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Let $ M $ be an aspherical closed manifold with an abelian fundamental group. The fundamental group of an aspherical manifold is torsion-free and the fundamental group of a closed manifold is finitely generated. So $$ \pi_1(M) \cong \mathbb{Z}^n $$ for some $ n $. Must it be the case that $ n=\text{dim}(M) $?

The answer is yes for $ \text{dim}(M) \leq 2 $.

I pretty sure the answer is yes for $ \text{dim}(M) =3 $.

I'm uncertain about $ \text{dim}(M) \geq 4 $.

Ian Gershon Teixeira
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  • Actually, much more is true: $M$ is homotopy-equivalent to $T^n$ (easy) and is homeomorphic to it (very hard). – Moishe Kohan Feb 19 '22 at 17:33
  • Indeed the argument from Lee Mosher shows that an aspherical closed manifold with abelian fundamental group has $ \pi_1(M) \cong \mathbb{Z}^n $ where $ n $ is the dimension of the manifold. From there it is immediate that $ M $ is an $ n $ dimensional closed manifold and a $ K(\mathbb{Z}^n,1) $ so by https://mathoverflow.net/questions/403202/what-happens-to-a-closed-manifold-to-ensure-it-is-homeomorphic-to-a-torus-tn?noredirect=1&lq=1 it must be homeomorphic to $ T^n $. – Ian Gershon Teixeira Feb 19 '22 at 21:59

2 Answers2

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Yes, this must be the case.

Two tools are needed for the proof: the Poincaré duality theorem for compactly supported cohomology; and group cohomology.

Let $m = \text{dim}(M)$. The universal cover $\tilde M$ is a contractible $m$-manifold, and so by the above version of Poincaré duality its compactly supported cohomology satisfies $$H^i_c(\tilde M;\mathbb Z) \approx \begin{cases} \mathbb Z & \quad\text{if $i=m$} \\ 0 &\quad\text{if $i \ne m$} \end{cases} $$ The relation to group cohomology is the isomorphism $$H^i_c(\tilde M;\mathbb Z) \approx H^i(\pi_1 M;\mathbb Z (\pi_1 M)) $$ where the right hand side is cohomology with group ring coefficients, twisted by the action of $\pi_1 M$ on its group ring. You can find this, for example, in Brown's book "Cohomology of Groups" (assuming that $M$ is a CW complex, which covers tremendously many cases, including the smooth case; I'm unsure of a reference without that assumption).

But then a simple calculation shows that $$H^i(\mathbb Z^n;\mathbb Z (\mathbb Z^n)) \approx \begin{cases} \mathbb Z &\quad\text{if $i=n$} \\ 0 &\quad\text{if $i \ne n$} \end{cases} $$ and therefore $m=n$.

Lee Mosher
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I have already accepted Lee Mosher's excellent and very direct proof.

Just wanted to mention here that I snooped around about the Borel conjecture and it turns out that the Borel conjecture is true for all solvable groups. So if $ \Gamma $ is any solvable group and $ M,N $ are any aspherical closed manifolds with $ \pi_1(M) \cong \Gamma \cong \pi_1(N) $ then we can conclude that $ M $ and $ N $ are homeomorphic.

Taking $ N=T^n $ and $ \Gamma=\mathbb{Z}^n $ we have desired result.

For more details on other types of groups for which the Borel conjecture holds see

https://mathoverflow.net/questions/416611/torus-bundles-and-compact-solvmanifolds?noredirect=1#comment1069392_416611

it seems to be all the groups that are good in the sense of Freedman.

Ian Gershon Teixeira
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