Timeline for Length and curvature for closed curves in negatively curved spaces
Current License: CC BY-SA 4.0
9 events
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| Nov 25, 2022 at 12:55 | vote | accept | Jean Raimbault | ||
| Nov 23, 2022 at 0:56 | comment | added | Otis Chodosh | I think this paper answers some versions of the question mathscinet.ams.org/mathscinet-getitem?mr=3415662 (not that $K \leq -1$ can be replaced by $K\leq 0$ if one makes the appropriate changes to the proof/statement). | |
| Nov 22, 2022 at 22:53 | history | edited | Anton Petrunin |
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| Nov 22, 2022 at 22:50 | answer | added | Anton Petrunin | timeline score: 4 | |
| May 22, 2020 at 13:39 | history | edited | Jean Raimbault | CC BY-SA 4.0 |
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| May 20, 2020 at 17:34 | comment | added | shurtados | For the hyperbolic plane, a horosphere has constant geodesic curvature 1 and a circle of radius $r$ has constant geodesic curvature $\frac{1}{\tanh{r}}$, so I guess that for $X = \mathbb{H^2}$ we have $f(\epsilon) = \infty$ if $\epsilon < 1$ and $f(\frac{1}{\tanh{r}}) = \text{length}_{\mathbb{H}^2}(\partial B_r) = 2\pi \sinh(r)$. See math.stackexchange.com/questions/2430495/…. | |
| May 20, 2020 at 10:27 | comment | added | YCor | Probably the right version should involve instead a punctual curvature condition given by some measure, so that for instance, it would encompass the fact that in the plane, the sum of $|\pi-\alpha_i|$ for $\alpha_i$ the angles of a closed polygonal path, is at least $2\pi$. The curvature is naturally bounded by some "measure", which in this case is supported by the vertices, while it it non-atomic for a $C^1$-curve. I'm not sure how to define this properly, but say for a curve parameterized with speed one I'd copy the curvature definition and take the derivative in distribution sense. | |
| May 20, 2020 at 10:23 | history | edited | YCor |
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| May 20, 2020 at 9:53 | history | asked | Jean Raimbault | CC BY-SA 4.0 |