Timeline for Ramified covers of 3-torus
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2009 at 23:27 | history | edited | Greg Kuperberg |
edited tags
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| Nov 15, 2009 at 22:46 | vote | accept | Dmitri Panov | ||
| Nov 15, 2009 at 21:09 | comment | added | Sam Nead | Ah, I don't think that having a connect summand with enough homology suffices. I give up. | |
| Nov 15, 2009 at 20:47 | answer | added | Allan Edmonds | timeline score: 13 | |
| Nov 15, 2009 at 17:04 | comment | added | Sam Nead | Ok - Instead of M having homology of rank at least three, lets assume that M has, as a connect summand, an irreducible manifold with homology of rank at least three. How is that? | |
| Nov 15, 2009 at 10:42 | history | edited | Dmitri Panov | CC BY-SA 2.5 |
added 267 characters in body
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| Nov 15, 2009 at 10:34 | comment | added | Dmitri Panov | 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. | |
| Nov 15, 2009 at 2:13 | comment | added | Ilya Nikokoshev | Could the en.wikipedia.org/wiki/Geometrization_conjecture help here by reducing to 8 Thurston geometries? | |
| Nov 15, 2009 at 2:04 | history | edited | Ilya Nikokoshev | CC BY-SA 2.5 |
retag, spelling, math
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| Nov 15, 2009 at 1:00 | comment | added | Sam Nead | (M assumed to be closed, orientable, connected.) | |
| Nov 15, 2009 at 0:59 | comment | added | Sam Nead | Conjecture: M is a branched cover of T^3 iff the first homology group of M has rank at least three. | |
| Nov 14, 2009 at 23:58 | comment | added | Dmitri Panov | 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 :) | |
| Nov 14, 2009 at 23:01 | comment | added | Anton Petrunin | Any conjectures? Clearly there is an onto homomorphism $\pi_1M\to Z^3$; can you do better? | |
| Nov 14, 2009 at 18:33 | history | asked | Dmitri Panov | CC BY-SA 2.5 |