Let $G$ be a discrete group and $BG$ some model for the classifying space of $G$. So $BG$ is an aspherical path-conected topological space.
Under which conditions is $BG$ a topological manifold or only homotopy equivalent to a topological manifold?
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Sign up to join this communityLet $G$ be a discrete group and $BG$ some model for the classifying space of $G$. So $BG$ is an aspherical path-conected topological space.
Under which conditions is $BG$ a topological manifold or only homotopy equivalent to a topological manifold?
Here is a more detailed answer.
Theorem. $K(G,1)$ is homotopy-equivalent to a (textbook) topological manifold if and only if $G$ is countable and has finite cohomological dimension (over ${\mathbb Z}$).
Sketch of the proof. One direction is clear, so suppose that $G$ is countable and has finite cohomological dimension (say, $n$). Then, by Eilenberg-Ganea theorem (see Theorem 1 in their paper "On the Lusternik-Schnirelmann category of abstract groups", see also Brown's book "Cohomology of Groups", Theorem 7.1), there exists a countable CW complex $X$ of dimension $m\le n+1$ which is $K(G,1)$. This theorem is usually stated without countability assumption/conclusion, but the same proof works in the countable context.
Now, by Whitehead's theorem (Theorem 13 from Whitehead's "Combinatorial Homotopy-I"), $X$ is homotopy-equivalent to an $m$-dimensional locally-finite CW complex $Y$. Without loss of generality, we can assume that $Y$ is simplicial. Then, by Whitney's embedding theorem (in the context of locally-finite simplicial complexes), there exists a PL embedding $Y\to {\mathbb R}^{2m+1}$. Next, take a suitable open regular neighborhood $N$ of $Y$ in ${\mathbb R}^{2m+1}$. Then $N$ is homotopy-equivalent to $X$ and, hence, provides a manifold which is $K(G,1)$.