Machine-translated from Chinese.
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
good situation
We recall the complete manifold for $\operatorname{Ric}\ge (n-1)K$, $p\in M,$
$\,\forall\,x\in S_p\setminus\{p\},$ We have:
$$
\Delta r(x)\le (n-1)\frac{sn_k'(r)}{sn_k(r)}.
$$
In the normal coordinate system,
$dV_g=\Theta (r,\theta)dr\wedge dS^{n-1},$
$$
\frac{\partial {}\Theta}{\partial {}r}=\Delta r\Theta.
$$
$$
\frac{\partial {}\log\Theta}{\partial {}r}=\Delta r\le \frac{\partial {}\log sn_k^{n-1}(r)}{\partial {}r}.
$$
$$
\frac{\partial {}\log \frac{\Theta}{\Theta_k} }{\partial {}r}\le 0.
$$
Therefore $\frac{\Theta}{\Theta_k}$ decreases simply,
$\lim_{r\rightarrow 0}\frac{\Theta}{\Theta_k}=1,$
Therefore $\Theta\le \Theta_k,$ is at $x\in \widehat{S}_p$.
When noticing $r_1<r_2$,
$\frac{\partial^\ast B_{r_1}^n}{r_1}\supset \frac{\partial^\ast B_{r_2}^n}{r_2}.$
$$
\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}=\int_{\partial^\ast B_{r_2}^n/r_2}\frac{\Theta(r_2,\theta)}{c_nsn_k^{n-1}(r_2)}dS^{n-1}\le \int_{\partial^\ast B_{r_2}^n/r_2}\frac{\Theta(r_1,\theta)}{c_nsn_k^{n-1}(r_1)}dS^{n-1}\le \frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_1}^g)}{S_k(r_1)}.
$$
Theorem 1. When $\operatorname{Ric}\ge (n-1)K$, $\frac{\mathcal{H}^{n-1}(\partial^\ast B_r^g(p))}{S_k(r)}$ decreases monotonically.
Corollary 2 (Myers’ Thm). When $\operatorname{Ric}\ge (n-1)K>0$, $\operatorname{diam}(M)\le \frac{\pi}{\sqrt{K} }$
The standard approach is the second variation method. Now we give the proof using the volume comparison theorem.
Let $\operatorname{diam}(M)>\frac{\pi}{\sqrt{K} },$ then $\,\exists\,p,$
Make$\partial^\ast B^g_{\frac{\pi}{\sqrt{K} } }(p)\neq \varnothing.$
$\mathcal{H}^{n-1}(\partial^\ast B^g_{\frac{\pi}{\sqrt{K} } }(p))>0.$
But at this time, the ratio of $S_k(\frac{\pi}{\sqrt{K} })=0,$ is $+\infty,$, which is inconsistent with the single subtractive property.
General situation
In fact, the general volume comparison theorem can be stated:
Theorem 3. $\operatorname{Ric}\ge (n-1)K,$ $\frac{\mathcal{H}^{n-1}(\partial B_r^g(p))}{S_k(r)}$ Decrease monotonically.
$$
\mathcal{H}^{n-1}(\partial B_r^g(p))=\int_{\partial^\ast B_r^n/r}\Theta(r,\theta)dS^{n-1}+\frac{1}{2}\int_{(\partial B_r^n\cap \widehat{N}_p)/r}\Theta(r,\theta)dS^{n-1}.
$$
$\Theta$ is well settled on $\widehat{C}_p$,
$$
\frac{\Theta(r_1,\theta)}{sn_k^{n-1}(r_1)}\le \frac{\Theta(r_2,\theta)}{sn_k^{n-1}(r_2)},\quad r_1>r_2, (r_1,\theta)\in \widehat{S}_p.
$$
The above formula is also correct when $(r_1,\theta)\in\widehat{N_p}$. Also note that:
$$
(\partial B_{r_2}^n\cap \widehat{N}_p)/r_2\subset \partial^\ast B_{r_1}^n/r_1,\quad r_1<r_2.
$$
Therefore, similarly, the conclusion can be obtained through integral scaling.
Theorem 4. $\operatorname{Ric}\ge (n-1)K,$ $\frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}$ Decrease monotonically.
On $T_pM\cong \mathbb{R}^n$,
Define $\widetilde{\Theta}|_{\widehat{S}_p}=\Theta,$ and take other parts to zero. Then
$$
\mathrm{vol}(B_r^g(p))=\int_{B_r^n}\widetilde{\Theta}drdS^{n-1}=\int_0^r\int_{S^{n-1} }\widetilde{\Theta}(r,\theta)dsdS^{n-1}\in AC_{loc}(0,+\infty)
$$
$$
\begin{aligned}
\frac{d {} }{d {}r}\frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}&=\frac{1}{V_k^2(r)}\cdot \left(\int_{\partial^*B^g_r/r}\Theta dS^{n-1}V_k(r)-\int_0^r\int_{\partial^*B^g_s/s}\Theta dS^{n-1}S_k(r)\right)\\
&=\frac{S_k(r)}{V_k(r)}\left(\frac{\mathcal{H}^{n-1}(\partial^* B_r^g)}{S_k(r)}-\frac{\int_0^r\frac{\mathcal{H}^{n-1}(\partial^*B_s^g)ds}{S_k(s)}S_k(s)}{\int_0^rS_k(s)ds}\right)\le 0
\end{aligned}
$$
The conclusion can be drawn from this.
The volume comparison theorem is generally used to estimate lower bounds rather than upper bounds.
Corollary 5. $\operatorname{Ric}\ge (n-1)K,$ $\operatorname{diam}M=1,$ There is a non-collapse condition $\mathrm{vol}(M)\ge v_0>0.$ then $\,\exists\,C_1=C_1(n,k,v_0),C_2=C_2(n,k,v_0)>0,$
$$
0<C_1\le \frac{\mathrm{vol}(B_r^g(p))}{r^n}\le C_2.
$$
According to the volume comparison theorem, we have
$$
0<\frac{\mathrm{vol}(M)}{V_k(1)}\le \frac{\mathrm{vol}(B_r^g(p))}{V_k(r)}\le 1.
$$
by
$$
\max_{[0,1]}\frac{V_k(r)}{r^n}<+\infty, \quad \inf_{[0,1]}\frac{V_k(r)}{r^n}>0
$$
There is a conclusion.
Corollary 6. $\operatorname{Ric}\ge (n-1)K,$ $r_1<r_2,$
$$
\frac{\mathrm{vol}(B_{r_2}^g(p)\setminus B_{r_1}^g(p))}{V_k(r_2)-V_k(r_1)}\le \frac{\mathrm{vol}(B_{r_2}^g(p))}{V_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}
$$
Take $r_1<r<r_2,$ and let $r\rightarrow r_2,$ get
$$
\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}
$$
Optional to $r_2'\in (r_1,r_2),$
Make$\mathcal{H}^{n-1}(\partial^\ast B_{r_2'}^g)=\mathcal{H}^{n-1}(\partial B_{r_2'}^g)$
$$
\frac{\mathcal{H}^{n-1}(\partial^\ast B_{r_2}^g)}{S_k(r_2)}\le \frac{\mathcal{H}^{n-1}(\partial B_{r_2}^g)}{S_k(r_2)}\le\frac{\mathcal{H}^{n-1}(\partial B_{r_2'}^g)}{S_k(r_2')} \le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}
$$
Corollary 7. $\operatorname{Ric}\ge (n-1)K,$ $r_1<r_2,$
$$
\frac{\mathcal{H}^{n-1}(\partial B_{r_2}^g(p))}{S_k(r_2)}\le \frac{\mathrm{vol}(B_{r_1}^g(p))}{V_k(r_1)}
$$
Corollary 8. $\operatorname{Ric}\ge (n-1)K.$ If $R$ satisfies $\frac{d {}\mathrm{vol}(B_t)}{d {}t}|_R=\mathcal{H}^{n-1}(\partial B_R),$ then
$$
\frac{\mathcal{H}^{n-1}(\partial B_{R}^g(p))}{S_k(R)}\le \frac{\mathrm{vol}(B_{R}^g(p))}{V_k(R)}
$$
Regarding the relationship between the sphere and the sphere, there are also the following commonly used forms: For $r<R,$
$$
\frac{\mathcal{H}^{n-1}(\partial B^g_R)}{S_k(R)}\le \frac{\mathrm{vol}(B_{R}^g(p)\setminus B_{r}^g(p))}{V_k(R)-V_k(r)}\le \frac{\mathcal{H}^{n-1}(\partial B^g_r)}{S_k(r)}
$$
Theorem 9. When $\operatorname{Ric}\ge 0$, $(M,g)$ is tight, or $\mathrm{vol}(B_R^g(p))\ge CR$ grows faster than linearly.
If $(M,g)$ is not compact, take $d(p,x_k)=k.$
$$
\frac{\mathrm{vol}(B_{k+1}(x_k)\setminus B_{k-1}(x_k))}{\omega_n((k+1)^n-(k-1)^n)}\le\frac{\mathrm{vol}(B_{k-1}(x_k))}{\omega_n(k-1)^n}
$$
$$
\mathrm{vol}(B_{2k(p)})\ge \mathrm{vol}(B_{k-1}(x_k))\ge Ck\mathrm{vol}(B_{k+1}(x_k)\setminus B_{k-1}(x_k))\ge Ck\mathrm{vol}(B_1(p))
$$
Note 10. If $\operatorname{Ric}\ge (n-1)K,$ If the ratio of any item of $B_r^g,\partial B_r^g,\partial^\ast B_r^g$ to the standard volume is $1,$, then $B_r^g(p)\cong B_r^k.$
The article was last updated on 2022-10-03 21:52:50
