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This question was originally posted to Math.StackExchange, but having got no response there, I'm reposting it here. I apologise if it is too elementary for this site. (Original post: https://math.stackexchange.com/questions/170310/calculating-hng-mathbbzg-as-co-homology-with-compact-support-of-a-prope )

I came across the following Exercise in Brown's book "Co-homology of Groups", and have been completely unable to solve it. If anyone could give me a hint that would point me in the right direction I would really appreciate it. I couldn't even think of a filtration of $X$ which might produce a useful spectral sequence.

Brown "Co-homology of Groups", Section VIII.7, p.209, Ex 4:

Prove the following generalisation of 7.5: Let $X$ be a contractible $G$-CW-complex with finite isotropy groups and only finitely many cells mod-$G$. Then $H^*(G, \mathbb{Z} G) \cong H_c^*(X, \mathbb{Z})$.

I presume that he means $H^*(G, \mathbb{Z} G) \cong H_c^*(X, \mathbb{Z})$ as $\mathbb{Z}G$-modules.

There is a hint:

First show, by a spectral sequence argument for instance, that $H^*(G, \mathbb{Z}G)$ can be computed from $\text{Hom}_G(C(X), \mathbb{Z}G)$.

Here $H_c^* $ is co-homology with compact support. Throughout Brown's book any action on a space is always rigid (set-wise and point-wise stabilisers of cells coincide). The "7.5" which he refers to is the statement that $H^*(G, \mathbb{Z}G) \cong H_c^*(X, \mathbb{Z})$ as $\mathbb{Z}G$ modules, where $X$ is a contractible, free, co-compact $G$-CW-complex.

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Perhaps the "spectral sequence argument" Brown is referring to is the cohomology analogue of Proposition 7.3 in Chapter VII. It seems that for any $G$-CW-complex $X$ and $G$-module $M$, there should be a spectral sequence $$ H^p(G;H^q(X;M))\Rightarrow H^{p+q}_G(X;M). $$ It then should follow (using Proposition 5.3 in the same chapter) that if $f\colon X\to Y$ is a map of $G$-CW-complexes which induces an isomorphism on ordinary integral homology, then $f$ induces an isomorphism $H^\ast_G(X;M)\cong H^\ast_G(Y;M)$. In particular, if $X$ is the complex in the exercise and $Y$ is a point, we have $H^\ast_G(X;\mathbb{Z}G)\cong H^\ast(G;\mathbb{Z}G)$.

So far the finite isotropy assumption hasn't come in. But I believe this should be used in the second step, that is, showing that $\operatorname{Hom}_G(C(X);\mathbb{Z}G)$ has the same homology as $\operatorname{Hom}_c(C(X),\mathbb{Z})$ (generalising Lemma 7.4 of Chapter VIII).

I haven't checked the details yet, let me know if it works (or not)!

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    $\begingroup$ Thanks! I've written out your argument and it seems to work. $\endgroup$ Commented Jul 19, 2012 at 14:23

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