Consider an extension\begin{equation}1\rightarrow N\rightarrow G\xrightarrow{\rho} K\rightarrow 1\end{equation}Let $K$ act on a $K$-module $A$ by $\phi_k: a\mapsto k\cdot a$. Define a $G$-action $a\mapsto g\cdot a:=\rho(g)\cdot a$; under this action, $A$ is a $G$-module. Now consider the Hochschild-Serre spectral sequence\begin{equation}E^2_{p,q}=H_p(K,H_q(N,A))\Rightarrow H_{p+q}(G,A)\end{equation}It would appear that the LHS, the page $E^2$, does not depend on the $K$-action $\phi_k$, while the RHS, the limit of the sequence, does depend on it. How does one resolve this apparent contradiction?

To be clear, I understand that, in general, $``H_p"$ is homology with local coefficients: $K$ acts on $H_q(N,A)$ according to $\rho$. I am comfortable with this.

I am concerned with the action $\phi_k$ of $K$ on $A$. Since $N\simeq\ker\rho$, $N$ always acts trivially on $A$, and so the effect of $\phi_k$ is erased on the second page. Is the data of $\phi_k$ encoded in the differentials?

I am primarily concerned with the case where $K=\mathbb{Z}_2=\{1,T\}$. I want to detect the difference between the trivial action $\phi^{triv}_T:a\mapsto a$ and the inversion action $\phi^{inv}_T:a\mapsto -a$. Presumably, these should yield different spectral sequences since they yield different limits $H_n(G,A)$.

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    $\begingroup$ I migrated this over at the request of the OP. $\endgroup$ Jun 4, 2014 at 12:28

1 Answer 1


The action of $K$ on $H_q(N;A)$ may be non-trivial, even though $A$ is a trivial $N$-module. This may even happen if the extension is central, so that the conjugation action of $K$ on $N$ is trivial. You have to take into account the coefficient homomorphisms $$ (\phi_k)_\ast : H_q(N;A)\to H_q(N;A),\quad k\in K. $$ Since $A$ is a non-trivial $K$-module, these may be non-trivial.

This is treated in section III.8 of Ken Brown's "Cohomology of groups" textbook, where he explains how to view group cohomology as a functor of two variables.


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