Timeline for Distinguishing 3-manifolds by homologies of covers
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2013 at 13:04 | vote | accept | Neil Hoffman | ||
| May 8, 2013 at 2:53 | answer | added | Ben Wieland | timeline score: 3 | |
| May 1, 2013 at 16:00 | comment | added | HJRW | Ben - in this case, the homology never gets bigger than $\mathbb{Z}$, so probably the action is trivial for every finite quotient. | |
| Apr 27, 2013 at 15:55 | comment | added | Ben Wieland | I am not familiar with the Sol example, but I would guess that the action of the finite groups on the homology groups is enough to distinguish them, as in Serre's example of $\mathbb Z^{p-1} \rtimes \mathbb Z/p$ of groups with isomorphic profinite completions. The points is that the actions are conjugate over every $\mathbb Z_p$, but not over $\mathbb Z$. | |
| Apr 26, 2013 at 16:47 | comment | added | Misha | Proposition 1.3 in Funar's paper provides the examples of Sol-manifolds where integer homology groups of all finite covers are isomorphic but manifolds are not homeomorphic. | |
| Apr 26, 2013 at 12:41 | comment | added | HJRW | (See, for instance, Corollary 1.4 of L. Funar, 'Torus bundles not distinguished by TQFT invariants'.) | |
| Apr 26, 2013 at 12:40 | comment | added | HJRW | Neil - there exist pairs of Sol manifolds with isomorphic profinite completions. I suspect they will answer your question in the integral case. | |
| Apr 26, 2013 at 12:39 | comment | added | HJRW | It's natural to include in your data the action of $\pi_1M_i$ on $H_{i,k}$, which factors through a finite quotient. If you do, then a positive answer to either question would imply that $\pi_1M_1$ and $\pi_1M_2$ have isomorphic profinite completions. For 3-manifolds with non-solvable fundamental group, this is Question 9.28 in our survey article on 3-manifold groups (arXiv:1205.0202v3). Without this data, it's not so clear to me, but it seems likely that the two questions are equivalent (ie you can reconstruct the action on homology). | |
| Apr 26, 2013 at 12:26 | comment | added | Neil Hoffman | Misha: Thank you for bringing up Sol-manifolds. I wrote the post with integral first homology groups in mind, and so I edited the question to reflect this. Since integral first homology can be often used to distinguish Sol-manifolds, do you know of a pair of Sol-manifolds that would answer the (refined) second question? | |
| Apr 26, 2013 at 12:12 | history | edited | Neil Hoffman | CC BY-SA 3.0 |
I specified I wanted integral homology groups.
|
| Apr 26, 2013 at 11:21 | comment | added | Misha | Neil: I do not think that with the present technology one can prove or disprove this for hyperbolic manifolds. For non-hyperbolic ones, all Sol-manifolds have isomorphic homology. | |
| Apr 26, 2013 at 10:51 | history | asked | Neil Hoffman | CC BY-SA 3.0 |