Timeline for Ricci flow descending from an universal cover
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2023 at 16:00 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Jan 21, 2017 at 17:07 | comment | added | Sylvain Maillot | Nice answer... to make it correct you need of course to assume all metrics are complete. | |
| Nov 23, 2012 at 19:40 | comment | added | Otis Chodosh | Oh, sorry I missed the update! | |
| Nov 23, 2012 at 18:19 | comment | added | malik | I specified it in my post above (see "EDIT"). In short my question was, why $g_p(t)(X,Y) := \tilde{g}_{f^{-1}(p)}(t)(df^{-1}X,df^{-1}Y)$ is well defined for $t>0$. But this was answered by Robert, so I updated my post. | |
| Nov 23, 2012 at 17:32 | comment | added | Otis Chodosh | Also just to be precise, you could use any ricci flow with bounded curvature although I do not see any point in passing to the universal cover. | |
| Nov 23, 2012 at 17:30 | comment | added | Otis Chodosh | Sorry, I don't understand "the question, how this is done still remains." Could you rephrase your exact question? | |
| Nov 23, 2012 at 9:35 | comment | added | malik | @Otis Thank you for the patience to summarize these results. So given a manifold of bounded curvature I could use any Ricci flow solution $(\tilde{M},\tilde{g}(t))$ with $\tilde{g}(0)=\tilde{g}$ on the universal cover $(\tilde{M},\tilde{g})$ to obtain a solution on my original manifold $(M,g)$. (The question, how this is done, still remains.) | |
| Nov 23, 2012 at 5:29 | history | answered | Otis Chodosh | CC BY-SA 3.0 |