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 2


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).

  • 5
    $\begingroup$ This is true in all dimensions. In dimension 3 you rely in Perelman and in dimension 4 you rely on Freedman and Quinn. $\endgroup$
    – mme
    Sep 4, 2021 at 17:19
  • 4
    $\begingroup$ Could you give me some specific references about dimension 3 and 4? $\endgroup$
    – 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.


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