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Suppose we have an acyclic group $G$ and let $X$ be a contractible CW-complex such that $G$ acts freely on $X$ (we do not suppose that the action is proper). Is there a way to understand the homology $\mathrm{H}_{\ast}(X/G, \mathbb{Z})$ ? We assume that the quotient space $X/G$ is Hausdorff.

For example if the action were free and proper then $$\mathrm{H}_{\ast}(X/G, \mathbb{Z})=\mathrm{H}_{\ast}(pt, \mathbb{Z})$$

Edit: Here is, I think, a more general question. Let $A$ be an abelian group. And let $G$ be a group such that $$\mathrm{H}_{\ast}(BG, A)=\mathrm{H}_{\ast}(pt, A)$$ Suppose that $X$ is a contractible CW-complex such that $G$ acts on $X$ freely. We assume that the orbit space $X/G$ is Hausdorff. What can we say about $\mathrm{H}_{\ast}(X/G, A)$ ?

Edit 2 (2019 January 5-th): May be the initial question sounds wild. I would be curious of an example where $G$ is an acyclic group acting freely on a contractible CW-complex and $$\mathrm{H}_{\ast}(X/G, \mathbb{Z})\neq\mathrm{H}_{\ast}(pt, \mathbb{Z})$$ with $X/G$ Hausdorff.

GSM
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  • Isn't $X/G$ just $BG$ in this case? – Exit path Dec 26 '18 at 22:23
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    @leibnewtz In the case when the action is free and proper, yes. in my question I'm assuming that the action is free but not proper... – GSM Dec 26 '18 at 22:28
  • Can you assume that $G$ permutes the cells of any CW-decomposition of $X$? Or is that equivalent to $G$ being proper? – Chris Gerig Dec 26 '18 at 23:46
  • @ChrisGerig No, I'm not assuming that it permutes the cells... – GSM Dec 26 '18 at 23:49
  • A tangential basic question: Is properness of G-action (or Hausdorffness of X/G, or both) equivalent to $X\to X/G$ being a cover (for a free G-action)? – Chris Gerig Dec 27 '18 at 04:05
  • You ask "what can be said?" but there are really at least two things you might mean by that. Are you more interested in (1) conditions under which $H_\ast(X/G,\mathbb Z) = H_\ast(pt, \mathbb Z)$ (as has come up in the comments), or (2) a better understanding of what happens when $H_\ast(X/G, \mathbb Z) \neq H_\ast(pt, \mathbb Z)$? Do you have an example where $H_\ast(X/G,\mathbb Z) \neq H_\ast(pt,\mathbb Z)$ that you're particularly interested in? – Tim Campion Dec 31 '18 at 14:05
  • @TimCampion I don't have a particular example! It seems to me that the second issue (2) is more interesting... – GSM Jan 01 '19 at 13:57
  • Why are we told that $X$ is a CW-complex? If the action of $G$ on $X$ preserves the cells and is free, then surely it will have to be proper. – IJL Jan 15 '23 at 11:26

1 Answers1

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As per the comments, there is currently a paradox in the first paragraph, somehow depending on $BG$ (where $G$ is a topological group) versus $BG^\delta$ (where $G^\delta$ has the discrete topology). I do not yet know the flaw/resolution.

I do not think properness is needed here, only that the (free) action is continuous. Since $X$ is a free $G$-space, there is a homotopy equivalence $EG\times_GX\simeq X/G$, and since $X$ is contractible there is a weak homotopy equivalence $EG\times_GX\simeq EG\times_G\lbrace pt\rbrace$. So in the notation of equivariant singular homology (the singular homology of the Borel construction) $$H_\ast(X/G)\cong H_\ast^G(X)\cong H_\ast^G(pt)\cong H_\ast(G)\cong 0$$ where the last isomorphism is due to acyclicity of $G$.

Here is a special case: Assume $G$ also permutes the cells of some CW decomposition of $X$ (called a $G$-CW-complex). Then I can work with equivariant cellular homology, the definition in Brown's book Cohomology of Groups (chapter VII.7) which uses chain complexes. Since $G$ acts freely, it is the equivariant homology $H_\ast(X/G)\cong H_\ast^G(X)$. Since $X$ is contractible, $H_\ast^G(X)=H_\ast^G(pt)=H_\ast(G)$. Since $G$ is acylic, $H_\ast(G)=0$. This is a special case of the Cartan-Leray spectral sequence, I believe. (It might be the case here that $X\to X/G$ is a regular cover.)

Chris Gerig
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    Something is wrong here. Consider the additive group $G=\mathbb{R}$. Setting $X = \mathbb{R}$, the group $G$ acts freely and continuously on the contractible CW complex $X$ (but of course does not permute the cells in any CW complex structure). The quotient $X/G$ is a single point, and thus all of its homology above degree $0$ vanish. However, $H_1(G) = \mathbb{R}$. – Andy Putman Dec 27 '18 at 19:17
  • @AndyPutman An acyclic group by definition has $H_k(G)=0$ for $k>0$. – Chris Gerig Dec 27 '18 at 19:20
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    Right, but in your first display you don't use the fact that $G$ is acyclic until the very last equality (and in my example the first and next to last items in your chain of equalities are not equal). – Andy Putman Dec 27 '18 at 19:21
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    Maybe I can say something more detailed/helpful later, but I think the issue here is that the homotopy equivalence $EG \times_G X \simeq X/G$ requires that the map $X\to X/G$ is a principal $G$-bundle (maybe you can get away with a little less than that?). In Andy's example, where $\mathbb{R}$ is considered with the discrete topology and $H_1 (\mathbb{R})$ means the group homology, i.e. the homology of $B\mathbb{R}^\delta$, the map is a (trivial) principal $\mathbb{R}$ bundle (over a point) instead of an $\mathbb{R}^\delta$-bundle. – Dan Ramras Dec 28 '18 at 01:18
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    Note that the oft-repeated statement "If $G$ acts freely and continuously on $X$, then $EG\times_G X \simeq X/G$" can't possibly be true without some extra hypotheses, since adding more open sets to the topology on $G$ won't destroy either freeness or continuity of the action, and won't change $X/G$, but can change $EG$ quite drastically. – Dan Ramras Dec 28 '18 at 01:22
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    (This is a pet peeve of mine, generalizing the pet peeve discussed here: https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/40070#40070) – Dan Ramras Dec 28 '18 at 01:27
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    @ChrisGerig I guess there is a confusion in your terminology, if the action of G on X is free,it does not mean that X is a free G-space. A free G-space (in my sense) is a cofibrant object in the model category of G-spaces (and in particular the action is free). If X were free G-space and contractible, then X/G is a model for BG. – GSM Dec 28 '18 at 11:15
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    @GSM Would it be possible to formulate it without mentioning any model structures? I guess something like $G\times X\to X\times X$ being a cofibration? But what are they? – მამუკა ჯიბლაძე Dec 31 '18 at 13:33
  • @მამუკაჯიბლაძე unfortunately, I don't know... – GSM Jan 01 '19 at 13:59