I know (using group cohomology, for example) that if a group $G$ admits a finite dimensional $K(G,1)$ complex, then $G$ needs to be torsion-free.

I'm interested in the converse of the statement.

Q: If a group $G$ is finitely generated torsion-free, then is it necessary that $G$ is geometrically finite (i.e. $G$ admits a finite dimensional $K(G,1)$ complex) (We need finitely generated $G$ otherwise we have examples like $\mathbb{Z}^\mathbb{Z}$)

I don't feel like it is true, but I can't come up with an example of it.

  • 1
    $\begingroup$ What do you mean by "geometrically finite?" $\endgroup$ May 28 at 22:20
  • $\begingroup$ @MoisheKohan added definition. $\endgroup$ May 28 at 22:49
  • $\begingroup$ One can verify that all groups given in answers to this question are torsion-free. Since they are also finitely generated and not finitely-presented, they provide many examples. If you want more examples (say finitely presentable but not geometrically finite), let me know. $\endgroup$ May 28 at 22:57
  • $\begingroup$ See also Holt's answer here. $\endgroup$ May 28 at 23:10
  • $\begingroup$ @MoisheKohan I think my formulation was not good. I would like K(G,1) to be finite dimensional, not a finite complex. I'll update the question, sorry for changing the question. $\endgroup$ May 28 at 23:17

1 Answer 1


Thompson's group F is an example of a torsion-free group of type $F_\infty$ (which is stronger than finite presentability and means existence of a classifying space with finitely many cells in each dimension). To see that Thompson's group F is torsion-free, note that it is realized as a subgroup of the (orientation-preserving) homeomorphism group of $[0,1]$. At the same time, this group has infinite cohomological dimension since it contains free abelian groups of arbitrarily high rank.


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