Given an aspherical 4-dimensional closed manifold $M$ with fundamental group $\mathbb{Z}^4$, it is homotopy-equivalent to $T^4 = S^1 \times \ldots S^1$, the 4-dimensional torus.

Question 1: Since I am no expert and could not dig out a reference I would be interested if it is open/known that under the circumstances above, $M$ must be homeomorphic (diffeomorphic) to $T^4$?

Question 2: Can one say more if one knows that $M$ is smoothly covered by $\mathbb{R} \times T^3$?

Remark: in all other dimensions it seems to be true (due to Wall et al.) that for dimensions $n \geq 5$ the manifold $M$ is finitely coverey by a manifold diffeomorphic to $T^n$ and for $n \leq 3$ it is even diffeomorphic to $T^3$.


Freedman and Quinn's "Topology of $4$-manifolds", Chapter 11.5, contains the following statement:

Let $f: M\to N$ be a homotopy equivalence of compact aspherical $4$-manifolds whose fundamental groups are poly-(finite or cyclic), restricting to a homeomorphism of boundaries. Then $f$ is homotopic rel $\partial M$ to a homeomorphism.

This implies a positive answer to (1) in the topological category. I would guess that the answer in the smooth category is unknown, since it appears to be unknown if there exist exotic $4$-tori, see Do there exist exotic 4-tori?

  • $\begingroup$ Thank you Mark for the reference. It would be interesting, though, if for the case of Question 2 or the fact that $M$ is a fibre bundle over the circle with toric fibers would give a positive answer in the smooth catefgory, too. $\endgroup$ – LCC1 Sep 2 '14 at 14:17

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