4

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.

Chris Gerig
  • 17,130
Alex Turzillo
  • 1,189
  • 1
    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? – user43326 Jun 04 '14 at 08:22
  • Yes, there is a Kunneth theorem for local coefficients (http://mathoverflow.net/questions/75472/kuenneth-formula-for-group-cohomology-with-nontrivial-action-on-the-coefficient), so I could easily compute the homology of $N\times\mathbb{Z}_2$ without using spectral sequences if I wanted to. – Alex Turzillo Jun 04 '14 at 08:34
  • I'm actually interested in a more general problem (http://math.stackexchange.com/questions/815901/hochschild-serre-spectral-sequence-and-non-trivial-action-on-coefficients), for which a Kunneth theorem is not always available. – Alex Turzillo Jun 04 '14 at 08:37
  • 1
    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? – user43326 Jun 04 '14 at 08:43
  • Yes, from the $N=1$ example, it appears that $E^2$ does depend on the $\mathbb{Z}_2$ action on $A$. So how does $\mathbb{Z}_2$ act on $H_q(N;A)$ in general? – Alex Turzillo Jun 04 '14 at 08:53
  • Sorry, I was already posting my answer while you were writing the comment... – user43326 Jun 04 '14 at 08:57
  • $\mathbb Z/2$ act on $H_q(N,A)$ simply by the action induced by $Z/2$ action on $A$. – user43326 Jun 04 '14 at 09:05
  • Can you please describe the induced action? – Alex Turzillo Jun 04 '14 at 09:07
  • 1
    @user151696: See Ken Brown's "Cohomology of groups", section III.8. – Mark Grant Jun 04 '14 at 09:10

1 Answers1

4

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.

user43326
  • 3,031