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 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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

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