TITLE: $\kappa$-free abelian groups and inaccessible cardinals
QUESTION [4 upvotes]: I found the following unproved statements in Shelah and Mekler's article "On the consistency strength of "every stationary set reflects"".

Let $\kappa$ be the least cardinal such that there exists a $\kappa$-free abelian group which is not $\kappa^+$-free. By a well known
  argument, $\kappa$ is either the successor of a singular cardinal or an inaccessible
  cardinal.
It is easy to see (and well known) that if every stationary set
  reflects in a regular cardinal then every $\kappa$-free abelian group is
  $\kappa^+$-free.

Can someone please give references for the above statements? In particulary I am deeply interested in the first one, but I couldn't find any introductory material. The wonderful book "Almost free modules" by Paul Eklof strangely doesn't mention the result (or I couldn't find it!).
Thank you in advance.

REPLY [2 votes]: The paper

MR1029909 (91b:03090). Mekler, Alan H.; Shelah, Saharon. The consistency strength of "every stationary set reflects". Israel J. Math. 67 (1989), no. 3, 353–366,

that you mention in the question actually provides the relevant references and explains the key idea of the argument. Note first that $\kappa$ is assumed regular.
They refer to

MR0379694 (52 #599). Eklof, Paul C. On the existence of $\kappa$-free abelian groups. Proc. Amer. Math. Soc. 47 (1975), 65–72.

In that paper, Eklof essentially shows that if $\kappa$ is regular and there is a stationary subset of $\kappa$ that does not reflect, then there is a $\kappa$-free not $\kappa^+$-free abelian group. This means that if we want the opposite, every stationary subset of $\kappa$ must reflect.
The point then is to note that $S^{\tau^+}_\tau$ is non-reflecting whenever $\tau$ is regular. Here $S^{\tau^+}_\tau$ is the set of ordinals smaller than $\tau^+$ of cofinality $\tau$. To see this, let $\alpha\in S^{\tau^+}_\tau$, fix a club subset $C$ of $\alpha$ of type $\tau$, and note that $C'$ cannot meet $S^{\tau^+}_\tau$ since no point of $C'$ has cofinality $\tau$.
This leaves us with $\kappa$ being inaccessible or the successor of a singular, since all other regular cardinals are of the form $\kappa^+$ for $\kappa$ regular.


TITLE: Finding vector for supremum norm of matrix
QUESTION [1 upvotes]: Is it possible to find a vector $w$ under the constraint $|w|=1$ such that the euclidean norm $|Aw|$ is maximized, where $A$ is a matrix?

REPLY [2 votes]: Let $\omega_* = \underset{\omega}{\arg\max} \|A\omega\|$ subject to $\|\omega\| =1$. The Lagrangian is $$L(\omega,\lambda) = \|A\omega\|^2 + \lambda(1 - \|\omega\|^2) \Rightarrow\dfrac{\partial L}{\partial \omega} = 2A^TA\omega - 2 \lambda \omega .$$ Setting $\left.\dfrac{\partial L}{\partial \omega}\right|_{\omega = \omega_*}=0$ gives
\begin{align}
A^TA\omega_* = \lambda \omega_*
\end{align}
Multiplying on the left by $\omega_*^T$, gives $\|A\omega_*\|^2 = \lambda \|\omega_*\|^2 = \lambda$.
Thus, $\omega_*$ is the eigenvector corresponding to the largest eigenvalue of $A^TA$.


TITLE: Finitely generated free modules of infinite rank.
QUESTION [2 upvotes]: We know that for general modules over a commutative ring with $1$, you can't always extract a basis from a generating set.
This makes me think that maybe there should be free modules of infinite rank which could be finitely generated. Do such things exist?

REPLY [6 votes]: Suppose we had such a horrible thing, a surjection: $$R^{\oplus n} \twoheadrightarrow R^{\oplus I}$$where $I$ is something infinite (or just finite and $> n$). Pick any maximal ideal of $R$ and tensor up with $R/\mathfrak{m}$, it's right exact so we still have: $$R^{\oplus n} \otimes R/\mathfrak{m} \twoheadrightarrow R^{\oplus I} \otimes R/\mathfrak{m}$$But these are isomorphic to: $$R/\mathfrak{m}^{\oplus n} \twoheadrightarrow R/\mathfrak{m}^{\oplus I}$$a surjection of vector spaces.
So, can't happen!


TITLE: Finding absolute extrema in a region
QUESTION [0 upvotes]: Given $-8x^3-6y^2+12xy=0$, find its absolute maximum and minimum in the region $x≥0$ and $y≥-1$.
So far, I got the critical points $(0,0)$ and $(1/2,1/2)$, and that $(1/2,1/2)$ yields the absolute maximum value. I don't know if I solved it correctly since its in an unbounded region (Or is it? I'm not sure, I only learned the ones with bounded regions). I feel like I may be overthinking this? Anyways, any help/hint would be appreciated!

REPLY [0 votes]: Considered the regularity of the objective function, any local or global extrema will occur either in stationary points in the interior of the region, or at points in the boundary.
Setting $y=-1$, you see that there is no hope of having a global minimum.
At the boundary, we have $f = -6y^2, y\ge -1$ or $f = -8 x^3 -12 x -6, x \ge 0.$
At the only local maximum in the interior, we have $f = \frac 12$.
Putting all this together, there is no global minimum and either the global maximum is $\frac 12$, or there is no global maximum. Can you proceed?


text:
\begin{definition}[Definition:Limit of Real Function/Left]
Let $\openint a b$ be an [[Definition:Open Real Interval|open real interval]].
Let $f: \openint a b \to \R$ be a [[Definition:Real Function|real function]].
Let $L \in \R$.
Suppose that:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: b - \delta < x < b \implies \size {\map f x - L} < \epsilon$
where $\R_{>0}$ denotes the [[Definition:Set|set]] of [[Definition:Strictly Positive/Real Number|strictly positive real numbers]].
That is, for every real strictly positive $\epsilon$ there exists a real strictly positive $\delta$ such that ''every'' real number in the [[Definition:Domain of Mapping|domain]] of $f$, less than $b$ but within $\delta$ of $b$, has an [[Definition:Image of Element under Mapping|image]] within $\epsilon$ of $L$.
:[[File:LimitFromLeft.png|400px]]
Then $\map f x$ is said to '''tend to the limit $L$ as $x$ tends to $b$ from the left''', and we write:
:$\map f x \to L$ as $x \to b^-$
or
:$\displaystyle \lim_{x \mathop \to b^-} \map f x = L$
This is voiced:
:'''the limit of $\map f x$ as $x$ tends to $b$ from the left'''
and such an $L$ is called:
:'''a limit from the left'''.
\end{definition}

config: wiki


TITLE: Prove that there exist arbitrarily long sequence of consecutive integers not of the form $x^k$ with $k > 1$. (Assuming that $x \in \mathbb Z$)
QUESTION [0 upvotes]: I'm not sure how to approach this question exactly, I've thought about assuming that the length of the sequence is finite and bounded above and somehow construct a contradiction.
Any hint on how to approach this question would be great, thanks.

REPLY [0 votes]: Any number of the form $4n+2$ can't be a perfect power (because it's divisible by $2$ but not by $4$). Similarly for any number of the forms $9n+3$, $25n+5$, $49n+7$, and so on. So all you have to do is show that for any $m$ there exists $r$ such that $r\equiv2\bmod4$ and $r+1\equiv3\bmod9$ and and and $r+m-1\equiv p_m\bmod{p_m^2}$.


text:
\begin{document}

\title{five-term exact sequence for Kac cohomology}

\author{C\'esar Galindo}
\address{Departamento de Matem\'aticas\\ Universidad de los Andes\\ Carrera 1 N.
18A - 10,  Bogot\'a, Colombia}
\email{cn.galindo1116@uniandes.edu.co}

\author{Yiby Morales}
\address{Departamento de Matem\'aticas\\ Universidad de los Andes\\ Carrera 1 N.
18A - 10,  Bogot\'a, Colombia}
\email{yk.morales964@uniandes.edu.co}

\thanks{C.G. is partially supported by Faculty of Science of  Universidad de los Andes, Convocatoria 2018-2019 para la Financiaci\'on de Programas de Investigaci\'on, programa ''Simetr\'{i}a $T$ (inversi\'on temporal) en
categor\'{i}as de fusi\'{o}n y modulares''.}

\dedicatory{Dedicated to Nicol\'as Andruskiewitsch on the occasion of his 60th birthday}
\begin{abstract}
We use relative group cohomologies to compute the Kac cohomology of matched pairs of finite groups. This cohomology naturally appears in the theory of abelian extensions of finite dimensional Hopf algebras. We prove that Kac cohomology can be computed using relative cohomology and relatively projective resolutions. This allows us to use other resolutions, besides the bar resolution, for computations. We compute, in terms of relative cohomology, the first two pages of a spectral sequence which converges to the Kac cohomology and its associated five-term exact sequence. Through several examples, we show the usefulness of the five-term exact sequence in computing groups of abelian extensions.
\end{abstract}
\maketitle

\section{Introduction}

Extension theory of groups plays a significant role in the construction and classification of finite groups. In the same way, the extension theory of Hopf algebras has led to results on the still wide open problem of construction and classification of finite-dimensional semisimple Hopf algebras, \cite{Kashina,Masuoka-p3,Natale1,Natale2,Natale-pqr}. The set of equivalence classes of extensions of a group $G$ by a $G$-module $M$ is an abelian group with the Baer product of extensions, which is isomorphic to the second cohomology group $H^2(G,M)$.  A generalization of this theory to Hopf algebras is obtained for the so-called \emph{abelian extensions}, that is,  cleft Hopf algebra extensions of a commutative Hopf algebra $K$ by a cocommutative Hopf algebra  $H$, see \cite{Hofstetter,Kac,Masuoka-Calculations,masuoka-notas,Masuoka-TAMS,Singer}.

In this paper, we deal with  $H = kF$, a group algebra, and $K = k^G$, the dual of such an algebra, where $F$ and $G$ are finite groups. In this case, each abelian extension has an associated matched pair, that is, a larger group $\Sigma$ such that $G$ and $F$ are subgroups of $\Sigma$ satisfying $G \cap F=\{e\}$ and $\Sigma = GF$. The set of equivalence classes of abelian extensions associated to a fixed matched pair, denoted by $\Opext(kF,k^G)$, is an abelian group that can be computed as the second total cohomology of certain double complex whose cohomology is called \emph{Kac cohomology}.

Obtaining a computation for the $\Opext(kF,k^G)$ of a matched pair of groups can be quite difficult. In fact, there are few general computations in the literature \cite{Masuoka-Calculations}. One obstacle for the computation of  $\Opext(kF,k^G)$ comes from the fact that it is defined as the cohomology of a very specific total complex, and the unique ``cocycle free" tool is the so-called \emph{Kac exact sequence}, (see \cite{Masuoka-Calculations} Corollary \ref{Kac exact sequence}). Perhaps one of first the results that provides a  cocycle free description and interpretation of the Kac cohomology is given in \cite[Proposotion 7.1]{Kac-cohomology}, where the authors describe the Kac cohomology as the \emph{singular cohomology} of the mapping cone $BG\sqcup BF \to B\Sigma$.

In this paper, we use two different kinds of relative cohomology groups: the Auslander relative cohomology and Hochschild relative cohomology (see Section \ref{Section:Kac Cohomomlogy}), in order to compute $\Opext(kF,k^G)$. We prove that $\Opext_{\rhd,\lhd}(kF,k^G)$ can be computed using Auslander relative cohomology and that  Auslander relative cohomology of a matched pair can be computed using relatively projective resolutions. This allows us to use other resolutions, besides the bar resolution, for computing $\Opext_{\rhd,\lhd}(kF,k^G)$.  In addition, we compute the first and second page of a spectral sequence which converges to the Kac cohomology. As a consequence, we compute, in terms of Hochschild relative cohomology, the associated five-term exact sequence, whose second term is $\Opext(kF,k^G)$. In the particular case of a semidirect product, the five-term exact sequence is described in terms of ordinary group cohomology. Finally, doing use of the five-term exact sequence and some non-standard resolutions, we compute  $\Opext(kF,k^G)$ for several families of matched pairs.

The organization of the paper is as follows: in Section \ref{Sec:Preliminaries} we discuss
preliminaries on group cohomology and abelian extensions of Hopf algebras. In Section \ref{Section:Kac Cohomomlogy} we recall the definitions of Auslander relative cohomology \cite{auslander1993relative} and Hochschild relative cohomology \cite{hochschild1956relative}. We prove that $\Opext(kF,k^G)$ can be computed using Auslander Relative cohomology and Auslander relative cohomology of matched pairs can be computed using relatively projective resolutions. In Section \ref{Section:Five-term exact sequence} we compute the first and second page of a spectral sequence which converges to the Kac cohomology. We also compute, in terms of Hochschild relative cohomology, the associated five-term exact sequence, whose second term is $\Opext(kF,k^G)$. Finally, in Section \ref{Section:Computations} we compute  $\Opext(kF,k^G)$ for several families of matched pairs.

\smallbreak\subsubsection*{Acknowledgements} We thank Paul Bressler and Bernardo Uribe for kindly answering some questions and for very useful conversations.
\section{Preliminaries}\label{Sec:Preliminaries}
\subsection{Cohomology of groups}
Let $G$ be a group and let $M$ be a $G$-module. The $n^{th}$-\emph{cohomology group} of $G$ with coefficients in $M$ is defined as
\begin{equation*}
H^n(G,M)=\Ext^n_{\mathbb{Z}G}(\mathbb{Z},M).
\end{equation*}
We will use occasionally the \emph{normalized bar resolution} $(\mathbb{Z}\stackrel{\epsilon}{\leftarrow}P_i,\delta)$ of $\mathbb{Z}$ as a trivial $G$-module. That is \begin{equation*}P_i=\Z G[G]^i:=\Z G[s_1|\cdots |s_i],\end{equation*} where $s_i\in G$, $s_i\neq e$ for all $i$. The differentials are given by

\begin{equation*}
\begin{aligned}
\delta([s_1|\ldots|s_{p+1}])=&s_1[s_2|\ldots|s_{p+1}]+\sum_{i=1}^{p}(-1)^i[s_1|\ldots|s_is_{i+1}|\ldots|s_{p+1}]\\
&+(-1)^{p+1}[s_1|\ldots|s_p].
\end{aligned}
\end{equation*}

For a finite cyclic group $C_n=\langle g\rangle$ of order $n$, we occasionally  use a periodic resolution of $\mathbb{Z}$, defined as follows:

\begin{equation}\label{CyclicResolution}
\cdots\stackrel{}{\to}\mathbb{Z}C_n\stackrel{g-1}{\to} \mathbb{Z}C_n\stackrel{N}{\to}\mathbb{Z}C_n\stackrel{g-1}{\to}\mathbb{Z}C_n\stackrel{\epsilon}{\to}\mathbb{Z},
\end{equation}
where, $N=\sum_{i=0}^{n-1} g^i$.  From this, we have that

\begin{equation}\label{CyclicCohomology}
H^m(C_n,M) = \left\{
\begin{array}{c l}
 M^{C_n}/\Img(N), & m =2k\\
 \Ker(N)/\Img(g-1) & m=2k+1.
\end{array}
\right.
\end{equation}

\subsubsection{Resolutions and cohomology for direct products}
Let $G=G_1\times G_2$ be a direct product of groups. Let $(\mathbb{Z}\stackrel{\epsilon}{\leftarrow}P_i)$ and $(\mathbb{Z}\stackrel{\epsilon}{\leftarrow}Q_i)$ be projective resolutions of $\mathbb{Z}$ as $G_1$-module and $G_2$-module respectively. Then, the total complex $\Tot(P_i\otimes Q_i)$ is a $G$-projective resolution of $\mathbb{Z}$. A useful description of the cohomology for direct products is the following. If $M$ is a trivial $G$-module, then it holds (see e.g. \cite{karpilovsky1985projective})
\begin{equation}\label{Karpilovski}
    H^2(G_1\times G_2,M)\cong H^2(G_1,M)\oplus H^2(G_2,M)\oplus P(G_1,G_2;M),
\end{equation}

where $P(G_1,G_2;M)$ is the group of bicharacters from $G_1\times G_2$ to $M$. An isomorphism is given by $\overline{\alpha}\mapsto (\overline{\alpha_1},\overline{\alpha_2},\phi_\alpha)$, where $\alpha_i$ is the restriction of $\alpha$ to $Q_i\times Q_i$ and $$\phi_{\alpha}(x,y)=\Alt(\alpha)=\alpha(x,y)-\alpha(y,x).$$
the inverse isomorphism is defined by  $(\overline{\alpha_1},\overline{\alpha_2},\phi)\mapsto \overline{\alpha}$ with
$$\alpha((x_1,y_1),(x_2,y_2))=\alpha_1(x_1,x_2)\alpha_2(y_1,y_2)\phi(x_1,y_2).$$

\subsubsection{Second cohomology group and skew-symmetric matrices}

Another useful description of the second cohomology group in the case that $V$ is a finite abelian group is provided by using the universal coefficient theorem. Let $k$ be a field and let us consider the group $k^\times$ of units  of $k$ as a trivial $V$-module. There is a short exact sequence

\begin{equation}\label{UnivCoef}
0\to \operatorname{Ext}(V,k^\times)\to H^2(V,k^\times)\overset{\operatorname{Alt}}{\to} \Hom(\wedge^2(V),k^\times)\to 0.\end{equation}
If $V$ has exponent $n$  and $(k^\times)^n=k^\times$, then
$\operatorname{Ext}(V,k^\times)=0$. Therefore, the map
  \begin{equation}\label{AltBilForms}
  \Alt:H^2(V,k^\times)\to \wedge^2\widehat{V},\end{equation}
where $\widehat{V}=\Hom(V,k^\times)$, defines an isomorphism $H^2(V,k^\times)\cong \wedge^2\widehat{V}$. \\

\subsection{Extensions of Hopf algebras}\label{HopfAlgExt}

Let $k$ be a field. A sequence of \emph{finite dimensional} Hopf algebras and Hopf algebra maps  \begin{equation*}
    (A):k\to K\overset{i}\rightarrow A\overset{\pi}\rightarrow H \to k
\end{equation*} is called an \emph{extension} of $H$ by $K$
if, $i$ is injective, $\pi$ is surjective, and $K= A^{\co H}$ (see \cite{Andrus-Devoto, Kac, majid1990physics}).

Two extensions $(A),(A')$ of $H$ by $K$ are said to be \emph{equivalent} if there is an homomorphism $f:A\to A'$ of Hopf algebras such that the diagram
\[\xymatrix{
&&A \ar[dd]^{f}\ar[rd]^{\pi}&&\\
k\ar[r]&K \ar[ru]^{i}\ar[rd]_{i'}&&H \ar[r]& k\\
&&A'\ar[ru]_{\pi'}&&
}\]

commutes.

\subsection{Matched pairs of groups}

Let us recall (see e.g., \cite{takeuchi1981matched})  that a \emph{matched pair of groups} is a collection
 $(G,F,\rhd,\lhd)$ where $G,F$ are groups and $\rhd,\lhd$ are permutation actions \[\xymatrix{G & \ar[l]_{\lhd} G\times F \ar[r]^{\rhd}& F}\] such that
\begin{align*}
s\rhd xy=(s\rhd x)((s\lhd x)\rhd y),&&
st\lhd x=(s\lhd(t\rhd x))(t\lhd x),
\end{align*}for all $s,t \in G, x,y\in F$.

Having a group G and F with a matched pair structure is equivalent to having a group $\Sigma$ with an exact factorization;
the actions $\rhd$ and $\lhd$ are determined by the relations \[sx =(s \rhd x)(s \lhd x),\] $x \in F, s \in G$.

The group $\Sigma$ associated to a matched pair of groups  will be denoted by $F\bowtie G$, and it is $F\times G$ with product given by \[(x,s)(y,t)=(x(s\rhd y),(s\lhd y)t).\]

It is easy to see that the following
conditions are equivalent:
\begin{itemize}
    \item[(i)]  The action $\rhd$ is trivial.
\item[(ii)] The action $\lhd: G \times F \to G$ is by group automorphisms.
\end{itemize}In this case, the associated group $\Sigma=F \ltimes G$ is a semi-direct product of groups.

\subsection{Abelian extensions}
Let $(G,F,\rhd,\lhd)$ be a matched pair of groups and let us consider $2$-cocycles  $\sigma \in Z^2(F,(k^G)^\times)$, $\tau \in Z^2(G,(k^F)^\times)$. On the vector space \[k^G \#_{\sigma,\tau}  k F:= \operatorname{Span}_k \{e_s\# x : s\in G, x\in F\},\]we can define a unital associative algebra and counital coassociate coalgebra structure by
\begin{align*}
(e_s\#x) (e_t\# y)&= \delta_{s\lhd x,t}\sigma(s;x,y)e_s\# xy,\\
\Delta(e_s\# x)&= \sum_{s=ab}\tau(a,b;x)e_a\# (b\rhd x)\otimes e_b\# x.
\end{align*}Here the 2-cocycles $\sigma$ and  $\tau$ are seen as functions
\begin{align*}
    \sigma: G\times F\times F  \to k^\times,&&
    (s,x,y) &\mapsto \sigma(s;x,y),\\
    \tau:G\times G \times F\to k^\times, && (s,t,x) &\mapsto \tau(s,t;x).
\end{align*}

The map $\Delta$ is an algebra map if and only if the 2-cocycles satisfy the following compatibility condition
\begin{align*}
\sigma(st;x,y)\tau(s,t;xy) =&\sigma(s;t\rhd x,(t\lhd x)\rhd y)\sigma(t;x,y)\times \\&\tau(s,t;x)\tau(s\lhd(t\rhd x),t\lhd x;y),   \notag
\end{align*}
 for all $x,y \in G, s,t \in F$. In case that the 2-cocycles are compatible, the sequence  \begin{equation*}
k\to k^G\stackrel{i}{\rightarrow}k^G\#_{\sigma,\tau}kF\stackrel{\pi}{\to} kF\to k,
\end{equation*}is a Hopf algebra extension,
where $i(e_s)=e_s\# e$ and $\pi(e_s\# x)=x$. This kind of extension are called \emph{abelian extensions} of Hopf algebras.

The set of equivalence classes of abelian extensions associated to a fixed matched pair $ (G,F,\rhd,\lhd)$ is an abelian group with the Baer product of extensions, and will be denoted by $\Opext_{\rhd,\lhd}(kF,k^G)$ (see \cite{masuoka2002hopf} for mor details).

\section{Kac cohomology and relative cohomology}\label{Section:Kac Cohomomlogy}

In this section, we recall the definitions of two different kinds of relative group group cohomology: the Auslander relative cohomology \cite{auslander1993relative} and the Hochschild relative cohomology \cite{hochschild1956relative}. Our aim  is to prove that $\Opext_{\rhd,\lhd}(kF,k^G)$ can be computed using Auslander Relative cohomology which, in the case of matched pairs, can be computed using relatively projective resolutions. This allows us to use other resolutions besides the bar resolution for computing $\Opext_{\rhd,\lhd}(kF,k^G)$.

\subsection{Auslander relative cohomology of groups}\label{Auslander}

Let $\Sigma$ be a group and $X$ a $\Sigma$-set. We will denote by  $\Lambda_X$ the kernel of the augmentation map
\begin{align}\label{aumentation map}
\epsilon_X:\mathbb{Z}[X] \to \mathbb{Z},&&
x\mapsto 1,
\end{align}
where $\mathbb{Z}[X]$ is the  $\Sigma$-module associated to $X$.

\begin{definition}
Given $\Sigma$-module $A$, the \emph{$n^{th}$-cohomology group of $\Sigma$ relative to $X$ with coefficients in $A$} is defined by
 \[H_\mathscr{A}^k(\Sigma,X;A):=\operatorname{Ext}_{\mathbb{Z}\Sigma}^{k-1}(\Lambda_X,A), \quad k\geq 1.\]
\end{definition}

Let $X$ be a $\Sigma$-set and $\mathcal{R}_X$ a set of representatives of the $\Sigma$-orbits in $X$. Using Shapiro's Lemma, we have that \[\Ext_\Sigma^k(\Z[X],A)=\prod_{x\in \mathcal{R}_X}\Ext^k_{\Sigma}(\Z[\mathcal{O}(x)],A)\cong  \prod_{x\in \mathcal{R}_X} \Ext^k_{\operatorname{St(x)}}(\Z,A),\] where $\operatorname{St(x)}$ denotes the stabilizer of $x\in X$. Hence, \[\Ext^k(\Z[X],A)=\prod_{x\in \mathcal{R}_X} H^k(\operatorname{St}(x),A).\]
If we apply the functor $\Ext_{\Sigma}(-,A)$ to the exact sequence of $\Sigma$-modules\[ 0\to \Lambda_X\to \Z[X]\to \Z\to 0,\] we obtain the well known long exact sequence of relative cohomology

\begin{equation}\label{long-sequence relative}
    \xymatrix@C-1em{\cdots H^k(\Sigma,A)\ar[r] & \prod_{x \in \mathcal{R}_X} H^k(\operatorname{St}(x),A)\ar[r] & H^{k+1}_{\mathscr{A}}(\Sigma,X,A)\ar[r] & H^{k+1}(\Sigma,A) \cdots
    }
\end{equation}

\subsection{Hochschild Relative cohomology of groups}\label{Poincare}

Relative cohomology of groups was originally defined by Hochschild \cite{hochschild1956relative} and Adamson \cite{adamson1954cohomology}. We follow  the descrition given in \cite{Alperin-book}.\\

Let  $U$ be a $G$-module  and $S$ a subgroup of $G$. We say that $U$ is \emph{relatively $S$-projective} if it satisfies the  following equivalent properties (see  \cite[Proposition 1, page 65]{Alperin-book}):
\begin{itemize}
    \item[(i)] If $\psi:U \twoheadrightarrow V$ is a surjective  $G$-homomorphism and $\psi$ splits as $S$-homomorphisms then $\psi$ splits as $G$-homomorphism.
    \item[(ii)] If $\psi:V \twoheadrightarrow W$ is a surjective $G$-homomorphism and $\phi: U\to W$ is a $G$-homomorphism, then there is a $G$-homomorphism $\lambda:U\to V$ with $\psi \lambda=\phi$, provided that  there is an $S$-homomorphisms $\lambda_0:U\to V$ with the same property.
    \item[(iii)] $U$ is a direct summand of $U\downarrow_S \uparrow_G.$
\end{itemize}Here, $\downarrow_S$ means the restriction to $S$, and $ \uparrow_G$ the induction to $G$.

A complex
\begin{equation*}
\mathscr{R}:\cdots\stackrel{}{\to}R_3\stackrel{\delta_3}{\to} R_2\stackrel{\delta_2}{\to}R_1\stackrel{\delta_1}{\to}R_0\stackrel{\epsilon}{\to}M
\end{equation*}
of $G$-modules is called a \emph{relatively $S$-projective resolution} if:
\begin{enumerate}
\item each $G$-module $R_i$ is relatively $S$-projective,
\item the sequence has a contracting homotopy as $S$-modules.
\end{enumerate}
\begin{remark}
\begin{itemize}
    \item Since the canonical map  $M\downarrow_S \uparrow_G \to M$  splits as an $S$-homomorphism, if $\mathscr{T}$ is a projective $S$-resolution of $M\downarrow_S$, then $\mathscr{T}\uparrow_G$ is a relatively $S$-projective resolution of $M$.
    \item If $S$ is the trivial subgroup of $G$,  the relatively $S$-projective resolutions of  $G$-module  are the same as projective resolutions of $G$-modules.
\end{itemize}\end{remark}

\begin{definition}
Given a relatively $S$-projective resolution $\mathscr{R}$ of $M$, the $n^{th}$-relative $S$-cohomology group of $G$ is defined by
\[H^m(G,S;M)=H^n(\Hom_{G}(\mathscr{R},M),\delta^*).\]

\end{definition}
As expected, this definition does not depend on the chosen relatively $S$-projective resolution of $M$, (see e.g., \cite{hochschild1956relative}).
\begin{example}[Standard complex \cite{snapper1964cohomology}]\label{Example1}  Let  $X$ be a transitive left $G$-set.  Let $C_i=\mathbb{Z}X^{ (i+1) }$ be the free $\mathbb{Z}$-module generated by all $i+1$-tuples of elements of $X$. The group $G$ acts diagonally on $C_i$ and the sequence
\begin{equation*}
C^X_*:=\cdots\stackrel{}{\to}C_3\stackrel{\delta_3}{\to} C_2\stackrel{\delta_2}{\to}C_1\stackrel{\delta_1}{\to}C_0\stackrel{\epsilon}{\to}\mathbb{Z},
\end{equation*}
where
\begin{equation}
    \delta_i(x_1,\ldots,x_{r+1})=\sum_{j=1}^{r+1}(-1)^{j+1}(x_1,\ldots,\widehat{x_j},\ldots,x_{r+1})
\end{equation}
and $\epsilon(x)=1$ for all $x\in X$, is a complex of $G$-modules.

If $F$ denotes the stabilizer subgroup of $x_0\in X$, the complex $C^X_*$ is relatively $F$-projective resolution of $\mathbb{Z}$ called the \emph{standard complex of} $(G,X)$.

\end{example}

\begin{proposition}\label{prop:resultion relativa matched}
Let $\Sigma=F\bowtie G$ be a matched pair and $Q:=(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow}Q_i,\delta_i')$ be the normalized right bar resolution of the trivial $G$-module $\mathbb{Z}$. Then the group $\Sigma$ acts on $Q_i$ by
\begin{align}\label{formula:action}
     [s_n|\ldots|s_2|s_1]\cdot (x,s)&=[s_n\lhd\big ( (s_{n-1}\dots s_1)\rhd x\big )|\ldots|s_2\lhd(s_1\rhd x)|s_1\lhd x]s,
\end{align}
and $Q$ is a relatively $F$-projective resolution of right $\Sigma$-modules .
\end{proposition}
\begin{proof}
Since $\Sigma$ is a matched pair, the right $\Sigma$-set of cosets $F \backslash \Sigma$ can be identified with the set $G$ and $\Sigma$-action $s\cdot (f,g)=(s\lhd f)g$. This $\Sigma$-set will be denoted by $X$.
Applying the construction of Example \ref{Example1} to $X$, we obtain a relatively $F$-projective resolution $C:=C_i\stackrel{\epsilon}{\to} \Z$, since $F$ is the stabilizer of $e\in G$. The resolution $C$ coincides with the standard $G$-free resolution of $\mathbb{Z}$ as a trivial $G$-module. A $G$-basis of $C_i$ (called bar basis) is given by  \[[s_i|\cdots|s_2|s_1]=(s_i\cdots s_1,\cdots ,s_2s_1,s_1,e).\]  The action of $\Sigma$ in this basis is given by \eqref{formula:action}. The normalized bar resolution is a quotient of the bar resolution, and it is easy to see that this is also  relatively $F$-projective.
\end{proof}

\begin{remark}\label{Remark about projectively resolution}
\begin{itemize}
    \item In Proposition \ref{prop:resultion relativa matched}, we may consider the normalized \emph{left bar resolution} $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow}P_i,\delta_i')$ for the trivial $G$-module $\mathbb{Z}$:
with action of $\Sigma$ given by
\begin{align}\label{formula:action2}
     (x,s)\cdot[x_1|x_2|\cdots|x_n]=&x[s\rhd x_1|(s\lhd x_1)\rhd x_2|(s\lhd x_1x_2)\rhd x_3|\cdots\\
     &\cdots|(s\lhd(x_1\cdots x_{n-1})\rhd x_n)],\notag.
\end{align}
This is a relatively $G$-projective resolution.
\item The formulas \eqref{formula:action} and \eqref{formula:action2} appear in \cite{Masuoka-Calculations}.
\end{itemize}

\end{remark}

\begin{theorem}\label{thm:non-standard res}
Let $\Sigma=F\bowtie G$ be a matched pair. Let $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i,\delta_i)$  and  $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i')$  be  a relatively  $F$-projective  and a relatively  $G$-projective $\Sigma$-resolutions of $\Z$, respectively. Then the total complex of the tensor product double complex \begin{equation*}
P_{i}\otimes Q_{j}\mbox{ for }i,j\geq 0.\end{equation*}is a projective $\Sigma$-resolution of $\Z$.
\end{theorem}
\begin{proof}
Since $P_i$ is relatively $F$-projective, $P_i\downarrow_F\uparrow^\Sigma=P_i\oplus P'_i$ as $\Sigma$-modules. Analogously,  $Q_j\downarrow_G\uparrow^\Sigma=Q_i\oplus Q'_i$.

Using the fact that  $(F,G)$ is an exact factorization of $\Sigma$ and the Mackey's tensor product theorem (see e.g. \cite[Theorem 10.18]{Curtis-book}), we have that
\begin{align*}
(P_i\otimes Q_j)\downarrow_{\{e\}}\uparrow^\Sigma & \cong P_i\downarrow_F\uparrow^\Sigma\otimes Q_j\downarrow_G\uparrow^\Sigma,\\
&= (P_i\oplus P_i')\otimes (Q_j\oplus Q_j')\\
&= P_i\otimes Q_j \oplus \big (  P_i\otimes Q'_j \oplus P'_i\otimes Q_j \oplus  P_i'\otimes Q_j\oplus P_i'\otimes Q_j'\big ).
\end{align*}Hence $P_i\otimes Q_j$ is direct summand of the $\Sigma$-free module $(P_i\otimes Q_j)\downarrow_{\{e\}}\uparrow^\Sigma$, that is, $P_i\otimes Q_j$ is a projective $\Sigma$-module.

Finally, since each  $P_i$ and $Q_j$ is a free $\Z$-module,  all rows (and columns) of $P_*\otimes Q_*$ are exact, then it follows from the Acyclic Assembly Lemma (see e.g., \cite[Lemma 2.7.3]{Weibel}) that $\operatorname{Tot}(P_*\otimes Q_*)$ is acyclic.
\end{proof}
Theorem \ref{thm:non-standard res} generalizes the results of \cite{Masuoka2,MR1957830} about the construction of non-standard free resolutions  of $\Z$ associated to a matched pairs of groups. In fact, taking the relatively projective resolutions of Proposition \ref{prop:resultion relativa matched} and Remark \ref{Remark about projectively resolution} we can obtain the resolutions in \cite{Masuoka2,MR1957830}.
\subsection{Kac cohomology as relative group cohomology}

Using the bijective maps,
\begin{align*}
    \Sigma/G\to F,&&  (f,g)G\mapsto f, \\
    F\backslash\Sigma\to G, && F(f,g) \mapsto g
\end{align*}
 we can endow the set $F$ with a left $\Sigma$-action $(f,g)x=f(g\rhd x)$ and $G$ with a right $\Sigma$-action $s(f,g)=(s\lhd f)g$. From now on, $X$ will denote the left $\Sigma$-set defined as the disjoint union $F\sqcup G$, where $G$ is considered a left $\Sigma$-set using the inverse.
We denote by $\Lambda_X$ the kernel of the augmentation map \eqref{aumentation map}.
\begin{proposition}\label{ResDelta}
Let $\Sigma=F\bowtie G$ be a matched pair. Let $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i,\delta_i)$  and  $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i')$  be  a relatively  $F$-projective  and a relatively  $G$-projective $\Sigma$-resolutions of $\Z$, respectively.  Let $D^{*,*}$ be the truncated tensor product double complex
\begin{equation}\label{TruncateComplex}
D^{i,j}:=P_{i+1}\otimes Q_{j+1}\mbox{ for }i,j\geq 0.\end{equation}
Then, the total complex $(\Tot(D^{*,*}),d_i)$ completed with the map
\begin{equation*}
P_0\otimes Q_0\stackrel{-\delta_1\otimes\delta'_1}{\longleftarrow}\Tot(D^{*,*}),
\end{equation*}

is a projective $\Sigma$-resolution of $\Lambda_X$.
\end{proposition}

\begin{proof}
Let $\Lambda$ be the kernel of the map $\epsilon:P_0\oplus Q_0\to \mathbb{Z}$ defined by $\epsilon(x\oplus y)=\epsilon_P(x)+\epsilon_Q(y)$.
Let us consider the total complex $(\Tot(D^{*,*}),d_i)$ completed with the maps

\begin{equation}\label{eq:res_delta_proof}
0\stackrel{}{\leftarrow}\Lambda \stackrel{\theta}{\leftarrow}P_0\otimes Q_0\stackrel{-\delta_1\otimes\delta'_1}{\longleftarrow}\Tot(D^{*,*}),
\end{equation}where $\theta(p\otimes q)=-p\epsilon_Q(q)\oplus \epsilon_P(p)q$.
Using the same argument of Theorem \ref{thm:non-standard res},  we have that $\Tot(D^{*,*})$ is an acyclic complex of  projective $\Sigma$-modules.

Let us see that \eqref{eq:res_delta_proof} is exact in  $P_1\otimes Q_1$. The map $d_1:P_2\otimes Q_1\oplus P_1\otimes Q_2\to P_1\otimes Q_1$ in $\Tot(D^{*,*})$ is defined by $d_1(a\oplus b)=(\delta_2\otimes \id)(a)-(\id\otimes \delta_2')(b)$, where $a\in P_2\otimes Q_1$ and $b\in P_1\otimes Q_2$. Hence, the composed map is given by
\[(-\delta_1\otimes\delta_1')\circ d_1(a,b)=(-\delta_1\otimes\delta'_1)(\delta_2\otimes \id)(a)+(\delta_1\otimes\delta'_1)(\id\otimes \delta_2')(b)=0.\]
Thus, $\Img(d_1)\subseteq \Ker(-\delta_1\otimes\delta'_1)$. Now, since $P_i$ and $Q_i$ are free $\Z$-modules, then
\begin{align*}
    \Ker(-\delta_1\otimes\delta_1')&=\Ker(\delta_1)\otimes Q_1+P_1\otimes \Ker(-\delta_1')\\
    &=\Img(\delta_2)\otimes Q_1-P_1\otimes \Img(\delta_2').
\end{align*}
so $\Ker(-\delta_1\otimes\delta'_1)\subseteq \Img(d_1)$. To see the exactness in $P_0\otimes Q_0$, note that
\[\theta\circ(\delta_1\otimes\delta_1')(p\otimes q)=\delta_1(p)\epsilon_Q(\delta_1'(q))\oplus\epsilon_P(\delta_1(p))\delta_1(q)=0.\]
Hence, $\Img(\delta_1\otimes\delta_1')\subseteq \Ker(\theta)$.
To see that $\Ker(\theta)\subseteq\Img(\delta_1\otimes\delta_1')$, let us consider the tensor product of the two chain complexes $0\leftarrow\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i$ and $0\leftarrow\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i$. That is,

\begin{equation}\label{Diagram3}
\xymatrix{
  &\mathbb{Z}\otimes Q_1 \ar[d]_{-id\otimes\delta'_1} &P_0\otimes Q_1\ar[d]^{id\otimes\delta'_1} \ar[l]_{\epsilon_P\otimes id} &P_1\otimes Q_1\ar[l]_{\delta_1\otimes id}\ar[d]^{-id\otimes\delta'_1}  \\
    &\mathbb{Z}\otimes Q_0\ar[d]_{-id\otimes\epsilon_Q} &P_0\otimes Q_0\ar[l]^{\epsilon_P\otimes id}\ar[d]^{id\otimes\epsilon_Q}& P_1\otimes Q_0\ar[l]^{\delta_1\otimes id}\ar[d]^{-id\otimes\epsilon_Q}\\
 &\mathbb{Z} &P_0\otimes\mathbb{Z} \ar[l]^{\epsilon_P\otimes id} &P_1\otimes\mathbb{Z}\ar[l]^{\delta_1\otimes id}.\\
}
\end{equation}
whose total complex is given by the exact sequence

\begin{equation}\label{sequece3}0\leftarrow\mathbb{Z}\otimes\mathbb{Z}\stackrel{\epsilon}{\longleftarrow}P_0\otimes\mathbb{Z}\oplus \mathbb{Z}\otimes Q_0\stackrel{d_1}{\longleftarrow}P_1\otimes\mathbb{Z}\oplus P_0\otimes Q_0\oplus \mathbb{Z}\otimes Q_1\leftarrow\cdots\end{equation}

which can be written as
\begin{equation}\label{sequence4}
\mathbb{Z}
\stackrel{\epsilon}{\longleftarrow}P_0\oplus Q_0\stackrel{d_1}{\longleftarrow}P_1\oplus P_0\otimes Q_0\oplus  Q_1\leftarrow\cdots
\end{equation}
Note that $\theta$ is the restriction of $d_1$ to $P_0\otimes Q_0$. Suppose that $c\in \Ker(\theta)$. Then, $(\id\otimes\epsilon_Q)(c)=(\epsilon_P\otimes\id)(c)=0$. Let $b_2,b_3$ be the preimages of $c$ under the vertical and horizontal differentials in \eqref{Diagram3} respectively. There is a tuple $b=(b_1,b_2,-b_3,-b_4)$ such that $d_2(b)=0$. Since the total complex of \eqref{Diagram3} is acyclic, there is a tuple $a=(a_1,a_2,a_3,a_4,a_5)$ such that $d_3(a)=b$, and, it can be verified that $a_3\in P_1\otimes Q_1$ satisfies $\theta(a_3)=c$. Therefore, $\Ker(\theta)\subseteq\Img(\delta_1\otimes\delta_1')$.

To see that $\theta$ is surjective, note that $\epsilon=\epsilon_P+\epsilon_Q$ in \eqref{sequence4},  and \[d_1(a\oplus b\oplus c)=(\delta_1(a)+(id\otimes \epsilon_Q)(b))\oplus ((\epsilon_P\otimes id)(b) - \delta'_1(c)).\] Let $p_0\in d_1(P_1)=\Ker(\epsilon_P)$ and let $q\in Q_0$ such that $\epsilon_Q(q)=1$. Then, $$d_1(p_0\otimes q)=(-id\otimes\epsilon_Q+\epsilon_P\otimes id)(p_0\otimes q)=p_0,$$ so $d_1(P_1)\subseteq d_1(P_0\otimes Q_0)$. Similarly, $d_1(Q_1)\subseteq d_1(P_0\otimes Q_0)$. Hence, \[\Lambda=\Ker(\epsilon)=\Img(d_1)=\spann\{d_1(P_1)\cup d_1(P_0\otimes Q_0)\cup d_1(Q_1)\}=d_1(P_0\otimes Q_0).\]
Also, the map $d_1$ restricted to $P_0\otimes Q_0$ is given by $-\id\otimes\epsilon_Q\oplus\epsilon_P\otimes \id=\theta$, then $\Lambda=\Img(\theta)$ and therefore the sequence \eqref{eq:res_delta_proof} is exact.

Finally, we see that $\Lambda$ and $ \Lambda_X$ are isomorhpic as $\Sigma$-modules. Let us take $P_*'=C^G_*$ and $Q_*'=C^F_*$, the standard resolutions  as in Example \ref{Example1}, were $F$ and $G$ are consider as $\Sigma$-sets. In this case, \[C_0^G\oplus C_0^F= \Z[G]\oplus \Z[F]\cong \Z[G\sqcup F]=\Z[X],\]
and $\epsilon$ is the augmentation map \eqref{aumentation map}. Then, for this resolution $\Lambda=\Lambda_X$.

If  $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i,\delta_i)$ and $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i)$ are relatively projective resolutions, there exists homotopy equivalences $$s: P_*\to C^G_*, \quad l:Q_*\to C^F_*$$ This implies that $P_0\otimes Q_0\stackrel{-\delta_1\otimes\delta'_1}{\longleftarrow}\Tot(D^{*,*})$ is homotopically equivalent to
$P_0'\otimes Q_0'\stackrel{-\delta_1\otimes\delta'_1}{\longleftarrow}\Tot(D^{*,*})$.
Hence,
\begin{align*}
\Lambda_X \cong \operatorname{Coker}(P_1'\otimes Q_1'\to P_0'\otimes Q_0')\cong \operatorname{Coker}(P_1\otimes Q_1\to P_0\otimes Q_0)\cong \Lambda.
\end{align*}
\end{proof}

\begin{theorem}\label{OpextRelativeCohomology}
Let $k$ be a field and  $\Sigma= F\bowtie G$ a matched pair of groups. Then,
\[\Opext_{\rhd,\lhd}(kF,k^G)\cong H^3_\mathscr{A}(\Sigma,X;k^\times),\]
where $k^\times$ is considered as a trivial $\Sigma$-module.
\end{theorem}

\begin{proof}
Let $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i,\delta_i)$ and $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i´')$ be the resolutions in Proposition \ref{prop:resultion relativa matched}. These are the resolutions used in \cite{Masuoka2} to compute $\Opext_{\rhd,\lhd}(kF,k^G)$: they consider the truncated tensor product $D^{i,j}$ of the two resolutions to get \[\Opext_{\rhd,\lhd}(kF,k^G)\cong H^1(\Tot(\Hom_\Sigma(D^{i,j}))).\]
If $\Lambda$ is the kernel of the map $\epsilon:P_0\oplus Q_0\to \mathbb{Z}$ defined by $\epsilon(x\oplus y)=\epsilon_P(x)+\epsilon_Q(y)$, (here, $P_0:=\mathbb{Z}F=\mathbb{Z}X$ and $Q_0:=\mathbb{Z}Y$ for the $\Sigma$-sets $X=F$ and $Y=G$) then, by Theorem \ref{ResDelta}, the total complex of $E_{i,j}:=P_{i+1}\otimes Q_{j+1}\mbox{ for }i,j\geq 0$ completed in the following way
\begin{equation*}
0{\longleftarrow}\Lambda\stackrel{\theta}{\longleftarrow}P_0\otimes Q_0\stackrel{-\delta_1\otimes\delta'_1}{\longleftarrow}\Tot(D_{*,*}),
\end{equation*}
is a resolution of $\Lambda$, and $\Lambda=\Lambda_X$. Therefore, if we apply $\Hom_\Sigma(-,k^\times)$ to this total complex, we get the relative cohomology groups $H_\mathscr{A}^n(\Sigma,X;A)$. That is

\begin{equation}\label{RelativeTot}
    H^k(\Hom_\Sigma(\Tot(D^{*,*})))\cong H_\mathscr{A}^{k+2}(\Sigma,X;A).
\end{equation}
 In particular, since $\Hom_\Sigma(\Tot(D^{*,*}))\cong \Tot(\Hom_\Sigma(D^{*,*}))$, then,
\begin{equation}\label{OpextRelativeCohomologyEq}\Opext_{\rhd,\lhd}(kF,k^G)\cong H^1(\Tot(\Hom_\Sigma(D^{*,*})))\cong H^3_\mathscr{A}(\Sigma,X;k^\times).\end{equation}
\end{proof}

As a consequence of Theorem \ref{OpextRelativeCohomology} and the long exact sequence \eqref{long-sequence relative} we obtain the Kac's exact sequence (see \cite{Masuoka-Calculations,Kac}).
\begin{corollary}[Kac's exact sequence]\label{Kac exact sequence}

For a fixed matched pair of groups $(G,F,\rhd,\lhd)$, we have a long exact sequence
\end{corollary}
\begin{align*}
       0&\to H^1(F\bowtie G,k^\times)\to H^1(F,k^\times)\oplus H^1(G,k^\times)\to H^3_\mathscr{A}(\Sigma,X;k^\times)\\
       &\to H^2(F\bowtie G,k^\times)\to H^2(F,k^\times)\oplus H^2(G,k^\times)\to\Opext_{\rhd,\lhd}(kF,k^G)\\
       &\to H^3(F\bowtie G,k^\times)\to H^3(F,k^\times)\oplus H^3(G,k^\times)\to H^4_\mathscr{A}(\Sigma,X;k^\times).
   \end{align*}

\section{The five-term exact sequence for Kac double complex}\label{Section:Five-term exact sequence}

The group $\Opext_{\rhd,\lhd}(kF,k^G)$ can be obtained, as described in \cite{Masuoka-Calculations}, as the first cohomology group of a double complex, which can be computed by means of a spectral sequence. We compute the first pages of the spectral sequence associated to the double complex $D^{**}$ in $\eqref{TruncateComplex}$, which is a particular case of the double complex of Kac. The five-term exact sequence for this spectral sequence will be useful for computing the group $\Opext_{\rhd,\lhd}(kF,k^G)$ for different kinds of matched pairs.
\subsection{Spectral sequence of a double complex}
Through this section we deal with a first quadrant double complex, that is, a double complex $M^{p,q}$ such that $M^{p,q}=\{0\}$ when $p,q< 0$. There is a spectral sequence associated to a first quadrant double complex, whose first pages are obtained taking vertical and horizontal cohomology of the double complex.\\

Let $M^{*,*}$ be a first quadrant double complex with vertical and horizontal differentials given by $\delta_v,\delta_h$. Let $\Tot(M^{*,*})$ be the total complex associated to $M^{*,*}$.
 There is a spectral sequence $(E_r^{*,*},d_r)$ with differentials $d_r^{p,q}:E_r^{pq}\to E_r^{p+r,q-r+1}$, which converges to $H^*(\Tot(M^{*,*}))$, whose first pages are given by
\begin{align*}
E_0^{p,q}=M^{p,q},&&
E_1^{p,q}=H^q(M^{p,*},d_0),&&
E_2^{p,q}=H^p(E_1^{*,q},d_1),
\end{align*}see \cite{McCleary} for more details.
The differentials for each page are given by
\begin{align}\label{Differentials}
d_0^{p,q}=d_v,&&
d_1^{p,q}=d_h',&&
d_2^{p,q}(\bar{\alpha})=\overline{d_h(\gamma)},
\end{align}
where $d_h'$ is the differential induced by $d_h$ on $H^q(M^{p,*},d_0)$ and $\gamma\in M^{p+1,q-1}$ is such that $d_h(\alpha)=d_v(\gamma)$.
 Associated to the spectral sequence $(E_r^{*,*},d_r)$, there is a five-term exact sequence
\begin{equation}\label{FTES0}
0{\rightarrow}E_2^{1,0}\stackrel{i}{\to} H^1(\Tot(M^{*,*}))\stackrel{p}{\to}E_2^{0,1}\stackrel{d_2^{0,1}}{\to}E_2^{2,0}\stackrel{i}{\to}H^2(\Tot(M^{*,*})),
\end{equation}
where $i$ is a restriction map, the map $p$ is a projection map.

\subsection{The five-term exact sequence}
Given a matched pair $(G,F,\rhd,\lhd)$ we compute a five-term exact sequence to calculate $\Opext_{\rhd,\lhd}(kF,k^G)$.

\begin{theorem}\label{FirstSecondPage}
Let $\Sigma=F\bowtie G$ be a matched pair and $A$ be a $\Sigma$-module. The first and second pages of the spectral sequence associated to the double complex $\Hom_\Sigma(D^{*,*},A)$, where $D^{i,j}:=P_{i+1}\otimes Q_{j+1}\mbox{ for }i,j\geq 0,$ is the double complex defined in Proposition \ref{ResDelta}, are given by
\begin{align*}
    E_1^{i,n}=H^n(\Sigma,G;\Hom(P_i,A)), && E_2^{n,m}=H^m(H^n(\Sigma,G;\Hom(P_*,A))),
\end{align*}

The first page does not depend on the resolution $Q_*$ and the second page does not depend neither on the resolution $P_*$ nor on the resolution $Q_*$.
\end{theorem}

\begin{proof}
The double complex $\Hom_\Sigma(D^{*,*},A)$ with only the vertical differentials is the $0^{th}$-page of the spectral sequence:
\begin{equation}\label{HomSigma}E^{i,j}_0:=\Hom_{\Sigma}(P_i\otimes Q_j,A)\cong \Hom_{\Sigma}(Q_j,\Hom_\mathbb{Z}(P_i,A)).\end{equation}
Since $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i')$ is a relatively projective resolution of $\mathbb{Z}$, the first page of the spectral sequence is

\[E_1^{i,n}=H^n(\Sigma, G, \Hom(P_i,A))=F_n(P_i).\]
where $F_n:\Sigma$-$Mod$ to $Ab$ is the functor given by $H_1^n(\Sigma,G;\Hom(-,A))$. Hence it does not depend on the resolution $Q_i$. The diagram for the first page is given below.
\begin{equation*}
\xymatrix{
&\cdots  &\cdots &\cdots  \\
&F_2(P_0) \ar[r]^{F_2(\delta_1)}   &F_2(P_1)
\ar[r]^{F_2(\delta_2)}&F_2(P_2) \ar[r]^{F_2(\delta_3)}& \cdots\\
&F_1(P_0) \ar[r]^{F_1(\delta_1)}  &F_1(P_1) \ar[r]^{F_1(\delta_2)} &F_1(P_2) \ar[r]^{F_1(\delta_3)} &\cdots\\
&F_0(P_0)\ar[r]^{F_0(\delta_1)} &F_0(P_1) \ar[r]^{F_0(\delta_2)} &F_0(P_1) \ar[r]^{F_0(\delta_3)} &\cdots}
\end{equation*}
where $F_k(\delta_i)$ is the induced differential in relative cohomology. Then, the second page is \[E_2^{n,m}=H^m(F_n(P_*)).\]
To see that the objects in the second page do not depend on the resolution, let $P_i'$ be another relatively $F$-projective resolution of $\mathbb{Z}$. Then, there exists a homotopy equivalence $f:P_i\to P'_i$ as $F$-modules, that is, there exists $g:P'_i\to P_i$ and $h_i:P_i\to P_{i-1}$ such that
\[\delta_i h+h\delta_i=fg-\id.\]
Since the functor $F_n$ is additive, we get
\[F_n(\delta_i)F_n(h_i)+F_n(h_i)F_n(\delta_i)=F_n(f)F_n(f^{-1})-F_n(\id)\]
for each $n$, so the map $F_n(f)$ is a homotopy equivalence between the resolutions $F_n(P_i)$ and $F_n(P_i')$. This means that the second page, which consist on the cohomology groups of the resolutions $F_n(P_i)$ and $F_n(P'_i)$ with the respective induced differentials, is isomorphic to the first one.
\end{proof}

\begin{theorem}\label{FTESKac}
Let $(G,F,\rhd,\lhd)$ be a matched pair where the action $\rhd$ is trivial and let $k$ be a field. The spectral sequence in Theorem \ref{FirstSecondPage} associated to the group $\Sigma=F\bowtie G$ has second page given by

\begin{align*}
E_2^{p,q}=&H^{p+1}(F,H^{q+1}(G,k^\times)),&&
E_2^{p,0}\cong H^{p+1}(F,\widehat{G}) \label{SecondPage}\\
E_2^{0,q}\cong& \Der(F,H^{q+1}(G,k^\times)),&&
E_2^{0,0}=\Der(F,\widehat{G})\notag
\end{align*}
for $p\geq 1, q\geq 1$.

Therefore, we have the five-term exact sequence:
\begin{equation}\label{FTES}
\begin{aligned}
0{\rightarrow}H^2(F,\widehat{G}))\stackrel{i}{\to} & \Opext_{\rhd,\lhd}(kF,k^G)\stackrel{\pi}{\to}\Der(F,H^2(G,k^\times))\\
&\stackrel{d_2}{\to}H^3(F,\widehat{G}){\to}H^4_\mathscr{A}(\Sigma,X,k^\times).
\end{aligned}
\end{equation}
\end{theorem}
\begin{proof}
According to \eqref{HomSigma}, the $0^{th}$-page is given by

\begin{equation*}E^{i,j}_0:=\Hom_{\Sigma}(P_i\otimes Q_j,k^\times)\cong \Hom_{\Sigma}(Q_j,\Hom_\mathbb{Z}(P_i,k^\times)).\end{equation*}
If we take $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow} P_i,\delta_i)$ and $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow} Q_i,\delta_i´')$ to be the resolutions in Proposition \ref{prop:resultion relativa matched}, then we have the group isomorphism
\begin{equation*}\label{HomMap}
\Hom_{\Sigma}(P_i\otimes Q_j,k^\times)\cong \Map_+(G^{q+1}\times F^{p+1},k^\times)\end{equation*}
and the vertical and horizontal differentials of the double complex of groups
$$\Map_+(G^{q+1}\times F^{p+1},k^\times)$$ are respectively given by

\begin{equation*}
\begin{aligned}\label{VDifferentials}
\delta_i f(s_{i+1},\cdots,s_1;x_1,\cdots,x_{p})^{(-1)^p}=&f(s_{i+1},\cdots,s_2;s_1\rhd x_1,(s_1\lhd x_1)\rhd x_2,\cdots\\
& \cdots ,(s_1\lhd x_1\cdots x_{p-1})\rhd x_p)\times \\
&\prod_{k=1}^{i}f(s_{i+1},\cdots,s_{i+1}s_i,\cdots,s_1;x_1,\cdots,x_p)^{(-1)^k}\times \\
&f(s_i,\cdots,s_1;x_1,\cdots,x_p)^{(-1)^{q+1}}.
\end{aligned}
\end{equation*}
 \begin{equation*}
\begin{aligned}\label{HDifferentials}
\delta_i' f(s_q,\cdots,s_1;x_1,\cdots,x_{i+1})=&f(s_q\lhd(s_{q-1}\cdots s_1\rhd x_1),\cdots\\
&\cdots ,s_2\lhd(s_1\rhd x_1),s_1\lhd x_1;x_2,\cdots,x_{p+1})\times\\
&\prod_{k=1}^{i}f(s_q,\cdots,s_1;x_1,\cdots,x_ix_{i+1},\cdots,x_{i+1})^{(-1)^k}\times \\
&\  \  \  \  \ \ f(s_q,\cdots,s_1;x_1,\cdots,x_i)^{(-1)^{i+1}}.
\end{aligned}
\end{equation*}
Since the action $\lhd$ is trivial, the vertical  differentials are given by

\begin{equation*}\label{VDifferentials2}
\begin{aligned}
\delta (f)(s_{q+1},\cdots,s_1;x_1,\cdots,x_{p})^{(-1)^p}=&f(s_{q+1},\cdots,s_2;x_1,\cdots,x_p)\\&\prod_{i=1}^{q}f(s_{q+1},\cdots,s_{i+1}s_i,\cdots,s_1;x_1,\cdots,x_p)^{(-1)^i}\\
&f(s_q,\cdots,s_1;x_1,\cdots,x_p)^{(-1)^{q+1}}.
\end{aligned}
\end{equation*}
We have that $\Map_+(G^{q}\times F^{p},k^\times)\cong C^q(G,C^p(F,k^\times))$, where $C^q(G,C^p(F,k^\times))$ denotes the group of functions $f:G^q\to C^p(F,k^\times)$ (with a normalization property) and the group $G$ acts trivially on the group $C^p(F,k^\times)$.

Taking vertical cohomology to the $0$th-page, we get $H^q(G,C^p(F,k^\times))$. Therefore, the first page $E_1^{*,*}$ of the spectral sequence is given by
\begin{align*}E_1^{p,q}&=H^{q+1}(G,C^{p+1}(F,k^\times))\cong C^{p+1}(F,H^{q+1}(G,k^\times))\mbox{   for }q\geq 1.\\
E_1^{p,0}&=\Der(G,C^{p+1}(F,k^\times))\cong C^{p+1}(F,\Der(G,k^\times)),\notag\end{align*}
where the isomorphisms hold since the vertical differential leaves every element of $F$ fixed. On the other hand, the horizontal differentials are given by

\begin{equation*}\label{HDifferentials2}
\begin{aligned}
\delta'(f)(s_q,\cdots,s_1;x_1,\cdots,x_{p+1})=&f(s_q\lhd x_1,\cdots,s_1\lhd x_1;x_2,\cdots,x_{p+1})\\
&\prod_{i=1}^{p}f(s_q,\cdots,s_1;x_1,\cdots,x_ix_{i+1},\cdots,x_{p+1})^{(-1)^i}\\
&f(s_q,\cdots,s_1;x_1,\cdots,x_p)^{(-1)^{p+1}},
\end{aligned}
\end{equation*}
by differentiating each row by the induced horizontal differentials,

\begin{align}
E_2^{p,q}=&H^{p+1}(F,H^{q+1}(G,k^\times)),&&
E_2^{p,0}\cong H^{p+1}(F,\widehat{G})\\
E_2^{0,q}\cong& \Der(F,H^{q+1}(G,k^\times)),&&
E_2^{0,0}=\Der(F,\widehat{G})\notag
\end{align}
Since $k^\times$ is a trivial $G$-module, the sequence \eqref{FTES0} turns into
\begin{equation*}
\begin{aligned}
0{\rightarrow}H^2(F,\Der(G,k^\times))\stackrel{i}{\to} &H^1(\Tot(\Hom_{\Sigma}(P_i\otimes Q_j,k^\times)))\stackrel{p}{\to}\Der(F,H^2(G,k^\times))\\
&\stackrel{d_2^{0,1}}{\to}H^3(F,\Der(G,k^\times))\stackrel{i}{\to}H^2(\Tot(\Hom_{\Sigma}(P_i\otimes Q_j,k^\times))),
\end{aligned}
\end{equation*}

From \eqref{RelativeTot} and \eqref{OpextRelativeCohomologyEq} we get the five-term exact sequence

\begin{equation*}
\begin{aligned}
0{\rightarrow}H^2(F,\widehat{G})\stackrel{}{\to} & \Opext_{\rhd,\lhd}(kF,k^G){\to}\Der(F,H^2(G,k^\times))\\
&\stackrel{d_2^{0,1}}{\to}H^3(F,\widehat{G}){\to}H^4(\Sigma,X,k^\times).
\end{aligned}
\end{equation*}
\end{proof}
Note that, in the case that $\rhd$ is a trivial action, the terms $E_2^{p,q}$ with $p\geq 1,q\geq 1$ of the second page of the spectral sequence associated to the semidirect product $\Sigma=F\ltimes G$ coincides with the second page of the Lyndon-Hochschild-Serre spectral sequence \cite{Evens}.
\begin{corollary} Let $(G,F,\rhd,\lhd)$ be a matched pair with trivial $\rhd$ action and let $k$ be a field. Then,
\begin{enumerate}
\item If $H^2(G,k^\times)=1$ then   $\Opext_{\rhd,\lhd}(kF,k^G)\cong  H^2(F,\widehat{G}).$

\item If $(|F|,|\widehat{G}|)=1$, then $\Opext_{\rhd,\lhd}(kF,k^G)\cong\Der(F,H^2(G,k^\times)).$
\item If $G$ is a perfect group, then $    \Opext_{\rhd,\lhd}(kF,k^G)=\Der(F,H^2(G,k^\times))$
\item If $|F|=2k+1$ and $G=S_n$ with $n\geq 4$, then, $\Opext_{\rhd,\lhd}(\mathbb{C}F,\mathbb{C}^G)=\Hom(F,\mathbb{Z}/2)$.
\end{enumerate}
\end{corollary}
\begin{proof}
Part $(1)$ is straightforward.
$(2)$:  Since $(|F|,|\widehat{G}|)=1$, then $H^n(F,\widehat{G})=H^n(F,\widehat{G})=\{1\}$ and the result holds.\\

$(3)$: Since the abelianization of $G$ is trivial, then $\widehat{G}\cong\{0\}$. Therefore, $H^2(F,\widehat{G})=H^2(F,\widehat{G})=\{0\}$, similarly $H^3(F,\widehat{G})=0$, then, $\Opext_{\rhd,\lhd}(kF,k^G)=\Der(F,H^2(G,k^\times))$.\\

$(4)$: Under the given conditions it holds $(|F|,|\widehat{G}|)=1$, as in $(3)$. So $\Opext_{\rhd,\lhd}(kF,k^G)=\Der(F,H^2(G,k^\times))$. Now, $H^2(S_n,k^\times)= \mathbb{Z}/2$, for $n\geq 4$ and $\Der(F,H^2(G,k^\times))=\Hom(F,H^2(G,k^\times))$, so the result holds.
\end{proof}

\section{Computations}\label{Section:Computations}

We compute some examples of the group $\Opext_{\rhd,\lhd}(kF,k^G)$ for different semidirect products: the right action $\rhd$ is trivial, so we denote the group $\Opext_{\rhd\lhd}(kF,k^G)$ by $\Opext_\lhd(kF,k^G)$). The first calculation generalizes one from Masuoka in \cite{Masuoka2}.
\begin{theorem}
Let $k$ be a field. Let $G$ be a  group and  $\mathbb{Z}/2\ltimes(G\times G)$ be the the semidirect product with $(a,b)\lhd 1=(b,a).$
  Then
  $$\Opext_\lhd(k\mathbb{Z}/2,k^{G\times G})\cong H^2(G,k^\times )\oplus P(G,G;k^\times).$$
  \end{theorem}
    \begin{proof}

It follows from \eqref{CyclicCohomology} that  $H^n(\mathbb{Z}/2,H^1(G\times G,k^\times))=0$ for $n\geq 1$. Then, the sequence \eqref{FTES} implies that

\[\Opext_\lhd(k\mathbb{Z}/2,k^{G\times G})\cong \Der(\mathbb{Z}/2,H^2(G\times G,k^\times)).\]

According to \eqref{Karpilovski}, we have $H^2(G\times G, k^\times)= H^2(G,k^\times)\oplus P(G,G,k^\times)\oplus H^2(G,k^\times)$. If $F$ is a projective $G$-resolution of $\Z$, the chain map \begin{align*}
(F\otimes F)_n:= \bigoplus_{i+j=n} F_i\otimes F_j &\to \bigoplus_{i+j=n} F_i\otimes F_j\\
f_i\otimes f_j & \mapsto (-1)^{ij}f_j\otimes f_i,
\end{align*} defines a compatible action of $\mathbb{Z}/2$ on $F\otimes F$. Then, the action of $\mathbb{Z}/2$ on $H^2(
G\times G, k^\times)= H^2(G,k^\times)\oplus P(G,G;k^\times)\oplus H^2(G,k^\times)$ is given by \begin{align*}
    \tau: H^2(G\times G, k^\times)&\to H^2(G\times G, k^\times)\\
    (a,b,c) &\mapsto (c,-b,a).
\end{align*}
Since $\Der(\mathbb{Z}/2,H^2(G\times G,k^\times))=\{s\in H^2(G\times G, k^\times): a+\tau(a)=0\},$ we have that \[\Der(\mathbb{Z}/2,H^2(G\times G,k^\times))=\{(a,b,-a):a\in  H^2(G,k^\times), b\in P(G,G;k^\times) \}.\]
In conclusion, we have that $$\Opext_\lhd(k\mathbb{Z}/2,k^{G\times G})\cong H^2(G,k^\times )\oplus P(G,G;k^\times).$$
\end{proof}
 The following example includes the previous one in the case that $b=c=1$, $a=0$.
  \begin{theorem}
Let $\mathbb{Z}/2\ltimes (\mathbb{Z}/n\oplus \mathbb{Z}/n)$ be the semidirect product where the action of $\mathbb{Z}/2$ on $(\mathbb{Z}/n\oplus \mathbb{Z}/n)$ is defined by the matrix
\[ A=\left( \begin{array}{cc}
a & b\\
c & -a \end{array} \right)\]  with $\Det(A)=-1$. Let $k$ be a field such that $k^\times/(k^\times)^{2n}=0$ Then, $$\Opext_\rhd(k\mathbb{Z}/2,k^{\mathbb{Z}/n\oplus \mathbb{Z}/n})\cong \frac{\operatorname{Ker}(A-I)}{\operatorname{Im}(A+I)}
      \oplus \mu_n(k),$$
      where $\mu_n(k)$ is the group of $n^{th}$-roots of unity in $k$.
\end{theorem}
\begin{proof}
In this case the sequence \eqref{FTES} is given by
\begin{equation*}
\begin{aligned}
0{\rightarrow}&H^2(\mathbb{Z}/2,\Der(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\stackrel{i}{\to}  \Opext_\lhd (k\mathbb{Z}/2,k^{\mathbb{Z}/n\oplus \mathbb{Z}/n})\\&\stackrel{\pi}{\to}\Der(\mathbb{Z}/2,H^2(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\stackrel{d_2^{0,1}}{\to}H^3(\mathbb{Z}/2,\Der(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\\ &{\to}H^4(\Sigma,X,A),
\end{aligned}
\end{equation*}

We will see that
\begin{itemize}
    \item[(i)]  $H^2(\mathbb{Z}/2,H^1(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\cong \frac{\operatorname{Ker}(A-I)}{\operatorname{Im}(A+I)}$.
    \item[(ii)] $\Der(\mathbb{Z}/2,H^2(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\cong \mu_n(k)$,

    \item[(iii)] $d_2^{0,1}=0$,

\end{itemize}
Therefore, $\Opext_\lhd (k\mathbb{Z}/2,k^{\mathbb{Z}/n\oplus \mathbb{Z}/n})$ fits in a short exact sequence
\begin{equation}\label{ShortExactSequence}
0{\rightarrow}\frac{\operatorname{Ker}(A-I)}{\operatorname{Im}(A+I)}\stackrel{i}{\to}\Opext_\lhd (k\mathbb{Z}/2,k^{\mathbb{Z}/n\oplus \mathbb{Z}/n})\stackrel{\pi}{\to}\mu_n(k)\to 0,
\end{equation}
moreover, we will see  \eqref{ShortExactSequence} is split, so

$$\Opext_\lhd (k\mathbb{Z}/2,k^{\mathbb{Z}/n\oplus \mathbb{Z}/n})\cong \frac{\operatorname{Ker}(A-I)}{\operatorname{Im}(A+I)}\oplus \mu_n(k).$$

(i) It follows immediately from\eqref{CyclicCohomology}.

(ii) We identify $\wedge^2(\mathbb{Z}/n\oplus\mathbb{Z}/n)$ with the abelian group of alternating $2\times 2$-matrices over the ring $\mathbb{Z}/n\mathbb{Z}$. Therefore, $\wedge^2(\mathbb{Z}/n\oplus\mathbb{Z}/n)\cong \mathbb{Z}/n$ and  \[H^2(\mathbb{Z}/n\oplus\mathbb{Z}/n,k^\times)\cong \Hom(\wedge^2(\mathbb{Z}/n\oplus\mathbb{Z}/n),k^\times)\cong \mu_n(k),\]where $\mu_n(k)$ is the group of $n$-th roots  unit. Since  $A^TMA=-M$  for all $M\in \wedge^2(\mathbb{Z}/n\oplus\mathbb{Z}/n)$ we have that \[\Der(\mathbb{Z}/2,H^2(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\cong \mu_n(k).\]

(iii) To compute \[d_2^{0,1}:\mu_n(k)=\Der(\mathbb{Z}/2,H^2(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))\stackrel{}{\to}H^3(\mathbb{Z}/2,\Der(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times)),\] we follow \eqref{Differentials}. Given $\zeta\in \mu_n(k)=\Der(\mathbb{Z}/2,H^2(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))$, we need to find $\gamma\in \Map_+((\mathbb{Z}/n\oplus \mathbb{Z}/n)\times (\mathbb{Z}/2)^{2},k^\times)\cong C^2(\mathbb{Z}/2,C^1(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))$ such that  \begin{equation}\label{DeltaGamma}\delta_h(\alpha_\zeta)=\delta_v(\gamma),\end{equation}where
\begin{align}\label{alpha-zeta}
    \alpha_\zeta\in C^2(\Z/n\oplus \Z/n, k^\times), && \alpha_\zeta(x,y)=\zeta^{x_1 y_2}
\end{align}
and then compute the cohomology class of $\delta_h(\gamma)$ in $H^3(\mathbb{Z}/2,\Der(\mathbb{Z}/n\oplus \mathbb{Z}/n,k^\times))$.

We have that

\begin{align*}
\delta_h(\alpha_\zeta)(1,1)(x,y) &=\alpha_\zeta(Ax,Ay)\alpha_\zeta(x,y)\\&= \zeta^{x^T\begin{pmatrix}ac&bc\\bc&-ab\end{pmatrix}y}.
\end{align*}
This is a bicharacter with associated quadratic form  \[\omega(x,y)=\zeta^{-acx^2-2bcxy+aby^2}.\] Therefore, the cochain $\gamma\in C^2(\Z/2,C^1(\Z/n\oplus\Z/n,k^\times ))$  defined by
\begin{equation}\label{gamma}
\gamma(1,1)= \zeta^{-\frac{acx^2}{2}-bcxy+\frac{aby^2}{2}}, \quad  \quad (x,y)\in \Z/n\oplus \Z/n,
\end{equation}
and $\gamma(0,1)=\gamma(1,0)=\gamma(0,0)=1,$ satisfies \eqref{DeltaGamma}.

Finally, the horizontal differential of $\gamma$ is given by
\begin{align*}
\label{Dif3Gamma}\delta_h(\gamma)(1,1,1)&=\gamma(1,1)\gamma(1,0)(^1(\gamma(1,1))\gamma(0,1))^{-1}\\
&=\gamma(1,1)(^1(\gamma(1,1))^{-1}=1. \notag
\end{align*}
Hence, $d_2(\zeta)=1$.

A section of $\pi$ in the exact sequence \eqref{ShortExactSequence} is given by cohomology of $s(\zeta)=(\alpha_\zeta,\gamma_\zeta)$, where $\alpha_\zeta$ and $\gamma_\zeta$ are given by \eqref{alpha-zeta} and \eqref{gamma}, respectively.  It is clear from the definition that $s$ is a group homomorphism, that is, \eqref{ShortExactSequence} splits.
\end{proof}
\begin{theorem}\label{thm:odd groups}
Let $F$ be an arbitrary group acting on a finite abelian group $V$ with odd order. Suppose that  $(k^\times)^n=k^\times$, where $n$ is the exponent of $V$. Then
 \begin{align*}
   \Opext_\lhd (k F,k^{V})\cong  & H^2(F, \widehat{V})
      \oplus \Der(F,\wedge^2 \widehat{V}),
  \end{align*}where $\widehat{V}=\Hom(V,k^\times)$.
  \end{theorem}

 \begin{proof}
By \eqref{UnivCoef}, we have $H^2(V,k^\times)\cong \wedge^2\widehat{V}$. The sequence \eqref{FTES} is given by:\\
  \begin{equation*}
\begin{aligned}
0{\rightarrow}H^2(F,\widehat{V})\stackrel{i}{\to} & \Opext_\lhd(kF,k^V)\stackrel{\pi}{\to}\Der(F,\wedge^2\widehat{V})\\
&\stackrel{d_2}{\to}H^3(F,\widehat{V}){\to}H^4(\Sigma,X,A),
\end{aligned}
\end{equation*}
where $\Sigma= V \rtimes F$. We will see that $d_2=0$ and, since the resulting short exact sequence splits, we get the result.

Let $\alpha\in \Der(F,\wedge^2 \widehat{V})$, that is, $\alpha:F\to \wedge^2\widehat{V}$ such that $\alpha(gh)=\ ^g\alpha(h)\alpha(g)$. By \eqref{AltBilForms}, $\alpha$ can be identified with a map $\alpha:F\to H^2(V,k^\times)$ which can be lifted to a map $\tilde{\alpha}:F\to Z^2(V,k^\times)$ considering that, since $V$ has odd order, the map
  \begin{align*}
      \Alt: \wedge^2\widehat{V}&\to \wedge^2\widehat{V}
  \end{align*}
  given by $\Alt(\phi)(x,y)=\frac{\phi(x,y)}{\phi(y,x)}=\phi(x,y)^2$ is an isomorphism, so we can define the lifting map by $\tilde{\alpha}:F\to Z^2(V,k^\times)$ by
\begin{equation*}
  \tilde{\alpha}(g)=\alpha(g)^{\frac{1}{2}}.
  \end{equation*}
  In order to compute $d_2(\alpha)$, we must find a function $\gamma\in C^2(F,C^1(V,k^\times))$ such that
  \begin{align*}
    \delta(\gamma(g,h))&=\frac{^g\widetilde{\alpha}(h)\widetilde{\alpha}(g)}{\widetilde{\alpha}(gh)}\\
    &=\frac{\widetilde{^g\alpha}(h)\widetilde{\alpha}(g)}{\widetilde{\alpha}(gh)}\\
    &=\left(\frac{^g\alpha(h)\alpha(g)}{\alpha(gh)}\right)^\frac{1}{2}=1.
      \end{align*}
Hence $\gamma$ can be taken to be the constant cochain and, therefore, $d_2(\alpha)=1$ for all $\alpha\in \Der(F,\wedge^2\widehat{V})$.
\end{proof}

   \begin{corollary}
	Let $F=C_m=\langle \sigma \rangle $ be a cyclic group of order $m$ acting on a finite abelian group $V$ with odd order. Suppose that  $(k^\times)^n=k^\times$, where $n$ is the exponent of $V$. Then

 \begin{align*}
   \Opext_\lhd (k F,k^{V})\cong  & \{\psi \in \widehat{V}: \sigma \psi=\psi\}/\{N_{\sigma} \psi: \psi \in \widehat{V}\} \\ & \oplus \{b \in \wedge^2 \widehat{V}: N_{\sigma}  b=0 \},
  \end{align*}where $N_\sigma= 1+ \sigma +\cdots +\sigma^{m-1}$.
\end{corollary}

\subsection{An example with non-trivial differential $d_2$}
The next example illustrates the fact that the hypothesis in Theorem \ref{thm:odd groups} stating that the order of the order of $V$ must be odd, can not be avoided since otherwise the differential $d_2$ can be not trivial.
\begin{remark}\label{Rmk:matrices alternantes}
\begin{enumerate}[leftmargin=*,label=\rm{(\alph*)}]
\item Let $G$ be an elementary abelian $p$-group of rank $n$. Once a basis of $G$ is fixed, using the isomorphism \eqref{AltBilForms}  we can identify $H^2(G,\mathbb{C}^\times)$ with alternating matrices  over $\mathbb{Z}/p$ . A representative $2$-cocycle $\alpha_M\in H^2(G,\mathbb{C}^\times)$ corresponding to a matrix $M$ is defined by
\begin{equation}\label{2CocycleToAlternatingMatrix}
    \alpha_M(\textbf{x},\textbf{y})=\exp\Big ( \frac{2\pi i}{n} {\bf{x}^T} \widetilde{M} {\bf{y}}\Big ),
\end{equation}
where $\widetilde{M}$ is the upper triangular part of $M$.

\item Let $F=\langle t_1\rangle \oplus \langle t_2\rangle$ be a product of cyclic groups and let $M$ be an left $F$-module. If  $(\mathbb{Z}\stackrel{\epsilon}{\leftarrow}P_i)$ and $(\mathbb{Z}\stackrel{\epsilon}{\leftarrow}P_i')$ are periodic resolutions as in \eqref{CyclicResolution} for the groups $\langle t_1\rangle$ and $\langle t_2\rangle$ respectively, then the total complex $\Tot(P\otimes P')$  is a free $F$-resolution of $\mathbb{Z}$. Therefore, given a $F$-module $M$, we can compute $H^*(F,M)$ as the   cohomology of total complex of
\begin{equation*}\label{MDiagram}
\xymatrix{
&\vdots  &\vdots  &\vdots  \\
&M \ar[r]^{t_1-1}\ar[u]^{t_2-1}   &M
\ar[u]^{t_2-1}\ar[r]^{N_{t_1}}&M\ar[r]^{t_1-1}\ar[u]^{t_2-1}& \cdots\\
&M \ar[r]^{t_1-1} \ar[u]^{N_{t_2}} &M\ar[u]^{N_{t_2}} \ar[r]^{N_{t_1}} &M\ar[u]^{N_{t_2}} \ar[r]^{t_1-1} &\cdots\\
&M\ar[r]^{t_1-1}\ar[u]^{t_2-1} &M\ar[u]^{t_2-1} \ar[r]^{N_{t_1}} &M\ar[u]^{t_2-1} \ar[r]^{t_1-1} &\cdots}
\end{equation*}

Since we are mainly interested in $H^2(F,M)$, we  write the second and third differentials $\delta_2:M\oplus M\to M\oplus M\oplus M$, $\delta_3:M\oplus M\oplus M\to M\oplus M\oplus M\oplus M$ of the total complex are given by
\begin{align}
\delta_2(A,B)&=(A+\ ^{g_1}A,\ A-\ ^{g_2}A-(B-\ ^{g_1}B),B+\ ^{g_2}B),\label{eq: diferencial 2 prod cyclic}\\
\delta_3(A,B,C)&=(\ ^{g_1}A-A,\ ^{g_2}A-A+B+\ ^{g_1}B,B+\ ^{g_2}B+\ ^{g_1}C-C,\ ^{g_2}C-C).\label{eq: diferencial 3 prod cyclic}
\end{align}

\end{enumerate}

\end{remark}

\begin{lemma}\label{TwoGroups}
Let $F=\langle t_1,t_2\rangle$, $G=\langle s_1,\ldots,s_4\rangle$ be elementary abelian $2$-groups of rank 2 and 4, respectively. Consider the (right) action of $F$ on $G$ determined by the matrices

\begin{align*}
F_1=\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
0 & 1 & 0 & 1\end{array} \right),&& F_2=\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
1 & 1 & 0 & 1\end{array} \right).
\end{align*}
and the induced left action of $F$ on $H^2(G,\mathbb{C}^\times)$. Then,
\begin{enumerate}
\item The group $\Der(F,H^2(G,\mathbb{C}^\times))$ is in correspondence with the set of pairs of matrices
\begin{align*}
    A=\left( \begin{array}{cccc}
0 & 0 & b & c \\
0 & 0 & d & e\\
b & d & 0 & f\\
c & e & f & 0\end{array} \right),&&  B=\left( \begin{array}{cccc}
0 & 0 & b' & c' \\
0 & 0 & d' & e'\\
b' & d' & 0 & f'\\
c' & e' & f' & 0\end{array} \right),
\end{align*} with entries in $\mathbb{Z}/p$, such that

\begin{align}\label{EquationsExample}

    c'+b'+d'&=0\\
    c+d+e&=0\notag\\
    b+e+c'+d'+e&=0.\notag
\end{align}
\item The group $H^2(F,H^1(G,\mathbb{C}^\times))$ is isomorphic to  $\Z/2^2.$
\end{enumerate}
\end{lemma}
\begin{proof}
(1) \ By \eqref{eq: diferencial 2 prod cyclic}, elements in $\Der(F,H^2(G,\mathbb{C}^\times))$ are in correspondence with pairs $(A, B)$ of alternanting $4\times 4$-matrices such that

\begin{align}\label{equationsAB}
    A+F_1AF_1^T=0\notag\\
    B+F_2BF_2^T=0\\
    A- F_2AF_2^T-B+F_1BF_1^T=0.\notag
\end{align}
The system \eqref{equationsAB} is equivalent to \eqref{EquationsExample}.

$(2)$ In order to compute $H^2(F,H^1(G,\mathbb{C}^\times))$ we use the canonical identification $H^1(G,\mathbb{C}^\times)=\operatorname{Hom}(G,\mathbb{C}^\times)\cong G$ as left $F$-modules. By \eqref{eq: diferencial 3 prod cyclic}, we have that $\Ker(\delta_3)$ is in correspondence with $4\times 3$-matrices $S=[n_a,n_b,n_c]$ over $\mathbb{Z}/2$ such that
\begin{align*}\label{eq:sistema lineal Ker}
    (F_1-\operatorname{I})n_a&=0, &&
(F_2-\operatorname{I})n_c=0,
\\
(\operatorname{I}-F_1)n_c&=(F_2+\operatorname{I})n_{b},&&
(\operatorname{I}+F_1)n_{b}=(\operatorname{I}-F_2)n_a.\notag
\end{align*}
Thus, the space $\Ker(\delta_3)$ corresponds with all $4\times 3$-matrices over $\mathbb{F}_2$ such that $S_{ij}=0$ for $1\leq i\leq 2$, $1\leq j \leq 3$. On the other hand, by \eqref{eq: diferencial 2 prod cyclic} we have
\[\Img(\delta_2)= \{(l_a+F_1l_a,l_a-l_b+F_1l_b-F_2l_a,l_b+F_2l_b): l_a, l_b\in N\},\]
that is, $\Img(\delta_2)$ is in correspondence with all matrices of the form

\[\left(\begin{array}{ccc}
0 & 0 & 0  \\
0 & 0 & 0 \\
x_1+x_2 &  x_1+y_1+y_2 & y_1 \\
x_2 &  x_1+x_2+y_2 & y_1+y_2  \end{array}\right )\]where $x_i, y_i \in \mathbb{Z}/2$. Hence $H^2(F,H^1(G,\mathbb{C}^\times))\cong\Z/2^2$.
\end{proof}

\begin{lemma}\label{Example2}
 Let $\Sigma=F\ltimes G$ be a semidirect product and let
\begin{equation}\label{Resol}
\cdots\stackrel{}{\to}R_3\stackrel{}{\to} R_2\stackrel{}{\to}R_1\stackrel{}{\to}R_0,
\end{equation}
be a free resolution of a right $F$-modules $M$. The action of $F$ on $R_i$ can be extended to an action of $\Sigma$ by $r\cdot (f,g)=r\cdot f$. With this action, the sequence \eqref{Resol} turns out to be a relatively $G$-projective resolution of the right $\Sigma$-module $M$.
\end{lemma}
\qed

\begin{theorem}Let $F=\langle t_1,t_2\rangle$, $G=\langle s_1,\ldots,s_4\rangle$ be elementary abelian $2$-groups of rank 2 and 4, respectively. Consider the (right) action of $F$ on $G$ determined by the matrices

\begin{align*}
F_1=\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
1 & 1 & 1 & 0\\
0 & 1 & 0 & 1\end{array} \right),&& F_2=\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0\\
1 & 0 & 1 & 0\\
1 & 1 & 0 & 1\end{array} \right).
\end{align*}
Then  $\Opext_\lhd(\mathbb{C}F,\mathbb{C}^G)\cong (\Z/2)^3\oplus (\Z/4)^2$.

\end{theorem}
\begin{proof}
The five-term exact sequence \eqref{FTES} for this case is
\begin{equation*}
\begin{aligned}
0{\rightarrow}H^2(F,\Der(G,\mathbb{C}^\times))\stackrel{i}{\to} & \Opext_\lhd(\mathbb{C}F,\mathbb{C}^G)\stackrel{\pi}{\to}\Der(F,H^2(G,\mathbb{C}^\times))\\
&\stackrel{d_2}{\to}H^3(F,\Der(G,\mathbb{C}^\times)){\to}H^2(\Tot(M^{*,*})).
\end{aligned}
\end{equation*}
For this computation  the relatively $F$-projective resolution used in Theorem \ref{FirstSecondPage} $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow}Q_i,\delta_i')$ will be as in Proposition \ref{prop:resultion relativa matched}, and the relatevely $G$-projective resolution $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow}P_i,\delta_i)$ will be the total complex of the tensor product of the cyclic resolutions for $\langle t_1\rangle$ and $\langle t_2\rangle$, considered as a relatively $G$-projective $\Sigma$-resolution of $\Z$ using Lemma $\ref{Example2}$.

The $0^{th}$-page of the spectral sequence in Theorem \ref{FirstSecondPage} is given by
\[E_0^{i,j}=\Hom_{\Z\Sigma}(P_{i+1}\otimes Q_{j+1},\mathbb{C}^\times)=\Hom_{\Z\Sigma}\left(\bigoplus_{k=1}^{i+2}\Z F\otimes \Z G [G]^{j+1},\mathbb{C}^\times\right)\]

The horizontal and vertical differentials $d_h$ and $d_v$ are induced by the differentials of the resolutions $(\mathbb{Z}\stackrel{\epsilon_P}{\leftarrow}{P_i})$ and $(\mathbb{Z}\stackrel{\epsilon_Q}{\leftarrow}{Q_i})$ respectively.\\
Each $\Sigma$-module $\Z F\otimes \Z G[G]^{j+1}$ is free with basis $\{e\otimes[g_1|\cdots|g_{j+1}];e\neq g_1,\cdots g_{j+1}\in N\}$.
Therefore, an element in $E_0^{i,j}$ is defined by a tuple $(h_1,\ldots,h_{i+2})$ with $h_k\in C^{j+1}(F,\mathbb{C}^\times)$, where $h_k(f_1,\cdots,f_{j+1})=1$ if any of the entries is the identity of $F$.
Similarly, the differentials $d_0^{k,0}:\Hom_\Sigma(P^{k+1}\otimes Q^1,\mathbb{C}^\times)$ of the $0^{th}$-page, induced by the vertical differentials of the double complex are given by
\[d_0(f)(e\otimes[g_1|g_2])=f(e\otimes(g_1[g_2]-[g_1g_2]+[g_1])).\]
Since we are considering $\mathbb{C}^\times$ as a trivial $\Sigma$-module, the elements in $E_1^{k,0}=\Ker(d_0^{k,0})$ are in correspondence to tuples $(\chi_1,\ldots, \chi_{i+2})$ with $\chi_i \in \widehat{G}$.

First, we will compute $\Ker(d_2)$. By Lemma \ref{TwoGroups}, the group $E_2^{0,1}=\Der(F,H^2(G,\mathbb{C}^\times))$ is in correspondence with pairs $(A, B)$ of alternating $4\times 4$-matrices  satisfying the equations in \eqref{EquationsExample}.
According to Remark \ref{Rmk:matrices alternantes}(a), a representative element for $(A,B)$ in $ E^{0,1}_0=\Hom_\Sigma((\Z F\oplus \Z F)\otimes\Z G[G]^2,\mathbb{C}^\times)$  is defined by $\alpha =(\alpha_A,\alpha_B)$,
where $\alpha_{A}, \alpha_{B}$ are  $2$-cocycles defined in \eqref{2CocycleToAlternatingMatrix}.

By \eqref{Differentials}, we have  that $d_2(A,B)=\overline{d_h(\gamma)}$ were \[\gamma\in E^{1,0}_0= \Hom_\Sigma((\Z F\oplus \Z F\oplus \Z F)\otimes\Z G[G]),\] satisfies $d_h(\alpha_A,\alpha_B)=d_v(\gamma)$.

By  \eqref{eq: diferencial 2 prod cyclic} we have that
\begin{align}\label{eq:dAlphadBeta}
d_h(\alpha_A,\alpha_B)=(b_{M_1},b_{M_2},b_{M_3})\in E^{1,1}_0= \Hom_\Sigma((\Z F\oplus \Z F\oplus\Z F)\otimes\Z G[G]^2,\mathbb{C}^\times),
\end{align}
where $b_{M_i}(x,y)=(-1)^{x^T M_i y}$ and

\begin{align*}
M_1=\widetilde{A}+F_1\widetilde{A}F_1^T=\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 & b+d & d\\
0 & 0 & d & e\end{array} \right)\\\notag
M_2=\tilde{A}-F_2\tilde{A}F_2^T-(\tilde{B}-F_1\tilde{B}F_1^T)=\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 & b+b'+d' & b+d+d'\\
0 & & b+d+d'   & c+e+e'\end{array}\right)\\\notag
M_3=\widetilde{B}+F_2\widetilde{B}F_2^T= \left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\\
0 & 0 &b' & c'\\
0 & 0 & c' & c'+e'\end{array} \right)
. \end{align*}

The cochain $\gamma=(\gamma_{M_1},\gamma_{M_2},\gamma_{M_3})\in E^{1,0}_0= \Hom_\Sigma((\Z F \oplus \Z F\oplus \Z F)\otimes\Z G[G]^2,\mathbb{C}^\times)$ defined by

\begin{align}\label{GammaABC}
    \gamma_{M_1}(\vect{x})&= \exp\Big (-\frac{\pi}{2}\left((b+d)x_3^2+2dx_3x_4+ex_4^2\right)\Big ) ,\notag\\
     \gamma_{M_2}(\vect{x})&= \exp\Big (-\frac{\pi}{2}\left((b+b'+d')x_3^2+2(b+d+d')x_3x_4+(c+e+e')x_4^2\right)\Big ),\notag\\
       \gamma_{M_3}(\vect{x})&=\exp\Big( -\frac{\pi}{2}\left( b'x_3^2+2c'x_3x_4+(c'+e')x_4^2\right)\Big),
\end{align}
satisfies \eqref{eq:dAlphadBeta}. Therefore,

\begin{align*}
\delta_h(\gamma_{M_1},\gamma_{M_2},\gamma_{M_3}) &=
    \Big ( \frac{^{t_1}\gamma_{M_1}}{\gamma_{M_1}},\ \frac{^{t_2}\gamma_{M_1}\gamma_{M_2}(\ ^{t_1}\gamma_{M_2})}{\gamma_{M_1}},\frac{\gamma_{M_2}(\ ^{t_2}\gamma{M_2})\ ^{t_1}\gamma_{M_3}}{\gamma_{M_3}},\ \frac{^{t_2}\gamma_{M_3}}{\gamma_{M_3}}\Big )\\
    &=(1,\gamma_{M_2}^2,\gamma_{M_2}^2,1)\in \ker (d_1: E_1^{2,0} \to E_1^{3,0}).
\end{align*}

Since $G$ is an elementary abelian 2-group, we will use the canonical identification of $\widehat{G}$ with $G$. Under this identification we have that $\gamma_{M_2}^2=(0,0,b+b'+d',c+e+e')$.

The pair $(A,B)$ belongs to $\Ker(d_2)$ if and only if $(0,\gamma_{M_2}^2,\gamma_{M_2}^2,0)$ belongs to the image of $d_1:E_1^{1,0}\to E^{2,0}_1$, if and only if there exists $(\mu_A,\mu_B,\mu_C)\in E_1^{1,0}=F^{\times 3}$ such that
$d_h(\mu_A,\mu_B,\mu_C)=(0,\gamma_{M_2}^2,\gamma_{M_2}^2,0)$.
This means that
\begin{align*}
(F_1-I)\mu_A&=0\\
(F_2-I)\mu_C&=0\\
(F_2-I)\mu_A+(F_1+I)\mu_{B}&=\gamma_{M_2}^2\\
(F_1-I)\mu_C+(F_2+I)\mu_{B}&=\gamma_{M_2}^2.
\end{align*}
From this equation we obtain
$b+b'+d'=c+e+e'=0$. Joining these two equations with  \eqref{EquationsExample} we get a system of equation with $5$ free variables, hence $\Ker(d_2)=(\Z/2)^5.$

Hence we have the exact sequence
\begin{equation}\label{ShortSequence}
0\to H^2(F,G)\stackrel{}{\to}\Opext_\lhd(\mathbb{C}F,\mathbb{C}^G)\stackrel{\pi}{\to} \operatorname{Ker}(d_2)\to 0.\end{equation}

An element in $\Ker(d_2)$ is represented by a pair of matrices A,B as in \eqref{TwoGroups}. Let us assign $c'=1$ and consider the remaining variables zero, and let us call the respective pair of matrices $(A_c', B_c')$. A section of the sequence \eqref{ShortSequence} send $(A_c,B_c)$ to the class of the extension \[(\alpha_c,\gamma_c)\in H^1(\Hom_\Sigma(\Tot (D^{*,*},\mathbb{C}^*))\cong \Opext_\lhd(\mathbb{C}F,\mathbb{C}^G)\] where $\alpha_c$ is the $2$-cocycle associated to $(A_c',B_c')$ and $\gamma_c$ is given according to \ref{GammaABC}, by

\begin{align*}
    \gamma_{M_1}(\vect{x})&= \exp\Big (-\frac{\pi}{2}\left(x_3^2+x_4^2\right)\Big ) ,\notag\\
     \gamma_{M_2}(\vect{x})&= \exp\Big (-\frac{\pi}{2}\left((x_3^2\right)\Big ),\notag\\
       \gamma_{M_3}(\vect{x})&=\exp\Big( -\frac{\pi}{2}\left(x_4^2\right)\Big).
\end{align*}

It can be verified that the class of $(\alpha_c,\gamma_c)$ has order 4 in $\Opext_{\rhd,\lhd}(\mathbb{C}F,\mathbb{C}^G)$. In the same way, if we take the variable $d'$ to be $1$ and consider the remaining  variables null we get an element $(\alpha_d,\gamma_d)$ of order 4. Any other element outside the subgroup $\langle(\alpha_c,\gamma_c),(\alpha_d,\gamma_d)\rangle\cong (\Z/4)^2$ has order $2$, otherwise the order of $\Opext_{\rhd,\lhd}(\mathbb{C}F,\mathbb{C}^G)$ could not be $2^7$. That is why $\Opext_\lhd(\mathbb{C}F,\mathbb{C}^G)\cong (\Z/2)^3\oplus (\Z/4)^2$.

\end{proof}

config: arxiv


TITLE: Using Laurent series to show the residue at a pole
QUESTION [1 upvotes]: Use the Laurent series to show that the residue at a pole of the order $m$ can be evaluated by:
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z))$$
I have no clue how to approach this problem. First, it does look extremely related to Cauchy's Integral Theorem, or at least some generalization of it. However, I'm not exactly sure what I should be expanding and manipulating to show this.
If it's just straight forward plug in, then how do I simplify?
The definition I've been given is
$$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k+\sum_{k=1}^{\infty}a_{-k}(z-z_0)^{-k}$$
which I know can just be written as:
$$f(z)=\sum_{k=-\infty}^\infty a_k(z-z_0)^k$$
Using this directly we get
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^m\sum_{k=-\infty}^\infty a_k(z-z_0)^k)\implies$$
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\sum_{k=-\infty}^\infty\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}(a_k(z-z_0)^{k+m})$$
But I don't know how to proceed beyond this point...

REPLY [1 votes]: It's a basic result of complex analysis, here an approach.
Suppose that $f$ is analytic insde and on a simple closed curve $\gamma$, so indeed $f$ can be written by Laurent series as
$$f(z)=\sum_{k=-\infty}^{+\infty}a_{k}(z-z_{0})^{k}$$
But we know that $f$ has a pole $z_{0}$ of order $m$ (we need the pole inside $\gamma$), so $f$ can be written as $$f(z)=\sum_{k=0}^{+\infty}a_{k}(z-z_{0})^{k}+\sum_{k=1}^{m}a_{-k}(z-z_{0})^{-k}$$ $$\implies (z-z_{0})^{m}f(z)=a_{-m}+a_{-m+1}(z-z_{0})+\cdots+a_{-1}(z-z_{0})^{m-1}+a_{0}(z-z_{0})^{m}+\cdots$$ $$\implies \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=\lim_{z\to z_{0}}(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies  \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}$$ $$\implies a_{-1}:={\rm Res}(f,z_{0})= \lim_{z\to z_{0}} \frac{1}{(m-1)!} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)$$
Therefore,
$$\boxed{{\rm Res}(f,z_{0})= \frac{1}{(m-1)!} \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right) }$$


TITLE: Suppose $P, N$ are sub-modules of a module $M$. Then there exist natural isomorphisms..
QUESTION [0 upvotes]: Suppose $P, N$ are sub-modules of a module $M$. Then there exist natural isomorphisms $$(1) \ \ \frac{P+N}{P} \approx \frac{N}{N \cap P}  \ \  \text{ and } \ (2)\ \  \frac{P+N}{N} \approx \frac{P}{N \cap P}.$$

First we define a map $g : P+N \to \frac{N}{N \cap P}$ as $g(x+y) = y + (N \cap P)$ for all $x \in P$ and $y \in N$. One can easily check that $g$ is well-defined, surjective and a homomorphism.
Also I have found that $\ker(g) =  \frac{N}{N \cap P}$.
Then I am trying to apply the epimorphism theorem...
But I cannot conclude from here......Help Needed!
Thanks..

REPLY [1 votes]: Do this in another direction:
$$f:N\rightarrow (P+N)/P,n\mapsto n+P$$
Check this is a well-defined module homomorphism. Check this is a surjection. Check the kernel is $N\cap P$. Then apply your epimorphism theorem (I think you are referring to the first isomorphism theorem. The result you are trying to prove is also called the second isomorphism theorem.)


TITLE: Simple inequality, proof
QUESTION [0 upvotes]: There is a following inequality
$$\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}{n}\geq\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}+2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{n^{2}}.$$
In my opinion, it is held for all $x_i>0$. It can be rewritten to the following form
\begin{eqnarray*}
\frac{n-1}{n}\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}{n} & \geq & \frac{2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{n^{2}},\\
x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & \frac{2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{\left(n-1\right)}.
\end{eqnarray*}
However, how to proof the last inequlity? I used
\begin{eqnarray*}
x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & \frac{2}{n-1}\frac{n!}{\left(n-2\right)!2!}E(x_{i}x_{j}),\\
x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & nE(x_{i}x_{j})
\end{eqnarray*}
which may not be correct. However, there must be a more rigorous and simple proof.

REPLY [0 votes]: Expand
$$ \sum_{i<j} (x_i - x_j)^2.$$


TITLE: PDF & values of a random variable
QUESTION [0 upvotes]: Let $X: \Omega \rightarrow \mathbb{R}$ be a random variable and $f_X(x)$ its probability density function.
Is it true that $\{x : f_X(x) = 0\} \cap X(\Omega) = \emptyset$?
For one, if $X \sim \text{Uniform}[a,b]$, then it’s pretty intuitive that $X$ shouldn’t take values from $\mathbb{R} \setminus [a,b]$.

REPLY [1 votes]: No.  Let $X$ be a random variable uniformly distributed on the interval $[0, 1]$, and let $Y$ be a random variable defined by $Y = X+1$ if $X = 1/2$, and $Y = X$ otherwise.  The PDFs of $X$ and $Y$ are identical, but only $Y$ can take on a value ($3/2$) for which its PDF is equal to zero.


TITLE: Computing eigenfunction of expectational operator
QUESTION [4 upvotes]: Let $A$ be an operator from $\textbf{L}^2$ to itself defined by $$Af(x) = \mathbb{E}[e^{-W \gamma}\cdot f(x\cdot e^W)]$$ where $W$ is a $\mathbb{N}(0,1)$ distributed random variable and $\gamma \in [3,10]$ is a parameter.
I believe there is no analytical method to compute the principal eigenfunction and eigenvalue of this operator.
By principal eigenfunction, I mean the positive function $f(x)$ such that $$Af(x) = \kappa f(x)$$ $\forall x \in \mathbb{R}$.
What kind of numerical methods are used for problems of this sort?
Also, if anyone can recommend some textbook covering this kind of problems that would be amazing.
Thank you!
Edit: the source for this kind of problem is an Economics publication on long run risk exposure (Hansen and Scheinkman (2009))
Edit2: Uranix gave great tips to approach this specific problem but also made me realize that there is not much hope of solving this analytically in general (when the factor multiplying $f(x\cdot e^W)$ is a general function of $W$). Does anyone know which numerical techniques can be used in this context?

REPLY [1 votes]: This is not a answer, but just some ideas.
So basically the operator in question acts as
$$
(Af)(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-W\gamma} f(x e^W) e^{-W^2/2} dW
$$
Let $g(x) = (Af)(x)$ and let's split the $f$'s domain into three pieces:
$$
f(y) = \begin{cases}
f_{-}(y), &y < 0\\
f_0, & y = 0\\
f_{+}(y), &y > 0
\end{cases}
$$
One can easily see that $g_(x)$ is completely determined by $f_{-}(y)$, similarly $g_{+}(x)$ is determined by $f_{+}(y)$ and
$$
g_0 = f_0 e^{\gamma^2/2}.
$$
In other words, unless $f_0$ is zero, the eigenvalue should be equal to $e^{\gamma^2/2}$.
Assume that $f(y), \kappa$ is the principal eigenpair. It is necessary that
$$
(Af)(x) = \kappa f(x)
$$
for $x > 0$. Also it is sufficient, since we can easily extend $f(x)$ is the following way (not the only possible):
$$
f_{-}(y) = f_{+}(y), \qquad f_0 = 0.
$$
So, assume now that $x > 0$, $f: \mathbb R_{+} \to \mathbb R$ and
$$
\int_0^{\infty} f^2(y) dy < \infty.
$$
Introduce $s = x e^W > 0$. By changing variables,
$$
(Af)(x) = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty} \left(\frac{x}{s}\right)^\gamma f(s) \exp\left(-\frac{(\log s - \log x)^2}{2}\right) \frac{ds}{s}
$$
which can be treated as an integral transform with kernel
$$
K(s,x) = \frac{x^\gamma}{\sqrt{2\pi}s^{\gamma+1}} \exp\left(-\frac{(\log s - \log x)^2}{2}\right)\\
(Af)(x) = \int_0^\infty K(s, x) f(s) ds.
$$
This kernel is a homogenous function:
$$
K(\lambda s, \lambda x) = \frac1\lambda K(s, x),\qquad \lambda > 0.
$$
Hence,
$$
(Af)(\lambda x) = \int_0^\infty K(s, \lambda x) f(s) ds =
\int_0^{\infty} K(\lambda u, \lambda x) f(\lambda u)
\lambda du = \int_0^\infty K(u, x) f(\lambda u) du = (Ah)(x)\\
h(u) \equiv f(\lambda u).
$$
If $f(x)$ is an eigenfunction, so is $h(x) = f(\lambda x)$, and they both have the same eigenvalue. Due to linearity of the transform, any linear combination also will be an eigenfunction. So should be
$$
k(x) = \int_{0}^{\infty} p(\lambda) f(\lambda x) d\lambda
$$
for any integrable function $p(\lambda)$.
IDEA. Treat
$$
\int_{0}^{\infty} f(\lambda x) p(\lambda) d\lambda = k(x) \equiv 1
$$
as an integral equation for $p(\lambda)$. If it has a solution, then $k(x) \equiv 1$ is an eigenfunction of the $A$ with same as $f(x)$ eigenvalue.
From now I strongly suspect that the only possible eigenfunction is $f(y) = \operatorname{const}$ with the eigenvalue equal to $e^{\gamma^2/2}$.


TITLE: Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
QUESTION [2 upvotes]: Find
  $$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$

My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?

REPLY [3 votes]: \begin{eqnarray}
\mathcal L &=& \lim_{x\to0}\frac{1-\cos x \sqrt{\cos 2x}}{x\sin x}=\\
&=& \lim_{x\to0}\frac{1-\cos^2 x \cos 2x}{x\sin x (1+\cos x\sqrt{\cos 2x})}=\\
&=& \lim_{x\to0}\frac{1-\cos^2x(2\cos^2x-1)}{2x\sin x}=\\
&=& \lim_{x\to0}\frac{1-2\cos^4x+\cos^2x}{2x\sin x}=\\
&=&\lim_{x\to0}\frac{2\left(\frac12 + \cos^2 x\right)\left(1-\cos^2 x\right)}{2x\sin x}=\\
&=&\lim_{x\to0} \frac32 \frac{1-\cos^2 x}{x\sin x}=\frac32.
 \end{eqnarray}


TITLE: Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
QUESTION [6 upvotes]: Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?

REPLY [1 votes]: Clearly, $x\ne0$
$$\sqrt{x-1/x} + \sqrt{1 - 1/x} = x$$
$$\implies \sqrt{x^2-1}+\sqrt{x-1}=x\sqrt x$$
$$\implies \sqrt{x-1}(\sqrt{x+1}+1)=x\sqrt x$$
$$\implies \sqrt{x-1}\frac{x+1-x}{\sqrt{x+1}-1}=x\sqrt x\text{ (rationalizing the numerator) }$$
$$\implies \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}-\sqrt x$$
$$\implies \sqrt x+ \sqrt{x-1}=\sqrt x(\sqrt{x+1}-1)=\sqrt{x^2+x}$$
Squaring we get, $$x+x-1+2\sqrt{x^2-x}=x^2+x\iff 2\sqrt{x^2-x}=x^2-x+1$$
Putting $x^2-x=a^2,a^2=2a-1\implies (a-1)^2=0\implies a=1$


TITLE: Are there certain conditions that $a_n$ must meet in order for this series to converge?
QUESTION [10 upvotes]: Suppose we have an alternating series of the form
$$\sum_{n=1}^\infty \frac{(-1)^{a_n}}{n}$$
We know this converges for some basic cases, like when $a_n=n$ (the sum evaluating to $-\ln2$) or for interesting alternating patterns like
$$\sum_{n=1}^\infty \frac{(-1)^{\lfloor\frac{n}{2}\rfloor}}{n}=1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots=\frac{\pi}{4}-\frac{\ln2}{2}$$
I also recall seeing a question on here about an alternation resembling a factorial pattern, something like
$$1-\underbrace{\left(\frac{1}{2}+\frac{1}{3}\right)}_{2!\text{ terms}}+\underbrace{\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)}_{3!\text{ terms}}-\cdots$$
which may or may not have converged, I cannot seem to find it...
Is there a generalization that can be made about the conditions on $a_n$ for these types of series?

REPLY [3 votes]: I don't think you're going to find a characterization that's not just a disguised version of "the series converges if and only if the series converges"; convergence of this series is a ding an sich (sp?), seems to me.
But the series does converge for "most" choices! If $(\epsilon_n)$ is a sequence of plus or minus ones chosen "at random" then the series $\sum \epsilon_n/n$ converges almost surely.
(I was about to give a more precise statement of that, decided that would be silly; anyone with the background to understand the precise statement should be able to formulate it for himself. Regarding proving it, look in some probability book in the neighborhood of the Law of Large Numbers...)


TITLE: Showing a set is a subspace
QUESTION [0 upvotes]: Let $X$ denote the set of function $[0,1] \to \mathbb{R} $ and $X$ is a real linear space. Define $A$ and $B$ by:
$A = \{x \in  X \mid x(0) = 0\}$
$B = \{y \in X \mid y(1) = 0 \}$
Show $A$ and $B$ are subspaces of $X$.

To show that a set is a subspace, we have to show that the set is closed under multiplication and addition. However, i'm not really sure how we can do that with the information given. Advice?

REPLY [1 votes]: Define $\phi_k:X \to \mathbb{R}$ by $\phi_k(f) = f(k)$. Note that $\phi_k$ is linear.
Then $A=\ker \phi_0, B= \ker \phi_1$.

REPLY [1 votes]: To show a set is a sub vector space, one must show that the set contains the element 0, is closed under addition, and closed under scalar multiplication.
Let's look at set $A$.
Does A contain the $0$ element?  The answer is yes because the zero function is one where $x(0)=0$.
Is it closed under scalar multiplication?  The answer is yes.  For any function $x\in X$, $(\alpha x)(0)=0$.  So $(\alpha x)\in X$.
Is it closed under vector addition?  The answer is yes.  Let $x_1,x_2\in X$.  $(x_1+x_2)(0)=0$, so $(x_1+x_2)\in X$.
End of proof.


TITLE: Why do Capacitors discharge?
QUESTION [0 upvotes]: I understand how capacitors charge and i know they discharge but i am so confused why they discharge. How do they suddenly know when they are full to discharge. I am doing a school report and really need to be able to explain why rather than just saying they do.

REPLY [0 votes]: Capacitors discharge to send their charges to ground, or to get zero electric potential. Because all charges whether they are positive or negative have the tendency to flow the ground so as you provide them a path to flow to ground or zero electric potential they will flow to the zero potential to minimize their electric potential energy.
Same as water flows from a water tank at height (high potential energy) to ground to lose potential energy.


TITLE: Equivariance of vector bundles over G/B
QUESTION [4 upvotes]: Let $G$ be a complex semisimple group, $B$ a Borel subgroup of $G$ and $X=G/B$ the flag variety of $G$. If $G$ is simply connected, then every line bundle $L$ on $X$ can be made $G$-equivariant (see the proof of Theorem 1 here). Is this also true if $G$ isn't simply connected? By identifying $X$ with the flag variety of the universal covering group $\widetilde{G}$ of $G$, we can at least get a $\widetilde{G}$-action on $L$, but I haven't been able to figure out if this action factors through to yield a $G$-action.
More generally, can an arbitrary vector bundle $V$ on $X$ be made $G$-equivariant? If so, in how many ways?

REPLY [5 votes]: A vector bundle does not have to be $G$-equivariant. For instance, Horrock-Mumford bundle on $P^4$ is equivariant only with respect to Heisenberg group and not full $SL_5$. Pull it back to the flag variety of $SL_5$ to get a bundle of rank 2, not $SL_5$-equivariant. I am sure someone will tell you easier examples than that.


TITLE: Integral closure of $\mathbb{Q}[X]$ in $\mathbb{Q}(X)[Y]$
QUESTION [9 upvotes]: Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be the finite extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[Y]$, where $Y^2-X=0$.
Let $O_{K}$ be the integral closure of $\mathbb Q[X]$ in $K$. Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $Y$, hence
$$ O_{K}\supseteq \mathbb{Q}[X][Y]$$
My guess is that actually "=" holds. How can be proved this?

REPLY [5 votes]: Let's view ${\bf Q}\rm (x)[y]/(y-x^2)$ as ${\bf Q}\rm(x)[x^{1/2}]$. Set ${\bf Q}\rm(x)[x^{1/2}]\subseteq{\bf Q}\rm(x^{1/2})$ and observe
$$\rm \frac{1}{p(x)+x^{1/2}q(x)}=\frac{p(x)-x^{1/2}q(x)}{p(x)^2-x~\,q(x)^2}\in{\bf Q}(x)[x^{1/2}].$$
Since every $\rm f(x^{1/2})\in{\bf Q}[x^{1/2}]$ may be written as $\rm p(x)+x^{1/2}q(x)$, the above establishes that any rational function in $\rm x^{1/2}$ is in ${\bf Q}(x)[x^{1/2}]$, so in fact we may say that $\rm{\bf Q}(x)[x^{1/2}]={\bf Q}(x^{1/2})$.
Thus, we want to find the integral closure of ${\bf Q}[x]$ within ${\bf Q}(x)[y]\cong{\bf Q}(x^{1/2})$. Given any polynomial in the latter $\rm a:=p(x)+x^{1/2}q(x)$ you can find a polynomial in the variable $T$ with coefficients from ${\bf Q}[x]$ of which $\rm a$ is a root (make a quadratic from the roots $\rm p(x)\pm x^{1/2}q(x)$); this shows the first inclusion $\rm{\bf Q}[x^{1/2}]\subseteq {\cal O}_{{\bf Q}(x^{1/2})}$, now you want the reverse inclusion.
Suppose we have an $\rm a(x^{1/2})/b(x^{1/2})\in{\bf Q}(x^{1/2})\setminus {\bf Q}[x^{1/2}]$, and let $\rm \pi(x^{1/2})$ be an irreducible factor of the denominator $\rm b(x^{1/2})$ with valuation $\rm e$ (i.e. $\rm\pi(x^{1/2})^e\mid b(x)$ but $\rm\pi(x^{1/2})^{e+1}\nmid b(x)$.) Now let $\rm f(T)$ be a polynomial in $\rm {\bf Q}[x][T]$ such that $\rm f(a/b)=0$; clear denominators to show that we have something which is $\rm\ne0\bmod \pi(x^{1/2})$ equal to $0$, impossible.
See here for the (arguably easier-to-digest) number field analogue of this line of reasoning  (via clearing denominators to reach a congruence contradiction). As an aside, both number fields and function fields enjoy partial fraction decomposition - in the former case it is encoded in the Prufer factorization ${\bf Q}/{\bf Z}\cong\bigoplus{\bf Z}(p^\infty)$ (a special case of $p$-primary decomposition for abelian groups), where the Prufer $p$-groups are ${\bf Z}(p^\infty)\cong{\bf Q}_p/{\bf Z}_p\cong{\bf Z}[p^{-1}]$ (i.e. rationals with $p$-power denominators modulo the integers under addition).


TITLE: For each of the following values of ($a,b$), find the largest number that is not of the form $ax+by$ with $x\geq 0$ and $y \geq 0$.
QUESTION [0 upvotes]: For each of the following values of ($a,b$), find the largest number that is not of the form $ax+by$ with $x\geq 0$ and $y \geq 0$.
$(i) (a,b) = (3,7)$
$(ii) (a,b) = (5,7)$
$(iii) (a,b) = (4,11)$
I know how to compute numbers of these forms (clear), but have no idea how to generate one that is not of that form.  Then, of those, which is the largest?   More importantly, how do I prove its the largest?

REPLY [2 votes]: For the first one, for example, you can see that if $n$ is an integer, $n$, $n-7$ or $n-14$ is a multiple of 3 (by looking what happens modulo 3). So, if $n\geq 14$, you can write $n-7.x=3.y$ with $x=0, 1,$ or $2$ and $y\geq 0$. Then, the number you look for is between $1$ and $13$ and you can look one by one and take the largest.
For the other cases, you can try the same method, there will just be more cases to consider, I think (because a is larger).
I have an other idea, I write it in a few minutes.
(After verification, the idea is correct, but it is too much difficult to be interesting for this question).

REPLY [1 votes]: To help you, let us have a look on how it works with the first example.
With $x\geq0$ and $y\geq0$, what are the smallest numbers $3x+7y$ that you can generate ?

Clearly 0
The next smallest one is 3, using $x=1$ and $y=0$
you can't generate 4 or 5 so the next one is $6=3\times2+7\times0$
the next one is obviously 7
8 is not possible (check it out) so next one is $9=3\times3+7\times0$
$10=3\times1+7\times1$
11 is not possible (you should now see why)
$12=3\times4+7\times0$, $13=3\times2+7\times1$ and $14=3\times0+7\times2$ are 3 consecutive numbers, therefore every number from this point can be generated (do you see why ? and in particular do you understand why it is important to find 3 consecutive numbers one can generate ?).

The answer for point (i) is thus 11. You should manage with (ii) and (iii) easily.


TITLE: Example of non-abelian groups with these properties
QUESTION [0 upvotes]: I am looking for examples of non-abelian groups of arbitrarily large size with the following properties

Have order $p^a$, where $a$ is a positive integer and $p$ is prime.
Contain an abelian subgroup of order $p^{a-2}$.

I know one example which is the quaternion group. I am looking for more examples of groups of arbitrarily large size.

REPLY [1 votes]: Take, for example, the direct product of a nonabelian group of order $p^3$ with an abelian group of order $p^{a-3}$.


TITLE: Need help with problem of Invariance Principle
QUESTION [0 upvotes]: This problem is from Arthur Engel's Problem Solving Strategies.
Problem:-

Starting with a point S  (a,b) of the plane with 0 < b < a, we
generate a sequence of points $(x_n, y_n)$ according to the rule
$x_0 = a, y_0 = b, x_{n+1} = \frac {x_n + y_n}{2}, y_{n+1} =\frac{2x_ny_n}{x_n + y_n}$

Solution/Explanation:-

Here it is easy to find an invariant. From $x_{n+1}y_{n+1} = x_ny_n$, for all n
we deduce $x_ny_n = ab$ for all n. This is the invariant we are looking for. Initially, we have $y_0 < x_0$. This relation also remains invariant. Indeed, suppose $y_n < x_n$ for some n. Then $x_{n+1}$ is
the midpoint of the segment with endpoints $y_n,x_n$. Moreover,
$y_{n+1} < x_{n+1}$ since the harmonic mean is strictly less than the
arithmetic mean. Thus,
$0 < x_{n+1}−y_{n+1} =\frac{x_n−y_n}{x_n + y_n} ·\frac{x_n−y_n}{2} < \frac{x_n−y_n}{2}$ for all n.

Could anyone please help me understand how the paragraph concluded the equation in the end?

REPLY [0 votes]: Inductively, $x_n$ is decreasing and $y_n$ is increasing in $n$. Thus they both converge, and since the difference $x_n-y_n \to 0$, they have the same limit $L$. Since $x_n y_n=ab$ we get $L^2=ab$. In other words, iterating arithmetic and harmonic means converges to the geometric mean.
https://mathworld.wolfram.com/Arithmetic-HarmonicMean.html


TITLE: unique permutations
QUESTION [1 upvotes]: Let $X$ be a set of permutations with repetitions of numbers from $1$ to $n$
Let $Y \subseteq X$ be unique if for all $\sigma, \pi \in Y$, $1 \leqslant i < j \leqslant n$ the fact that $\pi(i) = \sigma(i)$ and $\pi(j) = \sigma(j)$ implies $\pi = \sigma$.
The question is what is the maximum number of elements in $Y$ and more interesting how we can get $Y$? What is the algorithm?

REPLY [1 votes]: It's easy to see that $Y$ has at most $n^2$ elements. If you have more than $n^2$ ordered $n$-tuples, then by the pigeonhole principle two of them share the same first two components.
Whether you can always achieve $n^2$ is not clear to me. I suspect I'm overlooking some simple construction. Anyway, for $n=2$ as you note we can achieve 4. For $n=3$ there are lots of ways of achieving 9: $$\{{123,132,213,231,312,321,111,222,333\}}$$ is one way, another is $$\{{112,121,211,223,232,322,331,313,133\}}$$ and another is $$\{{111,122,133,212,223,231,313,321,332\}}$$
EDIT: here's a solution for $n=4$: $$\matrix{1111&1234&1342&1423\cr2222&2143&2431&2314\cr3333&3412&3124&3241\cr4444&4321&4213&4132\cr}$$ I found this by using the field of 4 elements, $F=\{{0,1,\alpha,\beta\}}$, and taking the two-dimensional subspace of $F^4$ spanned by $(1,1,1,1)$ and $(0,1,\alpha,\beta)$, and then renaming the elements of $F$, 1, 2, 3, 4. This ought to work if $n$ is the order of a finite field, that is, if $n$ is a prime power. But maybe there's a simple construction I'm not seeing that works for all $n$.