Aspherical manifold with abelian fundamental group 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 $.
 A: 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 $m=n$.
A: 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.
