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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 \begin{equation} t \mapsto \frac{{A}_p(t,\Theta)}{sn^{n-1}_{\kappa}(t)} \end{equation} 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.

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  • $\begingroup$ May I know where do you find this problem? $\endgroup$ Commented Aug 18, 2020 at 15:39
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    $\begingroup$ 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. $\endgroup$
    – sz3
    Commented Aug 18, 2020 at 15:53
  • $\begingroup$ Just curious, is it a online workshop? $\endgroup$ Commented Aug 18, 2020 at 15:59
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    $\begingroup$ @Arctic Char Yes, it is! $\endgroup$
    – sz3
    Commented Aug 18, 2020 at 17:14

1 Answer 1

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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.

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