Timeline for Which surfaces admit unbounded-length simple geodesics?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2017 at 0:32 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
deleted 28 characters in body
|
| May 12, 2017 at 0:17 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Update w relevant reference.
|
| May 11, 2017 at 21:51 | answer | added | Pengfei | timeline score: 2 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Nov 12, 2015 at 22:00 | comment | added | Anton Petrunin | @foliations, so it seems to be complete answer --- why not to write it as an asnwer :) | |
| Nov 12, 2015 at 21:57 | comment | added | foliations | @AntonPetrunin Agreed. | |
| Nov 12, 2015 at 21:55 | comment | added | foliations | @IgorRivin True, I misread the question. Nevertheless, it seems to me that if the surface is closed and oriented, then the following should be true (by using standard compactness theorems). Any sequence of closed simple geodesics whose length goes to $\infty$, possesses a subsequence that either a) converges to a geodesic lamination with at least one leaf of infinite length or b) converges to a foliation of the surface by closed simple geodesics. Case b) seems fairly restrictive (for instance it should force the surface to be a torus). | |
| Nov 12, 2015 at 21:55 | comment | added | Anton Petrunin | @foliations you should say semi-stable, meaning that a perturbation along a vector field does not decrease it's length in second order. | |
| Nov 12, 2015 at 16:01 | answer | added | Vladimir S Matveev | timeline score: 6 | |
| Nov 12, 2015 at 13:37 | answer | added | Igor Rivin | timeline score: 2 | |
| Nov 12, 2015 at 13:23 | comment | added | Igor Rivin | @foliations the OP is asking for arbitrarily long geodesics, not infinitely long ones, so it is not clear what you mean by "the original geodesic" | |
| Nov 12, 2015 at 13:02 | comment | added | foliations | Stable just means locally length minimizing. So the equator on a sphere is not stable but the neck of a hyperbola of revolution is. | |
| Nov 12, 2015 at 12:53 | comment | added | Joseph O'Rourke | @foliations: Thanks. Every surface has three simple, closed geodesics (Lusternik-Schnirelmann), but I am not sure what is a "stable" simple closed geodesic. | |
| Nov 12, 2015 at 12:39 | comment | added | foliations | Conversely, if the surface admits a two-sided closed strictly stable simple geodeisc, then it should be possible to construct an arbitrarily long simple geodesic by minimizing in the universal cover of the tubular neighborhood of the closed geodesic . | |
| Nov 12, 2015 at 12:36 | comment | added | foliations | If the surface is compact and oriented (and we really are using the fact that we are on a 2d surface here), then to admit an arbitrarily closed simple geodesic, the surface needs to admit a stable closed simple geodesic. This can be seeing by taking the set theoretic closure of the original geodesic, this closure has the structure of a geodesic lamination and one of the leaves should be the closed geodesic (which is also necessarily stable). | |
| Nov 12, 2015 at 12:29 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |