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It is known that every orientable 3-manfiold can be obtained as a ramified cover of S3 with a ramification (of some order) at a link in S3. I am curious if there is a reasonable characterization of 3-manifolds that cover 3-torus?

Added. Notice that such a manifold is enlargeble, so it does not admit a metric of positive scalar curvature, so for example a connected sum of n copies of S2 x S1 does not admit a ramified cover of T3 (as far as I understand).

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Any conjectures? Clearly there is an onto homomorphism $\pi_1M\to Z^3$; can you do better? –  Anton Petrunin Nov 14 '09 at 23:01
    
I don't really know anything about this question, and can not do better than the homeomorphism to Z^3. For every M^3 its connected sum with several T^3 admits a cover to T^3 :) –  Dmitri Nov 14 '09 at 23:58
    
Conjecture: M is a branched cover of T^3 iff the first homology group of M has rank at least three. –  Sam Nead Nov 15 '09 at 0:59
    
(M assumed to be closed, orientable, connected.) –  Sam Nead Nov 15 '09 at 1:00
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No, in fact I did not read the proof for S^3. Also the existence of onto homeo $\pi_1M\to Z^3$ as well having rank of first homology at lest 3 is no sufficient because covers of T^3 don't admit a metric of positive scalar curvature, so the conncted sum of 3 S^2xS^1 will be a counterexample. –  Dmitri Nov 15 '09 at 10:34
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up vote 11 down vote accepted

Note that a branched covering induces an injection of rational cohomology rings, by transfer considerations. Therefore the cohomology of a manifold that is a branched covering of $T^3$ must contain three classes of degree 1 whose triple cup product is nontrivial.

This condition on a manifold $M^3$ implies the existence of a map $M^3\to T^3$ of nonzero degree. Passing to a covering space of $T^3$ if necessary we obtain such a map that is also surjective on $\pi_1$. Assuming the resulting map is of degree $\ge 3$, the main result of [Edmonds, Allan L.Deformation of maps to branched coverings in dimension three. Math. Ann. 245 (1979), no. 3, 273--279.] implies that this map is homotopic to a branched covering.

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Thank you very much for the answer!!! This is neat. –  Dmitri Nov 15 '09 at 22:51
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