# What happens to a closed manifold to ensure it is homeomorphic to a torus $T^{n}$?

If $$M$$ is a smooth connected closed $$n$$-dimensional manifold, its universal covering space is homeomorphic to Euclidean space $$R^{n}$$, and its fundamental group is $$Z^{n}$$, then is it homeomorphic to a torus $$T^{n}$$? If not, why?

For dimensions $$n \geq 5$$, the answer is yes. First, note that $$M$$ is homotopy equivalent to a torus since it must be a $$K(\mathbb{Z}^n,1)$$. Second, Hsiang-Wall show in "On Homotopy Tori II" that in such dimensions, homotopy tori are in fact actual tori (i.e. homeomorphic to tori).

• This is true in all dimensions. In dimension 3 you rely in Perelman and in dimension 4 you rely on Freedman and Quinn.
– mme
Sep 4, 2021 at 17:19
• Could you give me some specific references about dimension 3 and 4?
– Thom
Sep 4, 2021 at 17:30

This answer is intended to give references of the cases for the case $$n \leq 4$$. In dimensions $$n \leq 2$$ this is covered in a first topology course so there are two interesting cases.

In dimension 3, one has the prime decomposition theorem: every 3-manifold is uniquely the connected sum of prime 3-manifolds. Because $$\Bbb Z^3$$ has no non-trivial decomposition as a free product, it follows that $$M \cong M_P \# M_S$$, where $$\pi_1 M_S = 0$$, so $$M_S$$ is a homotopy sphere. This is the one place where one must use the geometrization theorem. The geometrization theorem implies the Poincare conjecture implies $$M_S = S^3$$ implies $$M = M_P$$. In Hempel's book you will see this called the "Poincare associate" $$\mathcal P(M)$$ of $$M$$; it is prime.

The argument below shows $$M_P \cong T^3$$. The canonical reference for such pre-geometric techniques is Hempel's book "3-Manifolds".

Hempel's Theorem 11.10 says that if $$\pi_1 M$$ fits into a sequence $$1 \to N \to \pi_1 M \to \pi_1 B$$, where $$B$$ is a closed surface of nonpositive Euler characteristic, then $$N = \Bbb Z$$ and $$M_P$$ is a circle bundle over $$B$$. Here we have $$B = T^2$$; if $$Y_k$$ is the circle bundle over $$B$$ with Euler class $$k$$, then $$H_1(Y_k) = \Bbb Z^2 \oplus \Bbb Z/k$$. Because $$H_1(M_P) = \Bbb Z^3$$, it follows that $$M_P \cong Y_0 = T^3$$, as desired.

Hempel's proof of this statement is quite explicit and combinatorial. It is not analagous to the high-dimensional proofs.

You may also be interested in Hempel Chapter 13, which describes how to show that 3-manifolds and their automorphisms are essentially determined by their fundamental group in certain cases ($$P^2$$-irreducible and "Haken").

In dimension 4, you may invoke the Theorem of section 11.5 of Freedman and Quinn's book "Topology of 4-Manifolds". This states: "Suppose $$f: M \to N$$ is a homotopy equivalence of compact aspherical 4-manifolds with polyfinite or polycyclic fundamental groups, which restricts to a homeomorphism of boundaries. Then $$f$$ is homotopic relative to the boundary to a homeomorphism."

Finitely generated abelian groups are polycyclic and your manifold $$M$$ is aspherical. Thus the classifying map $$f: M \to K(\Bbb Z^4, 1) = T^4$$ for its fundamental group is a homotopy equivalence, and Freedman-Quinn's result implies that this map is homotopic to a homeomorphism.

This result is very nice but much more inexplicit than the 3-dimensional case. It requires a complicated infinite procedure, salvaging as much of Whitney's ideas as possible (simplifying handle decompositions by finding non-intersecting discs). Because this is no longer guaranteed by transversality, the procedure is much more complicated.

Oxford University Press has just published a new book on the disc embedding theorem, which will hopefully make its proof accessible to a wider audience.