Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question is related to Lyndon-Hochschild-Serre spectral sequence and cup products.

I have the followin result by J.S Milne in his book Arithmetic duality theorems pg 105.

Let $$0 \rightarrow C \rightarrow W \rightarrow G \rightarrow 0$$ is an exact sequence of groups corresponding to the canonical class $u \in H^{2}(G,C)$, and $C$ is of finite index in $W$

Let $M$ be a $W$ module on which $C$ acts trivially. Then by Hochschild-serre gives a spectral sequence $$0 \rightarrow H^{1}(G,M) \rightarrow H^{1}(W,M) \rightarrow H^{1}(C,M)^{G} \overset{\tau}\rightarrow H^{2}(G,M) $$ where $\tau$ is the transgression map. Now he goes on to prove that in this case $\tau(\alpha) = - \alpha \cup u$ (cup-product) for some $\alpha \in H^{0}(G,Hom(C,M))$.

Now consider the LHS for group homology, with $C,W,G,M$ as above ($C$ acting trivially on $M$)

$$ H_{2}(G,M) \overset{\tau'}\longrightarrow H_{1}(C,M)_{G} \overset{cor}\longrightarrow H_1(W,M) \overset{coinf}\longrightarrow H_{1}(G,M) \longrightarrow 0$$

Does it follow that $\tau'$ is induced by a cup product?

Thank you

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

The homology analogue of the result stated by Milne is that the transgression in the Lyndon-Hochschild-Serre spectral sequence for group homology $$ H_2(G;M)\to H_0(G;H_1(C;M)) = H_1(C;M)_G $$ is given (up to a sign) by cap product with the extension class $H^2(G;C)$. This makes sense since the cap product is a map $$ H^2(G;C)\otimes H_2(G;M)\to H_0(G;C\otimes M)=H_0(G;H_1(C; M)), $$ the final equality since $C$ is abelian.

This is supposedly well-known, see page 3 of this paper. I don't know a reference to an explicit proof, though.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.