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Suppose $G$ is a discrete group given by finitely many generators with finitely many relations. Can the homology groups $H_i(G, \mathbb{Q})$, or equivalently $H_i(BG, \mathbb{Q})$ (topological homology of the classifying space) be infinite-dimensional? Can they be nonzero for infinitely many $i$?

For any finitely presented groups I've seen, the answer is a surprising "no" (all finitely presented groups I know act on a finite-dimensional contractible space with finite stabilizers, and it follows that above the dimension of this space, homology vanishes). But it really should be the case that a "general" finitely-presented group has infinite homology... does anyone know of an example?

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    $\begingroup$ Just to make sure: the assertion that $H_i(G,\mathbb{Q})$ is finite-dimensional (for all $i>2$) is an assumption, yes? $\endgroup$
    – Alex Suciu
    Commented Aug 7, 2014 at 4:24
  • $\begingroup$ Also, "infinite homology" means "infinite-dimensional homology" (as $\mathbb{Q}$-vector space), right? $\endgroup$
    – Alex Suciu
    Commented Aug 7, 2014 at 4:26
  • $\begingroup$ @Alex: I thought any group homology of a finitely presented group is finite-dimensional. Are there examples where $H_2$ isn't? And yes, "infinite" means infinite-dimensional $\endgroup$ Commented Aug 7, 2014 at 8:07
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    $\begingroup$ Yes, $H_2$ of an fp group is finitely generated. But, say, $H_3$ needs not be finitely generated. The first such example was given by John Stallings, in a seminal paper, titled, sure enough, A finitely presented group whose 3-dimensional integral homology is not finitely generated, see here. $\endgroup$
    – Alex Suciu
    Commented Aug 7, 2014 at 8:27
  • $\begingroup$ @Dmitry: you should edit the second sentence of your question according to Alex' comments. $\endgroup$
    – YCor
    Commented Aug 7, 2014 at 9:54

2 Answers 2

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Thompson's group F is an example. It's finitely presented and, according to this paper of Ken Brown, the integral homology is free abelian of rank 2 in every positive dimension.

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This answers the other part of your question, not answered by Thompson's group. For each $i\geq 3$ there is a finitely presented group $G_i$ with the property that $H_i(G_i\mathbb{Q})$ is infinite dimensional. The first such examples were due to Stallings ($i=3$ and Bieri $i>3$). Take a direct product of $i$ copies of the free group $F$ on two generators, and define $G_i$ to be the kernel of the homomorphism $F^i\rightarrow \mathbb{Z}$ that sends each of the $2i$ standard generators to $1\in \mathbb{Z}$. These are the Bieri-Stallings groups. For $i=2$ the analogous group is finitely generated but not finitely presented, and for $i=1$ it is free of infinite rank. These groups can be viewed as special cases of the Bestvina-Brady construction.

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