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Let $G$ be a group acting on $X$, $X$ a cellular complex and $cd(G)$ the cohomological dimension of $G$.

2 things:

(1) I'm looking for a reference (or proof!) of this:

Suppose $X$ is acyclic. Then $cd(G) \leq max_{\sigma} \space cd(Stab(\sigma) + dim \space \sigma$, where $\sigma$ runs over the cells of $X$.

(2) If $X$ isn't acyclic, can anything be said about $cd(G)$?

flag	asked Sep 15 2010 at 16:33 Romeo 1,816●2●17

Presumably you want the action to be faithful? – Qiaochu Yuan Sep 15 2010 at 16:40

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@Qiaochu: Not necessarily; e.g. when X is a point the whole group is the stabilizer. – Tyler Lawson Sep 15 2010 at 17:43

Mark Grant · Answer 1 · 2010-09-15 21:11:30Z UTC

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If $X$ is a free contractible $G$-complex, this is the Eilenberg-Ganea Theorem

edited Sep 15 2010 at 21:11

answered Sep 15 2010 at 18:57

Mark Grant
9,061●1●14●35

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Your link seems to go through your university proxy, so I can't use it. Can you fix that? – Nikita Sep 15 2010 at 19:01

Oops, sorry. I've linked instead to Math Reviews, hope that works. – Mark Grant Sep 15 2010 at 21:14

If I've parsed the statement of the Eilenberg--Ganea theorem correctly, in their set up the action of $G$ on $X$ is free. – HW Nov 1 2010 at 23:13

Yes. Then (1) says $cd(G)\leq \dim(X)$, and (assuming $G$ is discrete) this says $cd(G)$ is bounded above by its geometric dimension, ie the smallest dimension of a $BG$...Ah, I guess I meant to say "this is (the trivial part of) the Eilenberg-Ganea Theorem"... – Mark Grant Nov 2 2010 at 10:04

Nikita · Answer 2 · 2010-09-15 18:22:40Z UTC

If you don't require $X$ to be acyclic, then you can't say anything. Indeed, every group acts freely on its Cayley graph, which is a 1-dimensional cell complex.

You're not even saved by assuming the $X$ is highly connected. By attaching cells to the Cayley graph in an equivariant manner, you can obtain a $k$-connected $(k+1)$-dimensional complex on which the group acts freely.

Bruce Ikenaga · Answer 3 · 2010-09-15 17:33:09Z UTC

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Reference: Serre, J-P.(1971) "Cohomologie des groupes discrets," Ann. Math. Studies 70, 77-169 (Proposition 11, page 93). Serre credits this to Quillen, and I've never succeeded in locating this in any of Quillen's papers. Does anyone know where it is?

Proof: Use the equivariant cohomology spectral sequence.

answered Sep 15 2010 at 17:33

Bruce Ikenaga
21●1

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The proof that I have in mind involves filtering the space $X\times_GEG$ in two ways -- by skeleta of $BG$ and by skeleta of $X$ -- and comparing the information that these two filtrations yield. Each of the two filtrations leads to a spectral sequence, I suppose (although on each side you can do the job without that tool). I don't know which of these two is what Bruce means by "the equivariant cohomology spectral sequence"! – Tom Goodwillie Sep 15 2010 at 18:26

Its when you filter by skeleta of $X$; the other is the Serre spectral sequence. – Oscar Randal-Williams Sep 15 2010 at 19:27

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Tom, does looking at $X \times _G EG→X \times _{Stab(\sigma)} EG$ end up bounding $cd(G)$ by dim$\Sigma ^n BStab(\sigma) = n+dim BStab(\sigma)$? (n = dim $\sigma$) – Romeo Sep 15 2010 at 20:08

Cohomological dimension of a group acting on a cellular complex

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