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.
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.