3
$\begingroup$

In the below picture ,how to show the inequation 1?

In fact,I'm not familiar with Hessian comparison.So, hope a detail answer , Thanks very much.

The below picture is form 194th page of here ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

enter image description here

enter image description here

$\endgroup$
8
  • $\begingroup$ I see what look like several useful references just googling "Hessian comparison". I think Petersen's Riemannian Geometry is also a good reference for this kind of comparison. $\endgroup$ Nov 2 '15 at 6:56
  • $\begingroup$ In case it's not clear to you the "Hessian comparison" is taking place in the estimate $|\nabla^2 s| \le C_2 / s + \sqrt M$. $\endgroup$ Nov 2 '15 at 7:35
  • $\begingroup$ @AnthonyCarapetis Thanks ,I try it . $\endgroup$
    – lanse2pty
    Nov 2 '15 at 9:46
  • $\begingroup$ @AnthonyCarapetis Sorry, at the first line of 1, why $\frac{1}{r}g''\frac{1}{r^2}4s^2\nabla s\cdot\nabla s+\frac{1}{r}g'2\nabla s\cdot \nabla s\leq\frac{C^1}{r}$? $\endgroup$
    – lanse2pty
    Nov 2 '15 at 13:06
  • 1
    $\begingroup$ The comparison theorem estimates the Hessian of the distance function $s$ in terms of the equivalent quantity in a constant-curvature space (here of curvature $-\sqrt M$ if I'm correct). You can get the other estimates by using the fact that $g'$, $g''$ are bounded, $s \le r$ and $|\nabla s| = 1$. $\endgroup$ Nov 2 '15 at 13:14
5
+50
$\begingroup$

The Hessian comparison theorem is:

If the sectional curvatures of a manifold are bounded below by $M$, then the distance function $s(x) = d(p,x)$ satisfies $\nabla^2 s \le \nabla^2_M s_M$, where $s_M$ is the corresponding distance function on the space of constant curvature $M$.

[see e.g. ON THE DISTRIBUTIONAL HESSIAN OF THE DISTANCE FUNCTION by Mantegazza, Mascellani & Uraltsev.]

We know $|R_{ijkl}|<M$ and thus our sectional curvatures $K(e_i,e_j) = R_{ijij}$ satisfy $K > -M$; so we can compare to the hyperbolic space with curvature $-M$. In polar coordinates this space has metric $$g_M = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2$$ where $d \Omega^2$ is the round metric on the unit $(n-1)$-sphere. The Hessian of the distance function is (see e.g. Petersen Chapter 2.3) $$\nabla^2s_M = \frac1{\sqrt M} \sinh(\sqrt M s) \cosh(\sqrt M s) d\Omega^2 = \sqrt M \coth(\sqrt M s) (g_M - ds^2).$$

If we throw away the $-ds^2$ we get the estimate $$|\nabla^2 s_M| \le \sqrt M \coth(\sqrt M s)|g_M|.$$

Since $|g_M| = \sqrt{g^{ij}g_{ij}} = \sqrt n$ depends only on the dimension $n$, the theorem (along with an upper bound for $\coth$ you can try proving) gives us the estimate $$|\nabla^2 s| \le |\nabla^2 s_M| \le \sqrt {nM} \coth \sqrt M s \le \sqrt n \left(\sqrt M + \frac1s\right).$$

It looks like the authors have a slightly better estimate here with $\sqrt M$ instead of $\sqrt {nM}$ - I'm not sure whether the mistake is mine or theirs. (Taking the $-ds^2$ in to account only improves the $\sqrt n$ to $\sqrt{n-1}$.) It doesn't matter anyway, since we can still choose a constant $C_3$ dependent only on dimension that makes it work.

$\endgroup$
11
  • $\begingroup$ Really thanks , if you don't tell me ,I think I still can't understand it after long time. $\endgroup$
    – lanse2pty
    Nov 2 '15 at 14:34
  • $\begingroup$ In $g_M = ds^2 + \frac1M \sinh^2(\sqrt M s) d\Omega^2$, the $s$ of $\sqrt M s$ is $s(x)$ ? How does it ? They are on different manifold . $\endgroup$
    – lanse2pty
    Nov 8 '15 at 1:49
  • $\begingroup$ @lanse2pty: the $s$ is just the radial coordinate on each manifold, I gave them the same name to make the comparison simpler. $\endgroup$ Nov 8 '15 at 2:15
  • $\begingroup$ If so , in the last inequality , $|\nabla^2 s| \le \sqrt n \left(\sqrt M + \frac1s\right).$ the first $s$ is not the second $s $, $\endgroup$
    – lanse2pty
    Nov 8 '15 at 2:26
  • $\begingroup$ @lanse2pty: the comparison theorem could be more precisely stated as $|\nabla^2 s(x)| \le |\nabla^2 s_M(p)|$ where $p$ is a point in the hyperbolic space such that $s_M(p) = s(x)$. You could also think of my equation for $g_M$ as defining a hyperbolic metric on the original manifold in geodesic polar coordinates, in which case they really are the same $s$. $\endgroup$ Nov 8 '15 at 2:30
0
$\begingroup$

I just using this Hessian comparison,which I am familiar with.

enter image description here

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.