Timeline for Number of disjoint simple closed geodesics
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
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| Mar 23, 2014 at 4:19 | vote | accept | Ali Taghavi | ||
| Dec 26, 2013 at 10:13 | comment | added | Ali Taghavi | @AntonPetrunin when we pass from a max to a min, the sign of gaussian curvature, changes. so it seems that we can not count the number of disjoint simple closed geodesics. other subject;was my question on $\gamma_{\infty}$ ,trivial?(two my previous comment) | |
| Dec 23, 2013 at 17:53 | comment | added | Anton Petrunin | @AliTaghavi, I think the example in Benoît's comment (which is the same as in Igor's answer) should answer your question; it should be easy to show equality if the necks are almost invisible (i.e., the sphere looks almost like cylinder with two spherical caps). | |
| Dec 23, 2013 at 15:34 | comment | added | Ali Taghavi | What about the remaining part of my question: for a fixed analytic metric g, is the invariant $m$ defined in my question, finite? For a natural number $n$, is there an analytic metric $g$ for which $m=n$? | |
| Dec 23, 2013 at 8:08 | comment | added | Ali Taghavi | @AntonPetrunin Can you explain why $\gamma_{\infty}$ can not have vertex? | |
| Dec 23, 2013 at 7:56 | vote | accept | Ali Taghavi | ||
| Dec 23, 2013 at 8:09 | |||||
| Dec 22, 2013 at 23:13 | comment | added | Anton Petrunin | @AliTaghavi, these are very basic questions, ask any geometer on your math department, he/she should be able to help, or take any book in Riemannian geometry and read about geodesics. | |
| Dec 22, 2013 at 21:33 | comment | added | Ali Taghavi | @AntonPetronin Thank you for the answer. I honestly admit that I need to some reference to underestand the detail of your proof. for example analytic extension, stability... could you please give a reference( with minimal necessary background). Ihave also some question: Is not possible that $\gamma_{\infty}$ would be a triangle (not necessarily closed geodesic)? What is the domain of $\ell$? Why $\ell$ can be well defined? Iapologize if my questions are elementary, but I need to a reference to underestand the details of your proof.Thanks again for your help | |
| Dec 22, 2013 at 20:36 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Dec 22, 2013 at 17:36 | comment | added | Anton Petrunin | @AliTaghavi, right, I proved what you asked, given a Riemannian metric on $\mathbb S^2$ there is at most finite number of disjoint closed simple geodesics. | |
| Dec 22, 2013 at 15:40 | comment | added | Ian Agol | Ok, that's a nice argument, much simpler than what I suggested. Note that the surface could be a Klein bottle too. | |
| Dec 22, 2013 at 6:35 | comment | added | Ali Taghavi | I emphasis on "real analytic" because my question is influenced by Ilyashenk Eckel theorem which says every analytic vector filed on $S^{2}$ has only a finit number of isolated closed orbit. In my question, closed geodesics play the role of closed orbit in the later theorem | |
| Dec 22, 2013 at 6:33 | comment | added | Ali Taghavi | @AntonPetrunin you wrote "according to Igor construction, there is no a universal upper bound" please look at my first question: We fix a reimannian metric (we do not change it) and we search for a uniform upper bound for the number of disjoint simple closed geodesic. In the Igor construction the geometry changes namely for each n, he give a riemannian metric on $S^{2}$ for which there are at least n disjoint simple closed geodesics. | |
| Dec 22, 2013 at 5:53 | comment | added | Anton Petrunin | @IanAgol, are you happy now? | |
| Dec 22, 2013 at 5:52 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Dec 22, 2013 at 5:00 | comment | added | Ian Agol | @AntonPetrunin: I'm aware that the geodesic is stable (in the sense of having no negative eigenvalues of the Jacobi operator). One may use a converging sequence of geodesics to form a Jacobi field. And analyticity should imply that this Jacobi field along the geodesic gives rise to a Jacobi vector field in a neighborhood (and therefore globally by analyticity). But I'm not sure how this argument is made precise, which is what I'd like you to expand on (if that's what you had in mind). | |
| Dec 22, 2013 at 4:23 | comment | added | Anton Petrunin | @IanAgol, I think your argument works perfectly. | |
| Dec 22, 2013 at 4:14 | comment | added | Ian Agol | If there were a sequence of geodesics converging, then by the maximum principle, they would have to be leaves of the foliation. But the curvature should be an analytic function of the area, which is therefore constant, so the constant curvature curves would be geodesics. But I haven't found such an argument (giving a CMC foliated neighborhood) in the literature. | |
| Dec 22, 2013 at 4:14 | comment | added | Ian Agol | Can you expand on the analytic family argument? I'm aware of results of Bohm-Tomi that prove a similar thing for minimal surfaces. But I haven't found the statement you're making in the literature. One way one could prove it is to show that any closed simple geodesic in a surface has a neighborhood foliated by constant curvature curves (these should solve some relative isoperimetric inequality). Intuitively, these curves should be bubbles obtained by blowing air into an annular region near the geodesic. | |
| Dec 22, 2013 at 3:42 | comment | added | Anton Petrunin | @RobertBryant, it is rewritten, now the answer is NO. That is the first time I used analyticity in my live; hopefully I did it right. | |
| Dec 22, 2013 at 3:41 | history | undeleted | Anton Petrunin | ||
| Dec 22, 2013 at 3:41 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| Dec 21, 2013 at 21:58 | history | deleted | Anton Petrunin | via Vote | |
| Dec 21, 2013 at 21:57 | comment | added | Anton Petrunin | @RobertBryant Ups, sorry I thought "distinct". | |
| Dec 21, 2013 at 21:15 | comment | added | Robert Bryant | That's a nice construction, but why would this have an infinite sequence of simple geodesics that are pairwise disjoint? | |
| Dec 21, 2013 at 19:31 | history | answered | Anton Petrunin | CC BY-SA 3.0 |