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Let $G$ be a group that acts on a space $X$, and the quotient $X/G$ is compact. Then any torsion-free subgroup of finite index of $G$ is the fundamental group of an acyclic space with only finitely many cells in each dimension. The homology of $G$ is then finitely generated in all dimensions. Why is it so?


Edit: apparently the general statement is false. Here's the context of the question:

Let $\text{Out}(F_n) = \text{Aut}(F_n)/\text{Inn}(F_n)$ be the outer automorphism group of the free group on $n$ generators. The spine of the outer space $K_n$ (defined by Vogtmann and Culler in their 1986 paper to study automorphism groups of free groups) is a contractible space that acts on $\text{Out}(F_n)$ freely. The quotient $K_n/\text{Out}(F_n)$ is finite and thus compact.

The statement in the paper is as follows: the fact that $K_n/\text{Out}(F_n)$ is compact implies immediately that any torsion-free subgroup of finite index of $\text{Out}(F_n)$ is the fundamental group of an acyclic space with only finitely many cells in each dimension. The homology of $\text{Out}(F_n)$ is then finitely generated in all dimensions.

It seems that the statement about the fundamental group follows easily from the fact that the quotient is compact, but I don't see why.

Thank you!

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    $\begingroup$ Welcome to math.stackexchange! Don't forget to check out the guide for asking questions $\endgroup$ – William Apr 15 at 14:03
  • $\begingroup$ Can you provide more context for your question, maybe the original source? As stated it is not true. For the first sentence you could take $X=pt$ and $G$ any group (or even $X=G$ if you need a free action). For the second sentence, we could then take $G = \mathbb{Z}$ with subgroup $H=2\mathbb{Z}$ so that $BH = S^1$ which is not acyclic; in fact only perfect groups can be the fundamental group of an acyclic space since $H_1$ is the abelianization of $\pi_1$. The third sentence then cannot be true in this generality since there are (finitely presented) groups with infinitely generated $H_3$. $\endgroup$ – William Apr 15 at 14:23
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    $\begingroup$ Thanks for the edit, that's much more specific. I don't have access to their paper, but I did find expository notes by Vogtmann, which on page 13 says "Since Outer space and its spine $K_n$ are contractible, and since $Out(F_n)$ acts with finite stabilizers, the quotient of $K_n$ by any torsion-free subgroup $Γ$ of finite index is an aspherical space". Do you mean "aspherical" instead of "acyclic"? In that case I think I can answer. $\endgroup$ – William Apr 15 at 21:51
  • $\begingroup$ The original statement can be found on page 4 of pi.math.cornell.edu/~vogtmann/papers/ICM/kvogtmann.pdf under 2.3 Finite generation of homology. I think it does mean acyclic instead of aspherical... the question you're talking about is quite different. $\endgroup$ – yshen Apr 15 at 22:28
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I think I can show that any torsion-free subgroup $\Gamma\subset Out(F_n)$ with finite index is the fundamental group of an aspherical finite CW complex (i.e. having a potentially non-trivial fundamental group but no higher homotopy groups).

$Out(F_n)$ acts on $K_n$ with finite stabilizers, so if $\Gamma$ is a torsion-free subgroup then $\Gamma$ acts freely on $K_n$. Since $K_n$ is contractible and $Out(F_n)$ (and hence $\Gamma$) is discrete, it follows that $K_n/\Gamma \simeq K(\Gamma,1)$, i.e. it is an aspherical space with fundamental group isomorphic to $\Gamma$. If $\Gamma$ has finite index in $Out(F_n)$ then the map $K_n/\Gamma \to K_n/Out(F_n)$ is a finite covering space, so if $K_n/Out(F_n)$ is a finite cell complex then $K_n/\Gamma$ is as well.

I think the statement "with finitely many cells in each degree" might be superfluous, because it seems like $K_n/\Gamma$ is actually a finite complex. In fact in Vogtmann's notes on page 12 I found the following:

Since the quotient of the spine $K_n$ by $Out(F_n)$ is finite, so is the quotient by any finite index subgroup $H$

so this should apply in particular to our torsion-free $\Gamma$ of finite index.

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