Timeline for Is a manifold with flat ends of bounded geometry?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2013 at 12:39 | vote | accept | Thomas Rot | ||
| Jan 15, 2013 at 23:44 | answer | added | Sergei Ivanov | timeline score: 15 | |
| Jan 15, 2013 at 17:18 | comment | added | Thomas Rot | @Misha: Thanks. I think I understand the idea in principle, but have to think a little more about the covers appearing in the classification. I would still be very much interested in a more elementary argument, which does depend on the classification. | |
| Jan 15, 2013 at 17:00 | comment | added | Thomas Richard | If you manage to make the "collapsing" argument work, I would be interested ! | |
| Jan 15, 2013 at 14:06 | comment | added | Igor Belegradek | Actually, as Misha says Eschenburg-Schroeder seems to do the job. | |
| Jan 15, 2013 at 14:01 | comment | added | Igor Belegradek | If the answer to the displayed question is no, then there is a sequence of asymptotically flat n-manifolds that collapses to an Alexandrov space of dimension $<n$ and nonnegative curvature. Analysis of this collapse can in principle lead to a contradiction (or counterexamples) but I do not know how to do this. For partial answers check survey of Greene library.msri.org/books/Book30/files/greene.pdf (see page 120-121), and a paper by Petrunin-Tuscmann mis.mpg.de/publications/preprints/1999/prepr1999-47.html. | |
| Jan 15, 2013 at 13:53 | comment | added | Misha | @Thomas Rot: If you use the isometric classification of flat ends quoted in that paper, I think you can answer your own question (affirmatively). The classification is due to Eschenburg and Schroeder. | |
| Jan 15, 2013 at 13:28 | comment | added | Thomas Rot | @Thomas Richard, This paper: math.sciences.univ-nantes.fr/~carron/flat_end.pdf claims that the the number of ends is finite (I think they implicitly assume that $M$ is connected). | |
| Jan 15, 2013 at 12:11 | comment | added | Thomas Richard | The question is interesting. My intuition screams yes but I don't have any better reason than the fact that all the examples I can think off do have positive injectivity radius. I think it would be enough to prove that : 1) there's a finite number of ends, 2) up to a finite cover, these ends are isometric to open sets either in $\mathbb{S}^{n}\times\mathbb{R}$ or in a flat metric cone over $\mathbb{S}^n$ (minus the origin). But I wouldn't be surprised if there is a simpler way. | |
| Jan 15, 2013 at 11:07 | history | edited | Thomas Rot | CC BY-SA 3.0 |
added 28 characters in body
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| Jan 15, 2013 at 10:58 | history | asked | Thomas Rot | CC BY-SA 3.0 |