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Set-up:

Consider the trivial extension, where $p$ is the projection onto the $\mathbb{Z}_2$ component,$$1\rightarrow N\rightarrow N\times\mathbb{Z}_2\xrightarrow{p}\mathbb{Z}_2\rightarrow 1$$ Define actions of $\mathbb{Z}_2=\{1,T\}$ on a $\mathbb{Z}_2$-module $A$ by$$\phi^1_T:a\mapsto a\ \text{ (trivial)}\ \ \ \text{ and }\ \ \ \phi_T^2:a\mapsto -a\ \text{ (sign) }$$ Let $N\simeq\ker p$ act trivially on $A$, so that $g\in N\times\mathbb{Z}_2$ determines maps $a\mapsto\phi^{1,2}_{p(g)}(a)$.

Now consider the Hochschild-Serre spectral sequence for the group homology, with coefficients in the module $A$, of the extension. (Equivalently, consider the Serre spectral sequence for homology, with coefficients in the local system $A$, of the fibration of Eilenberg-Maclane spaces.) In terms of group homology, the second page is given by $$E^2_{p,q}:=H_p(\mathbb{Z}_2,H_q(N;A))\Rightarrow H_{p+q}(N\times\mathbb{Z}_2;A)$$ Since the extension is a direct sum, $\mathbb{Z}_2$ acts trivially on the fiber homology $H_q(N;A)$.

Problem:

It appears that the second page $E^2_{p,q}$ does not depend on the choice of action (trivial or sign) on the coefficient module while the limit $H_{p+q}(N\times\mathbb{Z}_2;A)$ does depend on this choice.

Am I correct?

If so, the choice of $\phi^{1,2}$ must somehow be encoded in the differentials of the sequence.

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    $\begingroup$ What is the role of $N$ here? Since the local system only depends on $\mathbf Z/2$ and the extension splits, can't we deduce the general case from the case $N=1$ by some sort of Kunneth isomorphism? $\endgroup$
    – user43326
    Commented Jun 4, 2014 at 8:22
  • $\begingroup$ Yes, there is a Kunneth theorem for local coefficients (mathoverflow.net/questions/75472/…), so I could easily compute the homology of $N\times\mathbb{Z}_2$ without using spectral sequences if I wanted to. $\endgroup$
    – Alex Turzillo
    Commented Jun 4, 2014 at 8:34
  • $\begingroup$ I'm actually interested in a more general problem (math.stackexchange.com/questions/815901/…), for which a Kunneth theorem is not always available. $\endgroup$
    – Alex Turzillo
    Commented Jun 4, 2014 at 8:37
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    $\begingroup$ Anyhow, I think a good place to start is N=1 case. The $E_2$ term is simply $H_p(\mathbb Z/2,A)$. Doesn't this depend on the action of $\mathbb Z/2 $ on A? $\endgroup$
    – user43326
    Commented Jun 4, 2014 at 8:43
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    $\begingroup$ @user151696: See Ken Brown's "Cohomology of groups", section III.8. $\endgroup$
    – Mark Grant
    Commented Jun 4, 2014 at 9:10

1 Answer 1

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I guess here $\mathbb Z/2$-module means $\mathbb Z[\mathbb Z/2]$-module. So let's take $N=1$, $A=\mathbb Z$. $H_0(\mathbb Z/2,A)$ is the coinvariant $A/\mathbb Z/2$ so it is $A$ in the case of the trivial action, $\mathbb Z/2$ in the case of the sign action. So $E_2$ term depends on the action.

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  • $\begingroup$ Thanks. This was helpful. Yes, $A$ is a module over the group ring $\mathbb{Z}[\mathbb{Z}_2]$. $\endgroup$
    – Alex Turzillo
    Commented Jun 4, 2014 at 9:03

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