Timeline for A problem on Gauss--Bonnet formula
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 27, 2020 at 20:52 | vote | accept | Anton Petrunin | ||
| Jun 22, 2020 at 20:18 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
2nd curve
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| Jun 22, 2020 at 19:18 | answer | added | Anton Petrunin | timeline score: 1 | |
| Nov 16, 2018 at 17:40 | answer | added | RBega2 | timeline score: 2 | |
| Nov 15, 2018 at 19:55 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
added 430 characters in body
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| Nov 14, 2018 at 23:25 | answer | added | RaphaelB4 | timeline score: 4 | |
| Nov 14, 2018 at 4:37 | comment | added | Michael | @AntonPetrunin, you are right, I missed the positive curvature condition. | |
| Nov 13, 2018 at 23:10 | comment | added | Michael | I don't understand what would prevent these curves from being geodesics. Imagine a surface in $\mathbb{R}^3$ with $x,y$ on the plane where you drew $\gamma$ and $z=\text{smoothed squared distance on the plane from (x,y) to }\gamma$. Intuitively that would make $\gamma$ a geodesic. $(x,y)$ sufficiently far from $\gamma$ would produce $z$ that behaves nicely enough to compactify the surface to $\Sigma$. | |
| Nov 13, 2018 at 19:01 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
edited body
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| Nov 12, 2018 at 22:52 | comment | added | Sylvain JULIEN | The curves that are displayed have non trivial symmetry groups, hence my comment. | |
| Nov 12, 2018 at 21:23 | comment | added | Sylvain JULIEN | As Ricci flow preserves the symmetry, can it be of any use in this problem ? | |
| Nov 12, 2018 at 21:12 | history | asked | Anton Petrunin | CC BY-SA 4.0 |