Timeline for An inequality on manifolds with non-negative Ricci curvature
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 30 at 15:02 | vote | accept | Jooh | ||
| May 30 at 4:22 | answer | added | Willie Wong | timeline score: 2 | |
| May 29 at 16:40 | comment | added | Jooh | @WillieWong Yes $\mathrm{Area}(S^{n-1})$ means the area of the standard sphere. By Bishop's theorem, I know $\mathrm{Vol}(B_r(p))\leq \mathrm{Area}(S^{n-1})r^n/n$. So if I fix $$h(r):=\frac{\int_{B_r(p)} f ds^2}{\mathrm{Vol}(B_r(p))},$$ then it is enough to prove $h(0)=f(p)\geq h(r)$. To this end, I want to compute $dh(r)/dr$, but I'm not sure how to do this... Could you explain a little bit more about this? Thanks! | |
| May 29 at 2:53 | comment | added | Willie Wong | This last thing should follow from the Riemannian version of the Raychaudhuri equation. | |
| May 29 at 2:44 | comment | added | Willie Wong | Also, have you tried the obvious thing? That is, take geodesics normal coordinates around $p$, which allows you to write the metric as $g = dr^2 + r^2 \gamma(r)$ where $\gamma(r)$ is a Riemannian metric on $S^{n-1}$. Then your hypotheses imply that $$ 0 \geq r^{n-1} \frac{\partial}{\partial r} \int_{\partial B_r} u dS_r - r^{n-1} \int_{\partial B_r} u (\partial_r dS_r) $$ where $dS_r$ is the volume element of $\gamma(r)$. What you want would follow if the comparison theorem would tell you that $(\partial_r dS_r) \leq 0$. | |
| May 29 at 2:26 | comment | added | Willie Wong | What does Area($S^{n-1}$) mean here? The area of the standard sphere? | |
| May 28 at 17:28 | history | asked | Jooh | CC BY-SA 4.0 |