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It is well-known that classification of manifolds up to homemorphism is, in general, out of question. However, this task is sometimes tractable under some additional assumptions on manifolds one would like to classify. I want to ask about one specific example of this situation.

Let $T^n$ denote the $n$-dimensional torus $(S^1)^n$. Suppose that a closed manifold is finitely covered by $T^n$. I would like to know to what extent (and whether at all) it is possible to classify such manifolds $M$. If $n=1$, then $M=S^1$; if $n=2$, then $M=S^1 \times S^1$ or $M =K$, the Klein bottle. (But then again, this is not really interesting, because in these dimensions all manifolds are classified.) What happens if $n \geq 3$, or at least if $n=3$?

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What do you exactly mean by 'finitely covered'? I guess you don't mean that $T^n$ is a finite-sheeted covering space, do you? –  Fernando Muro Dec 29 '12 at 15:42
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In all dimensions you at least have the Bieberbach/Euclidean manifolds. These are the ones whose fundamental groups are the crystallographic groups Mosher mentions (in dimension 3) but this family exists in all dimensions. To what extent you get more than that in high dimensions I'm not certain off the top of my head. –  Ryan Budney Dec 29 '12 at 15:59
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2 Answers

People have already mentioned the Bieberbach theorems, which imply that your manifold is homotopy equivalent to a Euclidean manifold. In fact, it is homeomorphic to the Euclidean manifold, at least if the dimension is at least $5$. This is predicted by the Borel conjecture, which asserts that homotopy equivalent aspherical manifolds are homeomorphic in high dimensions. The necessary case of the Borel conjecture is proven in the following paper of Farrell and Jones (building on lots of earlier work of theirs):

F. T. Farrell and L. E. Jones. Topological rigidity for compact non-positively curved manifolds. In Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), pages 229– 274. Amer. Math. Soc., Providence, RI, 1993.


EDIT : The question inspired me to do a little more reading about the history of the Borel conjecture, and I learned that for virtually abelian fundamental groups like the ones in this question, the Borel conjecture was proven in the earlier paper

MR0704219 (84k:57017) Farrell, F. T.(1-MI); Hsiang, W. C.(1-PRIN) Topological characterization of flat and almost flat Riemannian manifolds Mn (n≠3,4). Amer. J. Math. 105 (1983), no. 3, 641–672.

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Assuming that you DO mean that $T^n$ is a finite sheeted covering space, at the very least one can say that $\pi_1(M)$ is a torsion free $n$-dimensional crystallographic group. This follows from the Bieberbach theorems regarding groups that contain $\mathbb{Z}^n$ with finite index.

If $n=3$, it follows that $M$ is homeomorphic to the closed Euclidean 3-manifold that one gets as the quotient of $\mathbb{E}^3$ by the 3-D crystallographic group isomorphic to $\pi_1(M)$; this follows from standard 3-manifold arguments. The complete list of these manifolds is known: amongst the 219 different 3-D crystallographic groups listed in http://en.wikipedia.org/wiki/Space_group, pick out the torsion free ones.

I don't know what happens beyond dimension 3, other than to say that $M$ is homotopy equivalent to the closed Euclidean $n$-manifold that one gets as the quotient of $\mathbb{E}^n$ by the $n$-dimensional crystallographic group isomorphic to $\pi_1(M)$.

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Oops, I did not see your comment, Ryan, before I made the latest edits. –  Lee Mosher Dec 29 '12 at 16:06
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No problem. In dimension 4 you have topological surgery at your disposal so the only issue is if these manifolds can have non-zero Kirby-Siebenmann invariant. I think the answer is likely yes but my head is somewhere else at the moment. In high dimensions it appears there's analogous surgery-theoretical obstructions. –  Ryan Budney Dec 29 '12 at 16:32
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