《几何分析》笔记(5)-Ricci比较定理

准备工作

令$M_k$为$n$维单连通黎曼流形, 截面曲率恒为$k,$ 即空间形式. 令$V_k(r)$为半径为$r$的球的体积, $S_k(r)$为其球面面积. 记

$$g=dr^2+sn_k^2(r)g_{S_{n-1} }$$ $$sn_k(r)=\left\{ \begin{aligned} &\frac{\sin\sqrt{k}r}{\sqrt{k} },&&k>0\\ &r,&&k=0\\ &\frac{\sin\sqrt{-k}r}{\sqrt{-k} },&&k<0 \end{aligned}\right.$$

$sn_k(r)$是方程$f’’+kf=0,$ $f(0)=0,$ $f’(0)=1$的解. 记$r(x)=d(x,p),$ 那么

$$\Delta r(x)=(n-1)\frac{sn_k'(r)}{sn_k(r)}.$$ $$dV_g=sn_k^{n-1}dr_ndS^{n-1}.$$

取$M$为$\operatorname{Ric}\ge (n-1)K$的流形, 那么对于其上的$r(x),$

$$\Delta r_k\ge \Delta r,\quad sn_{k}^{n-1}(r)\ge \Theta(r,\theta).$$

一些结论是, $r(x)\in W_{loc}^{1,p}\cap C^0,$ $\,\forall\,p.$ 这就是Ricci比较定理. 由Bishop在$S_p$上给出了证明. Gromov将其推广到整体上, 统称Bishop-Gromov定理. 它在证明Gromov-Hausdorff定理中起到了重要的作用.

引理 1. 记$\omega$为体积形式, $r=d(x,p).$ 那么,

$$L_{\partial r}\omega=\Delta r\omega,\quad \frac{\partial {}\Theta}{\partial {}r}=\Delta r\Theta,\quad x\in S_p.$$

取平行标准正交基$\{e_i\},$ $\nabla_ie_j=0.$ $\{\theta_i\}$为对偶基. 那么

$$\omega=\theta^1\wedge\cdots\wedge\theta^n.$$ $$L_{\partial_r}\omega=(L_{\partial_r}\theta^1)\wedge\theta^2\wedge\cdots\wedge \theta^n+\cdots+\theta^1\wedge\cdots\wedge \theta^{n-1}\wedge (L_{\partial_r}\theta^n).$$ $$\begin{aligned} (L_{\partial_r}\theta^i)(X)=&\frac{\partial {}\theta^i(X)}{\partial {}r}-\theta^i(L_{\partial_r}X)\\ =&\frac{\partial {}\theta^i(X)}{\partial {}r}-\theta^i(\nabla_{\partial_r}X)+\theta^i(\nabla_X\partial_r)\\ =&(\nabla_{\partial_r}\theta^i)(X)+\theta^i(\nabla_X\partial_r)\\ =&\theta^i(\nabla_X\partial_r) \end{aligned}$$

从而,

$$L_{\partial_r}\theta^i=\theta^i(\nabla_{e_k}\partial_r)\theta^k.$$ $$L_{\partial_r}\omega=\theta^k(\nabla_{e_k}\partial_r)\omega=\operatorname{tr}(Hess_r)\omega=\Delta r\omega.$$ $$\begin{aligned} Hess_f(X,Y)&=XY(f)-d_f(\nabla_XY)\\ &=X\left<{}Y,\nabla_gf\right>-\left<{}\nabla_gf,\nabla_XY\right>\\ &=\left<{}Y,\nabla_X\nabla_gf\right>. \end{aligned}$$

那么对于体积元,

$$\omega=\Theta dr\wedge dS^{n-1},\quad \Delta r\omega=L_{\partial r}\omega=\frac{\partial {}\Theta}{\partial {}r}dr\wedge dS^{n-1}=\frac{\partial {}\Theta}{\partial {}r}\frac{\omega}{\Theta}.$$

比较定理

对于$r(x)=d(x,p)\in C^\infty(S_p\setminus\{p\}),$ $|\nabla_gr(x)|^2\equiv 1.$ $\Delta_g|\nabla_gr|^2=0.$ 由Bochner公式,

$$2|Hess_f|^2+2\left<{}\nabla_g\Delta_g r,\nabla_g r\right>+2\operatorname{Ric}(\nabla_gr,\nabla_gr)=0,$$ $$\frac{\partial {}\Delta r}{\partial {}r}+|Hess_r|^2=-\operatorname{Ric}(\partial_r,\partial_r),$$

由于

$$Hess_r(\partial_r,X)=\left<{}\nabla_{\partial_r}\partial_r,X\right>=0,\quad |Hess_r|^2\ge \frac{(\operatorname{tr}Hess_r)^2}{n-1}=\frac{(\Delta_gr)^2}{n-1}.$$

结合$\operatorname{Ric}\ge (n-1)K,$ 我们有:

$$\frac{\partial {}\Delta r}{\partial {}r}+\frac{(\Delta r)^2}{n-1}\le -(n-1)K.$$

记$f(r)=\Delta_gr,$ 则

$$f'+\frac{f^2}{n-1}\le -(n-1)K.$$

考虑$r=0$附近的渐进形态. 记$r=|x|,$ 局部我们有:

$$\begin{aligned} \Delta_gr&=g^{ij}\left(\frac{\partial^2 {}r}{\partial {}x_i\partial {}x_j}-\Gamma_{ij}^k\frac{\partial {}r}{\partial {}x_k}\right)\\ &=\delta^{ij}\frac{\partial^2 {}r}{\partial {}x_i\partial {}x_j}+(g^{ij}-\delta^{ij})\frac{\partial^2 {}r}{\partial {}x_i\partial {}x_j}-\Gamma_{ij}^k\frac{\partial {}r}{\partial {}x_k}=\frac{1}{r}+O(r). \end{aligned}$$

接下来我们分类讨论:

$K>0$

$$f'\le -((n-1)K+\frac{f^2}{n-1}),\quad \frac{f'}{(n-1)^2K+f^2}\le -\frac{1}{n-1}.$$

两边积分, 利用$\arctan$即可得到,

$$f\le (n-1)\frac{\cos\sqrt{k}r}{\sin\sqrt{k}r}$$

$K=0$

$$f'+\frac{f^2}{n-1}\le 0,\quad f'\le 0.$$

取$T=\min\{t|f(t)=0\}\in (0,+\infty].$

$$f|_{[T,+\infty)}\le 0\quad\Rightarrow\quad f|_{[T,+\infty)}\le \frac{n-1}{r}=(n-1)\frac{sn_0'(r)}{sn_0(r)}.$$

所以只需分析$(0,T)$上的情况.

$$\frac{f'}{f^2}+\frac{1}{n-1}\le 0,\quad \left(\frac{1}{f}\right)'\ge \frac{1}{n-1},\quad \frac{1}{f(r)}\ge \frac{r-r_0}{n-1}+\frac{1}{f({r_0})}.$$

$K<0$

$$f'\le -(n-1)K-\frac{f^2}{n-1},\quad T=\min\{t|-(n-1)K-\frac{f^2}{n-1}=0\}.$$

那么

$$-(n-1)K-\frac{f^2}{n-1}|_{(0,T)}< 0$$

记

$$I=\{t|-(n-1)K-\frac{f^2}{n-1}< 0\}$$

那么在$I$上, 记$\varphi=(n-1)\frac{sn_k’(r)}{sn_k(r)},$ 使得有:

$$\frac{f'}{-(n-1)K-\frac{f^2}{n-1} }\ge 1=\frac{\varphi'}{-(n-1)K-\frac{\varphi^2}{n-1} }.$$

记$F’(t)=\frac{1}{-(n-1)K-\frac{t^2}{n-1} },$ 它在$I$上取负. 那么$F$单调递减,

$$(F(f))'\ge (F(\varphi))',\quad \lim_{r\rightarrow 0}F(f)=\lim_{r\rightarrow 0}F(\varphi)\in (0,+\infty).$$ $$F(f)\ge F(\varphi),\quad f\le \varphi,\quad in\,[0,T].$$

记$\alpha(r)=f(r)-\varphi(r).$ 设$\alpha(r_0)>0.$ 定义$r_1=\max\{r| r<r_0,\alpha(r)=0\}.$ 在$(r_1,r_0)$上,

$$-(n-1)K<\frac{\varphi^2}{n-1}<\varphi<f.$$

因此$-(n-1)K-\frac{f^2}{n-1}<0,$ $(r_1,r_0)\subset I.$ 由同样的方法可以分析出其上有$f(r)\le \varphi(r),$ 矛盾. 从而$\alpha(r)\le 0,$ 即$f\le \varphi.$

综上, 我们证明了定理:

定理 2. 对$\operatorname{Ric}\ge (n-1)K$的完备黎曼流形$M,$ $\,\forall\,x\in S_p\setminus\{p\},$ 我们有:

$$\Delta r(x)\le (n-1)\frac{sn_k'(r)}{sn_k(r)}.$$

文章最后更新于 2022-09-26 19:08:06