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.

  • $\begingroup$ Isn't $X/G$ just $BG$ in this case? $\endgroup$
    – Exit path
    Dec 26, 2018 at 22:23
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    $\begingroup$ @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... $\endgroup$
    – GSM
    Dec 26, 2018 at 22:28
  • $\begingroup$ Can you assume that $G$ permutes the cells of any CW-decomposition of $X$? Or is that equivalent to $G$ being proper? $\endgroup$ Dec 26, 2018 at 23:46
  • $\begingroup$ @ChrisGerig No, I'm not assuming that it permutes the cells... $\endgroup$
    – GSM
    Dec 26, 2018 at 23:49
  • $\begingroup$ 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)? $\endgroup$ Dec 27, 2018 at 4:05

1 Answer 1


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.)

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    $\begingroup$ 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, $H_1(G) = \mathbb{R}$. $\endgroup$ Dec 27, 2018 at 19:17
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    $\begingroup$ 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). $\endgroup$ Dec 27, 2018 at 19:21
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    $\begingroup$ 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. $\endgroup$
    – Dan Ramras
    Dec 28, 2018 at 1:22
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    $\begingroup$ (This is a pet peeve of mine, generalizing the pet peeve discussed here: mathoverflow.net/questions/23478/…) $\endgroup$
    – Dan Ramras
    Dec 28, 2018 at 1:27
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    $\begingroup$ @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. $\endgroup$
    – GSM
    Dec 28, 2018 at 11:15

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