$\DeclareMathOperator{\Tot}{Tot}$I have the following problem. Let $H$ and $G$ be groups such that $H$ acts on $G$, i.e., there exists a group homomorphism $H\to \mathrm{Aut}(G)$ and let $M$ be an abelian group treated as a trivial $G$– and $H$-module.
Further on, there is a double complex $C^{p,q}:= C^p(H,C^q(G,M))$ (I use here bar resolutions). To this double complex there are two spectral sequences associated, one of them is Lyndon–Hochschild–Serre spectral sequence. In particular, I know that both spectral sequences converge to $H^\ast(G\rtimes H, M)$. On the other hand, by the general construction of both spectral sequences, we know that they converge to $H^\ast(\Tot(C^{\ast \ast}))$.
My question is the following. If I have a specific cohomology class in $H^\ast(\Tot(C^{\ast \ast}))$, how can I decide to which class in $H^\ast(G\rtimes H, M)$ it corresponds? In particular, an answer to any of the following questions would do the job (I think):
- Is there an explicit map $H^\ast(\Tot(C^{\ast \ast}))\to H^\ast(G\rtimes H, M)$ giving the isomorphism? Or is there a cochain level map?
- Given a class in $H^2(\Tot(C^{\ast \ast}))$, how can I construct a group extension $1\to M \to E\to G\rtimes H\to 1$?
I am mostly interested in cohomology classes of degree 2 (as indicated by the second question). Also, reference would be enough.
Best Answer
Let me change the notation a little: let $\phi \colon H \to \operatorname{Aut}(N)$ be a group homomorphism, and consider $G = N \rtimes_\phi H$ (we will drop the subscript $\phi$ in the rest of this post). As usual, write $^hn$ for $\phi(h)(n)$ if $h \in H$ and $n\in N$.
The point is that the double complex $C^{\bullet\bullet}$ actually comes from a natural bi-cosimplicial abelian group, as follows:
Definition. Define the bisimplical set $X_{\bullet\bullet}$ via $X_{p,q} = N^{q+1}\times H^{p+1}$, where a morphism $(f,g) \colon ([p],[q]) \to ([p'],[q'])$ in $\Delta \times \Delta$ acts as \begin{align*} X_{p',q'} &\to X_{p,q}\\ \big(n_0,\ldots,n_{q'},h_0,\ldots,h_{p'}\big) &\mapsto \big(n_{f(0)},\ldots,n_{f(q)},h_{g(0)},\ldots,h_{g(p)}\big). \end{align*} Endow each $X_{p,q}$ with a $G$-action via \begin{align*} G \times X_{p,q} &\mapsto X_{p,q}\\ \big((n,h),(n_0,\ldots,n_q,h_0,\ldots,h_p)\big) &\mapsto \big(n\ {^hn_0},\ldots,n\ {^hn_q},hh_0,\ldots,hh_p\big)\\ \big((n,h),\big(\underline n,\underline h\big)\big) &\mapsto \big(n\ ^h\underline n,h\underline h\big). \end{align*} This is visibly functorial in $(p,q) \in \Delta^{\text{op}} \times \Delta^{\text{op}}$, so defines a bisimplicial $G$-set. Thus we can view $\mathbf Z[X_{\bullet\bullet}]$ naturally as a bisimplicial $\mathbf Z[G]$-module, and I claim that $$\operatorname{Hom}_{\mathbf Z[G]}\big(\mathbf Z[X_{\bullet\bullet}],M\big) \cong C^{\bullet\bullet}\label{1}\tag{1}$$ as double complexes, for any $G$-module $M$ (not necessarily trivial). Indeed, note that \begin{align*} \operatorname{Hom}_{\mathbf Z[G]}\big(\mathbf Z[X_{p,q}],M\big) &\cong \operatorname{Map}_G\big(X_{p,q},M\big), \\ C^{p,q} &\cong \operatorname{Map}_H\big(H^{p+1},\operatorname{Map}_N(N^{q+1},M)\big). \end{align*} We have a bijection $\operatorname{Map}(X_{p,q},M) \stackrel\sim\to \operatorname{Map}(H^{p+1},\operatorname{Map}(N^{q+1},M))$, taking a function $\phi \colon X_{p,q} \to M$ to $\psi \colon H^{p+1} \to \operatorname{Map}(N^{q+1},M)$ given by $$\psi\big(h_0,\ldots,h_p\big)\big(n_0,\ldots,n_q\big) = \phi\big(n_0,\ldots,n_q,h_0,\ldots,h_p\big).$$ Now $\psi$ has image in $\operatorname{Map}_N(N^{q+1},M)$ if and only if $$\phi\big(n\underline n,\underline h\big) = n\phi\big(\underline n,\underline h\big) \qquad \text{for all } \big(\underline n, \underline h\big) \in X_{p,q}, \text{ all } n \in N.\label{2}\tag{2}$$ The $G$-action on $\operatorname{Map}_N(N^{q+1},M)$ is given by $(gf)(\underline n) = gf(g^{-1} \cdot \underline n)$, which is trivial on $N$ by definition of $\operatorname{Map}_N$. As action of $H$, it can be written explicitly by $(hf)(\underline n) = hf(^{h^{-1}}\!\underline n)$. This means that $\psi \colon H^{p+1} \to \operatorname{Map}_N(N^{q+1},M)$ is $H$-equivariant if and only if $$\phi\big(\underline n,h\underline h\big) = h\phi\big({^{h^{-1}}}\!\underline n,\underline h\big) \qquad \text{for all } \big(\underline n,\underline h\big) \in X_{p,q}, \text{ all } h \in H.\label{3}\tag{3}$$ Combining \eqref{2} and \eqref{3} shows that $\psi \in C^{p,q}$ if and only if $\phi(g \cdot (\underline n,\underline h)) = g \phi(\underline n,\underline h)$ for all $(\underline n,\underline h) \in X_{p,q}$ and all $g$ in either $N$ or $H$. But this means exactly that $\phi \colon X_{p,q} \to M$ is $G$-equivariant, so $$\operatorname{Map}_G\big(X_{p,q},M\big) \stackrel\sim\to \operatorname{Map}_H\big(H^{p+1},\operatorname{Map}_N\big(N^{q+1},M\big)\big).$$ This identification is natural in $p$ and $q$, giving the isomorphism \eqref{1}. This also shows that $C^{\bullet\bullet}$ is naturally a bi-cosimplicial abelian group.
In particular, the Eilenberg–Zilber theorem (nlab) shows that the totalisation of $C^{\bullet\bullet}$ is naturally chain homotopic (!) to the diagonal $dC^\bullet$ (i.e. $dC^n = C^{n,n}$). But the diagonal of $X_{p,q}$ is just the bar resolution of $G$, so we conclude that the inclusion $dX_\bullet \hookrightarrow X_{\bullet\bullet}$ induces a chain homotopy equivalence $$\operatorname{Tot}C^{\bullet\bullet}\simeq C^\bullet(G,M).$$ Explicit descriptions of the map follow from sufficiently effective proofs of Eilenberg–Zilber. See for instance 8.5.4 in Weibel's An introduction to homological algebra. (For your direction, you need to use the Alexander–Whitney map on $\mathbf Z[X_{\bullet\bullet}]$, because the direction flips when applying $\operatorname{Hom}_{\mathbf Z[G]}(-,M)$.)
Concretely, the map $\operatorname{Tot}(C^{\bullet\bullet}) \to C^\bullet(G,M)$ takes a $(p,q)$-cocycle $f \colon X_{p,q} \to M$ to the $(r,r)$-cocycle $X_{r,r} \to M$ given by $$\big(n_0,\ldots,n_r,h_0,\ldots,h_r\big) \mapsto f\big(n_p,\ldots,n_r,h_0,\ldots,h_q\big),$$ where $r = p+q$. (Here I am working with inhomogeneous cocycles, i.e. $G$-equivariant maps $G^{r+1} \to M$.)