8
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 Aug 7 '14 at 4:24
  • $\begingroup$ Also, "infinite homology" means "infinite-dimensional homology" (as $\mathbb{Q}$-vector space), right? $\endgroup$ – Alex Suciu Aug 7 '14 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 $\endgroup$ – Dmitry Vaintrob Aug 7 '14 at 8:07
  • 8
    $\begingroup$ 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 Aug 7 '14 at 8:27
  • $\begingroup$ @Dmitry: you should edit the second sentence of your question according to Alex' comments. $\endgroup$ – YCor Aug 7 '14 at 9:54
18
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.