Timeline for Is every closed curve in 3D a geodesic on a genus-0 surface?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| May 26, 2015 at 5:31 | comment | added | Andy Putman | @JosephO'Rourke: Thanks for the compliment! My writing backlog is embarrassingly long right now, but I'll think about think about turning this into a brief paper at some point. | |
| May 25, 2015 at 17:09 | comment | added | Joseph O'Rourke | Your proof would make a nice paper for e.g., the American Mathematical Monthly or its equivalent, in case you are so inclined :-). | |
| May 25, 2015 at 11:48 | vote | accept | Joseph O'Rourke | ||
| May 25, 2015 at 2:24 | comment | added | Andy Putman | @JosephO'Rourke: By the way, here's one way to think about the conditions. The "local" condition ensures that we can find a surface with boundary $A$ (either an annulus or a Mobius band) containing $\gamma$ such that $\gamma$ is a geodesic on $A$. The remaining conditions ensure that $A$ can be extended to a sphere. Those conditions are not needed when the set $U$ I define is not dense since in that case there is freedom in choosing $A$, so we can choose it to extend to a sphere. | |
| May 25, 2015 at 2:21 | comment | added | Andy Putman | @JosephO'Rourke : I just edited the answer to include a summary. I give necessary and sufficient conditions for the sphere to exist. These conditions are not necessarily satisfied when $\gamma''$ never vanishes. | |
| May 25, 2015 at 2:20 | history | edited | Andy Putman | CC BY-SA 3.0 |
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| May 24, 2015 at 22:59 | comment | added | Joseph O'Rourke | Because (a) of the iterative nature of your answer, and (b) my lack of understanding of your linking-number obstruction (not your fault---my limitations), I am left unclear on the status of the answer. Is it the case that if $\gamma''$ never vanishes, it is a geodesic on a genus-zero surface? Has the (most generalized) question been answered completely? I.e., have you isolated necessary and sufficient conditions, or are there still pockets to explore? | |
| May 23, 2015 at 17:19 | comment | added | Joseph O'Rourke | :-) $\mbox{}\mbox{}$ | |
| May 23, 2015 at 17:19 | comment | added | Andy Putman | @JosephO'Rourke: Thanks for posting this lovely problem, by the way! Solving it was a fun way of procrastinating on some very boring administrative work... | |
| May 23, 2015 at 17:18 | comment | added | Andy Putman | @IgorRivin: In my edit, I explain briefly why this is not a problem. | |
| May 23, 2015 at 17:17 | comment | added | Andy Putman | @DavidSpeyer: Good point! I was very tired and thought I had an argument for this, but I was wrong. And in fact there are even more obstructions even if it is an annulus. I deleted the comment and edited things to discuss the new obstructions. | |
| May 23, 2015 at 17:16 | history | edited | Andy Putman | CC BY-SA 3.0 |
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| May 23, 2015 at 16:46 | comment | added | Igor Rivin | It is also not completely clear that the union of the two disks is an embedded surface (it seems plausible, but far from obvious, that you can separate them). | |
| May 23, 2015 at 11:29 | comment | added | Joseph O'Rourke | Very clever and surprising construction, arranging a discontinuity in the normals at a $\gamma''=0$ point. So it seems the answer is Yes if $\gamma''$ never vanishes. (Apologies for misspelling your name!) | |
| May 23, 2015 at 11:21 | comment | added | David E Speyer | Also, nice answer! | |
| May 23, 2015 at 11:20 | comment | added | David E Speyer | It seems to me that, even if the surface exists locally, it might be a Mobius strip rather than an annulus. Of course, in this case, $\gamma''$ must still vanish somewhere, as otherwise it would provide a choice of normal direction. For example, take a Mobius strip in $\mathbb{R}^3$ and take a geodesic on it realizing a generator of $\pi_1$. | |
| May 23, 2015 at 5:29 | history | answered | Andy Putman | CC BY-SA 3.0 |