《几何分析》笔记(6)-体积比较定理

好的情形

我们回忆对$\operatorname{Ric}\ge (n-1)K$的完备流形, $p\in M,$ $\,\forall\,x\in S_p\setminus\{p\},$ 我们有:

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

在法坐标系下, $dV_g=\Theta (r,\theta)dr\wedge dS^{n-1},$

$$\frac{\partial {}\Theta}{\partial {}r}=\Delta r\Theta.$$ $$\frac{\partial {}\log\Theta}{\partial {}r}=\Delta r\le \frac{\partial {}\log sn_k^{n-1}(r)}{\partial {}r}.$$ $$\frac{\partial {}\log \frac{\Theta}{\Theta_k} }{\partial {}r}\le 0.$$

因此$\frac{\Theta}{\Theta_k}$单减, $\lim_{r\rightarrow 0}\frac{\Theta}{\Theta_k}=1,$ 从而$\Theta\le \Theta_k,$ 在$x\in \widehat{S}_p$时.

注意到$r_1<r_2$时, $\frac{\partial^\ast B_{r_1}^n}{r_1}\supset \frac{\partial^\ast B_{r_2}^n}{r_2}.$

$$\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}=\int_{\partial^\ast B_{r_2}^n/r_2}\frac{\Theta(r_2,\theta)}{c_nsn_k^{n-1}(r_2)}dS^{n-1}\le \int_{\partial^\ast B_{r_2}^n/r_2}\frac{\Theta(r_1,\theta)}{c_nsn_k^{n-1}(r_1)}dS^{n-1}\le \frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_1}^g)}{S_k(r_1)}.$$

定理 1. $\operatorname{Ric}\ge (n-1)K$时, $\frac{\mathcal{H}^{n-1}(\partial^\ast B_r^g(p))}{S_k(r)}$单调递减.

推论 2 (Myers’ Thm). $\operatorname{Ric}\ge (n-1)K>0$时, $\operatorname{diam}(M)\le \frac{\pi}{\sqrt{K} }$

标准的做法是第二变分方法. 现在我们利用体积比较定理给出证明.

设$\operatorname{diam}(M)>\frac{\pi}{\sqrt{K} },$ 那么$\,\exists\,p,$ 使得$\partial^\ast B^g_{\frac{\pi}{\sqrt{K} } }(p)\neq \varnothing.$ $\mathcal{H}^{n-1}(\partial^\ast B^g_{\frac{\pi}{\sqrt{K} } }(p))>0.$ 但此时$S_k(\frac{\pi}{\sqrt{K} })=0,$ 比值为$+\infty,$ 与单减性矛盾.

一般情形

事实上, 可以说明一般的体积比较定理:

定理 3. $\operatorname{Ric}\ge (n-1)K,$ $\frac{\mathcal{H}^{n-1}(\partial B_r^g(p))}{S_k(r)}$单调递减.

$$\mathcal{H}^{n-1}(\partial B_r^g(p))=\int_{\partial^\ast B_r^n/r}\Theta(r,\theta)dS^{n-1}+\frac{1}{2}\int_{(\partial B_r^n\cap \widehat{N}_p)/r}\Theta(r,\theta)dS^{n-1}.$$

$\Theta$在$\widehat{C}_p$上良定,

$$\frac{\Theta(r_1,\theta)}{sn_k^{n-1}(r_1)}\le \frac{\Theta(r_2,\theta)}{sn_k^{n-1}(r_2)},\quad r_1>r_2, (r_1,\theta)\in \widehat{S}_p.$$

上式当$(r_1,\theta)\in\widehat{N_p}$时也对. 同时注意到有:

$$(\partial B_{r_2}^n\cap \widehat{N}_p)/r_2\subset \partial^\ast B_{r_1}^n/r_1,\quad r_1<r_2.$$

因此类似的, 通过积分放缩即可得到结论.

定理 4. $\operatorname{Ric}\ge (n-1)K,$ $\frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}$单调递减.

在$T_pM\cong \mathbb{R}^n$上, 定义$\widetilde{\Theta}|_{\widehat{S}_p}=\Theta,$ 其他部分取零. 那么

$$\mathrm{vol}(B_r^g(p))=\int_{B_r^n}\widetilde{\Theta}drdS^{n-1}=\int_0^r\int_{S^{n-1} }\widetilde{\Theta}(r,\theta)dsdS^{n-1}\in AC_{loc}(0,+\infty)$$ $$\begin{aligned} \frac{d {} }{d {}r}\frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}&=\frac{1}{V_k^2(r)}\cdot \left(\int_{\partial^*B^g_r/r}\Theta dS^{n-1}V_k(r)-\int_0^r\int_{\partial^*B^g_s/s}\Theta dS^{n-1}S_k(r)\right)\\ &=\frac{S_k(r)}{V_k(r)}\left(\frac{\mathcal{H}^{n-1}(\partial^* B_r^g)}{S_k(r)}-\frac{\int_0^r\frac{\mathcal{H}^{n-1}(\partial^*B_s^g)ds}{S_k(s)}S_k(s)}{\int_0^rS_k(s)ds}\right)\le 0 \end{aligned}$$

由此即可得到结论.

体积比较定理一般用来估计下界而非上界.

推论 5. $\operatorname{Ric}\ge (n-1)K,$ $\operatorname{diam}M=1,$ 有非塌缩条件$\mathrm{vol}(M)\ge v_0>0.$ 那么$\,\exists\,C_1=C_1(n,k,v_0),C_2=C_2(n,k,v_0)>0,$

$$0<C_1\le \frac{\mathrm{vol}(B_r^g(p))}{r^n}\le C_2.$$

由体积比较定理, 有

$$0<\frac{\mathrm{vol}(M)}{V_k(1)}\le \frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}\le 1.$$

由

$$\max_{[0,1]}\frac{V_k(r)}{r^n}<+\infty, \quad \inf_{[0,1]}\frac{V_k(r)}{r^n}>0$$

即有结论.

推论 6. $\operatorname{Ric}\ge (n-1)K,$ $r_1<r_2,$

$$\frac{\mathrm{vol}(B_{r_2}^g(p)\setminus B_{r_1}^g(p))}{V_k(r_2)-V_k(r_1)}\le \frac{\mathrm{vol}(B_{r_2}^g(p))}{V_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}$$

取$r_1<r<r_2,$ 令$r\rightarrow r_2,$ 即有

$$\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}$$

可选到$r_2’\in (r_1,r_2),$ 使得$\mathcal{H}^{n-1}(\partial^\ast B_{r_2’}^g)=\mathcal{H}^{n-1}(\partial B_{r_2’}^g)$

$$\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}\le \frac{\mathcal{H}^{n-1}(\partial B_{r_2}^g)}{S_k(r_2)}\le\frac{\mathcal{H}^{n-1}(\partial B_{r_2'}^g)}{S_k(r_2')} \le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}$$

推论 7. $\operatorname{Ric}\ge (n-1)K,$ $r_1<r_2,$

$$\frac{\mathcal{H}^{n-1}(\partial B_{r_2}^g(p))}{S_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}$$

推论 8. $\operatorname{Ric}\ge (n-1)K.$ 若$R$满足$\frac{d {}\mathrm{vol}(B_t)}{d {}t}|_R=\mathcal{H}^{n-1}(\partial B_R),$ 则

$$\frac{\mathcal{H}^{n-1}(\partial B_{R}^g(p))}{S_k(R)}\le \frac{\mathrm{vol}(B_{R}^g(p))}{V_k(R)}$$

关于球面和球体的关系式, 还有如下常用的形式: 对于$r<R,$

$$\frac{\mathcal{H}^{n-1}(\partial B^g_R)}{S_k(R)}\le \frac{\mathrm{vol}(B_{R}^g(p)\setminus B_{r}^g(p))}{V_k(R)-V_k(r)}\le \frac{\mathcal{H}^{n-1}(\partial B^g_r)}{S_k(r)}$$

定理 9. $\operatorname{Ric}\ge 0$时, $(M,g)$紧, 或$\mathrm{vol}(B_R^g(p))\ge CR$比线性增长快.

若$(M,g)$非紧, 取$d(p,x_k)=k.$

$$\frac{\mathrm{vol}(B_{k+1}(x_k)\setminus B_{k-1}(x_k))}{\omega_n((k+1)^n-(k-1)^n)}\le\frac{\mathrm{vol}(B_{k-1}(x_k))}{\omega_n(k-1)^n}$$ $$\mathrm{vol}(B_{2k(p)})\ge \mathrm{vol}(B_{k-1}(x_k))\ge Ck\mathrm{vol}(B_{k+1}(x_k)\setminus B_{k-1}(x_k))\ge Ck\mathrm{vol}(B_1(p))$$

注 10. 若$\operatorname{Ric}\ge (n-1)K,$ 若$B_r^g,\partial B_r^g,\partial^\ast B_r^g$任意一项与标准体积比取$1,$ 则$B_r^g(p)\cong B_r^k.$

文章最后更新于 2022-10-03 21:52:50