Suppose $\Gamma$ is a finitely presented group that has a torsion element. Can the classifying space $K(\Gamma,1)$ be homotopic to a finite-dimensional manifold? If yes, what is the simplest example?
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4$\begingroup$ The answer is no, because the presence of torsion forces $\Gamma$ to have infinite cohomological dimension. See e.g. Ken Brown's "Cohomology of groups", Chapter VIII.2. $\endgroup$– Mark GrantCommented Mar 29, 2017 at 15:57
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3$\begingroup$ Note that there is a version of classifying spaces $\underline{E}G$ where $G$ acts so that compact (i.e. finite) subgroups have (weakly) contractible fixed point sets. Sometimes groups with torsion can have finite-dimensional $\underline{E}G$. ams.org/mathscinet-getitem?mr=1292018 $\endgroup$– Ian AgolCommented Mar 29, 2017 at 17:40
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1 Answer
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This is not possible. Let $B$ be a space of type $K(\Gamma,1)$ and of dimension $d<\infty$, and let $E$ be the universal cover, which is contractible. Then for any $A\leq\Gamma$ we see that $E/A$ is a $K(A,1)$ and has finite dimension, so the group cohomology $H^i(A)$ vanishes for $i>d$. However, for $n>1$ the group cohomology of $C_n$ is nonzero in all even degrees, so $\Gamma$ cannot contain $C_n$, so $\Gamma$ must be torsion-free.
answered Mar 29, 2017 at 16:03