# Differential inequality regarding volume comparison

Let $$(M,g)$$ be a complete $$n$$-dimensional Riemannian manifold and let $$p \in M$$. Consider $$(t,\Theta)$$ , the geodesic spherical coordinates around $$p$$, where $$t \in (0,\text{conj}_p(\Theta))$$ and $$\Theta$$ is a unit vector in $$T_pM$$. Let $$A_p(t,\Theta)$$ be the density of the volume measure in these coordinates, i.e. $$\begin{equation*} d\operatorname{Vol} = A_p(t,\Theta) dt d\Theta \end{equation*}$$ A well-known theorem of Gromov states that if $$\operatorname{Ric}(M) \geqslant (n-1)\kappa$$, then the map $$$$t \mapsto \frac{{A}_p(t,\Theta)}{sn^{n-1}_{\kappa}(t)}$$$$ is non-increasing in $$t$$. As usual, $$sn_{\kappa}$$ is given by \begin{align*} sn_{\kappa}(t) = \begin{cases} \frac{\sin{\sqrt{k}t}}{\sqrt{k}} & k > 0\\ t & k = 0\\ \frac{\sinh{\sqrt{-k}t}}{\sqrt{-k}} & k < 0 \end{cases} \end{align*} Now I would like to prove a similar result when the sectional curvature of $$M$$ is bounded from above. That is, if $$\text{sec}(M) \leqslant \kappa$$, then

$$\begin{equation*} \frac{d^2}{dt^2}\left(\frac{A_p(t,\Theta)}{sn^{n-2}_{\kappa}(t)}\right) + \kappa \left(\frac{A_p(t,\Theta)}{sn^{n-2}_{\kappa}(t)}\right) \geqslant 0 \end{equation*}$$ I'm trying to mimic the argument given by Gromov, letting $$\varphi(t) = A_p(t,\Theta)^{\frac{1}{n-2}}$$ and calculate that $$(\log \varphi(t))' = \frac{1}{n-2}\text{tr}(\text{II}(t))$$, where $$\text{II}(t)$$ is the second fundamental form of $$\partial B(p,t)$$ . But since we are not proving a statement about monotonicity, I don't know how I can get rid of the power $$(n-2)$$. Differentiating such expression directly seems intimidating and tedious, and I believe there's a shortcut to the problem since it is very similar to the estimate of the norm of Jacobi fields. Any insight of the problem will be appreciated.

• May I know where do you find this problem? Commented Aug 18, 2020 at 15:39
• I'm attending a Riemannian geometry workshop and it's one of the practice problems. After one night of sleep and I woke up to a solution. I will post my answer after I re-checked my proof.
– sz3
Commented Aug 18, 2020 at 15:53
• Just curious, is it a online workshop? Commented Aug 18, 2020 at 15:59
• @Arctic Char Yes, it is!
– sz3
Commented Aug 18, 2020 at 17:14

So my professor gave me an idea of how to solve this problem. After we get $$\begin{equation*} \underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]'}_{\text{Part A}} + \underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]^2}_{\text{Part B}} + \kappa \geqslant 0 \tag{\star} \end{equation*}$$ We can use Riccati's equation to rewrite \begin{align*} \text{Part A} = &[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]' \\ \geqslant & -\text{tr}(\text{II}(t)^2)-(n-1)\kappa - (n-2)[-(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^{2} -\kappa]\\ =& -\text{tr}(\text{II}(t)^2) + (n-2)(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^{2}-\kappa \end{align*} And after expanding out $$\text{Part B}$$, $$\star$$ becomes \begin{align*} &\underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]'}_{\text{Part A}} + \underbrace{[\text{tr}(\text{II}(t))-(n-2)\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}]^2}_{\text{Part B}} + \kappa \\ \geqslant & \text{tr}(\text{II}(t))^2-\text{tr}(\text{II}(t)^2) -2(n-2)\text{tr}(\text{II}(t))\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}' + (n-1)(n-2)(\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\ \geqslant & \text{tr}(\text{II}(t))^2-\text{tr}(\text{II}(t)^2) -2(n-2)\text{tr}(\text{II}(t))\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}' + (\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\ = &\sum_{1,i\neq j,n-1}\lambda_{i}(t)\lambda_j(t) - [\lambda_{i}(t)+\lambda_{j}(t)]\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}' + (\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)})^2\\ =& \sum_{1,i\neq j,n-1}(\lambda_i(t)-\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}')(\lambda_j(t)-\frac{\text{sn}'_{\kappa}(t)}{\text{sn}_{\kappa}(t)}')\\ \geqslant & 0 \end{align*} where $$\lambda_{i}(t), i=1,\dots,n-1$$ are the eigenvalues of $$\text{II}(t)$$. The last inequality follows from Hessian comparison, which is indicated in Corollary 2.4 in Petersen's book.