new basis of $\mathbb Z S$, related to the basis of permutations via M\"obius inversion in the poset of the Bruhat weak order of the symmetric groups. In \cite{DHNT}, Duchamp, Hivert, Novelli and Thibon studied the Hopf algebra of permutations (denoted there $FQSYM$), and gave among others a faithful representation by noncommutative polynomials. The Hopf algebra of tableaux was used by J\"ollenbeck \cite{J}, and Blessenohl and Schocker \cite{BS}, to define their noncommutative character theory of the symmetric group. Moreover, Muge Taskin \cite{T} used the order on tableaux, induced by the weak order of permutations, to characterize the product of two tableaux, in a way reminiscent of the result of Loday and Ronco. The purpose of this Note is to find the primitive elements of $\mathbb Z T$, the Hopf algebra of tableaux with respect to the product and coproduct, following the approach of Aguiar and Sottile. A new basis for $\mathbb Z T$ is obtained by M\"obius inversion for the Taskin order of tableaux. The nice feature of the proofs here is that we manage to avoid {\em ``jeu de taquin''}, and use a simpler description through a shifted left and right concatenation product of tableaux. \section{Preliminaries on permutations} We denote by $S_n$ symmetric group on $\{1,\ldots,n\}$. We often represent permutations as words: $\sigma\in S_n$ is represented as the word $\sigma(1)\sigma(2)\cdots\sigma(n)$. By abuse of notation, we identify $\sigma$ and the corresponding word. A {\em word} in the sequel will always be on the alphabet of positive integers, also called {\em letters}. We denote by $|\sigma|$ the number of letters of $\sigma$. We denote by $\leq$ the {\em right weak order} of permutations. Recall that it is defined as the reflexive and transitive closure of the relation $u < v$, $u,v\in S_n$, $v=u\circ\tau$, for some adjacent transposition $\tau\in S_n$ such that $l(u)<l(v)$, where $l(u)$ denotes as usual the {\em length} of $u$ in the sense of Coxeter groups. Recall that this order may also be defined by comparing inversions sets: let $Inv(\sigma)$ be the set of inversions (`by values') of $\sigma$, that is, the set of pairs $(j,i)$ with $j>i$ and $\sigma^{-1}(j)<\sigma^{-1}(i)$; then $u\leq v$ if and only if $Inv(u)\subseteq Inv(v)$. Note that $(j,i)$ is an inversion of $\sigma$ if and only if $j>i$ and $j$ appears on the left of $i$ in the word representing $\sigma$. This definition applies also to any word. Given $\sigma \in S_n$, and a subset $I$ of $[n]$, $\sigma\mid I$ denotes the word obtained by removing in the word $\sigma$ the digits not in $I$ (whereas the restriction of $\sigma$ to $I$ is the subword $\sigma\mid_I$ of the images of the letters in $I$). For example, for $\sigma=2517643$, one has $2517643\mid\{2,3,6\}=263$ (while $\sigma\mid_{\{2,3,6\}}=514$). Moreover, $v$ being a word without repetition of letters, $st(v)$ denotes the {\em standard permutation} of the word $v$, obtained by replacing each letter in $v$ by its image under the unique increasing bijection form the set of letters in $v$ onto $\{1,2,\ldots,|v|\}$. For example, $st(5713)=3412$. \begin{lemma}\label{restr-order-perm} If $s\leq t$ are permutations in $S_n$ and $I$ an interval in $\{1,\ldots,n\}$, then $st(s\mid I)\leq st(t\mid I)$. \end{lemma} \begin{proof} Let $(j,i)$ be an inversion of $s\mid I$; then $i,j\in I$. Then $(j,i)$ is an inversion of $s$, hence of $t$. It follows that it is also an inversion of $t\mid I$. Since standardizing amounts to apply an increasing bijection, we deduce the lemma. \end{proof} Let $S=\bigcup_{n\geq 0} S_n$ be the disjoint union of all symmetric groups. There is a classical associative product on $S$, denoted by $\square$, which turns it into a free monoid \cite{PR}: let $u\in S_p$ and $v\in S_q$; let $\bar v$ be obtained by adding $p$ to each digit in $v$; then $u\square v$ is the concatenation $u\bar v$ of $u$ and $\bar v$ (or {\em right shifted concatenation}). For example, $231\square 12=23145$, with here $p=3$. The free generators of this free monoid are the {\em indecomposable} permutations, which have some importance in algebraic combinatorics; see \cite{C}. A variant of this product is as follows: given two permutations as above, $v\bigtriangleup u=\bar vu$ (which we refer to as {\em left shifted concatenation}). Example: $12\bigtriangleup 231=45231$. Clearly, the product $\square$ and the opposite product of $\tr$ are conjugate under the mapping $w\mapsto \tilde w$ which reverses words: $$ v\tr u=\widetilde{\tilde u\square \tilde v}. $$ It follows that $S$ with the product $\tr$ is also a free monoid, freely generated by the permutations which are indecomposable for this product, which are the reversals of the indecomposable permutations. For later use, we need \begin{lemma}\label{perm-comp} The weak order $\leq$ on permutations is compatible with the product $\tr$: $u\leq u', v\leq v'\Rightarrow v\tr u\leq v'\tr u'$. \end{lemma} \begin{proof} Let $u\in S_p$, $v\in S_q$. Suppose that $u\leq u'$, $v\leq v'$. We show that $v\tr u\leq v'\tr u'$. Let $(j,i)$ be an inversion of $v\tr u=\bar vu$. If $i,j\leq p$, then by construction of the product $\tr$, $(j,i)$ is an inversion of $u$, hence of $u'$ (since $u\leq u'$), hence of $v'\tr u'$. If $i,j> p$, then $(j-p,i-p)$ is an inversion of $v$, hence of $v'$ (since $v\leq v'$) and therefore $(j,i)$ is an inversion of $v'\tr u'$. If $i\leq p$ and $j>p$, then $(j,i)$ is an inversion of $v'\tr u'$. There is no other case, since $i<j$. \end{proof} \section{Hopf algebra of permutations}\label{ZS} Denote by $\mathbb Z S$ be the free $\mathbb Z$-module with basis $S$. We define on $\mathbb Z S$ a product, denoted by $*$ (called {\em destandardized concatenation}), and a coproduct (called {\em standardized unshuffling}), denoted by $\delta$, which turn it into a Hopf algebra (see \cite{MR}). If $ \alpha\in S_p$, $\beta\in S_q$, $\alpha * \beta$ is the sum of all permutations in $S_{p+q}$ of the form $uv$ (concatenation of $u$ and $v$), where $u,v$ are of respective lengths $p,q$ and $st(u)=\alpha$, $st(v)=\beta$; for example, $12*21=1243+1342+1432+2341+2431+3421$. Moreover, for $\sigma\in S_n$, $$ \delta(\sigma)=\sum_{0\leq i\leq n} \sigma\mid\{1,\ldots,i\}\otimes st( \sigma\mid\{i+1,\ldots,n\}).$$ An example of coproduct is $\delta(3124)=\epsilon\otimes st(3124)+1\otimes st(324)+12\otimes st(34)+312\otimes st(4)+3124\otimes \epsilon=\epsilon\otimes 3124 + 1\otimes 213+12\otimes 12+312\otimes 1+3124\otimes \epsilon$. Here $\epsilon$ is the empty permutation in $S_0$, the neutral element of the bialgebra $\mathbb Z S$. Following Aguiar and Sottile \cite{AS}, we define a new linear basis $\mathcal M_\sigma$ of $\mathbb Z S$, indexed by permutations, and called {\em monomial basis}. (Notice that Aguiar and Sottile deal with the isomorphic dual Hopf algebra, the isomorphism being $\sigma\mapsto\sigma^{-1}$, and also that they use the left weak order.) These elements are given by the formula $$ \sigma=\sum_{\sigma\leq w}\mathcal M_w,$$ which defines them recursively, via M\"obius inversion formula, since $\leq $ is an order. The following result is equivalent to a result due to Aguiar and Sottile (\cite{AS} Theorem 3.1). \begin{theorem}\label{AS} For any permutation $\sigma$, one has $$\delta (\mathcal M_\sigma)=\sum_{\sigma=v\tr u}\mathcal M_u\otimes \mathcal M_v.$$ \end{theorem} We give below the proof of this result; it may help to understand the proof of the similar result on the Hopf algebra of tableaux, which we give in Section \ref{prim}. We begin by a lemma, that will also have an analogue for tableaux. \begin{lemma}\label{sigma-a-b} Let $n=p+q$, $\sigma\in S_n$, $a=\sigma\mid\{1,\ldots,p\}$, $b=st(\sigma\mid\{p+1,\ldots,n\}$. Then, for $v\in S_q$, $u\in S_p$, $\sigma\leq v\tr u$ if and only if $a\leq u$ and $b\leq v$. \end{lemma} \begin{proof} 1. We show first that $\sigma\leq b\tr a$. We have $b\tr a=\overline{b}a$ (concatenation of words), where it is easily seen that $\overline{b}=\sigma\mid\{p+1,\ldots,n\}$ (add $p$ to each letter of $b$). Let $(j,i)$ be an inversion of $\sigma$. Then $i<j$ and $j$ is at the left of $i$ in $\sigma$. If $j,i\leq p$, then $(j,i)$ is an inversion of $a$, and therefore also of $b\tr a$. If $j,i>p$, then $(j,i)$ is an inversion of $\overline{b}$, hence of $b\tr a$. If $i\leq p$ and $j>p$, then $(j,i)$ is an inversion of $b\tr a$, since in this latter permutation, each letter $>p$ is at the left of each letter $\leq p$. There is no other case since $j>i$. Thus $(j,i)$ is in each case an inversion of $b\tr a$, hence $\sigma\leq b\tr a$. 2. Suppose that $a\leq u$ and $b\leq v$. Then clearly $b\tr a\leq v\tr u$, by Lemma \ref{perm-comp}. Hence by 1, $\sigma\leq v\tr u$. 3. Suppose that $\sigma\leq v\tr u$. If $(j,i)$ is an inversion of $a$, then it is an inversion of $\sigma$, hence of $v\tr u=\bar vu$; therefore it is an inversion of $u$ since $i,j\leq p$; thus $a\leq u$. Similarly, $b\leq v$. \end{proof} \begin{proof}[Proof of Theorem \ref{AS}] Define the $\Z$-linear mapping $\delta_1:\Z S\mapsto \Z S\otimes \Z S$ by $\delta_1(\mathcal M_\sigma)=\sum_{\sigma=v\tr u}\mathcal M_u\otimes \mathcal M_v$. It is enough to show that $\delta_1=\delta$. We have for any permutation $\sigma\in S_n$, $$ \delta_1(\sigma)=\delta_1(\sum_{\sigma\leq w}\mathcal M_w)= \sum_{\sigma\leq w}\sum_{w=v\tr u}\mathcal M_u\otimes \mathcal M_v= \sum_{\sigma\leq v\tr u}\mathcal M_u\otimes \mathcal M_v.$$ This is by Lemma \ref{sigma-a-b} equal to $$\sum_{0\leq i\leq n} \sum_{\sigma\mid\{1,\ldots,i\}\leq u \atop st(\sigma\mid\{i+1,\ldots,n\}\leq v} \mathcal M_u\otimes \mathcal M_v$$ $$=\sum_{0\leq i\leq n}(\sum_{\sigma\mid\{1,\ldots,i\} \leq u}\mathcal M_u)\otimes (\sum_{st(\sigma\mid\{i+1,\ldots,n\})\leq v} \mathcal M_v)$$ $$=\sum_{0\leq i\leq n} \sigma\mid\{1,\ldots,i\} \otimes st(\sigma\mid\{i+1,\ldots,n\}=\delta(\sigma),$$ by the definition of the basis $\mathcal M_\sigma$. \end{proof} In order to understand the equivalence between the previous theorem and the theorem of Aguiar and Sottile, it may be useful to use the notion of global descents, introduced by them. Recall that according to Aguiar and Sottile (\cite{AS}, Definition 2.12), a {\em global descent} of $\sigma\in S_n$ is a position $i\in\{1,\ldots,n-1\}$ such that for any $j\leq i$ and any $k>i$ one has $ \sigma(j)>\sigma(k)$. Then the permutations that are indecomposable for the product $\tr$ (which are the free generators of the free monoid $S$) are those which have no global descents. Moreover, if $\sigma=\sigma_1\tr\ldots\tr \sigma_k$ with indecomposable $\sigma_i$'s, then the global descents of $\sigma$ are the positions $|\sigma_1|, |\sigma_1|+| \sigma_2|,\ldots,|\sigma_1|+\ldots |\sigma_{k-1}|$. For example, $78465213=12\tr 132 \tr 213$ has 2 and 5 as global descents, and $12,132,213$ are indecomposable for $\tr$, equivalently, have no global descents. \begin{corollary} The submodule of primitive elements of $\Z S$ is spanned by the $\mathcal M_\sigma$ such that $\sigma$ has no global descent, or equivalently, $\sigma$ is indecomposable for the product $\tr$. \end{corollary} \section{Preliminaries on tableaux} For unreferenced results quoted here, see \cite{S}. Denote by $T_n$ denotes the set of standard Young tableaux (we say simply {\em tableaux}) whose entries are $1,\ldots,n$. We denote by $(P(\sigma),Q(\sigma))$ the pair of tableaux associated with $\sigma\in S_n$ by the Schensted correspondence. Let $T=\cup_{n\geq 0}T_n$ be the set of all standard tableaux. The {\em plactic equivalence} on $T$ is the smallest equivalence relation generated by the {\em Knuth relations} $xjiky\sim_{plax}xjkiy$, $xikjy\sim_{plax}xkijy$, $i<j<k$, for $i,j,k\in \mathbb{N}$ and $x,y \in \mathbb{N}^*$. By Knuth's theorem, one has $P(\sigma)=P(\tau)$ if and only if $\sigma\sim_{plax}\tau$. Thus we may identify tableaux and plactic classes. In the sequel, we use systematically this identification. Following Taskin \cite{T} , we define the {\em weak order} of tableaux as follows: let $U,V\in T_n$; let $u,v\in S_n$ be such that $P(u)=U, P(v)=V$. Define the relation $U\leq V$ if $u\leq v$; then $\leq$ is the transitive closure of this relation. In other words, the weak order on tableaux is the smallest order on $T$ such that the mapping $P:S\to T$ is increasing for the weak order. This order was introduced by Melnikov \cite{M} (called there {\em Duflo order}) and Taskin \cite{T} (denoted there $\leq_{weak}$). The difficulty here is to show that it is indeed an order. For two tableaux $A,U$, one has $A\leq U$ if and only if there exist $n\geq 1$ and permutations $\alpha_0,\ldots,\alpha_{n-1},\beta_1,\ldots,\beta_n$ such that \begin{equation}\label{<} P(\alpha_0)=A, \alpha_0\leq \beta_1\sim\alpha_1\leq\beta_2\sim\ldots \alpha_{n-1}\leq\beta_n,P(\beta_n)=U.\end{equation} The product $\tr$ of permutations induces a product of tableaux, still denoted by $\tr$. This follows from the next lemma. \begin{lemma}\label{plax-comp} The plactic equivalence is compatible with the product $\tr$, that is: $u\sim_{plax}u' \Rightarrow v\tr u\sim_{plax} v\tr u'$. \end{lemma} \begin{proof} Suppose indeed that $u,u'\in S_p,v\in S_q$ and for some letters $1\leq i<j<k\leq p$, and words $x,y$, one has $u=xjiky$, $u'=xjkiy$. Then $\bar vu=\bar vxjiky,\bar vu'=\bar vxjkiy$ and therefore $v\tr u\sim_{plax}v\tr u'$. There are other similar cases, left to the reader. \end{proof} Therefore, one has for any permutations $u,v$ $$ P(v\tr u)=P(v)\tr P(u), $$ which means that $P$ is a homomorphism from the monoid $S$ into the monoid $T$, both with the product $\tr$. Note that one may compute directly $V\tr U$ as follows: $V\tr U$ is the tableau obtained by letting fall $\bar V$ onto $U$, with $\bar V$ obtained by adding $p$ to each letter in $V$ (this follows from the dual Schensted correspondence, that is, column insertion). Thus $V\tr U$ is the tableau denoted $U/V$ in \cite{T}, p. 1109. For example: \ytableausetup{aligntableaux=center} $$U \ \ \begin{ytableau} 2\\ 1& 3 \end{ytableau}\ ,\ V \ \ \begin{ytableau} *(blue!40) 1& *(blue!40) 2 \end{ytableau} \ \ \ \ \rightarrow \ \ \ \ {\begin{ytableau} *(blue!40) 4& *(blue!40) 5 \end{ytableau} \atop \begin{ytableau} 2\\ 1& 3 \end{ytableau}} \ \ \ \ \rightarrow\ \ \ \ V\tr U \ \ \begin{ytableau} *(blue!40) 4\\ 2& *(blue!40) 5\\ 1&3 \end{ytableau} $$ \bigskip We need also the following. \begin{lemma}\label{weak-comp} The weak order $\leq$ on tableaux is compatible with the product $\tr$: $U\leq U', V\leq V' \Rightarrow V\tr U\leq V'\tr U'$. \end{lemma} \begin{proof} This follows from Lemma \ref{perm-comp}, and the characterization through Eq.(\ref{<}) of the order, using the fact that $P$ is an increasing surjective homomorphims $S\to T$. \end{proof} By Proposition 2.5 in \cite{LS}, p.133, the plactic equivalence is compatible with the restriction to intervals, and with standardization. It follows that the plactic equivalence is compatible with the composition of the two operations: if $u,v\in S_n$, $u\sim_{plax}v$, and $I$ is an interval of $[n]$, then $st(u\mid I)\sim_{plax} st(v\mid I)$. Thus, if $A=P(u)$, we may denote without ambiguity by $st(A\mid I)$ the tableau $P(st(u\mid I))$. By the work of Sch\"utzenberger, the corresponding tableau is obtained by jeu-de-taquin straightening of the skew tableau which is the restriction to $I$ of the tableau $P(u)$; but we do not need this fact. \begin{lemma}\label{restr-order} Let $A,B\in T_n$ such that $A\leq B$, and $I$ be an interval in $[n]$. Then $st(A\mid I)\leq st(B\mid I)$. \end{lemma} \begin{proof} This follows from the characterization through Eq.(\ref{<}) of the order, Lemma \ref{restr-order-perm}, the previous observation, and the fact that $P$ is increasing. \end{proof} \section{Main lemma} Recall the Taskin weak order on tableaux, denoted $\leq$. \begin{lemma}\label{main} Let $n=p+q$ and $\Sigma\in \mathcal T_n$. Let $A=st(\Sigma\mid\{1,\ldots,p\})$, and let $B=st(\Sigma\mid \{p+1,\ldots,n\})$. Then for $U\in \mathcal T_p$, $V\in \mathcal T_q$, one has: $\Sigma\leq V\tr U$ if and only if $A\leq U$ and $B\leq V$. \end{lemma} \begin{proof}[Proof of Lemma \ref{main}] 1. We show first that $\Sigma \leq B\tr A$. Let $\sigma \in S_n$ be such that $\Sigma=P(\sigma)$. Let $a=st(\sigma\mid \{1,\ldots,p\})$ and $b=st(\sigma\mid \{p+1,\ldots,n\})$. Then $\sigma \leq b\tr a$ by Lemma \ref{sigma-a-b}. It follows that $\Sigma \leq B\tr A$ by Lemmas \ref{plax-comp} and \ref{weak-comp}. 2. Suppose that $A\leq U$ and $B\leq V$. Then by 1. and Lemma \ref{weak-comp}, we have $\Sigma\leq B\tr A\leq V\tr U$. 3. Suppose now that $\Sigma\leq V\tr U$. Let $u\in S_p,v\in S_q$ be such that $U=P(u),V=P(v)$. Then by Lemma \ref{restr-order}, we have $A=st(\Sigma\mid \{1,\ldots,p\}) \leq st((V\tr U)\mid \{1,\ldots,p\})=P(st((v\tr u)\mid\{1,\ldots,p\}))=P(u)=U$. Moreover, by the same lemma, $$B=st(\Sigma\mid \{p+1,\ldots,n\})\leq st((V\tr U)\mid \{p+1,\ldots,n\}) $$ $$=P(st((v\tr u)\mid\{p+1,\ldots,n\}))=P(v)=V.$$ \end{proof} \section{Primitive elements in the Hopf algebra of tableaux}\label{prim} The free $\Z$-module $\Z T$, based on the set $T$ of tableaux, becomes a structure of Hopf algebra, quotient of the Hopf algebra $\Z S$ of Section \ref{ZS}, and whose product and coproduct are therefore also denoted by $*$ and $\delta$. The quotient is obtained by identifying plactic equivalent permutations. In other words, consider the submodule $I$ spanned by the elements $u-v$, $u\sim_{plax} v$; then $I$ is an ideal and a co-ideal of $\Z S$, and the quotient $\Z S/I$ is canonically isomorphic with $\Z T$. Moreover, the canonical bialgebra homomorphism $\Z S\to\Z T$ maps each permutation $\sigma$ onto $P(\sigma)$. See \cite{PR}, Th\'eor\`eme 3.4 and 4.3 (iv), where the product and coproduct are there denoted $*'$ and $\delta'$. Now we introduce a new basis of $\mathbb ZT$, following the method of Aguiar and Sottile \cite{AS}, replacing the weak order on permutations by the Taskin weak order on tableaux. The new basis (that we may call the {\em monomial basis}, following \cite{AS}) $\mathcal M_W$, $W\in T$, is completely defined by the identities $$ \Sigma=\sum_{\Sigma\leq W}\mathcal M_W,$$ for all tableau $\Sigma$, via M\"obius inversion on the poset $(T,\leq)$. \begin{theorem}\label{coproduct} Let $\Sigma\in \mathcal T_n$. Then $$\delta (\mathcal M_\Sigma)=\sum_{\Sigma=V\tr U}\mathcal M_U\otimes \mathcal M_V.$$ \end{theorem} \begin{proof} Define the $\Z$-linear mapping $\delta_1:\Z T\mapsto \Z T\otimes \Z T$ by $$\delta_1(\mathcal M_W)=\sum_{W=V\tr U}\mathcal M_U\otimes \mathcal M_V.$$ It is enough to show that $\delta_1=\delta$. We have $$\delta_1(\Sigma)=\delta_1(\sum_{\Sigma\leq W}\mathcal M_W)=\sum_{\Sigma\leq W}\sum_{W=V\tr U} \mathcal M_U\otimes \mathcal M_V=\sum_{\Sigma\leq V\tr U}\mathcal M_U\otimes \mathcal M_V$$ $$=\sum_{p+q=n} \sum_{\Sigma\leq V\tr U \atop V\in \mathcal T_q,U\in \mathcal T_p}\mathcal M_U\otimes \mathcal M_V. $$ This is by Lemma \ref{main}, and with its notations, equal to $$\sum_{p+q=n} (\sum_{st(\Sigma\mid\{1,\ldots,p\})\leq U}\mathcal M_U) \otimes (\sum_{st(\Sigma\mid\{p+1,\ldots,n\})\leq V}\mathcal M_V) $$ $$=\sum_{p+q=n}(st(\Sigma\mid\{1,\ldots,p\})) \otimes (st(\Sigma\mid\{p+1,\ldots,n\})).$$ Choose $\sigma$ such that $P(\sigma)=\Sigma$. Then by definition of $st(\Sigma\mid I)$, the latter quantity is equal to $$ \sum_{p+q=n}P(st(\sigma\mid\{1,\ldots,p\})) \otimes P(st(\Sigma\mid\{p+1,\ldots,n\}))$$ $$=(P\otimes P)(\sum_{p+q=n}st(\sigma\mid\{1,\ldots,p\}) \otimes st(\sigma\mid\{p+1,\ldots,n\}))$$ $$=(P\otimes P)(\delta(\sigma))=\delta(P(\sigma) =\delta(\Sigma)),$$ since $P$ is a homomorphism of bialgebra, as recalled at the beginning of Section \ref{prim}. \end{proof} \begin{corollary} The submodule of primitive elements of $\Z T$ is spanned by the $\mathcal M_\Sigma$ sucht that $\Sigma$ is indecomposable for the product $\tr$. \end{corollary} The dimensions of the graded components of the submodule of primitive elements is therefore the sequence of the numbers of tableaux indecomposable for the product $\tr$; it is denoted $a_n$ in \cite{PR}, p.88-89. For $n=1,2,\ldots,10$, they are the numbers $$ 1,1,1,3,7,23,71,255, 911,3535. $$ They appear as sequence A140456 in the Online Encyclopedia of Integer Sequences \cite{OEIS}, with other interpretations. \section{Further remarks} \subsection{Product formulas using $\Delta$} The product $\tr$, both for permutations and tableaux, plays a role in product formulas in the dual Hopf algebras of $\Z S$ and $\Z T$. First, one has to consider also the product $\square$ of permutations: let $a\in S_p$, $b\in S_q$; then $a\square b=a\bar b$, where $\bar b$ is obtained from $b$ by adding $p$ to each letter in $b$. Recall the {\em shifted shuffle product} of permutations, denoted $\overline{\shuffle}$: $a \overline{\shuffle} b$ is the shuffle of $a$ and $\bar b$. This product is the dual product of the coproduct $\delta$ of $\Z S$. On has \begin{theorem}(Loday-Ronco \cite{LR} Theorem 4.1) Let $a,b\in S$. Then $$ a \overline{\shuffle} b=\sum_{a\square b\leq \sigma\leq b\tr a} \sigma. $$ \end{theorem} In other words, $a \overline{\shuffle} b$ is the sum of all permutations in the interval $[a\square b, b\tr a]$ of the weak order. The product $\square$ is compatible with the plactic equivalence, hence $\square$ is well-defined on tableaux. Denote also by $\overline{\shuffle}$ the product which is the dual of $\delta$ in the dual coalgebra of $\Z T$. Then one has \begin{theorem}(Taskin \cite{T} Theorem 4.1) Let $A,B\in \mathcal T$. Then $$ A\overline{\shuffle} B=\sum_{A\square B\leq T\leq B\tr A} T. $$ \end{theorem} In other words, $A\overline{\shuffle} B$ is the sum of all tableaux belonging to the interval $[A\square B, B\tr A]$ of the weak order. Note that the product $A\square B$ is denoted $A\backslash B$ in \cite{T}, p. 1109. It corresponds to put the tableau $\overline{B}$ aside the tableau $A$, then push every row of $\overline{B}$ toward $A$. For example: \bigskip $A \ \ \ytableausetup{aligntableaux=bottom} \begin{ytableau} 2\\ 1& 3 \end{ytableau}\ ,\ B \ \ \begin{ytableau} *(blue!40) 3\\ *(blue!40) 2\\ *(blue!40)1& *(blue!40) 4 \end{ytableau}\ \ \rightarrow$ \ \ $\ytableausetup{aligntableaux=bottom} \begin{ytableau} 2\\ 1& 3 \end{ytableau} \ \ \begin{ytableau} *(blue!40) 6\\ *(blue!40) 5\\ *(blue!40)4& *(blue!40) 7 \end{ytableau}$\ \ $\ytableausetup{aligntableaux=bottom} \rightarrow\ \ A\square B \ \ \begin{ytableau} *(blue!40) 6\\ 2& *(blue!40) 5\\ 1&3 & *(blue!40) 4 &*(blue!40) 7\\ \end{ytableau} $ \bigskip \subsection{Multiplicative basis} Note that Theorem \ref{coproduct} means that in the dual Hopf algebra of $\Z T$, the dual basis $\mathcal M_T^*$ of the basis $\mathcal M_T$ is {\em multiplicative}, in the sense of \cite{DHNT}: one has for tableaux $A,B$, $$\mathcal M_A^*\overline{\shuffle}\mathcal M_B^*=\mathcal M_{B\tr A}^*.$$ Indeed, using the canonical pairing between $\Z T$ and its dual, $$\l \mathcal M_A^*\overline{\shuffle}\mathcal M_B^*,\mathcal M_\Sigma\r=\l\mathcal M_A^* \otimes \mathcal M_B^*,\delta(\mathcal M_\Sigma)\r=\l \mathcal M_A^* \otimes \mathcal M_B^*,\sum_{\Sigma=V\tr U}\mathcal M_U\otimes \mathcal M_V\r$$ $$=\sum_{\Sigma=V\tr U}\l \mathcal M_A^*,\mathcal M_U\r\l\mathcal M_B^*, \mathcal M_V\r. $$ This is 1 exactly when $\Sigma=B\tr A$, otherwise it is 0. Thus it is equal to $\l \mathcal M_{B\tr A}^*,\mathcal M_\Sigma\r$, which proves the formula. \subsection{A counter-example by Franco Saliola} Our basis $\mathcal M_\Sigma$ was inspired by the construction of Aguiar and Sottile in \cite{AS}. They prove further that the structure constants for multiplication are positive (Theorem 4.1 in their article). This is not true in the case of Poirier--Reutenauer Hopf algebra of tableaux, as shows a counter-example of Franco Saliola, that he kindly permitted us to reproduce here. Indeed, he computed that $$ \mathcal M_{P(123)}*\mathcal M_{P(123)}=\mathcal M_{P(123456)}$$ $$-\mathcal M_{P(241356)}-\mathcal M_{P(251346)} -\mathcal M_{P(261345)}-\mathcal M_{P(351236)}-\mathcal M_{P(361245)}-\mathcal M_{P(461235)}$$ $$+\mathcal M_{P(256134)}+\mathcal M_{P(346125)}+\mathcal M_{P(356124)} +2\mathcal M_{P(456123)}$$ $$+2\mathcal M_{P(362514)}-\mathcal M_{P(462513)} \mathcal -M_{P(543126)}. $$ \medskip {\em Acknowledgments}. We thank Franco Saliola, who allowed us to include a counter-example arising from his computations.

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TITLE: $45^\circ$ Rubik's Cube: proving $\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle? QUESTION [17 upvotes]: I've been working on a problem related to the 3x3x3 Rubik's Cube where you allow faces to be turned by $45^\circ$ instead of just the usual $90^\circ$. We know for the standard 3x3x3 the cube is sliced up into 27 regions and 26 of them are the external pieces of the puzzle (8 corners + 12 edges + 6 face centers). For the $45^\circ$ Rubik's Cube, when you turn a face by $45^\circ$ there have to be additional cuts added to enable the adjacent faces to turn. One interesting property though is that no matter how many cuts you add, it seems like there are still configurations where faces are be blocked unless more cuts are added. The twisty puzzle community calls this property jumbling when an infinite number of cuts are required. I'm trying to prove that the $45^\circ$ Rubik's Cube jumbles. Here is an example of a Rubik's Cube where a few cuts have been added to enable some $45^\circ$ turns before things get blocked. In this photo the left face with the yellow octagon center was twisted $45^\circ$ and then the right face with the red octagon center was twisted $45^\circ$. Now the yellow face can not be twisted any further without adding more cuts through the pieces that are blocking it: I call the above sequence [45F 45R]. To go about trying to prove that an infinite number of cuts are require, I set out to show something simpler that implies that. Namely, that a point on the puzzle gets rotated around some axis by an irrational angle if you repeat [45F 45R]. If repeating these two moves over and over rotates points by an irrational angle, then the order of the sequence (the period) is infinite and an infinite number of cuts must be added, implying that the puzzle jumbles. Here is an animation showing how the corner point moves when you do [45F 45R] repeatedly: So I build the rotation matrix for 45F and 45R and found the combined rotation matrix: $$\left( \begin{array}{ccc} \sqrt{2}/2 & \sqrt{2}/2 & 0 \\ -1/2 & 1/2 & -\sqrt{2}/2 \\ -1/2 & 1/2 & \sqrt{2}/2 \end{array} \right)$$ If you do some additional work on that matrix you'll find that the axis points are being rotated around is $$[-1,\ -\sqrt{2} + 1,\ 1]$$ and the angle points are rotated through is $$\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$$ Here is a plot of several different points through a few hundred applications of [45F 45R] and a plot of the axis computed from the rotation matrix: All that is left now is to prove that $\theta = \arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle. By irrational angle I mean that $\theta$ is not of the form $\frac{m}{n}\pi$. The strategy I've tried is to assume that $\theta = \frac{m}{n}\pi$ and then show that leads to a contradiction. This seems like a good way to go because $\cos(n\theta) = T_n(\cos(\theta))$ where $T_n$ is the $n^{\mathrm{th}}$ Chebyshev polynomial. This is simple because $n\theta = n\frac{m}{n}\pi = m\pi$ where $m$ is an integer. $\cos(m\pi) = \pm 1$ Since we know $\cos(\theta) = \frac{\sqrt{2}}{2} - \frac{1}{4}$ if we can show $T_n(\frac{\sqrt{2}}{2} - \frac{1}{4}) \ne \pm 1$ for any $T_n$ then $\theta$ can't be of the form $\frac{m}{n}\pi$ which is a contradiction and would prove that the angle is irrational. Unfortunately I'm stuck. I've tried expanding out various Chebyshev polynomials and substituting $y$ for $\sqrt{2}$ plugging in $\frac{y}{2} - \frac{1}{4}$ and it certainly seems like the odd powers of $y$ could never fully cancel out but I can't work out the details. So, is there a way to prove $\arccos ( \frac{\sqrt{2}}{2} - \frac{1}{4} )$ is an irrational angle? REPLY [11 votes]: book http://www.amazon.com/Irrational-Numbers-Carus-Mathematical-Monographs/dp/0883850389 page 37, Theorem 3.9, due to Lehmer: given integers $n > 2$ and $k,$ then $$ 2 \cos \left( \frac{2k \pi}{n} \right) $$ is an algebraic integer of degree $\phi(n)/2.$ Twice your cosine is $$ t = \sqrt 2 - \frac{1}{2}. $$ This satisfies $$ 4 t^2 + 4 t - 7 = 0 $$ and is an algebraic number but not an algebraic integer. See https://en.wikipedia.org/wiki/Algebraic_integer#Non-example

TITLE: subgroup generated by two subgroups QUESTION [7 upvotes]: Let $A$ and $B$ be two subgroups of the same group $G$. let $$AB=\{ab\,|\, a\in A,\, b\in B \}$$ and $$\langle A,B\rangle$$ the subgroup generated by $A$ and $B$. Are $AB$ and $\langle A,B\rangle$ the same as sets? My guess is no since the element $aba\in \langle A,B\rangle$ but it is not in $AB$, but I guess when $A$ and $B$ commute: $AB=BA$ then this is true. is this correct? REPLY [20 votes]: Yes, the set $AB$ is a subgroup of $G$ if and only if $AB = BA$, as can be found in many algebra texts, such as Herstein's "Topics in Algebra". It is certainly necessary that $AB = BA$, since $(ab)^{-1} = b^{-1}a^{-1}$ for $a \in A, b \in B$ and subgroups are closed under taking inverses. On the other hand, if we do have $AB = BA$, then for any $a,c \in A$ and $b,d \in B$, we have $(ab)(cd) = a(bc)d = a(c^{*}b^{*})d$ for some $c^{*} \in A$, $b^{*} \in B$, so that $(ab)(cd) = (ac^{*})(b^{*}d) \in AB$. Hence $AB$ is closed under taking inverses, and closed under the group operation, so $AB$ is a subgroup. Also $AB = BA$ implies that $\langle A,B \rangle \subseteq AB \subseteq \langle A,B \rangle$, so $\langle A,B \rangle = AB$. Conversely, if $\langle A,B \rangle = AB$ (as a set), then $AB$ is a subgroup, so $AB = BA$. By the way, if $AB = BA$, it is usual to say that the subgroups $A$ and $B$ are permutable, rather than that they commute. Note that permutability need not imply that all elements of $A$ commute with all elements of $B$. REPLY [11 votes]: Note that, regardless of whether $AB$ is a subgroup or not, if both $A$ and $B$ are finite, then the number of elements it has is $$|AB| = \frac{|A||B|}{|A\cap B|}.$$ To see this, consider the map $A\times B \to AB$ given by $(a,b)\mapsto ab$. The map is clearly onto. If $x\in A\cap B$, then for each $a\in A$ and $b\in B$ you have $ax\in A$, $x^{-1}b\in B$, so $(ax,x^{-1}b)\in A\times B$ has the same image as $(a,b)$. Thus, each element of $AB$ is the image of at least $|A\cap B|$-many elements. Conversely, if $(a,b)$ and $(a',b')$ have the same image under this map, then $ab=a'b'$, hence $(a')^{-1}a=b'b^{-1}\in A\cap B$, and $a = a'(a'^{-1}a)$ and $b=(a'^{-1}a)^{-1}b' = (bb'^{-1})b'$, so any two pairs that map to the same element arise from an element of $A\cap B$. Thus, each image occurs $|A\cap B|$ times, so $|AB||A\cap B| = |A\times B|=|A||B|$. In particular, take $G=S_3$, $A=\langle(1,2)\rangle$, and $B=\langle (1,3)\rangle$. Then $A\cap B = \{e\}$, so the number of elements in $AB$ is $|A||B|=4$. This cannot be a subgroup of $S_3$, by Lagrange's Theorem. So $AB$ is properly contained in $\langle A,B\rangle$ (as sets). REPLY [2 votes]: Let $G$ be $S_3$, the group of all permutations on $3$ letters. Let $A$ be the two-element subgroup generated by the transposition $(1,2)$, and $B$ the two-element subgroup generated by $(1,3)$. Then $AB$ consists of the identity, $(1,2)$, $(1,3)$, and their product $(12)(13)=(1,2,3)$. But $\langle A, B\rangle$ is all of $S_3$. We can verify this directly. Or else note that $\langle A, B\rangle$ has at least $4$ elements. Since the order of $\langle A, B\rangle$ must divide $3!$, $\langle A, B\rangle$ is of $S_3$.

\begin{document} \author {Anna Marie Bohmann and J. P. May} \address{Department of Mathematics, Northwestern University\\ 2033 Sheridan Road, Evanston IL 60208\\[5pt] Department of Mathematics, University of Chicago\\5734 S. University Avenue, Chicago IL 60637\\} \title{A presheaf interpretation of the generalized Freyd conjecture} \def\xypic{\hbox{\rm\Xy-pic}} \newcommand{\sA}{\scr{A}} \newcommand{\sB}{\scr{B}} \newcommand{\sC}{\scr{C}} \newcommand{\sD}{\scr{D}} \newcommand{\sM}{\scr{M}} \newcommand{\sP}{\scr{P}} \newcommand{\sS}{\scr{S}} \newcommand{\sT}{\scr{T}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bU}{\mathbb{U}} \newcommand{\bY}{\mathbb{Y}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\epz}{\varepsilon} \newcommand{\io}{\iota} \newcommand{\SI}{\Sigma} \newcommand{\com}{\circ} \newcommand{\iso}{\cong} \newcommand{\rtarr}{\rightarrow} \newtheorem{thm}{Theorem} \newtheorem{prop}{Proposition} \newtheorem{lem}{Lemma} \newtheorem{cor}{Corollary} \newtheorem{conj}{Conjecture} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \newtheorem{exmp}[thm]{Example} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \let\pf\proof \let\epf\endproof \maketitle \begin{abstract} We give a generalized version of the Freyd conjecture and a way to think about a possible proof. The essential point is to describe an elementary formal reduction of the question that holds in any triangulated category. There are no new results, but at least one known example drops out very easily. \end{abstract} In algebraic topology, the generating hypothesis, or Freyd conjecture, is a long-standing conjecture about the structure of the stable homotopy category. It was initially formulated in 1965 \cite{F} and remains open to this day. Because the original conjecture has proved difficult to analyze, recent work has turned to studying similar conjectures in categories that share many properties with the stable homotopy category in hopes of further understanding the types of categories in which such a conclusion holds. In this note, we state a version of the generating hypothesis for an arbitrary triangulated category and analyze conditions under which this hypothesis holds. We emphasize that we impose no additional conditions on our triangulated categories, so that our results show which {\em formal} properties of a category imply the Freyd conjecture. Therefore our results give conceptual insight into the kind of category in which the generating hypothesis can be expected to hold. \section{The generalized Freyd conjecture} Let $\sT$ be a triangulated category and write $[X,Y]$ for the abelian group of maps $X\rtarr Y$ in $\sT$. Let $\sB$ be a (small) full subcategory of $\sT$ closed under its translation (or shift) functor $\SI$ and let $\sC$ be the thick full subcategory of $\sT$ that is generated by $\sB$; write $\io\colon \sB\rtarr \sC$ for the inclusion. For emphasis, we often write $\sB(X,Y) = [X,Y]$ when $X,Y\in\sB$ and $\sC(X,Y) = [X,Y]$ when $X,Y\in\sC$. The category $\sB$ is pre-additive (enriched over $\sA\!b$), and $\sC$ is additive (has biproducts). Let $\sP\sB$ and $\sP\sC$ denote the categories of abelian presheaves defined on $\sB$ and $\sC$. They consist of the additive functors from $\sB^{op}$ or $\sC^{op}$ to $\sA\!b$ and the additive natural transformations between them. \begin{defn}\label{defn:Ffunctor} Define the Freyd functor $\bF\colon \sT\rtarr \sP\sB$ by sending an object $X$ to the functor $\bF X$ specified on objects and morphisms of $\sB$ by $\bF X(-) = [-,X]$ and sending a map $f\colon X\rtarr Y$ to the natural transformation $f_* =[-,f]$. Define $\bY\colon \sT\rtarr \sP\sC$ similarly. We are only interested in the restrictions of $\bF$ and $\bY$ to $\sC$, and then $\bY\colon \sC\rtarr \sP\sC$ is the standard Yoneda embedding. \end{defn} \begin{conj}[The generalized Freyd conjecture]\label{FC} The functor $\bF\colon \sC\rtarr \sP\sB$ is faithful. Equivalently, $\bF f =0$ if and only if $f=0$. We then say that the Freyd conjecture holds for the pair $(\sC,\sB)$. \end{conj} We hasten to add that the conjecture is false in this generality. Additional hypotheses are needed, but none will be relevant to our formal analysis. For example, we might as well assume that a map $f\colon X\rtarr Y$ in $\sC$ is an isomorphism if and only $f_*\colon \bF X\rtarr \bF Y$ is an isomorphism, as holds in the stable homotopy category. This condition is necessary but not sufficient for the Freyd conjecture to hold. Indeed, if $Z$ is the third term in an exact triangle with one map $f$, then $f$ is an isomorphism if and only if $Z=0$ and $\bF f$ is an isomorphism if and only if $\bF Z=0$. If $\bF Z = 0$ and the Freyd conjecture is true, then $Z=0$ since its identity map is the zero map. \begin{exmp}\label{FC1} Take $\sT = \text{Ho}\sS$ to be the stable homotopy category. Let $\sB$ consist of the sphere spectra $S^n$ for integers $n$. Then $\sC$ is the homotopy category of finite CW spectra. Freyd \cite{F} conjectured that a map $f$ in $\sC$ is zero if and only if it induces the zero homomorphism $f_*\colon \pi_*(X)\rtarr \pi_*(Y)$. By the following observation, this is a special case of our Conjecture \ref{FC}. \end{exmp} \begin{lem}\label{FO} In Example \ref{FC1}, the category $\sP\sB$ is isomorphic to the category $\sM$ of right modules over the ring $\pi_*$ of stable homotopy groups of spheres. Under this isomorphism, the Freyd functor $\bF$ coincides with the stable homotopy group functor $\pi_*\colon \text{Ho}\sS\rtarr \sM$. \end{lem} \begin{proof} We have $\sB(S^m,S^n) = \pi_m(S^n)\iso \pi_{m-n}$. For $T\in \sP\sB$, let $T_n = T(S^n)$. The additive contravariant functor $T$ gives homomorphisms \[ T\colon \pi_{m-n} \iso \sB(S^m,S^n)\rtarr \sA\!b(T_n,T_m).\] By adjunction, these give homomorphisms $T_n \otimes \pi_{m-n} \rtarr T_m$. The functoriality gives the formula $(tx)y = t(xy)$. No signs appear since we are taking right modules, as is dictated categorically by contravariance. Conversely, given a right $\pi_*$-module $M$, define $T(S^n)=M_n$ and, for $t\in M_n$ and $x\in \pi_{m-n}\iso \sB(S^m,S^n)$ define $T(x)(t) = tx\in M_m$. The module axioms ensure that $T$ is a functor. This specifies the isomorphism of categories, and the consistency of $\bF$ and $\pi_*$ is clear. \end{proof} The following example originally prompted us to take a presheaf perspective on the Freyd conjecture. \begin{exmp}\label{FCG} Let $G$ be a compact Lie group and take $\sT = \text{Ho}G\sS$ to be the equivariant stable homotopy category. Let $G\sB$ consist of the orbit $G$-spectra $$S^n[G/H]\equiv \SI^n\SI^{\infty}(G/H)_+$$ for integers $n$ and closed subgroups $H$ of $G$. The thick subcategory $G\sC$ generated by $G\sB$ is the category of retracts of finite $G$-CW spectra. The equivariant version of the Freyd conjecture asserts that a map $f$ in $G\sC$ is zero if and only if it induces the zero homomorphism $f_*\colon \pi_*^H(X)\rtarr \pi_*^H(Y)$ for all $H$, where \[ \pi_n^H(X) \equiv \pi_n(X^H) \iso [S^n[G/H],X]_G. \] Again, this is a special case of our Conjecture \ref{FC}. This example is the focus of \cite{B}. \end{exmp} \begin{rem}\label{MacF} The full subcategory $\sB_0$ of $\sB$ whose objects are the $S[G/H]\equiv S^0[G/H]$ is called the Burnside category. A Mackey functor (or $G$-Mackey functor) $M$ is by definition an object of $\sP G\sB_0$. When $G$ is finite, this agrees with the more usual algebraic definition (\cite[V\S9]{LMS} or \cite[IX\S4, XIX\S3]{EHCT}). A map of Mackey functors is a natural transformation, that is, a map of presheaves. \end{rem} \begin{defn} The graded Burnside category $G\pi_*$ has objects the $S[G/H]$. Its abelian group of maps of degree $n$ from $S[G/H]$ to $S[G/J]$ is $$\pi_n^H(S[G/J]) = [S^n[G/H],S[G/J]]_G.$$ Composition is induced by suspension and composition in $G\sB$ in the evident fashion. Define a right $G\pi_*$-module $M$ to be a graded presheaf, that is, a contravariant functor $G\pi_*\rtarr \sA\! b_*$, where $\sA\!b_*$ is the category of graded abelian groups. \end{defn} \begin{lem}\label{FOG} In Example \ref{FCG}, the category $\sP G\sB$ is isomorphic to the category $G\sM$ of right $G\pi_*$-modules. Under this isomorphism, the Freyd functor $\bF$ coincides with the equivariant stable homotopy group functor $\pi_*^{(-)}\colon \text{Ho}G\sS\rtarr G\sM$. \end{lem} The proof is the same as that of Example \ref{FC1}. In this case, the presheaf formulation of the Freyd conjecture appears more natural to us than the equivalent homotopy group reformulation. \begin{exmp} In the rational equivariant stable homotopy category, the Freyd conjecture is true if $G$ is finite \cite{GM} and is false if $G = S^1$ \cite{B}. \end{exmp} Here is another example where the presheaf formulation, and the use of many objects and their shifts rather than a single object and its shifts in $\sB$, may be more natural than the formulation of the Freyd conjecture studied so far. \begin{exmp} In several recent papers, Benson, Carlson, Chebolu, Christensen and Min\'a\v c \cite{BCCM, CCM2008, CCM2007} study the Freyd conjecture in the stable module category of a finite group $G$ over a field $k$ whose characteristic divides the order of $G$. This is a triangulated category $\mathrm{StMod}_{kG}$ obtained from the category of $kG$--modules by modding out by maps that factor through a projective module. They restrict to the thick subcategory generated by the trivial representation $k$ and ask whether the Tate cohomology functor is faithful on this subcategory. Since the Tate cohomology of a module $M$ is given by maps out of $k$ in the stable module category, \[ \hat{H}^i(G,M)=\mathrm{Stmod}_{kG}(\Omega^i k,M),\] their formulation is equivalent to our Conjecture \ref{FC} with $\sB=\{\Omega^i k\}$. In this context, they prove that the Freyd conjecture holds in the stable module category of $kG$--modules if and only if the $p$--Sylow subgroup of $G$ is either $C_2$ or $C_3$, where $p$ is the characteristic of $k$ \cite{CCM2008}. Their proof in fact shows that this variant of the Freyd conjecture holds if and only if the thick subcategory generated by $k$ consists of finite direct sums of suspensions of $k$. For $p$--groups, the trivial module is the only irreducible module over $kG$, and the thick subcategory generated by $k$ is the subcategory of compact objects in the stable module category. For non-$p$--groups, other irreducible modules exist. It is thus natural to generalize the Freyd conjecture to our presheaf context by letting $\sB$ be the category whose objects are the suspensions of the irreducible $kG$--modules. The thick subcategory $\sC$ generated by this $\sB$ is then the subcategory of compact objects in the stable module category and one can ask when Conjecture \ref{FC} holds. This presheaf version of the generating hypothesis in $\mathrm{StMod}_{kG}$ takes into account all the generating objects in the stable module category. \end{exmp} \section{A general line of argument} The presheaf perspective suggests a method of attack on the generalized Freyd conjecture. We have the forgetful functor \[ \bU = \io^*\colon \sP\sC\rtarr \sP\sB\] given by restricting presheaves defined on $\sC^{op}$ to the full subcategory $\sB^{op}$. The functor $\bU$ has a left adjoint prolongation functor \[ \bP = \io_!\colon \sP\sB\rtarr \sP\sC. \] For $T\in \sP\sB$ and $K\in \sC$, $\bP T(K)$ is the categorical tensor product \[ \bP T(K) = T\otimes_{\sB}\sC(K,-). \] See, for example, \cite[I\S3]{MMSS} or \cite[I\S2]{MM}. Since $\sB$ is a full subcategory of $\sC$, the unit $\Id\rtarr \bU\bP$ of the adjunction is a natural isomorphism \cite[I.3.2]{MMSS}. We focus attention on the counit $\epz\colon \bP \bU\rtarr \Id$. Observe that the Freyd functor $\bF\colon \sT\rtarr \sP\sB$ is the composite $\bU\bY$. This leads to the following observation. \begin{prop}\label{crit1} The Freyd conjecture holds for $(\sC,\sB)$ if $$ \epz\colon \bP\bF X = \bP\bU\bY X\rtarr \bY X $$ is an isomorphism for all $X\in \sC$. \end{prop} \begin{proof} Since the unit of the adjunction $(\bP,\bU)$ is an isomorphism, $\bU\epz$ is an isomorphism by one of the triangle identities. Thus $\bF \iso \bU \bP\bF$. Therefore, for a map $f\colon X\rtarr Y$ in $\sC$, $\bF f= 0$ if and only if $\bP\bF f =0$. By the Yoneda lemma, $f=0$ if and only if $\bY f =0$. If $\epz$ is an isomorphism, then $\bP\bF f =0$ if and only if $\bY f =0$. \end{proof} In fact, less is needed. Consider $\bY X(X) =[X,X]$. We will shortly prove the following result. \begin{prop}\label{crit2} The Freyd conjecture holds for $(\sC,\sB)$ if the identity map of $X$ is in the image of $\epz$ for all $X\in\sC$. \end{prop} We have the following starting point towards verification of the hypothesis. \begin{lem}\label{start} The map $\epz\colon \bP\bF X(J)\rtarr \bY X(J)$ is an isomorphism for all $X\in\sT$ and $J\in\sB$. \end{lem} \begin{proof} We have seen that $\bU\epz$ is an isomorphism, and by definition $\bU T(J) = T(J)$ for any $J\in\sB$ and any $T\in\sP\sC$. \end{proof} Now consider an exact triangle \begin{equation}\label{cof} \xymatrix@1{ K\ar[r] & L\ar[r] & M\ar[r] & \SI K\\} \end{equation} in $\sT$, where $K$, $L$, and $M$ are in $\sC$. We have the commutative diagram {\small \begin{equation}\label{lad} \xymatrix{ \cdots \ar[r] & \bP\bF X (\SI K) \ar[r] \ar[d]_{\epz} & \bP\bF X (M) \ar[r] \ar[d]^{\epz} & \bP\bF X (L) \ar[r] \ar[d]^{\epz} & \bP\bF X (K) \ar[r] \ar[d]^{\epz} & \cdots \\ \cdots \ar[r] & \bY X (\SI K) \ar[r] & \bY X (M) \ar[r] & \bY X (L) \ar[r] & \bY X (K) \ar[r] & \cdots. \\} \end{equation} } Since $\bY X (K) = [K,X]$, the lower row is exact. By definition, $\sC$ is the smallest subcategory of $\sT$ that contains $\sB$, is closed under retracts, and has the property that if two terms of an exact triangle are in $\sC$ then so is the third. By an easy retract argument and the five lemma, this gives the following conclusion. \begin{prop}\label{crit3} If the top row of Diagram (\ref{lad}) is exact for every exact triangle (\ref{cof}) and every $X\in\sC$, then $\epz\colon \bP\bF X \rtarr \bY X$ is an isomorphism for every $X\in\sC$ and the Freyd conjecture holds for $(\sC,\sB)$. \end{prop} There is a reinterpretation of the exactness hypothesis that makes it reminiscent of the standard result that the adjoint (if it exists) of an exact functor between triangulated categories is exact. For $K$ and $X$ in $\sC$, the abelian group $\bP\bF X(K)$ is the coequalizer in $\sA\!b$ displayed in the diagram \begin{equation}\label{PF} \xymatrix{ \sum_{I,J\in \sB} \sC(J,X)\otimes \sB(I,J)\otimes \sC(K,I) \ar@<1ex>[d] \ar@<-1ex>[d] \\ \sum_{J\in\sB} \sC(J,X)\otimes \sC(K,J) \ar[d]\\ \sC(-,X)\otimes_{\sB}\sC(K,-),\\} \end{equation} where the parallel arrows are given by composition in $\sC$. We use this to interpolate the proof of Proposition \ref{crit2}. The composition maps \[ \com\colon \sC(X,Y)\otimes \sC(J,X)\rtarr \sC(J,Y)\] induce a pairing \[ \com\colon \sC(X,Y)\otimes \bP \bF X(K)\rtarr \bP\bF Y(K) \] such that $f\com z = \bP \bF f(z)$ for $z\in \bP\bF X(K)$ and the following diagram commutes: \[ \xymatrix{ \sC(X,Y)\otimes \bP \bF X(K) \ar[r]^{\id\otimes\epz} \ar[d]_{\com} & \sC(X,Y)\otimes \bY X(K) \ar[d]^{\com}\\ \bP\bF Y(K) \ar[r]_{\epz} & \bY Y(K).\\} \] Take $K=X$ and suppose that $\epz(z) = \id_X$ (for the top map $\epz$). If $\bF f = 0$, then $\epz(f\com z) = \epz \bP\bF f(z) = 0$. By the diagram, this equals $f\com \epz(z) = f$ and so $f=0$. Let us write $\sP_d\sB$ and $\sP_d\sC$ for the categories of covariant additive functors on $\sB$ and $\sC$, and similarly write $\bU_d$, $\bP_d$, $\bF_d$, and $\bY_d$ for the corresponding functors. (The $d$ stands for dual.) We are just interchanging $\sB$ and $\sC$ with their opposite categories. Visibly, we again have $\bU_d\bY_d = \bF_d$ and again have an adjunction $(\bP_d,\bU_d)$ with $\bU_d\bP_d\iso\Id$. By symmetry, we have \begin{equation}\label{P'F'} \bP\bF X(K) = \bP_d\bF_d K(X). \end{equation} But in this dual reformulation, the exactness hypothesis on $K$ for fixed $X$ is now a levelwise exactness statement about the composite functor $\bP_d\bF_d\colon \sC\rtarr \sP_d\sC$. Since $\bF_dK(J) = \sC(K,J)$ for $J\in\sB$, $\bF_d$ clearly takes exact triangles in the variable $K$ to exact sequences for each fixed $J$. Thus a more general question to ask is whether or not $\bP_d\colon \sP_d\sB\rtarr \sP_d\sC$ preserves levelwise exactness. That is, is it true that if $T'\rtarr T\rtarr T''$ is a sequence of diagrams $\sB\rtarr \sA\!b$ such that the sequence $T'(J)\rtarr T(J)\rtarr T''(J)$ is exact for each $J\in \sB$, then the sequence $\bP T'(X)\rtarr \bP T(X)\rtarr \bP T''(X)$ is exact for all $X\in \sC$? Observe that we have not yet used any hypothesis on $\sB$, other than that it generates the thick subcategory $\sC$ of the triangulated category $\sT$. Thus all that we have done is to give a purely formal reduction of the general problem. \section{The derived category of ring} Our framework for understanding the Freyd conjecture leads to a transparent proof of the result of Lockridge \cite[3.9]{L} that the generalized Freyd conjecture holds in the derived category of a von Neumann regular ring. We simply observe that the hypotheses of Proposition \ref{crit3} hold in this case by one of the equivalent definitions of a von Neumann regular ring. However, our methods do not prove the converse, which is proven in \cite{HLP, L}. Using right $R$-modules for definiteness, let $\sD(R)$ be the derived category of a ring $R$ and let $\sB$ be the full subcategory of $\sD(R)$ whose objects are the shifts $\SI^i R$ of the chain complex that is $R$ concentrated in degree $0$. Then $\sC$ is the category of perfect chain complexes, namely those isomorphic in $\sD(R)$ to bounded chain complexes of finitely generated projective $R$-modules. The Freyd functor assigns the homology groups $\sB(\SI^iR,X) = H_i X$ to a chain complex $X$, and Conjecture \ref{FC} is the assertion that a map between perfect chain complexes is $0$ in $\sD(R)$ if it induces the zero map on homology. Defining $H^i(K) = \sC(K,\Sigma^iR)$, as usual, we have the following observation in this case. \begin{lem}\label{chaincxlem} For perfect chain complexes $K$ and $X$, $\bP\bF X(K)$ is isomorphic to the abelian group $\sum_i H_i(X)\otimes_R H^i(K)$. \end{lem} \begin{proof} By definition, $\bP\bF X(K)$ is the coequalizer displayed in the diagram \[ \xymatrix{ \sum_{i,j\in \bZ} \sC(\Sigma^jR,X)\otimes \sB(\Sigma^iR,\Sigma^jR)\otimes \sC(K,\Sigma^iR) \ar@<1ex>[d] \ar@<-1ex>[d] \\ \sum_{i\in\bZ} \sC(\Sigma^iR,X)\otimes \sC(K,\Sigma^iR) \ar[d]\\ \sC(-,X)\otimes_{\sB}\sC(K,-).\\} \] There are no maps $\Sigma^iR\to\Sigma^jR$ unless $i=j$, when $\sB(\Sigma^iR,\Sigma^iR)\cong R$. The composition maps \[ \sC(\Sigma^iR,X)\otimes \sB(\Sigma^iR,\Sigma^iR)\rtarr \sC(\Sigma^iR,X) \] specify the right action of $R$ on $H_i(X)$. Therefore our coequalizer diagram can be rewritten as \[ \xymatrix{ \sum_{i\in \bZ} H_i(X)\otimes R\otimes H^i(K) \ar@<1ex>[d] \ar@<-1ex>[d] \\ \sum_{i\in\bZ} H_i(X)\otimes H^i(K) \ar[d]\\ \sum_{i\in\bZ} H_i(X)\otimes_R H^i(K).} \] The conclusion follows. \end{proof} \begin{prop}\label{flathomology} If the homology $R$-modules $H_i(X)$ of any perfect chain complex $X$ are $R$-flat, then the Freyd conjecture holds for the pair $(\sC,\sB)$ in $\sD(R)$. \end{prop} \begin{proof} Let $K\to L\to M\to \Sigma K$ be an exact triangle in $\sD(R)$. Since the functor $\sC(-,\Sigma^iR)$ takes exact triangles to exact sequences, the sequence \[H^{i-1}(K) \to H^i(M) \to H^i(L)\to H^i(K)\] is exact. By our flatness hypothesis, this sequence remains exact on tensoring with each $H_i(X)$. By Lemma \ref{chaincxlem}, when we take the direct sum over $i$ of these sequences, we obtain the exact sequence \[\dotsb\rtarr \bP\bF X(\Sigma K)\rtarr \bP\bF X(M)\rtarr \bP\bF X(L) \rtarr \bP\bF X(K)\rtarr \dotsb\] The conclusion follows from Proposition 2.6. \end{proof} By one definition, the ring $R$ is von Neumann regular if every $R$-module is flat. \begin{cor} The Freyd conjecture holds for the derived category $\sD(R)$ of a von Neumann regular ring $R$. \end{cor} We have the following more general analogous result. \begin{cor} The Freyd conjecture holds for the derived category $\sD(R)$ if $R$ has weak dimension at most one and all finitely generated submodules of finitely generated projective $R$-modules are FP-injective. \end{cor} The ring $R$ has weak dimension at most one if all submodules of flat modules are flat. A module $M$ is FP-injective if $\Ext^1(F,M)=0$ for all finitely presented modules $F$. By \cite[Theorem 4.89]{Lam}, $M$ is FP-injective if and only if any short exact sequence of right modules \[0\to M\to L\to N\to 0\] is pure exact, meaning that it remains exact on tensoring with any left module. \begin{proof} Let $R$ satisfy the stated hypotheses. We show that the homology modules $H_i(X)$ of a perfect complex $X$ are flat. Proposition \ref{flathomology} then applies. Consider the short exact sequence \begin{align}\label{SES} 0\to B_i(X)\to Z_i(X)\to H_i(X)\to 0. \end{align} The module $Z_i(X)$ is flat since it is a submodule of the projective, hence flat, module $X_i$. By \cite[Corollary 4.86]{Lam}, it follows that $H_i(X)$ is flat if and only if the sequence (\ref{SES}) is pure exact. The module $B_i(X)$ is finitely generated since it is a quotient of the finitely generated module $X_{i+1}$, and it is a submodule of the finitely generated projective $R$-module $X_i$. By assumption, this implies that $B_i(X)$ is FP-injective. Therefore (\ref{SES}) is pure exact and $H_i(X)$ is flat. \end{proof} This elementary result comes close to one direction of the best possible result about $\sD(R)$, which is proven in \cite[Theorem 2.1]{HLP} and states that the Freyd conjecture holds for $\sD(R)$ if and only if $R$ has weak dimension at most one and all finitely presented $R$-modules are FP-injective.

TITLE: Finding the recursive formula for calculating the solution of the linear equation system $A$x = b. QUESTION [2 upvotes]: Let $K$ be a field and n $\in$ $\mathbb{N}$. It's given that $A$ = $ \begin{bmatrix} 0 & p_{1} & 0 & 0 & \dots & 0 \\ q_{2} & 0 & p_{2} & 0 & \dots & 0 \\ 0 & q_{3} & 0 & p_{3} & \dots & 0 \\ \vdots & & \ddots & \ddots & \ddots & \vdots \\ 0 & \dots & 0 & q_{n-1} & 0 & p_{n-1} \\ 0 & \dots & 0 & 0 & q_{n} & p_{n} \end{bmatrix} $ $\in$ $K^{n,n}$, $b$ = $ \begin{bmatrix} b_{1} \\ \vdots \\ b_{n} \\ \end{bmatrix} $ $\in$ $K^{n,1}$ with $p_{i}$,$q_{i}$ $\neq$ 0 for all i = 1, $\dots$ ,n. Find a recursive formula for calculating the solution of the linear equation system $A$x = b. My ideas and thoughts: I thought about writing the matrix as the following:$$ \left[\begin{array}{cccccc|c} 0 & p_{1} & 0 & 0 & \dots & 0 & b_{1} \\ q_{2}&0&p_{2}&0&\dots&0&\vdots\\ 0&q_{3}&0&p_{3}&\dots&0&\vdots\\ \vdots&&\ddots&\ddots&\ddots&\vdots&\vdots \\ 0 & \dots & 0 & q_{n-1}&0&p_{n-1}&\vdots \\ 0&\dots&0&0&q_{n}&p_{n}&b_{n} \end{array}\right] $$ so that \begin{equation} = \begin{cases} 0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1} \\ q_{2} \cdot x_{1} + 0 \cdot x_{2} + p_{2} \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{2} \\ 0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3} \\ \vdots \\ 0 \cdot x_{1} \dots 0 \cdot x_{n} + q_{n-1} \cdot x_{n+1} + 0 \cdot x_{n+2} + p_{n-1} \cdot x_{n+3} = b_{n-1} \\ 0 \cdot x_{1} \dots 0 \cdot x_{n} + 0 \cdot x_{n+1} + q_{n} \cdot x_{n+2} + p_{n} \cdot x_{n+3} = b_{n} \end{cases} \end{equation} Now I could calculate the solution for each row: (1) $0 \cdot x_{1} + p_{1} \cdot x_{2} + 0 \cdot x_{3} + 0 \cdot x_{4} \dots 0 \cdot x_{n} = b_{1}$ $\Leftrightarrow$ $ p_{1} \cdot x_{2} = b_{1}$ $\quad$ $\mid$ $\div$ $p_{1}$ $\Leftrightarrow$ $x_{2}$ = $\frac{b_{1}}{p_{1}}$ (3) $ 0 \cdot x_{1} + q_{3} \cdot x_{2} + 0 \cdot x_{3} + p_{3} \cdot x_{4} \dots 0 \cdot x_{n} = b_{3}$ $\Leftrightarrow$ $q_{3} \cdot x_{2} + p_{3} \cdot x_{4} = b_{3}$ $\Leftrightarrow$ $q_{3} \cdot \frac{b_{1}}{p_{1}} + p_{3} \cdot x_{4} = b_{3}$ $\quad$ $\mid$ $- q_{3} \cdot \frac{b_{1}}{p_{1}}$ $\Leftrightarrow$ $p_{3} \cdot x_{4} = b_{3} - q_{3} \cdot \frac{b_{1}}{p_{1}}$$\quad$ $\mid$ $\div p_{3}$ $\Leftrightarrow$ $x_{4}$ = $\frac{b_{3}-q_{3} \cdot \frac{b_{1}}{p_{1}}}{p_{3}}$ $\Leftrightarrow$ $\frac{p_{1} \cdot b_{3}-q_{3} \cdot b_{1}}{p_{1} \cdot p_{3}}$ And so on $\dots$ I don't know how to continue at this point to find a recursive formula for calculating the solution of the linear equation system. My ideas did not really lead me anywhere (I don't even think that they are correct) and im not sure if I even understand the matrix correctly. So how should I start instead? Any hints guiding me to the right direction I much appreciate. REPLY [1 votes]: So, a) solve the 1st as $x_2=b_1/p_1$; b) solve the 2nd in terms of $x_3=(b_2-q_2x_1)/p_2$, the $x_1$ is unknown: keep it as such and continue; c) all the following $x_{2k}$ will be a function of the precedent even-index and are therefore determined; d) all the following $x_{2k+1}$ will be a linear function of the precedent odd-indexed and thus a linear function of $x_1$; e) continue till the last two rows, which you are going to solve both in terms of $x_n$: one will refer to $x_{n-2}$ and one to $x_{n-1}$, thus one will contain the unknown and the other not, and you can solve for both $x_1$ and $p_n$; f) replace the found value of $x_1$ in the precedent results depending on it.

TITLE: Equivalence of matrix norms: $C_1 \cdot \|A\|_p \leq \|A\|_q \leq C_2 \|A\|_p ~~~\forall A \in \mathbb{R}^{m×n}$ QUESTION [3 upvotes]: Using : $$ \|x\|_p \leq \lambda_{pq}(n) \cdot\|x\|_q \ \ \ \ \ ,\ \ \ \lambda_{pq}(n):= \left\{\begin{matrix} 1, \ \ \ \ \ \ \ \ \ \ \ p \geq q \\ n^{(\frac{1}{p} - \frac{1}{q})}, \ \ p \geq q \end{matrix}\right.$$ and $$ \|A\|_p := \max_{\|x\|_p = 1} \|Ax\|_p = \max_{x \neq 0} \frac{\|A\|_p}{\|x\|_p} $$ show that $\exists \ 0 < C_1 \leq C_2$ such that : $$C_1 \cdot \|A\|_p \leq \|A\|_q \leq C_2 \|A\|_p \ \ \ \ \forall A \in \mathbb{R}^{m×n}$$ What I have done : $$\|A\|_p = \max_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p} \leq \max_{x \neq 0} \frac{\|A\|_p\cdot||x||_p}{\|x\|_p} \leq \max_{x \neq 0} \frac{\|A\|_p}{\|x\|_p}\cdot \lambda_{pq}(n)\cdot\|x\|_q$$ $$\text{if } p<q \text{ then } \leq \max_{x \neq 0} \frac{\|A\|_p}{\|x\|_p}\cdot n^{(\frac{1}{p} - \frac{1}{q})}\cdot\|x\|_q = n^{(\frac{1}{p} - \frac{1}{q})} \cdot \max_{x \neq 0} \frac{\|A\|_p \cdot\|x\|_q}{\|x\|_p} \text{ if } p<q \text{ then } \leq \max_{x \neq 0} \frac{\|A\|_p \cdot\|x\|_q}{\|x\|_p}$$ So I have : $$\|A\|_p\leq n^{(\frac{1}{p} - \frac{1}{q})}\cdot \max_{x \neq 0} \frac{\|A\|_p \cdot\|x\|_q}{\|x\|_p}$$ but I can't manage to have $$C_1 \cdot \|A\|_p\leq \max_{x \neq 0} \frac{\|Ax\|_q}{\|x\|_q}$$ REPLY [1 votes]: With OP idea For $x\in\Bbb R^d$, we have $$\|x\|_p \leq \lambda_{pq}(d) \cdot\|x\|_q \ \ \ \ \ ,\ \ \ \lambda_{pq}(d):= \begin{cases} 1, & p \geq q \\ d^{(\frac{1}{p} - \frac{1}{q})}, & p \geq q \end{cases} $$ Hence By Symmetry$$\lambda^{-1}_{pq}(d) \cdot\|x\|_q \le \|x\|_p \leq \lambda_{qp}(d) \cdot\|x\|_q \tag{II}$$ Then, for $x\in\Bbb R^n$ we have, $Ax\in \Bbb R^m$ and hence, $$\|Ax\|_p \overset{(II)}{\le}\lambda_{pq}(m) \cdot\|Ax\|_q \le \lambda_{pq}(m) \|A\|_q\cdot\|x\|_q \overset{(II)}{\le} \lambda_{pq}(m)\lambda_{qp}(n) \|A\|_q\cdot\|x\|_p$$ that is $$\|A\|_p=\max_{x\neq0}\frac{\|Ax\|_p}{\|x\|_p} \le \lambda_{pq}(m)\lambda_{qp}(n) \|A\|_q $$ That is $$ \color{blue}{ \|A\|_p \le \lambda_{pq}(m)\lambda_{qp}(n)\|A\|_q }$$ By symmetry we have, $$ \color{blue}{ \|A\|_q \le \lambda_{pq}(n)\lambda_{qp}(m) \|A\|_p }$$ That is $$ \color{blue}{ \lambda^{-1}_{pq}(n)\lambda^{-1}_{qp}(m) \|A\|_p \le \|A\|_q \le \lambda_{pq}(m)\lambda_{qp}(n) \|A\|_p }$$

Kontera ContentLink In-Text Ads Review By Reaper-X on Mar 16, 2007 in Advertisement, Income Sources, Internet, Making Money Today i just got accepted into Kontera In-text advertisement program, and to those who didn’t know or never heard about Kontera In-Text Ads known as ContentLink before, here’s the short description about Kontera ContentLink What is Kontera ContentLink ? A contextually relevant keyword that is discovered in real-time on a web page from within Kontera´s vast network of publishers, and is automatically turned into a link to the most relevant ad from among Kontera´s thousands of advertisers. With ContentLink publishers generate incremental revenue while advertisers reach their most targeted audience on a Cost-Per-Click basis What are the requirements to Join Kontera ? - Generate more than 500,000 page impressions per month - Be content-rich with more than 50 words per page on the majority of the site - Use English as the primary language - Kontera technology works in real time and therefore we work with publishers and advertisers in all content categories, and even with user generated content such as blogs, forums, social networking sites etc. If you saw the italic part of the above Kontera Requirements, you’ll notice that your website must have a minimum 500.000 page impression per month before you can be accepted as Kontera Publisher, of course for this is a really high requirement for most publisher even for this website Fortunately, the above requirements is no longer valid .. and this is all thanks to Joel Comm and this has been discussed over at Digital Point forums Can i use Kontera with Adsense ? Definitely, ever since Google updated their adsense terms of use, many publishers now can use other contextual advertising network with adsense as long as the ad format is different from the adsense format If the ad format is quite similar to adsense ads, then you can still use it too as long as you use different color scheme to distinguish between adsense ads and other contextual ads Can i make more money from Kontera ? That’s depend on your site traffic itself, basically Kontera is just like adsense, if you only get small number of traffic per day then you can be sure you’re not going to make money from it. So your best bet is to increase your site traffic :) My own opinion about Kontera Some people might / might not like in-text advertisement, but if it’s for me i find Kontera ContentLink non obtrusive. After all it’s just a normal link but in this case it was more like Snap Preview. Some says Snap preview is useful while the other says Snap is useless / obtrusive. But i'm also interested on hearing your own opinion about in-text advertisement :) Hello, My name is Vered and I'm from Kontera. Welcome aboard and thanks for the great review :). If you're interested in additional information about optimizing ContentLinks performance, I suggest you visit our official Kontera blog (blog.kontera.com). Vered Publisher Services Manager Hi, it's been almost a year since you started to use Kontera and I'm writing a review about the service and I would like to hear your experience. How much did you made, is it worth? and if possible some screen shots. If you can help please send me a email so we can talk about it. jdwbn chezdpsqm ewsluzf dxfvbpwtn eats riboxwum ceuyi

My Baby: A Life Worth Living? Director: BBC Year: 2006 Country: UK Rating: N/A Filed under: Eugenics and Genetic Enhancement, Documentary Candida Harris explores the emotional and practical dilemmas faced by couples desperate for children, yet all too aware of the implications of passing on their disability to their children. Disabled herself, with a 50 percent chance of passing the condition on to her child, Candida speaks to four different women and their families, who find themselves in a similar situation. One couple's only definite way of avoiding passing a condition on to their second child would be to get pregnant, then screen and abort if the condition is detected. The second family has a severe genetic condition which cannot be found during a screening process. Others have knowingly passed on disabilities to their offspring. The film also explores new developments in reproductive technology - in particular, one technique which can produce a baby free of a certain condition. (Aired BBC4, Wednesday 26 April 2006, 21.00; re-broadcast BBC Two, Wednesday 24 May)

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\begin{document} \makeatletter \input{myeqn.tex} \newcommand{\mybin}[2]{\text{$\Bigl(\begin{array}{@{}c@{}}#1\\#2 \end{array}\Bigr)$}} \newcommand{\mybinn}[2]{\text{$\biggl(\begin{array}{@{}c@{}} #1\\#2\end{array}\biggr)$}} \def\overtwo#1{\mbox{\small$\mybin{#1}{2}$}} \newcommand{\mybr}[2]{\text{$\Bigl\lfloor\mbox{ \small$\displaystyle\frac{#1}{#2}$}\Bigr\rfloor$}} \def\mybrtwo#1{\mbox{\mybr{#1}{2}}} \def\myfrac#1#2{\raisebox{0.2em}{\small$#1$}\!/\!\raisebox{-0.2em}{\small$#2$}} \def\myeqnlabel{\bgroup\@ifnextchar[{\@maketheeq}{\immediate \stepcounter{equation}\@myeqnlabel}} \def\@maketheeq[#1]{\def\theequation{#1}\@myeqnlabel} \def\@myeqnlabel#1{ {\edef\@currentlabel{\theequation} \label{#1}\enspace\eqref{#1}}\egroup} \def\rato#1{\hbox to #1{\rightarrowfill}} \def\arrowname#1{{\enspace \setbox7=\hbox{F}\setbox6=\hbox{ \setbox0=\hbox{\footnotesize $#1$}\setbox1=\hbox{$\to$} \dimen@\wd0\advance\dimen@ by 0.66\wd1\relax $\stackrel{\rato{\dimen@}}{\copy0}$} \ifdim\ht6>\ht7\dimen@\ht7\advance\dimen@ by -\ht6\else \dimen@\z@\fi\raise\dimen@\box6\enspace}} \def\namedarrow#1{{\enspace \setbox7=\hbox{F}\setbox6=\hbox{ \setbox0=\hbox{\footnotesize $#1$}\setbox1=\hbox{$\to$} \dimen@\wd0\advance\dimen@ by 0.66\wd1\relax $\stackrel{\rato{\dimen@}}{\copy0}$} \ifdim\ht6>\ht7\dimen@\ht7\advance\dimen@ by -\ht6\else \dimen@\z@\fi\raise\dimen@\box6\enspace}} \def\epsfs#1#2{{\epsfxsize#1\relax\epsffile{#2.eps}}} \author{A. Stoimenow\footnotemark[1]\\[2mm] \small Department of Mathematics,\\ \small University of Toronto,\\ \small Canada M5S 3G3,\\ \small e-mail: {\tt stoimeno@math.toronto.edu},\\ \small URL: {\hbox{\tt http://www.math.toronto.edu/stoimeno/}} } {\def\thefootnote{\fnsymbol{footnote}} \footnotetext[1]{Supported by a DFG postdoc grant.} } \title{\large\bf \uppercase{A property of the skein polynomial with}\\[2mm] \uppercase{an application to contact geometry}\\[4mm] {\small\it This is a preprint. I would be grateful for any comments and corrections!} } \date{\large Current version: \today\ \ \ First version: \makedate{19}{6}{2000}} \maketitle \makeatletter \let\point\pt \let\ay\asymp \let\pa\partial \let\al\alpha \let\be\beta \let\Gm\Gamma \let\gm\gamma \let\de\delta \let\dl\delta \let\eps\epsilon \let\lm\lambda \let\Lm\Lambda \let\sg\sigma \let\vp\varphi \let\om\omega \let\fa\forall \def\ssim{\stackrel{\ds \sim}{\vbox{\vskip-0.2em\hbox{$\scriptstyle *$}} }} \let\sm\setminus \let\tl\tilde \def\ncap{\not\mathrel{\cap}} \def\pr{\text{\rm pr}\,} \def\lra{\longrightarrow} \def\Lra{\leftrightarrow} \def\llra{\longleftrightarrow} \def\so{\Rightarrow} \def\So{\Longrightarrow} \let\ds\displaystyle \def\bt{\bar t_2} \def\Md{\max\deg} \def\md{\min\deg} \def\spn{{\mathop {\operator@font span}}} \def\mc{\min\cf} \def\cf{\mathop{\operator@font cf}} \let\reference\ref \long\def\@makecaption#1#2{ \vskip 10pt {\let\label\@gobble \let\ignorespaces\@empty \xdef\@tempt{#2} } \ea\@ifempty\ea{\@tempt}{ \setbox\@tempboxa\hbox{ \fignr#1#2} }{ \setbox\@tempboxa\hbox{ {\fignr#1:}\capt\ #2} } \ifdim \wd\@tempboxa >\captionwidth { \rightskip=\@captionmargin\leftskip=\@captionmargin \unhbox\@tempboxa\par} \else \hbox to\captionwidth{\hfil\box\@tempboxa\hfil} \fi} \def\fignr{\small\sffamily\bfseries} \def\capt{\small\sffamily} \newdimen\@captionmargin\@captionmargin2cm\relax \newdimen\captionwidth\captionwidth\hsize\relax \def\eqref#1{(\protect\ref{#1})} \def\proof{\@ifnextchar[{\@proof}{\@proof[\unskip]}} \def\@proof[#1]{\noindent{\bf Proof #1.}\enspace} \def\hint{\noindent Hint: } \def\problem{\noindent{\bf Problem.} } \def\@mt#1{\ifmmode#1\else$#1$\fi} \def\qed{\hfill\@mt{\Box}} \def\qqed{\hfill\@mt{\Box\enspace\Box}} \def\cA{{\cal A}} \def\cM{{\cal M}} \def\cF{{\cal F}} \def\cK{{\cal K}} \def\cU{{\cal U}} \def\cC{{\cal C}} \def\cP{{\cal P}} \def\tP{{\tilde P}} \def\tZ{{\tilde Z}} \def\cV{{\cal V}} \def\fg{{\frak g}} \def\tr{\text{tr}} \def\cZ{{\cal Z}} \def\cD{{\cal D}} \def\bR{{\Bbb R}} \def\cE{{\cal E}} \def\bZ{{\Bbb Z}} \def\bN{{\Bbb N}} \def\bQ{{\Bbb Q}} \def\bysame{\same[\kern2cm]\,} \def\br#1{\left\lfloor#1\right\rfloor} \def\BR#1{\left\lceil#1\right\rceil} \def\abstractname{} \@addtoreset {footnote}{page} \renewcommand{\section}{ \@startsection {section}{1}{\z@}{-1.5ex \@plus -1ex \@minus -.2ex} {1ex \@plus.2ex}{\large\bf} } \renewcommand{\@seccntformat}[1]{\csname the#1\endcsname . \quad} \def\bC{{\Bbb C}} \def\bP{{\Bbb P}} {\let\@noitemerr\relax \vskip-2.7em\kern0pt\begin{abstract} \noindent{\bf Abstract.}\enspace We prove a finiteness property of the values of the skein polynomial of homogeneous knots which allows to establish large classes of such knots to have arbitrarily unsharp Bennequin inequality (for the Thurston-Bennequin invariant of any of their Legendrian embeddings in the standard contact structure of $\bR^3$), and a give a short proof that there are only finitely many among these knots that have given genus and given braid index.\\[1mm] \noindent\em{Keywords:} homogeneous knots, Seifert surfaces, braid index, genus, Bennequin inequality, Legendrian knot.\\[1mm] \noindent\em{AMS subject classification:} 57M25 (primary), 53C15, 58A30 (secondary). \end{abstract} } \section{Introduction} In this paper, we will show the following result on the skein (or HOMFLY) polynomial \cite{HOMFLY}. \begin{theorem}\label{tm} The set \[ \{\,P_K\,:\,\spn_l P_K\le b,\,\tl g(K)\le g\,\} \] is finite for any natural numbers $g$ and $b$. \end{theorem} Here $\spn_l P_K$ if the span of the HOMFLY polynomial $P_K=P(K)$ of a knot $K$ in the (non-Alexander) variable $l$, that is, the difference between its minimal and maximal degree in $l$, $\md_l(P_K)$ and $\Md_l(P_K)$. By $\tl g(K)$ we denote the weak genus of $K$ \cite{gen1}. The main application of this theorem is to exhibit large families of knots to have arbitrarily unsharp Bennequin inequality for any of their realizations as topological knot type of a Legendrian knot, which simplifies and extends the main result of Kanda \cite{Kanda} and its alternative proofs given by Fuchs--Tabachnikov \cite{TabFuchs} and Dasbach-Mangum \cite{DasMan}. \begin{corr}\label{cr4} Bennequin's inequality becomes arbitrarily unsharp on any sequence of (Legendrian embeddings of distinct) properly obversed (mirrored) homogeneous knots. \end{corr} The main tool we use for the proof of theorem \reference{tm} is the result of \cite{gen1} and an analysis of the skein (HOMFLY) polynomial \cite{HOMFLY}. The proof of corollary \reference{cr4} uses the inequality, known from work of Tabachnikov \cite{Tab,TabFuchs}, relating the Thurston--Bennequin and Maslov (rotation) number of Legendrian knots to the minimal degree of their skein polynomial. This inequality suggests that one should look at knots behaving ``nicely'' with respect to their skein polynomial. The homogeneous knots introduced by Cromwell in \cite{Cromwell}, are, in some sense, the largest class of such knots. These are the knots having homogeneous diagrams, that is, diagrams containing in each connected component (block) of the complement of their Seifert (circle) picture only crossings of the same sign. This class contains the classes of alternating and positive/ negative knots. The other application we give of theorem \reference{tm} is also related to Bennequin's paper \cite{Bennequin} and the work of Birman and Menasco building on it. In their paper \cite{BirMen}, its referee made the observation that there are only finitely many knots of given genus and given braid index (theorem 2). This fact came as a bi-product of the work of the authors on braid foliations introduced in Bennequin's paper and bases on a rather deep theory. Here we will use our work in \cite{gen1} to give a simple, because entirely combinatorial, proof of a generalization of this result for homogeneous knots. In fact we show that the lower bound for the braid index coming from the inequality of Franks--Williams \cite{WilFr} and Morton \cite{Morton} gets arbitrarily large for homogeneous knots of given genus (corollary \reference{cry}). The result of \cite{BirMen} for homogeneous knots is then a formal consequence of ours. \begin{corr}\label{crBM} There are only finitely many homogeneous knots $K$ of given genus $g(K)$ and given braid index $b(K)$. \end{corr} We should remark that the corollary will straightforwardly generalize to links. The arguments we will give apply for links of any given (fixed) number of components, and clearly a link of braid index $n$ has at most $n$ components. Since this paper was originally written, more work was done on the subject, including by Etnyre, Honda, Ng, and in particular Plamenevskaya \cite{Plamenevskaya}. A recent survey can be found in \cite{Etnyre}. \section{Knot-theoretic preliminaries} The \em{skein (HOMFLY) polynomial}\footnote{Further names I have seen in the literature are: 2-variable Jones, Jones-Conway, LYMFHO, FLYPMOTH, HOMFLYPT, LYMPHTOFU, \dots} $P$ is a Laurent polynomial in two variables $l$ and $m$ of oriented knots and links and can be defined by being $1$ on the unknot and the (skein) relation \begin{eqn}\label{1} l^{-1}\,P\bigl( \diag{5mm}{1}{1}{ \picmultivecline{0.18 1 -1.0 0}{1 0}{0 1} \picmultivecline{0.18 1 -1.0 0}{0 0}{1 1} } \bigr)\,+\, l \,P\bigl( \diag{5mm}{1}{1}{ \picmultivecline{0.18 1 -1 0}{0 0}{1 1} \picmultivecline{0.18 1 -1 0}{1 0}{0 1} } \bigr)\,=\, -m\,P\bigl( \diag{5mm}{1}{1}{ \piccirclevecarc{1.35 0.5}{0.7}{-230 -130} \piccirclevecarc{-0.35 0.5}{0.7}{310 50} } \bigr)\,. \end{eqn} This convention uses the variables of \cite{LickMil}, but differs from theirs by the interchange of $l$ and $l^{-1}$. Let $[Y]_{t^a}=[Y]_a$ be the \em{coefficient} of $t^a$ in a polynomial $Y\in\bZ[t^{\pm1}]$. For $Y\ne 0$, let $\cC_Y\,=\,\{\,a\in\bZ\,:\,[Y]_a\ne 0\,\}$ and define \[ \md Y=\min\,\cC_Y\,,\quad \Md Y=\max\,\cC_Y\,,\mbox{\qquad and\qquad} \spn Y=\Md Y-\md Y\, \] to be the \em{minimal} and \em{maximal degree} and \em{span} (or \em{breadth}) of $Y$, respectively. Similarly one defines for $Y\in\bZ[x_1,\dots,x_n]$ the coefficient $[Y]_X$ for some monomial $X$ in the $x_i$. For a multi-variable polynomial the coefficient may be taken with respect only to some variables, and is a polynomial in the remaining variables, for example $[Y]_{x_1^k}\in \bZ[x_2,\dots,x_n]$. (Thus it must be clear as a monomial in which variables $X$ is meant. For example, for $X=x_1^k\in\bZ[x_1]$, the coefficient $[Y]_{X}=[Y]_{x_1^k}\in\bZ[x_2,\dots,x_n]$ is \em{not} the same as when regarding $X=x_1^k=x_1^kx_2^0\in\bZ[x_1,x_2]$ and taking $[Y]_{X}=[Y]_{x_1^kx_2^0}\in\bZ[x_3,\dots,x_n]$.) We call the three diagram fragments in \eqref{1} from left to right a \em{positive} crossing, a \em{negative} crossing and a \em{smoothed out} crossing (in the skein sense). The smoothing out of each crossing in a diagram $D$ leaves a collection of disjoint circles called \em{Seifert circles}. We write $c(D)$ for the number of crossings of $D$ and $s(D)$ for the number of its Seifert circles. The \em{weak genus} of $K$ \cite{gen1} is the minimal genus of all its diagrams, and the genus $g(D)$ of a diagram $D$ we will call the genus of the surface, obtained by applying the Seifert algorithm to this diagram: \[ \tl g(K)\,=\,\,\min\left\{\,g(D)=\frac{c(D)-s(D)+1}{2}\,:\,\mbox{$D$ is a diagram of $K$}\,\right\}\,. \] The \em{genus} $g(K)$ of $K$ is the minimal genus of all Seifert surfaces of $K$ (not necessarily coming from Seifert's algorithm on diagrams of $K$). The \em{slice genus} $g_s(K)$ of $K$ is the minimal genus of all smoothly embedded surfaces $S\subset B^4$ with $\partial S=K\subset S^3=\partial B^4$. Clearly $g_s(K)\le g(K)\le \tl g(K)$. $\tl g(K)$ coincides with the usual genus for many knots, in particular knots up to 10 crossings and homogeneous knots. The \em{braid index} $b(K)$ of $K$ is the minimal number of strings of a braid having $K$ as its closure. See \cite{BirMen,WilFr,Morton}. Recall, that a knot $K$ is \em{homogeneous}, if it has a diagram $D$ containing in each connected component of the complement (in $\bR^2$) of the Seifert circles of $D$ (called \em{block} in \cite[\S 1]{Cromwell}) only crossings of the same sign (that is, only positive or only negative ones). This notion was introduced in \cite{Cromwell} as a generalization of the notion of alternating and positive knots. \section{On the genus and braid index of homogeneous knots} The main tool we use for the proof of theorem \reference{tm} is the result of \cite{gen1}. \begin{theorem}\label{thm}(\cite{gen1}) Knot diagrams of given genus (with no nugatory crossings and modulo crossing changes) decompose into finitely many equivalence classes under flypes \cite{MenThis} and (reversed) applications of antiparallel twists at a crossing \begin{eqn}\label{move} \diag{7mm}{1}{1}{ \picmultivecline{0.12 1 -1.0 0}{0 0}{1 1} \picmultivecline{0.12 1 -1.0 0}{1 0}{0 1} }\llra\quad \diag{7mm}{3}{2}{ \picPSgraphics{0 setlinecap} \pictranslate{0.5 1}{ \picrotate{-90}{ \lbraid{0 -0.5}{1 1} \lbraid{0 0.5}{1 1} \lbraid{0 1.5}{1 1} \pictranslate{-0.5 0}{ \picvecline{0.03 1.95}{0 2} \picvecline{0.03 -.95}{0 -1} } } } } \,. \end{eqn} \end{theorem} This theorem allows to define for every natural number $g$ an integer $d_g$ as follows (see \cite{gen1} for more details): call 2 crossings in a knot diagram equivalent, if there is a sequence of flypes making them to form a clasp $ \diag{6mm}{2}{1}{ \picrotate{-90}{ \lbraid{-0.5 0.5}{1 1} \lbraid{-0.5 1.5}{1 1} \picvecline{-0.95 1.9}{-1 2} \picvecline{-0.05 0.1}{0 0} } }\,, $ in which the strands are reversely oriented. One checks that this is an equivalence relation. Then $d_g$ is the maximal number of equivalence classes of crossings of diagrams of genus $g$. The theorem ensures that $d_g$ is finite. It follows from the work of Menasco and Thistlethwaite \cite{MenThis} that $d_g$ can we expressed more self-containedly as \[ d_g\,=\,1+\sup\left\{\,i\in\bR\,:\,\limsup_{n\to\infty} \,\frac{a_{n,g}}{n^i}\,>\,0\,\right\}\,, \] where $a_{n,g}$ is the number of alternating knots of $n$ crossings and genus $g$. (Note that it is not \em{a priori} clear, whether the supremum on the right is integral or even finite.) \proof[of theorem \reference{tm}] It follows from theorem \reference{thm} that we can w.l.o.g. consider only one equivalence class $\cD$ of diagrams of genus $g$ modulo the move \eqref{move}. We will now argue that the skein relation for the HOMFLY polynomial implies that for a knot diagram $D$ in $\cD$ we have for its polynomial $P(D)=P_D$ \begin{eqn}\label{ff} (l^2+1)^{d_g}P(D)\,=\,\sum_{i=1}^{n_\cD}\,l^{t_i(D)}\, L_{i,\cD}\,, \end{eqn} where the number $n_\cD$ and the polynomials $L_{i,\cD}\in\bZ[l^{\pm 1}, m^{\pm 1}]$ depend on $\cD$ only, and the only numbers depending on $D$ are the $t_i$. (The reader may compare to \cite[proof of theorem 3.1]{gwg} for the case of the Jones polynomial, which is analogous.) An easy consequence of this skein relation \eqref{1} is the relation \begin{eqn}\label{g} P_{2n+1}\,=\,\frac{(il)^{2n}-1}{l+l^{-1}}m\,P_\infty+(il)^{2n}P_1\,, \end{eqn} where $L_{2n+1}$ is the the link diagram obtained by $n$ $\bt$ moves at a positive crossing $p$ in the link diagram $L_1$ (we call the tangle in $L_{2n+1}$ containing the $2n+1$ crossings so obtained a ``twist box''), $L_\infty$ is $L_1$ with $p$ smoothed out, and $P_i$ is the polynomial of $L_i$. (Compare also to the formula (8) of \cite{gwg}, but note that this formula has a misprint: the second term on the right must be multiplied by $t$.) To obtain \eqref{ff}, iterate \eqref{g} over the at most $d_g$ twist boxes of a genus $g$ diagram. The factor $(l^2+1)^{d_g}$ is used to get disposed of the denominators. {}From \eqref{ff} we obtain for a diagram $D$ of genus $g$ \begin{eqn}\label{eq1} \big|\,\bigl[\,(l^2+1)^{d_g}P(D)\,\bigr]_{l^pm^q}\,\big|\,\le\,C_{g} \end{eqn} for some constant $C_g$ depending on $g$ only. Morton showed in \cite{Morton} that $[P(D)]_{l^pm^q}=0$ if $q>2g(D)$, and, as well-known, the same is true for $q<0$ (we assume that $D$ is a \em{knot} diagram). Furthermore, it follows from the identity $P_D(l,-l-l^{-1})=1$ that $[P(D)]_{l^pm^q}\ne 0$ for some $|p|\le 2g(D)$. Thus, if $\spn_l P_D\le b$ and $g(D)\le g$, then $[P_D]_{l^pm^q}=0$ for $|p|+q>C'_{b,g}$ with $C'_{b,g}$ depending only on $b$ and $g$, and hence the same is true for $(l^2+1)^{d_g}P_D$. But we already saw that $(l^2+1)^{d_g}P_D$ has uniformly bounded coefficients \eqref{eq1}, so that \[ \{\,(l^2+1)^{d_g}P_D\,:\,\spn_l P_D\le b,\,g(D)\le g\,\} \] is finite. From this the theorem follows because multiplication with $(l^2+1)^{d_g}$ is injective (the polynomial ring is an integrality domain). \qed \begin{corr}\label{cry} There are only finitely many homogeneous knots $K$ of given genus $g(K)$ and given value of $\spn_l P_K$. \end{corr} We should point out that (trivially) a given knot may have infinitely many diagrams of given genus, and that even infinitely many different knots may have diagrams of given genus with the same HOMFLY polynomial \cite{Kanenobu}. This, not unexpectedly, shows that the combinatorial approach has its limits. \proof[of corollary \reference{cry}] Combine theorem \reference{tm} with the facts that for a homogeneous knot $K$ we have $g(K)=\tl g(K)$ \cite[corollary 4.1] {Cromwell}, that there are only finitely many homogeneous knots already of given Alexander polynomial \cite[corollary 3.5]{gen1}. \qed Finally, as the inequality $\max\deg_mP(K)\le 2\tl g(K)$ of Morton \cite{Morton} is known to be sharp in very many cases, we are led to conjecture more. \begin{conjecture} The set \[ \{\,P_K\,:\,\mbox{$K$ knot, } \spn_l P_K\le b,\,\max\deg_mP_K\le g\,\} \] is finite for any natural numbers $g$ and $b$. \end{conjecture} \section{The HOMFLY polynomial and Bennequin's inequality for Legendrian knots} A \em{contact structure} on a smooth 3-manifold is a 1-form $\al$ with $\al\wedge d\al\ne 0$ (which is equivalent to the non-integrability of the plane distribution defined by $\ker\al$). In the following we consider the $1$-form $\al=dx+y\,dz$ on $\bR^3(x,y,z)$, called the \em{standard contact space}. A \em{Legendrian knot} is a smooth embedding $\cK:S^1\to \bR^3$ with $\al\left(\frac{ \partial \cK}{\partial t}\right)\equiv 0$. Each such knot has its underlying topological knot type $K=[\cK]$ and two fundamental invariants in contact geometry known as the \em{Thurston--Bennequin number} $tb(\cK)$ and \em{Maslov index} $\mu(\cK)$. (See \cite{TabFuchs,Ferrand} for an excellent introductory account on this subject.) \begin{defi} The Thurston--Bennequin number $tb(\cK)$ of a Legendrian knot $\cK$ in the standard contact space is the linking number of $\cK$ with $\cK'$, where $\cK'$ is obtained from $\cK$ by a push-forward along a vector field transverse to the (hyperplanes of the) contact structure. The Maslov (rotation) index $\mu(\cK)$ of $\cK$ is the degree of the map \[ t\in S^1\,\mapsto\,\frac{\pr\,\frac{\partial \cK}{\partial t}(t)}{ \bigl |\pr\,\frac{\partial \cK}{\partial t}(t)\bigr |}\in S^1\,, \] where $\pr\,:\,\bR^3\to\bR^2\simeq \bC$ is the projection $(x,y,z)\mapsto(y,z)$. \end{defi} Both invariants $tb(\cK)$ and $\mu(\cK)$ can be interpreted in terms of a regular diagram of the (topological) knot $[\cK]$, and thus it was recently realized that the theory of polynomial invariants of knots and links in $\bR^3$, developed after Jones \cite{Jones}, can be applied in the context of Legendrian knots to give inequalities for $tb$ and $\mu$. In particular we have the inequality \begin{eqn}\label{tbm} tb(\cK)+|\mu(\cK)|\,\le\,\md_lP([\cK])-1\,. \end{eqn} This follows from the work of Morton \cite{Morton} and Franks--Williams \cite{WilFr}, and was translated to the Legendrian knot context by Tabachnikov and Fuchs \cite{TabFuchs}. See also \cite{Tab,CGM,GorHill,Ferrand}. On the other hand, a purely topological inequality was previously known for a while~-- Bennequin's inequality. In \cite{Bennequin}, Bennequin proved \begin{eqn}\label{bi} tb(\cK)+|\mu(\cK)|\,\le\,2g([\cK])-1\,. \end{eqn} This inequality was later improved by Rudolph \cite{Rudolph3} who showed \begin{eqn}\label{rbi} tb(\cK)+|\mu(\cK)|\,\le\,2g_s([\cK])-1\,, \end{eqn} where $g_s(K)$ is the slice (4-ball) genus of $K$. This improvement used the proof of the Thom conjecture by Kronheimer and Mrowka, achieved originally by gauge theory \cite{KroMro,KroMro2}, and later much more elegantly by Seiberg--Witten invariants \cite{KroMro3}. While the r.h.s. of \eqref{bi} and \eqref{rbi} are invariant w.r.t. taking the mirror image, the l.h.s. are strongly sensitive, so we have \begin{eqn} \tau'(L)\le \tau(L)\le 2g_s(L)-1\le 2g(L)-1 \end{eqn} for any topological knot type $L$, where \[ \tau'(L):=\max\{tb(\cK)+|\mu(\cK)|\,:\,[\cK]=L\,\} \] and \[ \tau(L):=\max\{tb(\cK)+|\mu(\cK)|\,:\,[\cK]\in\{L,!L\}\,\}\,=\, \max(\tau'(L),\tau'(!L))\,, \] and $!L$ is the obverse (mirror image) of $L$. In \cite{Kanda}, Kanda used an original argument and the theory of convex surfaces in contact manifolds developed mainly by Giroux \cite{Giroux} to show that the inequality $\tau'\le 2g-1$ can get arbitrarily unsharp, i.e. $\exists\{L_i\}\,:\,\tau'(L_i)-2g(L_i)\to-\infty$. (Here, and in the following, an expression of the form `$x_n\to\infty$' should abbreviate $\lim\limits_{n\to\infty}x_n=\infty$. Analogously `$x_{n_m}\to\infty$' should mean the limit for $m\to\infty$ etc.) In Kanda's paper, all $L_i$ are alternating pretzel knots, and hence of genus 1, so that for these examples in fact we also have $\tau'(L_i)-2g_s(L_i)\to-\infty$. It was realized (see the remarks on \cite[p. 1035]{TabFuchs}) that Kanda's result admits an alternative proof using \eqref{tbm} (whose proof in turn is also ``elementary'' in a sense discussed more detailedly in \cite{Ferrand}). Other examples (connected sums of two $(2,\,.\,)$-torus knots) were given by Dasbach and Mangum \cite [\S 4.3]{DasMan}, for which even $\tau-2g\to-\infty$. However, their examples do not apply for the slice version \eqref{rbi} of Bennequin's inequality. In \cite{Ferrand} it was observed that Kanda's result also follows from the work of Rudolph \cite{Rudolph, Rudolph2}. Here we give a larger series of examples of knots with $2g-\tau \to\infty$ containing as very special cases the previous ones given by Kanda and Dasbach--Mangum. These knots show that the inexactness of Bennequin's inequality is by far not an exceptional phenomenon. While arguments also use \eqref{tbm} (and hence are much simpler than the original proof of Kanda), they still apply in many cases also for the slice version \eqref{rbi} of Bennequin's inequality. Similar reasoning works for links of any fixed number of components, but for simplicity we content ourselves only with knots. From theorem \reference{tm}, the aforementioned application to the unsharpness of Bennequin's inequality is almost straightforward. We formulate the consequence somewhat more generally and more precisely than in the introduction. \begin{theorem}\label{cr3} Let $\{L_i\}$ be a sequence of knots, such that only finitely many of the $L_i$ have the same skein polynomial. Then $2\tl g(L_i)-\min(\tau'(L_i),\tau'(!L_i))\to\infty$. If additionally $\tl g(L_i)\le C$ for some constant $C$, then even $\min(\tau'(L_i),\tau'(!L_i))\to-\infty$. \end{theorem} The condition $g=\tl g$ is very often satisfied, but unfortunately this is not always the case, as pointed out by Morton \cite[remark 2] {Morton}. Worse yet, as shown in \cite{gwg}, there cannot be any inequality of the type $\tl g(K)\le f(g(K))$ for any function $f\,:\,\bN\to\bN$ for a general knot $K$. Nevertheless, by the results mentioned in the proof of corollary \ref{cry}, any sequence of homogeneous knots satisfies $g(L_i)=\tl g(L_i)$ and the condition of theorem \reference{cr3}. In particular, we have \begin{corr}\label{cr5} If $\{L_i\}$ are negative or achiral homogeneous knots, then $2g(L_i)-\tau'(L_i)\to \infty$. If $L_i$ are negative or additionally $g(L_i)\le C$ for some constant $C$, then even $2g_s(L_i)-\tau'(L_i)\to\infty$. \qed \end{corr} \begin{rem} \def\labelenumi{\theenumi)} Before we prove theorem \reference{cr3}, we make some comments on corollary \reference{cr5}. \begin{enumerate} \item Clearly for an achiral knot $L$ we have $\tau(L)=\tau'(L)$, so that in the case all $L_i$ are achiral (like the examples $T_{2,n} \#T_{2,-n}$, with $T_{2,n}$ being the $(2,n)$-torus knot, given in \cite{DasMan}) the stronger growth statement with $\tau'$ replaced by $\tau$ holds, $2g-\tau\to\infty$. \item Contrarily, the statement $2g-\tau\to\infty$ is not true in the negative case: Tanaka \cite[theorem 2]{Tanaka} showed that $\tau'=2g-1$ for positive knots. On the other hand, this means that for negative knots $2g-\tau'\to\infty$, and in fact $2g_s-\tau'\to\infty$, as by \cite{pos} $g=g_s$ for positive (and hence also for negative) knots. However, we have from \cite{beha} the stronger statement that $\tau'\to-\infty$, which also holds for almost negative knots (see \cite[\S 5]{apos}). \item The conditions can be further weakened. For example we can replace achirality by self-conjugacy of the HOMFLY polynomial (invariance under the interchange $l\Lra l^{-1}$) and `negative' by `$k$-almost negative' for any fixed number $k$, as the condition $g=\tl g$ in corollary \reference{cr3} can in fact be weakened to $\tl g\le f(g)$ for any (fixed) function $f\,:\,\bN\to\bN$. However, in latter case the assumption needs to be retained that only finitely many $L_i$ have the same polynomial. (This is known to be automatically true for $k\le 1$ \cite{beha,apos}, but not known for $k\ge 2$.) \item The fact that our collection of examples is richer than the one of Kanda can be made precise followingly: the number of all pretzel knots of at most $n$ crossings is $O(n^3)$, while it follows from \cite{ErnSum} and \cite{fib} that the number of achiral and positive knots of crossing number at most $n$, already among the 2-bridged ones, grows exponentially in $n$. \item The boundedness condition on the genus in the achiral case is essential (at least for this method of proof) as show the examples $T_{2,n}\#T_{2,-m}$ of Dasbach and Mangum, on which the skein polynomial argument fails for the slice genus. \end{enumerate} \end{rem} \proof[of theorem \reference{cr3}] If $\tl g(L_i)\le C$, then theorem \reference{tm} implies that $\Md_lP(L_i)-\md_lP(L_i)=\spn_l P (L_i)\to\infty$. Thus $\min(\md_lP(L_i),\md_lP(!L_i))\to-\infty$, and the assertions follow from \eqref{tbm} and \eqref{rbi}. Else there is a subsequence $\{L_{i_j}\}$ with $\tl g(L_{i_j})\to\infty$. Clearly, \[ \min(\md_lP(L_{i_j}),\md_lP(!L_{i_j}))\le 0\,, \] so that $2\tl g(L_{i_j})- \min(\tau'(L_{i_j}),\tau'(!L_{i_j}))\to\infty$, that is, $\{L_i\}$ always has a subsequence with the asserted property. Applying the argument on any subsequence of $\{L_i\}$ gives the property on the whole $\{L_i\}$. \qed As a final remark, there is another inequality, proved in \cite{ChmGor} and \cite{Tab}, involving the Kauffman polynomial $F$ (in the convention of \cite{Kauffman}), \begin{eqn} tb(\cK)\,\le\,-\Md_aF([\cK])-1\,. \end{eqn} It gives in general better estimates on $tb(\cK)$, but lacks the additional term $|\mu(\cK)|$ and also a translation to the transverse knot context (see \cite[remark at end of \S 6]{Ferrand}). Contrarily, \eqref{tbm} admits a version for transverse knots as well (in which case the term $|\mu(\cK)|$ is dropped; see \cite{GorHill} and \cite[theorem 2.4]{TabFuchs}), and so our results hold in the transverse case, too. Moreover, as remarked by Ferrand in \cite[\S 8]{Ferrand} (see also \cite[Problem, p.\ 3428]{Tanaka}), the inequality \begin{eqn}\label{PF} -\Md_aF(K)\le \md_lP(K) \end{eqn} is not always satisfied. Among the 313,230 prime knots of at most 15 crossings tabulated in \cite{KnotScape} there are 134 knots $K$ such that at least one of $K$ and $!K$ fails to satisfy \eqref{PF}. The simplest examples are two 12 crossing knots, one of them, $12_{1584}$, being quoted by Ferrand. \noindent{\bf Acknowledgement.} I wish to thank to E. Ferrand and S. Tabachnikov for some helpful conversations and for introducing me to the subject. I also wish to thank to K. Hulek for organizing and inviting me to the conference ``Perspectives in Mathematics'' and letting me there prepare a part of this manuscript. {\small \let\old@bibitem\bibitem \def\bibitem[#1]{\old@bibitem}

transmission speed The actual transfer rate of the USB3.0 ultra-fast interface is about 3.2Gbps (400MB/S). The theoretical maximum speed is 5Gbps (625MB/S). data transmission USB3.0 introduces full-duplex data transmission. Two of the five wires are used to send data, two are used to receive data, and one is ground. In other words, USB 3.0 can synchronize read and write operations at full speed. Previous USB versions did not support complete duplex data transfer. Power Supply The load on the power supply has been increased to 150 mA (USB 2.0 is around 100 mA), and configured devices can be raised to 900 mA. That’s 80 percent higher than USB 2.0 and faster to charge. In addition, the minimum operating voltage of USB 3.0 is reduced from 4.4 volts to 4 volts, which is even more power-saving. power management Instead of device polling, USB 3.0 uses the interrupt driver protocol. Thus, the standby device does not consume power until there is an interrupt request for data transmission. In short, USB 3.0 supports standby, sleep, and pause states. physical appearance The above specification will also be reflected in the physical appearance of USB 3.0. USB 3.0, however, has thicker cables because it has four more internal cables than USB 2.0. However, this port is a drawback of USB 3.0. It contains additional connected devices. END.

2014—In short excerpts from Reimagining India, Starbucks CEO Howard Schultz and Abbott Laboratories CEO Miles White reflect on how global companies can succeed in India.more January 2014—Two leading Indian executives—Unique Identification Authority of India chairman Nandan Nilekani and Biocon chairman Kiran Mazumdar-Shaw—explain how technology is transforming the country.more—Two of McKinsey’s India directors examine the success of companies seeking to expand internationally and recommend how to turn domestic strength into a global presence.more December 2013—The global CEO of Ogilvy Public Relations Worldwide discusses the difficulties in simultaneously promoting India as exotic and amazing to tourists, while assuring foreign investors of the country’s stability and potential.—Google’s executive chairman expresses optimism about India’s technology sector but urges companies to prioritize domestic excellence over global dominance.more December 2013—Khosla Ventures founding partner Vinod Khosla argues that the “leapfrogging” mind-set requires policies that foster innovation not imitation.more December 2013—The founder and chief executive of InMobi explains why changing the expectations and traditions of Indian society is critical to driving economic growth.more November 2013—MIT Sloan School of Management professor Yasheng Huang warns of the dangers of India’s reliance on the IT sector for economic growth.more November 2013—India can’t afford to emulate China. Mahindra Group chairman Anand Mahindra says the country must instead develop its own path to sustainable prosperity.more November 2013—The dean of the Lee Kuan Yew School of Public Policy at the National University of Singapore on the gap between the country’s potential and its actual performance.more November 2013—Is diversity an excuse for disunity? CNN’s Fareed Zakaria says Indians must embrace their common ambitions if the nation is to fulfill its tremendous potential.more An all-star cast of contributors consider Asia’s “other superpower” Learn more

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A new batch of names has been released linked with the embattled anti-aging clinic Biogeneis. Among the new names is New York Yankees catcher Francisco Cervelli, who is entering spring training battling for the job of starting catcher. Cervelli’s name appears on the same document as former MVP Ryan Braun of the Milwaukee Brewers and Baltimore Orioles third baseman Danny Valencia. However, it does not necessarily mean Cervelli purchased or used banned substances. For Braun, this is the second time has been linked to performance enhancing drugs. He reportedly failed a drug test prior to last season, but was able to have his suspension dropped due to a technicality in the testing process. Braun’s name appears next to monetary amounts in the range of $20-30,000. This is the first time Cervelli’s name has been mentioned and it comes on the heels of Alex Rodriguez being named in the same files. Rodriguez was specifically linked to banned substances, whereas Cervelli has not at this time. Major League Baseball is currently investigating all of the players linked to Biogenesis and its owner, Anthony Bosch. It is not known at this time whether or not suspensions are forthcoming. Bosch and Biogeneis are under federal investigation for distribution of PEDs. Cervelli is entering spring training trying to replace Russell Martin, who left via free agency. It is not known at this time what exactly Cervelli might have purchased, taken or when. However, his name appearing in the documents must come as a bit of a surprise for the Yankees, who are still dealing with the fallout from Rodriguez’s name coming up. Cervelli isn’t exactly someone whose name pops out as a possible user of PEDs. Then again, in this day and age, it shouldn’t surprise anyone anymore with athletes testing positive everywhere. If MLB decides to hand out suspensions and Cervelli is one of those suspended 50 games, the Yankees will have a serious problem. A suspension of Cervelli would leave Chris Stewart and Austin Romine as the only two catchers on the 40-man roster. That would mean they would have to either sign a catcher or promote Bobby Wilson to the 40-man roster. The Yankees catching situation wasn’t exactly ideal to begin with. Cervelli’s possible ties to Biogenesis puts the Yankees in an even more difficult situation.

Rose Hotel guests never complain about the options available to them for that great national pastime: shopping. From the multitude of independent shops right outside our front door along historic Main Street, to the sprawling Stoneridge Mall, to the colorful confines of Hacienda Crossings in neighboring Dublin, Calif., (to name a few excellent retail centers), those in hot pursuit of goods and services are never disappointed. And this situation is about to get immensely better with the Nov. 8 grand opening of Paragon Outlets, the Tri-Valley’s largest retail complex yet with 543,000 square feet of sharply designed retail space situated on 42 acres along Interstate 580. Think 120 retailers anchored by the likes of Bloomingdale’s, Neiman Marcus and Saks Fifth Avenue. There are many other brands discriminating shoppers will relish, such as Armani, Barney’s, Coach, Michael Kors and Prada. Check out this more complete list on the Paragon Outlets’ website. These factory outlet stores are housed in a modern, walkable, retail center in neighboring Livermore, Calif., just a short drive from The Rose Hotel. Besides easy freeway access, Paragon Outlets had made provisions for ample parking, and it features amenities that include miles of adjacent multi-use trails and a future public park. The $162 million development is the largest factory outlet center in Northern California, and it already has future expansion plans. But let’s not get ahead of ourselves. The Champagne and Confetti Grand Opening is at 10 a.m. Nov. 8. You’re invited to join Livermore city officials, tourism representatives and media members to officially launch Paragon Outlets at a ceremony near the retail center’s Food Court. Throughout the day there will be: We have rooms available at the hotel and can arrange transportation for our guests, if you act fast. We’re likely to run out of vacancy quickly because the Paragon Outlets’ staff will, of course, be in town for the Grand Opening celebration. Many of them have been staying at The Rose during the development and construction phases of the project. Don’t miss the excitement – or the deals. Give us a call at 925-846-8802 and we’ll make arrangements. It’s a perfect opportunity to do some early Christmas shopping. Written by Mike Consol

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\begin{document} \maketitle \begin{abstract} We construct the \emph{inverse partition semigroup} $\IP_X$, isomorphic to the \emph{dual symmetric inverse monoid} $\I^{\ast}_X$, introduced in~\cite{FL}. We give a convenient geometric illustration for elements of $\IP_X$. We describe all maximal subsemigroups of $\IP_X$ and find a generating set for $\IP_X$ when $X$ is finite. We prove that all the automorphisms of $\IP_X$ are inner. We show how to embed the symmetric inverse semigroup into the inverse partition one. For finite sets $X$, we establish that, up to equivalence, there is a unique faithful effective transitive representation of $\IP_n$, namely to $\IS_{2^n-2}$. Finally, we construct an interesting $\GH$-cross-section of $\IP_n$, which is reminiscent of $\IO_n$, the $\GH$-cross-section of $\IS_n$, constructed in~\cite{Cowan-Reilly}. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{proposition}{Proposition} \newtheorem{corollary}{Corollary} \newtheorem{definition}{Definition} \newtheorem{example}{Example} \newtheorem{remark}{Remark} \section{Introduction}\label{sec:intro} The \emph{dual inverse symmetric monoid} $\I^{\ast}_X$ was introduced in \cite{FL}. It consists of all \emph{biequivalences} on a set $X$, i.e. all the binary relations $\alpha$ on $X$ that are both \emph{full}, that is $X\alpha=\alpha X=X$, and \emph{bifunctional}, that is $\alpha\circ\alpha^{-1}\circ\alpha=\alpha$. The multiplication in $\I^{\ast}_X$ is given by: \begin{equation}\label{eq:FL} \alpha\beta=\alpha\circ\bigl(\alpha^{-1}\circ\alpha\vee\beta\circ\beta^{-1}\bigr)\circ\beta, \end{equation} for $\alpha,\beta\in\I^{\ast}_X$. In the present paper we introduce the \emph{inverse partition semigroup} $\IP_X$, isomorphic to $\I^{\ast}_X$ (see Theorem~\ref{th:Isomorphism}), and investigate some its properties. The main idea for considering the same semigroup under another point of view as in \cite{FL} (see definition of $\IP_X$ below) is to provide a convenient geometric realization for elements of this semigroup, which will enable us to handle them more easily. Besides, the semigroup $\IP_X$ naturally arises as an inverse subsemigroup of the \emph{composition semigroup} $\Ch_X$ (see Proposition~\ref{pr:IP_X-inverse-subsemigroup}), constructed below, a generalization of the semigroup $\Ch_n$, introduced in~\cite{Changchang}. The latter semigroup is close to, so called, \emph{Brauer-type semigroups}, which were investigated for different reasons and from different contexts. The first paper within these investigations, was the work of Brauer,~\cite{Brauer}, where he introduced the \emph{Brauer semigroup} $\B_n$ in connection with representations of orthogonal groups. One more work, where $\B_n$ was studied in connection with representation theory is \cite{Kerov}. Further work, dedicated to $\B_n$ are \cite{KMM}, \cite{Maltcev}, \cite{MM}. For example, in \cite{KMM} all the $\GL$- and $\GR$-cross-sections are described and in \cite{MM} a presentation for the singular part of $\B_n$ is given with respect to its minimal generating set. There are several generalizations of the Brauer semigroup: the \emph{partial Brauer semigroup} $\PB_n$, introduced in \cite{Mazorchuk-PB}; the \emph{composition semigroup} $\Ch_n$, appeared in \cite{Changchang}; the \emph{dual symmetric inverse monoid} $\I^{\ast}_X$, introduced in \cite{FL}; the finite \emph{inverse partition semigroup} $\IP_n$, appeared in \cite{Maltcev-IP} (which is isomorphic to $\I^{\ast}_n$); the \emph{partial inverse partition semigroup} $\PIP_X$, introduced in \cite{KM}. For other papers, dedicated to these semigroups we refer reader to \cite{F}, \cite{LF}, \cite{Maltcev-PB-C}, \cite{Mazorchuk-End}. The main purpose of this paper is to investigate some inner semigroup properties of $\IP_X$, as well as to establish some connections of $\IP_X$ with other semigroups. The paper is organized in the following way. In section~\ref{sec:definitionIP} we define $\IP_X$. After this, in section~\ref{sec:Isomorphism}, we prove that the constructed semigroup $\IP_X$ is isomorphic to $\I^{\ast}_X$. In section~\ref{sec:green} we characterize the Green's relations and the natural order in $\IP_X$. In section~\ref{sec:some-objects} we investigate maximal subsemigroups and ideals of $\IP_X$ and define the \emph{inverse type-preserving semigroup}. In section~\ref{sec:AutomorphismsIP_X} we describe the automorphism group $\Aut(\IP_X)$. In section~\ref{sec:Connections} we obtain a method how to embed the \emph{symmetric inverse semigroup} $\IS_X$ into the inverse partition one. In section~\ref{ef} we obtain that $\IP_X$ embeds into $\IS_{2^{\lmod X\rmod}-2}$ when $\lmod X\rmod\in\mathbb{N}\setminus\{1\}$. Finally, in section~\ref{IOP_n} we define the \emph{inverse ordered partition semigroup} $\IOP_n$, which behaves similar to the $\GH$-cross-section $\IO_n$ of $\IS_n$, studied in \cite{Cowan-Reilly}. Throughout this paper for $S$ a semigroup we denote by $E(S)$ the set of all idempotents of $S$. The natural order on an inverse semigroup $S$ will be denoted by $\leq$, i.e., $a\leq b$ for $a,b\in S$ if and only if there is an idempotent $e$ of $S$ such that $a=be$ (see~\cite{Howie}). We will also need the notion of the \emph{trace} $\tr(S)$ of an inverse semigroup $S$: the set $S$ together with the partial multiplication $\ast$, defined as follows: $a\ast b$ is defined precisely when $ab\in\GR_{a}\cap\GL_{b}$ and is equal then to $ab$ (see \cite{Meakin} and section XIV.2 of \cite{Petrich}). Finally, we recall one more definition. For any inverse semigroup $S$, the \emph{inductive groupoid} of $S$, or \emph{imprint} $\im(S)$ of $S$, is the triple $\bigl(\tr(S),\leq,\star\bigr)$, where $\leq$ is the natural partial order in $S$, and $\star$ is a partial product defined by: for $e\in E(S)$, $a\in S$, $e\leq aa^{-1}$, $e\star a=ea$ (see section XIV.3.4 of \cite{Petrich}). \section{Definition of the inverse partition semigroup $\IP_X$} \label{sec:definitionIP} Throughout all the paper let $X$ be an arbitrary set. We consider a map $':X\to X'$ as a fixed bijection and will denote the inverse bijection by the same symbol, that is $(x')'=x$ for all $x\in X$. We are going to construct a semigroup $\Ch_X$. Let $\Ch_X$ be the set of all partitions of $X\cup X'$ into nonempty blocks. If $X\cup X'=\bigcup\limits_{i\in I}^{\cdot}A_i$ is a partition of $X\cup X'$ into nonempty blocks $A_i$, $i\in I$, corresponding to an element $a\in\Ch_X$, then we will write $a=\bigl(A_i\bigr)_{i\in I}$. In the case when $I=\{i_1,\ldots,i_k\}$ is finite, we will also write $a=\bigl\{A_{i_1},\ldots,A_{i_k}\bigr\}$. For $a\in\Ch_X$ and $x,y\in X\cup X'$, we set $x\equiv_{a}y$ provided that $x$ and $y$ are at the same block of $a$. Clearly, we can realize $a\in\Ch_X$ as the equivalence relation $\equiv_{a}$. Thus in spite of the fact that elements of $\Ch_X$ will be partitions, we will sometimes treat with them as with the associated equivalence relations. Take now $a,b\in\Ch_X$. Define a new equivalence relation, $\equiv$, on $X\cup X'$ as follows: \begin{itemize} \item for $x,y\in X$ we have $x\equiv y$ if and only if $x\equiv_{a} y$ or there is a sequence, $c_1,\ldots,c_{2s}$, $s\geq 1$, of elements in $X$, such that $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c_{2s-1}\equiv_{b} c_{2s}$, and $c'_{2s}\equiv_{a} y$; \item for $x,y\in X$ we have $x'\equiv y'$ if and only if $x'\equiv_{b} y'$ or there is a sequence, $c_1,\ldots,c_{2s}$, $s\geq 1$, of elements in $X$, such that $x'\equiv_{b} c_1$, $c'_1\equiv_{a} c'_2$, $c_2\equiv_{b} c_3,\ldots,$ $c'_{2s-1}\equiv_{a} c'_{2s}$, and $c_{2s}\equiv_{b} y'$; \item for $x,y\in X$ we have $x\equiv y'$ if and only if $y'\equiv x$ if and only if there is a sequence, $c_1,\ldots$, $c_{2s-1}$, $s\geq 1$, of elements in $X$, such that $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c'_{2s-2}\equiv_{a} c'_{2s-1}$, and $c_{2s-1}\equiv_{b} y'$. \end{itemize} \begin{proposition}\label{pr:well-defined} $\equiv$ is an equivalence relation on $X\cup X'$. \end{proposition} \begin{proof} It follows immediately from the definition of $\equiv$ that this relation is reflexive and symmetric. Let now $x\equiv y$ and $y\equiv z$ for some $x,y,z\in X\cup X'$. We are going to establish that $x\equiv z$. In the rest of the proof we may assume that $y\in X$, the other case is treated analogously. We have four possible cases. \emph{Case 1}. $x,z\in X$. If $x\equiv_{a}y$ or $y\equiv_{a}z$ then since $\equiv_a$ is an equivalence relation, we immediately obtain from the definition of $\equiv$ that $x\equiv z$. Otherwise we have that there exist $c_1,\ldots,c_{2s},d_1,\ldots,d_{2t}$, elements of $X$, such that $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c_{2s-1}\equiv_{b} c_{2s}$, $c'_{2s}\equiv_{a} y$ and $y\equiv_{a} d'_1$, $d_1\equiv_{b} d_2$, $d'_2\equiv_{a} d'_3,\ldots,$ $d_{2t-1}\equiv_{b} d_{2t}$, $d'_{2t}\equiv_{a} z$. Now, using transitiveness of $\equiv_a$, we can write $c'_{2s}\equiv_a d'_1$ and hence $x\equiv z$. \emph{Case 2}. $x,z\in X'$. Then there are $c_1,\ldots$, $c_{2s-1}$, $d_1,\ldots$, $d_{2t-1}$, elements of $X$, such that $x\equiv_{b}c_{2s-1}$, $c'_{2s-1}\equiv_{a}c'_{2s-2}\ldots,$ $c'_3\equiv_{a}c'_2$, $c_2\equiv_{b}c_1$, $c'_1\equiv_{a}y$ and $y\equiv_{a} d'_1$, $d_1\equiv_{b} d_2$, $d'_2\equiv_{a} d'_3,\ldots,$ $d'_{2t-2}\equiv_{a} d'_{2t-1}$, $d_{2t-1}\equiv_{b} z$. Again, using transitiveness of $\equiv_{a}$, we obtain that $c'_1\equiv_{a}d'_1$, whence $x\equiv z$. \emph{Case 3}. $x\in X$ and $z\in X'$. There exist $d_1,\ldots$, $d_{2t-1}$, elements of $X$, such that $y\equiv_{a} d'_1$, $d_1\equiv_{b} d_2$, $d'_2\equiv_{a} d'_3,\ldots,$ $d'_{2t-2}\equiv_{a} d'_{2t-1}$, and $d_{2t-1}\equiv_{b} z$. If $x\equiv_{a}y$ then due to transitiveness of $\equiv_{a}$, we have $x\equiv z$. Otherwise there are $c_1,\ldots,c_{2s}$, elements of $X$, such that $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c_{2s-1}\equiv_{b} c_{2s}$, and $c'_{2s}\equiv_{a} y$. Then it remains to notice that $c'_{2s}\equiv_{a}d'_1$. \emph{Case 4}. $x\in X'$ and $z\in X$. Then, since $z\equiv y$ and $y\equiv x$, according to Case 3, we have that $z\equiv x$, whence $x\equiv z$. The proof is complete. \end{proof} Thus $\equiv$ defines a partition of $X\cup X'$ into disjoint blocks and so belongs to $\Ch_X$. Set this partition to be a product $a\cdot b$ in $\Ch_X$. One can easily show that $\bigl(\Ch_X,\cdot\bigr)$ is a semigroup. 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\drawline(143.00,66.00)(143.00,33.00) \drawline(143.00,33.00)(149.00,32.00) \drawline(149.00,32.00)(149.00,27.00) \drawline(149.00,27.00)(145.00,27.00) \drawline(145.00,27.00)(124.00,48.00) \drawline(124.00,48.00)(120.00,48.00) \put(122.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(57.50,35.00){\makebox(0,0)[cc]{$\cdot$}} \put(109.50,35.00){\makebox(0,0)[cc]{$=$}} \put(12.00,00.00){\makebox(0,0)[cc]{$1\rightarrow$}} \put(12.00,10.00){\makebox(0,0)[cc]{$2\rightarrow$}} \put(12.00,20.00){\makebox(0,0)[cc]{$3\rightarrow$}} \put(12.00,30.00){\makebox(0,0)[cc]{$4\rightarrow$}} \put(12.00,40.00){\makebox(0,0)[cc]{$5\rightarrow$}} \put(12.00,50.00){\makebox(0,0)[cc]{$6\rightarrow$}} \put(12.00,60.00){\makebox(0,0)[cc]{$7\rightarrow$}} \put(12.00,70.00){\makebox(0,0)[cc]{$8\rightarrow$}} \end{picture} \caption{Elements of $\Ch_8$ and their multiplication.}\label{fig:f1} \end{figure} Let $\IP_X$ be the subset of $\Ch_X$, containing those elements $\bigl(A_i\bigr)_{i\in I}\in\Ch_X$ such that $A_i\cap X\ne\varnothing$ and $A_i\cap X'\ne\varnothing$ for all $i\in I$. Since the construction of $\Ch_X$, we have that $\IP_X$ is closed under the multiplication in $\Ch_X$ and so $\IP_X$ is a subsemigroup of $\Ch_X$. Observe that $\IP_X$ has the zero element, namely $\{X\cup X'\}$. We will denote this element by $0$. Obviously, if $\mid\!X\!\mid=\mid\!Y\!\mid$ then $\Ch_X\cong\Ch_Y$ and $\IP_X\cong\IP_Y$. In the case when $X=\{1,\ldots,n\}$, it will be convenient to denote $\Ch_X$ and $\IP_X$ by $\Ch_n$ and $\IP_n$ respectively. Figures~\ref{fig:f1} and~\ref{fig:f2} illustrate the given notions for the case when $X=\{1,\ldots,8\}$, where we consider elements of semigroups as couples of vertical rows of points, divided into blocks. More precisely, the left vertical row corresponds to the set $X$ and the right one to $X'$. The multiplication $a\cdot b$ is just a gluing of elements $a$ and $b$ by dint of identifying the points of $X'$ from $a$ with the corresponding elements of $X$ from $b$. On Fig.~\ref{fig:f1} we present the equality \begin{multline} \bigl\{\{1,2,1'\},\{3,4\},\{5,2'\},\{3',4',5'\},\{6,7,6',7',8'\},\{8\}\bigr\} \cdot\\ \bigl\{\{1,1'\},\{2,3,4\},\{2',3'\},\{5,5'\},\{6,4'\},\{7\},\{6',7'\},\{8,8'\}\bigr\}=\\ \bigl\{\{1,2,1'\},\{3,4\},\{2',3'\},\{5,5'\},\{6,7,4',8'\},\{6',7'\},\{8\}\bigr\} \end{multline} and on Fig.~\ref{fig:f2} we present the following one: \begin{multline} \bigl\{\{1,2'\},\{2,3,1',4'\},\{4,3'\},\{5,6,5',6',7'\},\{7,8,8'\}\bigr\} \cdot\\ \bigl\{\{1,2'\},\{2,1',3'\},\{3,4,4'\},\{5,6',8'\},\{6,5'\},\{7,8,7'\}\bigr\}=\\ \bigl\{\{1,1',3'\},\{2,3,4,2',4'\},\{5,6,7,8,5',6',7',8'\}\bigr\}. \end{multline} \begin{figure} \special{em:linewidth 0.4pt} \unitlength 0.80mm \linethickness{0.4pt} \begin{picture}(150.00,75.00) \put(20.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(20.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(20.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(20.00,30.00){\makebox(0,0)[cc]{$\bullet$}} 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\drawline(149.00,03.00)(149.00,-03.00) \drawline(149.00,-03.00)(120.00,-03.00) \put(122.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(122.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(147.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(57.50,35.00){\makebox(0,0)[cc]{$\cdot$}} \put(109.50,35.00){\makebox(0,0)[cc]{$=$}} \put(12.00,00.00){\makebox(0,0)[cc]{$1\rightarrow$}} \put(12.00,10.00){\makebox(0,0)[cc]{$2\rightarrow$}} \put(12.00,20.00){\makebox(0,0)[cc]{$3\rightarrow$}} \put(12.00,30.00){\makebox(0,0)[cc]{$4\rightarrow$}} \put(12.00,40.00){\makebox(0,0)[cc]{$5\rightarrow$}} \put(12.00,50.00){\makebox(0,0)[cc]{$6\rightarrow$}} \put(12.00,60.00){\makebox(0,0)[cc]{$7\rightarrow$}} \put(12.00,70.00){\makebox(0,0)[cc]{$8\rightarrow$}} \end{picture} \caption{Elements of $\IP_8$ and their multiplication.}\label{fig:f2} \end{figure} Now we move to the proof of the fact that $\IP_X$ is isomorphic to $\I^{\ast}_X$. \section{$\IP_X$ is isomorphic to $\I^{\ast}_X$}\label{sec:Isomorphism} The main goal of this section is to prove the following \begin{theorem}\label{th:Isomorphism} $\IP_X\cong\I^{\ast}_X$. \end{theorem} \begin{proof} We begin with recalling one notion from~\cite{FL}. A \emph{block bijection} of $X$ is a bijection between two quotient sets $X/\sigma$ and $X/\tau$ for certain equivalence relations $\sigma$ and $\tau$ on $X$ such that $\lmod X/\sigma\rmod=\lmod X/\tau\rmod$. We will need the following statement, stated in~\cite{FL} (one might find it also in~\cite{Lawson}, Section 4.2). \begin{lemma}[Lemma 2.1 from~\cite{FL}]\label{lm:FL} If $\alpha$ is a biequivalence on $X$, then both $\alpha\circ\alpha^{-1}$ and $\alpha^{-1}\circ\alpha$ are equivalence relations on $X$. Moreover the map $\widetilde{\alpha}$ defined by $\widetilde{\alpha}:x(\alpha\circ\alpha^{-1})\mapsto x\alpha$ for $x\in X$ is a block bijection of $X/\alpha\circ\alpha^{-1}$ to $X/\alpha^{-1}\circ\alpha$. Conversely, given equivalence relations $\beta$ and $\gamma$ on $X$ together with a block bijection $\mu:X/\beta\to X/\gamma$, a unique biequivalence $\widehat{\mu}$ on $X$ inducing $\mu$ is given by: $x\widehat{\mu}y$ if and only if $x\beta\mapsto y\gamma$ under the block bijection $\mu$ (in which case $\beta=\widehat{\mu}\circ\widehat{\mu}^{-1}$ and $\gamma=\widehat{\mu}^{-1}\circ\widehat{\mu}$). Finally, the two processes are reciprocal: $\widehat{\widetilde{\alpha}}=\alpha$ and $\widetilde{\widehat{\mu}}=\mu$. \end{lemma} To define an isomorphism between $\IP_X$ and $\I^{\ast}_X$, we need some auxiliary notation. Let $a\in\IP_X$. Define the following relations $\rho_a$ and $\lambda_a$ on $X$ as follows: \begin{equation} x\rho_{a}y~\mbox{if and only if}~x\equiv_a y,~\mbox{and}~x\lambda_{a} y~\mbox{if and only if}~x'\equiv_a y', \end{equation} for $x,y\in X$. Since $\rho_{a}$ is a restriction of the relation $\equiv_{a}$ to $X$, we obtain that $\rho_{a}$ is an equivalence relation on $X$. From the definition of $\lambda_{a}$ and similar arguments it follows that $\lambda_{a}$ is an equivalence relation on $X$ as well. Remark that $a$ is not determined by $\lambda_{a}$ and $\rho_{a}$. Define a map $\pi:\IP_X\to\I^{\ast}_X$ as follows: for $a\in\IP_X$ we put $\pi(a)=\widehat{\mu_a}$, where $\mu_a$ is a block bijection from $X/\rho_a$ onto $X/\lambda_a$ such that the block $A$ of $\rho_a$ is mapped under $\mu_a$ to that block $B$ of $\lambda_a$, for which $A\cup B'$ is a block of $\equiv_a$. In view of our definition of $\IP_X$ and Lemma~\ref{lm:FL}, we obtain that $\pi$ is a bijection from $\IP_X$ onto $\I^{\ast}_X$. We are left to prove that $\pi$ is a morphism from $\IP_X$ to $\I^{\ast}_X$. Take $a,b\in\IP_X$. We need to prove that $\widehat{\mu_{ab}}=\widehat{\mu_a}\widehat{\mu_b}= \widehat{\mu_a}\circ\bigl(\widehat{\mu_a}^{-1}\circ\widehat{\mu_a}\vee \widehat{\mu_b}\circ\widehat{\mu_b}^{-1}\bigr)\circ\widehat{\mu_b}$. Notice that due to Lemma~\ref{lm:FL}, we have that $\widehat{\mu_b}\circ\widehat{\mu_b}^{-1}=\rho_b$ and $\widehat{\mu_a}^{-1}\circ\widehat{\mu_a}=\lambda_a$ and hence we must establish that $\widehat{\mu_{ab}}=\widehat{\mu_a}\circ\bigl(\lambda_a\vee \rho_b\bigr)\circ\widehat{\mu_b}$. Note also that for all $c\in\IP_X$ it follows immediately from the definition of $\mu_c$ that for all $x,y\in X$ one has $x\widehat{\mu_c}y$ if and only if $x\equiv_c y'$. Finally, we recall that for equivalence relations $\lambda$ and $\rho$ on $X$, the join $\lambda\vee\rho$ coincides with the transitive closure of the relation $\lambda\cup\rho$. Suppose firstly that $x\widehat{\mu_{ab}}y$, for some $x,y\in X$. Then $x\equiv_{ab}y'$ and so there exist $c_1,\ldots$, $c_{2s-1}$, $s\geq 1$, elements of $X$, such that $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c'_{2s-2}\equiv_{a} c'_{2s-1}$, and $c_{2s-1}\equiv_{b} y'$. Then we have $x\widehat{\mu_{a}}c_1$, $c_1\rho_{b} c_2$, $c_2\lambda_{a} c_3,\ldots,$ $c_{2s-2}\lambda_{a} c_{2s-1}$, and $c_{2s-1}\widehat{\mu_{b}}y$. Thus, we have $x\widehat{\mu_{a}}c_1$, $c_1(\lambda_a\vee\rho_b)c_{2s-1}$ and $c_{2s-1}\widehat{\mu_{b}}y$, whence $(x,y)\in\widehat{\mu_a}\circ\bigl(\lambda_a\vee \rho_b\bigr)\circ\widehat{\mu_b}$. Conversely, suppose that $(x,y)\in\widehat{\mu_a}\circ\bigl(\lambda_a\vee \rho_b\bigr)\circ\widehat{\mu_b}$. Then there exist $c,d\in X$ such that $x\widehat{\mu_{a}}c$, $c(\lambda_a\vee\rho_b)d$ and $d\widehat{\mu_{b}}y$. Then we have $x\equiv_a c'$ and $d\equiv_b y'$. Notice that if $c\lambda_a r$ then $x\equiv_a r'$ and if $t\rho_b d$ then $t\equiv_b y'$. Hence, taking to account $c(\lambda_a\vee\rho_b)d$, there exist $c_1,\ldots$, $c_{2s-1}$, $s\geq 1$, elements of $X$, such that $x\equiv_{a}c_1'$, $c_1\rho_{b} c_2$, $c_2\lambda_{a} c_3,\ldots,$ $c_{2s-2}\lambda_{a} c_{2s-1}$, and $c_{2s-1}\equiv_{b}y'$. These imply $x\equiv_{a} c'_1$, $c_1\equiv_{b} c_2$, $c'_2\equiv_{a} c'_3,\ldots,$ $c'_{2s-2}\equiv_{a} c'_{2s-1}$, and $c_{2s-1}\equiv_{b} y'$. Thus $x\equiv_{ab}y'$, whence $x\widehat{\mu_{ab}}y$. The proof of the theorem is complete. \end{proof} As a consequence of Theorem~\ref{th:Isomorphism} we obtain the following statement. \begin{proposition} \label{cor:IP-inverse} $\IP_X$ is an inverse semigroup. \end{proposition} \begin{proof} Follows from the fact that $\I^{\ast}_X$ is inverse, see~\cite{FL}. \end{proof} Due to what we have already obtained, we can now call $\IP_X$ the \emph{inverse partition semigroup} on the set $X$. \section{Green's relations and the natural order in $\IP_X$}\label{sec:green} We begin this section with description of Green's relations on $\IP_X$. But before we need some preparation. First notice that it follows immediately from the definition of multiplication in $\IP_X$ that \begin{equation}\label{eq:remark-for-ranks} \rho_{ab}\supseteq\rho_a~\mbox{and}~\lambda_{ab}\supseteq\lambda_b~\mbox{for all}~a,b\in\IP_X. \end{equation} Then we obtain that every $\rho_{ab}$-class is a union of some $\rho_a$-classes and that every $\lambda_{ab}$-class is a union of some $\lambda_b$-classes. Note also that the cardinal number of the set of all $\rho_a$-classes and the cardinal number of the set of all $\lambda_a$-classes coincide with the cardinal number of the set of all $\equiv_a$-classes. Denote this common number by $\rank(a)$. We will call this number the \emph{rank} of $a$. Due to~\eqref{eq:remark-for-ranks}, we have \begin{equation} \label{eq:ranks} \rank(ab)\leq\min\bigl\{\rank(a),\rank(b)\bigr\}~\mbox{for all}~a,b\in\IP_X. \end{equation} Note that if $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$ then $\rank(a)=\lmod I\rmod$. We denote the Green's relations in the standard way: $\GR$, $\GL$, $\GH$, $\GD$, and $\GJ$ (see~\cite{Howie}). \begin{theorem} \label{th:Green} Let $a,b\in\IP_X$. Then \begin{enumerate} \item\label{R} $a\GR b$ if and only if $\rho_a=\rho_b$; \item\label{L} $a\GL b$ if and only if $\lambda_a=\lambda_b$; \item\label{H} $a\GH b$ if and only if $\rho_a=\rho_b$ and $\lambda_a=\lambda_b$ hold simultaneously; \item\label{J=D} $a\GJ b$ if and only if $a\GD b$ if and only if $\rank(a)=\rank(b)$; \item\label{card-sem} $\mid\!\IP_n\!\mid=\sum\limits_{k=1}^{n}\bigl(s(n,k)\bigr)^2\cdot k!$, where $s(n,k)$ denotes the Stirling number of the second kind; \item\label{card-idem} $\mid\!E(\IP_n)\!\mid=\mathrm{B}_n$, where $\mathrm{B}_n$ denotes the Bell number. \end{enumerate} \end{theorem} \begin{proof} In view of Theorem~\ref{th:Isomorphism}, these statements are just reformulations of those of Theorem 2.2 from~\cite{FL}. \end{proof} Now we move to description of the group of units of $\IP_X$. Denote by $\mathcal{S}_X$ the \emph{symmetric group} on $X$. Set a map $\eta:\mathcal{S}_X\to\IP_X$ as follows: \begin{equation} \eta(g)=\bigl(\{x,g(x)'\}\bigr)_{x\in X}~\mbox{for all}~g\in\mathcal{S}_X. \end{equation} \begin{lemma} \label{lm:S_XembedsintoIP_X} The map $\eta$ is an injective homomorphism. \end{lemma} \begin{proof} That $\eta$ is a homomorphism, follows from the definition of the multiplication in $\IP_X$. If now $\eta(g_1)=\eta(g_2)$ for some $g_1,g_2\in\mathcal{S}_X$, then $g_1(x)=g_2(x)$ for all $x\in X$ and so $g_1=g_2$. This completes the proof. \end{proof} As a consequence of Lemma~\ref{lm:S_XembedsintoIP_X} we obtain that $\IP_X$ contains a subgroup $\eta(\mathcal{S}_X)$, isomorphic to $\mathcal{S}_X$. Let us identify this subgroup with $\mathcal{S}_X$. Clearly, the identity element $1$ of $\Sym_X$ is the identity element of $\IP_X$. Using Theorem~\ref{th:Green}, we obtain now the following corollary. \begin{proposition} \label{cor:Invertible-Elements} The group of all invertible elements of $\IP_X$ coincides with $\Sym_X$. \end{proposition} \begin{proof} Since the maximal subgroup of an arbitrary semigroup coincides with some $\GH$-class of this semigroup (see~\cite{Howie}), we obtain that an element $g$ is invertible in $\IP_X$ if and only if $g\GH 1$. Due to Theorem~\ref{th:Green}, this is equivalent to $g\in\Sym_X$. \end{proof} Let us now switch to the description of the natural order on $\IP_X$. But before, we need to describe the idempotents of $\IP_X$. \begin{lemma} \label{pr:idempotentIP} Let $e\in\IP_X$. Then $e$ is an idempotent if and only if there is a partition $X=\bigcup\limits_{i\in I}^{\cdot}E_i$ such that $e=\bigl(E_i\cup E_i'\bigr)_{i\in I}$. In addition, for idempotents $e$ and $f$ the elements $ef$ and $fe$ coincide with the minimum equivalence relation on $X\cup X'$, which contains $e$ and $f$. \end{lemma} \begin{proof} Let us prove firstly the first part of the statement. The sufficiency of it is obvious. Let now $e$ be an idempotent of $\IP_X$. Let $A\cup B'$ be some block in $e$. Suppose that $A\setminus B\ne \varnothing$. Then there is $a\in A$ such that $a\notin B$. Take an arbitrary $b$ of $B$. Take also $c\in X$ such that $c\equiv_{e}a'$. Then $c\notin A$. Indeed, otherwise we would have $a\equiv_{e}c\equiv_{e}a'$ which implies $a\in B$. Thus, $c\notin A$. Now due to $c\equiv_{e}a'$ and $a\equiv_{e}b'$, we obtain that $c\equiv_{e^2}b'$. But the latter gives us $c\in A$. We get a contradiction. Thus, $A\setminus B=\varnothing$ and so $A\subseteq B$. Analogously, $B\subseteq A$. Thus, every block of $e$ has the form $A\cup A'$ for certain $A\subseteq X$. This completes the proof of the first part of the statement. The second one now follows immediately from the definition of the multiplication in $\IP_X$. \end{proof} \begin{proposition}\label{pr:Omega} Let $a,b\in\IP_X$. Then $a\leq b$ if and only if $\equiv_a\supseteq\equiv_b$. \end{proposition} \begin{proof} Let $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$ and $b=\bigl(C_j\cup D_j'\bigr)_{j\in J}$. Suppose first that $\equiv_{b}\subseteq\equiv_{a}$. Then we have that for all $i\in I$, $A_i\cup B_i'$ is a union of some blocks $C_j\cup D_j'$, $j\in J$. Put $f=\bigl(B_i\cup B_i'\bigr)_{i\in I}$. Then we obtain that $a=bf$. It remains to note that, due to Lemma~\ref{pr:idempotentIP}, $f$ is an idempotent. Suppose now that there is an idempotent $e$ of $\IP_X$ such that $a=be$. Due to Lemma~\ref{pr:idempotentIP}, we have that $e=\bigl(E_k\cup E_k'\bigr)_{k\in K}$ for some partition $X=\bigcup\limits_{k\in K}^{\cdot}E_k$. Take now $(x,y)\in\equiv_b$. There is $z$ of $X$ such that $z'$ is $\equiv_b$--equivalent to $x$ and $y$. Then, since $z\equiv_e z'$, we obtain that $(x,y)\in\equiv_{be}$ or just that $(x,y)\in\equiv_{a}$. This completes the proof. \end{proof} Now we are able to characterize the trace of $\IP_X$. \begin{proposition}\label{pr:trace} Let $a,b\in\tr(\IP_X)$. The product $a\ast b$ is defined if $\lambda_{a}=\rho_{b}$ and in this case $\pi(a)\circ\pi(b)\in\I^{\ast}_X$ and $a\ast b=\pi^{-1}(\pi(a)\circ\pi(b))$. \end{proposition} \begin{proof} It is known that for $x,y\in\tr(S)$, where $S$ is an inverse semigroup, the product $x\ast y$ is defined if and only if $x^{-1}x=yy^{-1}$ (see~\cite{Meakin}). Note also that, using Lemma~\ref{pr:idempotentIP}, we have that for every $x\in\IP_X$ the condition $\rho_{x}=\lambda_{x}$ holds if and only if $x\in E(\IP_X)$. In addition, for $e,f\in E(\IP_X)$ we have that $\lambda_{e}=\lambda_{f}$ if and only if $e=f$. Hence, $a\ast b$ is defined if and only if $a^{-1}a=bb^{-1}$ if and only if $\lambda_{a^{-1}a}=\rho_{bb^{-1}}$. It remains to notice that since $a^{-1}a\GL a$ and $bb^{-1}\GR b$, using Theorem~\ref{th:Green}, we have $\lambda_{a^{-1}a}=\lambda_{a}$ and $\rho_{bb^{-1}}=\rho_{b}$. If now $a\ast b$ is defined then $\pi(a)\ast\pi(b)$ is defined in $\I^{\ast}_X$ and then $\pi(a)\ast\pi(b)=\pi(a)\circ\pi(b)$ (see~\cite{Lawson}). The statement follows. \end{proof} The following proposition is concerned with $\im(\IP_X)$, the imprint of $\IP_X$. \begin{proposition} Let $e\in E(\IP_X)$ and $a\in\IP_X$. The product $e\star a$ is defined if and only if $\rho_{a}\subseteq\rho_{e}$. \end{proposition} \begin{proof} By the definition of imprint, we have that $e\star a$ is defined if and only if $e\leq aa^{-1}$, which, in view of Proposition~\ref{pr:Omega}, holds if and only if $\equiv_{aa^{-1}}\subseteq\equiv_{e}$ which is equivalent to $\rho_{aa^{-1}}\subseteq\rho_{e}$. It remains to notice that $\rho_{a}=\rho_{aa^{-1}}$. \end{proof} \section{Generating set, ideals and maximal subsemigroups of $\IP_n$} \label{sec:some-objects} To begin this section, we put some auxiliary notations. Let $A\subseteq X$. Define an element $\tau_{A}$ of $\IP_X$ as follows: \begin{equation} \tau_A=\bigl\{A\cup A',\{x,x'\}_{x\in X\setminus A}\bigr\}. \end{equation} Clearly, $\tau_X$ is the zero element of $\IP_X$. If $x$ and $y$ are distinct elements of $X$, we will use the notation $\tau_{x,y}=\tau_{\{x,y\}}$. Suppose that $\mid\!X\!\mid\geq 3$. For pairwise distinct elements $x,y,z$ of $X$ define an element $\xi_{x,y,z}$ as follows: \begin{equation} \xi_{x,y,z}=\bigl\{\{x,y,x'\},\{z,y',z'\},\{t,t'\}_{t\in X\setminus \{x,y,z\}}\bigr\}. \end{equation} If necessary, we will write $\xi_{x,y,z}^{X}$ instead of $\xi_{x,y,z}$ to stress on that $\xi_{x,y,z}\in\IP_X$. \begin{lemma}\label{lm:XI-and-TAU} Let $\mid\!X\!\mid\geq 3$. Then \begin{multline} g^{-1}\xi_{x,y,z}g= \xi_{g(x),g(y),g(z)},~g^{-1}\tau_{x,y}g=\tau_{g(x),g(y)},\\ \xi_{x,y,z}^{2}=\tau_{\{x,y,z\}}~\mbox{and}~\xi_{x,y,z}\xi_{z,y,x}=\tau_{x,y} \end{multline} for all pairwise distinct $x,y,z\in X$ and $g\in\Sym_X$. \end{lemma} \begin{proof} Direct calculation. \end{proof} Now our local goal is to provide a generating set for $\IP_n$ (see Proposition~\ref{pr:Generating-SystemsIP}). In order to do this we will construct an inverse subsemigroup $\IT_n$ of $\IP_n$ (see below), which is interesting itself as a semigroup. In addition, the notion of $\IT_n$ will help us to describe all the maximal subsemigroups of $\IP_n$. So we are starting with putting some auxiliary notations. Let $n\geq 2$. Set $\IT_n=\langle \Sym_n,\tau_{1,2}\rangle$. Set also $\IT_1=\IP_1$. Let $\rho$ be some equivalence relation on $\{1,\ldots,n\}$. Define a \emph{type} of the relation $\rho$ as a tuple $(t_1,\ldots,t_n)$, where $t_i$ denotes the number of all $i$-element $\rho$-classes, $1\leq i\leq n$. The following proposition shows that $\IT_n$ is an inverse subsemigroup of $\IP_n$. But before, we give one more definition: an element $a$ of $\IP_n$ is said to be \emph{special} if \begin{equation} x\equiv_a y'~\mbox{implies}~\mid\!x\rho_a\!\mid=\mid\!y\lambda_a\!\mid~\mbox{for all}~x,y\in\{1,\ldots,n\}. \end{equation} \begin{proposition}\label{pr:IT_n} The following statements hold: \begin{enumerate} \item\label{1} $\IT_n$ is an inverse subsemigroup of $\IP_n$; \item\label{2} $\tau_A\in \IT_n$ for all $A\subseteq\{1,\ldots,n\}$; \item\label{3} the elements of $\IT_n$ are precisely all special elements of $\IP_n$; \item\label{4} if $a\in\IT_n$ then the types of $\rho_a$ and $\lambda_a$ coincide. \end{enumerate} \end{proposition} \begin{proof} We will assume that $n\geq 2$ as all the statements hold in the case when $n=1$. Since $\Sym_n$ is a subgroup of $\IP_n$ and $\tau_{1,2}$ is an idempotent in $\IP_n$, we obtain that $\IT_n$ is an inverse subsemigroup of $\IP_n$. This completes the proof of~\ref{1}). Note that, due to Lemma~\ref{lm:XI-and-TAU}, we have that $\tau_{x,y}\in\IT_n$ for all distinct $x$ and $y$ of $\{1,\ldots,n\}$. Now the statement~\ref{2}) follows from the equality $\tau_{\{x\}}=1$, for all $x\in\{1,\ldots,n\}$, and the fact that if $A=\{x_1,\ldots,x_k\}$, $k\geq 2$, then \begin{equation} \tau_{A}=\prod\limits_{i=1}^{k-1}\tau_{x_k,x_{k+1}}. \end{equation} Let us prove~\ref{3}). Let $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$ be an element of $\IP_n$ such that $x\equiv_a y'$ implies $\mid\!x\rho_a\!\mid=\mid\!y\lambda_a\!\mid$ for all $x,y\in\{1,\ldots,n\}$. Then $\mid\!A_i\!\mid=\mid\!B_i\!\mid$ for all $i\in I$ and so there exists $g\in\Sym_n$ such that $ga=\bigl(B_i\cup B_i'\bigr)_{i\in I}$. Now due to~\ref{2}), we have that \begin{equation} a=g^{-1}\cdot\prod\limits_{i\in I}\tau_{B_i}\in\IT_n. \end{equation} Conversely, suppose that $a\in\IT_n$. Note that $\tau_{1,2}$ is special and all the elements of $\Sym_n$ are special, too. Hence, to prove that $a$ is special, it is enough to prove that if $b\in\IP_n$ is special then $b\tau_{1,2}$ is special and $bg$ is special for all $g\in\Sym_n$. Suppose that $b=\bigl(C_i\cup D_i'\bigr)_{i\in K}\in\IP_n$ is special. Then, obviously, $bg$ is also special for all $g\in\Sym_n$. We have two cases. \emph{Case 1}. There is $i\in K$ such that $D_i\supseteq\{1,2\}$. Then $b\tau_{1,2}=b$ is special. \emph{Case 2}. There are distinct $i$ and $j$ of $K$ such that $1\in D_i$ and $2\in D_j$. Then $b\tau_{1,2}=\bigl\{\bigl(C_i\cup C_j\bigr)\bigcup\bigl(D_i\cup D_j\bigr)',\bigl(C_k\cup D_k'\bigr)_{k\in K\setminus\{i,j\}}\bigr\}$ is, obviously, special. This completes the proof of~\ref{3}). The statement~\ref{4}) follows immediately from~\ref{3}). \end{proof} As a consequence of~\ref{4}) of Proposition~\ref{pr:IT_n}, we can now call $\IT_n$ the \emph{inverse type-preserving semigroup} of degree $n$. We give an illustration of elements of $\IT_8$ on Fig.~\ref{fig:f3}. It also follows from Proposition~\ref{pr:IT_n} that $\IT_n=\Sym_nE(\IP_n)$, that is $\IT_n$ is the greatest factorizable inverse submonoid of $\IP_n$. Remark that $\IT_n$ (more precisely, $\pi(\IT_n)$, the greatest factorizable inverse submonoid of $\I^{\ast}_X$) appeared in~\cite{F},~\cite{FL} and~\cite{AO} under the name of the \emph{monoid of uniform block permutations}. \begin{figure} \special{em:linewidth 0.4pt} \unitlength 0.80mm \linethickness{0.4pt} \begin{picture}(150.00,75.00) \put(15.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \drawline(13.00,-03.00)(13.00,03.00) \drawline(13.00,03.00)(42.00,13.00) \drawline(42.00,13.00)(42.00,07.00) \drawline(42.00,07.00)(13.00,-03.00) \drawline(13.00,57.00)(13.00,73.00) \drawline(13.00,73.00)(42.00,73.00) \drawline(42.00,73.00)(42.00,57.00) \drawline(42.00,57.00)(13.00,57.00) \drawline(13.00,27.00)(13.00,53.00) \drawline(13.00,53.00)(17.00,53.00) \drawline(17.00,53.00)(42.00,43.00) \drawline(42.00,43.00)(42.00,17.00) \drawline(42.00,17.00)(38.00,17.00) \drawline(38.00,17.00)(13.00,27.00) \drawline(13.00,07.00)(13.00,23.00) \drawline(13.00,23.00)(38.00,53.00) \drawline(38.00,53.00)(42.00,53.00) \drawline(42.00,53.00)(42.00,48.00) \drawline(42.00,48.00)(36.00,48.00) \drawline(36.00,48.00)(36.00,03.00) \drawline(36.00,03.00)(42.00,03.00) \drawline(42.00,03.00)(42.00,-03.00) \drawline(42.00,-03.00)(38.00,-03.00) \drawline(38.00,-03.00)(13.00,07.00) \drawline(103.00,-03.00)(103.00,33.00) \drawline(103.00,33.00)(132.00,33.00) \drawline(132.00,33.00)(132.00,-03.00) \drawline(132.00,-03.00)(103.00,-03.00) \drawline(103.00,58.00)(103.00,63.00) \drawline(103.00,63.00)(107.00,63.00) \drawline(107.00,63.00)(132.00,43.00) \drawline(132.00,43.00)(132.00,38.00) \drawline(132.00,38.00)(128.00,38.00) \drawline(128.00,38.00)(103.00,58.00) \drawline(103.00,68.00)(103.00,73.00) \drawline(103.00,73.00)(107.00,73.00) \drawline(107.00,73.00)(132.00,53.00) \drawline(132.00,53.00)(132.00,48.00) \drawline(132.00,48.00)(128.00,48.00) \drawline(128.00,48.00)(103.00,68.00) \drawline(103.00,38.00)(103.00,52.00) \drawline(103.00,52.00)(128.00,73.00) \drawline(128.00,73.00)(132.00,73.00) \drawline(132.00,73.00)(132.00,58.00) \drawline(132.00,58.00)(107.00,38.00) \drawline(107.00,38.00)(103.00,38.00) \end{picture} \caption{Elements of $\IT_8$.}\label{fig:f3} \end{figure} The following proposition gives us an example of a generating system of $\IP_n$. But to prove this proposition, we need some auxiliary facts. \begin{lemma}\label{lm:BeforePrGeneratingSystems} Let $n\geq 3$, $a\in\IP_n$ and $\rank(a)=n-1$. Then either $a\in\xi_{x,y,z}\Sym_n$ or $a\in\tau_{x,y}\Sym_n$ for some pairwise distinct $x,y,z\in\{1,\ldots,n\}$. \end{lemma} \begin{proof} Straightforward. \end{proof} Take $n\in\mathbb{N}$. Set $\Pi_n=\bigl\{q\in\IP_{n+1}:~q~\mbox{contains the block}~\{n+1,(n+1)'\}\bigr\}$. \begin{lemma}\label{lm:Pi_n} Let $n\in\mathbb{N}$. Then the map $a\mapsto a\cup\{n+1,(n+1)'\}$, $a\in\IP_n$, is an isomorphism from $\IP_n$ onto $\Pi_n$, which maps $\xi_{1,2,3}^{\{1,\ldots,n\}}$ to $\xi_{1,2,3}^{\{1,\ldots,n+1\}}$. \end{lemma} \begin{proof} Obvious. \end{proof} \begin{proposition}\label{pr:Generating-SystemsIP} Let $n\geq 3$. Then $\IP_n=\langle \Sym_n, \xi_{1,2,3}\rangle$. Moreover, for $u\in\IP_n$, $\IP_n=\langle \Sym_n,u\rangle$ if and only if $u\in \Sym_n\xi_{1,2,3}\Sym_n$. \end{proposition} \begin{proof} We will prove the statement that $\IP_n=\langle \Sym_n, \xi_{1,2,3}\rangle$ for all $n\geq 3$ by the complete induction on $n$. First, let us verify that the basis of the induction, the case when $n=3$, holds. We are to prove that $\IP_3=\langle \Sym_3, \xi_{1,2,3}\rangle$. Note that, due to Lemma~\ref{lm:XI-and-TAU}, $0=\xi_{1,2,3}^2$. Thus, we are left to prove that every element $v$ of $\IP_3$ such that $\rank(v)=2$, belongs to $\langle \Sym_3, \xi_{1,2,3}\rangle$. But this follows from Lemmas \ref{lm:XI-and-TAU} and~\ref{lm:BeforePrGeneratingSystems}. Thus, the basis of induction holds. Assume now that the proposition of induction holds for all numbers $k$, $3\leq k\leq n$. We are going to prove now that $\IP_{n+1}=\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. Let $a\in \IP_{n+1}$. Then there is $g\in \Sym_{n+1}$ such that $b=ag$ contains a block $\bigl(E\cup \{n+1\}\bigr)\bigcup \bigl(F\cup {\{n+1\}}\bigr)'$ for certain subsets $E$ and $F$ of $\{1,\ldots,n\}$. Note that, due to Lemma~\ref{lm:XI-and-TAU}, $\tau_{x,y}$ and $\xi_{x,y,z}$ are both elements of $\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$ for all pairwise distinct $x,y,z\in\{1,\ldots,n\}$. Then taking to account Proposition \ref{pr:IT_n}, we obtain that $\IT_n\subseteq\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. In particular, $0\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. Thus, without loss of generality we may assume that $a\ne 0$, which implies $b\ne 0$. Suppose that all the blocks of $b$, except $\bigl(E\cup \{n+1\}\bigr)\bigcup \bigl(F\cup {\{n+1\}}\bigr)'$, are precisely $E_i\cup F_i'$, $1\leq i\leq k$. By the proposition of induction and Lemma~\ref{lm:Pi_n}, we obtain that \begin{equation} c=\bigl\{\bigl(E\cup E_1\bigr)\cup \bigl(F\cup F_1\bigr)',E_2\cup {F_2}',\ldots,E_k\cup {F_k}',\{n+1,(n+1)'\}\bigr\} \end{equation} is an element of $\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. We have four possibilities. \emph{Case 1}. $E=\varnothing$ and $F=\varnothing$. Then $b=c\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. \emph{Case 2}. $E=\varnothing$ and $F=\{f_1,\ldots,f_m\}\ne\varnothing$. Fix an element $f\in F_1$. Then $b=c\cdot\prod\limits_{i=1}^{m}\xi_{f,f_i,n+1}$ and so $b\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. \emph{Case 3}. $E=\{e_1,\ldots,e_l\}\ne\varnothing$ and $F=\varnothing$. Fix an element $e\in E_1$. Then $b=\prod\limits_{i=1}^{l}\xi_{n+1,f_i,e}\cdot c$, whence $b\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. \emph{Case 4}. $E\ne\varnothing$ and $F\ne\varnothing$. Put $d=\bigl\{E\cup F',E_1\cup {F_1}',\ldots,E_k\cup {F_k}',\{n+1,(n+1)'\}\bigr\}$. Due to proposition of induction and Lemma~\ref{lm:Pi_n}, we have that $d\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. Then $b=\tau_{E\cup\{n+1\}} d\tau_{F\cup\{n+1\}}\in\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$. In all these cases we obtained that $b$ belongs to $\langle \Sym_{n+1}, \xi_{1,2,3}\rangle$ and so does $a$. Thus, we have just proved that $\IP_n=\langle\Sym_{n}, \xi_{1,2,3}\rangle$ for all $n\geq 3$. This implies that $\IP_n=\langle \Sym_{n},u\rangle$ for all $u\in\Sym_{n}\xi_{1,2,3}\Sym_{n}$. Conversely, suppose that $\IP_n=\langle \Sym_{n},u\rangle$ for some $u\in\IP_n$. Then, due to \eqref{eq:ranks}, we obtain that $\rank(u)=n-1$. Now taking to account Lemmas~\ref{lm:BeforePrGeneratingSystems} and \ref{lm:XI-and-TAU}, we have that either $u\in\Sym_n\xi_{1,2,3}\Sym_n$ or $u\in\Sym_n\tau_{1,2}\Sym_{n}$. But $u\in\Sym_n\tau_{1,2}\Sym_{n}$ is impossible. Indeed, otherwise we would have $\langle\Sym_n,\xi_{1,2,3}\rangle=\IT_n$ and it remains to note that, due to~\ref{3}) of Proposition~\ref{pr:IT_n}, $\xi_{1,2,3}\notin\IT_n$ when $n\geq 3$. Hence, $u\in\Sym_n\xi_{1,2,3}\Sym_n$ holds, as was required. This completes the proof. \end{proof} Let $k\in\mathbb{N}$, $k\leq n$. Set $I_k=\bigl\{a\in\IP_n:~\rank(a)\leq k\bigr\}$. Note that \begin{equation}\label{eq:Ideal-Chain} \{0\}=I_1\subset I_2\subset\ldots\subset I_n=\IP_n. \end{equation} We will prove in the following proposition that these sets exhaust all the double-sided ideals (or just ideals) of $\IP_n$. \begin{proposition}\label{pr:Ideals} Let $I$ be an ideal of $\IP_n$ and $k\in\mathbb{N}$ such that $k\leq n$. Then \begin{enumerate} \item\label{Ideals-1} for all $b\in\IP_n$, $I_k=\IP_nb\IP_n$ if and only if $\rank(b)=k$; \item\label{Ideals-2} $I=I_m$ for some $m\in\mathbb{N}$, $m\leq n$; \item\label{Ideals-3} $I=\IP_na\IP_n$ for certain $a\in\IP_n$. \end{enumerate} \end{proposition} \begin{proof} Let us prove first that~\ref{Ideals-1}) holds. Take $b\in\IP_n$. Suppose that $I_k=\IP_nb\IP_n$. Then due to~\eqref{eq:ranks}, we obtain that $\rank(b)\geq k$. From the other hand, $b=1\cdot b\cdot 1\in I_k$ and so $\rank(b)\leq k$. Thus, $\rank(b)=k$. Conversely, suppose that $\rank(b)=k$. Then $b=\bigl(A_i\cup B_i'\bigr)_{1\leq i\leq k}$ for some partitions $\{1,\ldots,n\}=\bigcup\limits_{1\leq i\leq k}^{\cdot}A_i$ and $\{1,\ldots,n\}=\bigcup\limits_{1\leq i\leq k}^{\cdot}B_i$. Take $c\in I_k$ and let $\rank(c)=m\leq k$. Since \begin{multline} d=b\tau_{B_1\cup\ldots\cup B_{k+1-m}}=\bigl\{\bigl(A_1\cup\ldots\cup A_{k+1-m}\bigr)\cup\bigl(B_1\cup\ldots\cup B_{k+1-m}\bigr)',\\A_{k+2-m}\cup B_{k+2-m}',\ldots,A_k\cup B_{k}'\bigr\} \end{multline} is an element of the rank $m$, then due to~\ref{J=D}) of Theorem \ref{th:Green}, we obtain that there are $u,v\in\IP_n$ such that $c=udv=ub\tau_{B_1\cup\ldots\cup B_{k+1-m}}v\in\IP_nb\IP_n$. Thus, $I_k=\IP_nb\IP_n$ and the proof of~\ref{Ideals-1}) is complete. Let now $a$ be an arbitrary element of $I$ such that $\rank(a)$ has the maximum value among the numbers $\rank(x)$, $x\in I$. Then due to the statement~\ref{Ideals-1}), condition~\eqref{eq:Ideal-Chain} and the fact that $I=\bigcup\limits_{x\in I}\IP_nx\IP_n$, we have that $I=I_{\rank(a)}=\IP_na\IP_n$. Thus, statements~\ref{Ideals-2}) and~\ref{Ideals-3}) hold. \end{proof} As a corollary we obtain now the following proposition. \begin{proposition}\label{cor:Ideals} All the ideals of $\IP_n$ are principal and form the chain \eqref{eq:Ideal-Chain}. \end{proposition} \begin{proof} Follows from Proposition~\ref{pr:Ideals}. \end{proof} Set $\mathcal{D}_k=\bigl\{a\in\IP_n:~\rank(a)=k\bigr\}$ for all $k\in\mathbb{N}$, $1\leq k\leq n$. Due to~\ref{J=D}) of Theorem \ref{th:Green}, we have that all these sets exhaust all the $\GD$-classes of $\IP_n$. Now we are able to formulate a result on the structure of maximal subsemigroups of $\IP_n$. \begin{theorem}\label{th:MaximalSusemigroups} Let $n\geq 3$ and $S$ be a subset of $\IP_n$. Then the following statements are equivalent: \begin{enumerate} \item\label{MaxSubsemi-1} $S$ is a maximal subsemigroup of $\IP_n$; \item\label{MaxSubsemi-2} either $S=\IT_n\cup I_{n-2}$ or $S=G\cup I_{n-1}$ for some maximal subgroup $G$ of $\Sym_n$. \end{enumerate} In addition, every maximal subsemigroup of $\IP_n$ is an inverse subsemigroup of $\IP_n$. \end{theorem} \begin{proof} Let us prove first that~\ref{MaxSubsemi-2}) implies \ref{MaxSubsemi-1}). If $S$ coincides with the subsemigroup $G\cup I_{n-1}$ of $\IP_n$ for some maximal subgroup $G$ of $\Sym_n$ then since the condition~\eqref{eq:Ideal-Chain}, we have that $S$ is a maximal subsemigroup of $\IP_n$. Note that $\IT_n\cup I_{n-2}$ is a subsemigroup of $\IP_n$, as $\IT_n$ is a subsemigroup of $\IP_n$ and $I_{n-2}$ is an ideal of $\IP_n$. If now $\IT_n\cup I_{n-2}$ is a proper subsemigroup of $T$, where $T$ is a subsemigroup of $\IP_n$, then, due to Lemma~\ref{lm:BeforePrGeneratingSystems}, $T$ contains an element of $\Sym_n\xi_{1,2,3}\Sym_n$ and so, taking to account Proposition~\ref{pr:Generating-SystemsIP} and the fact that $\Sym_n\subseteq\IT_n$, we obtain that $T=\IP_n$. Thus, \ref{MaxSubsemi-2}) implies~\ref{MaxSubsemi-1}). Let now $S$ be a maximal subsemigroup in $\IP_n$. Note that $S\cup I_{n-2}$ is a subsemigroup of $\IP_n$. Besides, $S\cup I_{n-2}$ is a proper subset of $\IP_n$. Indeed, otherwise we would have $S\cup I_{n-2}=\IP_n$, whence $\Sym_n\cup \mathcal{D}_{n-1}\subseteq S$ and so due to Proposition~\ref{pr:Generating-SystemsIP}, we would obtain that $S=\IP_n$. Thus, $S\cup I_{n-2}=S$ and so $I_{n-2}\subseteq S$. Since $S\cup \{1\}$ is a proper subsemigroup of $\IP_n$, we have that $S=S\cup\{1\}$ and $G=S\cap \Sym_n\ne\varnothing$. Obviously, $G$ is a subgroup of $\Sym_n$. Now we have two possibilities. \emph{Case 1}. $G$ is a proper subgroup of $\Sym_n$. Then $S\subseteq G\cup I_{n-1}$ and due to the fact that $G\cup I_{n-1}$ is a proper subsemigroup of $\IP_n$, we obtain that $S=G\cup I_{n-1}$. It remains to note that the latter implies that $G$ is a maximal subgroup of $\Sym_n$. \emph{Case 2}. $G=\Sym_n$. Then $\Sym_n\cup I_{n-2}\subseteq S$. Since $\Sym_n\cup I_{n-2}$ is a proper subsemigroup of $\IT_n\cup I_{n-2}$, we have that $S$ contains an element $a$ of $\mathcal{D}_{n-1}$. Then due to Lemma~\ref{lm:BeforePrGeneratingSystems} and Proposition \ref{pr:Generating-SystemsIP}, we obtain that $S\subseteq\IT_n\cup I_{n-2}$. But $\IT_n\cup I_{n-2}$ is a maximal subsemigroup of $\IP_n$ and so $S=\IT_n\cup I_{n-2}$. This completes the proof of that~\ref{MaxSubsemi-1}) implies~\ref{MaxSubsemi-2}). That every maximal subsemigroup of $\IP_n$ is an inverse subsemigroup of $\IP_n$, follows from what we already have done and the fact that $\IT_n\cup I_{n-2}$ and $G\cup I_{n-1}$ are inverse subsemigroups of $\IP_n$ for all subgroups $G$ of $\Sym_n$. \end{proof} \section{Automorphism group $\Aut(\IP_X)$} \label{sec:AutomorphismsIP_X} Let $g\in\Sym_X$. Denote by $\varphi_g$ the map from $\IP_X$ to $\IP_X$, given by \begin{equation} \varphi_g(a)=g^{-1}ag~\mbox{for every}~a\in \IP_X. \end{equation} Clearly, $\varphi_g$ belongs to $\Aut(\IP_X)$, automorphism group of $\IP_X$. Throughout this section, denote by $\id$ the identity map of the set $X$ to itself. The main result of this section is the following theorem. \begin{theorem}\label{th:Automorphisms} Let $\varphi\in \Aut(\IP_X)$. Then $\varphi=\varphi_g$ for some $g\in\Sym_X$. In particular, $\Aut(\IP_X)\cong \Sym_X$ when $\mid\!X\!\mid\ne 2$ and $\Aut(\IP_2)=\{\id\}$. \end{theorem} We will divide the proof of this theorem into few lemmas. Naturally, $\varphi$ induces an automorphism $\chi=\varphi\mid_{E(\IP_X)}$ of the semilattice $E(\IP_X)$. Set $\zeta_{x}=\tau_{X\setminus\{x\}}$ for all $x\in X$. Set also $\Phi=\bigl\{\zeta_x\in\IP_X:~x\in X\bigr\}$. Recall that if $(E,\leq)$ is a semilattice with the zero element $0$, then an element $f$ of $E$ is said to be \emph{primitive} if $g\leq f$ implies either $g=f$ or $g=0$, for all $g\in E$. For all $n\geq 2$ set \begin{multline} \Theta_{\max}^n=\bigl\{\tau_{i,j}\in\IP_n:~i,j\in\{1,\ldots,n\},~i\ne j\bigr\}~\mbox{and}\\\Theta_{\mathrm{pr}}^n=\bigl\{\tau_{F}\tau_{\{1,\ldots,n\}\setminus F}\in\IP_n:F~\mbox{is a proper subset of}~\{1,\ldots,n\}\bigr\}=\\\mathcal{D}_2\cap E(\IP_n). \end{multline} Notice that $\Phi\subseteq\Theta_{\mathrm{pr}}^n$. \begin{lemma}\label{lm:Aut-Max-and-Pr} Let $n\geq 2$. Then the set of all primitive elements of the semilattice $E(\IP_n)$ coincides with $\Theta_{\mathrm{pr}}^n$. Also then the set of all maximal elements of the semilattice $E(\IP_n)\setminus\{1\}$ coincides with $\Theta_{\max}^n$. \end{lemma} \begin{proof} Follows from Proposition~\ref{pr:Omega}. \end{proof} \begin{lemma}\label{lm:Aut-Semilattice} Take $\theta\in\Aut\bigl(E(\IP_n)\bigr)$. Then there is $g\in\Sym_n$ such that $\theta(e)=g^{-1}eg$ for all $e\in E(\IP_n)$. \end{lemma} \begin{proof} Clearly, the statement holds when $n=1$. Thus, let us assume that $n\geq 2$. Obviously, $\theta(1)=1$. Then $\theta\bigl(E(\IP_n)\setminus\{1\}\bigr)=E(\IP_n)\setminus\{1\}$. Hence, due to Lemma~\ref{lm:Aut-Max-and-Pr}, we obtain that $\theta\bigl(\Theta_{\max}^n\bigr)=\Theta_{\max}^n$ and $\theta\bigl(\Theta_{\mathrm{pr}}^n\bigr)=\Theta_{\mathrm{pr}}^n$. Take $f=\tau_{F}\tau_{\{1,\ldots,n\}\setminus F}\in\Theta_{\mathrm{pr}}^n$. Set $\Lambda_f=\bigl\{a\in\Theta_{\max}^n:~fa=f\bigr\}$. Then $\theta(\Lambda_f)=\Lambda_{\theta(f)}$. If $f\notin\Phi$ then $2\leq\mid\!F\!\mid\leq n-2$. Thus, \begin{equation}\label{eq:Lambda_f} \mid\!\Lambda_f\!\mid=\binom{\mid\!F\!\mid}{2}+\binom{n-\mid\!F\!\mid}{2}, ~\mbox{if}~f\notin\Phi. \end{equation} Otherwise, we have the following: \begin{equation}\label{eq:Zeta} \mid\!\Lambda_f\!\mid=\binom{n-1}{2},~\mbox{if}~f\in\Phi. \end{equation} Let us prove now that for all $n\geq 4$ and for all $k$, $2\leq k\leq n-2$, the following holds: \begin{equation}\label{eq:Inequality} \binom{k}{2}+\binom{n-k}{2}<\binom{n-1}{2}. \end{equation} Indeed, the inequality \begin{multline} k(k-n)=(k^2-1)+1-kn=(k-1)(k+1)+1-kn<\\(k-1)n+1-kn=1-n~\mbox{implies} \end{multline} \begin{multline} \binom{k}{2}+\binom{n-k}{2}=\frac{1}{2}\bigl(k(k-1)+(n-k)(n-k-1)\bigr)=\\\frac{1}{2} \bigl(2k^2-2kn+n^2-n\bigr)=k(k-n)+\frac{1}{2} \bigl(n^2-n\bigr)<\\\frac{1}{2} \bigl(n^2-n\bigr)+1-n=\frac{1}{2}(n-1)(n-2)=\binom{n-1}{2}. \end{multline} Now due to~\eqref{eq:Lambda_f},~\eqref{eq:Zeta}, \eqref{eq:Inequality} and the equality $\theta(\Lambda_f)=\Lambda_{\theta(f)}$, we obtain that $\theta(\Phi)=\Phi$. Then there is an element $g$ of $\Sym_n$ such that $\theta(\zeta_{x})=\zeta_{g(x)}$ for all $x\in\{1,\ldots,n\}$. Take now distinct $x$ and $y$ of $\{1,\ldots,n\}$. Since $\zeta_{x}\tau_{x,y}=0$ and $\zeta_{y}\tau_{x,y}=0$, we have that $\zeta_{g(x)}\theta(\tau_{x,y})=0$ and $\zeta_{g(y)}\theta(\tau_{x,y})=0$. The latter, taking to account $\theta\bigl(\Theta_{\max}^n\bigr)=\Theta_{\max}^n$, implies that $\theta(\tau_{x,y})=\tau_{g(x),g(y)}=g^{-1}\tau_{x,y}g$. Let now $e=(E_i\cup E_i')_{i\in I}$ be a nonidentity idempotent element of $E(\IP_n)$. Then \begin{equation} e=\prod\bigl\{\tau_{x,y}:~x\ne y,~\{x,y\}\subseteq E_i~\mbox{for some}~i\in I\bigr\} \end{equation} implies \begin{multline} \theta(e)=\prod\bigl\{\tau_{g(x),g(y)}:~x\ne y,~\{x,y\}\subseteq E_i~\mbox{for some}~i\in I\bigr\}=\\\prod\bigl\{g^{-1}\tau_{x,y}g:~x\ne y,~\{x,y\}\subseteq E_i~\mbox{for some}~i\in I\bigr\}=g^{-1}eg. \end{multline} This completes the proof. \end{proof} Take distinct $x$ and $y$ of $X$. Define an element $\varepsilon_{x,y}$ of $\Sym_X$ as follows: \begin{equation} \varepsilon_{x,y}(x)=y,~\varepsilon_{x,y}(y)=x~\mbox{and}~\varepsilon_{x,y}(t)=t~\mbox{for all}~t\in X\setminus\{x,y\}. \end{equation} \begin{corollary}\label{cor:Aut-IP_6} Let $\mid\!X\!\mid=6$. Then there is $g\in\Sym_6$ such that $\varphi(h)=\varphi_g(h)$ for all $h\in\Sym_6$. \end{corollary} \begin{proof} If we put $\chi=\theta$ and $n=6$ in the statement of Lemma \ref{lm:Aut-Semilattice}, we will obtain that there is $g\in\Sym_6$ such that $\chi(e)=g^{-1}eg$ for all $e\in E(\IP_6)$. Take distinct $x$ and $y$ of $\{1,\ldots,6\}$. Then \begin{equation} g^{-1}\tau_{x,y}g= \varphi(\tau_{x,y})=\varphi(\tau_{x,y}\varepsilon_{x,y})=g^{-1}\tau_{x,y}g\varphi(\varepsilon_{x,y}), \end{equation} whence \begin{equation} \tau_{x,y}=\tau_{x,y}g\varphi(\varepsilon_{x,y})g^{-1}. \end{equation} The latter implies that either $g\varphi(\varepsilon_{x,y})g^{-1}=1$ or $g\varphi(\varepsilon_{x,y})g^{-1}=\varepsilon_{x,y}$. But since the order of $g\varphi(\varepsilon_{x,y})g^{-1}$ equals $2$, we have that $g\varphi(\varepsilon_{x,y})g^{-1}=\varepsilon_{x,y}$, which is equivalent to $\varphi(\varepsilon_{x,y})=g^{-1}\varepsilon_{x,y}g$. Now, taking to account the known fact that $\langle\varepsilon_{x,y}:~x\ne y\rangle=\Sym_n$ (see \cite{Kurosh}), we obtain that $\varphi(h)=\varphi_g(h)$ for all $h\in\Sym_6$. \end{proof} \begin{lemma}\label{lm:Aut-S_X} There is $g\in\Sym_X$ such that $\varphi(h)=\varphi_g(h)$ for all $h\in\Sym_X$. \end{lemma} \begin{proof} Due to Corollary~\ref{cor:Aut-IP_6}, we have that the statement holds when $\mid\!X\!\mid=6$. Assume now that $\mid\!X\!\mid\ne 6$. Since $\varphi$ preserves the set of all invertible elements of $\IP_X$, we have, due to Proposition~\ref{cor:Invertible-Elements}, that $\varphi(\Sym_X)=\Sym_X$. Hence, $\varphi$ induces an automorphism of $\Sym_X$. Then due to known fact, which claims that if $\mid\!X\!\mid\ne 6$ then every automorphism of $\Sym_X$ is inner (see \cite{Kurosh}), we have that there is $g\in\Sym_X$ such that $\varphi(h)=g^{-1}hg=\varphi_g(h)$ for all $h\in\Sym_X$. This completes the proof. \end{proof} Set now $\psi =\varphi\varphi_g$. Then $\psi$ is, obviously, an automorphism of $\IP_X$ and, due to Lemma~\ref{lm:Aut-S_X}, $\psi\mid_{\Sym_X}$ is the identity map of $\Sym_X$ to itself. For all $M\subseteq X$ set \begin{equation} \widetilde{\Sym}_M=\bigl\{h\in\Sym_X:~h(x)=x~\mbox{for all}~x\in X\setminus M\bigr\}. \end{equation} For all $a\in\IP_X$ set \begin{equation} \Fix_l(a)=\bigl\{h\in\Sym_X:~ha=a\bigr\}~\mbox{and}~\Fix_r(a)=\bigl\{h\in\Sym_X:~ah=a\bigr\}. \end{equation} \begin{lemma}\label{lm:Aut-C(e)} Let $a\in\IP_X$. Let also $X=\bigcup\limits_{i\in I}^{\cdot}A_i=\bigcup\limits_{i\in I}^{\cdot}B_i$. Then \begin{enumerate} \item $\Fix_l(a)=\bigoplus\limits_{i\in I}\widetilde{\Sym}_{A_i}$ if and only if $a=\bigl(A_i\cup U_i'\bigr)_{i\in I}$ for some partition $X=\bigcup\limits_{i\in I}^{\cdot}U_i$; \item $\Fix_r(a)=\bigoplus\limits_{i\in I}\widetilde{\Sym}_{B_i}$ if and only if $a=\bigl(V_i\cup B_i'\bigr)_{i\in I}$ for some partition $X=\bigcup\limits_{i\in I}^{\cdot}V_i$. \end{enumerate} \end{lemma} \begin{proof} Straightforward. \end{proof} \begin{corollary}\label{cor:Aut-H-classes} $a\GH\psi(a)$ for all $a\in\IP_X$. In particular, $\psi(e)=e$ for all $e\in E(\IP_X)$. \end{corollary} \begin{proof} That $a\GH\psi(a)$ for all $a\in\IP_X$ follows from Lemma \ref{lm:Aut-C(e)} and Theorem~\ref{th:Green}. Then $\psi(e)=e$ for all $e\in E(\IP_X)$, due to the fact that every $\GH$-class of an arbitrary semigroup contains at most one idempotent (see Corollary 2.2.6 from \cite{Howie}). \end{proof} \begin{lemma}\label{lm:Aut-n-geq-3} Let $a\in\IP_X$ and $\rank(a)\geq 3$. Then $\psi(a)=a$. \end{lemma} \begin{proof} Let $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$, $\mid\!I\!\mid\geq 3$. Due to Corollary~\ref{cor:Aut-H-classes}, we have that $a\GH\psi(a)$ and so $\psi(a)=\bigl(A_i\cup B_{\alpha(i)}'\bigr)_{i\in I}$ for some bijective map $\alpha:I\to I$. Due to Corollary \ref{cor:Aut-H-classes}, we also have that $ea\GH e\psi(a)$ for all $e\in E(\IP_X)$. Take arbitrary distinct $i$ and $j$ of $I$. Since $\tau_{A_i\cup A_j}a\GH\tau_{A_i\cup A_j}\psi(a)$, we have that \begin{multline} \bigl\{\bigl(A_i\cup A_j\bigr)\cup\bigl(B_i\cup B_j\bigr)',\bigl(A_l\cup B_l'\bigr)_{l\in I\setminus\{i,j\}}\bigr\}~\mbox{and}\\\bigl\{\bigl(A_i\cup A_j\bigr)\cup\bigl(B_{\alpha(i)}\cup B_{\alpha(j)}\bigr)',\bigl(A_l\cup B_l'\bigr)_{l\in I\setminus\{i,j\}}\bigr\} \end{multline} are $\GH$-equivalent, whence $\{i,j\}=\{\alpha(i),\alpha(j)\}$. Let now $k\in I$. Then $\alpha(k)=k$. Suppose the contrary. Then $\{k,m\}=\{\alpha(k),\alpha(m)\}$ for all $m\in I\setminus\{k\}$ implies that $\alpha(k)=m$ for all $m\in I\setminus\{k\}$. But $\mid\!I\!\mid\geq 3$ and we get a contradiction. Thus, $\alpha$ is an identity map of $I$ to itself, which is equivalent to $\psi(a)=a$. This completes the proof. \end{proof} Note that since $\IP_1$ is isomorphic to the unit group and since $\IP_2\cong\mathbb{Z}_2^0$, where $\mathbb{Z}_2^0$ denotes the group $\mathbb{Z}_2$ with adjoint zero, we have that $\Aut(\IP_X)=\{\id\}$ when $\mid\!X\!\mid\leq 2$. \begin{lemma}\label{lm:Aut-leq-2} Let $a\in\IP_X$ and $\rank(a)\leq 2$. Then $\psi(a)=a$. \end{lemma} \begin{proof} If $\rank(a)=1$ then $a=0$ and, obviously, $\psi(a)=a$. So let us suppose that $\rank(a)=2$. Assume that $a=\bigl\{A\cup B',C\cup D'\bigr\}$. Fix $x\in A$ and $y\in B$. Suppose that $\lmod A\rmod\geq 2$ and $\lmod B\rmod\geq 2$. Then $\psi(a)=a$. Indeed, we have that $A\setminus\{x\}\ne\varnothing$ and $B\setminus\{y\}\ne\varnothing$, so if $y_1\in B\setminus\{y\}$ then we can consider the equality $a=\bigl\{\{x,y'\},\bigl(A\setminus\{x\}\bigr)\bigcup\bigl(B\setminus\{y\}\bigr)',C\cup D'\bigr\}\cdot\tau_{y,y_1}$, whence, due to Corollary \ref{cor:Aut-H-classes} and Lemma~\ref{lm:Aut-n-geq-3}, we will have that $\psi(a)=a$. Analogously, if $\lmod C\rmod\geq 2$ and $\lmod D\rmod\geq 2$ then $\psi(a)=a$. Thus, we may assume that either $\lmod A\rmod=1$ or $\lmod B\rmod=1$, and that either $\lmod C\rmod=1$ or $\lmod D\rmod=1$. Without loss of generality we may suppose that $\lmod A\rmod=1$. Then we will have two possibilities. \emph{Case 1}. $\lmod C\rmod=1$. Then $\lmod X\rmod=2$ and we obtain $\psi=\id$. \emph{Case 2}. $\lmod D\rmod=1$. Then $\psi(a)=a$. Suppose the contrary. Then we would obtain that $\psi(a)=\bigl\{A\cup D',C\cup B'\bigr\}=\zeta_{x}h$ for some $h\in\Sym_X$. But $\zeta_{x}h=\psi(\zeta_{x}h)$ and so $a=\zeta_{x}h$, whence $B=D$, which leads to a contradiction. Thus, we proved that $\psi(a)=a$, which was required. \end{proof} As a consequence of that we have from Lemmas~\ref{lm:Aut-n-geq-3} and~\ref{lm:Aut-leq-2}, we have that $\psi=\id$, whence $\varphi=\varphi_g$. It remains to prove that $\Aut(\IP_X)\cong\Sym_X$ when $\lmod X\rmod\ne 2$. This follows from the following lemma. \begin{lemma}\label{lm:Aut-cong-Sym_X} Suppose that $\lmod X\rmod\geq 3$. Then a map $\vartheta:~\Sym_X\to\Aut(\IP_X)$, given by \begin{equation} \vartheta(h)=\varphi_h~\mbox{for all}~h\in\Sym_X, \end{equation} is an isomorphism from $\Sym_X$ onto $\Aut(\IP_X)$. \end{lemma} \begin{proof} We have already proved that $\vartheta$ is an onto homomorphism from $\Sym_X$ to $\Aut(\IP_X)$. But, besides, $\vartheta$ is an injective map. Indeed, $\vartheta(h_1)=\vartheta(h_2)$ implies that $h_1^{-1}hh_1=h_2^{-1}hh_2$ or just that $\bigl(h_1h_2^{-1}\bigr)^{-1}h\bigl(h_1h_2^{-1}\bigr)=h$ for all $h\in\Sym_X$ and it remains to note that $\Sym_X$ is a center-free group when $\lmod X\rmod\geq 3$ (see \cite{Kurosh}). Thus, $\vartheta$ is an isomorphism. \end{proof} The proof of theorem is complete. \section{Connections between $\IP_X$ and other semigroups}\label{sec:Connections} Set $\Upsilon=\bigl\{X,X'\bigr\}$. Then $\Upsilon\in\Ch_X$. The following proposition shows that $\IP_X\cup\{\Upsilon\}$ is a maximal inverse subsemigroup of $\Ch_X$ when $\lmod X\rmod\geq 2$. \begin{proposition}\label{pr:IP_X-inverse-subsemigroup} Let $\lmod X\rmod\geq 2$. Then $\IP_X\cup \{\Upsilon\}$ is a maximal inverse subsemigroup of $\Ch_X$. \end{proposition} \begin{proof} Since $\IP_X$ is an inverse subsemigroup of $\Ch_X$ and $a\Upsilon=\Upsilon a=\Upsilon$ for all $a\in\IP_X\cup \{\Upsilon\}$, we obtain that $\IP_X\cup \{\Upsilon\}$ is a proper inverse subsemigroup of $\Ch_X$. Suppose now that $S$ is an inverse subsemigroup of $\Ch_X$ such that $\IP_X\cup\{\Upsilon\}$ is a subsemigroup of $S$. Take $s\in S\setminus\IP_X$. Then there is a nonempty subset $A$ of $X$ such that either $s$ contains a block $A$ or $s$ contains a block $A'$. Without loss of generality we may assume that $s$ contains the block $A$. Let $t$ be the inverse of $s$ in $S$. Then $st$ is an idempotent in $S$ and so, due to the fact that idempotents of inverse semigroup commute, we obtain that $u=st\cdot\Upsilon=\Upsilon\cdot st$. The latter implies that $u$ contains both blocks $A$ and $X$, whence $A=X$. Then $s$ is an idempotent and due to equalities $s=\Upsilon s$ and $\Upsilon s=s\Upsilon$, we have that $s$ contains the block $X'$ and so $s=\Upsilon$. That is, $S=\IP_X\cup\{\Upsilon\}$. This implies that $\IP_X\cup\{\Upsilon\}$ is a maximal inverse subsemigroup of $\Ch_X$ which was required. \end{proof} Denote by $\IS_X$ the \emph{symmetric inverse} semigroup on the set $X$. Let $s\in\IS_X$. Denote by $\dom(s)$ and $\ran(s)$ the \emph{domain} and the \emph{range} of $s$ respectively. The following theorem shows how one can embed the symmetric inverse semigroup into the inverse partition one. \begin{theorem}\label{th:IS-embeds-into-IP} Let $\overline{x}\notin X$. Then $\IS_X$ isomorphically embeds into $\IP_{X\cup\{\overline{x}\}}$. \end{theorem} \begin{proof} For all $s\in\IS_X$, set \begin{equation}\label{eq:Omega} \Omega_s=\bigl(X\cup\{\overline{x}\}\setminus\dom(s)\bigr)\bigcup \bigl(X\cup\{\overline{x}\}\setminus\ran(s)\bigr)'. \end{equation} Set a map $\kappa:~\IS_X\to\IP_{X\cup\{\overline{x}\}}$ as follows: \begin{equation}\label{eq:kappa-definition} \kappa(s)=\bigl\{\Omega_s,\bigl(\{x,s(x)'\}\bigr)_{x\in\dom(s)}\bigr\}~\mbox{for all}~s\in\IS_X. \end{equation} Take an arbitrary $s$ of $\IS_X$. Then we have the following condition: \begin{equation}\label{eq:kappa} x\equiv_{\kappa(s)}\overline{x}\equiv_{\kappa(s)}{\overline{x}}'\equiv_{\kappa(s)}y'~\mbox{for all}~x\in X\setminus\dom(s)~\mbox{and}~y\in X\setminus\ran(s). \end{equation} Take $s,t\in\IS_X$. Then due to~\eqref{eq:Omega} and \eqref{eq:kappa}, we obtain that \begin{multline}\label{eq:dreadful} x\equiv_{\kappa(s)\kappa(t)}\overline{x}\equiv_{\kappa(s)\kappa(t)}\overline{x}' \equiv_{\kappa(s)\kappa(t)}y'~\mbox{for all}~x,y\in X~\mbox{such that}\\x\notin s^{-1}\bigl(\ran(s)\cap\dom(t)\bigr)~\mbox{and}~y\notin t\bigl(\dom(t)\cap\ran(s)\bigr). \end{multline} Notice that \begin{equation}\label{eq:S-T} s^{-1}\bigl(\ran(s)\cap\dom(t)\bigr)=\dom(st)~\mbox{and}~ t\bigl(\dom(t)\cap\ran(s)\bigr)=\ran(st). \end{equation} If now $x\in\dom(st)$ then $x\equiv_{\kappa(s)}s(x)'~\mbox{and}~s(x)\equiv_{\kappa(t)}st(x)'$, whence \begin{equation}\label{eq:last} x\equiv_{\kappa(s)\kappa(t)}st(x)'~\mbox{for all}~x\in\dom(st). \end{equation} The conditions~\eqref{eq:dreadful},~\eqref{eq:S-T} and \eqref{eq:last} imply that \begin{equation} \kappa(s)\kappa(t)=\bigl\{\Omega_{st},\bigl(\{x, st(x)'\}\bigr)_{x\in\dom(st)}\bigr\}=\kappa(st). \end{equation} Thus, $\kappa$ is a homomorphism from $\IS_X$ to $\IP_{X\cup\{\overline{x}\}}$. It remains to prove that $\kappa$ is an injective map. Suppose that $\kappa(s)=\kappa(t)$ for some $s,t\in\IS_X$. Then it follows from~\eqref{eq:kappa-definition} that $\dom(s)\subseteq\dom(t)$ and $\dom(t)\subseteq\dom(s)$, whence $\dom(s)=\dom(t)$. Then~\eqref{eq:kappa-definition} implies that $s(x)=t(x)$ for all $x\in\dom(s)=\dom(t)$. Hence, $s=t$ and so $\kappa$ is injective. The proof is complete. \end{proof} It follows immediately from Theorem~\ref{th:IS-embeds-into-IP} that $\IS_n$ embeds into $\IP_{n+1}$ for all $n\in\mathbb{N}$. Surprisingly, the following theorem shows that one can not construct an embedding map from $\IS_n$ to $\IP_n$. \begin{theorem}\label{th:IS-doesn't-embeds-into-IP} Let $n\in\mathbb{N}$. There is no an injective homomorphism from $\IS_n$ to $\IP_n$. \end{theorem} \begin{proof} Suppose the contrary. Then there is a subsemigroup $U$ of $\IP_n$ such that $U\cong\IS_n$. Then we have that $U$ is a regular subsemigroup of $\IP_n$, whence, due to Proposition 2.4.2 from \cite{Howie}, we obtain that $\GD^{U}=\GD\cap(U\times U)$, where $\GD^{U}$ denotes the Green's $\GD$-relation on $U$. Note that $\IP_n$ contains exactly $n$ different $\GD$-classes. This implies that $U$ contains at most $n$ different $\GD^{U}$-classes. But since $U\cong\IS_n$, we have that $U$ contains exactly $n+1$ different $\GD^{U}$-classes. We get a contradiction. This completes the proof. \end{proof} \section{$\IP_n$ embeds into $\IS_{2^n-2}$}\label{ef} Let $S$ be an inverse semigroup with the natural partial order $\leq$ on it. For $A\subseteq S$ denote by $[A]$ the order ideal of $S$ with respect to $\leq$, i.e., $[A]=\bigl\{b:~a\leq b~\mbox{for some}~a\in A\bigr\}$. Let also $H$ be a \emph{closed inverse subsemigroup} of $S$, i.e., $H$ is an inverse subsemigroup of $S$ and $[H]=H$ (see \cite{Howie}). Recall (see \cite{Howie}) that one can define the set of all \emph{right $\leq$-cosets} of $H$ as follows: \begin{equation} \mathcal{C}=\mathcal{C}_H=\bigl\{[Hs]:~ss^{-1}\in H\bigr\}. \end{equation} Further, one can define the \emph{effective transitive representation} $\phi_H:S\to\IS_{\mathcal{C}}$, given by \begin{equation} \phi_H(s)=\bigr\{\bigl([Hx],[Hxs]\bigr):~ [Hx],[Hxs]\in\mathcal{C}\bigl\}. \end{equation} Let now $K$ and $H$ be arbitrary closed inverse subsemigroups of $S$. For a definition of the \emph{equivalence} of representations $\phi_K$ and $\phi_H$, we refer reader to \cite{Howie}. But we note that due to Proposition IV.4.13 from \cite{Petrich}, one has that $\phi_K$ and $\phi_H$ are equivalent if and only if there exists $a\in S$ such that $a^{-1}Ha\subseteq K$ and $aKa^{-1}\subseteq H$. We will need the following well-known fact. \begin{theorem}[Proposition 5.8.3 from \cite{Howie}]\label{th:Howie} Let $H$ be a closed inverse subsemigroup of an inverse semigroup $S$ and let $a,b\in S$. Then $[Ha]=[Hb]$ if and only if $ab^{-1}\in H$. \end{theorem} The main result of this section is the following theorem. \begin{theorem} Let $n\geq 2$. Up to equivalence, there is only one faithful effective transitive representation of $\IP_n$, namely to $\IS_{2^n-2}$. In particular, $\IP_n$ isomorphically embeds into $\IS_{2^n-2}$. \end{theorem} We divide the proof of this theorem into lemmas. Throughout all further text of this section we suppose that $H$ is a closed inverse subsemigroup of $\IP_n$. \begin{lemma}\label{lm:Ef-Tr-1} $H=[G]$ for some subgroup $G$ of $\IP_n$. \end{lemma} \begin{proof} Since $\IP_n$ is finite, we have that $E(H)$ contains a zero element. It remains to use Proposition IV.5.5 from \cite{Petrich}, which claims that if the set of idempotents of a closed inverse subsemigroup contains a zero element, then this subsemigroup is a closure of some subgroup of the original semigroup. \end{proof} Denote by $e$ the identity element of $G$. \begin{lemma}\label{lm:Ef-Tr-2} If $e=0$ then $\phi_H$ is not faithful. \end{lemma} \begin{proof} We have $G=\{0\}$, whence $H=[0]=\IP_n$ and so $[Hx]\supseteq [0]=\IP_n$ for all $x\in\IP_n$. Thus, $[Hx]=\IP_n$ for all $x\in\IP_n$. Then $\lmod\phi_H(\IP_n)\rmod=1$, whence we obtain that $\phi_H$ is not faithful. \end{proof} \begin{lemma}\label{lm:Ef-Tr-3} Let $\rank(e)\geq 3$. Then $\phi_H$ is not faithful. \end{lemma} \begin{proof} Take $b\in\mathcal{D}_2$. Since $bb^{-1}\in\mathcal{D}_2$, we have that $bb^{-1}\notin H$ and so $[Hb]\notin\mathcal{C}$. The latter gives us that $\phi_H(b)$ equals the zero element of $\IS_{\mathcal{C}}$. Then, due to $\lmod\mathcal{D}_2\rmod\geq 2$, we obtain that $\phi_H$ is not faithful. \end{proof} \begin{lemma}\label{lm:Ef-Tr-4} Let $\rank(e)=2$ and $G\cong\mathbb{Z}_2$. Then $\phi_H$ is not faithful. \end{lemma} \begin{proof} Let $G=\{e,q\}$. We are going to prove that $\phi_H(e)=\phi_H(q)$. Let us prove first that $\dom\bigl(\phi_H(e)\bigr)=\dom\bigl(\phi_H(q)\bigr)$. Indeed, take $[Hx]\in\mathcal{C}$. Then, due to the equality $(xe)(xe)^{-1}=xex^{-1}=xqq^{-1}x^{-1}=(xq)(xq)^{-1}$, we obtain that $[Hxe]\in\mathcal{C}$ if and only if $(xe)(xe)^{-1}\in H$ if and only if $(xq)(xq)^{-1}\in H$ if and only if $[Hxq]\in\mathcal{C}$. Thus, $\dom\bigl(\phi_H(e)\bigr)=\dom\bigl(\phi_H(q)\bigr)$. Take now $x\in\dom\bigl(\phi_H(e)\bigr)$. Then $xex^{-1}\in H=[\{e,q\}]$. But since $xex^{-1}$ is an idempotent and $\rank(xex^{-1})\leq\rank(e)=2$, we obtain, taking to account Proposition~\ref{pr:Omega}, that $xex^{-1}=e$. Hence, $(xe)(xe)^{-1}=ee^{-1}$ and so, due to Proposition~2.4.1 from \cite{Howie}, we obtain that $xe\GR e$. But then we have that $\rank(xe)=\rank(e)$ and due to $\lambda_{xe}\supseteq\lambda_{e}$ (which follows, in turn, from \eqref{eq:remark-for-ranks}), we deduce that $\lambda_{xe}=\lambda_{e}$, whence due to Theorem~\ref{th:Green}, we have that $xe\GL e$. Thus, $xe\GH e$, whence $xe\in G$ and so $xq=xe\cdot q\in G$. But then $(xq)(xe)^{-1}\in G\subseteq H$, whence, due to Theorem~\ref{th:Howie}, we have that $[Hxe]=[Hxq]$. The latter implies that $\phi_H(e)(x)=\phi_H(q)(x)$. Thus, $\phi_H(e)=\phi_H(q)$ and so $\phi$ is not faithful. \end{proof} \begin{lemma}\label{ugly-lemma} Let $f\in\Theta_{\mathrm{pr}}^{n}$ and $T=[f]$. Take $[Tx]\in\mathcal{C}_T$. Then $\rank(fx)=2$ and $[Tx]=[fx]$. \end{lemma} \begin{proof} Clearly, $[Tx]\in\mathcal{C}_T$ is equivalent to $f\leq xx^{-1}$. Obviously, $\rank(fx)\leq\rank(f)=2$. But $\rank(fx)=1$ is impossible. Indeed, otherwise we would have $fx=0$, whence $0=fxx^{-1}=f$, which does not hold. Thus, $\rank(fx)=2$. Note that $[fx]\subseteq[Tx]$. It remains to prove that $[Tx]\subseteq[fx]$. Take $t\in T$. Then $f\leq t$ and due to the fact that the natural partial order on an arbitrary inverse semigroup is compatible (see \cite{Howie}), we obtain that $fx\leq tx$. That is, $tx\in [fx]$. Hence, $Tx\subseteq[fx]$, whence $[Tx]\subseteq\bigl[[fx]\bigr]=[fx]$. The proof is complete. \end{proof} \begin{lemma}\label{lm:Ef-Tr-5} Let $\rank(e)=2$ and $G=\{e\}$. Then $\phi_H$ is faithful. \end{lemma} \begin{proof} Note that $H=[e]$. Let $e=\tau_E\tau_{E_1}$, where $E$ and $E_1$ are nonempty subsets of $\{1,\ldots,n\}$ such that $\{1,\ldots,n\}=E\bigcup\limits^{\cdot}E_1$. Suppose that $\phi_H(s)=\phi_H(t)$ for some $s$ and $t$ of $\IP_n$. Let $A$ be an arbitrary $\rho_{t}$-class. Set $\overline{A}=\{1,\ldots,n\}\setminus A$. Suppose first that $s=0$. We are going to prove that $t=0$. Suppose the contrary. We have that $\rank(e\cdot xs)=1$ for all $x\in\IP_n$ such that $xx^{-1}\in [e]$. So, due to Lemma~\ref{ugly-lemma}, we obtain that $\dom\bigl(\phi_H(s)\bigr)=\varnothing$. Then, again by Lemma~\ref{ugly-lemma}, we have that $\rank(e\cdot xt)=1$, or just that $ext=0$, for all $x\in\IP_n$ such that $xx^{-1}\in [e]$. Put now $u=\bigl\{E\cup A',E_1\cup\overline{A}'\bigr\}$ (note that, due to assumption, $\overline{A}\ne\varnothing$). Then $uu^{-1}=e\in [e]$ and $eut\ne 0$. Thus, we get a contradiction and so $s=0$ implies $t=0$. Analogously, $t=0$ implies $s=0$. Assume now that $s\ne 0$, then $t\ne 0$ and so $\overline{A}\ne\varnothing$. Put again $u=\bigl\{E\cup A',E_1\cup\overline{A}'\bigr\}$. Due to Theorem~\ref{th:Howie} and the equality $\phi_H(s)=\phi_H(t)$, we have that $(xt)(xs)^{-1}\in H$ for all $x\in\dom\bigl(\phi_H(t)\bigr)$. Note that $u\in\dom\bigl(\phi_H(t)\bigr)$. Indeed, we have $uu^{-1}=e\in [e]$ and since $A$ is a $\rho_{tt^{-1}}$-class, we have that \begin{equation} (ut)(ut)^{-1}=utt^{-1}u^{-1}=e\in [e]. \end{equation} This implies that $u\cdot ts^{-1}\cdot u^{-1}\in [e]$. Moreover, since $\rank(uts^{-1}u^{-1})\leq\rank(u)=2$, we obtain that $\rank(uts^{-1}u^{-1})=2$, whence $(ut)(us)^{-1}=uts^{-1}u^{-1}=e$. In particular, we have that $us\ne 0$. But then $A$ is a union of some $\rho_s$-classes. Since $A$ was an arbitrary chosen $\rho_t$-class, we obtain that $\rho_s\subseteq\rho_t$. Analogously, one can prove that $\rho_t\subseteq\rho_s$. Thus, $\rho_s=\rho_t$. Further, if $s$ contains a block $A\cup B'$ then $us=\bigl\{E\cup B',E_1\cup \overline{B}'\bigr\}$, where $\overline{B}=\{1,\ldots,n\}\setminus B$. But $ut=\bigl\{E\cup A',E_1\cup \overline{A}'\bigr\}$ and so \begin{multline} \bigl\{E\cup E',E_1\cup E_1'\bigr\}=e=(ut)(us)^{-1}=\bigl\{E\cup A',E_1\cup \overline{A}'\bigr\}\cdot\bigl\{E\cup B',E_1\cup \overline{B}'\bigr\}^{-1}=\\\bigl\{E\cup A',E_1\cup \overline{A}'\bigr\}\cdot\bigl\{B\cup E',\overline{B}\cup E_1'\bigr\}. \end{multline} This implies $A=B$. Indeed, otherwise we would have $B\subseteq\overline{A}$ and so $A\subseteq\overline{B}$, whence $e=\bigl\{E\cup E_1',E_1\cup E'\bigr\}$, which is not true. Again, since $A$ was an arbitrary chosen $\rho_t$-class, we have that $\equiv_s=\equiv_t$. Thus, $s=t$. The proof is complete. \end{proof} \begin{lemma}\label{lm:Ef-Tr-6} Let $f\in\Theta_{\mathrm{pr}}^{n}$. Then $\lmod\mathcal{C}_{[f]}\rmod=2^n-2$. \end{lemma} \begin{proof} Take $[Hx]$ and $[Hy]$ of $\mathcal{C}_{[f]}$. Then due to Lemma~\ref{ugly-lemma}, we have that $[fx]=[fy]$ and $\rank(fx)=\rank(fy)$, whence $fx=fy$. Conversely, if $fx=fy$ then $[Hx]=[fx]=[fy]=[Hy]$. Thus, since $\rank(fx)=\rank(f)$ and $fx=f$ hold simultaneously if and only if $f\GL fx$, we obtain that $\lmod\mathcal{C}_{[f]}\rmod$ equals the cardinality of $\GL$-class, which contains $f$, which, in turn, equals the number of all partitions of $\{1,\ldots,n\}$ into two nonempty blocks. The latter number is equal to $2^n-2$. \end{proof} \begin{lemma}\label{lm:Ef-Tr-7} Let $f_1,f_2\in\Theta_{\mathrm{pr}}^{n}$. Then $\phi_{[f_1]}$ and $\phi_{[f_2]}$ are equivalent. \end{lemma} \begin{proof} Let $f_1=\tau_{F_1}\tau_{\{1,\ldots,n\}\setminus F_1}$ and $f_2=\tau_{F_2}\tau_{\{1,\ldots,n\}\setminus F_2}$ for certain proper subsets $F_1$ and $F_2$ of $\{1,\ldots,n\}$. Put $a=\bigl\{F_1\cup F_2',\bigl(\{1,\ldots,n\}\setminus F_1\bigr)\bigcup\bigl(\{1,\ldots,n\}\setminus F_2\bigr)'\bigr\}$. Then, taking to account Proposition \ref{pr:Omega}, we have that $a^{-1}[f_1]a=\{f_2\}\subseteq [f_2]$ and $a[f_2]a^{-1}=\{f_1\}\subseteq [f_1]$, whence $\phi_{[f_1]}$ and $\phi_{[f_2]}$ are equivalent. This completes the proof. \end{proof} Lemmas~\ref{lm:Ef-Tr-2}, \ref{lm:Ef-Tr-3}, \ref{lm:Ef-Tr-4}, \ref{lm:Ef-Tr-5}, \ref{lm:Ef-Tr-6}, \ref{lm:Ef-Tr-7} imply the statement of our theorem. We are done. \section{Definition of the ordered partition semigroup $\IOP_n$}\label{IOP_n} Let $n\in\mathbb{N}$. Consider the natural linear order on the set $\{1,\ldots,n\}$. Take $A\subseteq\{1,\ldots,n\}$. Denote by $\m_A$ the minimum element of $A$ with respect to this order. Denote by $\IOP_n$ the set of all elements $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$ of $\IP_n$ such that \begin{equation}\label{eq:IOP-definition} \m_{A_i}\leq\m_{A_j}\Rightarrow\m_{B_i}\leq\m_{B_j}~\mbox{for all}~i,j\in I. \end{equation} The following theorem shows that $\IOP_n$ is an inverse subsemigroup of $\IP_n$. \begin{theorem}\label{th:IOP_n} $\IOP_n$ is an inverse subsemigroup of $\IP_n$. \end{theorem} \begin{proof} That $a\in\IOP_n$ implies $a^{-1}\in\IOP_n$, follows immediately from \eqref{eq:IOP-definition}. It remains to prove that $\IOP_n$ is a subsemigroup of $\IP_n$. Take $a,b\in\IOP_n$. Set $c=ab$. Let $a=\bigl(A_i\cup B_i'\bigr)_{i\in I}$, $b=\bigl(C_j\cup D_j'\bigr)_{j\in J}$. Obviously, $0\in\IOP_n$, so we may assume that $c\ne 0$. Let also $c=\bigl(E_k\cup F_k'\bigr)_{k\in K}$ and set a linear order $\preceq$ on $K$, given by \begin{equation} \m_{E_k}\leq\m_{E_l}~\mbox{if and only if}~k\preceq l~\mbox{for all}~k,l\in K. \end{equation} Let now $K=\{k_1,\ldots,k_m\}$ and $k_1\preceq k_2\preceq\ldots\preceq k_m$. Set $P_i=E_{k_i}$ and $Q_i=F_{k_i}$ for all $i$, $1\leq i\leq m$. Then we have \begin{equation}\label{eq:wichtig} \m_{P_1}\leq\ldots\leq\m_{P_m}. \end{equation} Obviously, we have that $1\equiv_{a}1'$ and $1\equiv_{b}1'$. So $1\equiv_{c}1'$. Due to this fact, we obtain that $\{1,1'\}$ is a subset of the block $P_1\cup Q_1'$ of the element $c$. This implies that $\m_{Q_1}=1$. So $\m_{Q_1}\leq\m_{Q_2}$ and $\m_{Q_1}$ is the first number among the numbers $\m_{Q_1},\ldots,\m_{Q_m}$. Suppose now that $\m_{Q_1}\leq\ldots\leq\m_{Q_t}$ and that $\m_{Q_1},\ldots,\m_{Q_t}$ are the first $t$ numbers among the numbers $\m_{Q_1},\ldots,\m_{Q_m}$, for some $t$, $t<m$. Then $\m_{Q_t}\leq\m_{Q_{t+1}}$. Since $Q_1,Q_2,\ldots,Q_t$ are all $\lambda_{ab}$-classes, we obtain that each $Q_i$, $i\leq t$, is a union of some $\lambda_{b}$-classes and so $Z=Q_1\cup Q_2\cup\ldots\cup Q_t$ is a union of the sets $D_{r}$, $r\in R\subseteq J$. Further, we have that there is a subset $U$ of $I$ such that $\bigcup\limits_{u\in U}B_u=\bigcup\limits_{r\in R}C_r=W$. There is $r_0$ of $R$ such that $\m_{Q_t}\in D_{r_0}$. Then, obviously, $\m_{D_{r_0}}=\m_{Q_t}$. Since $\m_{Q_1},\ldots,\m_{Q_t}$ are the first $t$ numbers among $\m_{Q_1},\ldots,\m_{Q_m}$, we obtain that \begin{equation} \{1,\ldots,\m_{D_{r_0}}\}=\{1,\ldots,\m_{Q_1}\}\subseteq\bigcup\limits_{i=1}^{t}Q_i=\bigcup\limits_{r\in R}D_r. \end{equation} The latter implies that $\m_{D_r}$, $r\in R$, are the first $\lmod R\rmod$ numbers among the numbers $\m_{D_j}$, $j\in J$. Besides, $\m_{D_{r}}\leq\m_{D_{r_0}}$ for all $r\in R$. Then, taking to account that $b\in\IOP_n$, we obtain that $\{1,\ldots,\m_{C_{r_0}}\}\subseteq\bigcup\limits_{r\in R}C_r$ and $\m_{C_{r}}\leq\m_{C_{r_0}}$ for all $r\in R$. Then, taking to account $\bigcup\limits_{u\in U}B_u=\bigcup\limits_{r\in R}C_r$, we obtain that $\m_{B_u}$, $u\in U$, are the first $\lmod U\rmod$ numbers among the numbers $\m_{B_i}$, $i\in I$. Then, applying $a\in\IOP_n$, we obtain that $\m_{A_u}$, $u\in U$, are the first $\lmod U\rmod$ numbers among the numbers $\m_{A_i}$, $i\in I$. Note that $\bigcup\limits_{u\in U}A_u=\bigcup\limits_{i=1}^{t}P_t=Y$. Put $y=\m_{\{1,\ldots,n\}\setminus Y}$, $w=\m_{\{1,\ldots,n\}\setminus W}$ and $y=\m_{\{1,\ldots,n\}\setminus Z}$. Then due to what we have already obtained and due to~\eqref{eq:wichtig}, we have that $y=\m_{P_{t+1}}$. Suppose now that $z=\m_{Q_{g}}$, $g>t$. Then due to our assumption, we have that \begin{multline}\label{eq:final} \m_{Q_1},\ldots,\m_{Q_t},z~\mbox{are the first}~t+1~\mbox{numbers}\\ \mbox{among the numbers}~\m_{Q_1},\ldots,\m_{Q_m}. \end{multline} Due to $a,b\in\IOP_n$, we have that $y\equiv_a w'$ and $w\equiv_b z'$, whence $y\equiv_c z'$. This implies that $z\in Q_{t+1}$, whence $z=\m_{Q_{t+1}}$. Thus, due to~\eqref{eq:final}, we obtain that inductive arguments lead us to \begin{equation}\label{eq:one-more-last} \m_{Q_1}\leq\ldots\leq\m_{Q_m}. \end{equation} The conditions~\eqref{eq:wichtig} and~\eqref{eq:one-more-last} complete the proof. \end{proof} Thus, due to~Theorem \ref{th:IOP_n}, we can name $\IOP_n$ as the \emph{inverse ordered partition semigroup} of degree $n$. On Fig.~\ref{fig:f4} we give some examples of elements of $\IOP_8$. \begin{figure} \special{em:linewidth 0.4pt} \unitlength 0.80mm \linethickness{0.4pt} \begin{picture}(150.00,75.00) \put(15.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(15.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(40.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \put(105.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,00.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,10.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,20.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,30.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,40.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,50.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,60.00){\makebox(0,0)[cc]{$\bullet$}} \put(130.00,70.00){\makebox(0,0)[cc]{$\bullet$}} \drawline(13.00,-03.00)(13.00,03.00) \drawline(13.00,03.00)(17.00,03.00) \drawline(17.00,03.00)(17.00,28.00) \drawline(17.00,28.00)(13.00,28.00) \drawline(13.00,28.00)(13.00,33.00) \drawline(13.00,33.00)(42.00,43.00) \drawline(42.00,43.00)(42.00,38.00) \drawline(42.00,38.00)(38.00,38.00) \drawline(38.00,38.00)(17.00,-03.00) \drawline(17.00,-03.00)(13.00,-03.00) \drawline(13.00,07.00)(13.00,12.00) \drawline(13.00,12.00)(42.00,12.00) \drawline(42.00,12.00)(42.00,-03.00) \drawline(42.00,-03.00)(38.00,-03.00) \drawline(38.00,-03.00)(13.00,07.00) \drawline(13.00,18.00)(13.00,23.00) \drawline(13.00,23.00)(42.00,33.00) \drawline(42.00,33.00)(42.00,18.00) \drawline(42.00,18.00)(13.00,18.00) \drawline(13.00,47.00)(13.00,52.00) \drawline(13.00,52.00)(42.00,63.00) \drawline(42.00,63.00)(42.00,58.00) \drawline(42.00,58.00)(13.00,47.00) \drawline(13.00,38.00)(13.00,42.00) \drawline(13.00,42.00)(17.00,42.00) \drawline(17.00,42.00)(17.00,58.00) \drawline(17.00,58.00)(13.00,58.00) \drawline(13.00,58.00)(13.00,73.00) \drawline(13.00,73.00)(42.00,73.00) \drawline(42.00,73.00)(42.00,68.00) \drawline(42.00,68.00)(38.00,68.00) \drawline(38.00,68.00)(38.00,52.00) \drawline(38.00,52.00)(42.00,52.00) \drawline(42.00,52.00)(42.00,47.00) \drawline(42.00,47.00)(17.00,38.00) \drawline(17.00,38.00)(13.00,38.00) \drawline(103.00,-03.00)(103.00,03.00) \drawline(103.00,03.00)(107.00,03.00) \drawline(107.00,03.00)(128.00,22.00) \drawline(128.00,22.00)(132.00,22.00) \drawline(132.00,22.00)(132.00,-03.00) \drawline(132.00,-03.00)(103.00,-03.00) \drawline(103.00,08.00)(103.00,12.00) \drawline(103.00,12.00)(107.00,12.00) \drawline(107.00,12.00)(128.00,32.00) \drawline(128.00,32.00)(132.00,32.00) \drawline(132.00,32.00)(132.00,28.00) \drawline(132.00,28.00)(128.00,28.00) \drawline(128.00,28.00)(107.00,08.00) \drawline(107.00,08.00)(103.00,08.00) \drawline(103.00,18.00)(103.00,33.00) \drawline(103.00,33.00)(128.00,42.00) \drawline(128.00,42.00)(132.00,42.00) \drawline(132.00,42.00)(132.00,38.00) \drawline(132.00,38.00)(107.00,18.00) \drawline(107.00,18.00)(103.00,18.00) \drawline(103.00,48.00)(103.00,52.00) \drawline(103.00,52.00)(107.00,52.00) \drawline(107.00,52.00)(107.00,68.00) \drawline(107.00,68.00)(103.00,68.00) \drawline(103.00,68.00)(103.00,73.00) \drawline(103.00,73.00)(132.00,73.00) \drawline(132.00,73.00)(132.00,68.00) \drawline(132.00,68.00)(128.00,68.00) \drawline(128.00,68.00)(107.00,48.00) \drawline(107.00,48.00)(103.00,48.00) \drawline(103.00,38.00)(103.00,42.00) \drawline(103.00,42.00)(109.00,42.00) \drawline(109.00,42.00)(109.00,58.00) \drawline(109.00,58.00)(103.00,58.00) \drawline(103.00,58.00)(103.00,63.00) \drawline(103.00,63.00)(132.00,63.00) \drawline(132.00,63.00)(132.00,48.00) \drawline(132.00,48.00)(128.00,48.00) \drawline(128.00,48.00)(109.00,38.00) \drawline(109.00,38.00)(103.00,38.00) \end{picture} \caption{Elements of $\IOP_8$.}\label{fig:f4} \end{figure} Recall that a subsemigroup $T$ of a semigroup $S$ is said to be an \emph{$\GH$-cross-section} of $S$ if $T$ contains exactly one representative from each $\GH$-class of $S$. In the following proposition we show that $\IOP_n$ is an $\GH$-cross-section of $\IP_n$. \begin{proposition}\label{pr:IOP-cross-section} $\IOP_n$ is an $\GH$-cross-section of $\IP_n$. \end{proposition} \begin{proof} Follows from \eqref{eq:IOP-definition}, Theorem~\ref{th:Green} and Theorem~\ref{th:IOP_n}. \end{proof} As a consequence of Proposition~\ref{pr:IOP-cross-section}, we obtain the following corollary. \begin{corollary}\label{cor:IOP} Let $n\in\mathbb{N}$. Then $E(\IOP_n)=E(\IP_n)$. \end{corollary} \begin{proof} Recall that every maximal subgroup of an arbitrary semigroup $S$ coincides with some $\GH$-class of $S$, which contains an idempotent (see~\cite{Howie}). Then every $\GH$-cross-section of $\IP_n$ contains all the idempotents of $\IP_n$. In particular, $E(\IOP_n)=E(\IP_n)$, which was required. \end{proof} \section{Acknowledgments} The author is indebted to Professor Norman Reilly and to the two anonymous referees whose comments and suggestions contributed to a significant improvement of this paper.

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and a time for every purpose under heaven." Ecclesiastes 3:1 The past week has been a good one for me... My parents came to Oklahoma to visit me from Maryland and see what my "new life" is all about. I think they've really enjoyed their time here and I know I have certainly enjoyed their company! Last Saturday we went to Fort Worth, a town I am becoming ever increasingly fond of every time I go there, to go to the Fronteir Fort Days. It was amazing and I think my dad enjoyed seeing the old stockyards. We went to Oklahoma City yesterday to the Oklahoma National Stockyards - the world's largest feeder calf market - and then to the National Softball Hall of Fame. As many of you know, softball used to be my life. My biggest goal was to play Division I softball... and after I found the team that I loved, I shortly had the game taken from me. I tore my ACL twice before my sophomore season and was unable to play anymore. Softball is still a ghost I face everyday of my life, even though it has been nearly two years since I have last played. The Oklahoma City field is a place that every young girl that has a deep passion for the game dreams of playing - including me. I would be lying to say I didn't have a lump in my throat the entire time we were there yesterday, but it's also something I am so grateful I got to see... I am a strong believer that everything happens for a reason. While it is still a hard thing to swallow, I know that perhaps God used softball as a way to get me where He needed me to be. If I had not had the gift of softball, I would not have gone to the school I chose to play. I would never had been given the opportunities that I had in college at any other school, and I know that I wouldn't be here if it weren't for those opportunities. And if I had it all to do over again, I would want to end up right where I am... Right here. Sometimes it's so hard to remember all of our blessings when we have to face what we've lost. But sometimes we have to face what we've lost so we can more clearly see what all we've been blessed with....

: Medical Sleep, years in the works. About two years ago I found a new family doctor; general practice doctor? Anyways, my doctor growing up was in my hometown, and as much as I loved her, we really clicked, it was silly to drive 30 miles home, and back for an appointment when I could find someone closer. It took a bit, … Continue reading Sleep, years in the works.

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- . Dr. Zahid Hossain, P.E. Associate Professor of Civil Engineering Education - Doctor of Philosophy Civil Engineering The University of Oklahoma, Norman - Master of Science Computer Science The University of Oklahoma, Norman - Master of Science Civil Engineering The University of Oklahoma, Norman - Bachelor of Science Civil Engineering Khulna University of Eng. and Tech., Bangladesh Professional Credentials - Member, American Society of Civil Engineers (ASCE) - Member, Geo-Institute (GI) - Member, National Dean List - Trustee, Chi-Epsilon Honor Society - Member, Association for Computing Machinery (ACM) - Member, University of Oklahoma Database Group - Member, Institute of Engineers, Bangladesh Teaching Specialties Transportation Engineering, Construction Materials, Sustainable Engineering, Pavement Analysis and Design, Advanced Civil Engineering Materials, Engineering Mechanics, Geotechnical Engineering, Foundation Engineering, and Technical Communication. Courses Taught - ENGR 2411 Mechanics of Materials Lab, Spring 2015-2017, Fall 2015-2017 - CE 3223 Civil Engineering Materials, Fall 2013, 2014, 2015, 2016, 2017 - CE 429V Transportation Engineering, Spring 2013, 2014 - CE 4223 Transportation Engineering II, Fall 2015, 2016 - CE 5293 Advanced CE Materials, Spring 2015, Fall 2016, 2017 - ENGR 4463 Senior Design I (Group Supervision), Fall 2014, Spring 2016, 2017 - ENGR 4482 Senior Design II (Group Supervision), Spring 2015, Fall 2016, 2017 - ENGR 6163 Pavement Analysis and Design, Spring 2014, 2016, 2017 - ENGR 6053 Sustainable Engineering, Fall 2014, 2015 - ENGR 6693 Engineering Research, Fall 2016, Spring 2017, Summer 2017 - ENGR 629V Special Topics in Engineering, Summer 2014, 2015, 2016, 2017 - ENGR 689V Thesis Research Interests Transportation Engineering, Construction Materials, Sustainable Engineering, Pavement Analysis and Design, Advanced Civil Engineering Materials, Engineering Mechanics, Geotechnical Engineering, Foundation Engineering, and Technical Communication. Biography Dr. Hossain joined as a faculty member of Arkansas State University (ASU) in 2012 and serves as an Associate Professor of Civil Engineering. His research interests include energy conservation, recycling, nano- and bio-modifications, and intelligent system design of geotechnical and transportation materials for pavement applications. He is also faculty member of an interdisciplinary research team at the Center for Efficient and Sustainable Use of Resources (CESUR). Dr. Hossain focuses on the development and characterization of innovative geotechnical (stabilized subgrades and aggregates) and asphalt pavement materials and green technologies that will safeguard the environment and facilitate to build longer lasting pavements. Besides mechanistic performance evaluation, Dr. Hossain is interested in surface science and analytical chemistry (i.e., spectroscopy) approaches to evaluate these materials. Furthermore, Dr. Hossain's is interested in constitutive modeling, data mining and visualization, neural network modeling, lean construction, and molecular dynamics simulation of pavement materials. Dr. Hossain also holds "Green Belt" certification on Lean and Six/Sigma from the University of Oklahoma. Dr. Hossain a recipient of President's Gold Medal from Khulna University of Engineering and Technology (erstwhile Bangladesh Institute of Technology) for his outstanding performance in his undergraduate study. Contact Information P: 870-680-4299 mhossain@astate.edu Website Office Building: LSW Room: 246 Alternative Contact Timi Saffell P: 870-972-3565 tsaffell@astate.edu

) - this year we again (22 Sept 2018) Sunday (23 Sept 2018) Other possible topics: - Mobility under surveillance? - Collaboration software - What could be a joint EU-wide campaign in 2018? - Preparation of Monday meetings: Pad for meetings Monday (24 Sept 2018): free slot … -.

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Huzzah! I took my own advice & did this all reps routine the day after a rest day. I’d intended to practice yoga late yesterday afternoon, but ending up being downtown longer than expected & returning full, with the one meal I ate that day. Sometimes it’s cool to do a longer than usual fast before your first meal. It fills you up longer. My main & sometimes only meal of the day during ashtanga obsession was always lunch. “It’s way more than 1,000 reps; I’m telling you right now.” – Julia Just occurred to me today I should start tagging Julia’s workouts by who wrote them so I can get an idea of my own favorites! I always forget. I’m not going to go back through all my old workouts and tag them, but I will start doing that now. 1000 Rep Full Body Challenge by Daniela Part 1: 20 reps each, 1x through=220 reps Julia finishes just under 18min (including her very short introduction – she gets right to it which I love). My time = 22min. And of course I took a break after! - Diamond pushup burpee hands in diamond & diamond jump - Plank rows and frog hop (2 row + 1 hop = 1 rep) 2, 15lb weights 180tuck jump squats Squat & tuck jump I don’t like turning while doing tuck jumps. Need to be careful of 3rd Series knees! - Bulgarian LL + fwd lunge 2, 15lb weights / Do bulgarian & step same leg fwd for lunge Did this move 10x each leg, alt, for a total of 20reps ea leg - Bulgarian RL + fwd lunge - Jab cross 2x + heel grab 10 w left arm starting & switch - LL SL DL + hammer curl and shoulder press 13lbs ea hand Went light here b/c 15lbs would be still a bit high for me w/ upright rows, the next move. Legs must be even! lol - RL SL DL + upright row same weight as #7 - Yoga tricep pushups Chest goes close to ground, butt is up in air. Then push up into updog. J moves pretty fast here & stays up on toes. - Close grip chest press + skull crusher 2, 10lb weights I didn’t wanna challenge myself with 15s on the skull crusher! Also wouldn’t wanna hold 2 weights in each hand here, like I do when I pick up a 10 and a 3 for each hand. I’d do 12s if I had ’em. - Switch lunges 2=1 Part 2: Pyramid= 300 reps (520 total) My time = 17:36 - 100 squats 2, 15lb weights Went light! 50 holding weights on shoulders, 50 down by my side - 80 touchdowns 1=1 touching down in the middle - 60 hip thrusts 2, 20lb weights, on the floor J does sets of 20, alt weighted w/ unweighted & uses a shoulder elevation. I did the same, albeit w/ lesser weight. Someday I’ll have a sandbag! - 40 side lunges 2, 15lb weights / Toes turned out & feet planted wide. Alt side to side 1=1 - 20 back fly burpees with lat raise at the top 2, 10lbs No push-ups hooray! But still brutal. Jump back into plank & fly each arm. Jump forward & stand up straight for lat raise. Part 3: 20 reps each, 1x through=220 reps (740 total) My time = 24:17 And I thought this reps set went much faster than the 1st one! I guess it was the psychological ease of closing in on the end of the workout. - Tricep pushup plank tuck jump - SA clean & press and rev lunge (10x each arm) 15lbs Hold the weight up & then lunge back w/ same leg - Reverse grip pushup + bentover reverse grip row 2, 15lb weights / jump up & bentover row w/ both hands - Dive bombers - Swings Slams35lb kettlebell - SL box squat + fwd lunge (10 each leg) no added weight - weighted squat jumps 2, 10lb weights - Chest fly in bridge hold 2, 15lb weights on bench - SA burpee no push-up!! 10 ea arm obv - Sumo DL 2, 20lb weights - Good mornings 2, 15lb weights Part 4: pyramid 260 reps (1000 total) Forgot 2 reset stopwatch so don’t have time for this set - 90 SA curtsy swings (45/side) 10lbs Did 20/20/15/15/10/10 alt arms - 70 weighted punches 2, 3lb weights in sumo squat hold,, 2=1, 35 leading w/ LH 35 leading with right hand - 50 reverse curls on bench= - 30 ski squat + front raise 2, 10lb weight, 15 palms facing ea other / 15 palms down. Squat w/ feet together & then lift - 10 explosive pushups hands & feet leave the floor - 5 split lunge snatches RL 2, 15lb weights - 5 split lunge snatches LL Time = 1:26 including all breaks / better than expected Yoga About 15min. I should do an ashtangesque tomorrow.

Music Just Got Smarter with Logitech UEDecember 17, 2012 2 Comments on Music Just Got Smarter with Logitech UE What better way to get all of the juicy details on gifts from Logitech this season than to host, With the holiday season in full swing, many of us are baking treats for coworkers, hosting a potluck for friends, Our third video installment for Logitech UE tells the story of music going where few people have gone before: space. The 2012 Fall Tokyo Headphone Festival was held on Oct. 27-28 and is known to be one of the biggest Our second film highlighting the power of music, brings to life a time when government restrictions made it difficult for With today’s announcement of even more Apple products using the new Lightening connector, you may have concerns about the compatibility “If music has the power to stop war, if only for one night, it deserves to be heard in all As you know, over the past month Logitech UE announced a new line of music products, and we’ve held celebrations

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\section{Cocircuits of a threshold-linear network} In this section we identify an affine hyperplane arrangement associated with any threshold-linear network and show that the combinatorial equivalence class of this arrangement is sufficient to determine the fixed point supports of the network. Specifically, we show that the set of cocircuits of the arrangement determines $\operatorname{FP}(W,b)$. It follows that support bifurcations correspond to changes in the cocircuits of the arrangement. We establish a correspondence between network parameters and the geometry of the arrangement by connecting the quantities $b_i$, $b_iW_{ji}+b_j$ and $b_iW_{kj}-b_kW_{ij}$ with its cocircuits. Such a connection will allow an understanding of how modulation of network parameters leads to qualitative changes in the network dynamics. \par To obtain an affine arrangement from a threshold-linear network, consider the linear functionals: \begin{equation} \begin{split} h^*_i(x)&:=-x_i+\sum_{j\not=i}W_{ij}x_j+b_i\\ e^*_i(x)&:=x_i \end{split} \end{equation} for $i=1,...,n$. By taking the zero sets $E_i:=\{e^*_i(x)=0\}$ and $H_i:=\{h^*_i(x)=0\}$ we obtain an arrangement of $2n$ affine hyperplanes in the state space $\mathbb{R}^n$ which we will denote by $\mathcal{A}=\mathcal{A}(W,b)$. Note that the relationship with the linear boundaries is given by the equation: \begin{equation}\label{glueing} h^*_i(x)=-e^*_i(x)+l^*_i(x) \end{equation} for $i=1,...,n$. The equations of each linear region are constructed from a subset of $\{e^*_1,h^*_1,...,e^*_n,h^*_n\}$ and in particular, the fixed point $x^\sigma$ of the linear region $L^\sigma$ is a vertex of the arrangement $\mathcal{A}$. The arrangement $\mathcal{A}$ can be viewed as the smallest hyperplane arrangement that contains the nullclines of the threshold-linear network. To encode the geometry of this arrangement we consider its cocircuits. \begin{definition} The \textit{cocircuits of a threshold-linear network} we define to be the image of the map \begin{align*} \pi:\mathbb{R}^n&\to \{+,0,-\}^{2n}\\ x&\mapsto \big( \text{sgn}\:e^*_i(x),\text{sgn}\:h^*_i(x) \big)_{i=1,...,n} \end{align*} restricted to the vertices of the arrangement $\mathcal{A}$. \end{definition} For each vertex $x$ in the arrangement $\mathcal{A}$, the cocircuit of $x$ is a sign vector recording the position of $x$ with respect to each hyperplane in $\mathcal{A}$. In general, the collection of cocircuits of a hyperplane arrangement defines an oriented matroid (see, for instance, \cite{OMbook}) and the above definition is just a restriction to the case of arrangements arising from threshold-linear networks. Oriented matroids capture the combinatorial properties of an arrangement and two threshold-linear networks with the same set of cocircuits we will think of as combinatorially equivalent. The following lemma says that two distinct but combinatorially equivalent threshold-linear networks must have the same set of fixed point supports. \begin{lemma}\label{the_lemma} A subset of neurons $\sigma$ supports a fixed point if and only if the cocircuit $C^\sigma\in\{+,0,-\}^{2n}$ of $x^\sigma$ satisfies: \[ C^\sigma(E_i)=+ \text{ and } C^\sigma(H_j)=- \] for $i\in\sigma$ and $j\not\in\sigma$. \end{lemma} \begin{proof} By definition of $x^\sigma$, we have $h^*_i(x^\sigma)=0$ and $e^*_j(x^\sigma)=0$ for $i\in\sigma$ and $j\not\in\sigma$. From equation \eqref{glueing} with $x=x^\sigma$, we have the equalities: \[ l^*_i(x^\sigma)=e^*_i(x^\sigma) \text{ and } l^*_j(x^\sigma)=h^*_j(x^\sigma) \] for $i\in\sigma $ and $j\not\in\sigma$. By taking signs, we have that $x^\sigma$ lies in the linear cell $L^\sigma$ if and only if $\text{sgn }e^*_i(x^\sigma)=+$ and $\text{sgn }h^*_j(x^\sigma)=-$. \end{proof} If follows that support bifurcations correspond to changes in the cocircuits of the arrangement. Note that such a change occurs when the network parameters are varied in such a way that a hyperplane is pushed over a vertex. To illustrate this, we revisit the bifurcation in Figure~\ref{fig:switching} but from the perspective of the arrangement $\mathcal{A}$ and its cocircuits. \begin{example} \begin{figure}[h] \centering \includegraphics[width=5.5in]{Fig2.pdf} \caption{The support bifurcation from Figure~\ref{fig:switching} from the perspective of the hyperplane arrangement $\mathcal{A}$. The fixed points $x^{12}$ and $x^1$ colliding at the boundary $L_2$ and crossing can equivalently be viewed as the hyperplane $H_2$ being pushed over the vertex $x^1$ changing the cocircuits of the arrangement and hence its combinatorial equivalence class. } \label{fig:2d_bif} \end{figure} In the two dimensional case the arrangement $\mathcal{A}=\{E_1,H_1,E_2,H_2\}$ is an arrangement of four lines in the plane. From the definition of the linear functions $h^*_1$ and $h^*_2$, we see that the lines $\{H_1,H_2\}$ are oriented with the origin on their positive side (Figure~\ref{fig:2d_bif}). This information is sufficient to determine the cocircuits of the arrangement and we see that the bifurcation in Figure~\ref{fig:switching} corresponds to the change in cocircuits: \begin{align*} C^1:&\: (+,0,0,-)\to (+,0,0,+) \\ C^{12}:&\: (+,0,+,0) \to (+,0,-,0) \end{align*} where each cocircuit records the position of the fixed point with respect to the ordered set of hyperplanes $\{E_1,H_1,E_2,H_2\}$. In other words, $x^{12}$ and $x^1$ crossing the boundary $L_2$ can equivalently be viewed as $x^{12}$ crossing the hyperplane $E_2$ and $x^{1}$ crossing the hyperplane $H_2$. If we observe that the intersection of $H_2$ with $E_2$ is given by $x_1=-b_2/W_{21}$ and $x_2=0$, we see that this bifurcation can be realized by increasing the negative synaptic weight $W_{21}$ toward zero, for instance. \end{example} In this way, we view support bifurcations as changes in the combinatorial geometry of the arrangement arising from modulation of network parameters. In the following three lemmas we translate the network parameters into geometric properties of the hyperplane arrangement. In particular, we work out the relationship between network parameters and the restriction of the arrangement to each coordinate axis. Although the intersection of an arrangement with the coordinate axes does not uniquely determine the arrangement, in dimension three an understanding of this geometry is sufficient for a complete understanding of support bifurcations. \begin{lemma}\label{lemma1} The intersection of the hyperplane $H_i$ with the $x_i$-axis lies on the positive side of $E_i$ if and only if $b_i>0$. \end{lemma} \begin{proof} The intersection of $H_i$ with the $x_i$-axis is given by: \[ x^i:=H_i\cap \bigcap_{j\not=i}E_j \] Since $H_i$ is given by the zero set: \[ H_i:=\big\{-x_i+\sum_{j\not=i}W_{ji}x_j + b_i=0\big\} \] it follows that $x^i$ is defined by the equations $x_i=b_i$ and $x_j=0$ for $j\not=i$. Thus, $x^i$ lies on the positive side of the coordinate hyperplane $E_i$ if and only if $e^*_i(x^i)=b_i>0$. \end{proof} \begin{lemma}\label{lemma2} The intersection of $H_i$ with the $x_i$-axis lies on the positive side of $H_j$ if and only if $s^{ij}_j=b_iW_{ji}+b_j>0$. \end{lemma} \begin{proof} The position of $x^i$ with respect to the hyperplane $H_j$ is obtained by evaluating $h^*_j$ at $x_i$: \[ h^*_j(x^i) =\big( -x_j+\sum_{k\not= j}W_{jk}x_k+b_j\big )\bigg\rvert_{x^i} = b_iW_{ji}+b_j \] Thus, $x^i$ lies on the positive side of $H_j$ if and only if $b_iW_{ji}+b_j>0$. \end{proof} The above lemmas imply that a single neuron $i$ supports a fixed point only if the external input to that neuron is positive and $s^{ij}_j=b_iW_{ji}+b_j<0$ for all $j$. If we encode the signs of $s^{ij}_j=b_iW_{ji}+b_j$ in a directed graph, then we have that a single neuron supports a fixed point if and only if the external input to $i$ is positive and the node corresponding to $i$ is a sink. The bifurcation in Figure~\ref{fig:2d_bif} can be viewed from this perspective as varying the network parameters such that the edge $1\to 2$ is added to $G$. In dimension two, all bifurcations of competitive threshold-linear networks with uniform positive external input are obtained as changes to the underlying graph $G$,\cite{robust-motifs}. In dimension three this is no longer the case and it becomes necessary to consider higher order interactions between triples of neurons. \begin{lemma} The intersection of $H_i$ with the $x_j$-axis lies on the positive side of $H_k$ if and only if $\Delta^{ki}_j=b_iW_{kj}-b_kW_{ij}>0$. \end{lemma} \begin{proof} Let $p_{ij}$ denote the intersection of $H_i$ with the $x_j$ axis. We have \[ p_{ij}:=H_i\cap \bigcap_{l\not=j}E_l \] and so $p_{ij}$ is defined by $x_j=-b_i/W_{ij}$ and $x_l=0$ for $l\not=j$. The position of $p_{ij}$ with respect to $H_k$ is then obtained by evaluating: \[ h^*_k(p_{ij})=-x_k+\sum_{l\not=k}W_{kl}x_l+b_k\bigg\rvert_{p_{ij}}=W_{kj}(-b_i/W_{ij})+b_k \] Using our assumption of competitiveness and multiplying by $W_{ij}<0$, we have that $h^*_k(p_{ij})>0$ if and only if $b_iW_{kj}-b_kW_{ij}>0$. \end{proof} In the following example we illustrate the correspondences between network parameters and geometry established in the above lemmas by exhibiting some support bifurcations in dimension three. \begin{example}\label{Example1} Consider the threshold-linear network defined by the parameters \[ W=\small\begin{bmatrix} 0& -0.97 & -1.47\\ -0.65 & 0 & -0.57\\ -1.34 & -1.45 & 0\\ \end{bmatrix} \hspace{.25cm} b=\begin{bmatrix} 0.49\\ 0.40\\ 0.62 \end{bmatrix} \] In dimension three the arrangement $\mathcal{A}$ consists of six hyperplanes in $\mathbb{R}^3$. The arrangement for this particular network is shown Figure~\ref{Arrangements 1}. The hyperplanes $\{E_1,H_1\}$, $\{E_2,H_2\}$ and $\{E_3,H_3\}$ are colored blue, red and yellow, respectively, and $H_1$, $H_2$ and $H_3$ are all oriented with the origin on the positive side. \begin{figure}[h] \centering \includegraphics[width=6in]{Fig3.pdf} \caption{Varying the parameters of the network such that the quantity $b_3W_{23}+b_2$ changes sign changes the relative positions of the hyperplane $H_2$ and $H_3$ along the $x_3$-axis leading to the persistent support bifurcation $\{23,12,123\}\to \{3,12,123\}$. } \label{Arrangements 1} \end{figure} From the arrangement we can read off cocircuits and determine that $\operatorname{FP}(W,b)=\{12,123,23\}$. To see how modulation of network parameters leads to changes in this set consider flipping the sign of the quantity $b_3W_{23}+b_2$ by letting $W_{23}\to -0.8$. In the arrangement, this pushes the red hyperplane $H_2$ over the virtual fixed point $x^3$. Moreover, as this is done, the fixed point $x^{23}$ collides with $x^3$ and changes position with respect to $E_2$ and we have the persistent support bifurcation $\{23\}\to \{3\}$. If we encode the signs of $b_iW_{ji}+b_j$ in a directed graph, then this bifurcation corresponds to deleting an edge so that neuron three becomes a sink (Figure~\ref{graph_11}). \begin{figure}[h] \centering \includegraphics[width=3in]{Fig4.pdf} \caption{The directed graph of a threshold-linear network is determined by the signs of the quantities $b_iW_{ji}+b_j$. Such a graph captures pairwise asymmetries in the connectivity of the network. The bifurcation in the previous figure corresponds to deleting the edge $3\to 2$ in this graph. } \label{graph_11} \end{figure} Although this bifurcation is obtained by letting $W_{23}\to -0.8$, we can similarly consider varying either of the external inputs $b_2$ or $b_3$ in order to flip the sign of $b_3W_{23}+b_2$. Consider for instance, letting $b_2\to 0.25$. While this has the effect of flipping the sign of $b_3W_{23}+b_2$ and realizing the bifurcation $\{23\}\to \{3\}$, varying the external input in this way has a less localized effect on the network and in fact leads to several other bifurcations in addition. Specifically, decreasing the external input to neuron two in this way produces three persistent bifurcations $\{12\}\to \{1\}$, $\{123\}\to \{13\}$ and $\{23\}\to \{3\}$. In sum, the effect is to change the admissible sets from $\operatorname{FP}(W,b)=\{12,123,23\}$ to $\operatorname{FP}(W,b)=\{1,13,3\}$. The reason for this effect is that the parameter $b_2$ is involved in determining the intersection of $H_2$ with each of the coordinate axes. Thus a perturbation of $b_2$ moves each of these vertices simultaneously leading to less localized changes in the arrangement. In general, an understanding of how perturbations of network parameters leads to changes in the collection of admissible fixed points requires an understanding of the local perturbations of the hyperplane arrangement which we will develop in later sections. \begin{figure}[h] \centering \includegraphics[width=6in]{Fig5.pdf} \caption{Varying the external input leads to multiple support bifurcations. Here, the perturbation $b_2:0.4\to0.25$ contracts the hyperplane $H_2$ along all three coordinate axes simultaneously producing three persistent bifurcations $\{12\}\to \{1\}$, $\{123\}\to \{13\}$ and $\{23\}\to \{3\}$ changing $\operatorname{FP}(W,b)=\{12,123,23\}$ to $\operatorname{FP}(W,b)=\{1,13,3\}$. } \label{fig:A_2} \end{figure} \par Consider now the quantity $\Delta^{31}_2=b_1W_{32}-b_3W_{12}$. The sign of this quantity determines the position of the intersection of $H_1$ with the $x_2$-axis with respect to the hyperplane $H_3$. In other words, increasing this quantity away from zero increases the separation of the hyperplanes $H_1$ and $H_3$ along the $x_2$-axis. This can be accomplished by increasing $W_{12}$ toward zero and decreasing $W_{32}$ (Figure~\ref{fig: A_3}.) The effect of this is a non-smooth fold bifurcation $\{123,23\}\to \emptyset$ and a change in the admissible sets from $\operatorname{FP}(W,b)=\{12,123,23\}$ to $\operatorname{FP}(W,b)=\{12\}$. \begin{figure}[h] \centering \includegraphics[width=6in]{Fig6.pdf} \caption{The non-smooth fold bifurcation $\{123,23\}\to \emptyset$ can be obtained by varying the quantity $\Delta^{31}_2=b_1W_{32}-b_3W_{12}$ which corresponds to separating the hyperplanes $H_1$ and $H_3$ along the $x_2$ axis. This bifurcation can equivalently be viewed as flipping the simplicial cell bounded by the hyperplanes $\{H_1,H_2,H_3,E_1\}$. We will show in section four that all generic support bifurcations in dimension three correspond to flipping a simplicial cell. } \label{fig: A_3} \end{figure} Note that this bifurcation does not change the underlying directed graph as it does not change the position of $H_1$ or $H_3$ with respect to $H_2$ along the $x_2$-axis. Another important observation here is that this bifurcation corresponds to flipping a simplicial cell in the arrangement. If we consider the decomposition of the hyperplane arrangement $\mathcal{A}$ into cells, then an $n$-dimensional cell is called simplicial if it has a minimal number of vertices and faces. The bifurcation in Figure~\ref{fig: A_3} corresponds to flipping the simplicial cell bounded by the hyperplanes $H_1$, $H_2$ and $H_3$ and $E_1$. For an arrangement in general position, the smallest possible changes are in one-to-one correspondence with its simplicial cells. Thus, the set of support bifurcations obtainable by perturbing network parameters is constrained by the set of simplices in the arrangement. By considering the chirotope of the arrangement we will be able to express these constraints algebraically. \end{example}

TITLE: mean and variance formula for negative binomial distribution QUESTION [2 upvotes]: The equation below indicates expected value of negative binomial distribution. I need a derivation for this formula. I have searched a lot but can't find any solution. Thanks for helping :) $$ E(X)=\sum_{x=r}^\infty x\cdot \binom {x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} =\frac{r}{p} $$ I have tried: \begin{align} E(X) & =\sum _{x=r} x\cdot \binom{x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} \\[8pt] & = \sum_{x=r}^\infty x \cdot \frac{(x-1)!}{(r-1)! \cdot ((x-1-(r-1))!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt] & = \sum_{x=r}^\infty \frac{x!}{(r-1)!\cdot ((x-r)!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt] \Longrightarrow & \phantom{={}} \sum_{x=r}^\infty r\cdot \frac{x!}{r!\cdot (x-r)!}\cdot p^r \cdot (1-p{)}^{x-r} \\[8pt] & = \frac{r}{p} \cdot \sum_{x=r}^\infty \frac{x!}{r!\cdot (x-r)!}\cdot p^{r+1}\cdot (1-p)^{x-r} \end{align} If the power of $p$ in the last equation were not $r + 1,$ I can implement Newton Binomial. So It will be true. But I am stuck here. REPLY [1 votes]: Here is an alternative derivation to compute the expectation of a negative binomial distribution. Let $X_r$ be the number of trials needed to get $r$ successes, and $p$ be the probability of success on any given trial. If $r=1$, then $X_r$ has a geometric distribution, and so $\mathrm{E}(X_r)=1/p$. To prove this, let $S$ be the event that the first trial is success. By the law of iterated expectation, \begin{align} \DeclareMathOperator{\E}{\mathrm{E}} \DeclareMathOperator{\P}{\mathrm{P}} \E(X_1) &= \E(X_1 \mid S)\P(S)+\E(X_1\mid S')P(S') \\[4pt] &= (1 \cdot p)+(\E(X_1)+1)(1-p) \, , \end{align} and rearranging yields the desired result. Computing $\E(X_r)$ in general is not too difficult if we use linearity of expectation. Let $A_i$ be the number of additional trials needed to get the $i$-th success once you have had $i-1$ successes. Note that $X_r=\sum_{i=1}^{r}A_i$, and so $$ \E(X_r)=\E\left(\sum_{i=1}^r A_i\right)=\sum_{i=1}^r \E(A_i) \, . $$ But the expected number of trials needed to get the $i$-th success is no different to the expected number of trials needed to get the first success, and so $\E(A_i)=\E(A_1)=\E(X_1)=1/p$. Hence, $E(X_r)=r/p$.

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\begin{document} \title[Auslander-Buchweitz approximation theory]{Auslander-Buchweitz approximation theory for triangulated categories} \author{O. Mendoza, E. C. S\'aenz, V. Santiago, M. J. Souto Salorio.} \thanks{2000 {\it{Mathematics Subject Classification}}. Primary 18E30 and 18G20. Secondary 18G25.\\ The authors thank the financial support received from Project PAPIIT-UNAM IN101607 and MICINN-FEDER TIN2010-18552-C03-02.} \date{} \begin{abstract} We introduce and develop an analogous of the Auslander-Buchweitz approximation theory (see \cite{AB}) in the context of triangulated categories, by using a version of relative homology in this setting. We also prove several results concerning relative homological algebra in a triangulated category $\T,$ which are based on the behaviour of certain subcategories under finiteness of resolutions and vanishing of Hom-spaces. For example: we establish the existence of preenvelopes (and precovers) in certain triangulated subcategories of $\T.$ The results resemble various constructions and results of Auslander and Buchweitz, and are concentrated in exploring the structure of a triangulated category $\T$ equipped with a pair $(\X,\omega),$ where $\X$ is closed under extensions and $\omega$ is a weak-cogenerator in $\X,$ usually under additional conditions. This reduces, among other things, to the existence of distinguished triangles enjoying special properties, and the behaviour of (suitably defined) (co)resolutions, projective or injective dimension of objects of $\T$ and the formation of orthogonal subcategories. Finally, some relationships with the Rouquier's dimension in triangulated categories is discussed. \end{abstract} \maketitle \section*{Introduction.} The approximation theory has its origin with the concept of injective envelopes and it has had a wide development in the context of module categories since the fifties. In independent papers, Auslander, Reiten and Smalo (for the category $\mathrm{mod}\,(\Lambda)$ of finitely generated modules over an artin algebra $\Lambda$), and Enochs (for the category $\mathrm{Mod}\,(R)$ of modules over an arbitrary ring $R$) introduced a general approximation theory involving precovers and preenvelopes (see \cite{AR}, \cite{AS} and \cite{E}). Auslander and Buchweitz (see \cite{AB}) studied the ideas of injective envelopes and projective covers in terms of maximal Cohen-Macaulay approximations for certain modules. In their work, they also studied the relationship between the relative injective dimension and the coresolution dimension of a module. They developed their theory in the context of abelian categories providing important applications in several settings. Based on \cite{AB}, Hashimoto defined the so called ``Auslander-Buchweitz context" for abelian categories, giving a new framework to homological approximation theory (see \cite{H}). Recently, triangulated categories entered into the subject in a relevant way and several authors have studied the concept of approximation in both contexts, abelian and triangulated categories (see, for example, \cite{AM} \cite{B}, \cite{BR} and \cite{MS}). In this paper, an analogous theory of approximations in the sense of Auslander and Buchweitz (see \cite{AB}), is developed for triangulated categories. Throughout this paper, $\T$ denotes an arbitrary triangulated category and $\X$ a class of objects in $\T$. The main result (Theorem \ref{specialtrian}) deals with a pair $(\X,\omega)$ of classes of objects in $\T ,$ where $\X$ is closed under extensions, and $\omega$ satisfies a weak cogenerating condition with respect to the objects of $\X.$ Like Auslander and Buchweitz, we consider the class $\X^\wedge$ of objects of $\T$ admitting a finite resolution by objects of $\X.$ We prove that any object of $\X^\wedge$ admits two distinguished triangles: one giving rise to a right $\X$-approximation, and the other to a left $\omega^\wedge$-approximation. In the present paper, it is also introduced and discussed a notion of $\X$-resolution dimension, which is compared with other relative homological dimensions. The paper is organized as follows: In Section 1, we give some basic notions and properties of triangulated categories, that will be used in the rest of the work. In Section 2, we study the notion of $\mathcal{X}$-resolution dimension which allows us to characterize the triangulated subcategory $\Delta_\T(\X)$ of $\T,$ generated by a cosuspended subcategory $\X$ of $\T$ (see Theorem \ref{resBuan}). In Section 3, the properties of the $\mathcal{X}$-projective (respectively, $\mathcal{X}$-injective) dimension and its relation to the $\mathcal{X}$-resolution (respectively, coresolution) dimension are established. The main result of this section is Theorem \ref{pdyresdim} that relates different kinds of relative homological dimensions by using suitable subcategories of $\mathcal{T}$. In Section 4, we focus our attention to the notions of $\mathcal{X}$-injectives and weak-cogenerators in $\mathcal{X}.$ We relate these ideas to the concepts of injective and coresolution dimension. This leads us to characterize several triangulated subcategories; and moreover, in Theorem \ref{specialtrian} we establish the existence of $\mathcal{X}$-precovers and $\omega^\wedge$-preenvelopes. Finally, in Theorem \ref{tilde-id}, for a given pair $(\X,\omega)$ satisfying certain conditions, we give several characterizations of the triangulated subcategory $\Delta_{\X^\wedge}(\omega)$ of $\X^\wedge$ generated by $\omega.$ We remark that the results we get will be applied in a forthcoming paper \cite{MSSS2}, where a connection between Auslander-Buchweitz approximation theory in triangulated categories and co-$t$-structures (see \cite{Bo} and \cite{P}) is established. Finally, some relationship with other notions, as torsion theories (see Corollary \ref{torsion}) in the sense of Iyama-Yoshino \cite{IY} and Rouquier's dimension \cite{Ro}, are discussed (see Section 5). \section{Preliminaries} \ Throughout this paper, $\T$ will be a triangulated category and $[1]:\T\rightarrow\T$ its suspension functor. The term subcategory, in this paper, means a subcategory which is full, additive, and closed under isomorphisms. \ An important tool, which is a consequence of the octahedral axiom in $\T,$ is the so-called co-base change (see \cite{K}). That is, for any diagram in $\T$ $$\begin{CD} X @>>> Y\\ @VVV @.\\ Z @.\, \end{CD}$$ \vspace{.2cm} there exists a commutative and exact diagram in $\T$ $$\begin{CD} @. W[-1] @= W[-1] @.\\ @. @VVV @VVV @.\\ U[-1] @>>> X @>>> Y @>>> U\\ @| @VVV @VVV @|\\ U[-1] @>>> Z @>>> E @>>> U\\ @. @VVV @VVV @.\\ @. W @= W @.\, \\ {} \end{CD} $$ where exact means that the rows and columns, in the preceding diagram, are distinguished triangles in $\T.$ The base change, which is the dual notion of co-base change, also holds (see \cite{K}). Let $\X$ and $\Y$ be classes of objects in $\T.$ We put ${}^\perp\X:=\{Z\in\T\,:\,\Hom_\T(Z,-)|_{\X}=0\}$ and $\X^\perp:=\{Z\in\T\,:\,\Hom_\T(-,Z)|_{\X}=0\}.$ \noindent We denote by $\X*\Y$ the class of objects $Z\in\T$ for which exists a distinguished triangle $X\rightarrow Z\rightarrow Y\rightarrow X[1]$ in $\T$ with $X\in\X$ and $Y\in\Y.$ In case $\Y=\{Y\},$ we write $\X*Y$ instead of $\X*\Y.$\\ It is also well known that the operation $*$ is associative (see \cite[1.3.10]{BBD}). Furthermore, it is said that $\X$ is {\bf closed under extensions} if $ \X*\X\subseteq \X.$ \ Recall that a class $\X$ of objects in $\T$ is said to be {\bf suspended} (respectively, {\bf cosuspended}) if $\X[1]\subseteq\X$ (respectively, $\X[-1]\subseteq\X$) and $\X$ is closed under extensions. By the following lemma, it is easy to see, that a suspended (respectively, cosuspended) class $\X$ of objects in $\T,$ can be considered as a subcategory of $\T.$ \begin{lem}\label{xx=x} Let $\X$ be a class of objects in $\T.$ \begin{enumerate} \item[(a)] If $0\in \X$ then $\Y\subseteq\X*\Y$ and $\Y\subseteq\Y*\X$ for any class $\Y$ of objects in $\T.$ \item[(b)] If $\X$ is either suspended or cosuspended, then $0\in \X$ and $\X=\X*\X.$ \end{enumerate} \end{lem} \begin{dem} (a) If $0\in \X$ then we get $\Y\subseteq \X*\Y$ by using the distinguished triangle $0\rightarrow Y\stackrel{1_Y}{\to}Y \to 0$ for any $Y\in \Y.$ The other inclusion follows similarly. \ (b) Let $\X$ be cosuspended (the other case, is analogous). Then, it follows that $0\in \X$ since we have the distinguished triangle $X[-1]\rightarrow 0\rightarrow X \rightarrow X $ for any $X\in \X.$ Hence (b) follows from (a). \end{dem} \vspace{.2cm} Given a class $\X$ of objects in $\T,$ it is said that $\X$ is {\bf{closed under cones}} if for any distinguished triangle $A\rightarrow B\rightarrow C\rightarrow A[1]$ in $\T$ with $A,B\in \X$ we have that $C\in \X.$ Similarly, $\X$ is {\bf{closed under cocones}} if for any distinguished triangle $A\rightarrow B\rightarrow C\rightarrow A[1]$ in $\T$ with $B,C\in \X$ we have that $A\in \X.$ \ We denote by $\U_{\X}$ (respectively, ${}_{\X}\U$) the smallest suspended (respectively, cosuspended) subcategory of $\T$ containing the class $\X.$ Note that if $\X$ is suspended (respectively, cosuspended) subcategory of $\T,$ then $\X=\U_{\X}$ (respectively, $\X={}_{\X}\U$). We also recall that a subcategory $\U$ of $\T,$ which is suspended and cosuspended, is called {\bf{triangulated subcategory}} of $\T.$ A {\bf{thick}} subcategory of $\T$ is a triangulated subcategory of $\T$ which is closed under direct summands in $\T.$ We also denote by $\Delta_\T(\X)$ (respectively, $\overline{\Delta}_\T(\X))$ to the smallest triangulated (respectively, smallest thick) subcategory of $\T$ containing the class $\X.$ Observe that $\Delta_\T(\X)\subseteq\overline{\Delta}_\T(\X).$ For the following definition, see \cite{AR}, \cite{B}, \cite{BR} and \cite{E}. \begin{defi} Let $\X$ and $\Y$ be classes of objects in the triangulated category $\T.$ A morphism $f:X\to C$ in $\T$ is said to be an {\bf $\X$-precover} of $C$ if $X\in\X$ and $\Hom_\T(X',f):\Hom_\T(X',X)\to\Hom_\T(X',C)$ is surjective, $\forall X'\in\X.$ If any $C\in\Y$ admits an $\X$-precover, then $\X$ is called a {\bf precovering class} in $\Y.$ By dualizing the definition above, we get the notion of an {\bf $\X$-preenveloping} of $C$ and a {\bf preenveloping class} in $\Y.$ \end{defi} \ Finally, in order to deal with the (co)resolution, relative projective and relative injective dimensions, we consider the extended natural numbers $\overline{\mathbb{N}}:=\mathbb{N}\cup\{\infty\}.$ Here, we set the following rules: (a) $x+\infty=\infty$ for any $x\in\overline{\mathbb{N}},$ (b) $x<\infty$ for any $x\in\mathbb{N}$ and (c) $\mini(\emptyset):=\infty.$ \section{resolution and coresolution dimensions} Now, we define certain classes of objects in $\T$ which will lead us to the notions of resolution and coresolution dimensions. \begin{defi}\label{epsilon} Let $\X$ be a class of objects in $\T.$ For any natural number $n,$ we introduce inductively the class $\varepsilon^\wedge_n(\X)$ as follows: $\varepsilon^\wedge_0(\X):=\X$ and assuming defined $\varepsilon^\wedge_{n-1}(\X),$ the class $\varepsilon^\wedge_{n}(\X)$ is given by all the objects $Z\in\T$ for which there exists a distinguished triangle in $\T$ $$\begin{CD} Z[-1] @>>> W @>>> X @>>> Z \, \end{CD}$$ with $W\in\varepsilon^\wedge_{n-1}(\X)$ and $X\in\X.$ Dually, we set $\varepsilon^\vee_0(\X):=\X$ and supposing defined $\varepsilon^\vee_{n-1}(\X),$ the class $\varepsilon^\vee_{n}(\X)$ is formed for all the objects $Z\in\T$ for which there exists a distinguished triangle in $\T$ $$\begin{CD} Z @>>> X @>>> K @>>> Z[1]\, \end{CD}$$ with $K\in\varepsilon^\vee_{n-1}(\X)$ and $X\in\X.$ \end{defi} We have the following properties for $\varepsilon^\wedge_{n}(\X)$ (and the similar ones for $\varepsilon^\vee_{n}(\X)$). \begin{pro}\label{rkepsilon} Let $\T$ be a triangulated category, $\X$ be a class of objects in $\T,$ and $n$ a natural number. Then, the following statements hold. \begin{enumerate} \item[(a)] For any $Z\in \T$ and $n>0,$ we have that $Z\in\varepsilon^\wedge_{n}(\X)$ if and only if there is a family $\{K_j[-1]\to K_{j+1}\to X_j\to K_j\}_{j=0}^{n-1}$ of distinguished triangles in $\T$ with $K_0=Z,$ $X_j\in \X$ and $K_n\in\X.$ \item[(b)] $\varepsilon^\wedge_{n}(\X)=*_{i=0}^n\,\X[i]:=\X*\X[1]*\cdots*\X[n].$ \item[(c)] If $0\in \X$ then $\X[n]\subseteq \varepsilon^\wedge_{n}(\X)\subseteq\varepsilon^\wedge_{n+1}(\X)$ and $\varepsilon^\wedge_n(\X)[1]\subseteq \varepsilon^\wedge_{n+1}(\X)$ $\quad\forall\;n\in\N.$ \end{enumerate} \end{pro} \begin{dem} (a) If $n=1$ then the equivalence follows from the definition of $\varepsilon^\wedge_1(\X).$ Let $n\geq 2$ and suppose (by induction) that the equivalence is true for $\varepsilon^\wedge_{n-1}(\X).$ By definition, $Z\in \varepsilon^\wedge_{n}(\X)$ if and only if there is a distinguished triangle in $\T$ $$\begin{CD} Z[-1] @>>> K_1 @>>> X_0 @>>> Z \, \end{CD}$$ with $K_1\in\varepsilon^\wedge_{n-1}(\X)$ and $X_0\in\X.$ On the other hand, by induction, we have that $K_1\in\varepsilon^\wedge_{n-1}(\X)$ if and only if there is a family $\{K_j[-1]\to K_{j+1}\to X_j\to K_j\}_{j=1}^{n-1}$ of distinguished triangles in $\T$ with $X_j\in \X$ and $K_n\in\X;$ proving (a). \ (b) By definition, we have that $\varepsilon^\wedge_{n}(\X)=\X*\varepsilon^\wedge_{n-1}(\X)[1].$ So, by induction, it follows that $\varepsilon^\wedge_{n}(\X)=\X*(*_{i=0}^{n-1}\X[i])[1]=*_{i=0}^n\,\X[i].$ \ (c) Assume that $0\in \X.$ By (b), we know that $\varepsilon^\wedge_{n}(\X)=\varepsilon^\wedge_{n-1}(\X)*\X[n];$ and since $0\in\varepsilon^\wedge_{n-1}(\X),$ it follows from \ref{xx=x} (a) that $\X[n]\subseteq\varepsilon^\wedge_{n}(\X).$ Similarly, from the equalities $\varepsilon^\wedge_{n+1}(\X)=\varepsilon^\wedge_n(\X)*\X[n+1]$ and $\varepsilon^\wedge_{n+1}(\X)=\X*\varepsilon^\wedge_n(\X)[1],$ and the facts that $0\in \X[n+1]$ and $0\in\X,$ we get the other inclusions from \ref{xx=x} (a). \end{dem} \vspace{.2cm} Following \cite{AB} and \cite{B}, we introduce the notion of $\X$-resolution (respectively, coresolution) dimension of any class $\Y$ of objects of $\T.$ \begin{defi} Let $\X$ be a class of objects in $\T.$ \begin{enumerate} \item[(a)] $\X^\wedge:=\cup_{n\geq 0}\;\varepsilon^\wedge_n(\X)$ and $\X^\vee:=\cup_{n\geq 0}\;\varepsilon^\vee_n(\X).$ \item[(b)] For any $M\in\T,$ the $\X$-{\bf{resolution dimension}} of $M$ is $$\resdim_{\X}(M):=\mini\,\{n\in\mathbb{N} \;:\; M\in\varepsilon^\wedge_{n}(\X) \}.$$ Dually, the $\X$-{\bf{coresolution dimension}} of $M$ is $$\coresdim_{\X}(M):=\mini\,\{n\in\mathbb{N}:\; M\in\varepsilon^\vee_{n}(\X) \}.$$ \item[(c)] For any subclass $\Y$ of $\T$, we set $\resdim_{\X}(\Y):=\sup\,\{\resdim_{\X}(M)\,:\,M\in\Y\}.$ Similarly, we also have $\coresdim_{\X}(\Y).$ \end{enumerate} \end{defi} \begin{pro}\label{stardim} Let $\X$ and $\Y$ be classes of objects in $\T,$ $0\in\Y$ and $n\in\N.$ Then, the following statements hold. \begin{itemize} \item[(a)] $\resdim_\Y(\X)\leq n$ if and only if $\X\subseteq\varepsilon^\wedge_{n}(\Y)= *_{i=0}^n\,\Y[i].$ \item[(b)] If $\X$ is closed under extensions, then $\X* \X^{\wedge} \subseteq \X^{\wedge}.$ \item[(c)] If $\X$ is cosuspended, then $\varepsilon^\wedge_{n}(\X)=\X[n].$ \end{itemize} \end{pro} \begin{dem} (a) It follows by definition and \ref{rkepsilon} (b),(c). \ (b) It follows from \ref{rkepsilon} (b) since $\X*\X\subseteq\X.$ \ (c) Let $\X$ be cosuspended. since $0\in\X$ (see \ref{xx=x} (b)), we get from \ref{rkepsilon} (b), (c) that $\X[n]\subseteq\varepsilon^\wedge_{n}(\X)= *_{i=0}^n\,\X[i].$ On the other hand, using that $\X*\X\subseteq\X$ and $\X[-1]\subseteq \X,$ we conclude that $*_{i=0}^n\,\X[i]=(*_{i=0}^n\,\X[i-n])[n]\subseteq(*_{i=0}^n\,\X)[n]\subseteq\X[n].$ \end{dem} \vspace{.2cm} The following result will be useful in this paper. The item (a) already appeared in \cite{B}. We also recall that $\Delta_\T(\X)$ (respectively, $\overline{\Delta}_\T(\X))$ stands for the smallest triangulated (respectively, smallest thick) subcategory of $\T$ containing the class of objects $\X.$ \begin{teo}\label{resBuan} For any cosuspended subcategory $\X$ of $\T$ and any object $C\in\T,$ the following statements hold. \begin{itemize} \item[(a)] $\resdim_\X(C)\leq n$ if and only if $C\in\X[n].$ \item[(b)] $\X^\wedge=\cup_{n\geq 0}\,\X[n]=\Delta_\T(\X).$ \item[(c)] If $\X$ is closed under direct summands in $\T,$ then $\X^\wedge=\overline{\Delta}_\T(\X).$ \end{itemize} \end{teo} \begin{dem} (a) If follows from \ref{stardim} (a), (c) since $0\in\X$ (see \ref{xx=x} (b)). \ (b) From \ref{stardim} (c), we get $\X^\wedge=\cup_{n\geq 0}\,\X[n];$ and hence $\X^\wedge$ is closed under positive and negative shifts. We prove now that $\X^\wedge$ is closed under extensions. Indeed, let $X[n]\to Y\to X'[m]\to X[n][1]$ be a distinguished triangle in $\T$ with $X,X'\in\X.$ We assume that $n\leq m$ and then $X[n]=X[n-m][m]\in\X[m]$ since $n-m\leq 0$ and $\X[-1]\subseteq\X.$ Using now that $\X$ is closed under extensions, it follows that $Y\in\X[m]\subseteq \X^\wedge;$ proving that $\X^\wedge$ is closed under extensions. Hence $\X^\wedge$ is a triangulated subcategory of $\T$ and moreover it is the smallest one containing $\X$ since $\X^\wedge=\cup_{n\geq 0}\,\X[n].$ \ (c) It follows from (b). \end{dem} \begin{rk}\label{varios} $(1)$ Observe that a suspended class $\,\U$ of $\T$ is closed under cones. Indeed, if $A\rightarrow B\rightarrow C\rightarrow A[1]$ is a distinguished triangle in $\T$ with $A,B\in \U$ then $A[1],B\in \U;$ and so we get $C\in \U.$ Similarly, if $\,\U$ is cosuspended then it is closed under cocones. \ $(2)$ Let $(\Y,\omega)$ be a pair of classes of objects in $\T$ with $\omega\subseteq\Y.$ If $\Y$ is closed under cones (respectively, cocones) then $\omega^\wedge\subseteq\Y$ (respectively, $\omega^\vee\subseteq\Y$). Indeed, assume that $\Y$ is closed under cones and let $M\in\omega^\wedge.$ Thus $M\in\varepsilon^\wedge_n(\omega)$ for some $n\in\Bbb{N}.$ If $n=0$ then $M\in\omega\subseteq \Y.$ Let $n>0,$ and hence there is a distinguished triangle $M[-1] \to K\to Y\to M$ in $\T$ with $K\in\varepsilon^\wedge_{n-1}(\omega)$ and $Y\in\Y.$ By induction $K\in\Y$ and hence $M\in\Y$ since $\Y$ is closed under cones; proving that $\omega^\wedge\subseteq\Y.$ \ $(3)$ Note that $\X^\wedge\subseteq \U_{\X}$ (respectively, $\X^\vee \subseteq {}_{\X}\U$) since $\U_{\X}$ (respectively, ${}_{\X}\U$) is closed under cones (respectively, cocones) and contains $\X.$ \end{rk} Using the fact that the functor $\Hom$ is a cohomological one, we get the following description of the orthogonal categories. In particular, observe that ${}_{\X}\U^\perp$ (respectively, ${}^\perp\U_{\X}$) is a suspended (respectively, cosuspended) subcategory of $\T.$ \begin{lem}\label{susp-perp} For any class $\X$ of objects in $\T,$ we have that \begin{itemize} \item[(a)] ${}^\perp\U_{\X}=\{Z\in\T\;:\;\Hom_\T(Z,X[i])=0,\quad \forall i\geq 0, \forall X\in\X\},$ \item[(b)] ${}_{\X}\U^\perp=\{Z\in\T\;:\;\Hom_\T(X[i],Z)=0,\quad \forall i\leq 0, \forall X\in\X\}.$ \end{itemize} \end{lem} \begin{dem} It is straightforward. \end{dem} \begin{lem}\label{restri} Let $\Y$ and $\X$ be classes of objects in $\T,$ $n\geq 1$ and $Z\in \T.$ The following statements hold. \begin{itemize} \item[(a)] The object $Z$ belongs to $\Y*\Y[1]*\cdots*\Y[n-1]* \X[n]$ if and only if there exists a family $\{ K_i\rightarrow Y_i\rightarrow K_{i+1}\rightarrow K_i[1]\;:\; Y_i\in\Y\}_{i=0}^{n-1}$ of distinguished triangles in $\T$ with $K_0\in \X$ and $Z=K_n.$ \vspace{.2cm} \item[(b)] The object $Z$ belongs to $\X[-n] *\Y[-n+1]*\cdots*\Y[-1]*\Y$ if and only if there exists a family $\{K_{i+1}\rightarrow Y_i\rightarrow K_{i}\rightarrow K_{i+1}[1]\;:\; Y_i\in\Y\}_{i=0}^{n-1}$ of distinguished triangles in $\T$ with $K_0\in \X$ and $Z=K_n.$ \end{itemize} \end{lem} \begin{dem} (a) We proceed by induction on $n.$ If $n=1$ then (a) is trivial. Suppose that $n\geq 2$ and consider the class $$\Z_{n-1}:=\Y*\Y[1]*\cdots*\Y[n-2] * \X[n-1].$$ It is clear that $\Y*\Y[1]*\cdots*\Y[n-1]* \X[n]=\Y*\Z_{n-1}[1];$ and then, we have that $Z\in\Y*\Y[1]*\cdots*\Y[n-1]* \X[n]$ if and only if there is a distinguished triangle $$\begin{CD} K @>>> Y @>>> Z @>>> K[1]\end{CD}$$ in $\T$ with $Y\in\Y$ and $K\in\Z_{n-1}.$ On the other hand, by induction, we have that $K\in\Z_{n-1}$ if and only if there is a family $\{ K_i\rightarrow Y_i\rightarrow K_{i+1}\rightarrow K_i[1]\;:\; Y_i\in\Y\}_{i=0}^{n-2}$ of distinguished triangles in $\T$ with $K_0\in \X$ and $K=K_{n-1}.$ So the result follows by adding the triangle above to the preceding family of triangles. \ (b) It is similar to (a). \end{dem} \section{Relative homological dimensions} In this section, we introduce the $\X$-projective (respectively, injective) dimension of objects in $\T.$ Moreover, we establish a result that relates this relative projective dimension with the resolution dimension as can be seen in Theorem \ref{pdyresdim}. \begin{defi} Let $\X$ be a class of objects in $\T$ and $M$ an object in $\T.$ \begin{enumerate} \item[(a)] The {\bf{$\X$-projective dimension}} of $M$ is $$\pd_{\X}(M):=\mini\,\{n\in\mathbb{N}\; :\; \Hom_\T(M[-i],-)\mid_{\X}=0, \quad\forall i>n\}.$$ \item[(b)] The {\bf{$\X$-injective dimension}} of $M$ is $$\id_{\X}(M):=\mini\,\{n\in\mathbb{N}\; :\; \Hom_\T(-,M[i])\mid_{\X}=0, \quad\forall i>n\}.$$ \item[(c)] For any class $\Y$ of objects in $\T,$ we set $$\pd_{\X}(\Y):=\sup\,\{\pd_{\X}(C)\,:\,C\in\Y\}\quad\text{and}\quad\id_{\X}(\Y):=\sup\,\{\id_{\X}(C)\,:\,C\in\Y\}.$$ \end{enumerate} \end{defi} \begin{lem}\label{compdim} Let $\X$ be a class of objects in $\T.$ Then, the following statements hold. \begin{itemize} \item[(a)] For any $M\in\T$ and $n\in\Bbb{N},$ we have that \begin{itemize} \item[(a1)] $\pd_{\X}(M)\leq n$ if and only if $M\in{}^\perp\U_{\X}[n+1];$ \item[(a2)] $\id_{\X}(M)\leq n$ if and only if $M\in{}_{\X}\U^\perp[-n-1].$ \end{itemize} \item[(b)] $\pd_{\Y}(\X)=\id_{\X}(\Y)$ for any class $\Y$ of objects in $\T.$ \end{itemize} \end{lem} \begin{dem} (a) follows from \ref{susp-perp}, and (b) is straightforward. \end{dem} \begin{pro}\label{pdXresdim} Let $\X$ be a class of objects in $\T$ and $M\in\T.$ Then $$\pd_{\X}(M)=\resdim_{{}^\perp\U_{\X}[1]}(M)\;\text{ and }\;\id_{\X}(M)=\coresdim_{{}_{\X}\U^\perp[-1]}(M).$$ \end{pro} \begin{dem} Since ${}^\perp\U_{\X}$ is cosuspended (see \ref{susp-perp} (a)), the first equality follows from \ref{compdim} (a1) and \ref{resBuan} (a). The second equality can be proven similarly. \end{dem} \vspace{.2cm} Now, we prove the following relationship between the relative projective dimension and the resolution dimension. \begin{teo}\label{pdyresdim} Let $\X$ and $\Y$ be classes of objects in $\T.$ Then, the following statements hold. \begin{enumerate} \item[(a)] $\pd_{\X}(L)\leq \pd_{\X}(\Y) + \resdim_{\Y}(L),$ $\quad \forall L\in\T.$ \vspace{.2cm} \item[(b)] If $\Y\subseteq \U_{\X}\cap{}^{\perp}\U_{\X}[1]$ and $\Y$ is closed under direct summands in $\T,$ then $$\pd_{\X}(L)=\resdim_{\Y}(L), \quad \forall L\in\Y^\wedge.$$ \end{enumerate} \end{teo} \begin{dem} (a) Let $d:=\resdim_{\Y}(L)$ and $\alpha:=\pd_{\X}(\Y).$ We may assume that $d$ and $\alpha$ are finite. We prove (a) by induction on $d.$ If $d=0,$ it follows that $L\in\Y$ and then (a) holds in this case.\\ Assume that $d\geq 1.$ So we have a distinguished triangle $K\rightarrow Y\rightarrow L\rightarrow K[1]$ in $\T$ with $Y\in\Y$ and $K\in\varepsilon^\wedge_{d-1}(\Y).$ Applying the cohomological functor $\Hom_\T(-,M[j]),$ with $M\in\X,$ to the above triangle, we get and exact sequence of abelian groups $$\Hom_\T(K[1],M[j])\rightarrow \Hom_\T(L,M[j])\rightarrow \Hom_\T(Y,M[j]).$$ By induction, we know that $\pd_\X(K)\leq \alpha+d-1.$ Therefore $\Hom_\T(L,M[j])=0$ for $j>\alpha+d$ and so $\pd_\X(L)\leq \alpha+d.$ \ (b) Let $\Y\subseteq \U_{\X}\cap ^{\perp}\U_{\X}[1]$ and $\Y$ be closed under direct summands in $\T.$ Consider $L\in\Y^\wedge$ and let $d:=\resdim_{\Y}(L).$ By \ref{compdim} we have that $\pd_{\X}(\Y)=0$ and then $\pd_{\X}(L)\leq d$ (see (a)). We prove, by induction on $d,$ that the equality given in (b) holds. For $d=0$ it is clear.\\ Suppose that $d=1.$ Then, there is a distinguished triangle $$\eta:\quad Y_1\rightarrow Y_0\rightarrow L\stackrel{f}{\rightarrow} Y_1[1]\text{ in $\T$ with $Y_i\in\Y.$}$$ If $\pd_{\X}(L)=0$ then $L\in{}^{\perp}\U_{\X}[1]$ (see \ref{compdim}). Hence $f=0$ since $\Y\subseteq \U_{\X};$ and therefore $\eta$ splits giving us that $L\in\Y,$ which is a contradiction since $d=1.$ So $\pd_{\X}(L)>0$ proving (b) for $d=1.$\\ Assume now that $d\geq 2.$ Thus we have a distinguished triangle $K\rightarrow Y\rightarrow L\rightarrow K[1]$ in $\T$ with $Y\in\Y,$ $K\in\varepsilon^\wedge_{d-1}(\Y)$ and $\pd_\X(K)=d-1$ (by inductive hypothesis). Since $\pd_{\X}(L)\leq d,$ it is enough to see $\pd_{\X}(L)> d-1.$ So, in case $\pd_{\X}(L)\leq d-1,$ we apply the cohomological functor $\Hom_\T(-,X[d]),$ with $X\in\X,$ to the triangle $L\rightarrow K[1]\rightarrow Y[1]\rightarrow L[1].$ Then we get the following exact sequence of abelian groups $$\Hom_\T(Y[1],X[d])\rightarrow \Hom_\T(K[1],X[d])\rightarrow \Hom_\T(L,X[d]).$$ \vspace{.2cm} \noindent Therefore $\Hom_\T(K[1],X[d])=0$ contradicting that $\pd_\X(K)=d-1.$ This means that $\pd_{\X}(L)> d-1;$ proving (b). \end{dem} \begin{rk} Note that if $Y\neq 0$ and $Y\in \U_{\X}\cap{}^{\perp}\U_{\X}[1],$ then $Y[j]\notin \U_{\X}\cap{}^{\perp}\U_{\X}[1],$ $\forall j>0.$ \end{rk} The following technical result will be used in Section 4. \begin{lem}\label{pdim} Let $\X$, $\Y$ and $\Z$ be classes of objects in $\T.$ Then, the following statements hold. \begin{enumerate} \item[(a)] $\pd_{\Y}(\X^\vee)=\pd_{\Y}(\X).$ \vspace{.2cm} \item[(b)] If $\X\subseteq \Z\subseteq \X^\vee$ then $\pd_{\Y}(\Z)=\pd_{\Y}(\X).$ \end{enumerate} \end{lem} \begin{dem} To prove (a), it is enough to see that $\pd_{\Y}\,(\X^\vee) \leq\pd_{\Y}\,(\X).$ Let $M\in\X^\vee$. We prove by induction on $d:=\coresdim_{\X}\,(M)$ that $\pd_{\Y}\,(M) \leq\pd_{\Y}\,(\X)$. We may assume that $\alpha:=\pd_{\Y}\,(\X)<\infty.$ If $d=0$ then we have that $M\in\X$ and there is nothing to prove. \ Let $d\geq 1.$ Then we have a distinguished triangle $M\rightarrow X\rightarrow K\rightarrow M[1]$ in $\T$ with $X\in\X,$ $K\in\varepsilon^\vee_{d-1}(\X)$ and $\pd_{\Y}\,(K)\leq \alpha$ (by inductive hypothesis). Applying the cohomological functor $\Hom_\T(-,Y[i])$, with $Y\in\Y$, we get the exact sequence of abelian groups $$\Hom_\T(X,Y[i])\rightarrow \Hom_\T(M,Y[i])\rightarrow\Hom_\T(K,Y[i+1]).$$ Therefore $\Hom_\T(M,Y[i])=0$ for $i>\alpha$ since $\pd_{\Y}\,(K)\leq \alpha.$ So we get that $\pd_{\Y}\,(\X^\vee) \leq\pd_{\Y}\,(\X).$ \ Finally, it is easy to see that (b) is a consequence of (a). \end{dem} \vspace{.2cm} The following two lemmas resembles the so called ``shifting argument'' that is usually used for syzygies and cosyzygies in the $\Ext^n$ functor. \begin{lem}\label{corricores} Let $\X$ and $\Y$ be classes of objects in $\T$ such that $\id_{\X}(\Y)=0.$ Then, for any $X\in\X,$ $k>0$ and $K_n\in\Y*\Y[1]*\cdots*\Y[n-1]*K_0[n],$ there is an isomorphism of abelian groups $$\Hom_\T(X,K_0[k+n])\simeq\Hom_\T(X,K_n[k]).$$ \end{lem} \begin{dem} Let $X\in\X,$ $k>0$ and $K_n\in\Y*\Y[1]*\cdots*\Y[n-1]*K_0[n].$ By \ref{restri} (a), we have distinguished triangles $\eta_i:\quad K_i\rightarrow Y_i\rightarrow K_{i+1}\rightarrow K_i[1]$ with $Y_i\in\Y,\;0\leq i\leq n-1.$ Applying the functor $\Hom_\T(X[-k],-)$ to $\eta_i,$ we get the exact sequence of abelian groups $$(X[-k],Y_i)\rightarrow(X[-k],K_{i+1})\rightarrow(X[-k],K_i[1])\rightarrow(X[-k],Y_i[1]),$$ where $(-,-):=\Hom_\T(-,-)$ for simplicity. Since $\id_{\X}(\Y)=0,$ it follows that $\Hom_\T(X[-k],K_{i+1})\simeq\Hom_\T(X[-k],K_i[1]).$ Therefore, by the preceding isomorphism, we have\\ $\qquad\Hom_\T(X,K_{n}[k])\simeq\Hom_\T(X,K_{n-1}[k+1])\simeq\cdots\simeq\Hom_\T(X,K_0[k+n]).$ \end{dem} \begin{lem}\label{corrires} Let $\X$ and $\Y$ be classes of objects in $\T$ such that $\pd_{\X}(\Y)=0.$ Then, for any $X\in\X,$ $k>0$ and $K_n\in K_0[-n]*\Y[-n+1]*\cdots*\Y[-1]*\Y,$ there is an isomorphism of abelian groups $$\Hom_\T(K_0,X[k+n])\simeq\Hom_T(K_n,X[k]).$$ \end{lem} \begin{dem} The proof is similar to the one given in \ref{corricores} by using \ref{restri} (b). \end{dem} \section{relative weak-cogenerators and relative injectives} In this section, we focus our attention on pairs $(\X,\omega)$ of classes of objects in $\T.$ We study the relationship between weak-cogenerators in $\X$ and coresolutions. Also, we give a characterization of some special subcategories of $\T.$ \begin{defi} Let $(\X,\omega)$ be a pair of classes of objects in $\T.$ We say that \begin{itemize} \item[(a)] $\omega$ is a {\bf{weak-cogenerator}} in $\X,$ if $\omega\subseteq\X\subseteq\X[-1]*\omega;$ \item[(b)] $\omega$ is a {\bf{weak-generator}} in $\X,$ if $\omega\subseteq\X\subseteq \omega * \X[1];$ \item[(c)] $\omega$ is {\bf{$\X$-injective}} if $\id_{\X}(\omega)=0;$ and dually, $\omega$ is {\bf{$\X$-projective}} if $\pd_{\X}(\omega)=0.$ \end{itemize} \end{defi} The following result say us that an $\X$-injective weak-cogenerator, closed under direct summands, is unique (in case it exists). \begin{pro}\label{coinyec0} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ such that $\omega$ is $\X$-injective. Then, the following statements hold. \begin{enumerate} \item[(a)] $\omega^\wedge$ is $\X$-injective. \item[(b)] If $\omega$ is a weak-cogenerator in $\X,$ and $\omega$ is closed under direct summands in $\T,$ then $$\omega=\X\cap{}_{\X}\U^{\perp}[-1]=\X\cap\omega^\wedge.$$ \end{enumerate} \end{pro} \begin{dem} (a) It follows from the dual result of \ref{pdim} (a). \ (b) Let $\omega\subseteq\X\subseteq\X[-1]*\omega$ and $\omega$ be closed under direct summands in $\T.$ \ We start by proving the first equality. Let $X\in \X\cap{}_{\X}\U^{\perp}[-1]$. Since $\X\subseteq\X[-1]*\omega,$ there is a distinguished triangle $$\eta:\quad X\rightarrow W\rightarrow X'\stackrel{f}{\rightarrow} X[1]\text{ in $\T$ with $X'\in\X$ and $W\in\omega.$}$$ Moreover $X\in{}_{\X}\U^{\perp}[-1]$ implies that $\Hom_\T(-,X[1])|_\X=0$ (see \ref{susp-perp} (b)). Hence $\eta$ splits and so $X\in\omega;$ proving that $\X\cap{}_{\X}\U^{\perp}[-1]\subseteq\omega.$ The other inclusion follows from \ref{compdim} (a2) since $\omega\subseteq\X$ and $\id_{\X}(\omega)=0.$\\ On the other hand, it is easy to see that $\omega\subseteq \X\cap\omega^\wedge$ and since $\id_{\X}(\omega^\wedge)=0,$ it follows from \ref{compdim} (a2) that $\X\cap\omega^\wedge\subseteq \X\cap{}_{\X}\U^{\perp}[-1];$ proving (b). \end{dem} \begin{pro}\label{coresIdfin} Let $(\X,\omega)$ be a pair of classes of objects in $\T,$ and $\omega$ be closed under direct summands in $\T.$ If $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ then $$\X\cap\omega^\vee=\{X\in\X\,:\,\id_{\X}(X)<\infty\}.$$ \end{pro} \begin{dem} Let $M\in\X\cap\omega^\vee.$ We assert that $\id_\X(M)\leq d<\infty$ where $d:=\coresdim_\omega(M).$ Indeed, from \ref{rkepsilon} (a), dual version, and \ref{restri} (a), there is some $W_d\in\omega*\omega[1]*\cdots*\omega[d-1]*M[d]$ with $W_d\in\omega.$ So, by \ref{corricores} we get an isomorphism $\Hom_\T(X,M[k+d])\simeq\Hom_\T(X,W_d[k])$ for any $k>0$ and $X\in\X;$ and using that $\id_{\X}(\omega)=0,$ it follows that $\Hom_\T(X,M[k+d])=0$ for any $k>0,$ proving that $\id_\X(M)\leq d.$ \ Let $N\in\X$ be such that $n:=\id_\X(N)<\infty.$ Using that $\X\subseteq\X[-1]*\omega,$ we can construct a family $\{K_i\rightarrow W_i\rightarrow K_{i+1}\rightarrow K_i[1]\;:\; W_i\in\omega,\;0\leq i\leq n-1\}$ of distinguished triangles in $\T$ where $K_0:=N$ and $K_i\in\X,$ $\forall i\;0\leq i\leq n.$ Thus, by \ref{restri} (a), it follows that $K_n\in\omega*\omega[1]*\cdots*\omega[n-1]*N[n];$ and so by \ref{corricores} we get that $\Hom_T(X,K_n[k])\simeq \Hom_\T(X,N[k+n]),$ $\forall X\in\X,$ $\forall k>0.$ But $ \Hom_\T(X,N[k+n])=0,$ $\forall X\in\X,$ $\forall k>0$ because $\id_{\X}(N)=n.$ Therefore $\id_\X(K_n)=0$ and then $K_n\in\omega$ (see \ref{compdim} and \ref{coinyec0} (b)); proving that $N\in\X\cap\omega^\vee.$ \end{dem} \ Now, we are in condition to prove the following result. In the statement, we use the notions of precovering and preenveloping classes (see Section 1). \begin{teo}\label{specialtrian} Let $(\X,\omega)$ be a pair of classes of objects in $\T,$ $\X$ be closed under extensions and $\omega$ be a weak-cogenerator in $\X.$ Then, the following statements hold. \begin{enumerate} \item[(a)] For all $C\in\X^\wedge$ there exist two distinguished triangles in $\T:$\\ $\begin{CD} C[-1] @>>> Y_C @>>> X_C @>\varphi_C>> C \text{ with }Y_C\in\omega^\wedge\text{ and } X_C\in\X,\end{CD}$ \noindent $\begin{CD} C @>\varphi^C>> Y^C @>>> X^C @>>> C[1] \text{ with }Y^C\in\omega^\wedge\text{ and } X^C\in\X. \end{CD}$ \item[(b)] If $\omega$ is $\X$-injective, then \vspace{.2cm} \begin{itemize} \item[(b1)] $Y_C[1]\in\X^\perp$ and $\varphi_C$ is an $\X$-precover of $C,$ \vspace{.2cm} \item[(b2)] $X^C[-1]\in {}^\perp(\omega^\wedge)$ and $\varphi^C$ is a $\omega^\wedge$-preenvelope of $C.$ \end{itemize} \end{enumerate} \end{teo} \begin{dem} (a) Let $C\in\X^\wedge.$ We prove the existence of the triangles in (a) by induction on $n:=\resdim_\X(C).$ If $n=0,$ we have that $C\in\X$ and then we can consider $C[-1] \rightarrow 0\rightarrow C\stackrel{1_C}{\rightarrow} C$ as the first triangle; the second one can be obtained from the fact that $\X\subseteq\X[-1]*\omega.$ \ Assume that $n>0.$ Then, we have a distinguished triangle $C[-1]\rightarrow K_1\rightarrow X_0\rightarrow C$ in $\T$ with $X_0\in\X$ and $\resdim_\X(K_1)=n-1.$ Hence, by induction, there is a distinguished triangle $ K_1 \to Y^{K_1} \to X^{K_1} \to K_1[1] $ in $\T$ with $Y^{K_1}\in\omega^\wedge$ and $X^{K_1}\in\X.$ By the co-base change procedure applied to the above triangles, there exists a commutative diagram $$\begin{CD} @. X^{K_1}[-1] @= X^{K_1}[-1] @.\\ @. @VVV @VVV @.\\ C[-1] @>>> K_1 @>>> X_0 @>>> C\\ @| @VVV @VVV @|\\ C[-1] @>>> Y^{K_1} @>>> U @>>> C\\ @. @VVV @VVV @.\\ @. X^{K_1} @= X^{K_1} @.\\ {} \end{CD}$$ where the rows and columns are distinguished triangles in $\T.$ Since $X_0,X^{K_1}\in\X$ it follows that $U\in\X.$ By taking $X_C:=U$ and $Y_C:=Y^{K_1},$ we get the first triangle in (a). On the other hand, since $U\in\X$ and $\X\subseteq\X[-1]*\omega,$ there exists a distinguished triangle $X^C[-1]\rightarrow U\rightarrow W\rightarrow X^C$ in $\T$ with $X^C\in \X$ and $W\in\omega.$ Again, by the co-base change procedure, there exists a commutative diagram $$\begin{CD} @. Y^{K_1} @= Y^{K_1} @.\\ @. @VVV @VVV @.\\ X^C[-1] @>>> U @>>> W @>>> X^C\\ @| @VVV @VVV @|\\ X^C[-1] @>>> C @>>> Y^C @>>> X^C\\ @. @VVV @VVV @.\\ @. Y^{K_1}[1] @= Y^{K_1}[1] @.\\ {} \end{CD}$$ where the rows and columns are distinguished triangles in $\T.$ By the second column, in the diagram above, it follows that $Y^C\in\omega^\wedge.$ Hence the second row in the preceding diagram is the desired triangle. \ (b2) Consider the triangle $X^C[-1]\stackrel{g}{\rightarrow} C\stackrel{\varphi^C}{\to} Y^C\to X^C$ with $Y^C\in\omega^\wedge$ and $X^C\in\X.$ Since $\id_{\X}(\omega)=0$ we have by \ref{coinyec0} that $\id_{\X}(\omega^\wedge)=0.$ Thus $\Hom_\T(X[-1],-)|_{\omega^\wedge}=0$ for any $X\in\X;$ and so $X^C[-1]\in {}^\perp(\omega^\wedge).$ Let $f:C\to Y$ be a morphism in $\T$ with $Y\in\omega^\wedge.$ Since $\Hom_\T(X^C[-1],Y)=0,$ we have that $fg=0$ and hence $f$ factors through $\varphi^C;$ proving that $\varphi^C$ is a $\omega^\wedge$-preenvelope of $C.$ \ (b1) It is similar to the proof of (b2). \end{dem} \vspace{.2cm} The following result provides a characterization of the category $\X^\wedge,$ and will be applied in \cite{MSSS2} to deal with co-$t$-structures. \begin{cor}\label{igualresdim} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ such that $\X$ is closed under extensions and $\omega$ is a weak-cogenerator in $\X.$ Then, the following statements hold. \begin{enumerate} \item[(a)] If $0\in \omega$ then $\X^\wedge=\X*\omega^\wedge=\X*\omega^\wedge [1].$ \vspace{.2cm} \item[(b)] If $\X[-1]\subseteq\X $ then $\X^\wedge=\X*\omega^\wedge=\X*\omega^\wedge [1]=\X[-1]*\omega^\wedge .$ \end{enumerate} \end{cor} \begin{dem} We assert that $\X*\omega^\wedge \subseteq \X^\wedge.$ Indeed, since $\omega\subseteq\X$ it follows from \ref{rkepsilon} (b) that $\varepsilon^\wedge_n(\omega)\subseteq \varepsilon^\wedge_n(\X),$ giving us that $\omega^\wedge \subseteq \X^\wedge.$ Hence $\X*\omega^\wedge \subseteq \X*\X^\wedge$ and then $\X*\omega^\wedge \subseteq \X^\wedge $ by \ref{stardim} (b). \ (a) Let $0\in \omega.$ By \ref{specialtrian} (a) we have that $\X^\wedge\subseteq\X*\omega^\wedge[1],$ and therefore, by \ref{rkepsilon} (d) we get $\X^\wedge\subseteq \X*\omega^\wedge[1] \subseteq \X*\omega^\wedge.$ But $\X*\omega^\wedge \subseteq \X*\X^\wedge \subseteq \X^\wedge $ by \ref{stardim} (b), and then $\X^\wedge=\X*\omega^\wedge=\X*\omega^\wedge [1].$ \ (b) Let $\X[-1]\subseteq\X .$ By \ref{specialtrian} (a) and the assertion above, we have $\X^\wedge\subseteq \X [-1]*\omega^\wedge \subseteq \X*\omega^\wedge \subseteq \X^\wedge.$ On the other hand, from \ref{specialtrian} (a), it follows that $\X^\wedge\subseteq\X*\omega^\wedge[1].$ So, to prove (b), it is enough to see that $\X*\omega^\wedge[1]\subseteq \X^\wedge.$ Let $C\in\X*\omega^\wedge[1].$ Then there is a distinguished triangle $Y \to X \to C \to Y[1]$ in $\T$ with $X\in\X$ and $Y\in\omega^\wedge.$ Hence it follows that $C\in\X^\wedge$ since $\omega^\wedge \subseteq \X^\wedge;$ proving (b). \end{dem} \vspace{.2cm} We are now in position to prove that if $\omega$ is an $\X$-injective weak-cogenerator in a suitable class $\X,$ then the $\omega^\wedge$-projective dimension coincides with the $\X$-resolution dimension for every object of the thick subcategory of $\T$ generated by $\X.$ \begin{teo}\label{pd-resd} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ which are closed under direct summands in $\T.$ If $\X$ is closed under extensions and $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ then $$\pd_{\omega^\wedge}(C)=\pd_\omega(C)=\resdim_{\X}(C),\quad\forall C\in\X^\wedge.$$ \end{teo} \begin{dem} Let $C\in\X^\wedge.$ By \ref{compdim} (b) and the dual of \ref{pdim} (a), it follows that $\pd_\omega(C)=\id_{\{C\}}(\omega)=\id_{\{C\}}(\omega^\wedge)=\pd_{\omega^\wedge}(C).$ To prove the last equality, we proceed by induction on $n:=\resdim_\X(C).$ To start with, we have $\pd_\omega(\X)=\id_\X(\omega)=0.$ If $n=0$ then $C\in\X$ and so $\pd_\omega(C)=0=\resdim_{\X}(C).$ \ Let $n=1.$ Then, we have a distinguished triangle $X_1\to X_0\to C\to X_1[1]$ in $\T$ with $X_i\in\X.$ By \ref{specialtrian} (a), there is a distinguished triangle $Y_C \to X_C \stackrel{\varphi_C}{\to} C \to Y_C[1]$ in $\T$ with $Y_C\in\omega^\wedge$ and $X_C\in\X.$ By the base change procedure, there exists a commutative diagram\\ $$\begin{CD} @. Y_C @= Y_C @.\\ @. @VVV @VVV @.\\ X_1 @>>> E @>>> X_C @>>> X_1[1]\\ @| @VVV @V\varphi_C VV @|\\ X_1 @>>> X_0 @>\alpha >> C @>>> X_1[1]\\ @. @VVV @V\beta VV @.\\ @. Y_C[1] @= Y_C[1]\,, @.\, \end{CD}$$ \noindent ${}$ \vspace{.2cm} where the rows and columns are distinguished triangles in $\T.$ Since $X_1,X_C\in\X$ it follows that $E\in\X.$ On the other hand, since $\Hom_\T(X,Y[1])=0$ for any $X\in\X$ and $Y\in\omega^\wedge$ (see \ref{coinyec0} (a)), we get that $\beta\alpha=0$ and then the triangle $Y_C\to E\to X_0\to Y_C[1]$ splits getting us that $Y_C\in\X\cap\omega^\wedge=\omega$ (see \ref{coinyec0}). Using that $\pd_\omega(\X)=0$ and \ref{pdyresdim} (a), we have that $\pd_\omega(C)\leq\resdim_\X(C)=1.$ We assert that $\pd_\omega(C)>0.$ Indeed, suppose that $\pd_\omega(C)=0;$ and then $\Hom_\T(C,W[1])=0$ for any $W\in\omega.$ Since $Y_C\in\omega$ we get that $\beta=0$ and hence the triangle $Y_C\to X_C\to C\to Y_C[1]$ splits. Therefore $C\in\X$ contradicting that $\resdim_\X(C)=1;$ proving that $\pd_\omega(C)=1=\resdim_\X(C).$ \ Let $n\geq 2.$ From \ref{pdyresdim} (a), we have that $\pd_\omega(C)\leq\resdim_\X(C)=n$ since $\pd_\omega(\X)=0.$ Then, it is enough to prove that $\Hom_\T(C[-n],-)|_\omega\neq 0.$ Consider a distinguished triangle $K_1\to X_0\to C\to K_1[1]$ in $\T$ with $X_0\in\X$ and $\resdim_\X(K_1)=n-1=\pd_\omega(K_1).$ Applying the functor $\Hom_\T(-,W[n]),$ with $W\in\omega,$ to the triangle $C\to K_1[1]\to X_0[1]\to C[1]$ we get the exact sequence of abelian groups $$\Hom_\T(X_0[1],W[n])\to\Hom_\T(K_1[1],W[n])\to\Hom_\T(C,W[n]).$$ Suppose that $\Hom_\T(C[-n],-)|_\omega=0.$ Then $\Hom_\T(K_1[1],W[n])=0$ since $\id_{\X}(\omega)=0$ and $n\geq 2;$ contradicting that $\pd_\omega(K_1)=n-1.$ \end{dem} \begin{lem}\label{relid} Let $\X$ be a class of objects in $\T$ and $A\rightarrow B\rightarrow C\rightarrow A[1]$ a distinguished triangle in $\T.$ Then \begin{enumerate} \item[(a)] $\id_{\X}(B)\leq\maxi\,\{\id_{\X}(A),\id_{\X}(C)\};$ \vspace{.2cm} \item[(b)] $\id_{\X}(A)\leq\maxi\,\{\id_{\X}(B),\id_{\X}(C)+ 1 \};$ \vspace{.2cm} \item[(c)] $\id_{\X}(C)\leq\maxi\,\{\id_{\X}(B),\id_{\X}(A)-1 \}.$ \end{enumerate} \end{lem} \begin{dem} It is straightforward. \end{dem} \vspace{.2cm} The following result gives a relationship between the relative injective dimensions, attached to the pair $(\X,\omega),$ and the $\omega$-coresolution dimension. Such a result will be applied in \cite{MSSS2} to deal with co-$t$-structures. Observe that \ref{idXcores} is not the dual version of \ref{pd-resd}. \begin{pro}\label{idXcores} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ such that $\omega\subseteq{}_\X\U.$ If $\omega$ is closed under direct summands and $\X$-injective, then $$\id_\omega(C)=\id_\X(C)=\coresdim_\omega(C),\quad\forall\,C\in{}_\X\U\cap\omega^\vee.$$ \end{pro} \begin{dem} Assume that $\omega$ is closed under direct summands and $\id_{\X}(\omega)=0.$ Let $C\in{}_\X\U\cap\omega^\vee$ and $n:=\coresdim_\omega(C).$ By the dual of \ref{pdyresdim} (b), it follows $$(*)\quad\alpha:=\id_\omega(C)\leq\id_{\X}(C)=\coresdim_\omega(C)=n.$$ Moreover, since $C\in\omega^\vee$ there is a distinguished triangle $(\eta)\;:\;C\rightarrow W_0\rightarrow K_1\rightarrow C[1]$ in $\T$ with $W_0\in\omega$ and $\coresdim_\omega(K_1)=n-1.$ Furthermore, from \ref{rkepsilon} (a) we get that $K_1\in{}_\X\U$ since ${}_\X\U$ is closed under cocones and $\omega\subseteq{}_\X\U.$ Now, we prove the result by induction on $\alpha.$ \ Let $\alpha=0.$ We assert that $C\in\omega$ (note that if this is true, then the result follows). We proceed by induction on $n.$ If $n=0$ it is clear that $C\in\omega.$ So we may assume that $n>0,$ and then, applying \ref{relid} to $(\eta)$ it follows that $\id_\omega(K_1)=0.$ Hence by induction we get that $K_1\in\omega,$ and so $\Hom_\T(K_1,C[1])=0$ since $\id_\omega(C)=0.$ Therefore the triangle $(\eta)$ splits and then $C\in\omega;$ proving the assertion. \ Assume that $\alpha>0.$ Applying \ref{relid} to $(\eta),$ we get that $\id_\omega(K_1)\leq\alpha-1.$ Thus, by induction, it follows that $\id_\omega(K_1)=\id_\X(K_1)=\coresdim_\omega(K_1)=n-1.$ In particular, we obtain that $n-1\leq\alpha-1$ and hence by $(*)$ the result follows. \end{dem} \vspace{.2cm} The following result provides a characterization of $\omega^\wedge$ in terms of $\X.$ From this, we get some nice relationship between other subcategories. \begin{pro}\label{resomega} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ such that $\omega$ is closed under direct summands in $\T,$ $\X$ is closed under extensions and $\omega$ is an $\X$-injective weak-cogenerator in $\X.$ Then, the following statements hold. \begin{itemize} \item[(a)] ${}_\X\U^\perp[-1]\cap\X^\wedge=\omega^\wedge.$ \vspace{.2cm} \item[(b)] If $\X[-1]\subseteq\X$ then $\U_\omega=\omega^\wedge=\X^\perp[-1]\cap\X^\wedge.$ \end{itemize} \end{pro} \begin{dem} (a) Let $C\in{}_\X\U^\perp[-1]\cap\X^\wedge.$ In particular, from \ref{specialtrian} (a), there exists a distinguished triangle $Y_C\to X_C\to C\to Y_C[1]$ in $\T$ with $Y_C\in\omega^\wedge$ and $X_C\in\X.$ We assert that $\id_\X(X_C)=0.$ Indeed, it follows from \ref{relid} (a) since $\id_\X(C)=0=\id_\X(Y_C)$ (see \ref{compdim} and \ref{coinyec0} (a)). Therefore, $X_C\in\X\cap{}_\X\U^\perp[-1]$ and by \ref{coinyec0} (b), we get that $X_C\in\omega$ proving that $C\in\omega^\wedge.$ On the other hand, since $\id_\X(\omega^\wedge)=0,$ we have from \ref{compdim} that $\omega^\wedge\subseteq{}_\X\U^\perp[-1]\cap\X^\wedge.$ \ (b) Assume that $\X[-1]\subseteq\X.$ Hence, by \ref{xx=x} (b), we have that $\X$ is a cosuspended subcategory of $\T.$ Therefore, from (a), it follows that $\omega^\wedge=\X^\perp[-1]\cap\X^\wedge.$ Furthermore, since $\X^\perp[-1]$ is suspended and $\X^\wedge$ is triangulated (see \ref{resBuan}), we conclude that $\omega^\wedge$ is a suspended subcategory of $\T;$ and so $\U_\omega\subseteq\omega^\wedge.$ Finally, the equality $\U_\omega=\omega^\wedge$ follows from \ref{varios} (3). \end{dem} \begin{teo}\label{thicksub} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ which are closed under direct summands, $\X$ be cosuspended and $\omega$ be an $\X$-injective weak-cogenerator in $\X.$ Then, $$\varepsilon^\wedge_n(\X)=\X[n]=\X^\wedge\cap{}^\perp\U_\omega[n+1]=\X^\wedge\cap{}^\perp(\omega^\wedge)[n+1],\quad\forall n\geq 0.$$ \end{teo} \begin{dem} From \ref{resBuan}, we have that $\varepsilon^\wedge_n(\X)=\X[n]$ and $\X^\wedge=\cup_{n\geq 0}\,\X[n].$ On the other hand, by \ref{compdim} and \ref{pd-resd}, it follows that $$\X^\wedge\cap{}^\perp\U_{\omega^\wedge}[n+1]=\X^\wedge\cap{}^\perp\U_\omega[n+1]=\X[n]\cap\X^\wedge=\X[n].$$ Finally, since $\omega^\wedge$ is a suspended subcategory of $\T$ (see \ref{resomega} (b)), we have that ${}^\perp\U_{\omega^\wedge}={}^\perp(\omega^\wedge);$ proving the result. \end{dem} \vspace{.2cm} The previous results can be seen under the light of the so called torsion theories, in the sense of Iyama-Yoshino. Such torsion theories have been extensively studied in relation to the cluster theory (see \cite{IY}). \begin{defi}\cite[Definition 2.2]{IY} A pair $(\X,\Y)$ of subcategories of $\T$ is called a torsion theory in $\T,$ if the following conditions hold. \begin{itemize} \item[(a)] $\X$ and $\Y$ are closed under direct summands in $\T.$ \item[(b)] $\Hom_\T(\X,\Y)=0.$ \item[(c)] $\T=\X*\Y.$ \end{itemize} \end{defi} \begin{cor}\label{torsion} Let $(\X,\omega)$ be a pair of classes of objects in $\T,$ which are closed under direct summands in $\T,$ and such that $\X$ is cosuspended and $\omega$ is an $\X$-injective weak-cogenerator in $\X.$ Then, the pair $(\X,\omega^\wedge [1])$ is a torsion theory in the thick triangulated subcategory $\X^\wedge$ of $\T.$ \end{cor} \begin{dem} Since $\X$ is cosuspended and closed under direct summands, we know from \ref{resBuan} (c) that $\X^\wedge$ is a thick triangulated subcategory of $\T.$ On the other hand, from \ref{igualresdim}, it follows that $\X^\wedge=\X*\omega^\wedge [1];$ and furthermore, $\Hom_\T(\X,\omega^\wedge [1])=0$ since $\omega$ is $\X$-injective. On the other hand, by \ref{resomega}, we get that $\omega^\wedge$ is a subcategory closed under direct summands in $\T.$ \end{dem} \vspace{.2cm} \begin{defi} For a given class $\Y$ of objects in $\T,$ we set $\Y^\sim:=(\Y^\wedge)^\vee.$ \end{defi} \begin{lem}\label{Xsim} Let $\X$ be a class of objects in $\T.$ Then, the following statements hold. \begin{itemize} \item[(a)] If $\X^\wedge$ is closed under cocones then $\omega^\sim\subseteq\X^\wedge$ for any $\omega\subseteq\X.$ \vspace{.2cm} \item[(b)] $\X^\wedge$ is closed under cocones if and only if $\X^\wedge=\X^\sim.$ \vspace{.2cm} \item[(c)] If $\X^\wedge=\X^\sim$ then $\X^\wedge[-1]\subseteq\X^\wedge.$ \end{itemize} \end{lem} \begin{dem} (a) Let $\omega\subseteq\X$ and assume that $\X^\wedge$ is closed under cocones. Hence $\omega^\wedge\subseteq\X^\wedge$ and so by \ref{varios} (2), we conclude that $\omega^\sim\subseteq\X^\wedge.$ \ (b) Assume that $\X^\wedge$ is closed under cocones. It is clear that $\X^\wedge\subseteq\X^\sim.$ On the other hand, by (a) it follows that $\X^\sim\subseteq\X^\wedge.$ \ Suppose that $\X^\wedge=\X^\sim.$ Let $A\to B\to C\to A[1]$ be a distinguished triangle in $\T$ with $B,C$ in $\X^\wedge.$ Then $A\in\X^\sim=\X^\wedge$ and so $\X^\wedge$ is closed under cocones. \ (c) Let $\X^\wedge=\X^\sim$ and consider $X\in\X^\wedge.$ Since, we have the distinguished triangle $X[-1]\to 0\to X\stackrel{1_X}{\to} X$ and $0,X \in \X^{\wedge},$ it follows from (b) that $X[-1]\in\X^\wedge;$ proving the lemma. \end{dem} \begin{cor}\label{tilde-eq} Let $(\X,\omega)$ be a pair of classes of objects in $\T.$ If $\X$ is cosuspended and $\omega\subseteq\X,$ then $\omega^\sim\subseteq\X^\wedge=\X^\sim.$ \end{cor} \begin{dem} It follows from \ref{Xsim} and the fact that $\X^\wedge$ is triangulated (see \ref{resBuan}). \end{dem} In case $\omega$ is an $\X$-injective weak-cogenerator in a cosuspended subcategory $\X$ of $\T,$ both closed under direct summands, the thick subcategory $\overline{\Delta}_\T(\omega)$ of $\T$ can be characterized as follows. \begin{teo}\label{tilde-id} Let $(\X,\omega)$ be a pair of classes of objects in $\T,$ $\X$ be cosuspended and $\omega$ be closed under direct summands in $\T.$ If $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ the following statements hold. \begin{itemize} \item[(a)] $\omega^\sim=\{C\in\X^\wedge : \id_{\X}(C)<\infty\}=\X^\wedge\cap(\X^\perp[-1])^\vee.$ \vspace{.2cm} \item[(b)] $\omega^\sim$ is the smallest triangulated subcategory of $\X^\wedge$ containing $\omega,$ that is $\omega^\sim=\Delta_{\X^\wedge}(\omega).$ \vspace{.2cm} \item[(c)] If $\X$ is closed under direct summands in $\T,$ then $$\overline{\Delta}_\T(\omega)=\omega^\sim=\overline{\Delta}_\T(\X)\cap(\X^\perp[-1])^\vee.$$ \end{itemize} \end{teo} \begin{dem} Assume that $\omega\subseteq\X\subseteq \X[-1]*\omega$ and $\id_{\X}(\omega)=0.$ Let $\Y:=\{C\in\X^\wedge : \id_{\X}(C)<\infty\}.$ We start by proving that $\omega^\sim\subseteq\Y.$ By \ref{tilde-eq}, we know that $\omega^\sim\subseteq\X^\wedge.$ On the other hand, since $\id_{\X}(\omega^\wedge)=0$ (see \ref{coinyec0}(a)), we can apply the dual of $\ref{pdyresdim} (a),$ and then $\id_\X(C)\leq\coresdim_{\omega^\wedge}(C)<\infty$ for any $C\in\omega^\sim;$ proving that $\omega^\sim\subseteq\Y.$\\ Let $C\in\Y.$ By \ref{specialtrian} (a), there is a distinguished triangle $C\to Y^C\to X^C\to C[1]$ in $\T$ with $Y^C\in\omega^\wedge$ and $X^C\in\X.$ Hence, from \ref{relid} (c) we get that $\id_{\X}(X^C)<\infty$ and then, by \ref{coresIdfin} $X^C\in\omega^\vee \subseteq \omega^\sim;$ proving that $C\in\omega^\sim.$ Hence $\Y\subseteq\omega^\sim.$ In order to get the second equality in (a), we use \ref{compdim} and the fact that $\X={}_\X\U$ to obtain $$\{C\in\X^\wedge : \id_{\X}(C)<\infty\}=\X^\wedge\cap(\cup_{n\geq 0}\,\X^\perp[-n-1]).$$ On the other hand, since $\X^\perp[-1]$ is suspended, then by the dual of \ref{resBuan}, it follows that $(\X^\perp[-1])^\vee=\cup_{n\geq 0}\,\X^\perp[-n-1]$ and also that $(\X^\perp[-1])^\vee$ is a thick subcategory of $\T.$ In particular, by \ref{resBuan}, we get (b). Finally, (c) follows from (a) and \ref{resBuan}. \end{dem} \vspace{.2cm} The following result will be applied in \cite{MSSS2} to deal with co-$t$-structures. Also, it will be applied in Section 5, to get some connection with relative Rouquier's dimension. \begin{pro}\label{idwX} Let $(\X,\omega)$ be a pair of classes of objects in $\T,$ $\X$ cosuspended and $\omega$ closed under direct summands in $\T.$ If $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ then \begin{itemize} \item[(a)] $\id_\omega(C)=\id_\X(C)<\infty,\quad\forall C\in\omega^\sim;$ \vspace{.2cm} \item[(b)] $\omega^\sim\cap{}_\omega\U^\perp[-n-1]=\omega^\sim\cap\X^\perp[-n-1],\quad \forall n\geq 0.$ \end{itemize} \end{pro} \begin{dem} (a) By \ref{resBuan} and \ref{tilde-id}, we know that $\X^\wedge$ and $\omega^\sim$ are triangulated subcategories of $\T.$ Furthermore, from \ref{tilde-eq} it follows that $\omega^\sim\subseteq\X^\wedge.$ Let $C\in\omega^\sim.$ It is enough to prove that $\id_\X(C)\leq\id_\omega(C).$ In order to do that, we will use induction on $n:=\id_\omega(C).$ \ Since $C\in\X^\wedge,$ we have from \ref{specialtrian} the existence of a distinguished triangle $(\eta)\;:\;C\to Y^C\to X^C\to C[1]$ in $\T$ with $Y^C\in\omega^\wedge\subseteq\omega^\sim$ and $X^C\in\X.$ We assert that $X^C\in\X\cap\omega^\vee.$ Indeed, using that $\omega^\sim$ is triangulated we conclude that $X^C\in\X\cap\omega^\sim$ and hence $\id_\X(X^C)$ is finite (see \ref{tilde-id} (a)). Thus $X^C\in\X\cap\omega^\vee$ by \ref{coresIdfin}; proving the assertion. \ Let $n=0.$ Then $\id_\omega(X^C)=0$ since $\id_\omega(Y^C)=0$ (see \ref{coinyec0} and \ref{relid}). On the other hand, \ref{idXcores} gives the equalities $\coresdim_\omega(X^C)=\id_\omega(X^C)=0.$ Hence $X^C\in\omega$ and since $\id_\omega(C)=0,$ it follows that $\Hom_\T(X^C,C[1])=0.$ Therefore, the triangle $(\eta)$ splits giving us that $C$ is a direct summand of $Y^C,$ and hence $\id_\X(C)\leq\id_\X(Y^C)\leq\id_\X(\omega^\wedge)=0.$ \ Assume that $n>0.$ Since $\id_\X(Y^C)=0=\id_\omega(Y^C),$ it follows from \ref{relid} that $\id_\omega(X^C)\leq n-1.$ Hence, by induction $\id_\X(X^C)\leq\id_\omega(X^C)\leq n-1.$ Therefore, applying again \ref{relid} to the triangle $(\eta),$ we get that $\id_\X(C)\leq n=\id_\omega(C);$ proving the result. \ (b) By \ref{compdim}, the item (a) and the fact that ${}_\X\U=\X$ the result follows. \end{dem} \section{Some connection with Rouquier's dimension} In this section, we introduce some kind of ``relative Rouquier's dimension''; and relate it with the Rouquier's dimension and the other relative dimensions developed in this paper. \ Let $\X$ and $\Y$ be classes of objects in a triangulated category $\T.$ Consider the smallest subcategory $\left\langle \X\right\rangle$ of $\T$ containing $\X,$ closed under shifts, finite direct sums and direct summands, that is, $\left\langle \X\right\rangle:=\add\,(\cup_{i\in\Enteros}\,\X[i]).$ Let $\X\diamondsuit\Y:=\left\langle\X*\Y\right\rangle.$ Following R. Rouquier, we inductively define $\la \X\ra_0:=0$ and $\la \X\ra_n:=\la \X\ra_{n-1}\diamondsuit\la \X\ra$ for $n\geq 1.$ The objects of $\la \X\ra_{n}$ are the direct summands of the objects obtained by taking a n-fold extension of finite direct sums of shifts of objects of $\X.$ The following dimension for triangulated categories was introduced by R. Rouquier and has been extensively studied in \cite{Ro}. \begin{defi}\cite[Definition 3.2]{Ro} Let $\T$ be a triangulated category. The dimension of $\T$ is $$\dim\,(\T):=\mini\{n\in\N\;\;|\;\text{ there exists }X\in\T\text{ such that } \la X\ra_{n+1}=\T\}.$$ \end{defi} \begin{lem}\label{conexion} Let $\X$ be a class of objects in a triangulated category $\T.$ Then, for any $n\in\N,$ the following statements hold. \begin{itemize} \item[(a)] $\la \X\ra_{n+1}=\la\varepsilon_n^\wedge(\la \X\ra)\ra=\la\varepsilon_n^\vee(\la \X\ra)\ra.$ \item[(b)] $\la \X\ra_{n}\subseteq\la \X\ra_{n+1}.$ \end{itemize} \end{lem} \begin{dem} (a) By induction over $n,$ it can be seen that $\la \X\ra_{n+1}=\la *_{i=1}^{n+1}\la \X\ra\ra.$ On the other hand, from \ref{rkepsilon} (b), we have $\varepsilon_n^\wedge(\la \X\ra)=*_{i=0}^{n}\la \X\ra[i]=*_{i=1}^{n+1}\la \X\ra$ since $\la \X\ra[i]=\la \X\ra$ for any $i\in\Enteros;$ proving that $\la \X\ra_{n+1}=\la\varepsilon_n^\wedge(\la \X\ra)\ra.$ Similarly, by the dual of \ref{rkepsilon} (b), it follows that $\la \X\ra_{n+1}=\la\varepsilon_n^\vee(\la \X\ra)\ra.$ \ (b) It follows from (a) and \ref{rkepsilon} (c), since $0\in\la \X\ra.$ \end{dem} \vspace{.2cm} \vspace{.2cm} In what follows, we introduce the relative Rouquier's dimension as follows. \begin{defi} Let $\T$ be a triangulated category, $\X$ a class of objects in $\T$ and $M\in\T.$ The $\X$-dimension of $M$ is $$\dim_\X(M):=\mini\{n\in\N\text{ such that } M\in\la \X\ra_{n+1}\}.$$ \noindent For a class $\Y$ of objects in $\T,$ we set $\dim_\X(\Y):=\sup\,\{ \dim_\X(Y): \quad\forall\,Y\in\Y\}.$ \end{defi} \begin{lem}\label{XdimY} Let $\T$ be a triangulated category, and $\X,\,\Y$ be classes of objects in $\T.$ Then, the following statements hold. \begin{itemize} \item[(a)] For any $n\in\N,$ $\dim_\X(\Y)\leq n$ if and only if $\Y\subseteq\la \X\ra_{n+1}.$ \item[(b)] If $\X\subseteq\Y$ then $\dim_\Y(M)\leq\dim_\X(M)$ $\;\forall\;M\in\T.$ \end{itemize} \end{lem} \begin{dem} (a) Let $\dim_\X(\Y)\leq n.$ Hence, for all $Y\in\Y,$ we have $m(Y):=\dim_\X(Y)\leq n.$ Therefore, from \ref{conexion} (b), it follows that $Y\in\la \X\ra_{m(Y)+1}\subseteq\la \X\ra_{n+1};$ proving that $\Y\subseteq\la \X\ra_{n+1}.$ Finally, assume that $\Y\subseteq\la \X\ra_{n+1}.$ So, it follows directly that $\dim_\X(\Y)\leq n$ \ (b) Let $\X\subseteq\Y$ and $M\in\T.$ Thus $\la\X\ra\subseteq\la\Y\ra$ and hence $*_{i=1}^n\la\X\ra\subseteq *_{i=1}^n\la\Y\ra\subseteq\la *_{i=1}^n\la\Y\ra\ra=\la\Y\ra_n.$ Therefore $\la\X\ra_n\subseteq\la\Y\ra_n$ and so we get that $\dim_\Y(M)\leq\dim_\X(M).$ \end{dem} \vspace{.2cm} Now, the relative Rouquier's dimension is related to the Rouquier's dimension as follows. \begin{pro} Let $\T$ be a triangulated category. Then, $$\dim\,(\T)=\mini\,\{\dim_X(\T): \quad\forall\; X\in\T\}.$$ \end{pro} \textbf{Proof}. From \ref{XdimY} (a), we can write down the following equalities $$\begin{array}{lcl} \dim\,(\T)&=&\mini\{n\in\N\;\;|\;\text{ there exists }X\in\T\text{ such that } \la X\ra_{n+1}=\T\}\\ &=&\mini\{n\in\N\;\;|\;\text{ there exists }X\in\T\text{ such that } \dim_X(\T)\leq n\}\\ &=&\mini\{\dim_X(\T)\quad\forall\; X\in\T\}. \mbox {\hspace{.2cm} \Qed } \end{array} $$ \vspace{.2cm} The following are some relationships between relative Rouquier's dimension and the other relative dimensions as coresolution, resolution, relative projective and relative injective. \begin{pro}\label{dimrescores} Let $\T$ be a triangulated category, $\X$ a class of objects in $\T$ and $M\in\T.$ Then, we have that $$\dim_\X(M)\leq\maxi\,\{\resdim_{\la \X\ra}(M),\coresdim_{\la \X\ra}(M)\}.$$ \end{pro} \begin{dem} From \ref{conexion} (a), we have the inclusions $\varepsilon_n^\wedge(\la \X\ra) \subseteq \la \X\ra_{n+1}$ and $\varepsilon_n^\vee(\la \X\ra)\subseteq \la \X\ra_{n+1}.$ Hence the result follows. \end{dem} \begin{pro} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ which are closed under direct summands in $\T.$ If $\X$ is closed under extensions and $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ then $$\dim_\X(C)\leq\pd_\omega(C)<\infty,\quad\forall\; C\in\X^\wedge.$$ \end{pro} \begin{dem} It follows from \ref{dimrescores} and \ref{pd-resd}, since $\X\subseteq \la \X\ra.$ \end{dem} \begin{pro} Let $(\X,\omega)$ be a pair of classes of objects in $\T$ which are closed under direct summands in $\T.$ If $\X$ is cosuspended and $\omega$ is an $\X$-injective weak-cogenerator in $\X,$ then $$\dim_\omega(C)\leq\id_\X(C)=\id_\omega(C)<\infty,\quad\forall\; C\in\omega^\vee.$$ \end{pro} \begin{dem} By \ref{idwX}, we know that $\id_\omega(C)=\id_\X(C)<\infty$ for any $C\in\omega^\sim.$ On the other hand, from \ref{coinyec0} (b), it follows that $\omega={}_\X\U \cap {}_\X\U^\perp[-1]$ since $\X={}_\X\U.$ In particular, the dual of \ref{pdyresdim} (b) holds (take $\Y:=\omega$) and hence $\id_\X(C)=\coresdim_\omega(C)$ for any $C\in\omega^\vee.$ Observe that $\omega^{\vee}\subseteq\omega^{\sim}$ since $\omega^{\sim}$ is the smallest triangulated subcategory of $\T$ containing $\omega$ (see \ref{tilde-id} (b)). Therefore $\id_\omega(C)=\id_\X(C)<\infty$ for any $C\in\omega^\vee.$ Finally, from \ref{dimrescores}, we get that $\dim_\omega(C)\leq\coresdim_\omega(C)$ since $\omega\subseteq\la\omega\ra,$ proving the result. \end{dem} \vspace{.2cm} In what follows, $\T$ is assumed to be a $k$-linear, Hom-finite triangulated and Krull-Schmidt category over a fixed field $k;$ and $\mathcal{C}$ is a Krull-Schmidt subcategory of $\T$ which is closed under direct summands in $\T.$ It is said that $\mathcal{C}$ is $n$-cluster tilting (see \cite{KR}) if it is functorially finite and $\mathcal{C}=\cap_{i=1}^{n-1}\,\mathcal{C}[-i]^\perp=\cap_{i=1}^{n-1}\,{}^\perp\mathcal{C}[i].$ The following is an example of a triangulated category $\T$ having finite $\mathcal{C}$-dimension. \begin{pro} Let $\mathcal{C}$ be a $n$-cluster tilting subcategory of $\T.$ Then, $$\dim_\mathcal{C}(\T)\leq\resdim_\mathcal{C}(\T)\leq n-1.$$ \end{pro} \begin{dem} From \cite[Theorem 3.1]{IY}, we know that $\T=*_{i=0}^{n-1}\,\mathcal{C}[i].$ Therefore, by \ref{rkepsilon} (b), we conclude that $\resdim_\mathcal{C}(\T)\leq n-1.$ Finally, by \ref{dimrescores}, it follows that $\dim_\mathcal{C}(\T)\leq\resdim_\mathcal{C}(\T)$ since $\mathcal{C}\subseteq\la\mathcal{C}\ra;$ proving the result. \end{dem} \vspace{.4cm} \textbf{Acknowledgement} The authors are very grateful to the referee for the comments, corrections and suggestions.

section \<open>Auxiliary Lemmas for Coinductive Lists\<close> text \<open>Some lemmas to allow better reasoning with coinductive lists.\<close> theory MoreCoinductiveList imports Main Coinductive.Coinductive_List begin subsection \<open>@{term "lset"}\<close> lemma lset_lnth: "x \<in> lset xs \<Longrightarrow> \<exists>n. lnth xs n = x" by (induct rule: llist.set_induct, meson lnth_0, meson lnth_Suc_LCons) lemma lset_lnth_member: "\<lbrakk> lset xs \<subseteq> A; enat n < llength xs \<rbrakk> \<Longrightarrow> lnth xs n \<in> A" using contra_subsetD[of "lset xs" A] in_lset_conv_lnth[of _ xs] by blast lemma lset_nth_member_inf: "\<lbrakk> \<not>lfinite xs; lset xs \<subseteq> A \<rbrakk> \<Longrightarrow> lnth xs n \<in> A" by (metis contra_subsetD inf_llist_lnth lset_inf_llist rangeI) lemma lset_intersect_lnth: "lset xs \<inter> A \<noteq> {} \<Longrightarrow> \<exists>n. enat n < llength xs \<and> lnth xs n \<in> A" by (metis disjoint_iff_not_equal in_lset_conv_lnth) lemma lset_ltake_Suc: assumes "\<not>lnull xs" "lnth xs 0 = x" "lset (ltake (enat n) (ltl xs)) \<subseteq> A" shows "lset (ltake (enat (Suc n)) xs) \<subseteq> insert x A" proof- have "lset (ltake (eSuc (enat n)) (LCons x (ltl xs))) \<subseteq> insert x A" using assms(3) by auto moreover from assms(1,2) have "LCons x (ltl xs) = xs" by (metis lnth_0 ltl_simps(2) not_lnull_conv) ultimately show ?thesis by (simp add: eSuc_enat) qed lemma lfinite_lset: "lfinite xs \<Longrightarrow> \<not>lnull xs \<Longrightarrow> llast xs \<in> lset xs" proof (induct rule: lfinite_induct) case (LCons xs) show ?case proof (cases) assume *: "\<not>lnull (ltl xs)" hence "llast (ltl xs) \<in> lset (ltl xs)" using LCons.hyps(3) by blast hence "llast (ltl xs) \<in> lset xs" by (simp add: in_lset_ltlD) thus ?thesis by (metis * LCons.prems lhd_LCons_ltl llast_LCons2) qed (metis LCons.prems lhd_LCons_ltl llast_LCons llist.set_sel(1)) qed simp lemma lset_subset: "\<not>(lset xs \<subseteq> A) \<Longrightarrow> \<exists>n. enat n < llength xs \<and> lnth xs n \<notin> A" by (metis in_lset_conv_lnth subsetI) subsection \<open>@{term "llength"}\<close> lemma enat_Suc_ltl: assumes "enat (Suc n) < llength xs" shows "enat n < llength (ltl xs)" proof- from assms have "eSuc (enat n) < llength xs" by (simp add: eSuc_enat) hence "enat n < epred (llength xs)" using eSuc_le_iff ileI1 by fastforce thus ?thesis by (simp add: epred_llength) qed lemma enat_ltl_Suc: "enat n < llength (ltl xs) \<Longrightarrow> enat (Suc n) < llength xs" by (metis eSuc_enat ldrop_ltl leD leI lnull_ldrop) lemma infinite_small_llength [intro]: "\<not>lfinite xs \<Longrightarrow> enat n < llength xs" using enat_iless lfinite_conv_llength_enat neq_iff by blast lemma lnull_0_llength: "\<not>lnull xs \<Longrightarrow> enat 0 < llength xs" using zero_enat_def by auto lemma Suc_llength: "enat (Suc n) < llength xs \<Longrightarrow> enat n < llength xs" using dual_order.strict_trans enat_ord_simps(2) by blast subsection \<open>@{term "ltake"}\<close> lemma ltake_lnth: "ltake n xs = ltake n ys \<Longrightarrow> enat m < n \<Longrightarrow> lnth xs m = lnth ys m" by (metis lnth_ltake) lemma lset_ltake_prefix [simp]: "n \<le> m \<Longrightarrow> lset (ltake n xs) \<subseteq> lset (ltake m xs)" by (simp add: lprefix_lsetD) lemma lset_ltake: "(\<And>m. m < n \<Longrightarrow> lnth xs m \<in> A) \<Longrightarrow> lset (ltake (enat n) xs) \<subseteq> A" proof (induct n arbitrary: xs) case 0 have "ltake (enat 0) xs = LNil" by (simp add: zero_enat_def) thus ?case by simp next case (Suc n) show ?case proof (cases) assume "xs \<noteq> LNil" then obtain x xs' where xs: "xs = LCons x xs'" by (meson neq_LNil_conv) { fix m assume "m < n" hence "Suc m < Suc n" by simp hence "lnth xs (Suc m) \<in> A" using Suc.prems by presburger hence "lnth xs' m \<in> A" using xs by simp } hence "lset (ltake (enat n) xs') \<subseteq> A" using Suc.hyps by blast moreover have "ltake (enat (Suc n)) xs = LCons x (ltake (enat n) xs')" using xs ltake_eSuc_LCons[of _ x xs'] by (metis (no_types) eSuc_enat) moreover have "x \<in> A" using Suc.prems xs by force ultimately show ?thesis by simp qed simp qed lemma llength_ltake': "enat n < llength xs \<Longrightarrow> llength (ltake (enat n) xs) = enat n" by (metis llength_ltake min.strict_order_iff) lemma llast_ltake: assumes "enat (Suc n) < llength xs" shows "llast (ltake (enat (Suc n)) xs) = lnth xs n" (is "llast ?A = _") unfolding llast_def using llength_ltake'[OF assms] by (auto simp add: lnth_ltake) lemma lset_ltake_ltl: "lset (ltake (enat n) (ltl xs)) \<subseteq> lset (ltake (enat (Suc n)) xs)" proof (cases) assume "\<not>lnull xs" then obtain v0 where "xs = LCons v0 (ltl xs)" by (metis lhd_LCons_ltl) hence "ltake (eSuc (enat n)) xs = LCons v0 (ltake (enat n) (ltl xs))" by (metis ltake_eSuc_LCons) hence "lset (ltake (enat (Suc n)) xs) = lset (LCons v0 (ltake (enat n) (ltl xs)))" by (simp add: eSuc_enat) thus ?thesis using lset_LCons[of v0 "ltake (enat n) (ltl xs)"] by blast qed (simp add: lnull_def) subsection \<open>@{term "ldropn"}\<close> lemma ltl_ldrop: "\<lbrakk> \<And>xs. P xs \<Longrightarrow> P (ltl xs); P xs \<rbrakk> \<Longrightarrow> P (ldropn n xs)" unfolding ldropn_def by (induct n) simp_all subsection \<open>@{term "lfinite"}\<close> lemma lfinite_drop_set: "lfinite xs \<Longrightarrow> \<exists>n. v \<notin> lset (ldrop n xs)" by (metis ldrop_inf lmember_code(1) lset_lmember) lemma index_infinite_set: "\<lbrakk> \<not>lfinite x; lnth x m = y; \<And>i. lnth x i = y \<Longrightarrow> (\<exists>m > i. lnth x m = y) \<rbrakk> \<Longrightarrow> y \<in> lset (ldropn n x)" proof (induct n arbitrary: x m) case 0 thus ?case using lset_nth_member_inf by auto next case (Suc n) obtain a xs where x: "x = LCons a xs" by (meson Suc.prems(1) lnull_imp_lfinite not_lnull_conv) obtain j where j: "j > m" "lnth x j = y" using Suc.prems(2,3) by blast have "lnth xs (j - 1) = y" by (metis lnth_LCons' j(1,2) not_less0 x) moreover { fix i assume "lnth xs i = y" hence "lnth x (Suc i) = y" by (simp add: x) hence "\<exists>j>i. lnth xs j = y" by (metis Suc.prems(3) Suc_lessE lnth_Suc_LCons x) } ultimately show ?case using Suc.hyps Suc.prems(1) x by auto qed subsection \<open>@{term "lmap"}\<close> lemma lnth_lmap_ldropn: "enat n < llength xs \<Longrightarrow> lnth (lmap f (ldropn n xs)) 0 = lnth (lmap f xs) n" by (simp add: lhd_ldropn lnth_0_conv_lhd) lemma lnth_lmap_ldropn_Suc: "enat (Suc n) < llength xs \<Longrightarrow> lnth (lmap f (ldropn n xs)) (Suc 0) = lnth (lmap f xs) (Suc n)" by (metis (no_types, lifting) Suc_llength ldropn_ltl leD llist.map_disc_iff lnth_lmap_ldropn lnth_ltl lnull_ldropn ltl_ldropn ltl_lmap) subsection \<open>Notation\<close> text \<open>We introduce the notation \$ to denote @{term "lnth"}.\<close> notation lnth (infix "$" 61) end

Fossilised turtle arms reunited The arm of a sea turtle first discovered 160 years ago has been put back together as the second half of the broken bone has just been found. The broken arm bone of the turtle Atlantochelys mortoni was first found back in the 1840s and helped scientists map the breed, which existed 70-75 million years ago. However, a new dig has just found the corresponding half of the bone, creating a perfect match. This has helped further understanding of the Atlantochelys mortoni, with scientists now realising that the turtle was much bigger than first thought. To learn more about all the incredible animals on our planet, why not visit our AnimalAnswers website – from the makers of How It Works.

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\section{Proof of Lemma \ref{Lemma:BoundsPerturbation}} \label{Appendix:Proof_BoundsPerturbation} First, we establish the bound for $P[\bm{z}_{i-1}]$ in \eqref{Equ:Lemma:BoundsPerturbation:P_z}. Substituting \eqref{Equ:DistProc:invTF_wi_wi_prime} and \eqref{Equ:DistProc:relation_phi_w_wprime} into the definition of $\bm{z}_{i-1}$ in \eqref{Equ:DistProc:s_hat_phi_decomp_zdef_vdef} we get: \begin{align} P&[\bm{z}_{i-1}] \nn\\ &\preceq P\big[ s\left( \mathds{1} \!\otimes \! \bm{w}_{c,i-1} \!+\! (A_1^T U_L \!\otimes\! I_M) \bm{w}_{e,i-1} \right) \!-\! s(\mathds{1} \!\otimes\! \bm{w}_{c,i-1}) \big] \nonumber\\ &\overset{(a)}{\preceq} \lambda_U^2 \cdot P\left[ (A_1^T U_L \otimes I_M) \bm{w}_{e,i-1} \right] \nonumber\\ &\overset{(b)}{\preceq} \lambda_U^2 \cdot \left\| \bP[\mc{A}_1^T \mc{U}_L] \right\|_{\infty}^2 \cdot \mathds{1}\mathds{1}^T \cdot P[\bm{w}_{e,i-1}] \end{align} where step (a) uses the variance relation \eqref{Equ:VarPropt:s}, and step (b) uses property \eqref{Equ:VarPropt:Linear_ub}. Next, we prove the bound on $P[s(\mathds{1}\otimes \bm{w}_{c,i-1})]$. It holds that \begin{align} P&[s(\mathds{1}\otimes \bm{w}_{c,i-1})] \nonumber\\ &= P\Big[ \frac{1}{3}\cdot 3 \big( s(\mathds{1} \otimes \bm{w}_{c,i-1}) - s(\mathds{1}\otimes \bar{w}_{c,i-1}) \big) \nn\\ &\qquad + \frac{1}{3}\cdot 3 \big( s(\mathds{1}\otimes \bar{w}_{c,i-1}) - s(\mathds{1}\otimes w^o) \big) \nn\\ &\qquad + \frac{1}{3}\cdot 3 \cdot s(\mathds{1}\otimes w^o) \Big] \nonumber\\ &\overset{(a)}{\preceq} \frac{1}{3} \cdot P\big[ 3 \big( s(\mathds{1} \otimes \bm{w}_{c,i-1}) - s(\mathds{1}\otimes \bar{w}_{c,i-1}) \big) \big] \nonumber\\ &\quad + \frac{1}{3}\cdot P \big[ 3 \big( s(\mathds{1}\otimes \bar{w}_{c,i-1}) - s(\mathds{1}\otimes w^o) \big) \big] \nn\\ &\quad + \frac{1}{3}\cdot P \big[ 3 \cdot s(\mathds{1}\otimes w^o) \big] \nonumber\\ &\overset{(b)}{=} 3P\big[ s(\mathds{1} \otimes \bm{w}_{c,i-1}) \!-\! s(\mathds{1}\otimes \bar{w}_{c,i-1}) \big] \nn\\ &\quad + 3P \big[ s(\mathds{1}\otimes \bar{w}_{c,i-1}) \!-\! s(\mathds{1}\otimes w^o) \big] + 3P \big[ s(\mathds{1}\otimes w^o) \big] \nonumber\\ &\overset{(c)}{\preceq} 3\lambda_U^2 \cdot P\big[ \mathds{1} \otimes (\bm{w}_{c,i-1} - \bar{w}_{c,i-1}) \big] \nn\\ &\quad + 3\lambda_U^2 \cdot P[\mathds{1}\otimes(\bar{w}_{c,i-1}-w^o)] + 3P[s(\mathds{1}\otimes w^o)] \nonumber\\ &\overset{(d)}{=} 3\lambda_U^2 \cdot \|\check{\bm{w}}_{c,i-1}\|^2 \cdot \mathds{1} + 3\lambda_U^2 \cdot \|\bar{w}_{c,i-1} - w^o\|^2 \cdot \mathds{1} \nn\\ &\quad + 3P[s(\mathds{1}\otimes w^o)] \nonumber\\ &\overset{(e)}{\preceq} 3\lambda_U^2 \cdot \|\check{\bm{w}}_{c,i-1}\|^2 \!\cdot\! \mathds{1} \!+\! 3\lambda_U^2 \|\tilde{w}_{c,0}\|^2 \!\cdot\! \mathds{1} \!+\! 3P[s(\mathds{1}\otimes w^o)] \end{align} where step (a) uses the convexity property \eqref{Equ:Properties:PX_CvxComb}, step (b) uses the scaling property in Lemma \ref{Lemma:BasicPropertiesOperator}, step (c) uses the variance relation \eqref{Equ:VarPropt:s}, step (d) uses property \eqref{Equ:Properties:PX_Kron}, and step (e) uses the bound \eqref{Equ:Thm:ConvergenceRefRec:NonAsympBound} and the fact that $\gamma_c<1$. Finally, we establish the bounds on $P[\bm{v}_i]$ in \eqref{Equ:Lemma:BoundsPerturbation:P_v_E_Fiminus1}--\eqref{Equ:Lemma:BoundsPerturbation:P_v}. Introduce the $MN \times 1$ vector $\bm{x}$: \begin{align} \bm{x} \defeq \mathds{1} \otimes \bm{w}_{c,i-1} + \mc{A}_1^T \mc{U}_L \bm{w}_{e,i-1} \equiv \bm{\phi}_{i-1} \end{align} We partition $\bm{x}$ in block form as $\bm{x} = \col\{\bm{x}_1,\ldots,\bm{x}_N\}$, where each $\bm{x}_k$ is $M\times 1$. Then, by the definition of $\bm{v}_i$ from \eqref{Equ:DistProc:s_hat_phi_decomp_zdef_vdef}, we have \begin{align} \E \{ P[\bm{v}_i] | \mc{F}_{i-1} \} &= \E \{ P[\hat{\bm{s}}_i(\bm{x}) - s(\bm{x}) ] | \mc{F}_{i-1} \} \nonumber\\ &= \col\big\{ \E \left[ \| \hat{\bm{s}}_{1,i}(\bm{x}_1) - s_1(\bm{x}_1) \|^2 \big| \mc{F}_{i-1} \right], \nn\\ &\qquad \ldots, \E \left[ \| \hat{\bm{s}}_{N,i}(\bm{x}_N) - s_N(\bm{x}_N) \|^2 \big| \mc{F}_{i-1} \right] \big\} \nonumber\\ &\overset{(a)}{\preceq} \col\big\{ \alpha \cdot \| \bm{x}_1 \|^2 \!+\! \sigma_{v}^2,\; \ldots,\; \alpha \cdot \| \bm{x}_N \|^2 \!+\! \sigma_{v}^2 \big\} \nonumber\\ \label{Equ:Appendix:P_v_bound1} &= \alpha \cdot P[\bm{x}] + \sigma_v^2 \mathds{1} \end{align} where step (a) uses Assumption \eqref{Equ:Assumption:Randomness:RelAbsNoise}. Now we bound $P[\bm{x}]$: \begin{align} P[ \bm{x} ] &= P\left[ \mathds{1} \otimes \bm{w}_{c,i-1} + \mc{A}_1^T \mc{U}_L \bm{w}_{e,i-1} \right] \nonumber\\ &= P\Big[ \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes ( \w_{c,i-1} \!-\! \bar{w}_{c,i-1} ) \!+\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes (\bar{w}_{c,i-1} \!-\! w^o) \nn\\ &\qquad +\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \mA_1^T \mU_L \w_{e,i-1} \!+\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes w^o \Big] \nonumber\\ &= P\Big[ \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes \check{\w}_{c,i-1} \!+\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes \tilde{w}_{c,i-1} \nn\\ &\qquad +\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \mA_1^T \mU_L \w_{e,i-1} \!+\! \frac{1}{4} \!\cdot\! 4 \!\cdot\! \one \otimes w^o \Big] \nn\\ &\overset{(a)}{\preceq} \frac{1}{4} \cdot 4^2 \!\cdot\! P[ \one \otimes \check{\w}_{c,i-1} ] + \frac{1}{4} \cdot 4^2 \!\cdot\! P[ \one \otimes \tilde{w}_{c,i-1} ] \nn\\ &\quad + \frac{1}{4} \cdot 4^2 \!\cdot\! P[ \mA_1^T \mU_L \w_{e,i-1} ] + \frac{1}{4} \cdot 4^2 \!\cdot\! P[ \one \otimes w^o ] \nonumber\\ &\overset{(b)}{=} 4 \cdot \| \check{\bm{w}}_{c,i-1} \|^2 \cdot \mathds{1} + 4 \cdot \| \tilde{w}_{c,i-1} \|^2 \cdot \mathds{1} \nn\\ &\quad + 4 \cdot P[ \mc{A}_1^T \mc{U}_L \bm{w}_{e,i-1} ] + 4 \cdot \| w^o \|^2 \cdot \one \nonumber\\ &\overset{(c)}{\preceq} 4 \cdot \| \check{\bm{w}}_{c,i-1} \|^2 \!\cdot\! \mathds{1} \!+\! 4 \!\cdot\! \| \bP[ \mc{A}_1^T \mc{U}_L ] \|_{\infty}^2 \!\cdot\! \one\one^T \!\cdot\! P[ \bm{w}_{e,i-1} ] \nn\\ &\quad + 4 \cdot \|\tilde{w}_{c,0}\|^2 \cdot \mathds{1} + 4 \cdot \| w^o \|^2 \cdot \one \label{Equ:Appendix:P_v_bond2} \end{align} where step (a) uses the convexity property \eqref{Equ:Properties:PX_CvxComb} and the scaling property in Lemma \ref{Lemma:BasicPropertiesOperator}, step (b) uses the Kronecker property \eqref{Equ:Properties:PX_Kron}, step (c) uses the variance relation \eqref{Equ:VarPropt:Linear} and the bound \eqref{Equ:Thm:ConvergenceRefRec:NonAsympBound}. Substituting \eqref{Equ:Appendix:P_v_bond2} into \eqref{Equ:Appendix:P_v_bound1}, we obtain \eqref{Equ:Lemma:BoundsPerturbation:P_v_E_Fiminus1}, and taking expectation of \eqref{Equ:Lemma:BoundsPerturbation:P_v_E_Fiminus1} with respect to $\mF_{i-1}$ leads to \eqref{Equ:Lemma:BoundsPerturbation:P_v}.

This Fall, I collaborated with Envy Clothing Company to bring you outfit pairings for Back-To-School! When I was asked to do this collaboration, it brought me back to my college days when I had to figure out what was comfortable, but fashionable for studying. And within every college or university, you’ll definitely find different fashion personalities and styles, so I came up with 5 different styles you can see above (starting from the left): 1) Hobo Chic – It’s all about maxi skirts, layered necklaces, and draped sweaters. What is more boho chic than that? I am loving the addition of the wool hat with the feather detail and the leather Matt&Natt Crossbody. Play around with bodychains and layering chunky knits this season! 2) Street Style – One of my favorite looks, street chic is all about combining the hard and soft textiles. You need a staple pair of loose boyfriend jeans (size up!) and definitely some plaid and leather in your closet! Throw on some plaid or leather around your waist with an outfit you and your bf could possibly share, and a leather jacket; you’re good to go! Look for the perfect shades to give you that effortless, cool – look. 3) Prepped Up – This semester it’s all about blazers, denim crop jackets, flouncy skirts, and embellishment. Pair a comfortable high waisted skirt with a jeweled top and a cute jacket. Don’t forget to add on your staple ballet flats! 4) Trendy – In with the trends, you’re always rocking pieces from the fashion magazines! Put on that silky patterned jumpsuit with a loose vest, and you’re ready to hit the books! Don’t forget to save money in the bank for that new Canada Goose jacket you wanted for the cold Canadian winters! 5) Sports Inspired – Baseball T’s, cargo pants, bomber jackets, and leather caps are your go-to pieces! Another one of my favorites, make your outfit stand out with a statement baseball T from Maison Scotch and throw on a bomber around your waist in case it gets cold in the library! When I was coming up with the outfits for this semester, I thought it wouldn’t be college perfect without the right nighttime outfits to go along with it, which was when I came up with the theme: Study By Day, Party By Night. University/college is the perfect time for you to test new styles and figure out what works and what doesn’t work for you! It’s also where the craziest stories will come from, and why not look good and enjoy it while it lasts. So check out both Halifax Envy locations for my picks for Back-To-School fashion, and I’ll see you around! xoxo

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TITLE: Identifying superalgebras with fixed points under Cartan involution QUESTION [1 upvotes]: I am making my way through the "Foundations of the $AdS_5 x S^5$ Superstring: Part I" paper by Arutyunov/Frolov 2009 (https://arxiv.org/abs/0901.4937v2) and am hoping someone can help me bridge a logical breaking point. In the introduction, the authors assert that the type IIB Green-Schwarz superstring in the $AdS_5 x S^5$ background can be written as a non-linear sigma-model with the target space being $$ \frac{\text{PSU(2,2|4)}}{\text{SO(4,1) x SO(5)}}. $$ Logically, the next first section is a discussion of the superconformal algebra $\mathfrak{psu}(2,2|4)$. In order to get to this superalgebra, they start by introducing $\mathfrak{sl}(4|4)$. They then identify the superalgebra $\mathfrak{su}(2,2|4)$ with the fixed points $M^* =M$ of $\mathfrak{sl(4|4)}$ under Cartan involution $$ M^* = -H M^{\dagger} H^{-1}, $$ where M is the 8 x 8 block matrix $$ M= \left(\begin{array}{cccc} m&\theta\\ \eta & n\\ \end{array}\right) $$ and H is the 8 x 8 matrix $$ H= \left(\begin{array}{cccccccc} 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&-1&0&0&0&0&0\\ 0&0&0&-1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\\ \end{array}\right). $$ The point where I'm getting lost is when they identify the superalgebra $\mathfrak{su}(2,2|4)$ with the fixed points $M^* =M$ of $\mathfrak{sl(4|4)}$ under Cartan involution. If anybody could explain the significance of fixed points under Cartan involution and the logic behind this route of studying the superconformal algebra, I would greatly appreciate it. REPLY [0 votes]: It is just a reality condition on $M$. This will select a particular signature to your space-time and also impose reality conditions to the spinors. You see that as follows: the matrix $H$ can be written as $$ H=\begin{pmatrix}\eta^{(2,2)}&0\\0&\eta^{(4,0)}\end{pmatrix} $$ where $\eta^{(2,2)}$ is a metric with signature $(2,2)$ and $\eta^{(4,0)}$ a metric with signature $(4,0)$. If you write $M$ as $$ M=\begin{pmatrix}m&\theta\\\eta&m\end{pmatrix} $$ the reality condition $$ M^{*}=-HM^{T}H $$ will constraint $m$ to be a $u(2,2)$ generator, while $n$ to be a $u(4)$ generator, i.e. $$ m^{*} =-\eta^{(2,2)}m^{T}\eta^{(2,2)},\qquad n^{*} =-\eta^{(4,0)}n^{T}\eta^{(4,0)} $$ If $M$ is a $gl(4|4)$ generator the generators that survive under this reality condition are $$ u(2,2)\oplus u(4)=u(1)\oplus u(1)\oplus su(2,2)\oplus su(4) $$. Imposing the constraint $gl(4|4)\rightarrow sl(4|4)$ constraint $$ \text{tr}(m)-\text{tr}(n)=0 $$ killing one $u(1)$. The other $u(1)$ that survive is generated by $$ \text{tr}(m)+\text{tr}(n) $$ and is projected out by doing $sl(4|4)\rightarrow psl(4|4)$. With the reality conditions imposed we have $psu(2,2|4)$ Note that $su(2,2)\cong so(2,4)$ which is the correct signature for a conformal transformation in Minkowski space-time $\mathbb{R}^{1,3}$. The fermionic spinors $\theta$ and $\eta$ are in the fundamental (or anti-fundamental) of $SU(2,2)\oplus SU(4)$ so the reality condition consistent with the $SU(2,2)\oplus SU(4)$ transformations involve appropriate contractions with the metrics $\eta^{(2,2)}$ and $\eta^{(4,0)}$. The reality condition $M^{*}=-HM^{T}H$ does this automatically.

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\begin{document} \footnotesize \normalsize \newpage \thispagestyle{empty} \begin{center} \phantom. \bigskip {\hfill{\normalsize hep-th/0608152}\\ \hfill{\normalsize ITEP-TH-XX/06}\\ \hfill{\normalsize HMI-06-XX}\\ \hfill{\normalsize TCD-MATH-06-XX}\\ [10mm]\Large\bf Givental Integral Representation for Classical Groups \footnote{To be published in the Proceedings of the Satellite ICM 2006 conference: "Integrable systems in Applied Mathematics", Colmenarejo (Madrid, Spain), 7-12 September 2006.}} \vspace{0.5cm} \bigskip\bigskip {\large A. Gerasimov} \\ \bigskip {\it Institute for Theoretical and Experimental Physics, 117259, Moscow, Russia} \\ {\it School of Mathematics, Trinity College, Dublin 2, Ireland } \\ {\it Hamilton Mathematics Institute, TCD, Dublin 2, Ireland},\\ \bigskip {\large D. Lebedev\footnote{E-mail: lebedev@ihes.fr}} \\ \bigskip {\it Institute for Theoretical and Experimental Physics, 117259, Moscow, Russia} \\ {\it l'Institute des Hautes \'Etudes Scientifiques, 35 route de Chartres, Bures-sur-Yvette, France},\\ \bigskip {\large S. Oblezin} \footnote{E-mail: Sergey.Oblezin@itep.ru}\\ \bigskip {\it Institute for Theoretical and Experimental Physics, 117259, Moscow, Russia}\\ {\it Max-Planck-Institut f\"ur Mathematik, Vivatsgasse 7, D-53111 Bonn, Germany},\\ \end{center} \vspace{0.5cm} \begin{abstract} \noindent We propose integral representations for wave functions of $B_n$, $C_n$, and $D_n$ open Toda chains at zero eigenvalues of the Hamiltonian operators thus generalizing Givental representation for $A_n$. We also construct Baxter $Q$-operators for closed Toda chains corresponding to Lie algebras $B_{\infty}$, $C_{\infty}$, $D_{\infty}$, affine Lie algebras $B^{(1)}_n$, $C^{(1)}_n$, $D^{(1)}_n$ and twisted affine Lie algebras $A^{(2)}_{2n-1}$ and $A^{(2)}_{2n}$. Our approach is based on a generalization of the connection between Baxter $Q$-operator for $A_n^{(1)}$ closed Toda chain and Givental representation for the wave function of $A_n$ open Toda chain uncovered previously. \end{abstract} \vspace{1cm} \clearpage \newpage \normalsize \section{Introduction} A remarkable integral representation for the common eigenfunctions of $A_n$ open Toda chain Hamiltonian operators was proposed in \cite{Gi} (see also \cite{JK}). This representation is based on a flat degeneration of $A_n$ flag manifolds to a Gorenstein toric Fano variety (see \cite{L},\cite{Ba}, \cite{BCFKS} for details). This results in a purely combinatorial description of the integrand in the integral representation \cite{Gi}. An important application of the Givental integral representation so far was an explicit construction of the mirror dual of $A_n$ flag manifolds. Later it turns out that the representation introduced in \cite{Gi} is also interesting from another points of view. Thus it was shown in \cite{GKLO} that this integral representation has natural iterative structure allowing the connection of $A_{n-1}$ and $A_{n}$ wave functions by a simple integral transformation. It was demonstrated that thus defined integral transformation is given by a degenerate version of the Baxter $Q$-operator realizing quantum B\"acklund transformations in closed Toda chain \cite{PG}. Let us note that the torification of flag manifolds leads to a distinguished set of coordinates on its open parts. A group theory construction of these coordinates is also connected with a degenerate $Q$-operator and was clarified in \cite{GKLO}. Up to now the Givental integral representation was only generalized \cite{BCFKS} to the case of degenerate $A_n$ open Toda chains \cite{STS} leading to a construction of the mirror duals to partial flag manifolds $G/P$ for $G=SL(n+1,\mathbb{C})$ and $P$ being a parabolic subgroup. A natural approach to a generalization of the integral representation to other Lie algebras could be based on the relation with Baxter $Q$-operator. However no generalization of the Baxter $Q$-operator to other Lie algebras was known. In this note we solve both these problems simultaneously for all classical series of Lie algebras. We propose a generalization of the Givental integral representation to other classical series $B_n$, $C_n$, $D_n$ and construct $Q$-operators for affine Lie algebras $A_n^{(1)}$, $A^{(2)}_{2n}$, $A_{2n-1}^{(2)}$, $B_n^{(1)}$, $C_n^{(1)}$, $D_n^{(1)}$ and infinite Lie algebras $B_{\infty}$, $C_{\infty}$ and $D_{\infty}$. We also generalize the connection between $Q$-operators and integral representations of wave functions to all classical series. Let us stress that there is an important difference in the construction of the integral representations between $A_n$ and other classical series. The kernel of the integral operator providing recursive construction of the integral representation for $A_n$ has a simple form of the exponent of the sum of exponents in the natural coordinates. For other classical groups recursive operators of the same type exist but they relate Toda chain wave functions for different classical series (e.g. $C_{n}$ and $D_{n}$). Integral operators connecting Toda wave functions in the same series (e.g. $D_n$ and $D_{n-1}$) are given by compositions of the elementary integral operators. In this note we restrict ourselves by explicit constructions of the integral representations of eigenfunctions of the quadratic open Toda chain Hamiltonian operators at zero eigenvalues. The general case of all Hamiltonians and non-zero eigenvalues will be published elsewhere. Also we leave for another occasion the elucidation of a group theory interpretation of the obtained results. The plan of this paper is as follows. In Section 2 we summarize the results of \cite{GKLO}. In Section 3 we construct kernels of the elementary integral operators intertwining Hamiltonian operators of open Toda chains for (in general different) classical series of finite Lie algebras. In Section 4 using the results from Section 3 we give explicit integral representations for the wave functions of $B_n$, $C_n$ and $D_n$ open Toda chains. In Section 5 we describe a generalization of Givental diagrams to other classical series and remark on the connection with toric degeneration of $B_n$, $C_n$ and $D_n$ flag manifolds. In Section 6 we construct elementary integral operators intertwining Hamiltonian operators of closed Toda chains for (in general different) classical series of affine Lie algebras. In Section 7 we construct integral kernels for Baxter $Q$-operators for all classical series of (twisted) affine Lie algebras as appropriate compositions of the elementary intertwining operators. In Section 8 we construct $Q$ operators for $B_{\infty}$, $C_{\infty}$ and $D_{\infty}$ Toda chains. We conclude in Section 9 with a short discussion of the results presented in this note. We were informed by E.~Sklyanin that he also has some progress in the construction of Baxter $Q$-operators for Toda theories. {\em Acknowledgments}: The authors are grateful to S.~Kharchev for discussions at the initial stage of this project and to B.~Dubrovin and M.~Kontsevich for their interest in this work. The research of A.~Gerasimov was partly supported by the Enterprise Ireland Basic Research Grant. D.~Lebedev is grateful to Institute des Hautes \'Etudes Scientifiques for warm hospitality. S.~Oblezin is grateful to Max-Planck-Institut f\"ur Mathematik for excellent working conditions. \section{Recursive structure of Givental representation} In this section we recall a recursive construction of the Givental integral representation discussed in \cite{GKLO}. The solution of a quantum integrable system starts with the finding of the full set of common eigenfunctions of the quantum Hamiltonian operators of $A_n$ Toda chain. Note that the difference between wave functions for $\mathfrak{sl}_n$ and $\mathfrak{gl}_n$ manifests only at non-zero eigenvalues of the element of the center of $U\mathfrak{gl}_n$ linear over the generators. In the following we will consider only the wave functions corresponding to zero eigenvalues of all elements of the center. Thus we will always consider $\mathfrak{gl}_n$ Toda chain instead of $A_n$ Toda chain. The quadratic quantum Hamiltonian of $\mathfrak{gl}_n$ open Toda chain is given by \be H^{\mathfrak{gl}_n}(x)=-\frac{\hbar^2}{2}\sum\limits_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}+ \sum\limits_{i=1}^{n-1}g_i e^{x_{i+1}-x_{i}}\,. \ee In \cite{Gi} the following remarkable representation for a common eigenfunction of quantum Hamiltonians of $\mathfrak{gl}_n$ open Toda chain was proposed \be\label{intrep} \Psi(x_1,\ldots,x_{n})=\int_{\Gamma} e^{\frac{1}{\hbar}\mathcal{F}_n(x)}\bigwedge_{k=1}^{n-1} \bigwedge_{i=1}^k dx_{k,i}, \ee where $x_{n,i}:=x_i$, the function $\mathcal{F}_n(x)$ is given by \be\label{pot} \hspace{-0.5cm} \mathcal{F}_n(x)=\sum_{k=1}^{n-1}\sum_{i=1}^k\Big(e^{x_{k,i}-x_{k+1,i}} +g_ie^{x_{k+1,i+1}-x_{k,i}}\Big), \ee and the cycle $\Gamma$ is a middle dimensional submanifold in the $n(n-1)/2$- dimensional complex torus with coordinates $\{\exp\,{x_{k,i}},\, i=1,\ldots,k;\,k=1,\ldots, n-1\}$ such that the integral converges. The eigenfunction (\ref{intrep}) solves the equation \bqa H^{\mathfrak{gl}_n}(x)\,\Psi(x_1,\cdots, x_n)= 0. \eqa In the following we put $\hbar=1$ for convenience. The derivation of the integral representation (\ref{intrep}) using the recursion over the rank $n$ of the Lie algebra $\mathfrak{gl}_n$ was given in \cite{GKLO}. The integral representation for the wave function can be represented in the following form \bqa\label{recintrep} \Psi(x_{1},\ldots,x_{n})\,=\, \int\bigwedge_{k=1}^{n-1}\bigwedge_{i=1}^kdx_{k,i}\, \prod_{k=1}^{n-1}Q_{k+1,\,k}(x_{k+1,1},\ldots,x_{k+1,k+1}; \,x_{k,1},\ldots,x_{k,k}), \eqa with the integral kernel \be\label{QBAXTER} Q_{k+1,k}(x_{k+1,i};\, x_{k,i})= \exp\Big\{\, \sum_{i=1}^{k} \left(e^{x_{k,i}-x_{k+1,i}}+g_ie^{x_{k+1,i+1}-x_{k,i}}\right)\,\Big\}. \ee Here we have $x_i:=x_{n,i}$. The following differential equation for the kernel holds \bqa\label{intertw} H^{\mathfrak{gl}_{k+1}}(x_{k+1,i})Q_{k+1,k}(x_{k+1,i},\, x_{k,i})= Q_{k+1,k}(x_{k+1,i},\, x_{k,i})\,H^{\mathfrak{gl}_{k}}(x_{k,i}), \eqa where \be H^{\mathfrak{gl}_k}(x_i) =-\frac{1}{2}\sum\limits_{i=1}^{k}\frac{\partial^2}{\partial x_{k,i}^2}+ \sum\limits_{i=1}^{k-1}g_i e^{x_{k,i+1}-x_{k,i}}\, . \ee Here and in the following we assume that in the relations similar to (\ref{intertw}) the Hamiltonian operator on l.h.s. acts on the right and the Hamiltonian on r.h.s. acts on the left. Thus the integral operator with the kernel $Q_{k+1,k}$ intertwines Hamiltonian operators for $\mathfrak{gl}_{k+1}$ and $\mathfrak{gl}_k$ open Toda chains. The integral operator defined by the kernel (\ref{QBAXTER}) is closely related with a Baxter $Q$-operator realizing B\"{a}cklund transformations in a closed Toda chain corresponding to affine Lie algebra $\widehat{\mathfrak{gl}}_n$. Baxter $Q$-operator for zero spectral parameter can be written in the integral form with the kernel \cite{PG} \bqa\label{baff} Q^{\widehat{\mathfrak{gl}}_n}(x_i,y_i)=\exp\Big\{\, \sum_{i=1}^{n} \left(e^{x_{i}-y_{i}}+g_ie^{y_{i+1}-x_{i}}\right)\,\Big\}, \eqa where $x_{i+n}=x_i$ and $y_{i+n}=y_i$. This operator commutes with the Hamiltonian operators of the closed Toda chain. Thus for example for the quadratic Hamiltonian we have \bqa\label{intertwaff} \CH^{\widehat{\mathfrak{gl}}_n}(x_i)Q^{\widehat{\mathfrak{gl}}_n}(x_i,\, y_i)= Q^{\widehat{\mathfrak{gl}}_n} (x_i,\, y_i)\CH^{\widehat{\mathfrak{gl}}_n}(y_i), \eqa where \be \CH^{\widehat{\mathfrak{gl}}_n}= -\frac{1}{2}\sum\limits_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}+ \sum\limits_{i=1}^{n}g_i e^{x_{i+1}-x_{i}}\,. \ee Here we impose the conditions $x_{i+n}=x_i$. The recursive operator (\ref{QBAXTER}) can be obtained from Baxter operator (\ref{baff}) in the limit $g_{n}\rightarrow 0$, $x_n\rightarrow-\infty$. The main objective of this note is to generalize the representation (\ref{intrep}), (\ref{pot}) to other classical series $B_n$, $C_n$ and $D_n$ of finite Lie groups. Before we present the integral representations for $B_n$, $C_n$ and $D_n$ let us comment on the main subtlety in their constructions. As it was explained in \cite{GKLO} the variables $x_{k,i}$ in the integral representation for $A_n$ have a clear meaning of the linear coordinates on Cartan subalgebras of the intermediate Lie algebras entering recursion $A_n\to A_{n-1}\to \cdots \to A_{1}$. This is a consequence of the identity \be ({\rm dim}\, (\mathfrak{gl}_n)-{\rm rk}\,(\mathfrak{gl}_n)) -({\rm dim}\, (\mathfrak{gl}_{n-1})-{\rm rk}\,(\mathfrak{gl}_{n-1})) =2\,{\rm rk}\, \mathfrak{gl}_{n-1}. \ee However for other classical series there is no such simple relation. In general one finds more integration variables in the integral representation then those arising as linear coordinates on intermediate Cartan subalgebras. It turns out that the elementary integral operators intertwine Hamiltonians corresponding to Toda chains for {\it different} Lie algebras. The recursive operators are then constructed as appropriate compositions of the elementary intertwining operators. \section{Elementary intertwiners for open Toda chains} Let $\mathfrak{g}$ be a simple Lie algebra, $\mathfrak{h}$ be a Cartan subalgebra, $n=\dim \mathfrak{h}$ be the rank of $\mathfrak{g}$, $R\subset \mathfrak{h}^*$ be the root system, $W$ be the Weyl group. Let us fix a decomposition $R=R_+\cup R_-$ of the roots on positive and negative roots. Let $\alpha_1,\ldots,\alpha_n$ be the bases of simple roots. Let $(,)$ be a $W$-invariant bilinear symmetric form on $\mathfrak{h}^*$ normalized so that $(\alpha,\alpha)=2$ for a long root. This form provides an identification of $\mathfrak{h}$ with $\mathfrak{h}^*$ and thus can be considered as a bilinear form on $\mathfrak{h}$. Choose an orthonormal basis $e=\{e_1,\ldots,e_{n}\}$ in $\mathfrak{h}$. Then for any $x\in \mathfrak{h}$ one has a decomposition $x=\sum_{i=1}^n x_ie_i$. One associates with these data an open Toda chain with a quadratic Hamiltonian \bqa H^{R}(x_i)= -\frac{1}{2}\sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}\,+\, \sum_{i=1}^ng_ie^{\alpha_i(x)}\, .\eqa For the standard facts on Toda theories corresponding to arbitrary root systems see e.g. \cite{RSTS}. We start with explicit expressions for elementary intertwining operators for open Toda chains. The necessary facts on the root systems (including non-reduced ones) can be found in \cite{He}. \subsection{ $BC\leftrightarrow B$} Let $e=\{e_1,\ldots,e_{n}\}$ be an orthonormal basis in $\mathbb{R}^n$. Non-reduced root system of type $BC_n$ can be defined as \bqa \alpha_0=2e_1,\qquad \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1 \eqa and the corresponding Dynkin diagram is \bqa \frac{\alpha_0}{\alpha_1}\,\Longleftrightarrow\,\alpha_2 \longleftarrow\ldots\longleftarrow\alpha_n \eqa where the first vertex from the left is a doubled vertex corresponding to a reduced $\alpha_1=e_1$ and non-reduced $\alpha_0=2e_1=2\alpha_1$ roots. Quadratic Hamiltonian operator of the corresponding open Toda chain is given by \bqa H^{BC_n}(x_i)=- \frac{1}{2}\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}+ \frac{g_1}{2}\Big(e^{x_1}+g_1e^{2x_1}\Big)+ \sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}. \eqa Let us stress that the same open Toda chain can be considered as a most general form of $C_n$ open Toda chain (see e.g. \cite{RSTS}, Remark p.61). However in the following we will use the term $BC_n$ open Toda chain to distinguish it from a more standard $C_n$ open Toda chain that will be consider below. The root system of type $B_n$ can be defined as \bqa \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1 \eqa and the corresponding Dynkin diagram is \bqa \alpha_1\,\Longleftarrow\,\alpha_2 \longleftarrow\ldots\longleftarrow\alpha_n\,. \eqa Quadratic Hamiltonian operator of the corresponding open Toda chain is given by \bqa H^{B_n}(x_i)=- \frac{1}{2}\sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}+ g_1e^{x_1}+\sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}.\eqa An elementary operator intertwining open Toda chain Hamiltonians for $BC_n$ and $B_{n-1}$ can be written in the integral form with the kernel \be Q_{BC_n}^{\,\,\,\,B_{n-1}}(z_1,\ldots,z_n;\,x_1,\ldots,x_{n-1})= \exp\Big\{\,g_1e^{z_1}+\sum_{i=1}^{n-1}(e^{x_i-z_i}+ g_{i+1}e^{z_{i+1}-x_i})\,\Big\},\ee satisfying the following relation \bqa\label{BCB} H^{BC_n}(z)\,Q_{BC_n}^{\,\,\,\,B_{n-1}}(z,\, x)= Q_{BC_n}^{\,\,\,\,B_{n-1}}(z,\, x) \,H^{B_{n-1}}(x) . \eqa Similarly an elementary operator intertwining $B_n$ and $BC_{n}$ Hamiltonians has an integral kernel \be Q_{B_n}^{\,\,\,\,BC_{n}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n})=\\ \exp\Big\{\,g_1e^{z_1}+\sum_{i=1}^{n-1}\Big(e^{x_i-z_i}+ g_{i+1}e^{z_{i+1}-x_i}\Big)+e^{x_n-z_n}\,\Big\}.\ee \subsection{ $C \leftrightarrow D$} The root system of type $C_n$ can be defined as \bqa \alpha_{i}=e_{i+1}-e_i,\qquad \alpha_n=2e_n,\qquad 1\leq i\leq n-1, \eqa and the corresponding Dynkin diagram is \bqa \alpha_1\,\longleftarrow\,\ldots \longleftarrow \alpha_{n-1} \Longleftarrow\alpha_n\,. \eqa Quadratic Hamiltonian operator of the corresponding open Toda chain is given by \be H^{C_n}(x_i)=- \frac{1}{2}\sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}+\sum_{i=1}^{n-1}g_{i}e^{x_{i+1}-x_i}+ 2g_ne^{-2x_n}. \ee The root system of type $D_n$ is \bqa \alpha_{i}=e_{i+1}-e_i,\qquad \alpha_n=-e_{n-1}-e_{n},\qquad 1\leq i< n, \eqa and the corresponding Dynkin diagram is \bqa\begin{CD} \alpha_1 @>>> \ldots @>>> \alpha_{n-2} @>>> \alpha_{n-1}\\ @.@. @VVV @.\\ @.@. \alpha_{n} @.\end{CD}\eqa Quadratic Hamiltonian operator of the $D_n$ open Toda chain is given by\be H^{D_n}(x_i)=- \frac{1}{2}\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}+ \sum_{i=1}^{n-1}g_ie^{x_{i+1}-x_i}+g_{n-1}g_ne^{-x_n-x_{n-1}}. \ee An integral operator intertwining $C_n$ and $D_{n}$ Hamiltonians has the kernel \be Q_{C_n}^{\,\,\,\,D_{n}}(z_1,\ldots,z_n;x_1,\ldots,x_{n})=\\ \exp\Big\{\, \sum_{i=1}^{n-1}\Big(e^{x_i-z_i}+g_{i}e^{z_{i+1}-x_i}\Big)+ e^{x_n-z_{n}}+g_ne^{-x_n-z_{n}}\Big\}.\ee Similarly an integral operator with the kernel \be Q_{D_n}^{\,\,\,C_{n-1}}(x_1,\ldots, x_n;\, z_1,\ldots,z_{n-1})=\\= \exp\Big\{\sum_{i=1}^{n-1}\Big(e^{z_i-x_i}+g_ie^{x_{i+1}-z_i}\Big)+ g_ne^{-x_n-z_{n-1}}\Big\},\ee intertwines the following $D_n$ and $C_{n-1}$ quadratic Hamiltonians \bqa H^{D_n}(x_i)=- \frac{1}{2}\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}+ \sum_{i=1}^{n-1}g_ie^{x_{i+1}-x_i}+g_ne^{-x_n-x_{n-1}},\\ H^{C_{n-1}}(z_i)=-\frac{1}{2}\sum_{i=1}^{n-1}\frac{\partial^2}{\partial z_i^2}+\sum_{i=1}^{n-2}g_{i}e^{z_{i+1}-z_i}+ 2g_{n-1}g_ne^{-2z_{n-1}}.\eqa \section{Givental representation for wave functions} In the previous section explicit expressions for the kernels of elementary intertwining operators were presented. Now integral representations for eigenfunctions of open Toda chain Hamiltonians are given by a quite straightforward generalization of $A_n$ case. Below we provide integral representations for all classical series. For simplicity we put $g_i=1$ below. \subsection{$B_n$} The eigenfunction for $B_n$ open Toda chain is given by \be \Psi^{B_n}(x_1,\ldots,x_n)\,=\, \int\bigwedge_{k=1}^{n-1}\bigwedge_{i=1}^kdx_{k,i}\, \prod_{k=1}^{n-1}Q_{B_{k+1}}^{\,\,B_k}(x_{k+1,1},\ldots,x_{k+1,k+1}; \,x_{k,1},\ldots,x_{k,k}), \ee where $x_i:=x_{n,i}$ and the kernels $Q_{B_{k+1}}^{\,\,\,\,B_k}$ of the integral operators are given by the convolutions of the kernels $Q_{B_{k+1}}^{\,\,\,\,BC_{k+1}}$ and $Q_{BC_{k+1}}^{\,\,\,\,B_k}$ \bqa Q_{B_{k+1}}^{\,\,\,\,B_k}(x_{k+1,i};\,x_{k,j})=\int\bigwedge_{i=1}^kdz_{k,i} \, Q_{B_{k+1}}^{\,\,\,\,BC_{k+1}}(x_{k+1,1},\ldots,x_{k+1,k+1}; z_{k+1,1},\ldots,z_{k+1,k+1})\times \\ \nonumber \times Q_{BC_{k+1}}^{\,\,\,\,B_k} (z_{k+1,1},\ldots,z_{k+1,k+1};x_{k,1},\ldots,x_{k,k}).\eqa Notice that the wave function is given by the integral over a contour of the real dimension equal to a complex dimension of the flag manifold $X=G/B$, where $G=SO(2n+1,\mathbb{C})$ and $B$ is a Borel subgroup \be \sum_{k=1}^n (2k-1)=n^2=|R_+|. \ee \subsection{ $C_n$} The eigenfunction for $C_n$ open Toda chain is given by \be \Psi^{C_n}(z_1,\ldots,z_n)\,=\, \int\bigwedge_{k=1}^{n-1}\bigwedge_{i=1}^kdz_{k,i}\, \prod_{k=1}^{n-1}Q_{C_{k+1}}^{\,\,\,\,C_k}(z_{k+1,1},\ldots,z_{k+1,k+1}; \,z_{k,1},\ldots,z_{k,k}), \ee where $z_i:=z_{n,i}$ and the kernels $Q_{C_{k+1}}^{\,\,\,\,C_k}$ of the integral operators are given by the convolutions of the kernels $Q_{C_{k+1}}^{\,\,\,\,D_{k+1}}$ and $Q_{D_{k+1}}^{\,\,\,\,C_k}$ \bqa\label{QC} Q_{C_{k+1}}^{\,\,\,\,C_k}(z_{k+1,i};\,z_{k,j})=\int\bigwedge_{i=1}^kdx_{k,i} \, Q_{C_{k+1}}^{\,\,\,\,D_{k+1}}(z_{k+1,1},\ldots,z_{k+1,k+1}; x_{k+1,1},\ldots,x_{k+1,k+1})\times \\ \nonumber \times Q_{D_{k+1}}^{\,\,\,\,C_k} (x_{k+1,1},\ldots,x_{k+1,k+1};z_{k,1},\ldots,z_{k,k}).\eqa Thus the wave function is given by the integral over a contour of the real dimension equal to a complex dimension of the flag manifold $X=G/B$, where $G=Sp(n,\mathbb{C})$ and $B$ is a Borel subgroup \be \sum_{k=1}^n (2k-1)=n^2=|R_+|. \ee \subsection{$D_n$} The eigenfunction for $D_n$ open Toda chain is given by \be\label{WDN} \Psi^{D_n}(x_1,\ldots,x_n)\,=\, \int\bigwedge_{k=1}^{n-1}\bigwedge_{i=1}^kdx_{k,i}\, \prod_{k=1}^{n-1}Q_{D_{k+1}}^{\,\,\,\,D_k}(x_{k+1,1},\ldots,x_{k+1,k+1}; \,x_{k,1},\ldots,x_{k,k}),\ee where $x_i:=x_{n,i}$ and the kernels $Q_{D_{k+1}}^{\,\,\,\,D_k}$ of the integral operators are given by the convolutions of the kernels $Q_{D_{k+1}}^{\,\,\,\,C_{k}}$ and $Q_{C_{k}}^{\,\,\,\,D_k}$ \bqa Q_{D_{k+1}}^{\,\,\,\,D_k}(x_{k+1,i};\,x_{k,j})=\int\bigwedge_{i=1}^kdz_{k,i} \, Q_{D_{k+1}}^{\,\,\,\,C_k}(x_{k+1,1},\ldots,x_{k+1,k+1}; z_{k,1},\ldots,z_{k,k})\times \\ \nonumber \times Q_{C_{k}}^{\,\,\,\,D_k} (z_{k,1},\ldots,z_{k,k};x_{k,1},\ldots,x_{k,k}).\eqa Thus the wave function is given by the integral over a contour of the real dimension equal to a complex dimension of the flag manifold $X=G/B$, where $G=SO(2n,\mathbb{C})$ and $B$ is a Borel subgroup \be \sum_{k=1}^{n-1}\, 2k=n(n-1)=|R_+|. \ee \section{ Givental diagrams and Toric degenerations} In this section we describe combinatorial structure entering the integral representations presented above. This structure reflects flat toric degenerations of the corresponding flag manifolds (see \cite{BCFKS} for $A_n$ and \cite{B} for a general approach to mirror symmetry via degeneration). The combinatorial structure of the integrand readily encoded into the (generalized) Givental diagrams. Let us note that the diagrams for $B_n$, $C_n$ and $D_n$ can be obtained from those for $A_n$ by a factorization. This factorization is closely related with a particular realization \cite{DS} of classical series of Lie algebras as fixed point subalgebras of the involutions acting on an algebra $\mathfrak{gl}_N$ for some $N$. \subsection{ $A_n$ Diagram } Givental diagram \cite{Gi} for $A_n$ has the following form \be\label{AnDiag} \begin{CD}x_{n+1,n+1} @. @. @. @.\\ @AAb_{n,n}A @. @. @. @.\\ x_{n,n} @<<< x_{n+1,n} @. @. @.\\ @AAb_{n-1,n-1}A @AAb_{n,n-1}A @. @. @. @.\\ \vdots @. \vdots @. \ddots @.\\ @AAb_{2,2}A @AAb_{3,2}A @. @. @. @.\\ x_{2,2} @<a_{2,2}<< x_{3,2} @<a_{3,2}<< \ldots @<a_{n,2}<< x_{n+1,2} @.\\ @AAb_{1,1}A @AAb_{2,1}A @. @AAb_{n,1}A @.\\ x_{1,1} @<a_{1,1}<< x_{2,1} @<a_{2,1}<< \ldots @<a_{n-1,1}<< x_{n,1} @<a_{n,1}<< x_{n+1,1}\end{CD}\ee We assign variables $x_{k,i}$ to the vertexes $(k,i)$ and functions $e^{y-x}$ to the arrows $(x\longrightarrow y)$ of the diagram (\ref{AnDiag}). The potential function $\mathcal{F}(x_{k,i})$ (see (\ref{pot})) is given by the sum of the functions assigned to all arrows. Note that the variables $\{x_{k,i}\}$ naturally parametrize an open part $U$ of the flag manifold $X=SL(n+1,\mathbb{C})/B$. The non-compact manifold $U$ has a natural action of the torus and can be compactified to a (singular) toric variety. The set of the monomial relations defining this compactifiaction can be described as follows. Let us introduce the new variables $$a_{k,i}=e^{x_{k,i}-x_{k+1,i}} ,\,\,\,\,\,b_{k,i}=e^{x_{k+1,i+1}-x_{k,i}},\,\,\,\,\,\,\quad 1\leq k\leq n,\,\,1\leq i\leq k\,$$ assigned to the arrows of the diagram (\ref{AnDiag}). Then the following defining relations hold \be\label{defrelAn} a_{k,i}\cdot b_{k,i}\,=\, b_{k+1,i}\cdot a_{k+1,i+1},\qquad 1\leq k< n,\,\,1\leq i\leq k\\ a_{n,i}\cdot b_{n,i}=e^{x_{n,i+1}-x_{n,i}}\ee The defining relations of the toric embedding are given by the monomial relations for the variables associated with the paths on the diagram. They are given by a simple generalization of the relations (\ref{defrelAn}) (see \cite{BCFKS} for details). \subsection{$B_n$ Diagram} Diagram for $B_n$ has the following form $(n=3)$ \be\begin{CD} @. @. @. @Vb_{31}VV @. @.\\ @. @. @>a_{31}>> z_{31} @>c_{31}>> x_{31} @. @.\\ @. @. @Vb_{21}VV @Vd_{21}VV @Vb_{32}VV @.\\ @. @>a_{21}>> z_{21} @>c_{21}>> x_{21} @>a_{32}>> z_{32} @>c_{32}>> x_{32} @.\\ @. @Vb_{11}VV @Vd_{11}VV @Vb_{22}VV @Vd_{22}VV @Vb_{33}VV @.\\ @>a_{11}>> z_{11} @>c_{11}>> x_{11} @>a_{22}>> z_{22} @>c_{22}>> x_{22} @>a_{33}>> z_{33} @>c_{33}>> x_{33}\end{CD}\ee Here we use the same rules for assigning variables to the arrows of the diagram as in $A_n$ case. In addition we assign functions $e^x$ to the arrows $(\longrightarrow x)$. Note that the diagram for $B_n$ can be obtained by factorization of the diagram (\ref{AnDiag}) for $A_{2n}$ by the following involution \be \label{inv} \iota\,:\quad X\longmapsto w_0^{-1}X^T w_0, \ee where $w_0$ is the longest element of $A_{2n}$ Weyl group $\mathfrak{W}(A_{2n})$ isomorphic to a symmetric group $\mathfrak{S}_{2n+1}$ and $X^T$ denotes the standard transposition. Correspondingly the diagram for $B_n$ can be obtained from $A_{2n}$ diagram by the quotient \be \label{w0invBn} w_0\,:\qquad x_{k,i}\longleftrightarrow -x_{k,k+1-i}. \ee An analog of the monomial relations (\ref{defrelAn}) is as follows. Associate to the arrows of Givental diagram parameters \be \,a_{k,i}=e^{z_{k,i}-x_{k-1,i-1}},\,\,\,b_{k,i}=e^{z_{k,i}-x_{k,i-1}},\,\, \,c_{k,i}=e^{z_{k,i}-x_{k,i}},\,\,\,d_{l,j}=e^{x_{l,j}-z_{l+1,j}}\,\,\,\\ \,1\leq k\leq n,\,\,1\leq i\leq k, \quad 1\leq l\leq n-1,\,\, 1\leq j\leq l.\,\ee Then the following relations hold: \bqa a_{k,1}&=&b_{k,1},\qquad \qquad \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\leq k\leq n,\nonumber \\ d_{k,i}\cdot a_{k+1,i+1}\,&=&\,c_{k+1,i}\cdot b_{k+1,i+1},\,\,\,\qquad 1\leq k<n-1,\,\,1\leq i\leq k \\ b_{k,i}\cdot c_{k,i}\,&=&\, a_{k+1,i}\cdot d_{k,i},\qquad \qquad \,1\leq k<n-1,\,\,1\leq i\leq k\nonumber\\ b_{n,i}\cdot c_{n,i}&=&e^{x_{n,i}-x_{n,i-1}}\nonumber\eqa \subsection{ $C_n$ Diagram} Diagram for $C_n$ has the following form $(n=3)$ \be \label{diagrCn}\begin{CD} @. @. z_{33} @. @. @. @.\\ @. @. @| @. @. @. @. @. @.\\ @. z_{22} @= x_{33} @<c_{33}<< z_{33} @.\\ @. @| @Ad_{22}AA @Ab_{32}AA @. @.\\ z_{11} @= x_{22} @<c_{22}<< z_{22} @<a_{32}<< x_{32} @<c_{32}<< z_{32} @.\\ @| @Ad_{11}AA @Ab_{21}AA @Ad_{21}AA @Ab_{31}AA @. @.\\ x_{11} @<c_{11}<< z_{11} @<a_{21}<< x_{21} @<c_{21}<< z_{21} @<a_{31}<< x_{31} @<c_{31}<< z_{31} @.\end{CD}\ee where one assigns functions $e^{-z-x}$ to double arrows $(\begin{CD}x@= z\end{CD})$. The Lie algebra $C_n$ can be realized as a fixed point subalgebra of $A_{2n-1}$ using the involution \be \label{inv} \iota\,:\quad X\longmapsto w_0^{-1}X^T w_0, \ee where $w_0$ is the longest element of Weyl group $\mathfrak{W}(A_{2n-1})=\mathfrak{S}_{2n}$ and $X^T$ denotes the standard transposition. Correspondingly the diagram for $C_n$ can be obtained from $A_{2n-1}$ diagram by the quotient \be \label{w0inv} w_0\,:\qquad x_{k,i}\longleftrightarrow -x_{k,k+1-i}. \ee Note that diagram for $C_n$ can be also obtained by erasing the last row of vertexes and arrows on the right slope from the diagram for $D_{n+1}$ (see (\ref{D4Diag}) below). An analog of the monomial relations (\ref{defrelAn}) is as follows. Let us introduce the variables \be a_{l,j}=e^{z_{l-1,j}-x_{l,j}},\qquad\qquad\qquad\,\,\, \qquad 1<l\leq n,\,\,1\leq j\leq l,\,\, l\neq j,\\ b_{k,i}=e^{z_{k,i+1}-x_{k,i}},\,\,\,\qquad\qquad\qquad\qquad 1\leq k\leq n,\,\, 1\leq i\leq k,\,\,k\neq i,\\ c_{k,i}=e^{x_{k,i}-z_{k,i}},\qquad \qquad\qquad \qquad 1\leq k\leq n,\,\, 1\leq i\leq k,\\ d_{m,j}=e^{x_{m+1,j+1}-z_{m,j}},\qquad\qquad\qquad \qquad 1\leq m<n,\,\,1\leq j\leq m\ee where one assign the variables $b_{11}$, $a_{22}$, $b_{22}$, $a_{33}$, and $b_{33}$ to the left slope of the diagram (\ref{diagrCn}). Then the following relations hold \be c_{k,i}\cdot b_{k,i}\,=\,d_{k,i}\cdot a_{k+1,i+1},\\ a_{k,i}\cdot d_{k-1,i}\,=\,b_{k,i}\cdot c_{k,i+1},\\ c_{n,i}\cdot b_{n,i}=e^{z_{n,i+1}-z_{n,i}},\qquad a_{n,n}\cdot b_{n,n}=e^{-2z_{n,n}}.\ee \subsection{ $D_n$ Diagram} Diagram for $D_n$ has the following form $(n=4)$ \be\label{D4Diag}\begin{CD} \\ @. @. z_{33} @= x_{44} @. @. @.\\ @. @. @| @Ad_{33}AA @. @. @. @. @. @.\\ @. z_{22} @= x_{33} @<c_{33}<< z_{33} @<a_{43}<< x_{43}\\ @. @| @Ad_{22}AA @Ab_{32}AA @Ad_{32}AA @.\\ z_{11} @= x_{22} @<c_{22}<< z_{22} @<a_{32}<< x_{32} @<c_{32}<< z_{32} @<a_{42}<< x_{42}\\ @| @Ad_{11}AA @Ab_{21}AA @Ad_{21}AA @Ab_{31}AA @AAd_{31}A @.\\ x_{11} @<c_{11}<< z_{11} @<a_{21}<< x_{21} @<c_{21}<< z_{21} @<a_{31}<< x_{31} @<c_{31}<< z_{31} @<a_{41}<< x_{41} \end{CD}\ee Note that Lie algebra $D_n$ can be realized as a fixed point subalgebra of $A_{2n-1}$ using the involution \be \label{inv} \iota\,:\quad X\longmapsto w_0^{-1}X^T w_0, \ee where $w_0$ is the longest element of Weyl group $\mathfrak{W}(A_{2n-1})=\mathfrak{S}_{2n}$ and $X^T$ denotes the standard transposition. Correspondingly the diagram for $D_n$ can be obtain from that for $A_{2n-1}$ by the identification of the variables assigned to the vertexes of $A_{2n-1}$ diagram \be \label{w0inv} w_0\,:\qquad x_{k,i}\longleftrightarrow -x_{k,k+1-i}. \ee An analog of the monomial relations (\ref{defrelAn}) is as follows. Let us introduce new variables associate to the arrows of the diagram \be a_{l,j}=e^{z_{l-1,j}-x_{l,j}},\qquad\qquad\qquad\qquad\,\,\qquad\qquad\qquad \qquad 1<l\leq n,\,\,1\leq j\leq l,\\ b_{k,i}=e^{z_{k,i+1}-x_{k,i}}\,\,\,c_{k,i}=e^{x_{k,i}-z_{k,i}},\,\, \,d_{k,i}=e^{x_{k+1,i}-z_{k,i}},\qquad 1\leq k<n,\,\,1\leq i\leq k.\ee The following defining relations hold \be c_{k,i}\cdot b_{k,i}\,=\, d_{k,i}\cdot a_{k+1,i+1},\\ a_{k,i}\cdot d_{k-1,i}\,=\,b_{k,i}\cdot c_{k,i+1},\\ a_{k,i}\cdot d_{k-1,i}=e^{x_{k,i+1}-x_{k,i}}.\ee Finally let us note that it is easy to check that the potentials $\mathcal{F}(x,z)$ obtained from $B_n$, $C_n$ and $D_n$ diagrams by summing the function assigned to the arrows coincide with that entering the integral representations in Section 4. \section{Elementary intertwiners for closed Toda chains} In this section we generalize the construction of the elementary intertwiners to the classical series of affine Lie algebras. For the necessary facts in the theory of affine Lie algebras see \cite{K},\,\cite{DS}. Let us first recall the construction of the $Q$-operator for $A_n^{(1)}$ closed Toda chain~\cite{PG}. The integral kernel of the intertwining $Q$-operator in this case reads \be Q^{A_n^{(1)}}(x_1,\ldots,x_{n+1};\,y_1,\ldots,y_{n+1})=\exp\Big\{\, \sum_{i=1}^{n+1}(e^{x_i-y_i}+g_{i+1}e^{y_{i+1}-x_i})\,\Big\},\quad y_{n+2}=y_1,\ee The corresponding integral operator intertwines the following Hamiltonians operators for $A_n^{(1)}$ closed Toda chains \bqa \CH^{A_n^{(1)}}(x_i)=-\frac{1}{2}\sum_{i=1}^{n+1} \frac{\partial^2}{\partial x_i^2}+g_1e^{x_1-x_{n+1}}+ \sum_{i=1}^ng_{i+1}e^{x_{i+1}-x_i},\\ \CH^{A_n^{(1)}}(y_i)=-\frac{1}{2}\sum_{i=1}^{n+1} \frac{\partial^2}{\partial y_i^2}+g_1e^{y_1-y_{n+1}}+ \sum_{i=1}^ng_{i+1}e^{y_{i+1}-y_i}\,.\eqa Below we provide kernels of the integral operators intertwining Hamiltonians of the closed Toda chains corresponding to other classical series of affine Lie algebras. \subsection{ $A^{(2)}_{2n}\leftrightarrow BC_{n+1}^{(2)}$} Simple roots of the twisted affine root system $A^{(2)}_{2n}$ can be expressed in terms of the standard basis $\{e_i\}$ as follows \be \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-2e_n. \ee Corresponding Dynkin diagram is given by $$\alpha_1 \Longleftarrow\alpha_2\longleftarrow\ldots\longleftarrow\alpha_{n-1} \Longleftarrow\alpha_n. $$ Simple roots of the twisted affine non-reduced root system $BC^{(2)}_{n}$ are given by \be\alpha_0=2e_1,\quad\alpha_1=e_1,\quad \alpha_{i+1}=e_{i+1}-e_i,\,\,\, 1\leq i\leq n-1,\quad \alpha_{n+1}=-e_n-e_{n-1}\ee and the corresponding Dynkin diagram is as follows \bqa\begin{CD}\frac{\alpha_0}{\alpha_1} \Longleftrightarrow\alpha_2@<<< \ldots@<<< \alpha_{n-1} @<<< \alpha_n\\ @. @. @AAA @.\\ @. @. \alpha_{n+1} @.\end{CD}\eqa The integral operator with the following kernel \be \label{inttwA2even} Q_{A^{(2)}_{2n}}^{BC^{(2)}_{n+1}}(x_i,\,z_i)= \exp\Big\{\, g_1e^{z_1}+ \sum_{i=1}^n\Big( e^{x_i-z_i}+g_{i+1}e^{z_{i+1}-x_i}\Big)+ g_{n+2}e^{-z_{n+1}-x_n}\,\Big\},\ee intertwines Hamiltonian operators for $A_{2n}^{(2)}$ and $BC_{n+1}^{(2)}$ \bqa \CH^{A_{2n}^{(2)}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}+g_1e^{x_1}+ \sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}+2g_{n+1}g_{n+2}e^{-2x_{n}},\\ \CH^{BC^{(2)}_{n+1}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^{n+1} \frac{\partial^2}{\partial z_i^2}+ \frac{g_1}{2}\Big(e^{z_1}+g_1e^{2z_1}\Big)+ \sum_{i=1}^{n}g_{i+1}e^{z_{i+1}-z_i}+g_{n+2}e^{-z_{n+1}-z_n}.\eqa The integral kernel for the inverse transformation is given by \be Q^{\,\,\,\,A^{(2)}_{2n}}_{BC^{(2)}_{n+1}}(x_i,\,z_i)=Q_{A^{(2)}_{2n}} ^{\,\,\,\,BC^{(2)}_{n+1}}(z_i,\,x_i).\ee \subsection{ $A^{(2)}_{2n-1}\leftrightarrow A^{(2)}_{2n-1}$} Simple roots of the twisted affine root system $A^{(2)}_{2n-1}$ are given by \be \alpha_1=2e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-e_n-e_{n-1}, \ee and corresponding Dynkin diagram is $$ \begin{CD}\alpha_1 \Longrightarrow \alpha_2 @>>> \ldots @>>> \alpha_{n-1} @>>> \alpha_n\\ @. @. @VVV @.\\ @. @.\alpha_{n+1}@.\end{CD} $$ The integral operator represented by the following kernel \be\label{inttwA2odd} Q_{A^{(2)}_{2n-1}}^{\,\,\,\,A^{(2)}_{2n-1}}(x_i,\,z_i)=\\ =\exp\Big\{\, g_1e^{x_1+z_1}+ \sum_{i=1}^{n-1}\Big( e^{x_i-z_i}+g_{i+1}e^{z_{i+1}-x_i}\Big)+ e^{x_n-z_n}+g_{n+1}e^{-x_{n}-z_n}\,\Big\},\ee intertwines Hamiltonian operators for $A^{(2)}_{2n-1}$ closed Toda chains with different coupling constants \bqa \CH^{A_{2n-1}^{(2)}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}+2g_1e^{2x_1}+\sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}+ g_n g_{n+1}e^{-x_{n}-x_{n-1}},\\ \widetilde{\CH}^{A_{2n-1}^{(2)}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial z_i^2}+ g_1g_2e^{z_1+z_2}+ \sum_{i=1}^{n-1}g_{i+1} e^{z_{i+1}-z_i}+2g_{n+1}e^{-2z_{n}}.\eqa \subsection{$B^{(1)}_n\leftrightarrow BC^{(1)}_{n}$} Simple roots of the affine root system $B^{(1)}_{n}$ are given by \be \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-e_n-e_{n-1}, \ee and corresponding Dynkin diagram is as follows $$ \begin{CD}\alpha_1 \Longleftarrow \alpha_2 @<<< \ldots @<<< \alpha_{n-1} @<<< \alpha_n\\ @. @. @AAA @.\\ @. @.\alpha_{n+1}@.\end{CD} $$ Simple roots of the affine non-reduced root system $BC^{(1)}_{n}$ are \be \alpha_0=2e_1,\quad\alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-2e_n, \ee and corresponding Dynkin diagram is given by $$ \frac{\alpha_0}{\alpha_1} \Longleftrightarrow \alpha_2 \longleftarrow \ldots \longleftarrow \alpha_{n} \Longleftarrow\alpha_{n+1}$$ The integral operator represented by the following kernel \be\label{inttwBn} Q_{B^{(1)}_n}^{\,\,\,\,\,BC^{(1)}_{n}}(x_i,\,z_i)=\\ =\exp\Big\{\, g_1e^{z_1}+ \sum_{i=1}^{n-1}\Big( e^{x_i-z_i}+g_{i+1}e^{z_{i+1}-x_i}\Big)+ e^{x_n-z_n}+g_{n+1}e^{-x_{n}-z_n}\,\Big\},\ee intertwines Hamiltonians of $B^{(1)}_n$ and $BC^{(1)}_{n}$ closed Toda chains \bqa \CH^{B^{(1)}_n}(x_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}+g_1e^{x_1}+\sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}+ g_ng_{n+1}e^{-x_{n}-x_{n-1}},\\ \CH^{BC^{(1)}_{n}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial z_i^2}+ \frac{g_1}{2}\Big(e^{z_1}+g_1 e^{2z_1}\Big)+\sum_{i=1}^{n-1}g_{i+1} e^{z_{i+1}-z_i}+2g_{n+1}e^{-2z_{n}}.\eqa The integral kernel for the inverse transformation is given by \be Q^{\,\,\,\,B^{(1)}_{n}}_{BC^{(1)}_{n}}(x_i,\,z_i)=Q_{B^{(1)}_{n}} ^{\,\,\,\,BC^{(1)}_{n}}(z_i,\,x_i).\ee \subsection{ $C^{(1)}\leftrightarrow D^{(1)}$} Simple roots of the affine root system $C^{(1)}_{n}$ are \be \alpha_1=2e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-2e_n, \ee and corresponding Dynkin diagram is given by $$ \alpha_1\Longrightarrow\alpha_2\longrightarrow\ldots \longleftarrow\alpha_n\Longleftarrow\alpha_{n+1} $$ Simple roots of the affine root system $D^{(1)}_{n}$ are \be \alpha_1=e_1+e_2,\qquad \alpha_{i+1}=e_{i+1}-e_i,\qquad 1\leq i\leq n-1\qquad \alpha_{n+1}=-e_n-e_{n-1}, \ee and corresponding Dynkin diagram is given by $$ \begin{CD}\alpha_1 @<<< \alpha_3 @<<< \ldots @<<< \alpha_{n-1} @<<< \alpha_n\\ @. @AAA @. @AAA @.\\ @. \alpha_2 @. @. \alpha_{n+1} @.\end{CD} $$ The integral operator with the following kernel \be\label{inttwCn} Q_{C^{(1)}_{n}}^{\,\,\,\,D^{(1)}_{n+1}}(x_i,\,z_i)= \exp\Big\{ g_1e^{x_1+z_1}+\sum_{i=1}^{n}\Big( e^{x_i-z_i}+ g_{i+1}e^{z_{i+1}-x_i}\Big)+ g_{n+2}e^{-z_{n+1}-x_n}\Big\}\ee intertwines Hamiltonian operators for $C^{(1)}_n$ and $D^{(1)}_{n+1}$ closed Toda chains \bqa \CH^{C^{(1)}_{n}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}+2g_1e^{2x_1}+ \sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-x_i}+ 2g_{n+1}g_{n+2}e^{-2x_{n}},\\ \CH^{D^{(1)}_{n+1}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^{n+1} \frac{\partial^2}{\partial z_i^2}+g_1g_2e^{z_1+z_2}+ \sum_{i=1}^{n}g_{i+1}e^{z_{i+1}-z_i}+g_{n+2}e^{-z_{n+1}-z_{n}}.\eqa The integral operator with the kernel \be\label{inttwDn} Q_{D^{(1)}_{n}}^{\,\,\,\,C^{(1)}_{n-1}}(x_i,\,z_i)=\exp\Big\{\, g_1e^{x_1+z_1}+ \sum_{i=1}^{n-1}\Big(e^{z_i-x_i}+ g_{i+1}e^{x_{i+1}-z_i}\Big)+ g_{n+1}e^{-x_{n}-z_{n-1}}\,\Big\}\ee intertwines Hamiltonian operators for $D^{(1)}_n$ and $C^{(1)}_{n-1}$ closed Toda chains \bqa \CH^{D^{(1)}_{n}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^{n} \frac{\partial^2}{\partial x_i^2}+g_1g_2e^{x_1+x_2}+ \sum_{i=1}^{n-1}g_{i+1}e^{x_{i+1}-z_i}+g_{n+1}e^{-x_{n}-x_{n-1}},\\ \CH^{C^{(1)}_{n-1}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^{n-1} \frac{\partial^2}{\partial z_i^2}+2g_1e^{2z_1}+ \sum_{i=1}^{n-2}g_{i+1}e^{z_{i+1}-z_i}+ 2g_{n}g_{n+1}e^{-2z_{n-1}}.\eqa The integral kernel for the inverse transformation is given by \be Q^{\,\,\,\,D^{(1)}_{n}}_{C^{(1)}_{n}}(x_i,\,z_i)=Q_{D^{(1)}_{n}} ^{\,\,\,\,C^{(1)}_{n}}(z_i,\,x_i).\ee \section{Baxter $Q$-operators} Now we apply the results presented in the previous Sections to the construction of the Baxter integral $Q$-operators for all classical series of affine Lie algebras. Let us note that the elementary intertwining operators and recursive operators for finite Lie algebras can be obtained from the elementary intertwining operators and Baxter operators for affine Lie algebras by taking appropriate limits $g_i\rightarrow0$ in (\ref{inttwA2even})-(\ref{inttwDn}). This generalizes known relation between Baxter operator for $A_n^{(1)}$ and recursive operators for $A_n$. The integral kernels for $Q$-operators have the following form \bqa Q^{A^{(2)}_{2n}}(x_1,\ldots,x_n;\,y_1,\ldots,y_n)=\int \bigwedge_{i=1}^{n+1}dz_i Q_{A^{(2)}_{2n}}^{\,\,\,\,BC^{(2)}_{n+1}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n+1})\times \\ \nonumber \times Q_{BC^{(2)}_{n+1}}^{\,\,\,\,A^{(2)}_{2n}}(z_1,\ldots,z_{n+1};\,y_1,\ldots,y_n),\eqa \bqa Q^{A^{(2)}_{2n-1}}(x_1,\ldots,x_n;\,y_1,\ldots,y_n)=\int \bigwedge_{i=1}^{n}dz_i\, Q_{A^{(2)}_{2n-1}}^{\,\,\,\,A^{(2)}_{2n-1}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n})\times \\ \nonumber \times Q_{A^{(2)}_{2n-1}}^{\,\,\,\, A^{(2)}_{2n-1}}(y_1,\ldots,y_{n};\,z_1,\ldots,z_n),\eqa \bqa Q^{B^{(1)}_{n}}(x_1,\ldots,x_n;\,y_1,\ldots,y_n)=\int \bigwedge_{i=1}^{n}dz_i\, Q_{B^{(1)}_{n}}^{\,\,\,\,BC^{(1)}_{n}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n}) \times \\ \nonumber Q_{BC^{(1)}_{n}}^{\,\,\,\,B^{(1)}_{n}}(z_1,\ldots,z_{n};\,y_1,\ldots,y_n),\eqa \bqa Q^{C^{(1)}_{n}}(x_1,\ldots,x_n;\,y_1,\ldots,y_n)=\int \bigwedge_{i=1}^{n+1}dz_i\, Q_{C^{(1)}_{n}}^{\,\,\,\,D^{(1)}_{n+1}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n+1}) \times \\ \nonumber Q_{D^{(1)}_{n+1}}^{\,\,\,\,C^{(1)}_{n}}(z_1,\ldots,z_{n+1};\,y_1,\ldots,y_n),\eqa \bqa Q^{D^{(1)}_{n}}(x_1,\ldots,x_n;\,y_1,\ldots,y_n)=\int \bigwedge_{i=1}^{n-1}dz_i\, Q_{D^{(1)}_{n}}^{\,\,\,\,C^{(1)}_{n-1}}(x_1,\ldots,x_n;\,z_1,\ldots,z_{n-1}) \times \\ \nonumber Q_{C^{(1)}_{n-1}}^{\,\,\,\,D^{(1)}_{n}}(z_1,\ldots,z_{n-1};\,y_1,\ldots,y_n).\eqa \section{Baxter operators for $B_{\infty}$, $C_{\infty}$ and $D_{\infty}$} Similar approach can be applied to construct Baxter $Q$ operators for infinite root systems $B_{\infty}$, $C_{\infty}$, $BC_{\infty}$ and $D_{\infty}$. Simple roots and Dynkin diagrams for infinite Lie algebras $A_{\infty}$, $B_{\infty}$, $C_{\infty}$, $BC_{\infty}$ and $D_{\infty}$ are as follows \be A_{\infty}:\,\,\,\,\,\,\,\,\,\,\,\,\,\qquad\qquad\,\, \qquad \alpha_{i+1}=e_{i+1}-e_i, \qquad \quad i\in\mathbb{Z}, \ee \be\begin{CD} \ldots @>>> \alpha_{-1}\,@>>>\,\alpha_0 @>>> \alpha_1 @>>> \ldots \end{CD}\ee \be B_{\infty}:\,\,\,\,\,\,\,\, \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i, \qquad\qquad i\in\mathbb{Z}_{>0}, \ee \be\begin{CD} \alpha_1\,\Longleftarrow\,\alpha_2 @<<< \alpha_3 @<<< \ldots \end{CD}\ee \be C_{\infty}:\,\,\,\,\,\,\, \alpha_1=2e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i, \qquad \qquad i\in\mathbb{Z}_{>0}, \ee \be\begin{CD} \alpha_1\,\Longrightarrow\,\alpha_2 @>>> \alpha_3 @>>> \ldots \end{CD}\ee \be D_\infty:\,\,\,\,\,\,\ \alpha_1=e_1+e_2,\qquad \alpha_{i+1}=e_{i+1}-e_i, \qquad\quad i\in\mathbb{Z}_{>0}, \ee \be \begin{CD}\alpha_1 @>>> \alpha_3 @>>> \alpha_4 @>>> \ldots\\ @. @AAA @. @.\\ @. \alpha_2 @. @. \end{CD} \ee \be BC_{\infty}:\,\,\,\,\,\, \alpha_0=2e_1,\qquad \alpha_1=e_1,\qquad \alpha_{i+1}=e_{i+1}-e_i, \qquad i\in\mathbb{Z}_{>0}, \ee \be\begin{CD}\frac{\alpha_0}{\alpha_1}\Longleftrightarrow \alpha_2 @<<< \alpha_3 @<<<\ldots\end{CD}\ee Baxter $Q$-operator for $A_\infty$ infinte Toda chain is known (see \cite {T} for the classical limit). It is given by an integral operator with the kernel \be Q(x_i,\,y_i)=\exp\Big\{\,\sum_{i\in\mathbb{Z}}\left( e^{x_i-y_i}+g_ie^{y_{i+1}-x_i}\right)\,\Big\},\ee and intertwines $A_{\infty}$ Toda chain Hamiltonians \bqa \CH^{A_{\infty}}(x_i)&=&-\frac{1}{2}\sum_{i\in\mathbb{Z}}\frac{\partial^2}{\partial x_i^2}+\sum_{i\in \mathbb{Z}} g_i e^{x_{i+1}-x_i},\\ \CH^{A_{\infty}}(y_i)&=&-\frac{1}{2}\sum_{i\in\mathbb{Z}}\frac{\partial^2}{\partial y_i^2}+\sum_{i\in \mathbb{Z}}g_ie^{y_{i+1}-y_i}.\eqa Its generalization to other classical series relies on the construction of the integral operators intertwining different classical series. Thus we have the following set of integral operators. The integral operator with the kernel \be Q_{B_\infty}^{\,\,\,\,\,BC_\infty}(x_i,\,z_i)=\exp\Big\{\, g_1e^{z_1}+ \sum_{i>0} \left(e^{x_i-z_i}+g_{i+1}e^{z_{i+1}-x_i}\right)\,\Big\},\ee intertwines $B_\infty$ and $BC_\infty$ Toda chain Hamiltonian operators \bqa \CH^{B_{\infty}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial x_i^2} +g_1e^{x_1}+\sum_{i=1}^\infty g_{i+1}e^{x_{i+1}-x_i},\\ \CH^{BC_{\infty}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial z_i^2} +\frac{g_1}{2}\Big(e^{z_1}+g_1e^{2z_1}\Big)+\sum_{i=1}^\infty g_{i+1}e^{z_{i+1}-z_i}.\eqa $Q$-operator for $B_\infty$ is then obtained by composition of the intertwiner operators. For the integral kernel of the $Q$-operator for $B_{\infty}$ Toda chain we have $$Q^{B_\infty}(x_i;y_i)=\int \bigwedge_{i=1}^\infty dz_i\, Q_{B_\infty}^{\,\,\,\,BC_\infty}(x_i,\,z_i) \cdot Q_{BC_\infty}^{\,\,\,\,\,B_\infty}(z_i,\,y_i).$$ Similarly the integral operator with the kernel \be Q_{C_\infty}^{\,\,\,\,D_\infty}(x_i,\,z_i)= \exp\Big\{\,g_1e^{x_1+z_1}+ \sum_{i>0} \left(e^{x_i-z_i}+g_{i+1}e^{z_{i+1}-x_i}\right)\,\Big\},\ee intertwines $C_\infty$ and $D_\infty$ Toda chain Hamiltonian operators \bqa \CH^{C_{\infty}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial x_i^2} +2g_1e^{2x_1}+\sum_{i=1}^\infty g_{i+1}e^{x_{i+1}-x_i},\\ \CH^{D_{\infty}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial z_i^2} +g_1g_2e^{z_1+z_2}+\sum_{i=1}^\infty g_{i+1}e^{z_{i+1}-z_i}.\eqa Thus integral $Q$-operator for $C_\infty$ has as the kernel $$Q^{C_\infty}(x_i;y_i)=\int \bigwedge_{i=1}^\infty dz_i\, Q_{C_\infty}^{\,\,\,\,\,D_\infty}(x_i,\,z_i) \cdot Q_{D_{\infty}}^{\,\,\,\,C_{\infty}}(z_i,\,y_i).$$ The integral operator with the kernel \be Q_{D_\infty}^{\,\,\,\,C_\infty}(z_i,\,x_i)=\exp\Big\{\, g_1e^{x_1+z_1}+ \sum_{i>0}\left( e^{z_i-x_i}+g_{i+1}e^{x_{i+1}-z_i}\right)\,\Big\},\ee intertwines $D_{\infty}$ and $C_{\infty}$ Toda chain Hamiltonian operators \bqa \CH^{D_{\infty}}(x_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial x_i^2} +g_1g_2e^{x_1+x_2}+\sum_{i=1}^\infty g_{i+1}e^{x_{i+1}-x_i},\\ \CH^{C_{\infty}}(z_i)&=&-\frac{1}{2}\sum_{i=1}^\infty\frac{\partial^2}{\partial z_i^2} +2g_1e^{2z_1}+\sum_{i=1}^\infty g_{i+1}e^{z_{i+1}-z_i}.\eqa Therefore Baxter integral $Q$-operator for $D_\infty$ has the following kernel $$ Q^{D_\infty}(x_i;y_i)=\int \bigwedge_{i=1}^\infty dz_i\cdot Q_{D_\infty}^{\,\,\,\,C_\infty}(x_i,\,z_i) \cdot Q_{C_\infty}^{\,\,\,\,D_\infty}(z_i,\,y_i).$$ \section{Conclusions} In this note we provide explicit expressions for recursive operators and Baxter $Q$-operators for classical series of Lie algebras. This allows us to generalize Givental representation for the eigenfunctions of the open Toda chain quadratic Hamiltonians to all classical series. In this note we consider only the case of zero eigenvalues leaving a rather straightforward generalization to non-zero eigenvalues to another occasion\cite{GLO}. The proof of the eigenfunction property for the full set of Toda chain Hamiltonians follows the same strategy as in \cite{Gi},\,\cite{GKLO} and will be published separately. The results presented in this note provide a generalization to other classical series of only a part of \cite{GKLO}. The other part connected with the interpretation of elementary intertwining operators in representation theory framework will discussed elsewhere. Let us note that $Q$-operator was introduced by Baxter as a key tool to solve quantum integrable systems (see \cite{B}). Therefore one can expect that the Givental representation and its generalizations should play an important role in the theory of quantum integrable systems solved by the quantum inverse scattering method (see e.g. \cite{Fa}). Finally let us mention two possible applications of the obtained results. The generalization of the Givental integral representation and corresponding diagrams to other classical series allows to describe flat torifications of the corresponding flag manifolds. This provides interesting applications to the explicit description of the mirror symmetry for flag manifolds. In particular one expects the mirror/Langlands duality between Gromov-Witten invariants for flag manifolds and the corresponding period integrals for $B_n$ and $C_n$ series. Another interesting application connected with the theory of automorphic forms and corresponding $L$-functions. Note that open Toda chain wave functions are given (see e.g. \cite{STS}) by the Whittaker functions \cite{Ha} for the corresponding Lie groups. Taking into account the simplicity and the unified form of the arising constructions of Whittaker functions for all classical groups one can expect important computational advantages in using the proposed integral representations.

TITLE: Galois $\text{GF}(5)$, does $α^4=2$ make any sense? QUESTION [3 upvotes]: I just want to see if this is a misprint or am I making some really silly mistake. the question says let $α$ be a root of $f(x) = x^4 − 2$ over $\text{GF}(5)$. Obviously what one should do in such cases is take it s.t. $α^4=2$. However my difficulty arises through the fact that this doesn't seem to be true for any element of $\text{GF}(5):=\{0,1,2,3,4\}$. what am I doing wrong, because honestly I doubt its a misprint, and suspect it's much more likely that because I'm very new to galois fields that I'm doing something incorrectly. REPLY [2 votes]: I think there are some important things to remark concerning what you wrote above: To a). Your calculation is absolutely correct, but the fact that the question as you posted it in your comment speaks about explaining an "obvious" fact in "a few sentences", I would suspect that it aimed more for something like the following (I'm not entirely sure though, as the ordering of the given subtasks seems to disagree with my interpretation): Suppose we already know that $f(x)=x^4-2$ is irreducible over $\mathrm{GF}(5)$ (part d)). This would tell us that the degree of $\alpha$ over $\mathrm{GF}(5)$ is $4$ (part c)), so $K:=\mathrm{GF}(5)(\alpha)$ is a field with $5^4=625$ Elements. Its multiplicative group $K^\times=K\setminus\{0\}$ has order $\lvert K^\times \rvert=624$. So $a^{624}=1$ for any $a\in K^\times$, in particular $$a^{625}=a\cdot a^{624}=a\cdot 1=a,\quad\text{for any }a\in K^\times$$ Remark: Obviously, this equality holds for $a=0$ as well, so that every $a\in K$ is a root of $x^{625}-x\in\mathrm{GF}(5)[x]$. Considering that a polynomial of degree $n$ over a field can have only a maximum of $n$ roots (in some algebraic closure), one thus gets that every finite field $L$ of characteristic $p$ is a splitting field of $x^{\lvert L\rvert}-x$ over $\mathrm{GF}(p)$. To c). Here I am not sure what the first line of your solution is used for? An later on, I think it would have been more precise to explicitly give some $\mathrm{GF}(5)$-Basis of $\mathrm{GF}(5)(\alpha)$, you might for example prove that $\{1,\alpha,\alpha^2,\alpha^3\}$ is such a basis (your argument already shows it is generating, so you only had to use the irreducibility of $f$ to prove linear independence), hence $$[\mathrm{GF}(5)(\alpha):\mathrm{GF}(5)]=\dim_{\mathrm{GF}(5)}(\mathrm{GF}(5)(\alpha))=4.$$ To d). How exactly did you end up with only checking there is no $a\in \mathrm{GF}(5)$ with $a^2=2$ to ensure that $f$ has no irreducible factor of degree $2$? You were definitely right about checking for roots to exclude factors of degree $1$, but the point of excluding factors of degree $2$ could need some supplementation, I think. Basically, there are two natural ways to do this (unfortunately both are a bit lengthy): As $f$ is monic, you may without loss of generality assume both degree $2$ factors monic. So you could multiply out the assumption $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ compare the coefficients and try to solve for $a,b,c,d\in \mathrm{GF}(5)$, getting a contradiction. As $\mathrm{GF}(5)$ consists of only five elements, there are just $5^2=25$ monic polynomial of degree $2$ in $\mathrm{GF}(5)[x]$. By testing for roots in $\mathrm{GF}(5)$ you might find and list all irreducible ones, and then, for any of these, test if it divides $f$. To e). If I'm not mistaking what you wrote, this is an important misconception, I think! You could of course label $\alpha$ as $\sqrt[4]{2}$, which makes perfect sense, but the use of $i$ suggests you are thinking about complex numbers. However, the question happens in characteristic $5$! The procedure is similar, though. You already know one root of $f$ in $K=\mathrm{GF}(5)(\alpha)$, which is $\alpha$ (by definition). Now, what are the other roots? This is where the roots of unity come in. By definition, a $n$-th root of unity over a field $k$ (in some fixed algebraic closure of $k$) is nothing but a root of the polynomial $x^n-1\in k[x]$, that is, the roots of unity are the elements $\zeta $ with $\zeta ^n=1$. Now, if $\alpha$ is a root of $f$, that is $\alpha^4=2$, and $\zeta$ is a fourth root of unity over $\mathrm{GF}(5)$, then $$(\zeta\alpha)^4=\zeta^4\alpha^4=\alpha^4=2,$$ so $\zeta\alpha$ is a root of $f$ as well, for any fourth root of unity $\zeta$. Now $x^4-1\in\mathrm{GF}(5)[x]$ is separable (as the characteristic $5$ does not divide $4$), so there are four different 4th roots of unity over $\mathrm{GF}(5)$, call them $1,\zeta,\eta,\tau$ (actually, the group of roots of unity is cyclic as a finite subgroup of a multiplicative group of a field, so we might as well write them as $1,\zeta,\zeta^2,\zeta^3$), and the four elements $\alpha,\zeta\alpha,\eta\alpha,\tau\alpha$ are all roots of $f$. As $f$ is of degree $4$, these are of course all roots. Now, what are the fourth roots of unity over $\mathrm{GF}(5)$? As the multiplicative group $\mathrm{GF}(5)^\times$ has order $4$, we immediately have $a^4=1$ for all $a\in \mathrm{GF}(5)^\times$ - that means, the $4$-th roots of unity are nothing but $1,2,3,4\in\mathrm{GF}(5)$, and the roots of $f$ in $\mathrm{GF}(5)(\alpha)$ are thus $$\alpha,2\alpha,3\alpha,4\alpha.$$

According to the movies, America is currently in the midst of breast-beating and garment-rending over its own cruel immigration policy. It's tough to get to the land of plenty, but plenty are still willing to risk their lives to try. Latest entry in the timely film topic is Sin Nombre, a big winner at Sundance this year for director Cary Fukunaga -- who, incredibly, makes his feature debut with this stunning film. Sin Nombre is the story of two seemingly very different people, one a quiet Honduran girl and the other a violent Mexican gang member, who team up to survive the trek to the border of Texas. You can argue the ethics of using poverty and desperation as movie fodder on your own time. Sin Nombre is a spare, intelligent, visually transporting film, and despite some moments of somewhat shaky logic, it's also some gripping storytelling. Sin Nombre, which translates 'nameless', begins in Mexico with the life of Casper (Edgar Flores). He's a young man whose membership in the Mara Salvatrucha brotherhood is the centre of his existence, and he has recruited Smiley, a gang wannabe aged about 12. Smiley (Kristian Ferrer) gets viciously kicked by the whole gang as part of his initiation, and is later told he'll have to kill one of the gang's enemies as part of his initiation. The violence throughout Sin Nombre is quietly brutal and realistic, and endlessly disturbing. In a parallel story we meet Sayra (Paulina Gaitan) a teenage girl in Honduras who is going to try to get to the border of Texas with her uncle and her father. Her dad has been living in New Jersey with his new family, but was deported. He intends to return to the Garden State via illegal entry, and take his old family along with him. "Half of these people will not make it, but we will," Sayra's father predicts. The Hondurans make their way to Tapachula, Mexico, and wait in the train yards for a train that they can ride -- on the top, not inside -- to the U.S. border. That's where they encounter Casper and a few of the Mara; as the gang violence escalates, Casper has to go on the run from the rest of the brotherhood. Sayra attaches herself to the fierce Casper, and before long they are on the move together. Dodging immigration agents and vengeful gang members, they move toward the border of the United States. Sin Nombre is distinguished by elegant, understated performances, and better yet, they come from a cast that's more or less unknown to North American movie audiences -- the better to willingly suspend your disbelief, my dear. Both a drama and a thriller, Sin Nombre was nominated for the Grand Jury Prize at Sundance this year and won awards for director Cary Fukunaga and cinematographer Adriano Goldman. It won't disappoint you. That's rare, don't you think? The movie is in Spanish with English subtitles. (This film is rated 14-A)

\begin{document} \begin{abstract} We compute the expected number of commutations appearing in a reduced word for the longest element in the symmetric group. The asymptotic behavior of this value is analyzed and shown to approach the length of the permutation, meaning that nearly all positions in the reduced word are expected to support commutations. Finally, we calculate the asymptotic frequencies of commutations, consecutive noncommuting pairs, and long braid moves.\\ \noindent \emph{Keywords:} permutation, reduced word, commutation \end{abstract} \maketitle \section{Introduction} The symmetric group $\mathfrak{S}_n$ is generated by the simple reflections $\{s_1,\ldots, s_{n-1}\}$, where $s_i$ interchanges $i$ and $i+1$. These reflections satisfy the Coxeter relations \begin{eqnarray} \nonumber &s_i^2=1& \text{ for all $i$,}\\ \label{eqn:short} &s_is_j = s_js_i& \text{ for $|i - j| > 1$, and}\\ \label{eqn:long} &s_is_{i+1}s_i = s_{i+1}s_is_{i+1}.& \end{eqnarray} \begin{defn} Relations as in equation~\eqref{eqn:short} are called \emph{commutations} (also, \emph{short braid relations}), and those as in equation~\eqref{eqn:long} are \emph{long braid relations} (also, \emph{Yang-baxter moves}). \end{defn} There is a well-defined notion of ``length'' for a permutation, and it can be defined in terms of these simple reflections $\{s_i\}$. \begin{defn} Consider $w \in \mathfrak{S}_n$ written as a product $w = s_{i_1}s_{i_2}\cdots s_{i_{\ell(w)}}$ for $1 \le i_j \le n-1$, where $\ell(w)$ is minimal. This $\ell(w)$ is the \emph{length} of $w$, while the product $s_{i_1}s_{i_2}\cdots s_{i_l}$ is a \emph{reduced decomposition} for $w$ and the string of subscripts $i_1i_2\cdots i_l$ is a \emph{reduced word} for $w$. \end{defn} The Coxeter relations \eqref{eqn:short} and \eqref{eqn:long} have obvious analogues in reduced words. Namely, the letters $\{i,j\}$ commute when $|i - j| > 1$, and $i(i+1)i$ can be rewritten as $(i+1)i(i+1)$. We will abuse terminology slightly and refer to these phenomena in reduced words as commutations and long braid moves, respectively, as well. The structure of reduced words, and the influence of commutations and long braid moves, is of great interest. Some facts are known, such as those cited below, but many open questions remain. Additionally, there are objects defined naturally in terms of reduced words and Coxeter relations, including commutation classes and the graph of those classes (see, for example, \cite{elnitsky} and \cite{tenner rdpp}), for which much of the architecture is still largely obscure. \begin{defn} There is a \emph{longest element} $w_0 \in \mathfrak{S}_n$, and $\ell(w_0) = \binom{n}{2}$. In one-line notation, this $w_0$ is the permutation $n(n-1)(n-2)\cdots 321$. \end{defn} In \cite{reiner}, Reiner showed the following surprising fact about long braid moves in reduced words for $w_0 \in \mathfrak{S}_n$. \begin{thm}[{\cite[Theorem 1]{reiner}}]\label{thm:reiner} For all $n \ge 3$, the expected number of long braid moves in a reduced word for $w_0 \in \mathfrak{S}_n$ is $1$. \end{thm} Similar calculations were made for the finite Coxeter group of type $B$ in \cite{tenner expb}, one of which is restated here. \begin{thm}[{\cite[Theorem 3.1]{tenner expb}}]\label{thm:expb} For all $n \ge 3$, the expected number of long braid moves in a reduced word for the longest element in the Coxeter group of type $B_n$ is $2 - 4/n$. \end{thm} Recent work on random reduced words for the longest element appears under the guise of ``random sorting networks,'' as in \cite{ahrv, ah, agh, warrington}. In the present article, we compute the expected number of commutations in a reduced word for the longest element $w_0 \in \mathfrak{S}_n$. We obtain an explicit sum for this expectation, which is stated in Theorem~\ref{thm:main}. Unfortunately, occurrences of commutations are not as tidy as occurrences of long braid moves, and so our result is not independent of $n$ as in Reiner's work \cite{reiner}, nor of such straightforward form as the results of \cite{tenner expb}. In the final section of the paper, we discuss the asymptotic behavior of the enumeration from Theorem~\ref{thm:main}. Finally, we use that analysis to produce Corollary~\ref{cor:proportions}, giving the asymptotic frequencies of commutations, consecutive noncommuting pairs, and long braid moves in reduced words for the longest element. \section{Enumerating commutations} Throughout this section, fix a positive integer $n$ and consider $w_0 \in \mathfrak{S}_n$. Let $\ell = \ell(w_0) = \binom{n}{2}$. \begin{defn} For any integer $k \in [1,\ell-1]$, say that a reduced word $i_1 \cdots i_{\ell}$ for $w_0$ \emph{supports a commutation in position $k$} if $$|i_k - i_{k+1}| > 1.$$ \end{defn} We want to compute the expected number of commutations in a reduced word for the longest element $w_0 \in \mathfrak{S}_n$. In particular, we are interested in the following statistic. \begin{defn} Let $C_n$ be the random variable on a reduced word for $w_0 \in \mathfrak{S}_n$, which counts positions in the word that support commutations. \end{defn} To analyze $C_n$, we will actually look at the complementary event; that is, consecutive noncommuting pairs of symbols in a reduced word. \begin{defn} Consider a reduced word $i_1i_2\cdots i_{\ell}$ for $w_0$. Let $A_n^{(k,j)}$ be the indicator random variable for the event that $i_k = j$ and $i_{k+1} = j+1$, and let $$A_n = \sum_{k=1}^{\ell-1}\sum_{j=1}^{n-2}A_n^{(k,j)}.$$ Similarly, let $B_n^{(k,j)}$ be the indicator random variable for the event that $i_k = j+1$ and $i_{k+1} = j$, and let $$B_n = \sum_{k=1}^{\ell-1}\sum_{j=1}^{n-2}B_n^{(k,j)}.$$ \end{defn} Therefore $E(A_n)+E(B_n)$ computes the expected number of consecutive noncommuting pairs in a reduced word for $w_0$, and so \begin{equation}\label{eqn:complement} E(C_n) = \ell-1 - \big(E(A_n) + E(B_n)\big). \end{equation} In fact, the symmetry $s_{n-1-i}w_0s_i = w_0$ gives a correspondence between $A_n^{(k,j)}$ and $B_n^{(k,n-1-j)}$, and so we can simplify equation~\eqref{eqn:complement} to \begin{equation}\label{eqn:complement2} E(C_n) = \ell-1 - 2E(A_n). \end{equation} The proof of Theorem~\ref{thm:main} will employ similar techniques to those used by Reiner in \cite{reiner}. In particular, we will recognize that particular permutations are vexillary, and then use a result of Stanley to enumerate reduced words. \begin{defn} A permutation is \emph{vexillary} if it avoids the pattern $2143$. That is, $w(1)\cdots w(n)$ is vexillary if there are no indices $1 \le i_1 < i_2 < i_3 < i_4 \le n$ such that $w(i_2) < w(i_1) < w(i_4) < w(i_3)$. \end{defn} Stanley showed in \cite{stanley} that the reduced words for a vexillary permutation are enumerated by standard Young tableaux of a particular shape, which can be enumerated using the hook-length formula, as discussed in \cite{stanley ec2}. \begin{defn} Fix a permutation $w = w(1)\cdots w(n)$. For all integers $i \in [1,n]$, let $$r_i = \left|\{j : j < i \text{ and } w(j) > w(i)\}\right|.$$ The partition $\lambda(w) = (\lambda_1(w),\lambda_2(w),\ldots)$ is defined by writing $\{r_1,\ldots,r_n\}$ in nonincreasing order. \end{defn} \begin{thm}[{\cite[Corollary 4.2]{stanley}}]\label{thm:stanley} If $w$ is vexillary, then the number of reduced words for $w$ is equal to $f^{\lambda(w)}$, the number of standard Young tableaux of shape $\lambda(w)$. \end{thm} We are now able to compute the expected number of commutations in a reduced word for $w_0 \in \mathfrak{S}_n$. \begin{thm}\label{thm:main} For all $n \ge 3$, $$E(C_n) = \binom{n}{2} - 1 - \frac{1}{3\binom{n}{2} 2^{2n-7}} \sum_{j=1}^{n-2} \frac{(2j-1)!!}{(j-1)!}\cdot \frac{(2j+1)!!}{j!}\cdot \frac{(2n-2j-3)!!}{(n-j-2)!}\cdot \frac{(2n-2j-1)!!}{(n-j-1)!}.$$ \end{thm} \begin{proof} Fix an integer $n \ge 3$ and let $\ell = \binom{n}{2}$. To calculate the expected number of positions that support a commutation in a reduced word for $w_0 \in \mathfrak{S}_n$, we will analyze the complementary event and use equation~\eqref{eqn:complement2}. As noted in \cite{reiner}, the fact that $s_iw_0s_{n-i} = w_0$ for all $i \in [1,n-1]$ gives a cyclic symmetry to reduced words for $w_0$: $$i_1i_2\cdots i_{\ell}$$ is a reduced word for $w_0$ if and only if $$i_2i_3 \cdots i_{\ell}(n-i_1)$$ is as well. Therefore it is enough to consider the appearance of a commutation (or not) in the first position of a reduced word, using the identity $$A_n^{(k,j)} = A_n^{(1,j)}$$ for all $1 \le k \le \ell-1$. Thus equation~\eqref{eqn:complement2} can be rewritten as $$E(C_n) = \ell - 1 - 2(\ell - 1)\sum_{j=1}^{n-2} E(A_n^{(1,j)}).$$ The expected value $E(A_n^{(1,j)})$ is equal to the probability that the reduced word for $w_0$ is $$j(j+1)i_3i_4\cdots i_{\ell},$$ in which case $i_3i_4\cdots i_{\ell}$ is a reduced word for the permutation \begin{eqnarray*} a_n^{(j)} &=& s_{j+1}s_jw_0\\ &=& n(n-1)(n-2)\cdots(j+3)(j+1)j(j+2)(j-1)\cdots321, \end{eqnarray*} in one-line notation. The permutation $a_n^{(j)}$ is vexillary, meaning that we can enumerate its reduced words by counting standard Young tableaux as described in Theorem~\ref{thm:stanley}. The shape $\lambda(a_n^{(j)})$ is obtained from the staircase shape $\delta_n$ by deleting the corner squares from rows $j$ and $j+1$ as in Figure~\ref{fig:shapes}. Therefore, \begin{equation}\label{eqn:expectation in terms of shapes} E(C_n) = \ell - 1 - 2(\ell - 1)\sum_{j=1}^{n-2} \frac{f^{\lambda(a_n^{(j)})}}{f^{\delta_n}}. \end{equation} \begin{figure}[htbp] \begin{tikzpicture}[scale=.5] \fill[black!20] (0,5) rectangle (5,6); \fill[black!20] (0,6) rectangle (6,7); \fill[black!20] (4,7) rectangle (6,9); \foreach \x in {1,2,3,4,5,6,7,8} {\draw (0,\x) --+ (\x,0); \draw(\x,9) --+ (0,\x-9);} \draw (0,1) -- (0,9) -- (8,9); \draw (5,1) node {$\delta_9$}; \draw (6,1) node {\phantom{$\lambda(a_9^{(3)})$}}; \end{tikzpicture} \hspace{.5in} \begin{tikzpicture}[scale=.5] \fill[black!20] (0,5) rectangle (4,6); \fill[black!20] (0,6) rectangle (5,7); \fill[black!20] (4,7) rectangle (6,9); \foreach \x in {1,2,3,4,5,6,7,8} {\draw (0,\x) --+ (\x,0); \draw(\x,9) --+ (0,\x-9);} \draw (0,1) -- (0,9) -- (8,9); \draw (5,1) node {$\lambda(a_9^{(3)})$}; \fill[white] (4.05,4.8) rectangle (5.2,5.95); \fill[white] (5.05,5.8) rectangle (6.2,6.95); \foreach \x in {(4.5,5.5),(5.5,6.5)} {\fill[black] \x circle (4pt);} \end{tikzpicture} \caption{The staircase shape $\delta_9$ corresponding to $w_0 \in \mathfrak{S}_9$ and the shape $\lambda(a_9^{(3)})$. The shaded cells are the only ones in which the hook-lengths for the two shapes will differ.} \label{fig:shapes} \end{figure} We evaluate equation~\eqref{eqn:expectation in terms of shapes} using the hook-length formula. In fact, many hook-lengths of $\delta_n$ and $\lambda(a_n^{(j)})$ are the same, and thus cancel in the ratio $$\frac{f^{\lambda(a_n^{(j)})}}{f^{\delta_n}}.$$ The only non-identical hook-lengths in the shapes $\delta_n$ and $\lambda(a_n^{(j)})$ occur as indicated in Figure~\ref{fig:shapes}, and so equation~\eqref{eqn:expectation in terms of shapes} becomes $$E(C_n) = \ell - 1 - 2(\ell - 1)\sum_{j=1}^{n-2} \frac{(\ell-2)!}{\ell!}\cdot3 \cdot \left(\frac{2}{3}\right)^2 u_{j-1} u_j u_{n-j-2} u_{n-j-1},$$ where $$u_i = \frac{3\cdot5\cdots(2i+1)}{2\cdot4\cdots(2i)}.$$ Thus \begin{eqnarray*} E(C_n) &=& \ell - 1 - \frac{8}{3\ell}\sum_{j=1}^{n-2} u_{j-1}u_ju_{n-j-2}u_{n-j-1}\\ &=&\ell - 1 - \frac{1}{3\ell 2^{2n-7}} \sum_{j=1}^{n-2} \frac{(2j-1)!!}{(j-1)!}\cdot \frac{(2j+1)!!}{j!}\cdot \frac{(2n-2j-3)!!}{(n-j-2)!}\cdot \frac{(2n-2j-1)!!}{(n-j-1)!}, \end{eqnarray*} completing the proof. \end{proof} \section{Asymptotic behavior} In this last section of the paper, we analyze the expected value $E(C_n)$ as $n$ goes to infinity. In fact, we will draw conclusions about appearances of all three possibilities of consecutive symbols in a reduced word for the longest element: commutations, consecutive noncommuting pairs $i(i+1)$ or $(i+1)i$, and long braid moves. Set $$\sigma_n^{(j)} = \frac{1}{3\binom{n}{2} 2^{2n-7}} \cdot \frac{(2j-1)!!}{(j-1)!}\cdot \frac{(2j+1)!!}{j!}\cdot \frac{(2n-2j-3)!!}{(n-j-2)!}\cdot \frac{(2n-2j-1)!!}{(n-j-1)!},$$ and recall from Theorem~\ref{thm:main} that $$\sum_{j=1}^{n-2} \sigma_n^{(j)} = 2E(A_n).$$ In particular, this is the expected number of consecutive noncommuting pairs in a reduced word for $w_0 \in \mathfrak{S}_n$. Central binomial coefficients satisfy $$\binom{2x}{x} \le \frac{4^x}{\sqrt{\pi x}}.$$ Combining this with the fact that $$\frac{(2x-1)!!}{(x-1)!} = \binom{2x}{x}\cdot \frac{x}{2^x}$$ yields \begin{eqnarray*} \sigma_n^{(j)} &=& \frac{256}{3\pi^2} \cdot \frac{\sqrt{j(j+1)(n-j-1)(n-j)}}{n(n-1)}\\ &\approx& \frac{256}{\ 3\pi^2} \cdot\frac{j(n-j)}{n^2}. \end{eqnarray*} Similarly, $$\binom{2x}{x} \ge \frac{4^x}{\sqrt{\pi x}}\cdot \left(1 - \frac{1}{8x}\right)$$ yielding the lower bound \begin{eqnarray*} \sigma_n^{(j)} &\ge& \frac{256}{3\pi^2} \cdot \frac{\sqrt{j(j+1)(n-j-1)(n-j)}}{n(n-1)}\\ & & \hspace{.25in} \cdot \left(1 - \frac{1}{8j}\right) \cdot \left(1 - \frac{1}{8(j+1)}\right) \cdot \left(1 - \frac{1}{8(n-j-1)}\right) \cdot \left(1 - \frac{1}{8(n-j)}\right)\\ &\approx& \frac{256}{\ 3\pi^2} \cdot\frac{j(n-j)}{n^2} \left(1 - \frac{n}{8j(n-j)}\right)^2. \end{eqnarray*} Therefore $$\sigma_n^{(j)} = \frac{256}{\ 3\pi^2} \cdot\frac{j(n-j)}{n^2} + O\left(\frac{1}{n}\right).$$ From this we can compute the asymptotic behavior of the expected number of consecutive pairs of noncommuting symbols in a reduced word for $w_0 \in \mathfrak{S}_n$: \begin{eqnarray} \nonumber 2E(A_n) &=& \sum_{j=1}^{n-2}\sigma_n^{(j)}\\ \nonumber &=& \frac{256}{3\pi^2}\sum_{j=1}^{n-2}\frac{j(n-j)}{n^2} \ + \ O(1)\\ \nonumber &=& \frac{256}{\ 3\pi^2} \left(\frac{(n-1)(n-2)}{2n} - \frac{(n-1)(n-2)(2n-3)}{6n^2}\right) + O(1)\\ \nonumber &=& \frac{128}{\ 9\pi^2}\cdot \frac{(n-1)(n-2)(n+3)}{n^2} + O(1)\\ \label{eqn:asymp noncomm}&=& \frac{128}{\ 9\pi^2}n + O(1). \end{eqnarray} Combining equation~\eqref{eqn:asymp noncomm} with Theorem~\ref{thm:main} yields \begin{eqnarray} \nonumber E(C_n) &=& \binom{n}{2} - 1 - 2E(A_n)\\ \label{eqn:asymp comm} &=& \binom{n}{2} - \frac{128}{\ 9\pi^2}n + O(1). \end{eqnarray} It is interesting to compare equation~\eqref{eqn:asymp comm} with Theorem~\ref{thm:reiner}. These computations imply that appearances of commutations and appearances of long braid moves have vastly different behaviors in reduced words for the longest element. Namely, while only one long braid move is expected to appear, nearly all positions are expected to support commutations. This is, perhaps, not surprising because commutations are positions in a reduced word where a symbol $i$ is followed by anything other than $\{i-1,i,i+1\}$, and the proportion $(n-4)/(n-1)$ of ``acceptable'' successors to $i$ approaches $1$ as $n$ gets large. To conclude this article, we use equations~\eqref{eqn:asymp noncomm} and~\eqref{eqn:asymp comm} and Theorem~\ref{thm:reiner} to state the asymptotic expectation of the proportion of appearances of Coxeter-related behaviors in reduced words for $w_0 \in \mathfrak{S}_n$. \begin{cor}\label{cor:proportions} Consider the reduced words for the longest element $w_0 \in \mathfrak{S}_n$. \begin{itemize} \vspace{.1in} \item $\displaystyle{\frac{E(\text{number of commutations})}{\text{length}} \ \approx \ 1}$ \vspace{.1in} \item $\displaystyle{\frac{E(\text{number of consecutive noncommuting pairs})}{\text{length}} \ \approx \ \frac{256}{9\pi^2n}}$ \vspace{.1in} \item $\displaystyle{\frac{E(\text{number of long braid moves})}{\text{length}} \ \approx \ \frac{2}{n^2}}$ \end{itemize} \end{cor} \section*{Acknowledgements} I am indebted to Richard Stanley for helpful comments and enthusiasm about these results, and to Andrew Rechnitzer for asymptotic insight and Sage advice. I also appreciate the suggestions of the anonymous referees.

Ceiling Repairs Continue in Savin Apartment Kiana Quinonez, Managing Editor November 6, 2018 A potential mold outbreak, a falling ceiling, and concerned students greeted the residents of Apartment 31 in Savin Court Townhouses. The students said they were facing a unique situation, and an unresponsive facilities team. “It took facilities about 3 1/2 weeks to come and look at my ceiling...

From WENN.com The comic beat TV presenter Stephen Colbert, who had to settle with second place, while 30 Rock creator and star Tina Fey is third in Rolling Stone magazine's new giggle list. Jon Stewart lands fourth place and South Park creators Trey Parker & Matt Stone round out the top five. Chris Rock, Girls star Lena Dunham and Bridesmaids star Kristen Wiig also feature in the top 10.

TITLE: Number of "Overlapping" Cells Within a Hypercube QUESTION [1 upvotes]: I have a hypercube in $k$-space which is divided along each dimension into $n$ cells. Each cell in the hypercube is assigned a unique ordered set of coordinates as follows: $(a_{1}, a_{2}, a_{3}, ... , a_{k})$ Given a particular cell, I'd like to derive a closed form for the number of cells that "overlap," or share at least one coordinate with, that cell (inclusive). For example, for $n = 3$ and $k = 3$, the number of "overlaps" for any particular cell should be $20$. I know that number of possible $combinations$ where at least one coordinate is shared is: ${k\choose 1} * n^{k-1}$ , since there are $n$ possibilities for each coordinate that is not shared. Obviously this overstates the number of "overlapping" cells, since it double-counts cells that overlap in multiple dimensions. So how would I proceed to find a closed form from here (I know it probably involves inclusion-exclusion)? Thank you in advance. REPLY [0 votes]: There are $n^k$ total cells. For a cell to not overlap, we must change all the coordinates and have $n-1$ choices for each one. There are $(n-1)^k$ non-overlapping cells, so $n^k-(n-1)^k-1$ other cells that overlap. For $n=k=3$ that gives $18$

TITLE: Behavior of a non-linear differential equation QUESTION [1 upvotes]: Let us consider the following differential equation $$ \dot{x}(t)=a - b\sin(x(t)), \quad a,b\in\mathbb{R}. $$ My question. Suppose $a>|b|$ and $x(0)=x_0\in\mathbb{R}$. Can the solution to the above equation be written in the form $$ x(t) = at + r(a,t), $$ where the term $r(a,t)$ is such that $r(a,t)\to 0$ for $a\to \infty$? PS: Of course, the explicit solution can be computed via symbolic software tools, such as Wolfram Alpha. However, the symbolic expression (see here) looks quite messy and does not give much information on the behavior of the solution for $a\to\infty$... REPLY [2 votes]: Take the ODE $$ \tag{$*$} \dot{x}(t) = a - b \sin(x(t)) $$ on the one-dimensional torus $\mathbb{R}/2 \pi \mathbb{Z}$. The vector field has no zeros, so there is a unique periodic orbit, with period $$ \int\limits_{0}^{2 \pi} \frac{d\xi}{a - b \sin(\xi)} = \frac{2 \pi}{\sqrt{a^2 - b^2}}. $$ Return now to $(*)$ on the real line $\mathbb{R}$. We have $$ x\!\left(t + \frac{2 \pi}{\sqrt{a^2 - b^2}}\right) = x(t) + 2 \pi \quad \text{for all }t \in \mathbb{R}. $$ Put $$ h(t) := x(t) - \sqrt{a^2 - b^2} t. $$ $h$ is easily seen to be periodic, with period $2\pi/\sqrt{a^2 - b^2}$. Consequently, we have, for any solution $x(\cdot)$ to $(*)$, $$ x(t) = \sqrt{a^2 - b^2} t + h(t). $$ So, $$ r(a, t) = (\sqrt{a^2 - b^2} - a)t + h(t), $$ which, for a fixed $t$, converges to $h(t)$ as $a \to \infty$.

About and choices were stark. They were told his quality of life would be poor, he would need lots of difficult and risky open heart surgery, each involving risks of death, organ failure, brain injury. He would be blue and breathless. He would never run, probably have learning difficulties and faced a very uncertain future as surgery was considered palliative. They were offered a choice of termination, birth with no intervention which would allow him to die, or open heart surgery immediately after his birth and the heartbreaking consequences that could go with it, followed by at least two more huge ops in his early years to just keep him alive and re-route his heart. They returned home with those three impossible choices and cried. However, a nurse had put a leaflet in their hands from Little Hearts Matter. It was late evening by the time they got home and they had the weekend to make their decision, so they rang the number, not expecting anyone to answer. They did. They were preparing, coincidentally, for their annual Open Day, which would be held that weekend. They explained it was a chance to meet other families on the same journey, talk to the ex- cardiac sister who had set up the charity -and see the older children with the condition. They drove to Northampton the next day and they were given the warmest of welcomes. They saw children playing and got a chance to talk to their parents. They understood that they were seeing the lucky ones who’d survived the three high risk surgeries and that some were too ill to be there, but there was courage, resilience and warmth in the room that gave them the support to make their decision. Little Hearts Matter would support them whatever that decision was and ensured they were informed- and crucially- not alone. John’s journey has not been easy in many ways. He sustained neurological damage during one early heart surgery, which resulted in visual impairment, and hemiplegia, restricted mobility on one side plus a motor disorder. At one point they were told he was blind. Luckily, some sight came back but he is still partially-sighted. During the Fontan surgery they nearly lost him several times, he developed all sorts of complications and multi- resistant infections. At 4 years old he was traumatised by the painful hospital interventions, weeks of nil by mouth, collapsed lung, insertion of chest drains and endless blood tests. He pulled through and did well at school despite lots of absence for poor health, but at 11 years old they were told he had an autistic spectrum condition. This results in social difficulties that require support and understanding – another challenge in his teenage years. John is now 17 and has surprised everyone. He has about half-vision, so cannot drive, but is bright, articulate and funny. He passed 9 GCSEs and has an amazing interest in the world around him but often feels like school does not “get it”. The difficulties that go with his heart condition impact every day and in many ways. His family fought for an EHCP but schools can still forget that when he is exhausted he cannot work well, that if he is breathless he can’t concentrate, that he needs extra time to complete tasks but that extra time is also extra tiring. Classes can be at the top of stairs. Exams were back to back for days on end. He was expected to fit into a system that was not designed with his needs in mind. What would help? His family would like his school and future employers to understand that there is no easy fix for this, that he needs ongoing support and adjustments but in return they get a wonderful human being who can face adversity with humour and resilience – their gain if they can work with him and support his medical needs. John gets exhausted every day and needs a shorter day but part time days at school means missing out on lessons and friendships if school does not put in the right support. He doesn’t do PE but they expect him to sit in a room on his own during those lessons and don’t really help him overcome those barriers to inclusion. Often he would come home from school upset, go to his room and then sleep for 15 hours solid. University open days have been challenging as it’s exhausting for John to traipse around and try to keep up with walking tours. John is not sure if he will manage universtity or if it’s right for him but part time apprenticeships are not easy to find. He is bright but degree apprenticeships are usually 35 hours a week plus a commute. John is at risk of social isolation because he doesn’t have the same lifestyle as his peers: he can’t go out late, club all night, go on fast rides, drive a car, trek up a hill, swim, make specific plans requiring him to be well at a particular time, to name but a few things. He has found solace in online communities and is a bit of a tech-head. Emotional issues are ongoing. He has to cope with uncertainty and anxiety at a time when other young people are feeling optimistic, reckless, care-free.. He has an unknown life expectancy but is smart enough to understand the statistics. How do you concentrate on studies when you know you are lucky to make it to adulthood? That knowledge separates you from your peers and robs you of so many different possibilities in many ways. More funding is badly needed to support patients with single ventricle conditions – more access to counselling, more research into medical developments, more specialist staff and more possibilities for treatment. More part time rewarding courses, paths and careers. More understanding about transplants, more donor organs, more training for schools, more support for families. Little Hearts Matter supports a growing number of survivors, which is fantastic but needs extra funding. If we can raise awareness, we can open up possibilities.

\begin{document} \maketitle \begin{abstract} There is one generalization of fibered links in 3-manifolds, called homologically fibered links. It is known that the existence of a homologically fibered link whose fiber surface has a given homeomorphic type is determined by the first homology group and its torsion linking form of the ambient 3-manifold. In this paper, we interpret the existence of homologically fibered links with that of a solution of some equation, in terms of the first homology group and its torsion linking form or a surgery diagram of the ambient manifold. As an application, we compute the invariant ${\rm hc}(\cdot)$, defined through homologically fibered knots, for 3-manifolds whose torsion linkng forms represent a generator of linkings. \end{abstract} \section{Introduction} It is known that every connected orientable closed 3-manifold $M$ has a fibered link \cite{alexander}, i.e., a link $L$ in $M$ such that the complement of a small open neighborhood of $L$ admits a structure of a connected orientable compact surface bundle over $S^1$ and that each boundary component of a fiber surface is a longitude of a component of $L$. Moreover, it is known that every connected orientable closed 3-manifold $M$ has a fibered knot \cite{myers}. However, finding fibered links in a given 3-manifold is difficult in general. For example ${\rm op}(M)$, defined as the minimal genus of fiber surfaces of fibered knots which $M$ has, is a topological invariant which is difficult to calculate. There is one generalization of fibered links, called {\it homologically fibered links} \cite{goda_sakasai} (defined in Definition~\ref{homologycob} below). A homologically fibered link requests some Seifert surface such that the result of cut a 3-manifold along the surface is a homological product of a surface and an interval, whereas a fibered link requests some Seifert surface such that the result of cutting a 3-manifold along the surface is the product of a surface and an interval. Clearly, a fibered link is also a homologically fibered link, and ${\rm hc}(M)$ \cite{goda_sakasai}, defined as the minimal genus of homological fiber surfaces of homologically fibered knots which $M$ has gives a lower bound of ${\rm op}(M)$. Homological products of surfaces (of a fixed homeomorphic type) and intervals are not merely the homological constraint for fiber structures, but they form a monoid by stacking, which contains the mapping class group as a submonoid (see \cite{goda_sakasai}). This attracts attention as a generalization of the mapping class groups. Homological products of surfaces and intervals are studied using ``clasper theory''. As a consequence of clasper theory, it is known that the existence of a homologically fibered link whose fiber is a given homeomorphic type in $M$ depends on the first homology group of $M$ with its torsion linking form. In this paper, we interpret the existence of a homologically fibered link whose fiber is a given homeomorphic type with the existence of a solution for some equation as follows. \\ As a convention, for an $m\times m$-matrix $A$ and an $n\times n$-matrix $B$, $A\oplus B$ is an $(m+n)\times(m+n)$-matrix whose $(i,j)$-entry is the same as the $(i,j)$-entry of $A$ for $1\leq i,j\leq m$, the $(m+i',m+j')$-entry is the same as the $(i',j')$-entry of $B$ for $1\leq i',j'\leq n$, and the other entries are $0$.\\ Let $O_{r}$ be an $r\times r$-zero-matrix, $I_{r}$ an $r\times r$-identity matrix, and we regard $0\times 0$-matrix ($O_{0}$ and $I_{0}$) as the identity element for $\oplus$. \\ Let $P$, $Q$, $F^{k}_{0}$, $F^{k}_{1}$ and $G^{k}_{1}$ be matrices as follows, where $a\geq 0$ and $k\geq 0$: \begin{itemize} \item $P$ is a diagonal $a\times a$-matrix whose $(i,i)$-entry is $p_{i}\in \mathbb{Z}$ for $a>0$ or $I_{0}$ for $a=0$. \item $Q$ is a diagonal $a\times a$-matrix whose $(i,i)$-entry is $-q_{i}\in \mathbb{Z}$ for $a>0$ or $I_{0}$ for $a=0$. \item $F^{0}_{0}$, $F^{0}_{1}$ and $G^{0}_{1}$ to be $I_{0}$,\\ and $F^{k}_{0}= \left( \begin{array}{cc} 0 & 2^{k} \\ 2^{k} & 0 \\ \end{array} \right) $ , \hspace{1.0cm} $F^{k}_{1}= \left( \begin{array}{cc} 2^{k+1} & -2^{k} \\ -3\cdot2^{k} & 2^{k+1} \\ \end{array} \right)$ , \hspace{1.0cm} $G^{k}_{1}= \left( \begin{array}{cc} -1 & 0 \\ 0 & -3 \\ \end{array} \right)$ for $k>0$. \end{itemize} Also let $S$ and $T$ be matrices as follows, where $r\geq0$, $e_{0}\geq 0$, $e_{1}\geq 0$, $k_{0}=k'_{0}=0$, $k_{i}\geq 1$ for $i\neq0$, and $k'_{j}\geq 2$ for $j\neq0$: \begin{itemize} \item $S=O_{r}\oplus P\oplus (\bigoplus_{0\leq i\leq e_{0}}F^{k_{i}}_{0}) \oplus (\bigoplus_{0\leq j\leq e_{1}}F^{k'_{j}}_{1})$. \item $T=I_{r}\oplus Q\oplus I_{2e_{0}} \oplus (\bigoplus_{0\leq j\leq e_{1}}G^{k'_{j}}_{1})$. \end{itemize} As usual notation, $\Sigma_{g,n+1}$ denotes a connected orientable compact surface of genus $g$ with $n+1$ boundary components. \begin{thm} \label{thm} Let $M$ be a connected closed oriented 3-manifold whose free part of the first homology group with integer coefficient is isomorphic to $\mathbb{Z}^{r}$ and whose torsion linking form is equivalent to $(\bigoplus_{0\leq l\leq a}A^{p_l}(q_l))\oplus (\bigoplus_{0\leq i \leq e_0} E^{k_i}_{0}) \oplus (\bigoplus_{0\leq j\leq e_1}E^{k'_{j}}_1)$, where $a,e_0,e_1$ are non-negative integers and notations $A^{p_l}(q_l), E^{k_i}_{0}, E^{k'_{j}}_{1}$ represent linkings, defined in Subsection~\ref{linkingpair}. \\ Suppose $(g,n)\neq (0,0)$. Then $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$ if and only if there exist $(r+a+2e_{0}+2e_{1}) \times (2g+n)$-matrix of integer coefficients $X$ and $(2g+n) \times (2g+n)$- symmetric matrix of integer coefficients $Y$ satisfying the following equation: \begin{eqnarray} \label{condition} \left| \begin{array}{cc} S & TX \\ X^{t} & Y+(\mathcal{E}\oplus O_{n}) \\ \end{array} \right| = \pm 1, \end{eqnarray} where $\mathcal{E}$ is a $2g\times 2g$-matrix whose $(2i-1,2i)$-entry is $1$ for every $1\leq i\leq g$ and the others are $0$. \end{thm} \begin{rmk} For the case which is not covered by Theorem~\ref{thm}, it is known that a connected closed orientable 3-manifold $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,1}$ if and only if $M$ is an integral homology $3$-sphere. See the definitions in the Subsection~\ref{defs}. \end{rmk} In general, it is not easy to compute the torsion linking form of a 3-manifold. Sometimes surgery diagrams of a 3-manifold may be easy to handle. A statement in terms of surgery diagrams is given in Proposition~\ref{condition_surgery}. The rest of this paper is organized as follows: In Section~\ref{sec_homfiblinks}, we recall the definition of homologically fibered links and the fact of their dependence on torsion linking forms. In Section~\ref{sec_reps}, we recall three types of generators of linking forms and give rational homology 3-spheres whose torsion linking forms are the generators. In Section~\ref{surf_bands}, we give a correspondence between surfaces in 3-manifolds and tuples of annuli and bands for Theorem~\ref{thm}. In Section~\ref{proof}, we give a proof of Theorem \ref{thm}. In Section~\ref{sec_examples}, we compute ${\rm hc}(\cdot)$ for 3-manifolds whose torsion linking forms are two of three types of generators in Section~\ref{sec_reps}. \subsection*{Acknowledgements} The author would like to thank Takuya Sakasai and Yuta Nozaki for introducing him this subject and giving him many ingredients about it. He also is grateful to referee for his or her kindness, patience and correcting many mistakes in the proofs and the arguments. \section{Homologically fibered links} \label{sec_homfiblinks} In this section, we review the definition of homologically fibered links and the fact that the existence of homologically fibered links depends on the torsion linking forms. In this paper, the homology groups are with integer coefficients. \subsection{Definitions}\label{defs} \begin{defini} (homology cobordism, {\cite[Section~2.4]{garoufalidis}})\\ A {\it homology cobordism over} $\Sigma_{g,n+1}$ is a triad ($X,\partial_{+}X,\partial_{-}X$), where $X$ is an oriented connected compact 3-manifold and $\partial_{+}X \cup \partial_{-}X$ is a partition of $\partial X$, and $\partial_{\pm} X$ are homeomorphic to $\Sigma_{g,n+1}$ satisfying: \begin{itemize} \item $\partial_{+}X \cup \partial_{-}X = \partial X$. \item $\partial_{+}X \cap \partial_{-}X = \partial(\partial_{+}X)$. \item The induced maps $(i_{\pm})_{*}: H_{*}(\partial_{\pm} X) \rightarrow H_{*}(X)$ are isomorphisms, where $i_{\pm}: \partial_{\pm}X \rightarrow X$ are the inclusions.\\ \end{itemize} \end{defini} Note that the third condition is equivalent to the condition that $i_{\pm}$ induce isomorphisms on $H_{1}(\cdot)$. Moreover by using the Poincar$\rm{\acute{e}}$-Lefschetz duality, we see that it is sufficient to require only one of $i_{+}$ and $i_{-}$ to induce an isomorphism on $H_{1}(\cdot)$. \begin{defini}\label{homologycob} \cite{goda_sakasai} Let $L$ be a link in a closed orientable 3-manifold $M$. $L$ is called {\it homologically fibered link} if there exists a Seifert surface $F$ of $L$ such that $(M\setminus F, F_{+}, F_{-})$ is a homology cobordism over a surface homeomorphic to $F$, where $F_{\pm}$ are the cut ends. In this situation, we call $F$ a {\it homological fiber} of $L$. \end{defini} Clearly every fibered link is a homologically fibered link since the former requests that the manifold cut opened by some Seifert surface to be the product of an interval and a surface and the latter requests it to be the homological product of an interval and a surface. By the observations above the Definition~\ref{homologycob}, for a Seifert surface $F$ in a closed 3-manifold $M$ to be a homological fiber, it is enough to check that push-ups (or push-downs) of oriented simple closed curves on $F$ which form a basis of $H_{1}(F)$ also form a free basis of $H_{1}(M\setminus Int(F\times[0,1]))$. \begin{rmk}\label{lowerbound} By definition, when a connected closed orientable 3-manifold $M$ has a homologically fibered link with homological fiber $F$, $H_{1}(M)$ can be generated by $1-\chi(F)$ elements, where $\chi(F)$ is the Euler number of $F$. Therefore the minimal order of generating sets for $H_{1}(M)$ imposes some constraint on homeomorphic types of homological fibers in $M$. \end{rmk} \subsection{Dependence on the torsion linking forms} \begin{defini}(Torsion linking form)\\ Let $M$ be a connected closed oriented 3-manifold, $TH_{1}(M)$ the torsion part of $H_{1}(M)$. For every $a\in TH_{1}(M)$, there is a non-zero integer $n$ such that $na$ vanishes in $H_{1}(M)$, and we fix such an integer $n_a$ of minimal absolute value. This $a$ has a representative as oriented curves $L_a$ in $M$ such that the $n_a$-parallel copy of $L_a$ bounds a surface $S_{a}$ in $M$. The {\it torsion linking form} of $M$, $\psi_{M} : TH_{1}(M)\times TH_{1}(M) \longrightarrow \mathbb{Q}/\mathbb{Z}$ maps $(a,b)$ to $\frac{<S_{a}, L_{b}>}{n_a}\ {\rm mod} \ \mathbb{Z} $, where $<S_{a}, L_{b}>$ represents the algebraic intersection number of $S_a$ and $L_b$. It is known that $\psi_{M}$ is well-defined and is non-singular, symmetric bilinear pairing, and that $\psi_{-M}=-\psi_{M}$, where $-M$ denotes the manifold homeomorphic to $M$ with the opposite orientation. \end{defini} There is the operation of 3-manifolds preserving the first homology groups and the torsion linking forms, called the ``Borromean surgery'' introduced in \cite{matveev}. \begin{defini} Let $V$ be a standard handlebody of genus three in $S^3$ which contains $0$-framed link as in Figure \ref{borromeansurg}. Consider a compact orientable 3-manifold $M$ and an embedding $f : V\longrightarrow M$. Let $N$ be a manifold obtained as a result of the Dehn surgery on $M$ along the framed link in $f(V)$. This $N$ is called the manifold obtained from $M$ (and $f$) by a {\it Borromean surgery}. \end{defini} \begin{figure}[htbp] \begin{center} \includegraphics[width=30mm]{borromean.eps} \end{center} \caption{A standard handlebody of genus three containing a $0$-framed link} \label{borromeansurg} \end{figure} About Borromean surgeries, the following facts are known: \begin{fact}\label{borrom1}{\rm \cite{habiro}, \cite{massuyeau}} Let $X$ be a homology cobordism over $\Sigma_{g,n+1}$ and $Y$ a manifold obtained from $X$ by a Borromean surgery. Then $Y$ is also a homology cobordism over $\Sigma_{g,n+1}$. \end{fact} \begin{fact} \label{borrom2} {\rm \cite{matveev}} Let $M$ and $N$ be closed oriented 3-manifolds. Then the following are equivalent: \begin{itemize} \item There is an isomorphism between $H_1(M)$ and $H_1(N)$ such that $\psi_{M}$ and $\psi_{N}$ are equivalent under this isomorphism. \item $M$ is obtained from $N$ by a finite sequence of Borromean surgeries. \end{itemize} \end{fact} \vspace{0.5cm} By the facts above, we can see the dependence on the first homology groups and the torsion linking forms about the existence of homologically fibered links: \begin{prop}\label{dependence} Suppose that for connected closed oriented 3-manifolds $M$ and $N$, there exists an isomorphism between their first homology groups such that their torsion linking forms are equivalent under the isomorphism. Then $N$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$ if and only if $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$. \end{prop} \begin{proof} We prove the if part and note that it is enough. By Fact~\ref{borrom2}, there is a finite sequence of Borromean surgeries starting from $M$ and ending at $N$. Since it is enough to prove when $N$ is obtained from $M$ by one Borromean surgery, we assume that. Suppose $M$ has a homological fibered link with homological fiber $F$ which is homeomorphic to $\Sigma_{g,n+1}$. Take a spine $T$ of $F$ and assume that $F$ lies in a small neighborhood of $T$. Isotope $F$ so that it is disjoint from the surgery link of the Borromean surgery. This can be done since $T$ and the surgery link are graphs in a 3-manifold. Let $F'$ be a surface, which is the image of $F$ after the Borromean surgery. Since $F$ is disjoint from the surgery link, a triad $(N\setminus F', F'^{+},F'^{-})$ is obtained from $(M\setminus F, F^{+},F^{-})$ by the Borromean surgery. Then by Fact~\ref{borrom1}, $(N\setminus F', F'^{+},F'^{-})$ is a homology cobordism over $\Sigma_{g,n+1}$. This implies that $N$ has a homologically fibered link with homological fiber $F'$ which is homeomorphic to $\Sigma_{g,n+1}$. \end{proof} Note that if a connected closed oriented 3-manifold $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$, then $-M$ also has such a homologically fibered link. By Proposition~\ref{dependence}, for a connected closed oriented 3-manifold $M$ containing a homologically fibered link with homological fiber of a given homeomorphic type we can replace $M$ with another one whose first homology group with the torsion linking form is the same as arbitrary one of $M$ and $-M$. \section{Representatives with respect to the first homology groups and their torsion linking forms} \label{sec_reps} In this section, at first we recall a part of the result of \cite{wall}, which gives a generator of the semigroup of the linkings on finite abelian groups, and next we fix representatives of 3-manifolds with respect to the first homology groups and their torsion linking forms. In fact, representatives are given in \cite{kawauchi_kojima}. We give another representatives for our use. Note that they may coincide. \subsection{Linking pair}\label{linkingpair} A {\it linking} is a pair $(G,\psi)$ such that $G$ is a finite abelian group and $\psi$ is a non-singular, symmetric bilinear pairing $G\times G\longrightarrow \mathbb{Q}/\mathbb{Z}$. Sometimes $\psi$ is called a linking on $G$. We call two linkings $(G,\psi)$ and $(G',\psi')$ equivalent if there exists a group isomorphism between $G$ and $G'$ through which $\psi$ and $\psi'$ are equivalent. By fixing a basis of $G$, $\psi$ is represented as a non-singular, symmetric matrix with coefficients in $\mathbb{Q}/\mathbb{Z}$, whose $(i,j)$-entry is the image of the $i$-th element and the $j$-th element of the basis under $\psi$. For two likings $(G_1,\psi_1)$ and $(G_2,\psi_2)$, define the product as $(G_1\oplus G_2, \psi_1\oplus \psi_2)$, where $\psi_1\oplus \psi_2$ maps $\left( (g_1,g_2),(h_1,h_2)\right)$ to $\psi_1(g_1,h_1)+\psi_2(g_2,h_2)$ for every $g_{1}, h_{1}\in G_{1}$ and $g_{2}, h_{2}\in G_{2}$. Under this product, linkings form an abelian semigroup $\mathcal{R}$. For a connected closed orientable 3-manifold $M$, its torsion linking form, $(TH_1(M), \psi_{M})$ is an example of linkings. Moreover, for two connected closed oriented 3-manifolds $M$ and $N$, the torsion linking form of $M\# N$ is $\psi_{M}\oplus \psi_{N}$. In \cite{wall}, a generator of $\mathcal{R}$ is given as following: \begin{fact}\label{gen_of_r} {\rm \cite{wall}} \begin{itemize} \item Let $A^{p}(q)$ be a linking on $\mathbb{Z}/p\mathbb{Z}$ (with a generator $1\in \mathbb{Z}/p\mathbb{Z}$) which is represented by $\left( \begin{array}{c} \frac{q}{p} \\ \end{array} \right)$ for coprime integers $p>1$ and $q$, and \item let $E^{k}_{0}$ be a linking on $\mathbb{Z}/2^{k}\mathbb{Z} \oplus \mathbb{Z}/2^{k}\mathbb{Z}$ (with a basis $\{(1,0), (0,1)\}$) which is represented by $\left( \begin{array}{cc} 0 & 2^{-k} \\ 2^{-k} & 0 \\ \end{array} \right)$ for every integer $k>0$, and \item let $E^{k}_{1}$ be a linking on $\mathbb{Z}/2^{k}\mathbb{Z} \oplus \mathbb{Z}/2^{k}\mathbb{Z}$ (with a basis $\{(1,0), (0,1)\}$) which is represented by $\left( \begin{array}{cc} 2^{1-k} & 2^{-k} \\ 2^{-k} & 2^{1-k} \\ \end{array} \right)$ for every integer $k>1$. \end{itemize} Then $\mathcal{R}$ is generated by $A^{p}(q)$, $E^{k}_{0}$ and $E^{k}_{1}$. \end{fact} Moreover, the presentation of $\mathcal{R}$ is revealed by combining the results in \cite{wall} and \cite{kawauchi_kojima}. As a notation we regard $A^{1}(q)$, $E^{0}_{0}$ and $E^{0}_{1}$ as $\phi$, the identity element of $\mathcal{R}$.\\ Thus every linking is represented as $(\oplus_{0\leq l\leq a}A^{p_l}(q_l))\oplus (\oplus_{0\leq i \leq e_0} E^{k_i}_{0}) \oplus (\oplus_{0\leq j\leq e_1}E^{k'_{j}}_1)$, where $a\geq 0$, $p_{0}=1$, $p_{l}$ is an integer greater than $1$ for $l\neq 0$, $q_{l}$ is an integer prime to $p_{l}$, and $e_{0}\geq 0$, $k_{0}=0$, $k_{i}>0$ for $i\neq 0$, and $e_{1}\geq 0$, $k'_{0}=0$, $k'_{j}>1$ for $j\neq 0$. Note that this representation is not unique. \subsection{Representatives} We give 3-manifolds representing the generators in Fact~\ref{gen_of_r}. On showing that the torsion linking forms of these are the generators, we adopt a method of calculating torsion linking forms of rational homology 3-spheres by using their Heegaard splittings \cite{conway}. At first, we review the method. \subsubsection{Calculating torsion linking forms by using Heegaard splittings \cite{conway}} \label{cal_tor} Let $M$ be a rational homology 3-sphere and $M=V\cup W$ a Heegaard splitting i.e. $M$ is obtained from two handlebodies of same genus, say $g$, $V$ and $W$ by pasting their boundaries by some orientation reversing homeomorphism, say $f: \partial V\longrightarrow \partial W$. We give $V$ and $W$ orientations as standard handlebodies in $\mathbb{R}^{3}$, and we give $M$ an orientation coming from $V$. Take a symplectic basis $\{a^{V}_1, \dots a^{V}_g, b^{V}_1, \dots, b^{V}_g\}$ of $H_1(\partial V)$ such that $b^{V}_i$ is $0$ in $H_1(V)$ for every $i$, where symplectic basis means that the intersection form on $\partial V$ satisfies $\langle a^{V}_i, a^{V}_j \rangle=0$, $\langle b^{V}_i,b^{V}_j \rangle=0$, and $\langle a^{V}_i,b^{V}_j \rangle =\delta_{i,j}$ for every $1\leq i,j\leq g$. Similarly, take a symplectic basis $\{a^{W}_1, \dots a^{W}_g, b^{W}_1, \dots, b^{W}_g\}$ of $H_1(\partial W)$ such that $b^{W}_i$ is $0$ in $H_1(W)$ for every $i$. Denote by $\left( \begin{array}{cc} A & B \\ C & D \\ \end{array} \right)$ the matrix representing $f_{*}$, the map between the first homology group induced by $f$ with respect to these bases, where $A$, $B$, $C$ and $D$ are $(g\times g)$-matrices over $\mathbb{Z}$. \begin{fact} \label{cfh} {\rm \cite{conway}} In the situation above, $det(B)\neq 0$ and $H_1(M)$ is isomorphic to $\mathbb{Z}^{g}/B^{t}\mathbb{Z}^{g}$. Moreover, the torsion linking form $\psi_M$ is equivalent to $\mathbb{Z}^{g}/B^{t}\mathbb{Z}^{g} \times \mathbb{Z}^{g}/B^{t}\mathbb{Z}^{g} \longrightarrow \mathbb{Q}/\mathbb{Z}\ ; (v,w) \mapsto -v^{t}\left( B^{-1}A\right)w$. \end{fact} \begin{rmk} We review the procedure for the calculation in terms of Heegaard diagrams rather than gluing maps used in the following. Let $M=V\cup W$ be a genus $g$ Heegaard splitting of a closed oriented 3-manifold (not necessarily a rational homology three sphere), and $S$ the splitting surface. Give $S$ the orientation as $\partial V$. Take a family of pairwise disjoint oriented simple closed curves $\{b^{V}_{1},\dots,b^{V}_{g}\}$ on $S$ such that each $b^{V}_{i}$ bounds pairwise disjoint disks in $V$ and that these disks cut $V$ into a ball. Similarly, take a family of pairwise disjoint oriented simple closed curves $\{b^{W}_{1},\dots,b^{W}_{g}\}$ on $S$ such that each $b^{W}_{i}$ bounds pairwise disjoint disks in $W$ and that these disks cut $W$ into a ball. Note that the triplet $(S, \{b^{V}_{1},\dots,b^{V}_{g}\}, \{b^{W}_{1},\dots,b^{W}_{g}\})$ determines the homeomorphic type of $M=V\cup W$. It is well-known that we can compute $H_{1}(M)$ using the curves: $H_{1}(M)=\mathbb{Z}\langle x_{1},\dots ,x_{g}\rangle / \left( \Sigma^{g}_{i=1} \langle b^{V}_{i}, b^{W}_{j}\rangle_{S} \cdot x_{i} \ {\rm for }\ 1\leq j \leq g \right) $, where $\langle \cdot, \cdot \rangle_{S}$ is the algebraic intersection form on $S$. By this, we can see whether $M$ is a rational homology $3$-sphere or not. Suppose that $M$ is a rational homology $3$-sphere in the following. Choose a family of pairwise disjoint oriented simple closed curves $\{a^{V}_{1},\dots,a^{V}_{g}\}$ on $S$ such that $\langle a^{V}_i,b^{V}_j \rangle_{S} =\delta_{i,j}$ for every $1\leq i,j\leq g$. Similarly, choose a family of pairwise disjoint oriented simple closed curves $\{a^{W}_{1},\dots,a^{W}_{g}\}$ on $S$ such that $\langle a^{W}_i,b^{W}_j \rangle_{S} =-\delta_{i,j}$ for every $1\leq i,j\leq g$. Then the homology class of $\{a^{V}_{1},\dots,a^{V}_{g},b^{V}_{1},\dots,b^{V}_{g}\}$ in $S$ forms a symplectic basis of $H_{1}(\partial V)$ such that each $b^{V}_{i}$ vanishes in $H_{1}(V)$ , and the homology class of $\{a^{W}_{1},\dots,a^{W}_{g},b^{W}_{1},\dots,b^{W}_{g}\}$ in $S$ forms a symplectic basis of $H_{1}(\partial W)$ such that each $b^{W}_{i}$vanishes in $H_{1}(W)$. In this situation, we can compute the matrices $A$, $B$ using curves: The $(i,j)$-entry of $A$ is $\langle b^{W}_{i},a^{V}_{j}\rangle_{S}$, and the $(i,j)$-entry of $B$ is $\langle b^{W}_{i}, b^{V}_{j}\rangle_{S}$. \end{rmk} \subsubsection{Representatives for $A^{p}(q)$} Let $M$ be a lens space of type $(p,q)$ for $p>1$. $M$ has a surgery presentation as in Figure~\ref{a^p_q_surg} and a Heegaard splitting $V\cup W$ as in Figure~\ref{a^p_q}. In Figure~\ref{a^p_q}, $V$ is the inner handlebody and $W$ is the outer handlebody, and a box containing $\frac{-q}{p}$ represents curves in it, a result of resolving the intersection points of horizontal $|q|$ lines and vertical $|p|$ lines so that they twist in left-hand (or right-hand) side if $\frac{-q}{p}>0$ (or $<0$, respectively). We take the $(p,-q)$-curve as $b^{V}_{1}$ and the $(0,1)$-curve as $b^{W}_{1}$ on the splitting torus. Note that the orientation of $L(p,q)$ coming from the standard one of $S^3$ corresponds to that coming from $V$. By computation, we see that $M$ is a rational homology 3-sphere. Let $a^{V}_1$ be the $(r,s)$-curve for $r,s$ such that $-rq-sp=1$. Then the matrices in \ref{cal_tor} are $A=\left( -r \right)$ and $B=\left(-p\right)$ by computing the algebraic intersections $\langle a^{V}_1, b^{W}_1 \rangle$ and $\langle b^{V}_1, b^{W}_1 \rangle$ on $\partial W$. Note that $\partial V=-\partial W$. Thus by Fact~\ref{cfh}, $H_1(M)=\mathbb{Z}/p\mathbb{Z}$ and the torsion linking form is $\left( \begin{array}{c} -\frac{r}{p} \\ \end{array} \right)$. Let $x$ be a generator of $H_1(M)=\mathbb{Z}/p\mathbb{Z}$ corresponding to the $1\times 1$-matrix $\left( \begin{array}{c} -\frac{r}{p} \\ \end{array} \right)$. Note that $qx$ is also a generator of $H_1(M)=\mathbb{Z}/p\mathbb{Z}$ since $p$ and $q$ are coprime. Under this new generator, the matrix representation of the torsion linking form is $\left( \begin{array}{c} -\frac{rq^{2}}{p} \\ \end{array} \right)= \left( \begin{array}{c} \frac{q}{p} \\ \end{array} \right)$ since $rq \equiv -1$ mod $p$, the same as $A^{p}(q)$. Henceforth, $M(A^{p}(q))$ denotes $L(p,q)$. \begin{figure}[htbp] \begin{center} \includegraphics[width=25mm]{a_p_q_surg.eps} \end{center} \caption{A surgery description of $L(p,q)$} \label{a^p_q_surg} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=125mm]{a_p_q.eps} \end{center} \caption{left: A Heegaard splitting of $L(p,q)$\ \ \ \ right: Examples of boxes contain rational numbers} \label{a^p_q} \end{figure} \subsubsection{Representatives for $E^{k}_0$} Let $M$ be a connected closed orientable 3-manifold having a surgery description as in Figure~\ref{e^k_0_surg}. It has a Heegaard splitting $V\cup W$ as in Figure~\ref{e^k_0}, so-called a vertical Heegaard splitting. The convention of boxes containing rational numbers is as in Figure~\ref{a^p_q}. Note that this Heegaard diagram is not minimally intersecting. We regard $V$ as the inner handlebody and $W$ as the outer handlebody. Note that the orientation coming from the standard one of $S^3$ corresponds to that coming from $V$. By computation, we see that $M$ is a rational homology 3-sphere. Let $a^{V}_1, a^{V}_2$ be curves as in Figure~\ref{e^k_0_a}, where $b^{V}_1, b^{V}_2$ is the same as in Figure~\ref{e^k_0}, which we abbreviate. Then the matrices in \ref{cal_tor} are $A=\left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 2^{k} & 0 \\ -2^{k} & 2^{k} \\ \end{array} \right)$. Thus by Fact \ref{cfh}, $H_1(M)=\mathbb{Z}\langle x, y \rangle/(2^{k}x-2^{k}y, 2^{k}y)$ and $\psi_{M}=-B^{-1}A=\left( \begin{array}{cc} 0 & -2^{-k} \\ -2^{-k} & 2^{1-k} \\ \end{array} \right)$ under the basis $\{x,y\}$. By changing a basis, $H_{1}(M)=\left( \mathbb{Z}\langle x+y\rangle/(2^{k}(x+y)) \right) \oplus \left( \mathbb{Z} \langle -x\rangle /(2^{k}(-x)\right)$ and $\psi_{M}=\left( \begin{array}{cc} 0 & 2^{-k}\\ 2^{-k} & 0 \\ \end{array} \right)$ under the basis $\{x+y,-x\}$, the same as $E^{k}_0$. Henceforth, $M(E^{k}_0)$ denotes this $M$. \begin{figure}[htbp] \begin{center} \includegraphics[width=130mm]{e_k_0_surg.eps} \end{center} \caption{Surgery descriptions of $M$} \label{e^k_0_surg} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=80mm]{e_k_0.eps} \end{center} \caption{A Heegaard splitting of $M$} \label{e^k_0} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=80mm]{e_k_0_a.eps} \end{center} \caption{Curves $a^{V}_1, a^{V}_2, b^{W}_{1}, b^{W}_{2}$} \label{e^k_0_a} \end{figure} \subsubsection{Representatives for $E^{k}_{1}$} Let $M$ be a connected closed orientable 3-manifold having a surgery description as in Figure~\ref{e^k_1_surg}. It has a Heegaard splitting $V\cup W$ as in Figure~\ref{e^k_1}, so-called a vertical Heegaard splitting. The convention of boxes containing rational numbers is as in Figure~\ref{a^p_q}. We regard $V$ as the inner handlebody and $W$ as the outer handlebody. Note that the orientation coming from the standard one of $S^3$ corresponds to that coming from $V$. By computation, we see that $M$ is a rational homology 3-sphere. Let $a^{V}_1, a^{V}_2$ be curves as in Figure~\ref{e^k_1_a}, where $b^{V}_1, b^{V}_2$ are the same as in Figure~\ref{e^k_1}, which we abbreviate. Then the matrices in \ref{cal_tor} are $A=\left( \begin{array}{cc} 0 & -1 \\ -3 & 3 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 2^{k} & 2\cdot2^{k} \\ -2^{k} & -3\cdot 2^{k} \\ \end{array} \right)$. Thus by Fact~\ref{cfh}, $H_1(M)=\mathbb{Z}\langle x, y\rangle/(2^{k}x-2^{k}y, 2\cdot2^{k}x-3\cdot2^{k}y)$ and $\psi_{M}=-B^{-1}A=\left( \begin{array}{cc} 3\cdot 2^{1-k} & -3\cdot 2^{-k} \\ -3\cdot 2^{-k} & 2^{1-k} \\ \end{array} \right)$ under the basis $\{x,y\}$. By changing a basis, $H_{1}(M)=\left( \mathbb{Z}\langle x+y\rangle/(2^{k}(x+y)) \right) \oplus \left( \mathbb{Z} \langle -y\rangle /(2^{k}(-y)\right)$ and $\psi_{M}=\left( \begin{array}{cc} 2^{1-k} & 2^{-k}\\ 2^{-k} & 2^{1-k} \\ \end{array} \right)$ under the basis $\{x+y,-y\}$, the same as $E^{k}_1$. Henceforth, $M(E^{k}_1)$ denotes this $M$. \begin{figure}[htbp] \begin{center} \includegraphics[width=130mm]{e_k_1_surg.eps} \end{center} \caption{Surgery descriptions of $M$} \label{e^k_1_surg} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=80mm]{e_k_1.eps} \end{center} \caption{A Heegaard splitting of $M$} \label{e^k_1} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=100mm]{e_k_1_a.eps} \end{center} \caption{Curves $a^{V}_1, a^{V}_2, b^{W}_{1}, b^{W}_{2}$} \label{e^k_1_a} \end{figure} \vspace{1.0cm} From the above, we have representatives for connected closed oriented 3-manifolds with respect to the first homology groups and their torsion linking forms. \begin{prop} \label{representative} Suppose that a connected closed oriented 3-manifold $M$ whose free part of the first homology group is isomorphic to $\mathbb{Z}^{r}$ and whose torsion linking form is equivalent to $(\bigoplus_{0\leq l\leq a}A^{p_l}(q_l))\oplus (\bigoplus_{0\leq i \leq e_0} E^{k_i}_{0}) \oplus (\bigoplus_{0\leq j\leq e_1}E^{k'_{j}}_1)$ for integers $a, p_{l}, q_{l}, e_{0}, k_{i}, e_{1}, k'_{j}$ as in the end of Subsection \ref{linkingpair}. \\ Then a 3-manifold $\left(\#^{r}S^2\times S^1\right)\#\left(\#_{0\leq l\leq a}M\left(A^{p_l}(q_l)\right)\right)\# (\#_{0\leq i \leq e_0} M(E^{k_i}_{0})) \# (\#_{0\leq j\leq e_1}M(E^{k'_{j}}_1))$ has isomorphic first homology group to that of $M$ and has equivalent torsion linking forms to that of $M$, where as a notation we regard $\#^{0}S^2\times S^1$, $M(A^{1}(q_0))$, $M(E^{0}_{0})$ and $M(E^{0}_{1})$ as $S^3$. \end{prop} We use this in Section~\ref{proof}. \section{Thickened surface and its spine bands} \label{surf_bands} In this section, we explain how surfaces (with their spines) in a given 3-manifold correspond to pairs of annuli and bands in the manifold. All surfaces and 3-manifolds in this section are oriented, and we can use the terminology ``front'' and ``back'' sides of surfaces.\\\\ We state the statement as a proposition: \begin{prop}\label{correspondence} Let $M$ be an oriented 3-manifold. Fix non-negative integers $g$ and $n$.\\ There is a bijection between the set of isotopy classes of oriented surfaces in $M$, each of which is homeomorphic to $\Sigma_{g,n+1}$ with the spine as in Figure \ref{spine} and the set of oriented annuli $A_{1}, \dots,A_{2g+n}$ with oriented core curves and the oriented bands (regarded as small rectangles) $B_{1},\dots, B_{g}, C_{1}, \dots ,C_{g+n-1}$ satisfying the following condition (see Figure~\ref{bands} for example): \begin{itemize} \item One side of $\partial B_{i}$, which consists of four sides, is on the back side of $A_{2i-1}$ and is a properly embedded essential arc on $A_{2i-1}$, and the opposite side of $\partial B_{i}$ is on the front side of $A_{2i}$ and is a part of the core curve of $A_{2i}$ for $1\leq i\leq g$. \item For $1\leq i\leq g$, near the intersection of $B_i$ with $A_{2i-1}$, the front side of $B_i$ is on the positive direction with respect to the orientation of the core curve of $A_{2i-1}$, and near that of with $A_{2i}$, the front side of $B_{i}$ is the ``right'' side of the core curve of $A_{2i}$, where we look the intersection (this is on the front side of $A_{2i}$) so that the core curve of $A_{2i}$ runs from the bottom to the top. \item For $1\leq i \leq g-1$, $C_{i}$ connects the ``left'' boundary of $A_{2i}$ and the ``right'' boundary of $A_{2i+1}$ so that the front sides are attached, and for $g\leq j\leq g+n-1$, $C_{j}$ connects the ``left'' boundary of $A_{j+g}$ and the ``left'' boundary of $A_{j+g+1}$ so that the front sides are attached, where we look the front side of $A_{k}$ so that the core curve runs from the bottom to the top. \end{itemize} \end{prop} We call the condition for annuli and bands in Proposition~\ref{correspondence} the condition $\spadesuit$. We construct oriented annuli and bands from a surface with its spine in Subsection~\ref{fromsurf}, and construct a surface with its spine from oriented annuli and bands in Subsection~\ref{fromknots}. We can see that these operations are the inverses of each other. Thus Proposition~\ref{correspondence} holds. \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{spine.eps} \end{center} \caption{A spine of $\Sigma_{g,n+1}$} \label{spine} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{bands.eps} \end{center} \caption{Annuli and bands} \label{bands} \end{figure} \subsection{Annuli and bands obtained from surfaces}\label{fromsurf} Let $F$ be a surface homeomrphic to $\Sigma_{g,n+1}$ in $M$. Fix an embedding $\iota : \Sigma_{g,n+1}\times[-1,1] \longrightarrow M$ such that $\iota(\Sigma_{g,n+1}\times \{0\})=F$ and identify $\Sigma_{g,n+1}$ with $F$ via $\iota$. Take a spine $(K_{1}\cup K_{2})\cup w_{1}\cup \cdots \cup w_{g-1}\cup (K_{2g-1}\cup K_{2g})\cup w_{g}\cup K_{2g+1} \cup w_{g+1} \cup \cdots \cup w_{g+n-1} \cup K_{2g+n}$ of $F$ as in Figure~\ref{spine}. We assign an orientation to $K_i$ for each $1\leq i\leq 2g+n$. Denote the point $K_{2i-1}\cap K_{2i}$ for $1\leq i\leq g$ by $p_i$. Let $U_i$ be the square centered at $p_i$ and $V_j$ the rectangle around $w_j$ as in Figure~\ref{charts}. Take charts $\phi_{i} : [-1,1]^{2} \longrightarrow U_i$ such that $\phi([-1,1]\times \{0\})$ is a part of $K_{2i-1}$, $\phi(\{0\}\times [-1,1])$ is a part of $K_{2i}$, and $\psi_{j} : [-1,1]^{2} \longrightarrow V_{j}$ such that $\psi([-1,1]\times \{0\})$ is a part of $w_j$ and $\{-1\}\times [-1,1]$ is near $K_{2j}$ for $1\leq j\leq g$, near $K_{j+g}$ for $g+1\leq j\leq g+n-1$. \begin{figure}[htbp] \begin{center} \includegraphics[width=70mm]{charts.eps} \end{center} \caption{Charts} \label{charts} \end{figure} Denote the annulus whose core curve is $K_i$ by $A'_i$ for $1\leq i\leq 2g+n$ as in Figure~\ref{annuli}. Consider $F\times [-1,1]$, see Figure \ref{thicken1}. We regard that it is in $M$ via $\iota$. Let $A_{2i-1}$, $A_{2i}$ and $A_{j}$ be $A'_{2i-1}\times \{1\}$, $A'_{2i}\times \{-1\}$ and $A'_{j}\times \{1\}$, respectively for $1\leq i\leq g$ and $2g+1\leq j\leq 2g+n$. These annuli can be regarded as framed knots in $M$. Let $B_{i}$, $C_{i}$ and $C_{j}$ be $\phi_{i} \left( \{0\}\times[-1,1] \right) \times[-1,1]$, $\left\{\psi_{i}(\{t\}\times [-1,1])\times \{t\}| t\in [-1,1] \right\}$ and $\psi_{j}([-1,1]^{2})\times \{1\}$, respectively for $1\leq i\leq g$ and $g+1\leq j\leq g+n-1$. These are bands in $M$. We get an oriented annuli $A_{k}$'s and bands $B_{i}$'s and $C_{j}$'s (these are rectangles) in $M$ as in Figure~\ref{bands}. Observe that these satisfy the condition $\spadesuit$ with appropriate orientations. \begin{figure}[htbp] \begin{center} \includegraphics[width=80mm]{annuli.eps} \end{center} \caption{Annuli} \label{annuli} \end{figure} \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{thicken1.eps} \end{center} \caption{Thickened $F$} \label{thicken1} \end{figure} \subsection{Surfaces from annuli and bands} \label{fromknots} Let $A_{1}, \dots, A_{2g+n}$ be oriented annuli and $B_{1},\dots, B_{g}, C_{1}, \dots ,C_{g+n-1}$ be bands satisfying the condition $\spadesuit$ in $M$. These look like some embeddings of the union of annuli and bands as in Figure \ref{bands}. Thicken these (denoted by $N$) and consider curves with colors as in Figure \ref{thicken2}. In this figure, a part of the core curve of $A_{2i}$, which is also one edge of the band $B_{i}$ is replaced with the other three edges of $B_{i}$ for each $1\leq i\leq g$. A small neighborhood $F$ of these curves in $\partial N$ is homeomorphic to $\Sigma_{g,n+1}$ with a spine as in Figure~\ref{spine}. Let $K_{2i-1}$ denote the red curve coming from $A_{2i-1}$ and $K_{2i}$ the blue curve coming from $A_{2i}$ and $B_{i}$ for $1\leq i\leq g$, and $w_{j}$ the green curve coming from $C_{j}$ for $1\leq j\leq g+n-1$. For each $1\leq i \leq 2g+n$, fix an essential arc $\alpha _{i}$ in $A_{i}$ such that it is disjoint from $w_{j}$ for $1\leq j\leq g+n-1$ and $K_{l}$ for $l\neq i$, and let $D_{i}$ be an essential disk in $N$ obtained by thickening $\alpha_{i}$. Then each $\partial D_{i}$ intersects $\partial F$ twice with the opposite orientations and $D_{i}$'s cut $N$ into a ball. Therefore, $F$ gives $N$ the structure of $\Sigma_{g,n+1}\times[-1,1]$ so that $F$ is $\Sigma_{g,n+1}\times \{1\}$. The colored curves play a role of a spine of $\Sigma_{g,n+1}$. We get a surface corresponding to $\Sigma_{g,n+1}\times \{0\}$ with a spine obtained by projecting the curves along the product structure. In other word, the surface $\Sigma_{g,n+1}\times \{0\}$ with its spine is isotopic to $F$ with its spine in $M$. We can easily show that the operations in Subsections~\ref{fromsurf} and \ref{fromknots} are inverses of each other. \begin{rmk} Let $m_{A_{2i-1}}$ be a meridian curve of $A_{2i-1}$ (i.e. an oriented circle on the boundary of a small neighborhood of $A_{2i-1}$ which bounds a disk in the neighborhood intersecting the core of $A_{2i-1}$ once positively) for $1\leq i\leq g$. We assume $m_{A_{2i-1}}$ is in a small neighborhood of $A_{2i-1}$ and disjoint from $A_{2i-1}$. Note that the curve obtained from $A_{2i}$ by the above procedure represents $[A_{2i}]-[m_{A_{2i-1}}]$ in $H_{1}(M\setminus N)$, where $[A_{2i}]$ and $[A_{2i-1}]$ denote the elements represented by push-ups of the core curves of $A_{2i}$ and $A_{2i-1}$, respectively. \end{rmk} \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{thicken2.eps} \end{center} \caption{Thickened annuli and bands} \label{thicken2} \end{figure} \section{A proof of Theorem~\ref{thm}} \label{proof} In this section, we give a proof of Theorem~\ref{thm}. At first, we consider the condition for a connected closed orientable 3-manifold having a homologically fibered link whose homological fiber is given homeomorphic type using a surgery diagram. In order to define the linking number $lk(\cdot,\cdot)$ among knots in the surgery link, we have to give the link an orientation. We fix any one. However, this choice does not matter for the existence of a homologically fibered link whose fiber is the given homeomorphism type, see Remark \ref{choice}. \begin{prop} \label{condition_surgery} Let $M$ be a connected closed orientable 3-manifold. Suppose that $M$ is obtained from $S^3$ by the surgery along a link $L=L_{1}\sqcup \cdots \sqcup L_{m}$, where each $L_{i}$ is an oriented knot and its coefficient is $\frac{p_i}{q_i}$.\\ Then $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$ if and only if there exist $m \times (2g+n)$-matrix of integer coefficients $X$ and $(2g+n) \times (2g+n)$-symmetric matrix of integer coefficients $Y$ satisfying the following equation: \begin{eqnarray} \label{surg} \left| \begin{array}{cc} \Phi & \Psi X \\ X^{t} & Y+(\mathcal{E}\oplus O_{n}) \\ \end{array} \right| = \pm 1, \end{eqnarray} where $\Phi$ is an $m\times m$-matrix whose $(i,i)$-entry is $p_i$ and $(i,j)$-entry is $q_{i}lk(L_{i},L_{j})$ for $i\neq j$, $\Psi$ is an $m\times m$-diagonal matrix whose $(i,i)$-entry is $q_{i}$, and $\mathcal{E}$ is a $2g\times 2g$-matrix whose $(2i-1,2i)$-entry is $1$ for every $1\leq i\leq g$ and the others are $0$. \end{prop} \begin{proof} Suppose $M$ has a homologically fibered link whose homological fiber $F$ is homeomorphic to $\Sigma_{g,n+1}$. We will construct a solution of (\ref{surg}). By choosing a spine of $F$, and using the construction in Section \ref{surf_bands}, we get oriented annuli $A_{1},\dots,A_{2g+n}$ and oriented bands $B_{1},\dots,B_{g}$, $C_{1},\dots, C_{n}$ satisfying the condition $\spadesuit$. We regard these annuli and bands are in the surgery diagram on $S^3$ missing the surgery link. Let $T_{i}$ be the canonical longitude (regarded in $S^3$) of the core curve of $A_{i}$ and $m_{A_{i}}$ the meridian of (the core curve of) $A_{i}$ such that $lk(T_{i},m_{A_{i}})=1$ and $t_{i}$ the framing number of $A_{i}$ regarded as a framed knot. Note that $H_{1}(M\setminus F)\cong H_{1}\left( M\setminus \left( (\bigcup_{1\leq i \leq 2g+n}A_{i})\cup(\bigcup_{1\leq j \leq g}B_{j}))\cup(\bigcup_{1\leq l \leq n}C_{l}) \right) \right) \cong H_{1}\left( M\setminus(\bigcup_{1\leq i \leq 2g+n}A_{i}) \right)$, and that the push-up of $K_{i}$, which is in spines of $F$ as in Figure \ref{spine} represents\\ $\begin{cases} [T_{2i'-1}]+t_{2i'-1}[m_{A_{2i'-1}}]\ \ {\rm if} \ 1\leq i \leq 2g \ {\rm and}\ i=2i'-1,\\ [T_{2i'}]+t_{2i'}[m_{A_{2i'}}]-[m_{A_{2i'-1}}]\ \ {\rm if} \ 1\leq i \leq 2g \ {\rm and}\ i=2i',\\ [T_{i}]+t_{i}[m_{A_{i}}]\ \ {\rm if} \ 2g+1\leq i \leq 2g+n\\ \end{cases}$\\ in $H_{1}\left( M\setminus(\bigcup_{1\leq i \leq 2g+n}A_{i})\right)$. \\ Let $m_{L_{i}}$ be the meridian of $L_{i}$ and set $l_{i,j}$ to be $lk(L_{i},L_{j})$ and $\tilde{l}_{i,j}$ to be $lk(L_{i},T_{j})$ and $\bar{l}_{i,j}$ to be $lk(T_{i},T_{j})$. Note that $H_{1}\left( S^{3}\setminus \left(L\cup (\bigcup_{1\leq i \leq 2g+n}A_{i})\right)\right)\cong \mathbb{Z}^{m+2g+n}$ is generated by $\{ [m_{L_{1}}],\dots,[m_{L_m}], [m_{A_1}],\dots,[m_{A_{2g+n}}]\}$, and that $\frac{p_i}{q_i}$-slope of $L_i$ represents an element $p_{i}[m_{L_{i}}]+q_{i}\{ \Sigma_{\substack{1\leq j\leq m\\ i\neq j}} l_{i,j}[m_{L_{j}}]+\Sigma_{1\leq k \leq 2g+n} \tilde{l}_{i,k}[m_{A_{k}}]\}$ in $H_{1}\left( S^{3}\setminus \left(L\cup (\bigcup_{1\leq i \leq 2g+n}A_{i})\right)\right)$, denoted by $\mathbb{A}_{i}$. Note also that the push-up of $K_{i}$, which is in a spine of $F$ as in Figure \ref{spine} represents \\ $\begin{cases} t_{2i'-1}[m_{A_{2i'-1}}]+\Sigma_{1\leq j\leq m}\tilde{l}_{j,i}[m_{L_j}]+\Sigma_{\substack{1\leq k \leq 2g+n\\ k\neq i}} \bar{l}_{i,k}[m_{A_{k}}]\ \ {\rm if} \ 1\leq i \leq 2g \ {\rm and}\ i=2i'-1,\\ t_{2i'}[m_{A_{2i'}}]-[m_{A_{2i'-1}}]+\Sigma_{1\leq j\leq m}\tilde{l}_{j,i}[m_{L_j}]+\Sigma_{\substack{1\leq k \leq 2g+n\\ k\neq i}} \bar{l}_{i,k}[m_{A_{k}}]\ \ {\rm if} \ 1\leq i \leq 2g \ {\rm and}\ i=2i',\\ t_{i}[m_{A_i}]+\Sigma_{1\leq j\leq m}\tilde{l}_{j,i}[m_{L_j}]+\Sigma_{\substack{1\leq k \leq 2g+n\\ k\neq i}} \bar{l}_{i,k}[m_{A_{k}}]\ \ {\rm if} \ 2g+1\leq i \leq 2g+n\\ \end{cases}$\\ under the basis $\{ [m_{L_{1}}],\dots,[m_{L_m}], [m_{A_1}],\dots,[m_{A_{2g+n}}]\}$ of $H_{1}\left(S^{3}\setminus \left(L\cup (\bigcup_{1\leq i \leq 2g+n}A_{i})\right)\right)$, denoted by $\mathbb{B}_{i}$. Since $F$ is homological fiber, $H_{1}(M\setminus F)$ must be $\mathbb{Z}^{2g+n}$ and the push-ups of $K_{i}$'s form a free basis of $H_{1}(M\setminus F)$. This implies that $\{ [m_{L_{1}}],\dots,[m_{L_m}], [m_{A_1}],\dots,[m_{A_{2g+n}}]\}/\langle \mathbb{A}_{1},\dots ,\mathbb{A}_{m} \rangle$ is isomorphic to $\mathbb{Z}^{2g+n}$, which also implies that $\mathbb{A}_{1},\dots ,\mathbb{A}_{m}$ are linearly independent in $H_{1}\left(S^{3}\setminus \left(L\cup (\bigcup_{1\leq i \leq 2g+n}A_{i})\right) \right)$, and $\mathbb{B}_{j}$'s modulo $\langle \mathbb{A}_{1},\dots ,\mathbb{A}_{m} \rangle$ form a basis of it. Thus $\{ \mathbb{A}_{1},\dots,\mathbb{A}_m, \mathbb{B}_{1},\dots,\mathbb{B}_{2g+n}\}$ is a basis of $\mathbb{Z}^{m+2g+n}$ freely generated by $\{ [m_{L_{1}}],\dots,[m_{L_m}], [m_{A_1}],\dots,[m_{A_{2g+n}}]\}$. Thus considering the change of basis matrix, we get a solution of (\ref{surg}) by setting $X$ equal to a matrix whose $(i,j)$-entry is $\tilde{l}_{i,j}$, $Y$ equal to a symmetric matrix whose $(k,k)$-entry is $t_{k}$, $(2i-1,2i)$-entry and $(2i,2i-1)$-entry for $i\leq g$ are $\bar{l}_{2i-1,2i}-1$ and other $(i,j)$-entry is $\bar{l}_{i,j}$. \\ Conversely, suppose we have a solution $X$, $Y$ for (\ref{surg}). Then we can construct oriented framed knots $K_{1},\dots,K_{2g+n}$ in the surgery diagram on $S^3$, which will be the core curves of annuli $A_{1},\dots , A_{2g+n}$, such that $lk(L_{i},K_{j})$ is the $(i,j)$-entry of $X$, the framing of $K_{i}$ is the $(i,i)$-entry of $Y$, $lk(K_{2i-1},K_{2i})$ for $i\leq g$ is equal to $1$ plus ${\rm the}\ (2i-1,2i)$-entry of $Y$, and $lk(K_{i},K_{j})$ for other pair $(i,j)$ is the $(i,j)$-entry of $Y$. Take bands $B_{1},\dots, B_{g}, C_{1},\dots ,C_{2g+n-1}$ satisfying the condition $\spadesuit$. Note that we can choose bands arbitrarily since these do not affect on the homology. We get a surface which is homeomorphic to $\Sigma_{g,n+1}$ by the construction in Section~\ref{surf_bands}. By reversing the argument in the previous paragraph, we see that this surface is a homological fiber. \end{proof} \begin{rmk}\label{choice} Suppose there is a solution $X$, $Y$ for the equation (\ref{surg}) in Proposition~\ref{condition_surgery} for some orientation of the surgery link. If we reverse the orientation of the $i$-th component of $m$-component surgery link, then the signs of the $i$-th row and $i$-th column of $\Phi$ except for the $(i,i)$-entry and the $(i,i)$-th entry of $\Psi$ in the left hand side of the equation (\ref{surg}) are switched. Then a pair of the matrix obtained from $X$ by switching the signs of the $i$-th row and $Y$ is a solution of the new equation. Therefore the existence of a solution of the equation (\ref{surg}) is preserved. \end{rmk} \begin{rmk} For a given surgery diagram and a solution of (\ref{surg}), we can concretely construct a homologically fibered link as in the latter part of the proof of Proposition~\ref{condition_surgery}. However we have many choices in construction: There are many links with linking each others in the given number and linking the surgery link in the given number, and we can arbitrary take bands. These homologically fibered links corresponding to one solution are not always different, i.e. they may be isotopic. Moreover, two homological fibered links corresponding to two different solution may be isotopic. \end{rmk} \begin{proof} [Proof of Theorem~\ref{thm}] Suppose that $M$ is a connected closed oriented 3-manifold whose free part of the first homology group is isomorphic to $\mathbb{Z}^{r}$ and whose torsion linking form is equivalent to $(\bigoplus_{0\leq l\leq a}A^{p_l}(q_l))\oplus (\bigoplus_{0\leq i \leq e_0} E^{k_i}_{0}) \oplus (\bigoplus_{0\leq j\leq e_1}E^{k'_{j}}_1)$. \\ Then $\tilde{M}= \left(\#^{r}S^2\times S^1\right)\#\left(\#_{0\leq l\leq a}M\left(A^{p_l}(q_l)\right)\right)\# (\#_{0\leq i \leq e_0} M(E^{k_i}_{0})) \# (\#_{0\leq j\leq e_1}M(E^{k'_{j}}_1))$ has the isomorphic first homology group to $M$ and has a torsion linking form equivalent to that of $M$ by Proposition~\ref{representative}. By Proposition~\ref{dependence}, $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{g,n+1}$ if and only if $\tilde{M}$ has such homologically fibered link. We have a surgery diagram on $S^3$ for $\tilde{M}$ which consists of $r$ copies of an unknot with surgery slope $0$ (for $\#^{r}S^2\times S^1$), $a$ copies of an unknot with surgery slope $\frac{p_l}{-q_l}$ for $0\leq l\leq a$ (for $\#_{0\leq l\leq a}M\left(A^{p_l}(q_l)\right)$), $e_0$ copies of $2$-component link as in the right of Figure \ref{e^k_0_surg} for $0\leq i \leq e_0$ (for $\#_{0\leq i \leq e_0} M(E^{k_i}_{0})$), and $e_1$ copies of $2$-component link as in the right of Figure \ref{e^k_1_surg} for $0\leq j \leq e_1$ (for $\#_{0\leq j\leq e_1}M(E^{k'_{j}}_1)$). Applying Proposition~\ref{condition_surgery} with this surgery diagram with appropriate order, we get the equality (\ref{condition}). \end{proof} Almost the same statement as the next corollary was given by Nozaki. \begin{cor} Let $M$ be a connected closed orientable 3-manifold. Suppose that $M\#(S^{2}\times S^1)$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,n+1}$ for $n\geq 1$ (resp. $\Sigma_{1,1}$). Then $M$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,n}$ (resp. $\Sigma_{0,2}$). \end{cor} \begin{proof} At first, for the case where $M\#(S^2\times S^1)$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,2}$, we can see that $H_{1}(M\#(S^2\times S^1))\cong H_{1}(M)\oplus H_{1}(S^2\times S^1)$ is generated by one element by Remark \ref{lowerbound}. Thus $M$ is an integral homology $3$-sphere and has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,1}$. In the following, for the case where $M\#(S^2\times S^1)$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,n+1}$, we assume that $n\geq 2$.\\ Fix a surgery diagram $\mathcal{D}$ of $M$, and let $m-1$ be the number of knots in $\mathcal{D}$. A surgery diagram $\mathcal{D}'$ of $M\#(S^{2}\times S^1)$ is obtained by adding disjoint unknot $U$ with surgery slope $0(=\frac{0}{1})$ to $\mathcal{D}$. We give an order among the link in $\mathcal{D}'$ so that $U$ is the first component. Then by Proposition \ref{condition_surgery}, we have a solution of (\ref{surg}) for $\Sigma_{0,n+1}$ (resp. $\Sigma_{1,1}$) and fix it. Let $\mathbb{X}$ denote the matrix in the left hand side of (\ref{surg}) and let $\mathbb{X}_{i,j}$ denote the $(i,j)$-entry of $\mathbb{X}$. Note that $\mathbb{X}_{1,i}=\mathbb{X}_{i,1}=0$ for $1\leq i\leq m$, and that $\mathbb{X}_{1,i}=\mathbb{X}_{i,1}$ for $m+1\leq i\leq m+n$ (resp. $m+1 \leq i\leq m+2$). \\ Since (\ref{surg}) holds, $\mathbb{X}_{1,m+1}, \mathbb{X}_{1,m+2},\dots,\mathbb{X}_{1,m+n}$ (resp. $\mathbb{X}_{1,m+1}, \mathbb{X}_{1,m+2}$) are coprime. This implies that the (ordered) set $\{ \mathbb{X}_{1,m+1}, \mathbb{X}_{1,m+2},\dots,\mathbb{X}_{1,m+n}\}$ (resp. $\{ \mathbb{X}_{1,m+1}, \mathbb{X}_{1,m+2}\}$) can be changed into $n$-element set (resp. two-element set) such that the first element is $1$ or $-1$ and the other elements are $0$ in finitely many steps, at each of which the $l$-times of the $i$-th element is added to the $j$-th element for some integer $l$ and $1\leq i\neq j\leq n$ (resp. $1\leq i\neq j\leq 2$). Fix one of such sequence of steps. Along this sequence, at each step, say the $l$-times of the $i$-th element is added to the $j$-th element, we change $\mathbb{X}$ as follows: Add $l$ times the $(m+i)$-th column to the $(m+j)$-th column and then add $l$ times the $(m+i)$-th row to the $(m+j)$-th row. Note that at any time in the sequence, each matrix obtained from $\mathbb{X}$ is also a ``solution'' of (\ref{surg}). A solution for a homologically fibered link whose homological fiber is homeomorphic to neither $\Sigma_{0,n+1}$ nor $\Sigma_{1,1}$ is not necessarily preserved under this operation. At the end of the sequence we get a matrix $\mathbb{X}'$ obtained from $\mathbb{X}$ whose $(1,i)$-entry is $0$ for $i\neq m+1$ and $\pm1$ for $i=m+1$. Note that the $(i,1)$-entry of $\mathbb{X}'$ is $0$ for $i\neq m+1$ and $\pm1$ for $i=m+1$. \\ Expand the determinant of $\mathbb{X}'$ along the first row (the only one cofacter survives) and expand the determinant of the cofacter along the first column (the only one cofacter survives). The cofacter finally obtained is a solution for $\Sigma_{0,n}$ (resp. $\Sigma_{0,2}$) in a 3-manifold which has a surgery diagram $\mathcal{D}$, representing $M$. \end{proof} \section{Examples}\label{sec_examples} Though we have Theorem~\ref{thm}, it is difficult in general to find a solution of (\ref{condition}) for a given manifold and homeomorphic type of a surface. Nozaki \cite{nozaki} proved that $L(p,q)$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{1,1}$ for every pair $(p,q)$ by solving some equations, which is equivalent to (\ref{condition}), using the density theorem. The author \cite{sekino} proved that $L(p,q)$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{0,3}$ for every pair $(p,q)$ by solving (\ref{condition}) following the Nozaki's argument. Thanks to Nozaki \cite{nozaki}, we see that ${\rm hc}(M(A^{p}(q)))=1$ for every $(p,q)$. Moreover, he also proved in his thesis that ${\rm hc} \left ((\#^{2m}S^2\times S^1)\#M(A^{p}(q))\right) =m+1$, and ${\rm hc}\left( (\#^{2m+1}S^2\times S^1)\#M(A^{p}(q))\right) =m+1$ if $q$ or $-q$ is quadratic residue modulo $p$ and ${\rm hc}\left( (\#^{2m+1}S^2\times S^1)\#M(A^{p}(q))\right) =m+2$ otherwise for every non-negative integer $m$. Thus we know ${\rm hc}(\cdot)$ for all 3-manifolds whose torsion linking form is $A^{p}(q)$. In this section, we determine ${\rm hc}(\cdot)$ for 3-manifolds whose torsion linking forms are the other generators for linkings, $(\#^{r}S^2\times S^1)\#M(E^{k}_{0})$ and $(\#^{r}S^2\times S^1)\#M(E^{k}_{1})$ for a non-negative integer $r$, we state as a proposition, proved at Subsections~\ref{hcek0} and~\ref{hcek1}: \begin{prop}In the following, $\lceil n \rceil$ and $\lfloor n \rfloor$ for a real number $n$ represent the minimal integer greater than or equal to $n$ and the maximal integer less than or equal to $n$, respectively. \begin{itemize} \item For $r\geq 0$, ${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{k}_{0})\right) =\lceil \frac{r}{2} \rceil+1$ if $k=1$ or $k\geq 3$. \item ${\rm hc}(M(E^{2}_{0}))=2$, and for $r\geq 1$ ${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{2}_{0})\right) =\lceil \frac{r}{2} \rceil+1$. \item For $r\geq 0$, ${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{2}_{1})\right) =\lceil \frac{r}{2} \rceil+1$. \item For $r\geq 0$, ${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{k}_{1})\right) =\lfloor \frac{r}{2} \rfloor+2$ if $k\geq 3$. \end{itemize} \end{prop} As a preparation, we review two observations: Firstly, the invariant ${\rm hc}(\cdot)$ is subadditive under the connected sum i.e. ${\rm hc}(M_{1}\#M_{2})\leq {\rm hc}(M_{1})+{\rm hc}(M_{2})$ since we get a homological fiber by the plumbing of two homological fibers, see \cite{sekino} for example. Secondly, it is known that $S^2\times S^1$ has a fibered link whose fiber surface is an annulus. By the plumbing it and a Hopf annulus in $S^3$, and by the plumbing two fibered annuli of $S^2\times S^1$, we get genus one fibered knots in $S^2\times S^1$ and $\#^{2}S^2\times S^1$. Since they are not integral homology $3$-spheres, we see that ${\rm hc}(S^2\times S^1)=1$ and ${\rm hc}\left( \#^{2}S^2 \times S^1\right) =1$. \subsection{ ${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{k}_{0})\right) $}\label{hcek0} We divide the argument into three cases, where $r$ is zero, where $r$ is positive even, and where $r$ is odd. \subsubsection{${\rm hc}\left( M(E^{k}_{0})\right) $}\label{hcek0_r0} We will compute ${\rm hc} \left( M(E^{k}_0)\right)$. Since $M(E^{k}_0)$ is not an integral homology 3-sphere, ${\rm hc}\left( M(E^{k}_0)\right) \geq 1$. Moreover, it is known that $M(E^{k}_0)$ has a fibered link whose fiber surface is homeomorphic to $\Sigma_{0,3}$ as in Figure~\ref{e^k_0_fib}: If we ignore the green curve, the right of Figure~\ref{e^k_0_fib} represents a fiber surface of a fibered link whose fiber surface is homeomorphic to $\Sigma_{0,3}$ in $L(2^{k},1)\#L(2^{k},1)$, which is obtained by the plumbing of fibered annuli in each prime components in appropriate way. The green curve is on this fiber, and note that the canonical framing of this curve is identical with the surface framing. Then $\left (-\frac{1}{2^{k}}\right)$-surgery (with respect to the canonical framing) along the green curve preserves the fiber structure. By the plumbings of two Hopf annuli, we get a genus two fibered knot in $M(E^{k}_0)$. This implies ${\rm hc}\left( M(E^{k}_0)\right) \leq2$. Thus it is enough to show whether $M(E^{k}_{0})$ has a genus one homologically fibered knot or not.\\ \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{e_k_0_fib.eps} \end{center} \caption{Surgery link on a planar fiber surface} \label{e^k_0_fib} \end{figure} By Theorem~\ref{thm}, $M(E^{k}_0)$ has a genus one homoloically fibered knot if and only if there exist integers $x,y,z,w,\alpha,\beta$ and $\gamma$ satisfying \begin{eqnarray} \label{hc_mek0} \pm1 &=& \left| \begin{array}{cccc} 0 & 2^{k} & x & y \\ 2^{k} & 0 & z & w \\ x & z & \alpha & \beta +1\\ y & w & \beta & \gamma \\ \end{array} \right| \nonumber \\ &=& -2^{2k}\{ \alpha \gamma - \beta(\beta +1)\} +2^{k+1}\{x(z\gamma -w\beta)-y(z\beta -w\alpha)\}-2^{k}(xw+yz)+(xw-yz)^{2} \end{eqnarray} \vspace{0.5cm} (i) $k=1$\\ In this case, it is known that $M(E^{1}_{0})$ has a genus one fibered knot as in Figure~\ref{e^1_0_fib}: If we ignore the green curve, the right of Figure~\ref{e^1_0_fib} represents a fiber surface of a genus one fibered knot in $L(2,1)\#L(2,1)$, which is obtained by the plumbing of fibered annuli in each prime components in appropriate way. The green curve is on this fiber, and note that the surface framing of this curve is the $(-1)$-slope with the canonical framing. This implies that the $\left (-\frac{1}{2}\right)$-slope of the green curve with respect to the canonical framing is the $\left( -\frac{1}{2}\right)$-slope with respect to the surface framing. Then the surgery along the green curve preserves the fiber structure. Thus $M(E^{1}_{0})$ has a genus one homologically fibered knot. In fact, $x=3,y=1,z=0,w=1,\alpha=-1,\beta=0$ and $\gamma=0$ is one of the solutions. This solution may correspond to a non-fibered knot. \begin{figure}[htbp] \begin{center} \includegraphics[width=100mm]{e_1_0_fib.eps} \end{center} \caption{Surgery link on a fiber surface of genus one} \label{e^1_0_fib} \end{figure} (ii) $k=2$\\ In this case, there exist no solutions. If there was, $(xw-yz)$ must be odd since the other terms in the right hand side of (\ref{hc_mek0}) are even. If $(xw-yz)$ is odd, then the right hand side of (\ref{hc_mek0}) is congruent to $5$ modulo $8$ by noting that $xw+yz=xw-yz+2yz$ and that the square of every odd integer is congruent to $1$ modulo $8$. This cannot be $\pm1$. (iii) $k\geq3$\\ In this case, $x=2^{k-1}, y=1, z=1, w=1, \alpha=2^{k-3}+1, \beta=0$ and $\gamma=0$ is one of the solutions of (\ref{hc_mek0}). \subsubsection{${\rm hc}\left( (\#^{2m}S^2 \times S^1)\# M(E^{k}_{0})\right) $ for $m\geq 1$}\label{hcek0_evenr} Since $H_{1}\left( (\#^{2m}S^2 \times S^1)\# M(E^{k}_{0}) \right) \cong \mathbb{Z}^{2m}\oplus \mathbb{Z}/2^{k}\mathbb{Z}\oplus \mathbb{Z}/2^{k}\mathbb{Z}$ requires at least $2m+2$ elements for generating, ${\rm hc}\left( (\#^{2m}S^2 \times S^1)\# M(E^{k}_{0})\right) \geq m+1$ by Remark~\ref{lowerbound}. By the plumbing of two fibered annuli of $S^2 \times S^1$ to a fiber surface of $M(E^{k}_{0})$ in the right of Figure~\ref{e^k_0_fib}, we get a genus two fibered knot in $(\#^{2}S^2 \times S^1)\# M(E^{k}_{0})$. Thus \begin{eqnarray*} \label{hc_mek0} {\rm hc}\left( (\#^{2m}S^2 \times S^1)\# M(E^{k}_{0})\right) &\leq& (m-1){\rm hc}( \#^{2}S^2 \times S^1) + {\rm hc}\left( (\#^{2}S^2 \times S^1)\# M(E^{k}_{0})\right) \\ &\leq& m+1. \end{eqnarray*} Therefore, ${\rm hc}\left( (\#^{2m}S^2 \times S^1)\# M(E^{k}_{0})\right) =m+1$ for $m\geq 1$. \subsubsection{${\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{0})\right) $ for $m\geq 0$} \label{hcek0_oddr} Since $H_{1}\left( \left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{0})\right) \right) \cong \mathbb{Z}^{2m+1}\oplus \mathbb{Z}/2^{k}\mathbb{Z}\oplus \mathbb{Z}/2^{k}\mathbb{Z}$ requires at least $2m+3$ elements for generating, ${\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{0})\right) \geq m+2$ by Remark~\ref{lowerbound}. By the plumbing of two fibered annuli, one is in $S^2 \times S^1$ and the other is in $S^3$, to a fiber surface of $M(E^{k}_{0})$ in the right of Figure~\ref{e^k_0_fib}, we get a genus two fibered knot in $(S^2 \times S^1)\# M(E^{k}_{0})$. Thus \begin{eqnarray*} \label{hc_mek0} {\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{0})\right) &\leq& m\cdot {\rm hc}( \#^{2}S^2 \times S^1) +{\rm hc}\left( (S^2 \times S^1)\# M(E^{k}_{0})\right)\\ &\leq& m+2. \end{eqnarray*} Therefore, ${\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{0})\right) =m+2$ for $m\geq 0$. \subsection{${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{k}_{1})\right)$}\label{hcek1} We divide the argument into four cases, where $r$ is zero, where $k$ is two, where $k$ is not two and $r$ is even, and where $k$ is not two and $r$ is odd. \subsubsection{${\rm hc}\left( M(E^{k}_{1})\right) $} We will compute ${\rm hc}(M(E^{k}_{1}))$ for $k\geq2$. Since $M(E^{k}_1)$ is not an integral homology 3-sphere, ${\rm hc}\left( M(E^{k}_{1})\right) \geq 1$. Moreover, $M(E^{k}_1)$ has a genus two fibered knot as in Figure~\ref{e^k_1_fib}: If we ignore the red curve and the blue curve, two of the bottom of Figure~\ref{e^k_1_fib} represent fiber surfaces of genus two fibered knots in $L(2^{k},1)\#L(2^{k},1)\#L\left( -\frac{2}{3} \{2^{k-1}-(-1)^{k-1}\}, 1\right)\#L\left(2,(-1)^{k-1}\right)$ for odd $k$ and even $k$, which are obtained by the plumbings of fibered annuli in each prime components in appropriate way. The red curve and blue curve are on this fiber. Note that the $0$-slope of the red curve with respect to the canonical framing is the $(-1)$-slope with respect to the surface framing, and that the $\left((-1)^{k}\cdot2\right)$-slope of the blue curve with respect to the canonical framing is the $1$-slope with respect to the surface framing. Then the surgery along the red curve and the blue curve preserves the fiber structure. Therefore we have ${\rm hc}\left( M(E^{k}_1)\right) \leq 2$. Thus it is enough to show whether $M(E^{k}_{1})$ has a genus one homologically fibered knot or not.\\ \begin{figure}[htbp] \begin{center} \includegraphics[width=120mm]{e_k_1_fib.eps} \end{center} \caption{Surgery link on a fiber surface of genus two} \label{e^k_1_fib} \end{figure} By Theorem~\ref{thm}, $M(E^{k}_1)$ has a genus one homologically fibered knot if and only if there exist integers $x,y,z,w,\alpha,\beta$ and $\gamma$ satisfying \begin{eqnarray} \label{hc_mek1} \pm1 &=& \left| \begin{array}{cccc} 2^{k+1} & -2^{k} & -x & -y \\ -3\cdot2^{k} & 2^{k+1} & -3z & -3w \\ x & z & \alpha & \beta +1\\ y & w & \beta & \gamma \\ \end{array} \right| \nonumber \\ &=& 2^{2k}\{ \alpha \gamma - \beta(\beta +1)\} +2^{k+1}\{\gamma(x^{2}+3xz+3z^{2})+\alpha(y^{2}+3yw+3w^{2})-(2\beta+1)(xy+3zw)\} \nonumber\\ &\ & \hspace{2.0cm} -3\cdot2^{k}(2\beta+1)(xw+yz)+3(xw-yz)^{2} \end{eqnarray} \vspace{0.5cm} (i) $k=2$\\ In this case, we get a genus one fibered knot as in Figure~\ref{e^2_1_fib}: If we ignore the green curve, the right of Figure~\ref{e^2_1_fib} represents a fiber surface of a genus one fibered knot in $L(4,1)\#L(4,1)$, which is obtained by the plumbing of fibered annuli in each prime components in appropriate way. The green curve is on this fiber, and note that the surface framing of this curve is the $(-1)$-slope with the canonical framing. This implies that the $\left (-\frac{3}{4}\right)$-slope of the green curve with respect to the canonical framing is the $\left( {1}{4}\right)$-slope with respect to the surface framing. Then the surgery along the green curve preserves the fiber structure. Thus $M(E^{2}_{1})$ has a genus one homologically fibered knot. In Fact, $x=-1,y=1,z=1,w=0,\alpha=1,\beta=0$ and $\gamma=0$ is one of the solutions. This solution may correspond to a non-fibered knot. Thus we have ${\rm hc}\left( M(E^{2}_{1})\right) =1$. \\ \begin{figure}[htbp] \begin{center} \includegraphics[width=150mm]{e_2_1_fib.eps} \end{center} \caption{Surgery link on a fiber surface of genus one} \label{e^2_1_fib} \end{figure} (ii) $k\geq3$\\ In this case, there exist no solutions of (\ref{hc_mek1}). In fact, all terms but the last in the right hand side of (\ref{hc_mek1}) are divisible by $8$ and the last term is congruent to neither $1$ nor $-1$ modulo $8$. Thus we have ${\rm hc}\left( M(E^{k}_{1})\right) \geq2$. We conclude ${\rm hc}\left( M(E^{k}_{1})\right) =2$. \subsubsection{${\rm hc}\left( (\#^{r}S^2\times S^1)\#M(E^{2}_{1})\right) $ for $r\geq 0$} Since $H_{1}\left( \left( (\#^{r}S^2 \times S^1)\# M(E^{2}_{1})\right) \right) \cong \mathbb{Z}^{r}\oplus \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$ requires at least $r+2$ elements for generating, ${\rm hc}\left( (\#^{r}S^2 \times S^1)\# M(E^{2}_{1})\right) \geq \lceil \frac{r}{2} \rceil +1$ by Remark~\ref{lowerbound}. On the other hand, \begin{eqnarray*} \label{hc_mek0} {\rm hc}\left( (\#^{r}S^2 \times S^1)\# M(E^{2}_{1})\right) &\leq& {\rm hc}( \#^{r}S^2 \times S^1) + {\rm hc}\left( M(E^{2}_{1})\right) \\ &\leq& \lceil \frac{r}{2} \rceil +1. \end{eqnarray*} Therefore, ${\rm hc}\left( (\#^{r}S^2 \times S^1)\# M(E^{2}_{1})\right) =\lceil \frac{r}{2} \rceil +1$ for $r\geq 0$. \subsubsection{${\rm hc}\left( (\#^{2m}S^2\times S^1)\#M(E^{k}_{1})\right) $ for $m\geq 0$ and $k\geq 3$} Note that \begin{eqnarray*} \label{hc_mek0} {\rm hc}\left( (\#^{2m}S^2\times S^1)\#M(E^{k}_{1})\right) &\leq& m\cdot {\rm hc}(\#^{2}S^2\times S^1)+ {\rm hc}\left( M(E^{k}_{1})\right) \\ &=& m+2. \end{eqnarray*} We will show that $(\#^{2m}S^2\times S^1)\#M(E^{k}_{1})$ has no homologically fibered links whose homological fibers are homeomorphic to $\Sigma_{m+1,1}$. This implies that ${\rm hc}\left( (\#^{2m}S^2\times S^1)\#M(E^{k}_{1})\right)=m+2$.\\ Suppose that $(\#^{2m}S^2\times S^1)\#M(E^{k}_{1})$ had a homologically fibered links whose homological fibers are homeomorphic to $\Sigma_{m+1,1}$. By applying Theorem~\ref{thm}, we have a solution $X$, $(2m+2) \times (2m+2)$-integer matrix, and $Y$, $(2m+2) \times (2m+2)$-symmetric integer matrix, for an equation below. \begin{eqnarray} \label{abcdef} \left| \begin{array}{cc} O_{2m} \oplus \left( \begin{array}{cc} 2^{k+1} & -2^{k} \\ -3\cdot 2^{k} & 2^{k+1} \\ \end{array} \right) & \left( I_{2m} \oplus \left( \begin{array}{cc} -1 & 0 \\ 0 & -3 \\ \end{array} \right) \right)X \\ X^{t} & Y+\mathcal{E}\\ \end{array} \right| = \pm 1 \end{eqnarray} Note that the left hand side is congruent to $\left| X^{t}\left( I_{2m} \oplus \left( \begin{array}{cc} -1 & 0 \\ 0 & -3 \\ \end{array} \right) \right) X \right| =3 \left| X\right| ^{2}$ modulo $8$. This cannot be $\pm1$, and this leads a contradiction. \subsubsection{${\rm hc}\left( (\#^{2m+1}S^2\times S^1)\#M(E^{k}_{1})\right) $ for $m\geq 0$ and $k\geq 3$} First, we show that $M(E^{k}_{1})$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{1,2}$. By Theorem~\ref{thm}, the existence of such a link is equivalent to that of a solution $a,b,c,d,e,f,v,w,x,y,z$ of the equation below. \begin{eqnarray} \label{qwerty} \left| \begin{array}{ccccc} 2^{k+1} & -2^{k} & -x & -y & -u \\ -3\cdot 2^{k} & 2^{k+1} & -3z & -3w & -3v \\ x & z & a & b+1 & d \\ y & w & b & c & e \\ u & v & d & e & f\\ \end{array} \right| = \pm 1 \end{eqnarray} By substituting $u=1, v=d=e=f=0$, the left hand side of (\ref{qwerty}) becomes equal to $\left| \begin{array}{ccc} 2^{k+1} & -3z & -3w \\ z & a & b+1 \\ w & b & c \\ \end{array} \right|$. Note that the existence of $a,b,c,z,w$ such that the determinant becomes $\pm 1$ is equivalent to that of a genus one homologically fibered knot in $L(2^{k+1},3)$ by Theorem~\ref{thm}, and the existence is guaranteed by Nozaki. Thus $M(E^{k}_{1})$ has a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{1,2}$. Moreover, we get a homologically fibered link whose homological fiber is homeomorphic to $\Sigma_{2,1}$ in $(S^2\times S^1)\#M(E^{k}_{1})$ by the plumbing with a fibered annulus in $S^2\times S^1$. \\ Since $H_{1}\left( \left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{1})\right) \right) \cong \mathbb{Z}^{2m+1}\oplus \mathbb{Z}/2^{k}\mathbb{Z}\oplus \mathbb{Z}/2^{k}\mathbb{Z}$ requires at least $2m+3$ elements for generating, ${\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{1})\right) \geq m+2$ by Remark \ref{lowerbound}. On the other hand, \begin{eqnarray*} \label{hc_mek0} {\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{1})\right) &\leq& m\cdot {\rm hc}( \#^{2}S^2 \times S^1)+ {\rm hc}\left( (S^2 \times S^1)\# M(E^{k}_{1})\right)\\ &\leq& m+2. \end{eqnarray*} Therefore, ${\rm hc}\left( (\#^{2m+1}S^2 \times S^1)\# M(E^{k}_{1})\right) =m+2$ for $m\geq 0$.

For some reason, this hit a nerve with me. Robin Williams who is best known for his work in Jumaji, Aladdin, Goodwill Hunting and Mrs Doubtfire was a massive video game fan. So much so, that he named his daughter Zelda after Princess Zelda herself. Zelda Williams is a massive gamer and from time to time streams video games to raise awareness for charity. She has no announced that she’ll be live streaming The Legend of Zelda Breath of the Wild later this week. She hasn’t announced exactly what time yet but you can follow her Twitch channel here.

An SME’s Quick 10-Step Guide To Cyber Security 29 September 2017 00:00 There's many ways of losing sensitive data or having it stolen, and with them come many repercussions and punishments - especially for SMEs. Nowadays, the cyber security strength of SMEs is no stranger to scrutiny. Many of the articles and blogs you'll come across that relate to SME cyber security will be littered with phrases such as "woefully underprepared", "easy targets" and "growing threat". But, for these smaller to medium-sized enterprises, there are still huge responsibilities to protect personal information that your business and it's employees collect and use. Sorry to be the bore, but breaching this means financial, operational and reputational turmoil - which many businesses find it difficult to ever overcome. And not just for the SME themselves; For smaller businesses acting in the supply chain for larger organisations, a potential breach can have catastrophic effects on your clients (if you're not aware of this sort of threat, then this blog is definitely worth a read!). So, dig into this guide to get a step-by-step view of how to practice cyber security for SMEs (with credit to the ICO). 1. Assess The Threats And Risks To Your Business The Risk: In order to establish what level of security is suitable for your organisation, it’s important to sit down and review the personal data you hold, and what risks they pose. Consider the processes involved when collecting, storing, using and disposing of personal data. Also, consider just how valuable the sensitive and personal information in your ranks might be, and what repercussions or distress this could cause to the relevant stakeholders should this data be compromised in a breach. Once you’ve gathered this information, you’ll have a clear view of what security measures are appropriate for your company's needs - and you’ll be ready to start putting them into place. 2. Get Certified With Cyber Essentials The Risk: Unfortunately, there isn’t one single product that will be able to give you a complete guarantee of security for your business. Perhaps the best approach is to use a set of security controls that complement each other - although ongoing support is hugely important for a good level of security. What To Do: This is where the ‘Cyber Essentials Scheme’ comes into play. This provides a clear outlook on five key controls for keeping information secure. These are: - Boundary Firewalls and Internet Gateways - Secure Configuration - Access Control - Malware Protection - Patch Management and Software Updates Obtaining a Cyber Essentials certificate and following these key steps can help provide certain security assurances and help protect personal data in your IT systems. 3. Secure Your Data On The Move And In The Office The Risk: Often overlooked, the physical security of company equipment can provide just as many risks as that of the network infrastructure. Personal data on these devices can become easily compromised due to loss of theft of laptops, mobiles, tablets and USBs. There’s also the risk of printed documents containing sensitive information that can be left lying around long after they’ve served their purpose. Allowing untrusted devices to connect to your company network or connecting company devices to untrusted networks, is another security issue to consider. Often, employees working on the move connect to open Wi-Fi sources in hotels, cafes and conferences - putting personal and company data at risk. What To Do: You can strengthen the physical security of your office by storing your servers in a separate room with added protection. Any devices, USBs or CDs that are used for backup should be locked away when not in use. For further protection, either ensure that personal data is not stored on the device or that is has been appropriately secured and cannot be accessed when lost or stolen. Good access control and encryption are important here, with different forms of encryption being: - Full Disk Encryption - All data on the device is encrypted - File Encryption - Individual files are encrypted Make sure you know exactly what protection you are applying to your data, as some software only offers password protection to stop people making changes to the data (which might not stop a criminal from reading the data). For mobile devices, there can sometimes be an option to enable a remote to disable or wipe facility. This allows you to send a signal to a lost or stolen device in order to locate it and, if needed, delete its data (you will most likely need to pre-register the device to use this type of service). If employees are authorised to connect their own devices to your company’s network, then the security risks will inevitably be heightened. Here’s a small article on how you can address Bring-Your-Own-Device (BYOD) security risks. 4. Secure Your Data In The Cloud The Risk: In an age where smartphones and tablets are seen as business necessities, there is a host of online services that require users to transfer data to remote computing facilities - otherwise known as ‘the cloud’. The risks of processing data in the cloud arise due to the fact of personal data, which you are responsible for, will leave your network and be processed in those systems managed by the cloud provider. This means you’ll need to assess the security measures that the cloud provider has in place to ensure they are appropriate. What To Do: It’s important to know what data is being stored in the cloud. Most modern devices can have cloud backup or sync services switched on by default. Enabling two-factor authentication (2FA) for remote access to your data cloud is also a key consideration. 5. Backup Your Data The Risk: Disasters such as floods, fires or even vandalism can happen. For businesses, especially those that are smaller sized, you need to be up and running as quickly as possible for financial and reputational purposes. As well as this, loss of data can also put you at risk of breaching the Data Protection Act. The next risk is one you’ll be sure to have heard of; Malware, specifically ransomware, can disrupt the availability of access to your data. Ransomware can encrypt your data until you pay up to have it decrypted. The only problem is, most cyber criminals have no intention of ever allowing you to regain access, regardless of whether you’ve paid. What To Do: You need to have a strong data backup strategy implemented in order to protect against disasters and malware. Avoid storing backups in a way that is permanently visible to the rest of the network. This can help dodge the risk of falling victim to malware, as well as having files accidentally deleted. Also, store at least one of your backups off-site. 6. Train Your Staff The Risk: Another criminally overlooked method of protection, employee security awareness can act as your front line of defence. In an astounding statistic, over 90% of all successful data breaches contain elements or are caused by, human error. A lack of cyber security education is at fault for many of these cases. One of the main traps for end users is when falling victim to emails containing malicious links or attachments, otherwise known as ‘phishing attacks'. There’s also a high proportion of human-prone errors that involve simply sending them an email to the wrong recipient, containing sensitive information. What To Do: Employees at all levels, including the c-suite, need to be aware of their role and responsibilities in cyber security. Security awareness training should be conducted on a regular basis, covering a wide range of topics from social media, to working remotely. A cyber security culture within your business is proven to significantly reduce the risks of employee-caused breaches or loss of data. Also, keeping up-to-date on the latest threats and risks is vital. Subscribing to security related-blogs is a good way of ensuring this (we offer a free subscription to our weekly roundup, which you can sign up for here). 7. Keep An Eye Out For Problems The Risk: Most breaches go undiscovered for months, long after the damage has been caused. But when being attacked by cyber criminals and malware, the warning signs can sometimes be in clear view. What To Do: Regularly checking your monitoring services (i.e., software messages, access control logs and other reporting systems) for any alerts is vital. Also, make sure you can check what software or services are running on your network. Ensure you can identify any warning signs of what’s there that shouldn't be. Scan your system for known vulnerabilities with regular scans and penetration testing. Address any vulnerabilities that you may find. 8. Know What You Should Be Doing The Risk: Any risks you find within your business need to be addressed in a consistent manner - a good company policy will allow you to do just that. Some businesses fail in their bid for protection as they’re not currently using the security they already have in place and are not always able to spot and flag a problem. You should consider what actions need to be put in place should you suffer a breach. Effective incident management can reduce the damage and distress caused to stakeholders. What To Do: Review the personal data you have and what methods of protection you have. Ensure you are compliant with any industry guidance or other legal requirements. Document the controls you have in place and find where you improvements are needed. Once these have been located and improvements are in place, monitor the controls and adjust them where necessary. Be proactive and minimise the damage of a potential data breach by considering the risks for each type of personal data you hold. Use an acceptable use policy and training materials for employees in order for them to understand their responsibilities in data protection. 9. Minimise Your Data The Risk: Some (or a lot) of the data you hold may be out-of-date, inaccurate or no longer of use. This happens in most companies, as large amounts of information get collected over time. But for protection (and because the DPA says so), personal data should be accurate, up-to-date, and no longer kept if not needed. What To Do: Make a decision as to whether the data is still needed. If it is, make sure it’s stored in the right place.Move any wanted data to a secure location in order to prevent unauthorised access. If you have data that you no longer need, it’s best to delete it. But make sure this is in line with your data retention/ disposal policies. Specialist software or assistance might need to do this in a fully secure manner. 10. Make Sure Your IT Contractor Is Doing What They Should Be The Risk: For an SME, outsourcing your IT requirements to a third party is common. But, you need to be satisfied that they are putting in at least the same level of security with your data as you are. What To Do: Ask for a security audit of the systems containing your data, which can help highlight vulnerabilities that need to be addressed. Also, review copies of the security assessments of your provider and visit them in person if also be held responsible if personal data gathered by you is extracted from your old IT equipment when it is resold.

\begin{document} \bigskip \centerline{\bf { SHUKLA COHOMOLOGY AND ADDITIVE TRACK THEORIES} } \bigskip \bigskip \bigskip \bigskip \centerline{\bf { Hans-Joachim BAUES and Teimuraz PIRASHVILI}} \bigskip \bigskip \bigskip \bigskip \bigskip \section{Introduction} It is well known that the Hochschild cohomology for associative algebras has good properties only for algebras which are projective modules over the ground ring. For general algebras behavior of Hochschild cohomology is more pathological, for example there is no long cohomological exact sequence corresponding to a short exact sequence for coefficients, etc. In early 60-s Shukla \cite{sh} developed a cohomology theory for associative algebras with nicer properties than Hochschild theory. Quillen in \cite{Q} indicated that the Shukla cohomology fits in his general framework of homotopical algebra. The approach of Quillen is based on simplicial methods, which are usually quite hard to deal with. The aim of this work is to give the foundation of Shukla cohomology based on chain algebras. We also give an application to the problem of strengthening additive track theories, which is based on the comparison homomorphism between Shukla and Mac Lane cohomology theories \cite{PW}. We believe that our approach is much simpler than one used in \cite{Q} or \cite{sh}. Let us recall that a track category is a category enriched in groupoids. A track category $\ta$ is called \emph{abelian} if for any arrow $f$ the group of automorphisms of $f$ is abelian. A track theory is an abelian track category with finite lax products. If it admits strong products then it is called strong track theory. The main result of \cite{BJP} asserts that any track theory is equivalent to a strong one. An additive track theory is a track theory which moreover possesses a lax zero object and finite lax coproducts such that the natural map from the lax coproduct to the lax product is a homotopy equivalence and the corresponding homotopy category is an additive category. An additive track theory is called very strong if it possesses a strong zero object and strong products which are also strong coproducts. By the result of \cite{BJP}, any additive track theory is equivalent to one which possesses strong products and lax coproducts or strong coproducts and lax products. We show that in general it is impossible to get both strong products and coproducts. However this is possible if certain obstructions vanish. In particular this is possible if hom's of the corresponding homotopy category are vector spaces over a field. The contents of the sections below are as follows. In Section \ref{hochcoh} we recall basics on Hochschild cohomology theory and especially relationship between abelian extensions which are split over ground ring, and elements of the second Hochschild cohomology. In Section \ref{crossed} we introduce crossed bimodules and crossed extensions. We recall the relationship between crossed extensions which are split over the ground ring, and elements of the third Hochschild cohomology. This section also contains a new interpretation of the classical obstruction theory in terms of crossed extensions (see Theorem \ref{jvarediancinaag}). We also discuss a different generalization of the relationship between different sort of extensions and higher cohomology. In Section \ref{shuklacoh} we define Shukla cohomology as a kind of derived Hochschild cohomology on the category of chain algebras and we prove basic properties of the Shukla cohomology including relationship with crossed bimodules. In the original paper Shukla used an explicit cochain complex for the definition of Shukla cohomology. Unfortunately this complex is very complicated to work with. Quillen instead used closed model category structure on the category of simplicial algebras. We use the closed model category structure on the category of chain algebras, which is developed in the Appendix. The Section \ref{shgam} is devoted to some computations of Shukla cohomology when the ground ring is the ring of integers or $\Z/p^2\Z$; we also consider the relationship between the Shukla cohomology over integers and over $\Z/p^2\Z$. In this direction we prove the following result. Let $A$ be an algebra over $\F_p$ and let $M$ be a bimodule over $A$, then the base change morphism $$\Sh^i(A/\K,M)\to \Sh^i(A/\Z,M),\ \ \ \K=\Z/p^2\Z$$ is always an epimorphism. It is an isomorphism in dimensions $0,1$ and $2$. We also prove that in dimension three the kernel of this map is isomorphic to $\H^0(A,M)$. The Section \ref{relh} solves the problem of constructing a canonical cochain complex for computing the Shukla cohomology in the important case when the ground ring is an algebra over a field. Our cochain complex consists of tensors, unlike the one proposed by Shukla. The Section \ref{hmlsh} recalls basics of Mac Lane cohomology \cite{mac} and relationship with Shukla cohomology. It is well known that these two theories are isomorphic up to dimension two. It turns out that for algebras over fields they are also isomorphic in dimension three. The section \ref{ota} continues the study of track theories which was started in \cite{BJP}. In this section we show that the straightforward version of the strengthening result for additive track theories is not true and we construct the corresponding obstruction. This obstruction is defined using the exact sequence relating third Shukla and Mac Lane cohomology and is a main application of the theory considered in the previous sections. The Appendix contains the basic definitions on closed model categories. It contains also a proof of the fact that chain algebras over any ground ring form a closed model category. This fact is used in Section \ref{shuklacoh}. At the end of the Appendix we introduce a closed model category structure on the category of crossed bimodules over any ground ring. In a forthcoming paper we introduce the notion of a {\it strongly additive track theory} and we will prove that any additive track category is equivalent to a strong one. The notion of strongly additive track theory is based on theory of square rings \cite{square}. The second author is indebted to Mamuka Jibladze for the idea to modify classical obstruction theory in terms of crossed bimodules. \section{ Preliminaries on Hochschild Cohomology}\label{hochcoh} Here we recall the basic notion on Hochschild cohomology theory and refer to \cite{HC} and \cite{homology} for more details. In this section $\K$ denotes a commutative ring with unit, which is considered as a ground ring, except for the section \ref{relh}. \subsection{Definition} Let $R$ be a $\K$-algebra with unit and let $M$ be a bimodule over $R$. Consider the module $$ C^n(R,M):= \Hom(R^{\otimes n},M) $$ (where $ \otimes =\otimes _K$ and $\Hom=\Hom_K$). The {\it Hochschild coboundary} is the linear map $d: C^n(R,M)\to C^{n+1}(R,M)$ given by the formula \begin{multline*} d(f)(r_1,...,r_{n+1})=r_1f(r_2,...,r_{n+1})+\\ +\sum_{i=1}^n(-1)^if(r_1,...,r_ir_{i+1},...,r_{n+1})+\\ (-1)^{n+1}f(r_1,...,r_n)r_{n+1}. \end{multline*} Here $f\in C^n(R,M)$ and $r_1,\cdots,r_{n+1}\in R$. By definition the {\it $n$-th Hochschild cohomology group} of the algebra $R$ with coefficients in the $R$-bimodule $M$ is the $n$-th homology group of the Hochschild cochain complex $C^*(R,M)$ and it is denoted by $\H^*(R,M)$. Sometimes these groups are denoted by $$\H^*(R/\K,M)$$ in order to make clear that the ground ring is $\K$. We are especially interested in cases $\K=\Z,\F_p,\Z/p^2\Z$. It is clear that for such a $\K$ one has $$\H^i(\F_p/\K, \F_p)=0, \ i\geq 1.$$ In the following sections we consider two modifications of Hochschild cohomology, known as Shukla and Mac Lane cohomology. It should be noted that in both theories the algebra $\F_p$ has nontrivial cohomology over the ground ring $\K=\Z$ or $\K=\Z/p^2\Z$. \subsection{$\K$-split exact sequences} Let $$\xymatrix{0\ar[r]&M_1\ar[r]^{\mu}&M\ar[r]^{\sigma}&M_2\ar[r]&0}$$ be an exact sequence of bimodules over $R$. It is called \emph{$\K$-split} if there exists a $\K$-linear map $u:M_2\to M$ such that $\sigma\circ u=\id_{M_2}$. This condition is equivalent to the following one: there is a $\K$-linear map $v:M\to M_1$ such that $v\circ \mu=\id_{M_1}$. Let $f:M\to N$ be a morphism of bimodules over $R$. It is called $\K$-split, if the following exact sequences $$0\to \Ker(f)\to M\to {\sf Im}(f)\to 0,$$ and $$ 0\to {\sf Im}(f)\to N\to {\sf Coker}(f)\to 0$$ are $\K$-split. If $\xymatrix{0\ar[r]&M_1\ar[r]^{\mu}&M\ar[r]^{\sigma}&M_2\ar[r]&0}$ is a $\K$-split exact sequence, then $$ 0\to C^*(R,M_1)\to C^*(R,M)\to C^*(R,M_2)\to 0$$ is exact in the category of cochain complexes and therefore yields the long cohomological exact sequence: $$\cdots \to \H^n(R,M_1)\to \H^n(R,M)\to \H^n(R,M_2)\to \H^{n+1}(R,M_1)\to \cdots$$ \subsection{Induced bimodules} The category of bimodules over $R$ is equivalent to the category of left modules over the ring $R^e:=R\t R^{op}$, where $R^{op}$ is the opposite ring of $R$, which is isomorphic to $R$ as a $\K$-module via the map $r\mapsto r^{op}$, $R\to R^{op}$, while the multiplication structure in $R^{op}$ is given by $r^{op}s^{op}=(sr)^{op}$. The multiplication map $R\t R^{op}\to R$ is an algebra homomorphism. We always consider $R$ as a bimodule over $R$ via this homomorphism. If $A$ and $B$ are left $R$-modules, then $\Hom(A,B)$ is a bimodule over $R$ by the following action $$(rfs)(a)=rf(sa), \ \ r,s\in R, a\in M, f\in \Hom(A,B)$$ A bimodule is called {\it induced} if it is isomorphic to $\Hom(R,A)$ for an $R$-module $A$. It is well-known \cite{homology} that the Hochschild cohomology vanishes in positive dimensions on induced bimodules. For a bimodule $M$ the map $$\mu:M\to \Hom(R,M)$$ given by $\mu(m)(r)=mr$ is a homomorphism of bimodules, which is also $\K$-split monomorphism, hence one has a $\K$-split short exact sequence $$0\to M\to \Hom(R,M)\to N\to 0$$ where $N={\sf Coker}(\mu)$, which yields the isomorphism \begin{equation}\label{induced} \H^{i+1}(R,M)\cong \H^{i}(R,N), \ \ i>0 \end{equation} This shows that there is a natural isomorphism \cite{homology} $$\H^*(R,M)\cong \Ext^*_{R^{e},\K}(R,M)$$ where subscript $\K$ indicates that $\Ext$-groups in question are defined in the framework of relative homological algebra, where the proper class consists of $\K$-split exact sequences. If $R$ is projective as a $\K$-module, then one can take the usual $\Ext$-groups $\Ext^*_{R^{e}}(R,M)$ instead of the relative $\Ext$-groups. In particular, the Hochschild cohomology vanishes in positive dimensions on injective bimodules, provided $R$ is projective as a $\K$-module. \subsection{Hochschild cohomology in dimension $0$} For $n=0$ one has $$ \H^0(R,M)=\left\{ m\in M \mid rm=mr \ {\rm for \ any }\ r\in R\right\} . $$ In particular $\H^0(R,R)$ coincides with the center ${\sf Z}(R)$ of the algebra $R$. \subsection{Hochschild cohomology in dimension $1$} For $n=1$ a 1-cocycle is a linear map $D:R\to M$ satisfying the identity $$ D(xy)=xD(y)+D(x)y, \ \ x,y\in R. $$ Such a map is called a {\it derivation} from $R$ to $M$ and the $\K$-module of derivations is denoted by $\der(R,M)$. A derivation $D:R\to M$ is a coboundary if it has the form $$\ad_m(r)=rm-mr$$ for some fixed $m\in M$; $\ad_m$ is called an {\it inner derivation}. Therefore $$ \H^1(R,M)=\der(R,M)/\lbrace {\rm Inner\ derivations} \rbrace . $$ In particular one has the exact sequence $$\xymatrix{0\ar[r]&\H^0(R,M)\ar[r]&M\ar[r]^-{\ad}\ &\ \der(R,M)\ar[r]& \H^1(R,M)\ar[r]&0}$$ \subsection{Hochschild cohomology in dimension $2$} It is clear that a $2$-cocycle of $C^*(R,M)$ is a linear map $f:R\t R\to M$ satisfying $$xf(y,z)-f(xy,z)+f(x,yz)-f(x,y)z=0, \ \ x,y,z\in R.$$ For any linear map $g:R\to M$ the formula $f(x,y)=xg(y)-f(xy)+f(x)y$ defines a cocycle, all such cocycles are called coboundaries. We let ${\sf Z}^2(R,M)$ and ${\sf B}^2(R,M)$ be the collections of all $2$-cocycles and coboundaries. Hence $$\H^2(R,M)={\sf Z}^2(R,M)/{\sf B}^2(R,M).$$ We recall the relation of $\H^2(R,M)$ to abelian extensions of algebras. An {\it abelian extension} (sometimes called also {\it a singular extension}) of associative algebras is a short exact sequence $$\xymatrix{ 0\ar[r]& M\ar[r] &E\ar[r]^p& R\ar[r] &0} \leqno (E) $$ where $R$ and $E$ are associative algebras with unit and $p:E\to R$ is a homomorphism of algebras with unit and $M^2=0$, in other words the product in $E$ of any two elements from $M$ is zero. For an elements $m\in M$ and $r\in R$ we put $mr:=me$ and $rm=:em$. Here $e\in E$ is an element such that $p(e)=r$. This definition does not depend on the choice of $e$. Therefore $M$ has a bimodule structure over $R$. An abelian extension $(E)$ is called $\K$-split if there exists a linear map $s:R\to E$ such that $ps=\id_R$. Assume we have a bimodule $M$ over an associative algebra $R$, then we let ${\cal E}(R,M)$ be the category, whose objects are the abelian extensions $(E)$ such that the induced $R$-bimodule structure on $M$ coincides with the given one. The morphisms $(E)\to (E')$ are commutative diagrams $$ \xymatrix{0\ar[r] &M\ar[r]\ar[d]^{\id} &E\ar[r]\ar[d]^{\phi} &R\ar[r]\ar[d]^{\id}&0\\ 0\ar[r] &M\ar[r] &E'\ar[r] &R\ar[r]&0} $$ where $\phi$ is a homomorphism of algebras with unit. Moreover, we let ${\cal E}_{\K}(R,M)$ be the category, whose objects are $\K$-split singular extensions. It is clear that the categories ${\cal E}(R,M)$ and ${\cal E}_{\K}(R,M)$ are groupoids, in other words all morphisms in ${\cal E}(R,M)$ and ${\cal E}_{\K}(R,M)$ are isomorphisms. We let ${\sf Extalg}(R,M)$ and ${\sf Extalg}_{\K}(R,M)$ be the classes of connected components of these categories. Clearly ${\sf Extalg}_{\K}(R,M)\subset{\sf Extalg}(R,M)$. According to \cite{homology} there is a natural bijection \begin{equation}\label{abgaf} \H^2(R,M)\cong {\sf Extalg}_{\K}(R,M). \end{equation} We also recall that the map $\H^2(R,M)\to {\sf Extalg}_{\K}(R,M)$ is given as follows. Let $f:R\t R\to M$ be a $2$-cocycle. We let $M\rtimes _f R$ be an associative $\K$-algebra which is $M\oplus R$ as a $\K$-module, while the algebra structure is given by $$(m,r)(n,s)=(ms+rn+f(r,s),rs).$$ Then $$\xymatrix{0\ar[r]&M\ar[r]^-i \ & \ M\rtimes _fR\ \ar[r]^-p\ &R\ar[r]&0}$$ is an object of ${\cal E}_{\K}(R,M)$. Here $i(m)=(m,0)$ and $p(m,r)=r$. \subsection{Cohomology of tensor algebras.} \cite{HC}, \cite{homology}. Let $V$ be a $\K$-module. For the tensor algebra $R=T^*(V)$ one has $H^i(R,-)=0$ for all $i\geq 2$. An algebra is called {\it free} if it is isomorphic to $T(V)$, where $V$ is a free $\K$-module. \subsection{Cup-product in Hochschild cohomology} For any associative algebra $R$ the cohomology $\H^*(R,R)$ is a graded commutative algebra under the cup-product, which is defined by $$(f\cup g)(r_1,\cdots ,r_{n+m}):=f(r_1,\cdots,r_n)g(r_{n+1},\cdots,r_{n+m})$$, for $f\in C^n(R,R)$ and $g\in C^m(R,R)$ (see \cite{gerstenhaber}). This product corresponds to the Yoneda product under the isomorphism $\H^*(R,R)\cong \Ext^*_{R^{e},\K}(R,R)$. \section{Crossed bimodules and Hochschild cohomology }\label{crossed} \subsection{Crossed bimodules} Let us recall that a chain algebra over $\K$ is a graded algebra $C_*=\bigoplus_{n\geq 0} C_n$ equipped with a boundary map $\partial: C_*\to C_*$ of degree $-1$ satisfying the Leibniz identity $$\partial (xy)=\partial (x)+(-1)^{|x|}x \partial(y).$$ \begin{De} A {\it crossed bimodule} is a chain algebra which is trivial in dimensions $\geq 2$. \end{De} Thus a crossed bimodule consists of an algebra $C_0$ and a bimodule $C_1$ over $C_0$ together with a homomorphism of bimodules $$C_1 \buildrel \partial \over \to C_0$$ such that $$\partial (c)c'=c\partial (c'), \ c,c'\in C_1,$$ Indeed, since $C_2=0$ the last condition is equivalent to the Leibniz identity $0=\partial(cc')=\partial(c)c'-c\partial(c')$. It follows that the product defined by $$c*c':=\partial (c)c'$$ where $c,c'\in C_1$ gives an associative non-unital $\K$-algebra structure on $C_1$ and $\partial:C_1\to C_0$ is a homomorphism of non-unital $\K$-algebras. The equivalent but less economic definition goes back at least to Dedecker and Lue \cite{delu}. The notion of crossed bimodules is an associative algebra analogue of crossed modules introduced by Whitehead \cite{hanry} in the group theory framework, which plays a major role in homotopy theory of spaces with nontrivial fundamental groups \cite{B1}, \cite{jll}. We let ${\sf Xmod}$ and ${\sf Xmod}_R$ be the category of crossed bimodules and crossed $R$-bimodules respectively. We have also a category ${\sf Bim/Alg}$, whose objects are triples $(C_0,C_1,\d)$, where $C_0$ is an associative algebra, $C_1$ is a bimodule over $C_0$ and $\d:C_1\to C_0$ is a homomorphism of bimodules over $C_0$. It is clear that ${\sf Xmod}$ is a full subcategory of ${\sf Bim/Alg}$ and the inclusion ${\sf Xmod}\subset {\sf Bim/Alg}$ has a left adjoint functor, which assigns $\d:{\wt C_1}\to C_0$ to $C_1\to C_0$. Here ${\wt C_1}$ is the quotient of $C_1$ under the equivalence relation $x\d(y)-\d(x)y\sim 0$, $x,y\in C_1$. We let ${\sf Mod/Alg}$ be the category whose objects are triples $(V,C,\d)$, where $C$ is an associative algebra, $V$ is a $\K$-module and $\d:V \to C$ is a linear map. One has the forgetful functor ${\sf Bim/Alg}\to {\sf Mod/Alg}$, which has a left adjoint functor sending $(V, C,\d)$ to the triple $(M,d,C)$, where $M=C\t V\t C$ and $d$ is the unique homomorphism of bimodules which extends $\d$. As a consequence we see that the forgetful functor ${\sf Xmod}\to {\sf Mod/Alg}$ also has a left adjoint. Of special interest is the case when $C$ is a free associative algebra and $V$ is a free $\K$-module on $X \subset V$. In this case the corresponding crossed bimodule is called {\it free crossed bimodule}. \subsection{Hochschild cohomology in the dimension $3$ and crossed extensions} Here we recall the relation between Hochschild cohomology and crossed bimodules (see Exercise E.1.5.1 of \cite{HC} or \cite{baumin}). \smallskip Let $\partial: C_1\to C_0$ be a crossed bimodule. We put $M=\ker (\partial)$ and $R=\Cok (\partial)$. Then the image of $\partial$ is an ideal of $C_0$. We have also $MC_1=0=C_1M$ and $M$ has a well-defined bimodule structure over $R$. Let $R$ be an associative algebra with unit and let $M$ be a bimodule over $R$. A {\it crossed extension} of $R$ by $M$ is an exact sequence $$0\to M\to C_1 \buildrel \partial \over \to C_0\to R\to 0$$ where $\partial:C_1\to C_0$ is a crossed bimodule, such that $C_0\to R$ is a homomorphism of algebras with unit and an $R$-bimodule structure on $M$ induced from the crossed bimodule structure coincides with the prescribed one. A crossed extension of $R$ by $M$ is {\it $\K$-split}, if all arrows in the exact sequence $$0\to M\to C_1 \buildrel \partial \over \to C_0\to R\to 0$$ are $\K$-split. For fixed $R$ and $M$ one can consider the category ${\bf Crossext}(R,M)$ whose objects are crossed extensions of $R$ by $M$. Morphisms are maps between crossed modules which induce the identity on $M$ and $R$. \begin{Le}\label{crospullback} Assume $(\d)$ is a crossed extension of $R$ by $M$ and a homomorphism $f:P_0\to C_0$ of unital $\K$-algebras is given. Let $P_1$ be the pull-back of the diagram $$\xymatrix{&P_0\ar[d]\\ C_1\ar[r]&C_0.}$$ Then there exists a unique crossed module structure on $P_1\to P_0$ such that the diagram $$\xymatrix{0\ar[r]&M\ar[d]^{\id}\ar[r]&P_1\ar[r]\ar[d]&P_0\ar[d]^f \ar[r]&R\ar[r]\ar[d]^{\id}&0\\ 0\ar[r] & M\ar[r] &C_1\ar[r]&C_0\ar[r] &R\ar[r] &0}$$ defines a morphism of crossed extensions. \end{Le} \begin{Co}\label{crosfiltri} In each connected component of ${\bf Crossext}(R,M)$ there is a crossed extension $$0\to M\to P_1 \to P_0\to R\to 0 \leqno(P)$$ with free algebra $P_0$ and for any other object $(\d)$ in this connected component there is a morphism $(P)\to (\d)$. Thus $(\d)$ and $(\d')$ are in the same component of ${\bf Crossext}(R,M)$ iff there exists a diagram of the form $(\d)\leftarrow (P)\to (\d')$. \end{Co} We let ${\bf Crossext}_{\K}(R,M)$ be the subcategory of $\K$-split crossed extensions. Morphisms are such morphisms from ${\bf Crossext}(R,M)$ that all maps involved are $\K$-split. Let ${\bf Cros}(R,M)$ and ${\bf Cros}_{\K}(R,M)$ be the set of components of the category of crossed extensions and the category of $\K$-split crossed extensions respectively. Then there is a canonical bijection: \begin{equation}\label{bauh3} \H^3(R,M)\cong {\bf Cros}_{\K}(R,M) \end{equation} (see for example Exercise E.1.5.1 of \cite{HC} or \cite{baumin}). A similar isomorphism for group cohomology was proved by Loday \cite{jllst}, see also \cite{MW}. We recall only how to associate a 3-cocycle to a $\K$-split crossed extension: $$0\to M\to C_1 \buildrel \partial \over \to C_0\buildrel \pi \over \to R\to 0$$ of $R$ by $M$. We put $V:={\sf Im}(\partial)$ and consider $\K$-linear sections $p:R\to C_0$ and $q:V\to C_1$ of $\pi: C_0\to R$ and $\partial:C_1\to V$ respectively. Now we define $m:R\t R\to V$ by $m(r,s):=q(p(r)p(s)-p(rs))$. Finally we define $f:R\t R\t R\to M$ by $$f(r,s,t):=p(r)m(s,t)-m(rs,t)+m(r,st)-m(r,s)p(t).$$ Then $(f,g,h)\in {\sf Z}^3(R/\K,M)$ and the corresponding class in $\H^3(R/\K,M)$ depends only on the connected component of a given crossed extension and in this way one gets the expected isomorphism (see \cite{baumin}). \subsection{Obstruction theory} Now we explain a variant of the classical obstruction theory in terms of crossed extensions (compare with Sections IV.8 and IV.9 of \cite{homology}). Let $$0\to M\to C_1 \buildrel \partial \over \to C_0\to R\to 0 $$ be a crossed extension of $R$ by $M$. {\it A $\d$-extension of $C_1$ by $R$} is a commutative diagram with exact rows $$\xymatrix{&0\ar[r]&C_1\ar[d]^{\id} \ar[r]^{\mu}&S \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]&R\ar[r]&0}$$ where $S$ is a unital $\K$-algebra and $\si$ is a homomorphism of unital $\K$-algebras. Furthermore the equalities $\mu(x)s=\mu(x\xi(s)))$ and $s\mu(x)=\mu(\xi(s)x)$ hold, where $x\in C_1$, $s\in S$. It follows then that product in $C_1$ induced from $S$ coincides with the $*$-product: $x*y=\d(x)y=x\d(y)$. Moreover one has the exact sequence $$\xymatrix{0\ar[r]& M\ar[r]^{\mu} & S\ar[r]^{\xi}& C_0\ar[r]& 0.}$$ It is clear that $\d$-extensions of $C_1$ by $R$ form a groupoid, whose set of components will be denoted by $_\d{\Ext}(R,C_1)$. Now we assume that $\d$ is a $\K$-split crossed extension. A $\d$-extension of $C_1$ by $R$ is called $\K$-split if $\xi$ is a $\K$-split epimorphism. Of course in this case $\si$ is $\K$-split as well. We let $_\d\Ext_{\K}(R,C_1)$ be the subset of $_\d{\Ext}(R,C_1)$ consisting of $\K$-split $\d$-extensions. \begin{The}\label{jvarediancinaag} The class of a $\K$-split crossed extension $$0\to M\to C_1 \buildrel \partial \over \to C_0\to R\to 0 \leqno(\d)$$ is zero in $\H^3(R,M)$ iff $_\d\Ext_\K(R,C_1)$ is nonempty. If this is the case then the group $\H^2(R,M)$ acts transitively and effectively on $_\d\Ext_\K(R,C_1)$. \end{The} {\it Proof}. For a crossed extension $\d$ one considers sections $p:R\to C_0$ and $q:V\to C_1$, $V= {\sf Im}(\d)$ as above. We may and we will assume that $p(1)=1$. Then the class of $(\d)$ in $\H^3$ is given by the cocycle $f(r,s,t):=p(r)m(s,t)-m(rs,t)+m(r,st)-m(r,s)p(t)$ where $m(r,s)=q(p(r)p(s)-p(rs))$. Given a $\d$-extension of $C_1$ by $R$: $$\xymatrix{&0\ar[r]&C_1\ar[d]^{\id} \ar[r]^{\mu}&S \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]&R\ar[r]&0}$$ choose a $\K$-linear section $v:C_0\to S$ such that $v(1)=1$. One puts $u=vp:R\to S$. Then $\si u= {\sf Id}_R$. One defines $n:R\t R\to C_1$ by $\mu (n(r,s))=u(r)u(s)-u(rs)$. We claim that \begin{equation}\label{cinakocikli} p(r)n(s,t)-n(rs,t)+n(r,st)-n(r,s)p(t)=0 \end{equation} Indeed, $$p(r)n(s,t)=u(r)n(s,t)=u(r)u(s)u(t)-u(r)u(st).$$ Similarly $n(r,s)p(t)=u(r)u(s)u(t)-u(rs)u(t)$. Thus $$p(r)n(s,t)-n(rs,t)+n(r,st)-n(r,s)p(t)=u(r)u(s)u(t)-u(r)u(st)-u(rs)u(t)$$ $$+u(rst)+u(r)u(st)-u(rst)-u(r)u(s)u(t)+u(rs)u(t)=0.$$ Since $m(r,s)=q\d n(r,s)$, it follows that $g(r,s)=m(r,s)-n(r,s)$ lies in $M$. Thus we obtain a well-defined linear map $g:R\t R\to M$. Then it follows from the equation (\ref{cinakocikli}) that $$f(r,s,t)=rg(s,t)-g(rs,t)+g(r,st)-g(r,s)t,$$ which shows that the class of $\d$ in $\H^3$ is zero. Given any normalized $2$-cocycle $h:R\t R\to M$, one can define a new $\d$-extension $S_h$ of $R$ by $C_1$ by putting $S_h=C_1\os R$ with the following multiplication: $$(x,r)(y,s)=(x*y+p(r)y+xp(s)+n(r,s)+h(x,y),xy).$$ This construction yields a transitive and effective action of $\H^2\!(R,\!M)$ on $_\d\!\Ext_\K\!(R,\!C_1)$. Conversely, assume that the class of $0\to M\to C_1 \buildrel \partial \over \to C_0\to R\to 0$ is zero in $\H^3(R,M)$. Thus there is a linear map $g:R\t R\to M$ such that $f(r,s,t)=rg(s,t)-g(rs,t)+g(r,st)-g(r,s)t.$ One can define $n: R\t R\to C_1$ by $n(r.s)=m(r,s)-g(r,s)$. Then $p(r)n(s,t)-n(rs,t)+n(r,st)-n(r,s)p(t)=0$ and therefore $S=R\os C_1$ with the product $(x,r)(y,s)=(x*y+p(r)y+xp(s)+n(x,y),xy)$ defines a $\d$-extension. \subsection{ Abelian and crossed $n$-fold extensions}\label{baumin-n} {\it An abelian twofold extension} of an algebra $R$ by an $R$-$R$-bimodole $M$ is an exact sequence $$\xymatrix{0\ar[r] &M\ar[r]^{\aa}& N\ar[r]^{\mu} &S\ar[r]^{\pi}& R\ar[r] &0}$$ where $N$ is a bimodule over $R$ and $\aa$ is a bimodule homomorphism. Moreover, $S$ is an associative algebra with unit and $\pi$ is a homomorphism of algebras with unit, such that $\Ker(\pi)$ is a square zero ideal of $S$. Furthermore, for any $s\in S$ and $n\in N$ one has $$\mu(n\pi(s))=\mu(n)s, \ \ \mu(\pi(s)n)=s\mu(n).$$ We let ${\cal E}^2(R,M)$ be the category of abelian twofold extensions of $R$ by $M$, whose connected components are denoted by ${\bf Extalg}^2(R,M)$. As usual we have also a $\K$-split variant ${\cal E}^2_\K(R,M)$ of the category ${\cal E}^2(R,M)$: Objects of ${\cal E}^2_\K(R,M)$ are $\K$-split twofold abelian extensions (i.e. $\aa$, $\mu$ and $\pi$ are $\K$-splits), and the morphisms in ${\cal E}^2_\K(R,M)$ are $\K$-splits, accordingly we let ${\bf Extalg}^2_\K(R,M)$ be the connected components of ${\cal E}^2_\K(R,M)$. Let us note that for any abelian twofold extension $$\xymatrix{0\ar[r] &M\ar[r]^{\aa}& N\ar[r]^{\mu} &S\ar[r]^{\pi}& R\ar[r] &0}$$ the morphism $\mu:N\to S$ is a crossed bimodule, where the action of $S$ on $N$ is given via $\pi$. It is clear that the induced $*$-product on $N$ is trivial. Thus one obtains the functor ${\cal E}^2(R,M)\to {\bf Crossext}(R,M)$, which takes the subcategory ${\cal E}^2_\K(R,M)$ to the category ${\bf Crossext}_\K(R,M)$. \begin{Le}\label{2abgaf} The natural map ${\bf Extalg}^2_\K(R,M)\to {\bf Cros}_\K(R,M)$ is a bijection and therefore $${\bf Extalg}^2_\K(R,M)\cong\H^3(R,M)$$ \end{Le} {\it Proof}. We just construct the inverse map $$\xi:\H^3(R,M)\to {\bf Extalg}^2_\K(R,M).$$ Consider the $\K$-split short exact sequence $$0\to M\to \Hom(R,M)\to N\to 0$$ and the corresponding isomorphisms (\ref{induced}), (\ref{abgaf}) $$\H^3(R,M)\cong \H^2(R,N)\cong {\bf Extalg}(R,N).$$ Take now an element $a\in \H^3(R,M)$. It corresponds under these isomorphisms to an abelian extension $0\to N\to S\to R\to 0$. By gluing it with $0\to M\to \Hom(R,M)\to N\to 0$ one obtains an abelian twofold extension $$0\to M\to \Hom(R,M)\to S\to R\to 0$$ In this way one obtains the expected map $\xi$. It is clear now how to introduce the notion of abelian $n$-fold extension for all $n\geq 2$ and get the same sort of isomorphism in higher dimensions. Following the earlier work of Huebschmann \cite{hueb}, recently Baues and Minian \cite{baumin} obtained another interpretation of Hochschild cohomology in dimensions $\geq 4$. They introduced the notion of crossed $n$-fold extension and proved that $n$-fold extensions classify $(n+1)$-dimensional Hochschild cohomology for all $n\geq 2$. For $n=2$ this is an isomorphism (\ref{bauh3}). Here we give a sketch how to deduce the case $n>2$ from the case $n=2$ and from the classical results of Yoneda \cite{yoneda}. This argument gives also a new proof of Lemma \ref{2abgaf}. Let $T$ be an additive functor from the category of bimodules over $R$ to the category of $\K$-modules. Objects of the category ${\cal E}^n(T)$ are pairs $(E,x)$, where $$0\to M\to E_1\to \cdots \to E_n\to 0$$ is a $n$-fold extension of $E_n$ by $M$ in the category of $R$-$R$-bimodules and $x\in T(E_n)$. Morphisms in ${\cal E}^n(T)$ are defined in an obvious way. Let ${\bf E}^n(T)$ be the set of components of the category ${\cal E}^n(T)$. A result of Yoneda asserts that one has a natural isomorphism: $${\bf E}^n(T)\cong S^n T(M)$$ where $S^nT$ is the $n$-th satellite of $T$ \cite{CE}. Comparing with the definition of abelian twofold extension we see that $${\bf Exalg}^2(R,M)\cong {\bf E}^1(T)$$ where $T={\bf Extalg}(R,-)$. To show how to deduce Lemma \ref{2abgaf} from the Yoneda isomorphism, we consider the case when $\K$ is a field. Since $T\cong \H^2(R,-)$ the result of Yoneda yields $${\bf Extalg}^2(R,M)\cong S^1T(M)\cong \H^3(R,M).$$ This argument works also for general $\K$: we have to use a straightforward generalization of the Yoneda isomorphism in the framework of relative homological algebra. Let $n\geq 2$. A {\it crossed $n$-fold extension of $R$ by $M$} \cite{baumin} is an exact sequence $$\xymatrix{0\ar[r]&M\ar[r]^-{f}&M_{n-1}\ar[r]^{\d_{n-1}}&\cdots \ar[r]^{\d_2}& M_1\ar[r]^{\d_1}& A\ar[r]^{\pi}& R\ar[r]&0 } $$ of $\K$-modules with the following properties: i) $(M_1,R,\d_1)$ is a crossed bimodule with cokernel $R$; ii) $M_i$ is a bimodule over $R$ for $1<i\leq n-1$ and $\d_i$ and $f$ are maps of bimodules over $R$. Note that $\Ker(\d_1)$ is naturally a bimodule over $R$ and therefore it makes sense to require $\d_2$ to be a map of bimodules over $R$. We let ${\bf Cros}^n(R,M)$ denote the set of connected components of the category of crossed $n$-fold extensions of $R$ by $M$. Observe that $${\bf Cros}^n(R,M)={\bf E}^{n-2}(T)$$ where $T={\bf Cros}(R,-)$. Now, as in \cite{baumin} for simplicity we assume that $\K$ is a field. Theorem 4.3 of \cite{baumin} claims that there is a natural isomorphism \begin{equation}\label{baun} {\bf Cros}^n(R,M)\cong \H^{n+1}(R,M) \end{equation} For $n=2$ this is the isomorphism (\ref{bauh3}) and for $n>2$ it is an immediate corollary of Yoneda's isomorphism: \begin{multline*} {\bf E}^{n-2}(T)=S^{n-2}T=S^{n-2}\H^3(R,-)(M)=\\ S^{n-2}S^3\H^0(R,-)(M)\cong S^{n+1}\H^0(R,-)(M)=\H^{n+1}(R,M) \end{multline*} Here we used the isomorphism $T\cong H^3(R,-)$ and the classical fact that $\H^n(R,M)=\Ext_{R\t R^{op}}^n(R,M)=S^n\H^0(R,-)(M)$ see \cite{CE}. For general $\K$ one needs to work in the framework of relative homological algebra \cite{homology}. The corresponding class of proper exact sequences consists of $\K$-split exact sequences. Then the corresponding results hold for arbitrary $\K$. As we can see the results in this section strongly depend on the vanishing of Hochschild cohomology on (relative) injective modules. \section{Shukla Cohomology}\label{shuklacoh} As we already saw the Hochschild cohomology in dimensions two and three classifies $\K$-split abelian and crossed extensions respectively. However, there is a variant of Hochschild cohomology due to Shukla in the early 60-s which classifies all abelian and crossed extensions. We will present these results. Our approach to Shukla cohomology is based on chain algebras and especially on the possibility of extension of Hochschild cohomology to chain algebras. Actually there are two ways for such extension. First is a very naive: one replaces $\otimes $ and $\Hom$ in the definition of Hochschild cohomology by the tensor product and Hom of complexes to arrive at a cosimplicial cochain complex and then one takes the homology of the total complex. However, this definition does not respect weak equivalences of chain algebras. The second definition (called derived Hochschild cohomology) is a kind of Quillen's derivative of the Hochschild cohomology and uses the closed model category structure on the category of chain complexes introduced in the Appendix. Since the category of algebras is the full subcategory of the category of chain algebras, the derived Hochschild cohomology restricts to a cohomology theory of algebras, which is by definition the Shukla cohomology. \subsection{Hochschild cohomology for chain algebras} In this section we give a naive definition of the Hochschild cohomology for chain algebras. Let us recall that a chain algebra is a graded algebra $R_*=\bigoplus_{n\geq 0} R_n$ equipped with a differential $d:R_n\to R_{n-1}$ satisfying the Leibniz identity: $$d(xy)=d(x)y+(-1)^{n}xd(y), \ \ x\in A_n, y\in A_m.$$ Let ${\bf DGA}$ be the category of chain algebras. A morphism of chain algebras is a weak equivalence if it induces isomorphism in homology. An $R_*$-bimodule is a chain complex $M_*$ of $\K$-modules, equipped with actions from both sides: $R_*\t M_*\to M_*$ and $M_*\t R_*\to M_*$, satisfying usual axioms. However, for our purposes we restrict ourselves to the case when $M$ is concentrated in degree zero. In this case $R_*$-bimodule means simply a bimodule over $H_0(R_*)$. In particular $xm=0=mx$ as soon as $m\in M$ and $|x| \geq 1$. For a chain algebra $R_*$ and a $H_0(R_*)$-bimodule $M$ we denote by $C_*(R_*,M)$ the total complex of the following cosimplicial cochain complex. The $n$-th component of this cosimplicial object is the cochain complex $$ C^n(R_*,M):= \Hom(R^{\otimes n}_*,M). $$ Here $ \otimes$ denotes the tensor product of chain complexes. The coface operations are given via Hochschild coboundary formula: $$ d^0(f)(r_1,...,r_{n+1})=(-1)^{nk}r_1f(r_2,...,r_{n+1}), \ \ f:R^{\otimes n}_* \to M, \ \ r_1\in R_k$$ (actually this expression is zero provided $k>0$) $$ d^i(f)(r_1,...,r_{n+1})= f(r_1,...,r_ir_{i+1},...,r_{n+1}), \ \ 0<i<n+1 $$ $$d^{n+1}(f)(r_1,...,r_{n+1})=f(r_1,...,r_n)r_{n+1}.$$ The homology of $C_*(R_*,M)$ is denoted by $\H^*(R_*,M)$. The spectral sequences of a bicomplex in our situation have the following form: $$E_{pq}^1=H^q(\Hom(R_*^{\t p},M))\then \H^{p+q}(R_*,M)$$ $$F_{pq}^1=\H^q(| R_*|,M)\then\H^{p+q}(R_*,M)$$ Here $| R_*|$ denotes the underlying graded algebra of the chain algebra $(R_*,\d)$. \begin{Le}\label{hhisinvariantoba} Let $f:R_*\to S_*$ be a weak equivalence of chain algebras and let $M$ be a bimodule over $H_0(S)$. Then the induced homomorphism $$\H^*(S_*,M)\to \H^*(R_*,M)$$ s an isomorphism provided $R_*$ and $S_*$ are projective $\K$-modules. \end{Le} {\it Proof}. It is well known that any weak equivalence between degreewise projective bounded below chain complexes is a homotopy equivalence. Thus $f$ is a homotopy equivalence in the category of chain complexes of $\K$-modules. Therefore the induced map $R_*^{n \t}\to S_*^{n \t}$ is also a homotopy equivalence. It follows that the induced map of cosimplicial cochain complexes is a homotopy equivalence in each degree and therefore it induces a weak equivalence on the total complex level thanks to the spectral sequence argument associated to the the bicomplex. \subsection{The complex $\Der(| R_*|,M)$}\label{dercompleksi} Let $R_*$ be a chain algebra and $M$ be an $H_0(R_*)$-bimodule. We can take the derivations $| R_*|\to M$ from the underlying graded algebra to $M$. Since $| R_*|$ is graded, the space of derivations $\Der(| R_*|,M)$ is also graded. Since $M$ is concentrated only in dimension zero, we see that the $0$-th component is the space of all derivations $R_0\to M$, while in dimensions $n>0$ we get the space of linear maps $f:R_n\to M$ satisfying the conditions $$f(xy)=xf(y), \ \ f(yx)=f(x)y, \ \ x\in R_0, y\in R_n, n>0$$ $$f(uv)=0, \ \ u\in R_i, v \in R_j, i+j=n, i>0, j>0.$$ The boundary map $\partial: R_n\to R_{n-1}$ in $R_*$ yields a cochain complex structure on $\Der(| R_*|,M)$. In what follows $\Der(| R_*|, ,M)$ is always considered with this cochain complex structure. \Lem{Let $R_*$ be a quasi-free algebra, meaning that the underlying algebra structure is free, and let $M$ be an $H_0(A_*)$-bimodule. Then the Hochschild cohomology $\H^n(A_*,M)$ is isomorphic to the $(n-1)$-st homology of the cochain complex $\Der(| R_*|,M)$ provided $n>0$. } {\it Proof}. This is a direct consequence of the spectral sequence related to the bicomplex $C^*(R_*,M)$ together with the fact that the Hochschild cohomology of a free algebra is zero in dimensions $>1$. We also recall the K\"unneth formula for Hochschild cohomology \begin{Le}\cite{homology}\label{kiuneti} Let $\K$ be a field and let $R_*$ and $S_*$ be chain algebras. Assume that for each $n$, $R_n$ and $S_n$ are finite dimensional vector spaces. Then for any $R_*$-bimodule $M$ and $S_*$-bimodule $N$ one has the following isomorphism $$\H^n(R\t S, M\t N)\cong \bigoplus_{i+j=n}\H^i(R,M)\t \H^j(S,N)$$ \end{Le} \subsection{Derived Hochschild cohomology and Shukla cohomology} In this section we use the closed model category structure on chain algebras described in the Appendix. Let us recall that weak equivalences in this model category are usual ones and a morphism of chain algebras is a fibration if it is surjective in all positive dimensions. We also need the fact that any cofibrant object is a retract of a quasi-free algebra. It follows from Lemma \ref{hhisinvariantoba} that for any weak equivalence $f:R_*\to S_*$ of cofibrant chain algebras and any $H_0(S)$-bimodule $M$ the induced homomorphism $\H^*(S_*,M)\to \H^*(R_*,M)$ is an isomorphism. We can use this fact to define the derived Hochschild cohomology as follows. Let $R_*$ be a chain algebra. Thanks to the properties of closed model categories there exists a chain algebra morphism $f:R_*^c\to R_*$ which is a weak equivalence and $R_*^c$ is a cofibrant. For any $R$-bimodule $M$ the groups $\H^*(R^c_*,M)$ do not depend on the cofibrant replacement and they are called {\it the derived Hochschild cohomology of $R_*$ with coefficients in $M$} and are denoted by ${\bf H}^*(R_*,M)$. Thus $${\bf H}^*(R_*,M):=\H^*(R^c_*,M)$$ This definition has expected functorial properties: for any morphism $f:R_*\to S_*$ of chain algebras and any $H_0(S)$-bimodule $M$ there is a well-defined homomorphism ${\bf H}^*(S_*,M)\to {\bf H}^*(R_*,M)$ which depends only on the homotopy class of $f$. Moreover it is an isomorphism provided $f$ is a weak equivalence. One has also a natural homomorphism $\H^*(R_*,M)\to {\bf H}^*(R_*,M)$ which is induced by the chain algebra homomorphism $R_*^c\to R_*$. The following fact is a direct consequence of Lemma \ref{hhisinvariantoba}. \begin{Le} If $R_*$ is projective as a $\K$-module then $\H^*(R_*,M)\to {\bf H}^*(R_*,M)$ is an isomorphism. \end{Le} Since the category of algebras is a full subcategory of the category of chain algebras we can consider the restriction of the derived Hochschild cohomology ${\bf H}^*$ on the category of algebras. The resulting theory is called {\it the Shukla cohomology.} Thus for any algebra $R$ and any $R$-bimodule $M$ the Shukla cohomology of an algebra $R$ with coefficients in $M$ is defined by $$\Sh^*(R,M):={\bf H}^*(R,M)\cong H^*(\Der(R^c_*,M))$$ where $R^c_*\to R$ is a weak equivalence from a quasi-free chain algebra $R^c_*$. The natural transformation $$\H^n(R,M)\to \Sh^n(R,M), \ \ n\geq 0$$ is an isomorphism in dimensions $n=0,1$ and it is an isomorphism in all dimensions provided $R$ is projective as a $\K$-module. For example we have $\Sh^i(A,-)=0$ provided $A$ is a free algebra and $i\geq 2$. The cup-product in Hochschild cohomology yields a (commutative graded) algebra structure on $\Sh^*(A,A)$. \subsection{Shukla cohomology and extensions}\label{gafshu} The following properties of Shukla cohomology are of special interests. They are non-$\K$-split analogues of the isomorphisms (\ref{abgaf}) and (\ref{bauh3}). \begin{The}\label{shugaf} Let $A$ be an associative algebra and let $M$ be an $A$-bimodule. Then there are natural isomorphisms $$\Sh^2(A,M)\cong {\bf Extalg}(A,M)$$ $$\Sh^3(A,M)\cong {\bf Cros}(A,M).$$ \end{The} The first isomorphism is well known (see Theorem 4 of \cite{sh}). However we give an independent proof. {\it Proof}. i) Let $$ 0\to M\to E\to A\to 0 \leqno (E) $$ be a singular extension of algebras. Define the chain algebra $E_*$ as follows: $$E_0=E, \ \ E_1=M, \ \ E_n=0, \ \ n\geq 2$$ The only nontrivial boundary map is induced by the inclusion $M\to E$. Then one has a map of chain algebras $E_*\to A$ which is an acyclic fibration. Let $A_*\to A$ be a weak equivalence with quasi-free $A_*$. Since $A_*$ is cofibrant there exists a lifting $f_*:A_*\to E_*$. We consider now the first component $f_1: A_*\to E_1=M$ of $f_*$. Since $f_*$ is a homomorphism of algebras it follows that $f_1\in {\sf Der}(A_*,M)$ is a 1-cocycle of $ {\sf Der}(A_*,M)$ and therefore it gives a class $e(E)\in \Sh^2(A,M)$. If $g_*:A_*\to E_*$ is another lifting, then the values of $h=f_0-g_0:A_0\to E$ lie in $M$. Thus $h\in {\sf Der}(A_0,M)$ and $f_1-g_1= \partial ^*(h)$, which shows that the class $e(E)$ depends only on the isomorphism class of $(E)$. Conversely, if $f\in {\sf Der}(A_*,M)$ is a 1-cocycle, then one can form an abelian extension according to the following diagram: $$ \xymatrix{\cdots \ar[r] &\ A_2\ar[r]\ar[d] &A_1 \ar[r]\ar[d]^{f}&A_0 \ar[r]\ar[d] &A \ar[r]\ar[d]^{Id} &0\\ &0\ar[r] & M\ar[r] &E\ar[r] &A\ar[r] &0.} $$ In this way we obtain the isomorphism i). ii) Let $$0\to M\to C_1 \buildrel \partial \over \to C_0\to A\to 0$$ be a crossed extension. The algebra $C_0$ acts on $M$ via the projection to $A$. Moreover $$C_*=(\cdots \to0\to M\to C_1 \buildrel \partial \over \to C_0)$$ can be considered as a chain algebra as follows. In dimensions $0$ and $1$ it is already defined. In the dimension two one puts $C_2=M$, and $C_i=0$ for $i>2$. The pairing $C_i\t C_j\to C_{i+j}$ is the given one if $i=0$ or $j=0$, while the pairing $C_1\t C_1\to C_2$ as well as all other pairings are zero. Then $C_*\to A$ is an acyclic fibration. Therefore we have a lifting $f_*:A_*\to C_*$, where $A_*\to A$ is a weak equivalence with quasi-free $A_*$. It is clear that $f_2\in {\sf Der}(A_*,M)$ is a 2-cocycle in ${\sf Der}(A_*,M)$ and therefore gives rise to an element in $\Sh^3(A,M)$. Conversely, starting with a 2-cocycle $f\in {\sf Der}(A_*,M)$ one can construct the corresponding crossed extension using the diagram $$ \xymatrix{\cdots \ar[r]&A_3 \ar[r] \ar[d] &\ A_2\ar[r]\ar[d]^f &A_1 \ar[r]\ar[d]&A_0 \ar[r]\ar[d]^{Id} &A \ar[r]\ar[d]^{Id} &0\\ &0\ar[r] & M\ar[r] &C_1\ar[r]&C_0\ar[r] &A\ar[r] &0.} $$ The following theorem is the non-$\K$-split analogue of Theorem \ref{jvarediancinaag}. \begin{The} \label{nulisgamocnoba} The class of a crossed extension $$0\to M\to C_1 \buildrel \partial \over \to C_0\buildrel \pi\over \to R\to 0 \leqno(\d)$$ is zero in $\Sh ^3(R,M)$ iff $_\d\Ext(R,C_1)$ is nonempty. If this is the case then the group $\Sh ^2(R,M)$ acts transitively and effectively on $_\d\Ext(R,C_1)$. \end{The} {\it Proof}. It is clear that the crossed extension $$\xymatrix{ 0\ar[r]&M \ar[r]^{\id}&M\ar[r]^{0}&R\ar[r]^{\id}&R\ar[r]&0}$$ represents the zero element of ${\bf Cros}(R,M)$. Assume $_\d\Ext(R,C_1)$ is nonempty and let $$\xymatrix{&0\ar[r]&C_1\ar[d]^{\id} \ar[r]^{\mu}&S \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]^\pi&R\ar[r]&0}$$ be an object of the category $_\d\Ext(R,C_1)$. Then $d:M\oplus C_1\to S$ is a crossed bimodule, where $d(m,c_1)=\mu(c_1)$ and the action of $S$ on $M\oplus C_1$ is given by $s(m,c_1)=(\xi(s)m,\xi(s)c_1)$ and $(m,c_1)s=(m\xi(s),c_1\xi(s))$. Then one has the following commutative diagram in ${\bf Cross}(R,M)$: $$\xymatrix{ 0\ar[r]&M \ar[r]^{\id}&M\ar[r]^{0}&R\ar[r]^{\id}&R\ar[r]&0\\ 0\ar[r]&M\ar[d]^{\id} \ar[u]^{\id}\ar[r]^{i_1}&M\oplus C_1\ar[d]^{p_2}\ar[u]^{p_1}\ar[r]^{\mu}&S\ar[u]^{\si} \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]\ar[u]^{\id}&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]^\pi&R\ar[r]&0}$$ which shows that the class of $(\d)$ in $\Sh ^3(R,M)$ is zero. Here $p_1$ and $p_2$ are standard projections from the direct sums to summands and $i_1$ and $i_2$ are corresponding injections. Conversely, assume the class of $(\d)$ in $\Sh ^3(R,M)$ is zero. It follows from Corollary \ref{crosfiltri} that there exists a commutative diagram of crossed extensions: $$\xymatrix{ 0\ar[r]&M \ar[r]^{\id}&M\ar[r]^{0}&R\ar[r]^{\id}&R\ar[r]&0\\ 0\ar[r]&M\ar[d]^{\id} \ar[u]^{\id}\ar[r]^{i}&P_1 \ar[d]^{\ee}\ar[u]^{p}\ar[r]^{\mu}&P_0\ar[u]^{\si} \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]\ar[u]^{\id}&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]^\pi&R\ar[r]&0.}$$ It follows that the restriction of $\mu$ to $\Ker(p)$ is a monomorphism and therefore we have the following commutative diagram with exact rows: $$\xymatrix{ & 0\ar[r]&\Ker(p) \ar[d]^{\ee}\ar[r]^{\mu}&P_0 \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]^\pi&R\ar[r]&0.}$$ One defines the $\K$-algebra $S$ via the exact sequence $$\xymatrix{ 0\ar[r]&\Ker(p)\ar[r]^{(-\ee,\mu)}&C_1\oplus P_0\ar[r]&S\ar[r]&0.}$$ Here the product on $C_1\oplus P_0$ is given by $$(c,x)(c',x'):=(c*c'+c\xi(x')+\xi(x)c',xx').$$ One easily checks that $\Ker(p)$ is really an ideal of $C_1\oplus P_0$ and therefore $S$ is well-defined. Now it is clear that $$\xymatrix{&0\ar[r]&C_1\ar[d]^{\id} \ar[r]&S \ar[d]\ar[r] &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]^\pi&R\ar[r]&0}$$ is an object of $_\d\Ext(R,C_1)$ and the proof is finished. \ \ \noindent {\bf Remark}. One cannot get non-$\K$-split versions of results of Section \ref{baumin-n}. In other words for $n>2$ neither ${\bf Extalg^n}(R,M)$ nor ${\bf Cros}^n(R,M)$ are isomorphic to $\Sh^{n+1}(R,M)$ in general. This is because for such $n$ both groups ${\bf Extalg^n}(R,M)$ and ${\bf Cros}^n(R,M)$ vanish on injective bimodules, while Shukla cohomology does not. Indeed, if $\K=\Z$ and $R=\F_p$, then any bimodule over $R$ is injective, while the computation in Section \ref{shugaf}) shows that $\Sh^{2i}(\F_p/\Z,F_p)=\F_p$ for all $i$. By the same reason the groups ${\bf Extalg^2}(R,M)$ and $\Sh^2(R,M)$ are different. On the other hand we have another interpretation of the higher Shukla cohomology using chain algebras. Indeed, the above argument can be easily modified to get the following extension of Theorem \ref{shugaf} to higher dimensions. A chain algebra $A_*$ is called of length $\leq n$ if $A_i=0$ for all $i>n$. Let $R$ be an algebra and $M$ be a bimodule over $R$. For any $n\geq 1$ we let ${\bf Crosext}^{n}(R,M)$ be the category of triples $(A_*,\alpha, \beta)$ where $A_*$ is a chain algebra of length $\leq n$ with property $H_i(A_*)=0$ for all $0<i<n$. Moreover $\alpha:H_0(A_*)\to R$ is an isomorphism of algebras and $\beta:M\to H_n(A_*)$ is an isomorphism of $R$-bimodules, where the $R$-bimodule structure on $H_n(A_*)$ is induced via $\alpha^{-1}$. It is clear that for $n=1$ the category ${\bf Crosext}^{1}(R,M)$ and ${\bf Crosext}(R,M)$ are equivalent. The argument given in the proof of part ii) of Theorem \ref{shugaf} shows that connected components of the category ${\bf Crosext}^{n}(R,M)$ are in one-to-one correspondence with elements of the group $\Sh^{n+2}(R,M)$. Furthermore, for a given object $X=(A_*,\alpha, \beta)$ of the category ${\bf Crosext}^{n}(R,M)$ one can define the category $_X\Ext(R;A_n)$ of objects $(C_*,\gamma, \eta)$, where $C_*$ is a chain algebra of length $\leq n$ with the property $H_i(C_*)=0$ for all $i>0$, $\gamma:H_0(C_*)\to R$ is an isomorphism of algebras and $\eta:C_*\to A_*$ is a chain algebra homomorphism such that the diagram $$\xymatrix{H_0(C_*)\ar[r]^{\gamma}\ar[d]^{\eta}&R\ar[d]^{\id}\\ H_0(A_*)\ar[r]^{\alpha}&R} $$ commutes and $\eta_n:C_n\to A_n$ is an isomorphism. Then the category $_X\Ext(R;A_n)$ is nonempty iff the class of $X$ in $\Sh^{n+2}(R,M)$ is zero. If this is so, then the group $\Sh^{n+1}(R,M)$ acts transitively and effectively on the set of components of the category $_X\Ext(R;A_n)$. Duskin in \cite{duskin} introduced higher torsors to obtain an interpretation of elements of the cohomology groups in very general context. For associative algebras his approach also gives the interpretation of $\H^3$ via crossed bimodules, but in higher dimensions his approach is totally different from one indicated here. \subsection{Shukla cohomology via free crossed bimodules} Let $R$ be an associative algebra. We claim that there is a free crossed module $\d:F_1\to F_0$ with ${\sf Coker}(\d)=R$. Indeed, first we take a surjective homomorphism of rings $\pi:F_0\to R$, where $F_0$ is a free $\K$-algebra. Then we choose a free $\K$-module $V$ together with an epimorphism $V\to \ker(\pi)$. Finally we take $\d:F_1\to F_0$ to be the free crossed bimodule generated by $V\to F_0$. Then $\d$ has the expected property. \begin{Pro} Let $R$ be an associative algebra and let $M$ be a bimodule over $R$. Let $$0\to E\buildrel j \over \to F_1\buildrel \d \over \to F_0\to R\to 0\leqno(F)$$ be a crossed extension with free crossed bimodule $\d:F_1\to F_0$. Then there is an exact sequence $$\Hom_{F_0^e}(F_1,M)\buildrel j* \over \to \Hom_{R^e}(E,M)\to \Sh^3(R,M)\to 0$$ where $j^*(h)=hj$, for $h\in \Hom_{F_0^e}(F_1,M)$. \end{Pro} {\it Proof}. The crossed extension $(F)$ defines an element $ e\in \H^3(R,E)$. The homomorphism $e_*:\Hom_{R^e}(E,M)\to \Sh^3(R,M)$ sends an element $f\in \Hom_{R^e}(E,M)$ to $f_*(e)\in \Sh^3(R,M)$. Take any crossed extension $$0\to M\to C_1\to C_0\to R\to 0$$ Since $F_0$ is a free algebra and $\d:F_1\to F_0$ is a free crossed bimodule, there exists a morphism of crossed extensions $$\xymatrix{0\ar[r] &E\ar[r]\ar[d]^f& F_1\ar[r]\ar[d]& F_0\ar[r]\ar[d]^\id&R\ar[r]\ar[d]^\id& 0\\ 0\ar[r] &M\ar[r]& C\ar[r]& F_0\ar[r]&R\ar[r]& 0} $$ which shows that $e_*:\Hom_{R^e}(E,M)\to \Sh^3(R,M)$ is an epimorphism. We claim that $j_*(e)=0$. Indeed, $j_*(e)$ is represented by the bottom crossed extension in the following diagram: $$\xymatrix{0\ar[r] &E\ar[r]^j\ar[d]^j& F_1\ar[r]\ar[d]& F_0\ar[r]\ar[d]^\id&R\ar[r]\ar[d]^\id& 0\\ 0\ar[r] &F_1\ar[r]& X\ar[r]& F_0\ar[r]&R\ar[r]& 0} $$ Obviously $F_1\to X$ has a retraction, hence the claim. Take any $h\in \Hom_{F_0^e}(F_1,M)$. Then we have $$e_*j^*(h)=(hj)_*(e)=h_*j_*(e)=0$$ Thus it remains to show that if $f\in \Hom_{R^e}(E,M)$ satisfies $f_*(e)=0$, then $f=hj$ for some $h\in \Hom_{F_0^e}(F_1,M)$. If $f_*(e)=0$, then we can use Theorem \ref{nulisgamocnoba} to obtain a diagram $$\xymatrix{0\ar[r] &E\ar[r]^j\ar[d]^f& F_1\ar[r]^\d\ar[d]^g& F_0\ar[r]^{\pi}\ar[d]^\id& R\ar[r]\ar[d]^\id& 0\\ 0\ar[r] &M\ar[r]^{j'}& C\ar[r]^{\delta}& F_0\ar[r]^{\pi}&R\ar[r]& 0\\ & 0\ar[r]& C\ar[u]^\id\ar[r]^i & S\ar[u]_t\ar[r]^p&R\ar[r]\ar[u]_\id&0} $$ Since $F_0$ is a free $\K$-algebra, the homomorphism $t$ has a section $s:F_0\to S$. So we have $ts=\id_{F_0}$. Since $p=\pi t$, we obtain $ps=\pi t s=\pi$. It follows that $ps\d=\pi \d=0$, thus there exists a unique $r:F_1\to C$ such that $s\d=ir$. Then we have $irj=s\d j=0$ and therefore $rj=0$. On the other hand $$ \delta (g-r)=\delta g-\delta r=\d-tir=\d-ts\d=0.$$ Therefore there exists a unique $h:F_1\to M$ such that $g=r+j'h$. Since $j'f=gj=rj+j'hj=j'hj$ we obtain $f=hj$ and we are done. \section{Some computations of Shukla cohomology}\label{shgam} \subsection{The case $\K=\Z$} Let $\K=\Z$ and $R=\Z/n\Z$, $n\geq 2$. Consider the exterior algebra $\Lambda _{\Z}^*(x)$ on a generator $x$ of degree $1$ over $\Z$. We put $\partial (x)=n$. Then $\Lambda _{\Z}^*(x)$ is a chain algebra, which is weakly equivalent to $\Z/n\Z$. It is clear that the normalized Hochschild cochain complex of $\Lambda _{\Z}^*(x)$ with coefficients in $\Z/n\Z$ has a bicomplex structure, which is $\Z/n\Z$ in bidegree $(i,i)$, $i\geq 0$ and is zero elsewhere. Thus $$\Sh^*(R/\Z,R)=R[\xi],$$ where $$\xi\in \Sh^2({R/\Z,R})$$ has degree $2$. Based on the interpretation of the second Shukla cohomology via abelian extensions (see Section \ref{gafshu}) one easily sees that $\xi$ represents the following extension: $$\xi=(0\to \Z/nZ \to \Z/n^2\Z\to \Z/n\Z\to 0)\in {\sf Extalg}_{\Z}(R,R)$$ This example can be generalized as follows. Let $A$ be an algebra over $R=\Z/n\Z$. We will assume that $A$ is free as a module over $\Z/n\Z$ ( of course this holds automatically if $n=p$ is a prime). A ring $A_0$ is called {\it a lifting of $A$ to $\Z$} if there exists an isomorphism of rings $A_0/nA_0\cong A$ and additionally $A_0$ is free as an abelian group. \begin{Pro} Let $A$ be an algebra over $R= {\Z/n\Z}$, which is free as an $R$-module. If $A$ has a lifting to $\Z$ then $$\Sh^*(A/\Z,A)\cong \H^*(A/R,A)[\xi]$$ \end{Pro} {\it Proof}. Let $A_*$ be a chain algebra over $\Z$ defined as follows. As a graded algebra $A_*$ is the tensor product $A_*=\Lambda _{\Z} ^*(x)\t A_0$ where $x$ has degree one. The boundary homomorphism is defined by $\partial(x)=n$. Thus as a chain complex $A_*$ looks as follows: $$\cdots \to 0\to A_0 \buildrel n \over \to A_0$$ in particular $A_*\to A$ is a weak equivalence and the K\"unneth Theorem \ref{kiuneti} for Hochschild cohomology implies $$\Sh^*(A/\Z,A)\cong \H^*(A/R,A)\t \Sh^*(R/\Z,R)\cong \H^*(A/R,A)[\xi]$$ \subsection{The case $\K=\Z/p^2\Z$} Let $p$ be a prime and $\K=\Z/p^2\Z$ and $R=\Z/p\Z$. Consider the commutative chain algebra $$\Lambda^*_{\Z/p^2\Z}(x)\t \Gamma^*_{\Z/p^2\Z}(y),$$ where $x$ is of degree $1$ and $y$ is of degree $2$. Here $\Gamma ^*$ denotes the divided power algebra. Now we put $\partial (x)=p$ and $\partial (y)=px$. One easily checks that in this way one obtains a chain algebra compatible with divided powers. Since the augmentation $$\Lambda^*_{\Z/p^2\Z}(x)\t \Gamma^*_{\Z/p^2\Z}(y) \to \Z/p\Z$$ is a weak equivalence, one can use this chain algebra to compute the Shukla cohomology. It is clear that $$C^*(\Lambda^*_{\Z/p^2\Z}(x)\t \Gamma^*_{\Z/p^2\Z}(y),\Z/p\Z)\cong C^*(\Lambda^*_{\Z/p\Z}(x)\t \Gamma^*_{\Z/p\Z}(y), \Z/p\Z)$$ where $ \Lambda^*_{\Z/p\Z}(x)\t \Gamma^*_{\Z/p\Z}(y)$ is a chain algebra with zero boundary map. Then the K\"unneth theorem for Hochschild cohomology \cite{homology} implies that $$\Sh^*(R/\K,R)\cong R[\sigma x, \sigma y, \sigma y^{[2]},,\cdots \sigma y^{[2^n]},\cdots], \ \ {\rm if} \ \ p=2$$ where $|\sigma z| =1+|z|$. Similarly, if $p$ is odd, then $$\Sh^*(R/\K,R)\cong \Lambda^*(\sigma y, \cdots \sigma y^{[p^n]},\cdots)\t \Z/p\Z[\sigma x, \sigma^2 y, \cdots \sigma^2 y^{[p^n]},\cdots]\ \textrm{if $p$ is odd}$$ Here we use the fact that one has an isomorphism of algebras: $$\Gamma _{\Z/p\Z}^*(z)\cong \Z/p\Z[z]/(z^p)\t \Z/p\Z[z]/(z^{p^2})\t \Z/p\Z[z]/(z^{p^3})\t \cdots$$ The element $\sigma x$ is still represented by the following abelian extension of algebras $$(0\to \Z/pZ \to \Z/p^2\Z\to \Z/p\Z\to 0)\in {\bf Extalg}_{\Z/p^2\Z}({\Z/p\Z,\Z/p\Z})$$ while $\sigma y$ is represented by the crossed extension of algebras: $$(0\to \Z/pZ \to \Z/p^2\Z \buildrel p \over \to\Z/p^2\Z\to \Z/p\Z\to 0)\in {\bf Cros}_{\Z/p^2\Z}({\Z/p\Z,\Z/p\Z}).$$ More generally, let $A$ be an algebra over $\Z/p\Z$. A ring $A_0$ is called {\it a lifting of $A$ to $\Z/p^2\Z$} if there exists an isomorphism of algebras $A_0/pA_0\cong A$ and additionally $A_0$ is free as a $\Z/p^2\Z$-module. \begin{Pro} Let $A$ be an algebra over $\F_p$. If $A$ has a lifting to $\K=\Z/p^2\Z$, then $$\Sh^*(A/\K,A)\cong \H^*(A/\F_p,A)\t \Sh^*(\F_p/\K,\F_p)$$ where $$\Sh^*(\F_2/\K,\F_2)\cong \F_2[\sigma x, \sigma y, \sigma y^{[2]},,\cdots \sigma y^{[2^n]},\cdots]$$ and $$\Sh^*(\F_p/\K,\F_p)\cong \Lambda^*(\sigma y, \cdots \sigma y^{[p^n]},\cdots)\t \F_p[\sigma x, \sigma^2 y, \cdots \sigma^2 y^{[p^n]},\cdots]$$ if $p$ odd. \end{Pro} {\it Proof}. Let $A_*$ be a chain algebra over $\Z/p^2$ given as the tensor product of chain algebras: $$A_*=A_0\t \Lambda^*_{\Z/p^2\Z}(x)\t \Gamma^*_{\Z/p^2\Z}(y)$$ By the K\"unneth theorem \ref{kiuneti} $A_*\to A$ is a weak equivalence and hence $$\Sh^*_{\Z/p^ 2}(A,A)\cong \H^*_{\Z/p\Z}(A,A)\t \Sh^*_{\Z/p^2\Z}({\Z/p\Z,\Z/p\Z})$$ \ \ Let us observe that if a $\F_p$-algebra $A$ has a lifting to $\Z$ then it has also a lifting to $\Z/p^2\Z$. It is clear that group algebras (or more generally monoid algebras), truncated polynomial algebras have lifting to $\Z$. It is also known that any smooth commutative algebra has lifting to $\Z$ \cite{arabia}. It is also clear that the class of algebras having lifting to $\Z$ (or $\Z/p^2\Z$) is closed under tensor product. It is also closed under finite cartesian products. \subsection{On relationship between Shukla cohomology over $\Z$ and $\Z/p^2\Z$ up to dimension three}\label{sh3} In this section $\K=\Z/p^2\Z$ and $\H^*$ denotes the Hochschild cohomology over $\F_p$. Let $M$ be a bimodule over an $\F_p$-algebra $A$. Since $A$ is also an algebra over $\Z$ and $\K=\Z/p^2\Z$, we can consider not only the Hochschild cohomology $\H^*(A,M)$, but also the Shukla cohomologies $\Sh^*(A/\K,M)$ and $\Sh^*(A/\Z,M)$. The ring homomorphisms $\Z\to \K\to\F_p$ yield the natural transformations $$\H^i(A,M)\to \Sh^i(A/\K,M)$$ and $$b^i:\Sh^i(A/\K,M)\to \Sh^i(A/\Z,M)$$ which are obviously isomorphisms for $i=0,1$. For $i=2$, the groups in question classify abelian extensions of $A$ by $M$, respectively in the category of algebras over $\F_p$, $\K$ and $\Z$. Let us observe that if $X\to Y\to Z$ is a short exact sequence of abelian groups and $pX=0=pZ$, then $p^2Y=0$. Thus any abelian extension of $A$ by $M$ in the category of all rings lies in the category of algebras over $\K$. It follows that for $i=2$, the first map $\H^2(A,M)\to \Sh^2(A/\K,M)$ is a monomorphism, while the second homomorphism is an isomorphism: $$b^2:\Sh^2(A/\K,M)\cong \Sh^2(A/\Z,M)$$ In higher dimensions we have \begin{Le} For all $n$ the homomorphism $$b^n:\Sh^n(A/\K,M)\to \Sh^n(A/\Z,M)$$ is an epimorphism and it has a natural splitting. \end{Le} {\it Proof}. We have only to consider the case $n\geq 3$. We have to construct the homomorphism $d^n:\Sh^n(A/\Z,M)\to \Sh^n(A/\K,M)$, which is a right inverse of $b^n$. We consider more carefully the case $n=3$ and then we indicate how to modify the argument for $n>3$. In terms of crossed extensions, $b=b^3:\Sh^3(A/\K,M)\to \Sh^3(A/\Z,M)$ sends the class of a crossed extension $$0\to M\to C_1\to C_0\to A\to 0$$ of $\Z/p^2\Z$-algebras to the same crossed extension but now considered as algebras over $\Z$. Now we take any element from $\Sh^3(A/\Z,M)$, which is represented by the following crossed extension of $A$ by $M$ in the category of rings: $$0\to M\to D_1\to D_0\to A\to 0.$$ Thanks to Lemma \ref{crospullback} and Corollary \ref{crosfiltri} without loss of generality one can assume that $D_0$ is free as an abelian group (this follows also from Section \ref{cmcxmod}). Thus $V:={\sf Im}(\d)$ is also free as an abelian group and $0\to M\to D_1\to V\to 0$ splits as a sequence of abelian groups. It follows that $0\to M\to D/pD\to V/pV\to 0$ is exact. On the other hand $pV$ is a two-sided ideal in $D_0$ and therefore one has an exact sequence $0\to V/pV\to D_0/pV\to A$. It follows that $D_0/pV$ is a $\Z/p^2\Z$-algebra. By gluing these data we get a crossed extension $$\xymatrix{ 0\ar[r]& M\ar[r]& D_1/pD_1\ar[r]& D_0/pV\ar[r]& A\ar[r] &0} $$ and therefore an element in $\Sh^3(A/\K,M)$. In this way we obtain the homomorphism $d=d^3:\Sh^3(A/\Z,M)\to\Sh^3(A/\K,M)$. The commutative diagram $$\xymatrix{0\ar[r] &M \ar[r]\ar[d]_\id& D_1\ar[r]\ar[d]&D_0\ar[r]\ar[d] &A\ar[r]\ar[d]^{\id} & 0\\ 0\ar[r]& M\ar[r]& D_1/pD_1\ar[r]& D_0/pV\ar[r]& A\ar[r] &0} $$ shows that $bd=\id$ and the case $n=3$ is done. Assume now $n>3$. According to Remark at the and of Section \ref{gafshu} we know that elements of $\Sh^n(A/\Z,M)$ are equivalence classes of chain algebras $X_*$ of length $\leq n-2$ which are acyclic in all but the extreme dimensions: $$0\to M\to X_{n-2}\to \cdots \to X_0\to A\to 0$$ Without loss of generality one can assume that $X_0,\cdots, X_{n-3}$ are free as abelian groups (use Section \ref{cmctrun}, or modify the argument in Lemma \ref{crospullback} and Corollary \ref{crosfiltri}). By repeating the previous argument we can construct a diagram of the form $$\xymatrix{0\ar[r] &M \ar[r]\ar[d]_\id& X_{n-2} \ar[r]\ar[d]& \cdots \ar[r]& X_0\ar[r]\ar[d] &A\ar[r]\ar[d]^{\id} & 0\\ 0\ar[r]& M\ar[r]& X_{n-2}/pX_{n-2}\ar[r]& \cdots \ar[r]& X_0/pV\ar[r]& A\ar[r] &0} $$ where $V:= \Ker(X_0\to A)$ and we are done. Now we analyze the kernel of the homomorphism $$b=b^3:\Sh^3(A/\K,M)\to \Sh^3(A/\Z,M)$$ \begin{Pro} \label{langwail} Let $A$ be an algebra over $\F_p$ and let $M$ be a bimodule over $A$. Then one has a natural isomorphism $$\Sh^3(A/\K,M)\cong \Sh^3(A/\Z,M)\oplus \H^0(A,M)$$ where $\K=\Z/p^2\Z$. \end{Pro} {\it Proof} consists of several steps. We already defined the homomorphism $ d=d^3:\Sh^3(A/\Z)\to \Sh^3(A/\K)$ with $bd=\id$. Now we define the homomorphisms $$e: \Sh^3(A/\K)\to \H^0(A,M), \ \ c: \H^0(A,M)\to \Sh^3(A/\K)$$ with $$ ed=0, \ \ ec=\id, \ \ bc=0$$ and we prove that $(b,e):\Sh^3(A/\K)\to \H^0(A,M)\oplus \Sh^3(A/\Z)$ is a monomorphism. From these assertions the result follows. \ {\it First step. The homomorphism $e:\Sh^3(A/\K,M)\to \H^0(A,M)$}. Let $$0\to M\to C_1 \buildrel \partial \over \to C_0\buildrel \pi \over \to A\to 0\leqno(\d)$$ be a crossed extension, where $C_0$ and $C_1$ are $\K$-algebras. Since $A$ is an algebra over $\F_p$, one has $\pi(p1)=0$, where $1\in C_0$ is the unit of $C_0$. Therefore one can write $p1=\d([P])$ for a suitable $[P]$ in $C_1$. Now we put: $$e((\d))=p[P]\in M$$ it is easy to check that $e$ is a well-defined homomorphism. Let us observe that $e(\d)=0$ if $pC_1=0$. It follows that $ed=0$. \ {\it Second step. The canonical class $(\sigma)_A\in \Sh^3(A/\K,A)$}. Let $X$ be an abelian group. We let $\Z[X]$ be the free abelian group generated by $X$ modulo the relation $[0]=0$. Here $[x]$ denotes an element of $\Z[X]$ corresponding to $x\in X$. Then we have a canonical epimorphism $\eta:\Z[X]\to X$, $\eta([x])=0$ which gives rise to {\it the canonical free resolution of $X$}: $$0\to R(X)\to \Z[X]\to X\to 0$$ For any $x,y\in X$ we put $$\brk{x,y}:=[x]+[y]-[x+y]\in R(X)$$ We now assume that $pX=0$, that is $X$ is a vector space over $\F_p$. By applying the functor $(-)\t \Z/p^2\Z$ to the canonical free resolution we obtain the following exact sequence $$0\to X \buildrel i\over \to R(X)/p^2R(X) \buildrel \sigma \over \to \Z/p^2\Z[X] \buildrel \eta \over \to X\to 0 \leqno(\sigma)_X$$ Here we used the well-known isomorphism $V\cong \Tor_1(V,\Z/p^2Z)$ for any $\F_p$-vector space $V$ considered as an abelian group (the $\Tor$ and $\t$ are taken of course over $\Z$ and not over $\K=\Z/p^2\Z$). The homomorphism $i$ has the following form $$i(x)=\sum_{j=0}^{p-1}p\brk{jx,x} \ \ {\sf mod} (p^2R(X))$$ Let us turn back to our situation. We can take $X=A$. The multiplicative structure on $A$ can be extended linearly to $\Z[A]$ to get an associative algebra structure on it. Then not only $\eta$ is a ring homomorphism, but the exact sequence $(\sigma)_A$ is a crossed extension and therefore we obtain an element $$(\sigma)_A=( 0\to A \to R(A)/p^2R(A) \buildrel \sigma \over \to \Z/p^2\Z[A] \to A\to 0)\in \Sh^3(A/\K,A)$$ It is clear that $A\mapsto (\sigma)_A$ is a functor from $\F_p$-algebras to the category of crossed extensions of $\Z/p^2\Z$-algebras. Since $$\sigma(\sum_{j=0}^{j=p-1}\brk{j,1})=p[1]$$ one has $$e((\sigma)_A)=1\in \H^0(A,A)\subset A.$$ On the other hand $p^2R(A)$ is an ideal of $\Z[A]$. Thus we have a commutative diagram $$\xymatrix{&0\ar[r] & R(A)/p^2R(A)\ar[r]\ar[d]^{\id}& \Z[X]/p^2R(A)\ar[r]\ar[d]&A\ar[r]\ar[d]^{\id}&0\\ 0\ar[r]&A\ar[r] &R(A)\ar[r] &\Z/p^2\Z[A] \ar[r]&A\ar[r]&0 }.$$ It follows from Theorem \ref{nulisgamocnoba} that the class $(\sigma)_A$ has the following important property: $$0=b((\sigma)_A) \in \Sh^3(A/\Z,A).$$ \ {\bf Third step. The homomorphism $c:\H^0(A,M)\to \Sh^3(A/\K,M)$}. Using the class $(\sigma)_A$ we now define the homomorphism $c:\H^0(A,M)\to \Sh^3(A/\K,M)$ by $$c(m)=f_m^*((\sigma)_A).$$ Here $m\in \H^0(A,M)$ and $f_m:A\to M$ is the unique bimodule homomorphism with $f_m(1)=m$ and $f_m^*:\Sh^3(A/\K,A)\to \Sh^3(A/\K,M)$ is the induced homomorphism in cohomology. Since $e$ and $b$ are natural transformations of functors it follows that for any $m\in M$ we have $ec(m)=ef_m^*((\sigma)_A)=f_m^*e((\sigma)_A)=f_m(1)=m$ and $ bc(m)= bf_m^*((\sigma)_A)=f_m^*b((\sigma)_A)=0$. Thus $$ec=\id \ \ {\rm and } \ \ bc=0$$ \ {\it Fourth step.} It remains to show that $$(b,e):\Sh^3(A/\K)\to \H^0(A,M)\oplus \Sh^3(A/\Z)$$ is a monomorphism. Let $$0\to M\to C_1\to C_0\to A\to 0$$ be a crossed extension of $\Z/p^2\Z$-algebras which lies in $\Ker(b,e)$. Since it goes to zero in $\Sh^3(A/\Z,M)$ one has the following diagram $$\xymatrix{&0\ar[r]&C_1\ar[d]^{\id} \ar[r]^{\mu}&S \ar[d]^{\xi}\ar[r]^{\si} &R\ar[d]^{\id}\ar[r]&0\\ 0\ar[r]&M\ar[r]&C_1\ar[r]^\d&C_0\ar[r]&R\ar[r]&0}$$ where $S$ is a ring. Since $\xi$ is a homomorphisms of algebras with unit we have $[P]=p1_S$, where $1_S$ is the unit of $S$. Therefore $e(\d)=p^21_=0$, because $(\d)$ goes also to zero under the map $e$. It follows that $S$ is an algebra over $\Z/p^2\Z$. Theorem \ref{nulisgamocnoba} shows that the class of $0\to M\to C_1\to C_0\to A\to 0$ in $\Sh^3(A/\K,M)$ is zero and the proof is finished. \section{A bicomplex computing Shukla cohomology}\label{relh} \subsection{Construction of a bicomplex} In this section following \cite{relshukla} we construct a canonical bicomplex which computes the Shukla cohomology in the special case, when the ground ring $\K$ is an algebra over a field $k$. In this section, contrary to other parts of the paper the tensor product $\t$ denotes $\t_k$ and not $\t _\K$. The same is for $\Hom$. Let $R$ be a $\K$-algebra and let $M$ be a bimodule over $R$, where $\K$ is a commutative algebra over a field $k$. Thus $R$ is also an algebra over $k$. We let $C^*(R,M)$ be the Hochschild cochain complex of $R$ considered as an algebra over $k$. Similarly, we let $C^*(R/\K ,M)$ be the Hochschild cochain complex of $R$ considered as an algebra over $\K$. Accordingly $H^*(R,M)$ and $H^*(R/\K ,M)$ denotes the Hochschild cohomology of $R$ with coefficients in $M$ over $k$ and $\K$ respectively. We let $K^{**}(\K,R,M)$ be the following bicosimplicial vector space: $$ K^{pq}(\K ,R,M)= \Hom(\K ^{\t pq}\t R^{\t q},M)$$ The $q$-th horizontal cosimplicial vector space structure comes from the identification $$K^{*q}(\K ,R,M)=C^*(\K ^{\t q}, C^q(R,M)),$$ where $C^q(R,M))=\Hom(R^{\t q},M)$ is considered as a bimodule over $\K ^{\t q}$ via $$((a_1,\cdots,a_q)f(b_1,\cdots,b_q))(r_1,\cdots,r_q):=a_1\cdots a_qf(b_1r_1,\cdots,b_qr_q).$$ Here $f\in \Hom(R^q,M)$ and $a_i,b_j\in \K $, $r_k\in R$. The $p$-th vertical cosimplicial vector space structure comes from the identification $$ K^{p*}(\K ,R,M)= C^*(\K ^{\t p}\t R,M)$$ where $M$ is considered as a bimodule over $\K ^{\t p}\t R$ via $$(a_1\t\cdots \t a_p\t r)m(b_1\t \cdots \t b_p\t s):=(a_1\cdots a_pr)m(b_1\cdots b_ps).$$ We allow ourselves to denote the corresponding bicomplex by $ K^{**}(\K ,R,M)$ as well. Thus $ K^{**}(\K ,R,M)$ looks as follows: $$\xymatrix@C=1em{M\ar[r]^{0}\ar[d]^{\delta}& M\ar[r]^{\ Id}\ar[d]^{\delta}&M\ar[r]^{0}\ar[d]^{\delta}&\cdots\\ \Hom(R,M)\ar[r]^-d\ar[d]^{\delta}& \Hom(\K\t R,M)\ar[r]^-d\ar[d]^{\delta} &\Hom(\K\t\K\t R,M)\ar[r]\ar[d]^{\delta}&\cdots\\ \Hom(R\t R,M)\ar[r]^-d\ar[d]^{\delta} & \Hom(\K^{\t 2}\t R^{\t 2},M)\ar[r]^d\ar[d]^{\delta} &\Hom(\K^{\t 4}\t R^{\t 2},M)\ar[r]\ar[d]^{\delta}&\cdots\\ \vdots & \vdots & \vdots &}$$ Therefore for $f:\K ^{\t pq}\t R^{\t q}\to M$ the corresponding linear maps $$d(f): \K ^{\t (p+1)q}\t R^{\t q}\to M\ \ {\rm and } \ \ \delta(f):\K ^{\t p(q+1)}\t R^{\t (q+1)}\to M$$ are given by $$df(a_{01},...,a_{0q},a_{11},..., a_{1q},..., a_{p1},..., a_{pq},r_1,...,r_q)=$$ $$a_{01}...a_{0q}f(a_{11},..., a_{1q},..., a_{p1},..., a_{pq},r_1,...,r_q)+$$ $$+\sum_{0\leq i<p}(-1)^{i+1}f(a_{01},... ,a_{0q},...,a_{i1}a_{i+1,1},...,a_{iq}a_{i+1,q},..., a_{p1},..., a_{pq},r_1,...,r_q)+$$ $$(-1)^{p+1}f(a_{01},... ,a_{0q},...,a_{p-1,1},... ,a_{p-1,q},a_{p1}r_1,...,a_{pq}r_q).$$ and $$\delta(f)(a_{10},...,a_{1q},..., a_{p0},...,a_{pq},r_0,...,r_q)=$$ $$(-1)^pa_{10}...a_{p0}r_0f(a_{11},...,a_{1q},a_{p1},...,a_{pq},r_1,...,r_q)+$$ $$\sum_{0\leq i<q}(-1)^{i+p+1}f(a_{10},...,a_{1i}a_{1,i+1},... ,a_{pi}a_{pi+1},..., a_{pq},r_0,...,r_ir_{i+1},...,r_q)+$$ $$(-1)^{q+p+1}f(a_{10},...,a_{1q-1},..., a_{p0},...,a_{p,q-1},r_0,...,r_{q-1})a_{1q}... a_{pq}r_q$$ We let $H^*(\K ,R,M)$ be the homology of the bicomplex $ K^{**}(\K ,R,M)$. We also consider the following subbicomplex $\bar{K}^{**}(\K ,R,M)$ of $ K^{**}(\K ,R,M)$: $$\xymatrix@C=1em{M\ar[r]\ar[d]^{\delta}& 0\ar[r] \ar[d]& 0\ar[r]\ar[d]&\cdots \\ \Hom(R,M)\ar[r]^-d\ar[d]^{\delta}& \Hom(\K \t R,M)\ar[r]^-d\ar[d]^{\delta} &\Hom(\K \t \K \t R,M)\ar[r]\ar[d]^{\delta}&\cdots\\ \Hom(R\t R,M)\ar[r]^-d\ar[d]^{\delta} & \Hom(\K ^{\t 2}\t R^{\t 2},M)\ar[r]^-d\ar[d]^{\delta} &\Hom(\K ^{\t 4}\t R^{\t 2},M)\ar[r]\ar[d]^{\delta}&\cdots\\ \vdots & \vdots & \vdots &}.$$ It is clear that $H^*(\K ,R,M)\cong H^*(\bar{K}^{**}(\K ,R,M))$. \subsection{The homomorphism $\alpha$} It follows from the definition that $$\Ker(d:K^{*0}\to K^{*1})\cong C^*(R/\K ,M).$$ Therefore one has the canonical homomorphism $$\alpha^n:H^n(R/\K ,M)\to H^n(\K ,R,M), \ n\geq 0.$$ \begin{The} {\rm i)} The homomorphisms $\alpha^0$ and $\alpha^1$ are isomorphisms. The homomorphism $\alpha^2$ is a monomorphism. {\rm ii)} If $R$ is projective over $\K $, then $$\alpha^n:H^n(R/\K ,M)\to H^n(\K ,R,M)$$ is an isomorphism for all $n\geq 0$ {\rm iii)} The groups $H^*(\K ,R,M)$ are canonically isomorphic to $\Sh^*(R/\K,M)$ \end{The} {\it Proof}. i) is an immediate consequence of the definition of the bicomplex\linebreak $\bar{K}^{**}(\K ,R,M)$. ii) The bicomplex gives rise to the following spectral sequence: $$E^{pq}_1=H^q(\K ^{\t p},C^p(R,M))\Longrightarrow H^{p+q}(\K ,R,M).$$ Let us recall that if $X$ and $Y$ are left modules over an associative algebra $S$, then $\Ext_{S}^*(X,Y)\cong \H^*(S,\Hom(X,Y))$ \cite{CE}, where $\Hom(X,Y)$ is considered as a bimodule over $S$ via $(sft)(x)=sf(tx)$. Here $x\in X$, $s,t\in S$ and $f:X\to Y$ is a lineal map. Having this isomorphism in mind, we can rewrite $E^{pq}_1\cong \Ext^q_{\K ^{\t p}}(R^{\t p},M)$. By our assumptions $R^{\t p}$ is projective over $\K ^{\t p}$. Therefore the spectral sequence degenerates and we get $H^*(\K ,R,M)\cong H^*(C(R/\K ,M))=H^*(R/\K ,M)$. Here we used the obvious isomorphism $$\Hom_{\K \t \K\t \cdots \t \K }(R\t R\t \cdots \t R,M)= \Hom_{\K }(R\t_\K R\t_\K \cdots \t_\K R ,M).$$ iii) We let ${\bar K}^*(\K ,R,M)$ denote the total cochain complex associated to the bicomplex ${\bar K}^{**}(\K ,R,M)$. Then this construction has an obvious extension to the category of chain $\K$-algebras. Unlike Lemma \ref{hhisinvariantoba}, for any weak equivalence $R_*\to S_*$ of chain $\K$-algebras the induced map ${\bar K}^*(\K, S_*,M) \to {\bar K}^*(\K,R_*,M)$ is a weak equivalence. This is because the definition of ${\bar K}^*(\K ,R,M)$ involves the tensor products and hom's over the field $k$ and not over $\K$. Furthermore, by ii) $H^*(\K,R_*,M)$ is isomorphic to the Hochschild cohomology, provided $R_*$ is degreewise projective over $\K$. In particular this happens when $R_*$ is cofibrant. Now we take any $\K$-algebra $R$ and a cofibrant replacement $R^c_*$ of $R$. Then one has $$H^*(\K,R,M)\cong H^*(\K,R_*^c, M)\cong \H^*(R_*^c/\K,M)\cong \Sh^*(R/\K,M).$$ \begin{Co} {\rm i)} There is a natural bijection $${\bf Extalg}(\K,R,M)\cong \H^2(\K ,R,M).$$ {\rm ii)} There is a natural bijection $${\bf Cros}(\K,R,M)\cong \H^3(\K ,R,M).$$ \end{Co} Our next aim is to describe directly the cocycles of $H^*(\K,R,M)$ corresponding to abelian and crossed extensions. We have $$H^2(\K ,R,M)=Z^2(\K ,R,M)/B^2(\K ,R,M)$$ where $Z^2(\K ,R,M)$ consists of pairs $(f,g)$ such that $f:R\t R\to M$ and $g:\K \t R\to M$ are linear maps and the equalities $$ag(b,r)-g(ab,r)+g(a,br)=0$$ $$abf(r,s)-f(ar,bs)=arg(b,s)-g(ab,rs)+g(a,r)bs$$ $$rf(s,t)-f(rs,t)+f(r,st)-f(r,s)t=0$$ hold. Here $a,b\in \K $ and $r,s,t\in R$. Moreover, $(f,g)$ belongs to $B^2(A,R,M)$ iff there exists a linear map $h:R\to M$ such that $f(r,s)=rh(s)-h(rs)+h(r)s$ and $g(a,r)=ah(r)-h(ar).$ Starting with $(f,g)\in Z^2(\K ,R,M)$ we construct an abelian extension of $R$ by $M$ by putting $S=M\oplus R$ as a vector space. A $\K $-module structure on $S$ is given by $a(m,r)=(am+g(a,r),ar)$, while the multiplication on $S$ is given by $(m,r)(n,s)=(ms+rn+f(r,s),rs)$. Conversely, given an abelian extension $$ 0\to M\to S \to R\to 0$$ we choose a $k$-linear section $h:R\to S$ and then we put $f(r,s):=h(r)h(s)-h(rs)$ and $ g(a,r):=ah(r)-h(ar)$. One easily checks that $(f,g)\in Z^2(\K ,R,M)$ and one gets i). Similarly, we have $H^3(\K ,R,M)=Z^3(\K ,R,M)/B^3(\K ,R,M)$. Here $Z^3(\K ,R,M)$ consists of triples $(f,g,h)$ such that $f:R\t R\t R\to M$, $g: \K \t \K \t R\t R\to M$ and $h: \K\t \K \t R\to M$ are linear maps and the following relations hold: $$r_1f(r_2,r_3,r_3)-f(r_1r_2,r_3,r_4)+f(r_1,r_2r_3,r_4)-f(r_1,r_2,r_3r_4)+f(r_1,r_2,r_3)r_4=0$$ $$abc\!f(r,\!s,t)-f\!(ar\!,\!bs,ct)\!=\!arg(b,c\!,y,z)-g(ab,c,\!xy,z)+g(a,bc,x,yz)-g(a,b,x,\!y)cz$$ $$abg(c,\!d\!,x,\!y\!)-g(ac,\!bd,\!x,\!y)+g(a,b,cx,\!dy)\!=\!acxh(b,\!d,\!y)-h(ab,cd,\!xy)+h(a,c,x)bdy$$ $$ah(b,c,x)-h(ab,c,x)+h(a,bc,x)-h(a,b,cx)=0.$$ Moreover, $(f,g,h)$ belongs to $B^3(\K ,R,M)$ iff there exist linear maps $m:R\t R\to M$ and $n:\K \t R\to M$ such that $$f(r,s,t)=rm(s,t)-m(rs,t)+m(r,st)-m(r,s)t$$ $$g(a,b,r,s)=abm(r,s)-m(ar,bs)-arn(b,s)+n(ab,rs)-n(a,x)bs$$ $$h(a,b,r)=an(b,r)-n(ab,r)+n(a,br).$$ Let $$0\to M\to C_1 \buildrel \partial \over \to C_0\buildrel \pi \over \to R\to 0$$ be a crossed extension. We put $V:={\sf Im} (\partial)$ and consider $k$-linear sections $p:R\to C_0$ and $q:V\to C_1$ of $\pi: C_0\to R$ and $\partial:C_1\to V$ respectively. Now we define $$m:R\t R\to V \ \ {\rm and} \ \ n:A\t R\to V$$ by $m(r,s):=q(p(r)p(s)-p(rs))$ and $n(a,r):=q(ap(r)-p(ar))$. Finally we define $f:R\t R\t R\to M$, $g:\K^{\t 3} \t R\t R\to M$ and $h:\K\t \K \t R\to M$ by $$f(r,s,t):=p(r)m(s,t)-m(rs,t)+m(r,st)-m(r,s)p(t)$$ $$g(a,b,r,s):=p(as)n(b,s)-n(ab,rs)+bn(a,x)p(y)-abm(r.s)+m(ax,by)$$ $$h(a,b,r):= an(b,r)-n(ab,r)+n(a,bx).$$ Then $(f,g,h)\in Z^3(A,R,M)$ and the corresponding class in $\H^3(A,R,M)$ depends only on the connected component of a given crossed extension. Thus we obtain a well-defined map $ {\bf Cros}(A,R,M)\to \H^3(\K ,R,M)$ and a standard argument (see \cite{baumin}) shows that it is an isomorphism. \section{Applications to Mac Lane cohomology}\label{hmlsh} In this section we are working with rings. So our ground ring is the ring of integers $\K=\Z$. \subsection{Eilenberg-MacLane $Q$-construction and Mac Lane cohomology} The definition of the Mac Lane cohomology \cite{mac} of a ring $R$ with coefficients in an $R$-bimodule $M$ is based on the work of Eilenberg and Mac Lane on Eilenberg-Mac Lane spaces \cite{EM}. Namely, for any abelian group $A$ Eilenberg and Mac Lane constructed a chain complex $Q_*(A)$ whose homology is the stable homology of Eilenberg- Mac Lane spaces $$H_q(Q_*(A))\cong H_{n+q}(K(A,n), \ \ \ n>q.$$ In low dimensions $Q_*(A)$ is defined as follows \cite{EM}, \cite{mac}, \cite{macobs}, \cite{HC}. The group $Q_0(A)=\Z[A]$ is the free abelian group generated by elements $[a]$, $a\in A$ modulo the relation $[0]=0$. The group $Q_1(A)$ is the free abelian group generated by pairs $[a,b]$, $a,b\in A$ modulo the relations $[a,0]=0=[0,a]$, $a\in A$, while the group $Q_2(A)$ is the free abelian group generated by $4$-tuples $[a,b,c,d]$ modulo the relations $$[a,b,0,0]=[0,0,c,d]=[a,0,c,0]=[0,b,0,d]=[a,0,0,d]=0$$ in general $Q_n(A)$ is generated by $2^n$-tuples modulo some relations \cite{EM},\cite{mac},\cite{JP2}. The boundary map is given by $$d[a,b]=[a]+[b]-[a+b]$$ $$d[a,b,c,d]=[a,b]+[c,d]-[a+c,b+d]-[a,c]-[b,d]+[a+b,c+d].$$ For any $a\in A$, the element $\gamma(a):=[0,a,a,0]\in Q_2(A)$ is a two-dimensional cycle and $\gamma$ yields an isomorphism (see \cite{EM},\cite{mac}) $$\gamma:A/2A\cong H_2(Q_*(A)).$$ Moreover for any abelian groups $A$ and $B$ there is a natural pairing $$Q_*(A)\otimes Q_*(B)\to Q_*(A\t B)$$ (see for example \cite{mac}, \cite{JP2} or \cite{HC}). For any ring $R$, this pairing allows us to put a chain algebra structure on $Q_*(R)$. For example, in very low dimensions we have $$[x][y]=[xy],\ \ [x][y,z]=[xy,xz], \ \ [x,y][z]=[xz,yz],$$ $$[x][y,z,u,v]=[xy,xz,xu,xv], \ \ [x,y,z,t][u]=[xu,yu,zu,tu]$$ $$[x,y][u,v]=[xu,xv,yu,yv]$$ By definition the Mac Lane cohomology $\HML^*(R,M)$ is defined as the Hochschild cohomology of $Q_*(R)$ with coefficients in $M$. One can also introduce the dual objects -- Mac Lane homology. It was proved in \cite{PW} that Mac Lane homology is isomorphic to the topological Hochschild homology of B\"okstedt \cite{bo}. It is also isomorphic to the stable $K$-theory thanks to a result of Dundas and McCarthy \cite{dum}. \subsection{Relation with Shukla cohomology in low dimensions} Since $H_0(Q_*(R))$ $\cong$ $R$ we have a natural augmentation $\ee:Q_*(R)\to R$. Since $Q_*(R)$ is free as an abelian group the chain algebra $$V_*(R)=(\cdots \to 0 \to \ker (\ee)\to Q_0(R))$$ is $\Z$-free and $V_*(R)\to R$ is a weak equivalence. Hence $V_*(R)$ can be used to compute the Shukla cohomology. Thus the morphism of chain algebras $$ \xymatrix{\cdots \ar[r]&Q_2(R)\ar[r]\ar[d]&Q_1(R)\ar[r]\ar[d]& Q_0(R)\ar[r]\ar[d]^{\id} &R\ar[r]\ar[d]^{\id}&0\\ &0\ar[r]& \Ker(\ee)\ar[r]&Q_0(R)\ar[r]&R\ar[r]&0} $$ yields the natural transformation \begin{equation}\label{shumac} \Sh^i(R/\Z,M)\to \HML^i(R,M) \end{equation} which is an isomorphism in dimensions 0,1 and 2. Thus $\HML^2(R,M)$ classifies singular extensions of $R$ by $M$ in the category of rings, see also \cite{mac}. According to Theorem 9 of \cite{macobs} in the dimension $3$ one has the following exact sequence (see also Theorem \ref{zustimimdevrobebi}) \begin{equation}\label{HMLorigr} 0\to \Sh^3(R/\Z,M)\to \HML^3(R,M)\to \H^0(R, \ _2M) \end{equation} The connecting map $\HML^3(R,M)\to \H^0(R, \ _2M)$ is defined via $\gamma$ (see \cite{macobs}). \begin{Pro}\label{HMLpalg} Let $R$ be an algebra over $\F_p$ and $M$ be an $R$-bimodule. Then the natural map $$\Sh^3(R/\Z ,M)\to \HML^3(R,M)$$ is an isomorphism. \end{Pro} {\it Proof}. If $p\not =2$ then this is an immediate consequence of the exact sequence (\ref{HMLorigr}), because $_2M=0$. So we have to consider only the case $p=2$. For any $\F_2$-algebra $R$ we have the canonical homomorphism $\F_2\to R$, which yields the following commutative diagram $$\xymatrix{0\ar[r]&\Sh^3(R/\Z,M)\ar[r]\ar[d]&\HML^3(R,M)\ar[r]\ar[d]&\H^0(R,M)\ar[d] \\ 0\ar[r]&\Sh^3(\F_2/\Z ,M)\ar[r]&\HML^3(\F_2,M)\ar[r]&\H^0(\F_2,M)=M } $$ It is well known that $\HML^3(\F_2,M)=0$ see for example \cite{fls} or \cite{bo}. Since the last vertical arrow is a monomorphism we are done. Based on Proposition \ref{HMLpalg} and Proposition \ref{langwail} we obtain the following \begin{Co}\label{hjb} Let $A$ be an algebra over $\F_p$ and let $M$ be an $A$-bimodule. Then one has a split exact sequence $$0\to \H^0(A,M)\to \Sh^3(A/\K,M)\to \HML^3(A,M)\to 0$$ where $\K=\Z/p^2\Z$. \end{Co} {\bf Remark}. The homomorphism $\Sh^3(R/\Z ,M)\to \HML^3(R,M)$ in general is not an isomorphism. For example, if $R=\Z$, then $\Sh^i(\Z,-)=0$ for all $i\geq 1$, thanks to Lemma \ref{hhisinvariantoba}. On the other hand $\HML^*(\Z,-)$ is quite nontrivial (see \cite{bo}, \cite{FP}) and in particular $\HML^3(\Z,\F_2)=\F_2$. More about $\HML^*(\Z,-)$ see at the end of Section \ref{ssshmac}. \subsection{Relation with Shukla cohomology in higher dimensions}\label{ssshmac} The relationship between Shukla cohomology $\Sh^*(A/\Z,M)$ and Mac Lane cohomology $\HML^*(A,M)$ in higher dimensions is more complicated. Let us first consider the crucial case $A=\Z/p^k\Z$. We already saw $\Sh^i(A/\Z,M)$, $A=\Z/p^k\Z$, is $M$ if $i$ is even and is zero otherwise. Unlike the Shukla cohomology, the behavior of $\HML^*(A,M)$ depends on whether $k=1$ or $k>1$. If $k=1$, then similarly to Shukla cohomology the group $\HML^i(A,M)$ is $M$ if $i$ is even and is zero otherwise. However the natural map $$\Sh^i(\F_p/\Z, M)\to \HML^i(\F_p, M)$$ is an isomorphism only for $i=0, \cdots, 2p-1,$ and it is zero for $i>2p-2$. This follows from the fact that $\Sh^*(\F_p, \F_p)$ is a polynomial algebra on the generator $x$ of dimension two and $\HML^*(\F_p, \F_p)$ is a divided power algebra on the same generator $x$ \cite{fls}. If $k>1$, then situation with Mac Lane cohomology is more complicated. A computation made in \cite{P4} shows that $$\HML^{2n}({\Z/p^k\Z}, \F_p)= (\F_p)^{t}, \ \ \ \HML^{2n-1}({\Z/p^k\Z},\F_p)=(\F_p)^s,$$ where $t=1+[\frac{n}{p}]$ and $s=[\frac{n+1}{p}]$. The full computation of $\HML_*({\Z/p^k\Z},-)$ was obtained by Brun \cite{brun}. The relationship between Mac Lane cohomology and Shukla cohomology for general rings in all dimensions is given by the following theorem proved in \cite{PW} (see also \cite{P3}). \begin{The}\label{zustimimdevrobebi} Let $R$ be a ring. Then for any $R$-bimodule $M$ there is a spectral sequence $$E^{pq}_2(\K)=\Sh^p(R/\K,\HML^q(\K,M))\Longrightarrow \HML^{p+q}(R,M)$$ which is natural in $R$ and $M$. The spectral sequence in low dimensions gives rise to the exact sequence:\def\sto{{\scriptstyle\to}} $$0\sto\Sh^3(R/\Z,\!M)\sto\!\HML^3(R,\!M)\sto\H^0(R,_2\!M)\sto\!\Sh^4(R/\Z,M)\sto\!\HML^4(R,M).$$ \end{The} For the proof of the first part we refer to \cite{PW} and \cite{P3}. The second part first was proved in \cite{JP1}. It is an immediate consequence of the existence of the spectral sequence together with the following computation due to B\"okstedt \cite{bo} (see also \cite{fls} and \cite{FP} ). $$\HML^{2n}({\Z},M)= M/nM, \ \ \ \HML^{2n-1}({\Z},M)= \ _nM, \ n>0.$$ $$\HML^{2n}({\F_p}, M)= M, \ \ \ \HML^{2n-1}({\F_p},M)=0.$$ \subsection{Mac Lane cohomology and cohomology of small categories}\label{bwjp} In this section we recall the relationship between Mac Lane cohomology and cohomology of small categories \cite{JP2}. We assume that the reader is familiar with definition of cohomology of small categories with coefficients in a natural system \cite{BW}, \cite{BJP}. Let us recall that any bifunctor gives rise to a natural system, and therefore we can talk about the cohomology of small categories with coefficients in a bifunctor. For a ring $R$ we let $R\mod$ be the category of finitely generated free $R$-modules. Actually we will assume that objects of $R\mod$ are natural numbers and morphisms from $\bf n$ to $\bf m$ are the same as $R$-linear maps $R^n\to R^m$, or $m\times n$-matrices over $R$. Let $M$ be a bimodule over $R$. There is a bifunctor $$\hom(-,M\t_R-):R\mod^{op}\times R\mod\to \sf Ab$$ given by $$\hom(-,M\t_R-)(X,Y)=\Hom_R(X,M\t _RY).$$ Therefore one can consider the cohomology $H^*(R\mod, \hom(-,M\t_R-))$ of the category $R\mod$ with coefficients $\hom(-,M\t_R-)$ in the sense of Baues and Wirshing \cite{BW} (see also \cite{BJP}). A result of \cite{JP2} asserts that one has an isomorphism: \begin{equation}\label{jibpira} \HML^*(R,M)\cong H^*(R\mod, \hom(-,M\t_R-)). \end{equation} Comparing this isomorphism with the natural homomorphism $\Sh^*(R,M)\to \HML^*(R,M)$ one obtains the homomorphism \begin{equation}\label{shbw} \Sh^i(R,M)\to H^i(R\mod, \hom(-,M\t_R-)), \ \ i\geq 0. \end{equation} Now we recall the description of this homomorphism in terms of extensions for $i=2.$ Let $$0 \to M\buildrel i\over\to S\buildrel p\over \to R\to 0$$ be an abelian extension of rings. Then $$0\to \hom(-,M\t_R-)\to S\mod \buildrel p_*\over \to R\mod \to 0$$ is a linear extension of categories \cite{BW, BJP}, where the functor $p_*$ is given by $p_*(A)= A\t_SR$, $A\in S\mod$ (having in mind the identification of $R\mod$ as the category of natural numbers and matrices, the functor $p_*$ is the identity on objects and is given by applying $p$ on matrices). Let us recall that for fixed $R$ and $M$ the equivalence classes of abelian extensions of $R$ by $M$ form a group ${\sf Extal}(R,M)$, which is isomorphic to the second Shukla cohomology of $R$ with coefficients in $M$ (see Theorem \ref{shugaf}), while linear extensions are classified using the second cohomology of small categories \cite{BW},\cite{BJP}, thus we obtain the homomorphism $$\Sh^2(R,M)\to \H^2(\mod(R),\hom(-,M\t_R-))$$ which is an isomorphism according to isomorphisms (\ref{jibpira}) and (\ref{shumac}). One easily shows that any biadditive bifunctor $D$ on $R\mod$ is of the form $D=\hom(-,M\t_R-)$, where $M=D(R,R)$. Thus one can conclude that any extension of the category $R\mod$ by a biadditive bifunctor is also of the form $S\mod$, for some ring $S$. In particular it is an additive category, more generally any linear extension of an additive category by biadditive functor is an additive category. This fact is an immediate consequence of Lemma 5.1.2 of \cite{BJP}. \section{Applications to strengthening of additive track theories}\label{shstrict} \subsection{Additive and very strongly additive track theories}\label{ota} Let us recall that a \emph{track category} $\ta$ is a category enriched in groupoids \cite{BJP}. Thus $\ta$ consists of objects and for each pair of objects $X$, $Y$ of $\ta$ there is given the \emph{Hom-groupoid} $\hog{X,Y}$, whose objects are termed maps, while 2-arrows --- tracks. To any track category $\ta$ there is an associated category $\ta_\ho$ with the same objects as $\ta$, while for objects $A$ and $B$ of $\Ob(\ta)$ the set of morphisms $[A,B]$ in $\ta_\ho$ is the set of connected components of the groupoid $\hog{X,Y}$. A track category is \emph{abelian} if for any 1-arrow $f:X\to Y$, the group $\Aut(f)$ of tracks from $f$ to itself is abelian. Any abelian track category defines a natural system $D=D_{\ta}$ on $\ta_\ho$ and a canonical class $\Ch(\ta)\in H^3(\ta_\ho,D)$ --- see section 2.3 of \cite{BJP}. Conversely for any category $\bf C$, any natural system $D$ on $\bf C$ and any element $a\in H^3({\bf C},D)$ there exists an abelian track category $\ta=\ta_{{\bf C},D,a}$ unique up to equivalence such that $\ta_\ho=\bf C$ and $\Ch(\ta) =a$ (see \cite{BJP}). In fact for a given natural system $D$ on a category $\bf C$ there is a category ${\sf Trext}({\bf C},D)$ whose objects are abelian track categories $\ta$ with $\ta_\ho=\bf C$ and $D_{\ta}=D$ and the set of connected components of ${\sf Trext}({\bf C},D)$ is isomorphic to the third dimensional cohomology \cite{P1},\cite{P2}: \begin{equation}\label{mompares} \pi_0({\sf Trext}({\bf C},D))\cong H^3({\bf C},D) \end{equation} A {\it lax coproduct} $A\vee B$ in a track category $\ta$ is an object $A\vee B$ equipped with maps $i_1:A\to A\vee B$, $i_2:B\to A\vee B$ such that the induced functor $$(i_1^*,i_2^*):\hog{A\vee B,X}\to \hog{A,X}\x \hog{B,X}$$ is an equivalence of groupoids for all objects $X\in \ta$. The coproduct is {\it strong} if the functor $(i_1^*,i_2^*)$ is an isomorphism of groupoids. By duality we have also the notion of {\it lax product} and {\it strong product}. A {\it lax zero object} in a track category $\ta$ is an object $0$ such that the categories $\hog{0,X}$ and $\hog{X,0}$ are equivalent to the trivial groupoid for all $X\in\ta$. Let us recall that a \emph{trivial groupoid} has only one object and one arrow. A {\it strong zero object} in a track category $\ta$ is an object $0$ such that the categories $\hog{0,X}$ and $\hog{X,0}$ are trivial groupoids. A \emph{theory} is a category possessing finite products. A \emph{track theory} (resp. {\it strong track theory}) is a track category $\ta$ possessing finite lax products (resp. strong products) \cite{BJP}. If $\ta$ is a track theory, then $\ta_\ho$ is a theory. In this case the corresponding natural system on $\ta_\ho$ is a so called \emph{cartesian natural system}, meaning that it is compatible with finite product in an appropriate sense \cite{BJP}. Conversely, if $\ta$ is a track category, with property that $\ta_\ho$ is a theory and corresponding natural system is a cartesian natural system then $\ta$ is a track theory \cite{BJP}. Morphisms of track theories are enriched functors which are compatible with lax products. An equivalence of track theories is a track theory morphism which is a weak equivalence \cite{BJP} and two track theories are called \emph{equivalent} if they are made so by the smallest equivalence relation generated by these. Two track theories $\ta$ and $\ta'$ are equivalent iff there is an equivalence of categories $\ta_{\ho}\cong \ta'_\ho$ and after identification of these categories one should have $D_{\ta}=D_{\ta'}$ and ${\sf Ch}(\ta)={\sf Ch}(\ta')$. The main result of \cite{BJP} is the so called \emph{strengthening theorem}, which asserts that any abelian track theory is equivalent to a strong one. An \emph{additive track theory} is a track category $\ta$ such that $\ta_\ho$ is an additive category and the corresponding natural system is a biadditive bifunctor. We are going now to give an equivalent definition, but let us before that discuss the definition of an additive category. Let $\A$ be a category with zero object $0$ which possesses also finite coproducts and finite products. For objects $A$ and $B$ we have canonical inclusions $i_1=(\id,0):A\to A\x B$ and $i_2=(0,\id):B\to A\x B$ and therefore also the canonical morphism $\kappa:A\vee B\to A\x B$. The category $\A$ is called \emph{semi-additive} if the canonical morphism $\kappa:A\vee B\to A\x B$ is an isomorphism for all $A$ and $B$. If $\A$ is a semi-additive category and $f,g:A\to B$ are morphisms in $\A$, we let $f+g:A\to B$ be the following composite: $$\xymatrix{A\ar[r]^{\Delta}&A\x A\ar[r]^{(f,g)}&B\x B\ar[r]^{\kappa^{-1}}&B \vee B\ar[r]^\nabla&B}$$ where $\Delta=(\id,\id)$ is the diagonal and $\nabla$ is the codiagonal. Thus in a semi-additive category hom's are commutative monoids and the composition law is biadditive. If these monoids are abelian groups then a semi-additive category is called \emph{additive}. This happens iff the identity morphism $\id_A$ admits the additive inverse $-\id_A$, for each object $A$. Now we pass to the 2-world. Let $\ta$ be a track theory with lax zero object. Then for any objects $A$ and $B$ of $\ta$, there is a map $i_1:A\to A\x B$ and tracks $p_1i_1\then \id_A$, $p_2i_1\then 0$. Similar meaning has $i_2:B\to A\x B$. A {\it semi-additive track category} is an additive track theory with strong zero object, such that for any two objects $A$ and $B$ the lax product $A\x B$ is also lax coproduct via $i_1:A\to A\x B$ and $i_2:A\to A\x B$. It is clear that the homotopy category $\ta_\ho$ of a semi-additive track theory is a semi-additive category. One can prove that a track category $\ta$ is an additive track theory iff it is a semi-additive track category and additionally the semi-additive category $\ta_{\ho}$ is an additive category. An additive track category is called {\it very strong} if it admits strong zero object $0$, strong finite products and for any two objects $A$ and $B$ the strong product $A\x B$ is also the strong coproduct by $i_1:A\to A\x B$ and $i_2:A\to A\x B$. As we said a strengthening theorem of \cite{BJP} asserts that any track theory is equivalent to a strong one. In particular, any additive track category is equivalent to one which possesses strong products. Since the dual of an additive track category is still a track theory, we see that it is also equivalent to one which possesses strong coproducts. Can we always get strong products and coproducts simultaneously? In other words, is every additive track category $\ta$ equivalent to a very strong one? We will see that the answer is negative in general, but positive provided the corresponding homotopy category $\ta_\ho$ (which is an additive category in general) is $\F_2$-linear, or $2$ is invertible in $\ta_\ho$ (meaning that all $\Hom$'s are modules over $\Z[\frac{1}{2}])$. More precisely the following is true: \begin{The} \label{verystongstrengthening} Let $\ta$ be a small additive track theory with the homotopy category ${\bf C}=\ta_\ho$ and a canonical bifunctor $D=D_{\ta}$. Let $_2D$ be the two-torsion part of $D$. Then there is a well-defined element $o(\ta)\in H^0({\bf C},\ _2D)$, which is nontrivial in general and such that $o(\ta)=0$ iff $\ta$ is equivalent to a very strongly additive track theory. The class $o(\ta)$ is zero provided hom's of the additive category ${\bf C}$ are modules either over $\Z[\frac{1}{2}])$ or over $\F_2$. \end{The} The reader should compare Theorem \ref{verystongstrengthening} with the exact sequence (\ref{HMLorigr}) and Proposition \ref{HMLpalg}. The similarity of these results is not accidental. Indeed, let us give a quick proof of the Theorem \ref{verystongstrengthening} in the the key case when ${\bf C}=R\mod$ is the category of finitely generated free modules over a ring $R$. The proof of Theorem \ref{verystongstrengthening} in the general case is a repetition of the proof given below in the special case, except that one has to use ringoids instead of rings and we leave it as an exercise to the interested reader. {\it Proof of Theorem \ref{verystongstrengthening}. The case ${\bf C}=R\mod$}. For any biadditive bifunctor $D$ on $R\mod$ one has an isomorphism $D\cong \hom(-,M\t_R-)$, where $M=D(R,R)$. Here we used the notations of Section \ref{bwjp}. By Proposition \ref{verysh} $\Sh^3(R,M)$ classifies all very strong additive track categories (up to equivalence) $\ta$ with $\ta_\ho=R\mod$ and $D(R,R)=M$, where $D$ is the canonical bifunctor associated with $M$. On the other hand the isomorphism (\ref{mompares}) and the isomorphism (\ref{jibpira}) show that $\HML^3(R,M)$ classifies all additive track categories $\ta$ (up to equivalence) with $\ta_\ho=R\mod$ and $D(R,R)=M$. Let $\ta$ be an additive track category, then up to isomorphism (\ref{jibpira}) one can assume that ${\sf Ch}(\ta)\in \HML^3(R,M)$. Thanks to the exact sequence (\ref{HMLorigr}) we can take $o(\ta)$ to be the image of ${\sf Ch}(\ta)$ in $\H^0(R, \ _2M)$. Now Theorem \ref{verystongstrengthening} is a consequence of the exact sequence (\ref{HMLorigr}) and Proposition \ref{HMLpalg}. The example $R=\Z$ and $M=\F_2$ shows that the map $\HML^3(R,M)\to \H^0(R,M)$ is not trivial in general. It follows that the function $o$ is not trivial in general. {\bf Remarks}. 1) The following example introduces a well-known example from topology \cite{B1} of a track category $\ta$, which represents the generator of $\HML^3(\Z,\F_2)=\F_2$. Following \cite{B1} we consider the track category ${\sf Top}^*$ of compactly generated Hausdorff spaces with basepoint $*$. Maps in ${\sf Top}^*$ are pointed maps. A track $\aa:f\then g$ between pointed maps $f,g:A\to B$ is a homotopy class of a homotopy relative to $A\times \partial I$. Now we take ${\cal S}^k$ to be the full subcategory of ${\sf Top}^*$ consisting of finite one-point unions of spheres $S^k$, $k\geq 2$. Then ${\cal S}^k$ is an abelian track category and $({\cal S}^k)_\ho$ is equivalent to $\Z\mod$. For $k\geq 3$ the corresponding bifunctor is $\hom(-,\F_2\t -)$ and therefore ${\cal S}^k$ is an additive track theory, whose class in $$H^3(\Z\mod, \hom(-,\F_2\t -))\cong \HML^3(\Z,\F_2)=\F_2$$ is nontrivial. \noindent 2) One can describe the function $o$ in Theorem \ref{verystongstrengthening} as follows. Let $\ta$ be an additive track theory. Let $\vee$ denote the weak coproduct in $\ta$ and let $0$ be the weak zero object. For objects $X,Y$ one has therefore ``inclusions'' $i_1:X \to X\vee Y$ and $i_2:Y \to X\vee Y$. Since $X\vee Y$ is also a weak product of $X$ and $Y$ in $\ta$ it follows that one has also projection maps $p_1:X\vee Y\to X$ and $p_2:X\vee Y\to Y$. For each $X$ we choose maps $i_X:X\to X\vee X$ and $t_X:X\vee Y\to Y\vee X$ in such a way that classes of $i_X$ and $t_X$ in $\ta_\ho$ are the codiagonal and twisting maps in the additive category $\ta_\ho$. It follows that there is a unique track $$\aa_X:i_X\then t\circ i_X$$ such that $p_{i*}(\aa_X)=0$ for $i=1,2$. Now, let $(1,1):X\vee X\to X$ be a map which lifts the codiagonal map in $\ta_\ho$. Then $(1,1)_*\aa_X$ is a track $\id_X\to \id_X$ and therefore it differs from the trivial track by an element $o(X)\in D(X,X)$. One can prove that the assignment $X\mapsto o(X)$ is an expected one. 3) Corollary \ref{HMLpalg} shows that if $\ta$ is an additive track theory such that $\ta_\ho$ is an $\F_p$-linear category then $\ta$ is equivalent to a strong additive track theory with $\Z/p^2\Z$-linear hom's. This fact for a special track theory arising in the theory of the ``secondary Steenrod algebra'' was proved by the first author by completely different methods and was a starting point of this work. 4) Based on quadratic categories and square rings \cite{square} we in the fortcomming paper we introduce the notion of strongly additive track theories and we prove that any additive track category is equivalent to srong one. \subsection{Crossed bimodules and very strongly additive track theories} Let us recall that for a category $\bf C$ and a bifunctor $D$ there is a category ${\sf Trext}({\bf C},D)$ such that $\pi_0({\sf Trext}({\bf C},D))\cong H^3({\bf C},D)$. The objects of ${\sf Trext}({\bf C},D)$ are abelian track categories $\ta$ with $\ta_\ho\cong \sf C$ and $D_{\ta}=D$. If additionally $\bf C$ is an additive category and $D$ is a biadditive bifunctor, then any such $\ta$ is an additive track theory. We let ${\sf Strext}({\bf C},D)$ be the full subcategory of ${\sf Trext}({\bf C},D)$ whose objects are very strongly additive track theories. \begin{Pro}\label{verysh} Let $R$ be a ring and let $M$ be a bimodule over $R$. There is a functor $${\bf Crosext}(R,M)\to {\sf Strext}(R\mod,\hom(-,M\t _R-))$$ which is an equivalence of categories. \end{Pro} {\it Proof}. Let $\d:C_1 \to C_0$ be a crossed bimodule. We let $\ta=\ta(\d)$ be the following track category. The objects of $\ta$ are the same as the objects of $R\mod$, i.~e. natural numbers. For any natural numbers $n$ and $m$ the maps from $n$ to $m$ (which is the same as objects of the groupoid $\ta(n,m)$) are $m\times n$-matrices with coefficients in $C_0$. For $f,g\in Mat_{m\times n}(S)$ the set of tracks $f\to g$ (which is the same as the set of morphisms from $f$ to $g$ in the groupoid $\ta(n,m)$) is given by $$\Hom_{\ta(n,m)}(f,g)=\left\{h \in Mat_{m\times n}(C_1)\ \mid\ \d (h)=f-g\right\}.$$ The composition of 1-arrows is given by the usual multiplication of matrices, while the composition of tracks is given by the addition of matrices. One easily checks that in this way one really obtains a very strongly additive track theory $\ta(\d)$. It is clear that $\ta_\ho=R\mod$, where $R={\sf Coker}(\d)$ and the bifunctor associated to $\ta$ is $D=\hom(-,M\t_R-)$. Thus we obtain a functor $${\bf Crosext}(A,M)\to {\sf Strext}(R\mod,\hom(-,M\t _R-)).$$ Now we construct the functor in the opposite direction. Let $\ta$ be an object of ${\sf Strext}(R\mod,\hom(-,M\t _R-))$. Let $\ta_0$ be the category with the same objects as $\ta$ and with maps (i.e. 1-arrows) of $\ta$ as morphisms. Since $\ta$ is a strongly additive track theory, we see that $\ta_0$ is an additive category and therefore it is equivalent to $S\mod$, where $S={\sf End}_{\ta_0}(1)$. The restriction of the quotient functor $\ta\to \ta_0$ yields the homomorphism of rings $S\to R$. One defines $X$ to be the set of pairs $(h,x)$, where $x\in \Hom_{\ta_0}(1.1)$ and $h:x\then 0$ is a track in the groupoid $\ta(1,1)$. Moreover we put $\partial=\partial_{\ta}(h,x)=x$. Then $X$ carries a structure of a bimodule over $S$, and $$0\to M\to X\buildrel \partial \over \to S\to R\to 0$$ is a crossed extension. Then $\ta\mapsto \partial_{\ta}$ yields the functor $$ \mathsf{Strext}(R\mod,\hom(-,M\t _R-))\to\mathbf{Crosext}(A,M). $$ One easily checks that these two functors yield the expected equivalence of categories. \appendix \section{Closed model category structure on chain algebras and crossed bimodules }\label{cmc} \subsection{Closed model categories} We recall the definition of a closed model category introduced by Quillen \cite{qu}. We refer the reader to \cite{dwyerspalinski} for the basic facts on the closed model category theory. Let $\c$ be a category. A morphism $f$ is a \emph{retract} of a morphism $g$ if there exists a commutative diagram of the form $$ \xymatrix{A\ar[r]\ar[d]_f &C\ar[d]^g\ar[r]& A\ar[d]^f \\ B\ar[r] & D\ar[r] &A}$$ in which the horizontal composites are identities. Let $i:A\to B$ and $p:X\to Y$ be morphisms in $\c$. Then $i$ has \emph{left lifting property} with respect to $p$ and $p$ has \emph{right lifting property} with respect to $i$, if for every commutative diagram $$ \xymatrix{A\ar[r]^{ g'}\ar[d]_i &X\ar[d]^f \\ B\ar[r]_{ g} & Y }$$ there exists a commutative diagram $$ \xymatrix{A\ar[r]^{ g'}\ar[d]_i &X\ar[d]^f \\ B\ar[r]_{ g}\ar[ru]_h & Y }$$ Then $h$ is called a {\it lifting}. \begin{De} A closed model category consists of a category ${\cal C}$ together with three distinguished classes of morphisms called respectively weak equivalences, cofibrations and fibrations, so that the following 5 axioms hold. \smallskip \noindent {\bf CM 1}. $\cal C$ has all finite limits and colimits. All 3 classes form a subcategory. \smallskip \noindent {\bf CM 2}. If $f$ and $g$ are composable arrows in $\cal C$ and two of the three morphisms $f,g,gf$ are weak equivalences, then so is the third. \smallskip \noindent {\bf CM 3}. A retract of a fibration (resp. cofibration, weak equivalence) is still a fibration (resp. cofibration, weak equivalence). \smallskip \noindent {\bf CM 4}. Fibrations have the right lifting property with respect to acyclic cofibrations and cofibrations have left lifting property with respect to acyclic fibrations. Here a map is called an \emph{acyclic fibration} (resp. \emph{acyclic cofibration}) if it is both a fibration (resp. cofibration) and a weak equivalence. \smallskip \noindent {\bf CM 5}. Any arrow $f:A\to B$ has factorizations $f=pi$ and $f=qj$, where $i$ and $j$ are cofibrations, $p$ and $q$ are fibrations and $p$ and $j$ are weak equivalences too. \end{De} Here is more language corresponding to closed model categories. An object $X$ is called {\it cofibrant} if $ \emptyset\to X$ is a cofibration. An object $Y$ is called {\it fibrant} if $ Y\to *$ is a fibration. Here $ \emptyset$ and $*$ are respectively initial and terminal objects in $\cal C$. For any object $X$ there are weak equivalences $X\to X^f$ and $X^c\to X$ with fibrant $X^f$ and cofibrant $X^c$. This is an easy consequence of CM 5. Any such $X^c$ (resp. $X^f$) is called a cofibrant replacement (resp. fibrant replacement). It follows from the axioms that a map $i$ is a cofibration iff it has the left lifting property with respect to acyclic fibrations. Moreover $i$ is an acyclic cofibration iff it has the left lifting property with respect to fibrations. Therefore fibrations and weak equivalences completely determine cofibrations. The dual properties hold for fibrations. \bigskip Let $\cal C$ be a closed model category. We let $\cal W$ be the class of all weak equivalences. Then there exists a category ${\cal H}o:={\cal C}[{\cal W}^{-1}]$ together with a functor ${\cal C}\to {\cal H}o$ which takes all morphisms from ${\cal W}$ to isomorphisms and which is universal with respect to this property. Clearly the category ${\cal H}o$ is determined uniquely up to equivalence of categories. It has the following description: objects of ${\cal H}o$ are the same as those of $\cal C$, while morphisms are given by $$\Hom_{{\cal H}o}(X,Y):=\Hom _{\cal C}(X^c,Y^f)/\sim ,$$ where $\sim$ is an appropriate homotopy relation, which is defined as follows. Let $f,g:A\to B$ be two maps. Then $f\sim g$ if there exists a map $h:IA \to B$ such that $f=h\circ i_1$ and $g=h\circ i_2$. Here $IA$ and the maps $i_1,i_2:A\to IA$ satisfy the following conditions: the canonical map $(id,id):A\coprod A\to A$ is a composite $A\coprod A{\buildrel (i_1,i_2) \over\longrightarrow} IA \to A$, where the first map is a cofibration and the second one is an acyclic fibration. It turns out that this relation is an equivalence relation on $\Hom_{\cal C}(A,B)$ if $A$ is cofibrant and $B$ is fibrant. Moreover it is compatible with the composition law in $\cal C$ and the category ${\cal H}o$ is well defined. \subsection{Cofibrantly generated model categories} Suppose $\cal C$ is a category with all colimits. Let $I$ be a class of maps in $\cal C$. Following \cite{hov} we call a morphism {\it $I$-injective} (resp. {\it $I$-projective}) if it has the right (resp. left) lifting property with respect to every morphism in $I$. The class of $I$-injective and $I$-projective morphisms are denoted $I$-\emph{inj} and $I$-\emph{proj} respectively. A morphism is called an {\it $I$-cofibration} (resp. {\it $I$-fibration}) if it has the left (resp. right) lifting property with respect to every morphism in $I$-\emph{inj} (resp. $I$-\emph{proj}). The class of $I$-cofibrations and $I$-fibrations are denoted $I$-\emph{cof} and $I$-\emph{fib} respectively. Assume now $I$ is a set of morphisms. A morphism $f:A\to B$ is called a {\it relative $I$-cell complex} if there is an ordinal $\lambda$ and a $\lambda$-sequence $$X_0\to X_1\to \cdots \to X_{\beta}\to \cdots, $$ $ \beta\leq \lambda$, with $A=X_0$ and $B=colim X_{\beta}$ such that for all $\beta$ with $\beta + 1<\lambda$ there is a pushuot diagram $$\xymatrix{C_{\beta}\ar[r]^{g_{\beta}}\ar[d] & D_{\beta}\ar[d]\\ X_{\beta}\ar[r]& X_{\beta +1}}$$ such that $g_{\beta}\in I$. The class of relative $I$-cell complexes is denoted $I$-\emph{cell}. An object $A$ is called an {\it $I$-cell complex} if $0\to A$ is a relative $I$-cell complex. We will say that an object $A$ is {\it small} relative to a class of morphisms $I$ if there exists a cardinal $\kappa$ such that for each $\kappa$-filtered ordinal $\lambda$ and a $\lambda$-sequence $X_0\to X_1\to \cdots \to X_{\beta} \to \cdots$ one has $$colim \Hom_{\cal C}(A, X_{\beta})\cong \Hom_{\cal C}(A, colim X_{\beta}).$$ If $A$ is small with respect of $\cal C$ then $A$ is called {\it small}. The following result is well-known (see for example Theorem 2.1.19 of \cite{hov}). \begin{Pro}\label{Ho} Suppose $\cal C$ is a category with all colimits and limits. Suppose $W$ is a subcategory of $\cal C$ and $I$ and $J$ are two sets of morphisms of $\cal C$ such that the following conditions hold: \begin{itemize} \item[(i)] The subcategory $W$ is closed under retracts and satisfies the CM2 axiom. \item[(ii)] The domains of $I$ (resp. $J$) are small relative to $I$-\emph{cell} (resp. $J$-\emph{cell}). \item[(iii)] $J$-\emph{cell} $\subset W\bigcap I$-\emph{cof}. \item[(iv)] $I$-\emph{inj} $= W\bigcap J$-\emph{inj}. \end{itemize} Then there is a close model category structure on $\cal C$ with $W$ as the subcategory of weak equivalences, $I$-\emph{cof} as the class of cofibrations, $J$-\emph{inj} as the class of fibrations. Moreover $I$-\emph{inj} is the class of acyclic fibrations and $J$-\emph{cof} is the class of acyclic cofibrations. \end{Pro} The closed model categories obtained in this way are called {\it cofibrantly generated model categories}. \subsection{Chain algebras} We fix a commutative ring $\K$ and all algebras in what follows in this section are $\K$-algebras. Let us recall that a chain algebra is a graded algebra $A=\bigoplus_{n\geq 0} A_n$ equipped with a differential $d:A_n\to A_{n-1}$ satisfying the Leibniz identity: $$d(xy)=d(x)y+(-1)^{n}xd(y), \ \ x\in A_n, y\in A_m.$$ Let ${\bf DGA}$ be the category of chain algebras. \begin{The} Define a morphism of chain algebras to be \begin{itemize}\label{CMCDGA} \item[(i)] a weak equivalence if it induces isomorphism in homology \item[(ii)] a fibration if it is a surjection in positive dimensions \item[(iii)] a cofibration if it has the left lifting property with respect to all maps which are fibrations and weak equivalences \end{itemize} Then with these choices $\bf DGA$ is a cofibrantly generated closed model category. \end{The} To prove the theorem, we first introduce two classes of chain algebras. They play the role of discs and spheres. For $n\geq 1$ we let $D(n)$ be the following chain algebra. As graded algebra it is freely generated by elements $x$ and $dx$ of degree $n$ and $n-1$ respectively. The boundary map assigns $dx$ to $x$. For $n=0$ we let $D(0)$ be the algebra freely generated by an element $x$ of degree $0$ (of course $d(x)=0$ in this case). Moreover we define $S(n)$ to be the trivial algebra $\K$ if $n=-1$ and the algebra freely generated by an element $y$ of degree $n$ with zero boundary $d(y)=0$ provided $n\geq 0$. Then for all $n\geq 0$ we have a canonical homomorphism $S(n-1) \to D(n)$ which takes the generator $y$ to $dx$. We let $\coprod$ denote the coproduct in $\bf DGA$. One has the following isomorphism of chain complexes \begin{equation}\label{copr} A_*\coprod D(n)\cong A_*\oplus (A_*\t C_*\t A_*)\oplus (A_*\t C_*\t A_*\t C_*\t A_*)\oplus \cdots. \end{equation} Here $C_*$ is a chain complex, which is zero in all dimensions except for dimensions $n$ and $n-1$, where it is $\K$ and the unique nontrivial boundary map is the identity. Therefore the inclusion $A_*\to A_*\coprod D(n)$ is a weak equivalence, provided $n>0$. One observes that for any chain algebra $A_*$ one has the isomorphisms \begin{equation}\label{Dn}\Hom_{\bf DGA}(D(n),A_*)\cong A_n, \end{equation} \begin{equation}\label{Sn} \Hom_{\bf DGA}(S(n),A_*)\cong \Ker(d:A_n\to A_{n-1}). \end{equation} We let $W$ be the subcategory of all weak equivalences in $\bf DGA$. Moreover we put $$J:=\{\K\to D(n)\}_{n\geq 1},$$ $$I:=J\bigcup \{S(n-1)\to D(n)\}_{n\geq 0}.$$ Then the conditions i) and ii) of Proposition \ref{Ho} hold, because $D(n)$ and $S(n)$ are small thanks to isomorphisms \ref{Dn} and \ref{Sn}. We will show that all conditions of Proposition \ref{Ho} hold as well. To this end we need some preparations. Since $\K$ is the initial object in $\bf DGA$ a morphism $f:X_*\to Y_*$ is in $J$-\emph{inj} iff for any diagram $$ \xymatrix{& X_*\ar[d]^f \\ D(n)\ar[r]_{g} & Y_*}$$ there exists a morphism $h:D(n)\to X$ such that $f=gh$. Now the isomorphism (\ref{Dn}) gives that $f:X_*\to Y_*$ is in $J$-\emph{inj} iff $f_n$ is surjective for all $n>0$. Thus we proved the following \begin{Le}\label{fib} A map $f:X_*\to Y_*$ is a $J$-$inj$ iff it is fibration. \end{Le} \begin{Le}\label{iinj} Let $f:X_*\to Y_*$ be a morphism in $\bf DGA$. Then the following conditions are equivalent: \begin{itemize} \item[(i)] $f$ is $I$-injective \item[(ii)] $f$ is fibration and weak equivalence \item[(iii)] $f_n$ is surjective for all $n\geq 0$ and $\Ker \, f$ is acyclic, that is $H_*(\Ker \, f)=0$. \end{itemize} \end{Le} \noindent {\it Proof}. Lemma \ref{fib} and the homology exact sequence show that iii) $\Longrightarrow $ ii). Thanks to the isomorphism (\ref{Sn}) a morphism $f$ lies in $I$-\emph{inj} iff $f$ is a fibration with the following property: for all $x\in X_{n-1}$ and $y\in Y_n$ with $dx=0$ and $fx=dy$ there exists $z\in X_n$ such that $dz=x$ and $fz=y$. If the last condition holds, then $f_0$ is surjective and $\Ker \, f$ is acyclic. Thus by Lemma \ref{fib} we have i) $\Longrightarrow $ iii). Conversely, assume iii) holds. Suppose $x\in X_{n-1}$ and $y\in Y_n$ are given with $dx=0$ and $fx=dy$. Then there is $u\in X_n$ such that $fu=y$. Since $f(x-du)=0$ and $d(x-du)=0$ it follows that $x-du=dv$ for some $v\in X_n$ and therefore $x=d(u+v)$ which shows that iii) $\Longrightarrow $ i). To show ii) $\Longrightarrow $ iii) it suffices to show that $X_0\to Y_0$ is surjection. But this follows from the commutative diagram \bigskip $$ \xymatrix{& \cdots \ar[r] & X_1\ar[r]\ar[d] &X_0\ar[r]\ar[d] &H_0(X_*)\ar[r]\ar[d]_{\cong} &0\\ &\cdots \ar[r] & Y_1\ar[r]\ar[d] &Y_0\ar[r] &H_0(Y_*)\ar[r] &0\\ &&0&} $$\rdg \begin{Co}\label{ijinj} We have $I$-\emph{inj}$=W\bigcap J$-\emph{inj}. \end{Co} Recall that a morphism $f:X_*\to Y_*$ belongs to $I$-\emph{cof} if it has the left lifting property with respect to all maps from $I$-\emph{inj}. Thanks to Lemma \ref{iinj} this happens iff $f$ is a cofibration. A chain algebra $A_*$ is called $\K$-{\it projective} if each $A_n$ is projective as a $\K$-module. A chain algebra $A_*$ is called {\it quasi-free} if its underlying algebra is free. Let us recall that a graded algebra is free if it is isomorphic to the tensor algebra $T(V)$ of a graded $\K$-module $V_*$, which is free as a $\K$-module. A map of chain algebras $f:A_*\to B_*$ is called {\it quasi-free} if $B_*$ as a graded algebra is a coproduct $A_*\coprod X_*$ where $X_*$ is a free algebra. \begin{Le}\label{quasifree}{Quasi-free maps are cofibrations.}\end{Le} {\it Proof}. Let $$\xymatrix{A_*\ar[r]^{g}\ar[d]^i& X_*\ar[d]_p \\ B_*\ar[r]_h & Y_* }$$ be a commutative diagram of chain algebras, in which $i$ is quasi-free and $p$ is an acyclic fibration. We have to prove that there is a chain map $f:B_*\to X_*$ such that $g=fi$ and $h=pf$. By assumptions we have an isomorphism of algebras $B_*\cong A_*\coprod C_*$, where $C_*$ is a free algebra. Let $E$ be the set of free generators of $C_*$. Then $E$ is the union of subsets $E_n$ of degree $n$ elements, $n\geq 0$. In order to define $f$ one needs to specify elements $f(x)$ for $x\in E_n$, $n\geq 0$ with two properties \begin{itemize} \item[a)] $\partial f(x)= f(\partial x)$, \item[b)] $pf(x)= h(x)$. \end{itemize} We will work by induction on $n$. First consider the case $n=0$. Since $p$ is surjective, there exists $f(x)\in X_0$ such that $pf(x)=h(x)$. Consider now the case $n>0$. Suppose for all $m<n$ we already defined $f(x)$ for all $x\in E_m$ such that a) and b) holds for all $x\in E_j$, $1\leq j<n$. Take now $x\in E_n$. Since $p$ is surjective we can choose an element ${\bar f}(x)\in X_n$ such that $p{\bar f}(x)= g(x)$. Since $\partial x$ lies in the subalgebra generated by $A_*$ and $E_j$, $j<n$, the element $f(\partial x)$ is already defined. Set $z= \partial ({\bar f}(x))-f(\partial (x))$. Then $\partial (z)= 0$ and $p(z)=0$. Therefore $z=\partial (u)$ for some element $u\in \Ker (p)$. Now we put $f(x)= {\bar f}(x)-u$. It is clear that $f(x)$ satisfies properties a) and b). Thus induction step is finished and hence the lemma. \rdg \Cor{ The canonical maps $k\to D(n)$, $k\to S(n)$ and $S(n-1)\to D(n)$ are cofibrations.} {\bf Proof of Theorem \ref{CMCDGA}.} As was mentioned already the conditions i) and ii) of Proposition \ref{Ho} hold. By Corollary \ref{ijinj} the condition iv) holds as well. Thus we have to show that $J$-\emph{cell}$\subset W\bigcap I$-\emph{cof}. We have $J$-\emph{cell}$\subset I$-\emph{cof} because $J\subset I$. Since the domain of all maps from $J$ is $\K$, which is an initial object, we see that all pushouts in the definition of a relative $J$-cell complex are coproducts with $D(n)$ for some $n>0$. It follows that all such morphisms are weak equivalences and quasi-free maps and the result follows from Lemma \ref{quasifree}. Let us note that a similar theorem for cochain algebras was proved by Jardine \cite{jar}. Moreover our argument is merely a variant of the one given there (compare also with \cite{bg}). \subsection{Truncated chain algebras}\label{cmctrun} Let us fix a natural number $m\geq 1$. We let ${\bf DGA}_m$ be the full subcategory of $\bf DGA$ which consists of objects $X_*$ such that $X_i=0$ for all $i>m$. For any chain complex $(X_*,d)$ we let $\tam(X_*)$ be the following chain complex: $$(\tam(X_*))_i=X_i, \ {\rm if } \ i<m$$ $$(\tam(X_*))_m=X_m/d(X_{m+1}) \ \ $$ $$(\tam(X_*))_i=0, \ {\rm if } \ i>m$$ The quotient map $X_*\to \tam(X_*)$ is a chain map. Moreover $H_i(\tam(X_*))\cong H_i(X_*)$ if $i\leq m$ and $H_i(\tam(X_*))=0$ provided $i>m$. It is also clear that, if $X_*$ is a chain algebra, then there is a unique chain algebra structure on $\tam(X_*)$ such that the quotient map $X_*\to \tam(X_*)$ is a chain algebra homomorphism. Thus $$\tam:{\bf DGA}\to {\bf DGA}_m$$ is a well-defined functor, which is the left adjoint to the inclusion functor ${\bf DGA}_m\subset {\bf DGA}$. \begin{The}\label{CMCDGAM} Define a map in ${\bf DGA}_m$ to be a weak equivalence (resp. fibration) if it is a weak equivalence (resp. fibration) in $\bf DGA$. Define a map in ${\bf DGA}_m$ to be a cofibration if it has left lifting property with respect to all acyclic fibrations. Then this defines a cofibrantly generated model structure on ${\bf DGA}_m$. \end{The} {\it Proof}. We introduce two classes of morphisms in ${\bf DGA}_m$: $$J_m:=\{\K\to \tam D(n)\}_{ n\geq 1 },$$ $$I_m:=J_m\bigcup \{\tam S(n-1)\to \tam D(n)\}_{n\geq 0}.$$ We have to show that all assertions of Proposition \ref{Ho} hold. Conditions i) and ii) are clear. Formal argument with adjoint functors shows that a morphism $f:X_*\to Y_*$ in ${\bf DGA}_m$ considered as a morphism of $\bf DGA$ lies in $J$-\emph{inj} (resp. $I$-\emph{inj}) iff it is in $J_m$-\emph{inj} (resp. $I_m$-\emph{inj}). Therefore $f$ is a fibration (resp. acyclic fibration) iff it is in $J_m$-\emph{inj} (resp. $I_m$-\emph{inj}) and the condition iv) holds. We also have $J_m$-\emph{cell}$\subset I_m$-\emph{cof} because $J_m\subset I_m$. Thus it remains to show that $J_m$-\emph{cell}$\subset W$. Comparing the definitions we see that any morphism from $J_m$-\emph{cell} can be written as $\tam(g)$, where $g\in J$-\emph{cell}. In particular $g\in W$. Since $\tam$ preserves weak equivalences we are done. \subsection{A closed model category structure on crossed bimodules}\label{cmcxmod} Of the special interest is the case, when $m=1$. In this case Theorem \ref{CMCDGAM} gives the closed model category structure on the category $\sf Xmod$ of crossed bimodules. A map of crossed bimodules $$\xymatrix{C_1\ar[r]^\d\ar[d]^f&C_0\ar[d]^g\\ C_1' \ar[r]^{\d'}&C_0'} $$ is a fibration if $f$ is a surjective homomorphism. Moreover, $(f,g): \d\to \d'$ is a weak equivalence if induced maps $\ker(\d)\to \ker(\d')$ , ${\sf Coker}(\d)\to \ {\sf Coker}(\d')$ are isomorphisms. It follows that if $(f,g)$ is an acyclic fibration, then $g$ is a surjection and the induced map $\ker(f)\to \ker(g)$ is an isomorphism, in other words one has the following commutative diagram with exact rows and columns: $$\xymatrix@=1em{&& 0\ar[d]& 0\ar[d]&&\\ &&\ker(f)\ar[d]\ar[r]^{\cong}&\ker(g)\ar[d]&&\\ 0\ar[r]&\ker(\d)\ar[d]^\cong \ar[r]&C_1\ar[d]^f\ar[r]^\d&C_0\ar[r]\ar[d]^g&{\sf Coker}(\d)\ar[r]\ar[d]^\cong & 0\\ 0\ar[r]&\ker(\d') \ar[r]&C_1'\ar[r]^{\d'}\ar[d]&C_0\ar[r]\ar[d]&{\sf Coker}(\d')\ar[r]& 0\\ &&0&0&&} $$ Thus we proved the following \begin{Le} If $(f,g):\d\to \d'$ is an acyclic fibration in $\sf Xmod$, then $g$ is surjective and $$\xymatrix{ C_1\ar[d]^f\ar[r]^\d&C_0\ar[d]^g\\ C_1'\ar[r]^{\d'}&C_0'} $$ is a pullback diagram. \end{Le} \begin{Le} A crossed bimodule $\delta :R_1\to R_0$ is a cofibrant objects in $\sf Xmod$ provided $R_0$ is a free algebra. \end{Le} {\it Proof}. Let $(f,g):\d\to \d'$ be an acyclic fibration of crossed bimodules and let $(a',b'):\delta\to \d'$ be a morphism of crossed bimodules. We have to lift it to a morphism $(a,b):\delta\to \d$. Since $g$ is a surjective homomorphism of $\K$-algebras and $R_0$ is a free $\K$-algebra, we can lift $b'$ to a homomorphism $b:R_0\to C_0$ of $\K$-algebras. Then we have the following commutative diagram $$ \xymatrix@=.2em{R_1\ar[dddddr]_{a'}\ar[drrrrr]^{b\partial }\\ & C_1\ar[dddd]^f\ar[rrrr]_\d&&&&C_0\ar[dddd]^g\\ \\ \\ \\ &C_1'\ar[rrrr]^{\d'}&&&&C_0'} $$ and we have the unique homomorphism $a:R\to C_1$ which fits in the diagram. It is now clear that $(a,b)$ is an expected lifting.

In message <4077FE63.4010602 at cronyx.ru>, Roman Kurakin writes: >Poul-Henning Kamp wrote: >>How about we recognize that different users needs different levels of >>randomness ? >> >>Couldn't we provide at three levels of random bits: >> >> 1. "Random enough for games" >> >> 2. "Random enough for money" >> >> 3. "Random enough for lives" >> >How about computations? Well, depends what kind of computations you are talking about obviously. I specifically didn't list that because I have seen so many examples of monte carlo analysis done with rand() that I could cry. People who do stochastic simulation should think about randomization not just belive in some random (pun intended).

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