\documentclass[12pt]{article} \usepackage{amsmath,amssymb,palatino,verbatim,a4wide,amsthm,euler} \usepackage[matrix,arrow]{xy} \newtheorem{definition}{Definition} \newtheorem{claim}[definition]{Claim} \newtheorem{lemma}[definition]{Lemma} \newtheorem{theorem}[definition]{Theorem} \newtheorem{observation}[definition]{Observation} \newtheorem{trouble}[definition]{Troubling Observation} \renewcommand{\qedsymbol}{\rule{0.5em}{0.5em}} \title{Cohomology Products in \textsf{GAP}, Explained in not Unbearable Detail, but Still Bad Enough to Require Being Seated while Reading} \author{Marcus Bishop} \newcounter{savedenumii} \begin{document} \maketitle The purpose of this document is to explain the implementation of cohomology products in the \textsf{crime} package for \textsf{GAP} including the Massey $n$-fold product. In this document, a composition of two functions $g\circ f$ is the function obtained by applying $f$ first and then $g$. The symbol $\circlearrowright$ is used in diagrams to indicate that a polygon either commutes or anticommutes. Let $G$ be a finite $p$-group for some prime $p$ and let $k=\mathbb{F}_p$. Also write $k$ for the trivial $kG$-module. We assume that we can calculate a $kG$-projective resolution $P_\ast$ of $k$, that is, for $n$ as large as we need, we can compute the integers $\left\{b_m:0\le m\le n\right\}$, the maps $\left\{\partial_m:1\le m\le n\right\}$ and the map $\epsilon$ such that \begin{equation}\label{proj} \xymatrix{P_n\ar[r]^{\partial_n} &P_{n-1}\ar[r]&\dots\ar[r] &P_1\ar[r]^{\partial_1} &P_0\ar[r]^{\epsilon}&k}\end{equation} is exact, where $P_m=\left(kG\right)^{\oplus b_m}$. Later, we will assume moreover that $P_\ast$ is {\em minimal}, that is, that $\partial_m\left(P_m\right)\le \mathrm{Rad}\left(P_{m-1}\right)$ for all $m\ge 1$. \section{Cohomology Products}\label{cp} The following construction is taken from \cite{carlson}. We begin with two cocycles $f:P_i\to k$ and $g:P_j\to k$, that is, that $f\circ\partial_{i+1}=g\circ\partial_{j+1}=0$. We want to compute the cup product $fg:P_{i+j}\to k$. We first convert $f$ into an chain map, resulting in the following commutative diagram. \begin{equation}\label{ff}\xymatrix{ P_m\ar[r]^{\partial_m}\ar[d]^{f_m} &P_{m-1}\ar[d]^{f_{m-1}}\ar[r]^{\partial_{m-1}} &P_{m-2}\ar[d]^{f_{m-2}}\ar[r] &\dots\ar[r] &P_{i+2}\ar[r]^{\partial_{i+2}}\ar[d]^{f_{i+2}} &P_{i+1}\ar[r]^{\partial_{i+1}}\ar[d]^{f_{i+1}} &P_i\ar[d]^{f_i}\ar[dr]^f \\ P_{m-i}\ar[r]_{\partial_{m-i}} &P_{m-i-1}\ar[r]\ar[r]_{\partial_{m-i-1}} &P_{m-i-2}\ar[r] &\dots\ar[r] &P_2\ar[r]_{\partial_2} &P_1\ar[r]_{\partial_1} &P_0\ar[r]_\epsilon&k\ar[r]&0 }\end{equation} \begin{enumerate} \item Define $f_i$ such that $\epsilon\circ f_i=f$. This is possible by projectivity of $P_i$. \item Define $f_{i+1}$ such that $\partial_1\circ f_{i+1} =f_i\circ\partial_{i+1}$. This is possible by projectivity of $P_{i+1}$ since \[\mathrm{im}\left(f_i\circ\partial_{i+1}\right) \le\mathrm{im}\left(\partial_1\right) =\ker\left(\epsilon\right)\] as $\epsilon\circ\left(f_i\circ\partial_{i+1}\right) =f\circ\partial_{i+1}=0$. \item Define $f_{i+2}$ such that $\partial_2\circ f_{i+2} =f_{i+1}\circ\partial_{i+2}$. This is possible by projectivity of $P_{i+2}$ since \[\mathrm{im}\left(f_{i+1}\circ\partial_{i+2}\right) \le\mathrm{im}\left(\partial_2\right) =\ker\left(\partial_1\right)\] as $\partial_1\circ\left(f_{i+1}\circ\partial_{i+2}\right) =f_i\circ\partial_{i+1}\circ\partial_{i+2}=0$. \item Define $f_m$ for $m>i+2$ by recursion such that $\partial_{m-i}\circ f_m =f_{m-1}\circ\partial_m$. This is possible by projectivity of $P_m$ since \[\mathrm{im}\left(f_{m-1}\circ\partial_m\right) \le\mathrm{im}\left(\partial_{m-i}\right) =\ker\left(\partial_{m-i-1}\right)\] as $\partial_{m-i-1}\circ\left(f_{m-1}\circ\partial_m\right) =f_{m-2}\circ\partial_{m-1}\circ\partial_m=0$. \end{enumerate} Then the product $fg$ is calculated as $g\circ f_{i+j}$. The process above is used to compute the multiplication table used by the \verb!CohomologyRing! command and is used to find generators by the \verb!CohomologyGenerators! command. \section{The Yoneda Cocomplex} My understanding of the purpose of the Yoneda Cocomplex is the following. The definition of the Massey product below requires a cocomplex having an associative product. The product defined above, however, is defined only for $f$ and $g$ cocycles in $\mathrm{Hom}\left(P_\ast,k\right)$. The Yoneda cocomplex $Y$, on the other hand, has the same cohomology as $\mathrm{Hom}\left(P_\ast,k\right)$, but has an associative product defined for all cochains, namely composition. Moreover, we will show that via the isomorphism $\Phi:\mathrm{H}^\ast\left(G,k\right)\to \mathrm{H}^\ast\left(Y\right)$, composition in $Y$ agrees with the product defined in Section \ref{cp} up to the factor $\left(-1\right)^{\deg f\deg g}$, that is, \[\Phi\left(fg\right) =\left(-1\right)^{\deg f\deg g}\Phi\left(g\right) \circ\Phi\left(f\right).\] The following construction comes from \cite{borge}. \begin{definition} For $i\ge 0$, define \[Y^i=\prod_{m\ge i}\mathrm{Hom}_{kG}\left(P_m,P_{m-i}\right).\] Then an element $f\in Y^i$ is a collection of $kG$-homomorphisms $\left\{f_m:P_m\to P_{m-i}:m\ge i\right\}$ as in the following diagram. \begin{equation}\label{f}\xymatrix{ P_n\ar[r]^{\partial_n}\ar[d]^{f_n} &P_{n-1}\ar[r]\ar[d]^{f_{n-1}} &\dots\ar[r] &P_m\ar[d]^{f_m}\ar[r]^{\partial_m} &P_{m-1}\ar[d]\ar[d]^{f_{m-1}}\ar[r]^{\partial_{m-1}} &P_{m-2}\ar[d]^{f_{m-2}}\ar[r] &\dots\ar[r] &P_{i+1}\ar[r]^{\partial_{i+1}}\ar[d]^{f_{i+1}} &P_i\ar[d]^{f_i} \\ P_{n-i}\ar[r]_{\partial_{n-i}} &P_{n-i-1}\ar[r] &\dots\ar[r]\ar[r] &P_{m-i}\ar[r]_{\partial_{m-i}} &P_{m-i-1}\ar[r]_{\partial_{m-i-1}} &P_{m-i-2}\ar[r] &\dots\ar[r] &P_1\ar[r]_{\partial_1} &P_0 }\end{equation} \end{definition} Diagram (\ref{f}) is not required to commute. \begin{definition} Define $\displaystyle{Y=\bigoplus_{i\ge 0}Y^i}$. $Y$ is called the {\em Yoneda cocomplex} of $P_\ast$. We write $\deg\left(f\right)=~i$ for $f\in Y^i$. Let $f=\left\{f_m:m\ge i\right\}\in Y^i$ and define \begin{align*} \partial:Y^i&\to Y^{i+1}\\ f&\mapsto\left\{f_{m-1}\circ\partial_m -\left(-1\right)^i\partial_{m-i}\circ f_m :m\ge 1\right\}. \end{align*} \end{definition} We observe that cocycles in $Y$ are those elements $f$ for which (\ref{f}) commutes if $\deg f$ is even and anticommutes if $\deg f$ is odd. \begin{lemma} $Y$ with differentiation $\partial$ is a cocomplex, that is, $\partial^2=0$. \end{lemma} \begin{proof} Let $f\in Y^i$. We will show that $\partial^2 f=0$ at the point $P_m$ in (\ref{f}) for $m\ge i+2=\deg\left(\partial^2 f\right)$. Follow along in the picture. \begin{eqnarray*} \left(\partial\left(\partial f\right)\right)_m &=&\left(\partial f\right)_{m-1}\circ\partial_m -\left(-1\right)^{i+1} \partial_{m-i-1}\circ \left(\partial f\right)_m\\ &=&\left(f_{m-2}\circ\partial_{m-1} -\left(-1\right)^i\partial_{m-i-1}\circ f_{m-1} \right) \circ\partial_m\\ &&-\left(-1\right)^{i+1} \partial_{m-i-1}\circ \left(f_{m-1}\circ\partial_m -\left(-1\right)^i\partial_{m-i}\circ f_m\right)\\ &=&f_{m-2}\circ\partial_{m-1}\circ\partial_m -\partial_{m-i-1}\circ\partial_{m-i}\circ f_m\\ &=&0 \end{eqnarray*} \end{proof} \begin{theorem} The cohomology groups of $Y$ are $\mathrm{H}^\ast\left(G,k\right)$. \end{theorem} \begin{proof} We will define a group isomorphism $\Phi:H^i\left(G,k\right)\to H^i\left(Y\right)$. \begin{enumerate} \item Let $f:P_i\to k$ be a cocycle in $\mathrm{Hom}_{kG}^i\left(P_\ast,k\right)$, that is, assume $f\circ\partial_{i+1}=0$. Define $\Phi\left(f\right)= \left\{f_m:m\ge i\right\}\in Y^i$ as follows. The element $\Phi\left(f\right)$, together with $f$, is pictured in the following diagram. \begin{equation}\label{pff}\xymatrix{ P_m\ar[r]^{\partial_m}\ar[d]^{f_m} &P_{m-1}\ar[d]^{f_{m-1}}\ar[r]^{\partial_{m-1}} &P_{m-2}\ar[d]^{f_{m-2}}\ar[r] &\dots\ar[r] &P_{i+2}\ar[r]^{\partial_{i+2}}\ar[d]^{f_{i+2}} &P_{i+1}\ar[r]^{\partial_{i+1}}\ar[d]^{f_{i+1}} &P_i\ar[d]^{f_i}\ar[dr]^f \\ P_{m-i}\ar[r]_{\partial_{m-i}} &P_{m-i-1}\ar[r]\ar[r]_{\partial_{m-i-1}} &P_{m-i-2}\ar[r] &\dots\ar[r] &P_2\ar[r]_{\partial_2} &P_1\ar[r]_{\partial_1} &P_0\ar[r]_\epsilon&k\ar[r]&0 }\end{equation} \begin{enumerate} \item\label{first} Define $f_i$ such that $\epsilon\circ f_i=f$. This is possible by projectivity of $P_i$. \item\label{firstsecond} Define $f_{i+1}$ such that $\partial_1\circ f_{i+1} =\left(-1\right)^i f_i\circ\partial_{i+1}$. This is possible by projectivity of $P_{i+1}$ since \[\mathrm{im}\left(\left(-1\right)^i f_i\circ\partial_{i+1}\right) \le\mathrm{im}\left(\partial_1\right) =\ker\left(\epsilon\right)\] as $\epsilon\circ\left(\left(-1\right)^i f_i\circ\partial_{i+1}\right) =\left(-1\right)^i f\circ\partial_{i+1}=0$. \item Define $f_{i+2}$ such that $\partial_2\circ f_{i+2} =\left(-1\right)^i f_{i+1}\circ\partial_{i+2}$. This is possible by projectivity of $P_{i+2}$ since \[\mathrm{im}\left(\left(-1\right)^i f_{i+1}\circ\partial_{i+2}\right) \le\mathrm{im}\left(\partial_2\right) =\ker\left(\partial_1\right)\] as $\partial_1\circ\left(\left(-1\right)^i f_{i+1}\circ\partial_{i+2}\right) =f_i\circ\partial_{i+1}\circ\partial_{i+2}=0$. \item\label{last} Define $f_m$ for $m>i+2$ by recursion such that $\partial_{m-i}\circ f_m =\left(-1\right)^i f_{m-1}\circ\partial_m$. This is possible by projectivity of $P_m$ since \[\mathrm{im}\left(\left(-1\right)^i f_{m-1}\circ\partial_m\right) \le\mathrm{im}\left(\partial_{m-i}\right) =\ker\left(\partial_{m-i-1}\right)\] as $\partial_{m-i-1}\circ\left(\left(-1\right)^i f_{m-1}\circ\partial_m\right) =f_{m-2}\circ\partial_{m-1}\circ\partial_m=0$. \setcounter{savedenumii}{\value{enumii}} \end{enumerate} This completes the definition of $\Phi$. The maps $\left\{f_m:m\ge i\right\}$ defined in Steps \ref{firstsecond}-\ref{last} above satisfy \[\partial_{m-i}\circ f_m=\left(-1\right)^i f_{m-1}\partial_m.\] In other words, $\left(\partial \Phi\left(f\right)\right)_{m+1}=0$ for all $m\ge i+1$ so that $\partial\Phi\left(f\right)=0$. Thus, $\Phi\left(f\right)$ is a cocycle by construction. \item\label{otherf} We claim than any other choice of maps $\left\{f'_m:m\ge i\right\}$ satisfying the conditions in \ref{first}-\ref{last} above will be equivalent to $\left\{f_m:m\ge i\right\}$ in $\mathrm{H}^i\left(Y\right)$. More precisely, if $f$ and $f'$ both satisfy conditions \ref{first}-\ref{last}, then will define a map $\theta\in Y^{i-1}$ such that $\partial\theta=f-f'$. Write $g_m=f_m-f'_m$ for $m\ge i$. \begin{equation} \begin{xy}{\xymatrix{ P_m\ar[r]^{\partial_m}\ar[d]_{g_m} &P_{m-1}\ar[r]^{\partial_{m-1}}\ar[d]_{g_{m-1}} \ar[dl]_{\theta_{m-1}} &P_{m-2}\ar[r]^{\partial_{m-2}}\ar[d]_{g_{m-2}} \ar[dl]_{\theta_{m-2}} &P_{m-3}\ar[r]\ar[d]_{g_{m-3}} \ar[dl]_{\theta_{m-3}} &\dots\ar[r] &P_{i+2}\ar[r]^{\partial_{i+2}}\ar[d]_{g_{i+2}} &P_{i+1}\ar[r]^{\partial_{i+1}}\ar[d]_{g_{i+1}} \ar[dl]_{\theta_{i+1}} &P_i\ar[d]_{g_i}\ar[dl]_{\theta_i}\ar[r]^{\partial_{i}} &P_{i-1}\ar[dl]_{\theta_{i-1}} \\ P_{m-i}\ar[r]_{\partial_{m-i}} &P_{m-i-1}\ar[r]_{\partial_{m-i-1}} &P_{m-i-2}\ar[r]\ar[r]_{\partial_{m-i-2}} &P_{m-i-3}\ar[r] &\dots\ar[r] &P_2\ar[r]_{\partial_2} &P_1\ar[r]_{\partial_1} &P_0\ar[r]_\epsilon &k }}; "2,8";"1,7"**{}; ?(.35)*{\circlearrowright} \end{xy} \end{equation} \begin{enumerate} \setcounter{enumii}{\value{savedenumii}} \item\label{secondfirst} Take $\theta_{i-1}=0$. \item\label{secondsecond} Since $\epsilon\circ f_i=\epsilon\circ f'_i=f$, we have $\mathrm{im}\left(g_i\right)\le \ker\left(\epsilon\right)=\mathrm{im}\left(\partial_1\right)$. Define $\theta_i$ such that $\partial_1\circ\theta_i=\left(-1\right)^i g_i$. This is possible by projectivity of $P_i$. We rewrite the condition on $\theta_i$ for future reference as follows. \begin{equation}\label{t1} \left(\partial\theta\right)_i= 0-\left(-1\right)^{i-1}\partial_1\circ\theta_i=g_i \end{equation} \item By (\ref{secondsecond}), we have \[\partial_1\circ\theta_i\circ\partial_{i+1} =\left(-1\right)^i g_i\circ\partial_{i+1} =\partial_1\circ g_{i+1} \] so that \[\mathrm{im}\left(g_{i+1}-\theta_i\circ\partial_{i+1}\right) \le\ker\left(\partial_1\right)=\mathrm{im}\left(\partial_2\right).\] Define $\theta_{i+1}$ such that \[\partial_2\circ\theta_{i+1} =\left(-1\right)^i \left(g_{i+1}-\theta_i\circ\partial_{i+1}\right),\] and again, for future reference, we rewrite this as follows. \begin{equation}\label{t2} \left(\partial\theta\right)_{i+1}=\theta_i\circ\partial_{i+1} -\left(-1\right)^{i-1}\partial_2\circ\theta_{i+1}=g_{i+1} \end{equation} \item Assume by recursion that we have computed $\theta_{m-2}$ and $\theta_{m-3}$ such that \[\partial_{m-i-1}\circ\theta_{m-2} =\left(-1\right)^i\left(g_{m-2}-\theta_{m-3}\circ\partial_{m-2}\right).\] Then $\partial_{m-i-1}\circ\theta_{m-2}\circ\partial_{m-1} =\left(-1\right)^i g_{m-2}\circ\partial_{m-1} =\partial_{m-i-1}\circ g_{m-1}$ so that \[\mathrm{im}\left(g_{m-1}-\theta_{m-2}\circ\partial_{m-1}\right) \le\ker\left(\partial_{m-i-1}\right)=\mathrm{im}\left(\partial_{m-i}\right).\] Define $\theta_{m-1}$ such that \[\partial_{m-i}\circ\theta_{m-1} =\left(-1\right)^i\left(g_{m-1}-\theta_{m-2}\circ\partial_{m-1}\right),\] and again, for future reference, we rewrite this as follows. \begin{equation}\label{tm} \left(\partial\theta\right)_{m-1}=\theta_{m-2}\circ\partial_{m-1} -\left(-1\right)^{i-1}\partial_{m-i}\circ\theta_{m-1}=g_{m-1} \end{equation} \setcounter{savedenumii}{\value{enumii}} \end{enumerate} This completes the definition of $\theta$. Then $\theta$ satisfies $\partial\theta=f-f'$ by (\ref{t1}), (\ref{t2}), and (\ref{tm}). \item\label{dg} Suppose now that $f=\partial g$ for some cochain $g:P_{i-1}\to k$. Write $\Phi\left(g\circ\partial_i\right)=\left\{ g_m:m\ge i\right\}$. We will construct $\theta$ such that $\Phi\left(\partial g\right)= \partial\theta$ for some $\theta\in Y^{i-1}$ as in the following diagram. \[\begin{xy}{\xymatrix@+5pt{ P_{m+1}\ar[r]^{\partial_{m+1}}\ar[d]_{g_{m+1}} &P_m\ar[r]^{\partial_m}\ar[d]_{g_m}\ar[dl]_{\theta_m} &P_{m-1}\ar[r]^{\partial_{m-1}} \ar[d]_{g_{m-1}}\ar[dl]_{\theta_{m-1}} &P_{m-2}\ar[r]\ar[d]_{g_{m-2}}\ar[dl]_{\theta_{m-2}} &\dots\ar[r] &P_{i+1}\ar[r]^{\partial_{i+1}}\ar[d]_{g_{i+1}} &P_i\ar[d]_{g_i}\ar[r]^{\partial_i}\ar[dl]_{\theta_i} &P_{i-1}\ar[d]_g\ar[dl]_{\theta_{i-1}} \\ P_{m-i+1}\ar[r]_{\partial_{m-i+1}} &P_{m-i}\ar[r]_{\partial_{m-i}} &P_{m-i-1}\ar[r]_{\partial_{m-i-1}} &P_{m-i-2}\ar[r] &\dots\ar[r] &P_1\ar[r]_{\partial_1} &P_0\ar[r]_{\epsilon} &k}}; "2,8";"1,7"**{}; ?(.35)*{\circlearrowright} \end{xy}\] \begin{enumerate} \setcounter{enumii}{\value{savedenumii}} \item Define $\theta_{i-1}$ such that $\epsilon\circ\theta_{i-1}=g$. This is possible by projectivity of $P_{i-1}$. \item Since $\epsilon\circ\theta_{i-1}\circ\partial_i =g\circ\partial_i=\epsilon\circ g_i$, we have that \[\mathrm{im}\left(g_i-\theta_{i-1}\circ\partial_i\right) \le\ker\left(\epsilon\right) =\mathrm{im}\left(\partial_1\right).\] Thus, by projectivity of $P_i$, we have $\theta_i$ such that \[\partial_1\circ\theta_i =\left(-1\right)^i\left(g_i-\theta_{i-1}\circ\partial_i\right).\] Then \[\left(\partial\theta\right)_i= \theta_{i-1}\circ\partial_i-\left(-1\right)^{i-1}\partial_1\circ\theta_i=g_i.\] \item Assume by recursion that we have computed the maps $\theta_{m-1}$ and $\theta_{m-2}$ such that \[\theta_{m-2}\circ\partial_{m-1} -\left(-1\right)^{i-1}\partial_{m-i}\circ\theta_{m-1} = g_{m-1}.\] Then \[\partial_{m-i}\circ g_m=\left(-1\right)^i g_{m-1}\circ\partial_m =\partial_{m-i}\circ\theta_{m-1}\circ\partial_m\] so that \[\mathrm{im}\left(g_m-\theta_{m-1}\circ\partial_m\right) \le\ker\left(\partial_{m-i}\right)=\mathrm{im} \left(\partial_{m-i+1}\right).\] Define $\theta_m$ such that \[\partial_{m-i-1}\circ\theta_m= \left(-1\right)^i\left(g_m-\theta_{m-1}\circ\partial_m\right).\] Then \[\left(\partial\theta\right)_m= \theta_{m-1}\circ\partial_m-\left(-1\right)^{i-1} \partial_{m-i+1}\circ\theta_m=g_m.\] \end{enumerate} This completes the definition of $\theta$. Then $g=\partial\theta$ by construction. \item\label{hom} We will now show that $\Phi$ is a $k$-module homomorphism. Let $f,g:P_i\to k$ be cocycles and let $\alpha,\beta\in k$. Write $h=\alpha f+\beta g$. We want to show that $\Phi\left(h\right)=\alpha\Phi\left(f\right)+\beta\Phi\left(g\right)$. But $\epsilon\circ h_0=\epsilon\circ\left(\alpha f_0+\beta g_0\right) =\alpha f+\beta g$, so that we are in the situation of Step \ref{otherf} above. Thus, $\Phi\left(h\right)$ and $\alpha\Phi\left(f\right)+\beta\Phi\left(g\right)$ are equivalent elements of $Y$. \item By Steps \ref{dg} and \ref{hom}, we have that if $f$ and $f'$ are equivalent in $H^\ast\left(G,k\right)$, then $\Phi\left(f\right)$ and $\Phi\left(f'\right)$ are equivalent in $H^\ast\left(Y\right)$. This together with \ref{otherf} shows that $\Phi$ is a well-defined $k$-module homomorphism. \item Finally, $\Phi$ is a bijection, having inverse given by \[\left\{f_m:m\ge i\right\}\mapsto\epsilon\circ f_i.\] \end{enumerate} \end{proof} \section{Products in $Y$} Consider the following product $Y^i\otimes Y^j\to Y^{i+j}$ on $Y$. Let $f\in Y^i$ and $g\in Y^j$ and consider the composition of the individual component maps of $f$ with those of $g$ such that legitimate compositions are obtained, as in the following diagram. \begin{equation}\label{comp}\xymatrix{ P_n\ar[r]^{\partial_n}\ar[d]^{f_n} &P_{n-1}\ar[d]^{f_{n-1}}\ar[r] &\dots\ar[r] &P_m\ar[r]^{\partial_m}\ar[d]^{f_m} &P_{m-1}\ar[r]\ar[d]^{f_{m-1}} &\dots\ar[r] &P_{i+j+1}\ar[r]^{\partial_{i+j+1}}\ar[d]^{f_{i+j+1}} &P_{i+j}\ar[d]^{f_{i+j}} \\ P_{n-i}\ar[r]^{\partial_{n-i}}\ar[d]^{g_{n-i}} &P_{n-i-1}\ar[r]\ar[d]^{g_{n-i-1}} &\dots\ar[r] &P_{m-i}\ar[r]^{\partial_{m-i}}\ar[d]^{g_{m-i}} &P_{m-i-1}\ar[r]\ar[d]^{g_{m-i-1}} &\dots\ar[r] &P_{j+1}\ar[r]^{\partial_{j+1}}\ar[d]^{g_{j+1}} &P_j\ar[d]^{g_j} \\ P_{n-i-j}\ar[r]^{\partial_{n-i-j}} &P_{n-i-j-1}\ar[r] &\dots\ar[r] &P_{m-i-j}\ar[r]^{\partial_{m-i-j}} &P_{m-i-j-1}\ar[r] &\dots\ar[r] &P_1\ar[r]^{\partial_1} &P_0 }\end{equation} Observe that we have thrown away the maps $\left\{f_m:i\le m\le i+j-1\right\}$. I suppose that the natural symbol for the object in (\ref{comp}) would be $g\circ f$, to emphasize the fact that we're talking about the component-wise composition of two elements of $Y$ and {\em not} a cohomology product. \begin{observation}\label{differential} $\displaystyle{\partial\left(g\circ f\right) =g\circ\partial f +\left(-1\right)^{\deg f}\partial g\circ f}$. \end{observation} \begin{proof} Write $i=\deg\left(f\right)$ and $j=\deg\left(g\right)$ as in (\ref{comp}). We will show the claim at the point $P_m$ in (\ref{comp}) for $m\ge i+j+1=\deg\left(\partial\left(g\circ f\right)\right)$. Follow along in the picture. \begin{eqnarray*} \left(g\circ\partial f+\left(-1\right)^i \partial g\circ f\right)_m &=& g_{m-i-1}\circ \left(f_{m-1}\circ\partial_m -\left(-1\right)^i\partial_{m-i}\circ f_m \right)\\ &&+\left(-1\right)^i\left( g_{m-i-1}\circ\partial_{m-i} -\left(-1\right)^j \partial_{m-i-j}\circ g_{m-i}\right) \circ f_m\\ &=&g_{m-i-1}\circ f_{m-1}\circ\partial_m -\left(-1\right)^{i+j} \partial_{m-i-j}\circ g_{m-i}\circ f_m\\ &=&\left(\partial\left(g\circ f\right) \right)_m \end{eqnarray*} \end{proof} \begin{claim} Composition in $Y$ induces via $\Phi$ an associative binary operation \[\mathrm{H}^i\left(G,k\right) \otimes\mathrm{H}^j\left(G,k\right) \to\mathrm{H}^{i+j}\left(G,k\right)\] making $\mathrm{H}^\ast\left(G,k\right)$ into a ring with 1. \end{claim} \section{Relationships among products on $\mathrm{H}^\ast\left(G,k\right)$} Let $f\in\mathrm{H}^i\left(G,k\right)$ and $g\in\mathrm{H}^j\left(G,k\right)$. Consider the following products on $\mathrm{H}^\ast\left(G,k\right)$. \begin{enumerate} \item The {\em cup product}\label{cup} $fg$ defined in Section \ref{cp} \item\label{yp} The product induced from composition in $Y$ \[\left(f,g\right)\stackrel{\Phi}{\mapsto} \big(\Phi\left(f\right),\Phi\left(g\right)\big) \stackrel{\circ}{\mapsto} \Phi\left(g\right)\circ\Phi\left(f\right) \stackrel{\Phi^{-1}}{\mapsto} \epsilon\circ\big(\Phi\left(g\right)\circ\Phi\left(f\right)\big)_{i+j} \] \item The {\em Massey 2-fold product} $\left\langle f,g\right\rangle$, defined more generally in Section \ref{mp} below, \[\left(f,g\right)\stackrel{\Phi}{\mapsto} \big(\Phi\left(f\right),\Phi\left(g\right)\big) \stackrel{\left\langle\cdot\right\rangle}{\mapsto} \left(-1\right)^i \Phi\left(g\right)\circ\Phi\left(f\right) \stackrel{\Phi^{-1}}{\mapsto} \left(-1\right)^i \epsilon\circ \big(\Phi\left(g\right)\circ\Phi\left(f\right)\big)_{i+j} \] \end{enumerate} The cup product is calculated as $g\circ f_{i+j}$, where $f_{i+j}$ is as in (\ref{ff}), whereas product \ref{yp} is calculated as $g\circ f_{i+j}$, where $f_{i+j}$ is as in (\ref{pff}). Comparing (\ref{ff}) and (\ref{pff}), we see that the two $f_i$'s are the same, the $f_{i+1}$'s differ by $\left(-1\right)^i$, the $f_{i+2}$'s differ by $\left(-1\right)^{2i}$, and in general, the $f_{i+m}$'s differ by $\left(-1\right)^{im}$. Thus, products \ref{cup} and \ref{yp} differ by $\left(-1\right)^{ij}$, that is, \begin{align*} &\Phi^{-1}\big(\Phi\left(g\right)\circ\Phi\left(f\right)\big)= \left(-1\right)^{ij} fg\\ \intertext{so that} &\Phi\left(fg\right)=\left(-1\right)^{ij}\Phi\left(g\right)\circ \Phi\left(f\right) \intertext{and therefore} &\Phi\left(fg\right)=\left(-1\right)^{i\left(j+1\right)} \left\langle f,g\right\rangle. \end{align*} We observe that product \ref{cup} is associative (see \cite{carlson}), and that product \ref{yp} is also associative, consisting of composition of functions. The Massey product, however, is not associative in general. \section{Massey Products}\label{mp} The idea of the Massey product is to extend the cohomology product to an $n$-fold product for $n\ge 2$. The following definition is adapted from \cite{kraines}. \begin{definition}\label{massey} For $k\ge 2$, let $f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}$ be cocycles in $Y$. The {\em Massey $k$-fold product} $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}\right\rangle$ is defined provided that for each pair $\left(i,j\right)$ with $1\le i<j\le k$ other than $\left(1,k\right)$, the lower-degree product $\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle$ is defined and vanishes as an element of $H^\ast\left(Y\right)$, that is, if for each qualifying $\left(i,j\right)$, there exists $u^{i,j}\in Y$ such that $\partial u^{i,j} =\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle$. In this situation, the value of $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}\right\rangle$ is defined to be \[\sum_{t=1}^{k-1} u^{t+1,k}\circ \overline{u^{1,t}} \] where the symbols $u^{1,1}$ and $u^{k,k}$ are taken to be $f^{\left(1\right)}$ and $f^{\left(k\right)}$ respectively and $\overline{u}=\left(-1\right)^{\deg\left(u\right)}u$. \end{definition} Observe that in the case $k=2$, the condition on $\left(i,j\right)$ is vacuously satisfied, so that $\left\langle f,g\right\rangle=g\circ\overline{f}$. Traditionally, one organizes the information in Definition \ref{massey} in an array, such as the following, \[\begin{array}{cccc} f^{\left(1\right)}&u^{1,2}&u^{1,3}\\ &f^{\left(2\right)}&u^{2,3}&u^{2,4}\\ &&f^{\left(3\right)}&u^{3,4}\\ &&&f^{\left(4\right)}\end{array}\] and traces the top row with one hand while tracing the rightmost column with the other hand as $t$ runs from 1 to 3. In this case, we have \[\left\langle f^{\left(1\right)},f^{\left(2\right)}, f^{\left(3\right)},f^{\left(4\right)}\right\rangle =u^{2,4}\circ \overline{f^{\left(1\right)}} + u^{3,4}\circ \overline{u^{1,2}} + f^{\left(4\right)}\circ \overline{u^{1,3}}.\] \begin{lemma}\label{cocycle} $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}\right\rangle$ is a cocycle in $Y$. \end{lemma} The reason for the sign appearing in Definition \ref{massey} becomes apparent is the following proof. \begin{proof} We begin by making a general observation about $Y$. Suppose $f\in Y^i$ and that $g=\partial\theta$ for some $\theta\in Y^{j-1}$ as in the following diagram. \[\xymatrix@+15pt{ &P_{i+j+m+1}\ar[r]^{\partial_{i+j+m+1}}\ar[d]^{f_{i+j+m+1}} &P_{i+j+m}\ar[d]^{f_{i+j+m}} \\ &P_{j+m+1}\ar[r]^{\partial_{j+m+1}} \ar[d]^{g_{j+m+1}}\ar[dl]^{\theta_{j+m+1}} &P_{j+m}\ar[d]^{g_{j+m}}\ar[dl]^{\theta_{j+m}} \\ P_{m+2}\ar[r]_{\partial_{m+2}} &P_{m+1}\ar[r]_{\partial_{m+1}} &P_m }\] Then by Observation \ref{differential}, we have \begin{eqnarray*} \left(g\circ f\right)_{i+j+m+1} &=&g_{j+m+1}\circ f_{i+j+m+1}\\ &=&\theta_{j+m}\circ\partial_{j+m+1}\circ f_{i+j+m+1} -\left(-1\right)^{j-1} \partial_{m+2}\circ\theta_{j+m+1}\circ f_{i+j+m+1}\\ &=&\theta_{j+m}\circ\partial_{j+m+1}\circ f_{i+j+m+1} -\left(-1\right)^{j-1} \partial_{m+2}\circ\theta_{j+m+1}\circ f_{i+j+m+1}\\ && -\left(-1\right)^i\theta_{j+m}\circ f_{i+j+m}\circ\partial_{i+j+m+1} +\left(-1\right)^i\theta_{j+m}\circ f_{i+j+m}\circ\partial_{i+j+m+1}\\ &=&-\left(-1\right)^i \left(\theta\circ\left(\partial f\right)\right)_{i+j+m+1} +\left(-1\right)^i \partial\left(\theta\circ f\right)_{i+j+m+1} \end{eqnarray*} so that as elements of $\mathrm{H}^\ast\left(Y\right)$, we have \begin{equation} \partial\theta\circ f =-\left(-1\right)^i\theta\circ\partial f.\end{equation} Now we compute the derivative of $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}\right\rangle$. \begin{eqnarray*} \partial\left(\sum_{t=1}^{k-1} \left(-1\right)^{\left(\deg u^{1,t}\right)} u^{t+1,k}\circ u^{1,t}\right) &=&\sum_{t=1}^{k-1}\left( \left(-1\right)^{\left(\deg u^{1,t}\right)} u^{t+1,k}\circ \partial u^{1,t} +\partial u^{t+1,k}\circ u^{1,t} \right)\\ &=&\sum_{t=1}^{k-1}\left( -\partial u^{t+1,k}\circ u^{1,t} +\partial u^{t+1,k}\circ u^{1,t} \right)\\ &=&0 \end{eqnarray*} \end{proof} \begin{observation}\label{degree} The condition $\partial u^{i,j} =\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle$ forces \begin{align*} \deg\left(u^{i,j}\right) &=\sum_{t=i}^j \deg\left(f^{\left(t\right)}\right) +i-j\\ \text{and}\qquad\deg\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle &=\sum_{t=i}^j \deg\left(f^{\left(t\right)}\right) +i-j+1. \end{align*} \end{observation} \begin{trouble} $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}\right\rangle$ is not uniquely defined, unless for each $\left(i,j\right)$ the condition $\partial u^{i,j}=\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle$ is satisfied by exactly one cochain $u^{i,j}$. \end{trouble} Suppose that we are given cocycles $f^{\left(1\right)}, f^{\left(2\right)},\dots,f^{\left(k\right)}$ and we want to compute the map $u^{i,j}$ for some $\left(i,j\right)$ with $1\le i<j\le k$ other than $\left(1,k\right)$. Assume that recursively, we have computed all of the maps in the following array. \[\begin{array}{ccccc} f^{\left(i\right)}&u^{i,i+1} &\dots&u^{i,j-1}&\\ &f^{\left(i+1\right)}&&u^{i+1,j-1} &u^{i+1,j}\\ &&&&\vdots\\ &&&f^{\left(j-1\right)}&u^{j-1,j}\\ &&&&f^{\left(j\right)} \end{array}\] The map $u^{i,j}$ will be such that \begin{equation}\label{ij} \partial u^{i,j}= \left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots, f^{\left(j\right)}\right\rangle =\sum_{t=i}^{j-1} u^{t+1,j}\circ \overline{u^{i,t}} \end{equation} where $u^{i,i}=f^{\left(i\right)}$ and $u^{j,j}=f^{\left(j\right)}$. Write $g$ for the map on the right-hand side of (\ref{ij}). Write \[d=\deg\left(g\right) =\sum_{t=i}^j\deg\left(f^{\left(t\right)}\right)+i-j+1.\] The relevant maps are all pictured below. \begin{equation}\label{masseypic}\tiny \begin{xy}{\xymatrix@+13pt{ P_m\ar[r]^{\partial_m}\ar[d]_{g_m} &P_{m-1}\ar[r]^{\partial_{m-1}}\ar[d]_{g_{m-1}}\ar[dl]_{u^{i,j}_{m-1}} &P_{m-2}\ar[r]^{\partial_{m-2}}\ar[d]_{g_{m-2}}\ar[dl]_{u^{i,j}_{m-2}} &P_{m-3}\ar[r]\ar[d]_{g_{m-3}}\ar[dl]_{u^{i,j}_{m-3}} &\dots\ar[r] &P_{d+2}\ar[r]^{\partial_{d+2}}\ar[d]_{g_{d+2}} &P_{d+1}\ar[r]^{\partial_{d+1}}\ar[d]_{g_{d+1}}\ar[dl]_{u^{i,j}_{d+1}} &P_d\ar[d]_{g_d}\ar[r]^{\partial_d}\ar[dl]_{u^{i,j}_d} &P_{d-1}\ar[dl]_{u^{i,j}_{d-1}} \\ P_{m-d}\ar[r]_{\partial_{m-d}} &P_{m-d-1}\ar[r]_{\partial_{m-d-1}} &P_{m-d-2}\ar[r]_{\partial_{m-d-2}} &P_{m-d-3}\ar[r] &\dots\ar[r] &P_2\ar[r]_{\partial_2} &P_1\ar[r]_{\partial_1} &P_0\ar[r]_\epsilon&k }}; "2,8";"1,7"**{}; ?(.35)*{\circlearrowright} \end{xy} \end{equation} We assume now that $P_\ast$ is minimal, that is, that $\partial_m\left(P_m\right)\le\mathrm{Rad}\left(P_{m-1}\right)$ for all $m\ge 1$. This implies that $\partial f=0$ for {\em any} cochain $f$, that is, we have $\partial_{i+1}\circ f=0$ for any $kG$-homomorphism $f:P_i\to k$. The map $u^{i,j}\in Y^{d-1}$ is constructed as follows. \begin{enumerate} \item\label{zero} We take $u^{i,j}_{d-1}=0$. \item The assumption that $\left\langle f^{\left(i\right)}, f^{\left(i+1\right)},\dots,f^{\left(j\right)}\right\rangle=g$ vanishes as an element of $\mathrm{H}^d\left(Y\right)$ tells us that $\epsilon\circ g_d$ vanishes as an element of $\mathrm{H}^d\left(G,k\right)$. But since $P_\ast$ is {\em minimal}, this means that $\epsilon\circ g_d$ is actually the zero map. Then by projectivity of $P_d$, there exists $u^{i,j}_d$ such that $\partial_1\circ u^{i,j}_d=\left(-1\right)^d g_d$. Observe that this means \[\left(\partial u^{i,j}\right)_d =0-\left(-1\right)^{d-1}\partial_1\circ u^{i,j}_d=g_d.\] \item The map $g$ is a cocycle by Lemma \ref{cocycle}. This means that the {\em rectangles} in (\ref{masseypic}) either commute or anticommute, depending on whether $d$ is even or odd. Thus, \[\partial_1\circ \left(g_{d+1}-u_d^{i,j}\circ\partial_{d+1}\right) =\partial_1\circ g_{d+1} -\left(-1\right)^d g_d\circ\partial_{d+1}=0\] so that \[\mathrm{im}\left(g_{d+1}- u_d^{i,j}\circ\partial_{d+1}\right)\le\ker\left(\partial_1\right) =\mathrm{im}\left(\partial_2\right).\] Thus, there exists $u^{i,j}_{d+1}$ such that \[\partial_2\circ u^{i,j}_{d+1}= \left(-1\right)^d\left(g_{d+1}- u_d^{i,j}\circ\partial_{d+1}\right).\] Observe that this means \[\left(\partial u^{i,j}\right)_{d+1} =u^{i,j}_d\circ\partial_{d+1}-\left(-1\right)^{d-1} \partial_2\circ u^{i,j}_{d+1}=g_{d+1}.\] \item Assume by recursion that we have constructed that maps $u^{i,j}_{m-2}$ and $u_{m-3}^{i,j}$ such that \[\partial_{m-d-1}\circ u^{i,j}_{m-2}= \left(-1\right)^d\left(g_{m-2} -u_{m-3}^{i,j}\circ\partial_{m-2}\right).\] Thus \[\partial_{m-d-1}\circ \left(g_{m-1}-u_{m-2}^{i,j}\circ\partial_{m-1}\right) =\partial_{m-d-1}\circ g_{m-1} -\left(-1\right)^d g_{m-2}\circ\partial_{m-1}=0\] so that \[\mathrm{im}\left(g_{m-1}- u_{m-2}^{i,j}\circ\partial_{m-1}\right)\le\ker\left(\partial_{m-d-1}\right) =\mathrm{im}\left(\partial_{m-d}\right).\] Thus, there exists $u^{i,j}_{m-1}$ such that \[\partial_{m-d}\circ u^{i,j}_{m-1}= \left(-1\right)^d\left(g_{m-1}- u_{m-2}^{i,j}\circ\partial_{m-1}\right).\] Observe that this means \[\left(\partial u^{i,j}\right)_{m-1}= u^{i,j}_{m-2}\circ\partial_{m-1}-\left(-1\right)^{d-1} \partial_{m-d}\circ u^{i,j}_{m-1}=g_{m-1}.\] \end{enumerate} This completes the construction of $u^{i,j}$. By construction, we have $\partial\left(u^{i,j}\right)=g$. Finally, observe that in the last step in the calculation of $\left\langle f^{\left(1\right)}, f^{\left(2\right)},\dots, f^{\left(k\right)}\right\rangle$, which is actually the {\em first} step, as this is a recursive process, it is only necessary to calculate $u^{1,k-1}$, but none of the maps $u^{1,m}$ for $2\le m\le k-2$, and none of the maps $u^{m,k}$ for $2\le m\le k-1$. In effect, the sum \[\sum_{t=1}^{k-1} u^{t+1,k}\circ \overline{u^{1,t}} =\sum_{t=1}^{k-2} u^{t+1,k}\circ \overline{u^{1,t}} +f^{\left(k\right)}\circ\overline{u^{1,k-1}} \] appearing in Definition \ref{massey} is calculated as \[ \sum_{t=1}^{k-2} \stackrel{!}{\boxed{u^{t+1,k} _{\deg u^{t+1,k}}}} \circ \overline{u^{1,t}_{\deg u^{t+1,k}+\deg u^{1,t}}} +f^{\left(k\right)}_{\deg f^{\left(k\right)}} \circ\overline{u^{1,k-1}_{\deg f^{\left(k\right)}+\deg u^{1,k-1}}}, \] But $u^{t+1,k}_{\deg u^{t+1,k}}=0$ by construction (see Step \ref{zero} above), so the sum reduces to a single term. This is not the case with the intermediate maps $u^{i,j}$ with $j-i\le k-2$. \bibliographystyle{plain} \bibliography{crimebib.xml} \end{document}