Timeline for Riemannian manifolds with small geodesics and bounded curvature
Current License: CC BY-SA 3.0
12 events
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| Apr 26, 2013 at 21:46 | history | edited | Malte | CC BY-SA 3.0 |
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| Apr 26, 2013 at 21:21 | history | edited | Malte | CC BY-SA 3.0 |
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| Apr 26, 2013 at 17:41 | answer | added | Anton Petrunin | timeline score: 2 | |
| Apr 26, 2013 at 17:24 | comment | added | Benoît Kloeckner | Why should $\gamma$ disconnect your surface? think of a torus, or a hyperbolic surface with small non-separating systole. | |
| Apr 26, 2013 at 16:41 | history | edited | Malte | CC BY-SA 3.0 |
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| Apr 26, 2013 at 15:44 | comment | added | Rbega | In light of horse with no name's examples I guess you need to assume either that $\gamma$ is separating or $\Omega_0$ has positive genus for what I wrote to work. | |
| Apr 26, 2013 at 15:26 | comment | added | Asaf | For negative curvature, there is the thin-thick decomposition by Margulis' lemma... | |
| Apr 26, 2013 at 14:49 | answer | added | horse with no name | timeline score: 6 | |
| Apr 26, 2013 at 13:29 | comment | added | Rbega | I guess you could rescale the metric so you should be able to prove that $Area(\Omega_0)\geq 2\pi i_g^{-2}$. Perhaps I'm missing something here. | |
| Apr 26, 2013 at 12:45 | comment | added | Malte | Thank you, Rbega. This, of course, doesn't take the length of $\gamma$ into account. (One should expect the volume of $\Omega_i$ to increase as $i_g \rightarrow 0$. | |
| Apr 26, 2013 at 12:24 | comment | added | Rbega | If your on an oriented surface then doesn't Gauss-Bonnet together with your curvature bound tell you that $Area(\Omega_0)\geq 2\pi$? Here $\Omega_0$ is a component of $M\backslash \gamma$. I guess you could do better if the genus of $\Omega_0$ was known to be bigger than 1. | |
| Apr 26, 2013 at 11:55 | history | asked | Malte | CC BY-SA 3.0 |