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?

## 2 Answers

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 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.