# The homology of the orbit space

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.

• Isn't $X/G$ just $BG$ in this case? Dec 26, 2018 at 22:23
• @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, 2018 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? Dec 26, 2018 at 23:46
• @ChrisGerig No, I'm not assuming that it permutes the cells...
– GSM
Dec 26, 2018 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)? Dec 27, 2018 at 4:05

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.)

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