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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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I have edited the question in asking only for homotopy equivalence. –  berl13 Mar 8 '12 at 14:03
The following page provides some useful information: –  Xiaolei Wu Mar 8 '12 at 15:17
It all depends on your definition of a topological manifold. If you use the textbook definition, then the result is: $G$ admits a manifold $K(G,1)$ iff $G$ is countable and of finite cohomological dimension. Proof is a combination of a theorems by Eilenberg-Ganea, Whitehead (Theorem 13 from "Combinatorial Homotopy-I"), and Whitney's embedding theorem for locally finite CW complexes. However, if you insist on your manifold being closed, then you are facing Wall's conjecture that most topologists do not believe in. –  Misha Mar 8 '12 at 15:47
Thanks for your answer. I do not insist on the manifold being closed. What is the Eilenberg-Ganea theorem? I know only the conjecture which was proven to be false for groups of cohomological dimension larger than 2. –  berl13 Mar 8 '12 at 15:55
Eilenberg-Ganea theorem states that if $G$ had cohomological dimension $n$ then geometric dimension $gd(G)$ of $G$ is at most $n+1$. Furthermore, unless $n=2$, they proved that $gd(G)=n$. Hence, the conjecture. –  Misha Mar 8 '12 at 16:35

2 Answers 2

up vote 10 down vote accepted

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 $K(G,1)$.

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What do you mean by "(textbook) topological manifold"? –  Greg Friedman Mar 9 '12 at 1:37
Hausdorff, 2nd countable, locally homeomorphic to ${\mathbb R}^n$. –  Misha Mar 9 '12 at 1:46
The proof I gave is well-known (among topologists) provided that $G$ admits $K(G,1)$ which is a finite CW-complex. The only tricky issue in general is to get local finiteness for which the only reference I know is the original Whitehead's paper. Does somebody know a modern reference? –  Misha Mar 9 '12 at 2:00
Igor Belegradek has found a very nice modern reference for Whitehead's result on the last page of This is given as an exercise with a hint and the proof (via a mapping telescope) is much easier and nicer than the original one. –  Misha Mar 11 '12 at 23:44
That's interesting. Do you mean for the rationals $Q$ we can find a manifold whose fundamental group is $Q$? –  yeshengkui Mar 21 '12 at 16:26

This question here may be interesting for you.

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I have seen this discussion. But I am not so much interested in the smooth structure. So my question is purely topological. –  berl13 Mar 8 '12 at 14:02

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