# How to show the inequation by using Hessian comparison.

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

• 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. Nov 2 '15 at 6:56
• In case it's not clear to you the "Hessian comparison" is taking place in the estimate $|\nabla^2 s| \le C_2 / s + \sqrt M$. Nov 2 '15 at 7:35
• @AnthonyCarapetis Thanks ,I try it . Nov 2 '15 at 9:46
• @AnthonyCarapetis Sorry, at the first line of 1, why $\frac{1}{r}g''\frac{1}{r^2}4s^2\nabla s\cdot\nabla s+\frac{1}{r}g'2\nabla s\cdot \nabla s\leq\frac{C^1}{r}$? Nov 2 '15 at 13:06
• 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 $|\nabla s| = 1$. 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 $$M$$.

We know $$|R_{ijkl}| and thus our sectional curvatures $$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.

• Really thanks , if you don't tell me ,I think I still can't understand it after long time. Nov 2 '15 at 14:34
• 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 . Nov 8 '15 at 1:49
• @lanse2pty: the $s$ is just the radial coordinate on each manifold, I gave them the same name to make the comparison simpler. Nov 8 '15 at 2:15
• If so , in the last inequality , $|\nabla^2 s| \le \sqrt n \left(\sqrt M + \frac1s\right).$ the first $s$ is not the second $s$, Nov 8 '15 at 2:26
• @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 $s$. Nov 8 '15 at 2:30

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