Timeline for Derivation of yamabe flow
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2014 at 19:41 | vote | accept | Slm2004 | ||
| May 7, 2014 at 4:27 | answer | added | Otis Chodosh | timeline score: 4 | |
| Apr 28, 2014 at 19:00 | comment | added | Deane Yang | When choosing what flow to use, the main thing you want is one for which you can obtain "nice" formulas and desired estimates. This usually means that you can prove curvature bounds using the maximum principle. It should be emphasized that before Hamilton proved his spectacular theorem about positively curved 3-manifolds using the Ricci flow, nobody had studied geometric flows (except the harmonic map flow) in any setting, not even for curves. After that, the lesson learned was to find the simplest possible intrinsic flow that has curvature bounds using the maximum principle. | |
| Apr 27, 2014 at 19:39 | comment | added | Slm2004 | That does not help either. $L^2(g_0)$ will also give you strange power of $u$. @MarkPeletier | |
| Apr 27, 2014 at 19:38 | comment | added | Slm2004 | sounds reasonable. but I do want to have a way to explian why people study this kind of equation other than else. @DeaneYang | |
| Apr 27, 2014 at 15:05 | comment | added | Deane Yang | Also, there is no reason to believe that the $L^2$ gradient flow is a good flow to use here both because you are not looking for the minimum of the energy and because the energy is not an "$L^2$" energy. | |
| Apr 27, 2014 at 14:41 | comment | added | Deane Yang | Here are my imprecise impressions: the Ricci flow is also not the gradient flow of the total scalar curvature. The exact flow used is mostly due to pragmatic reasons, namely you get an easier PDE to work with. In particular, it is the heat operator plus lower order terms. The original motivation for the Ricci flow was not that it minimized any energy but just that it should head towards the solution of the time-independent solution. The Yamabe flow was studied only after Hamilton's work on Ricci flow, and it appears that the exact flow chosen was also for pragmatic reasons. | |
| Apr 27, 2014 at 9:56 | comment | added | Mark Peletier | Could the difference be whether the gradient flow is with respect to the $L^2(g_0)$ metric or the $L^2(g)$ metric? | |
| Apr 27, 2014 at 3:52 | comment | added | Deane Yang | Yes your formula appears to be correct. | |
| Apr 27, 2014 at 0:39 | comment | added | Slm2004 | I am pretty sure my calculation is right. I first saw the $E'$ on simon brendle's paper " a generalization of the yamabe flow for manifolds with boundary"page 629. I also checked by my hand. I think the problem may be we should not use $L^2$ gradient flow. For example, may be the $H^1$ gradient flow works. I don't know. Anyway thank you for your replying. @OtisChodosh | |
| Apr 26, 2014 at 22:04 | comment | added | Deane Yang | As you do your calculation, keep track of how your equations transform if you rescale either $u$ or space by a constant factor. Both sides must scale the same. This will help you figure out where your error is. | |
| Apr 26, 2014 at 21:08 | comment | added | Otis Chodosh | The computation for $E'$ should not be that disgusting. So maybe you're taking a long route and somewhere there's a mistake? The computation of the gradient of the Yamabe functional can be found here: projecteuclid.org/euclid.bams/1183553962, but its not so explicit. But it might help you get on the right track. If you can't figure it out, you should post your computation so that someone can help you out, otherwise its impossible to tell what went wrong. | |
| Apr 26, 2014 at 18:31 | history | edited | Slm2004 | CC BY-SA 3.0 |
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| Apr 26, 2014 at 18:30 | comment | added | Slm2004 | Yes. I notice that. It is just a typo. Thank your very much.@OtisChodosh | |
| Apr 26, 2014 at 15:22 | comment | added | Otis Chodosh | I can't promise this will fix everything, but I think that your expression for $V(t)$ is wrong. The power of $u$ should be $\frac{2n}{n-2}$ rather than $\frac{2n}{n-1}$. (but maybe this was just a typo in your post). | |
| Apr 26, 2014 at 0:55 | review | Close votes | |||
| Apr 28, 2014 at 18:45 | |||||
| Apr 26, 2014 at 0:30 | history | asked | Slm2004 | CC BY-SA 3.0 |