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回顾
对于$\operatorname{Cut}_p=N_p\cup Q_p\cup L_p,$
$(Q_p\cup L_p)$维数小于等于$n-2,$且它是闭的. 设$x_k\rightarrow x,$
若$x\notin Q_p,$ 那么$x$处有微分同胚, 从而周围的$x_k$只能在$L_p$中,
这将导致三条线段的(子列)收敛, $x\in L_p.$
记$\partial^\ast B_r^g(p):=\partial B_r^g(p)\cap S_p,$
是$\partial B_r^g(p)$中好的部分,
$\partial B_r^g(p)\cap N_p$是主要的坏的部分. 对几乎处处的$r,$
它是$n-2$维子流形; 对所有的$r,$ 它是一个$n-1$维varifold.
积分公式
对$\exp_p:\widehat{S}_p\rightarrow S_p,$ 取法坐标系
$$
\widehat{g}=\exp_p^\ast (g)=dr\otimes dr+\phi_{ij}(r,\theta)d\theta^i\otimes d\theta^j,
$$
$$
dV_{\widehat{g} }=\theta(r,\theta)dr\wedge dS^{n-1}.
$$
事实上$\Theta$就是$\phi_{ij}$的行列式.
$$
\mathcal{H}^{n-1}(\partial^\ast B_r^g(p))=\int_{(\partial B_r^n(0)\cap \widehat{S}_p)/r}\Theta(r,\theta)dS^{n-1},
$$
$$
\mathcal{H}^{n-1}(\partial B_r^g(p)\cap N_p)=\frac{1}{2}\int_{(\partial B_r^n(0)\cap \widehat{N}_p)/r}\Theta(r,\theta)dS^{n-1}.
$$
对几乎处处的$r,$ $\partial B_r^g(p)$只有第一个式子, 而对坏的$r,$
测度要加上第二行的式子.
应用
对$\dim M=2,$ 记$\operatorname{Cut}_p=N_p\cup R_p,$
$R_p:=L_p\cup Q_p$维数为零.
$$
E:=\{r\in (0,+\infty)|\partial B_r^g(p)\cap R_p\}\neq \varnothing.
$$
命题 1. $\dim E=0.$
由于$\dim R_p=0,$ 固定$\tau>0,$ $\widehat{B}_\infty^\tau(R_p)=0.$
从而$\,\exists\,x_i,r_i,$ 使得$x_i\in R_p,$
$R_p\subset \cup \overline{B}_{r_i}(x_i),$
$\sum_{i=1}^\infty r_i^\tau<\varepsilon.$
$\,\forall\,r\in E,$ $\,\exists\,x\in R_p\cap \partial B_r^g(p),$
$\,\exists\,i,$ 使得$x\in \overline{B}_{r_i}(x_i).$ 记$a_i:=d(p,x_i),$
$$
|r-a_i|=|d(p,x)-a_i|\le d(x,x_i)\le r_i.
$$
从而$r\in \cup [a_i-r_i,a_i+r_i]\supset E.$
$$
\widehat{B}^\tau_{\infty}(E)\le \sum \omega(\tau)r_i^\tau\le C\varepsilon.
$$
令$\varepsilon\rightarrow 0$即得$\widehat{B}^\tau_{\infty}(E)=0.$
由$\tau$任意性即有结论.
称$r\in \mathbb{R}_+$是正则的, 若$r\notin E,$
$\partial B_r^g(p)$与$N_p$横截相交. 此时$\partial B_r^g(p)$是分段光滑的.
正则的$r$是几乎处处的, 因此这样的性质也是.
文章最后更新于 2022-09-22 23:12:19
review
For $\operatorname{Cut}_p=N_p\cup Q_p\cup L_p,$
The dimension of $(Q_p\cup L_p)$ is less than or equal to $n-2,$ and it is closed. Let $x_k\rightarrow x,$
If $x\notin Q_p,$ then there is a diffeomorphism at $x$, so the surrounding $x_k$ can only be in $L_p$,
This will result in a (sub-column) convergence of three line segments, $x\in L_p.$
Remember$\partial^\ast B_r^g(p):=\partial B_r^g(p)\cap S_p,$
It is the good part of $\partial B_r^g(p)$,
$\partial B_r^g(p)\cap N_p$ is the main bad part. For almost everywhere $r,$
It is a $n-2$-dimensional submanifold; for all $r,$ it is a $n-1$-dimensional varifold.
integral formula
Get the normal coordinate system for $\exp_p:\widehat{S}_p\rightarrow S_p,$
$$
\widehat{g}=\exp_p^\ast (g)=dr\otimes dr+\phi_{ij}(r,\theta)d\theta^i\otimes d\theta^j,
$$
$$
dV_{\widehat{g} }=\theta(r,\theta)dr\wedge dS^{n-1}.
$$
In fact $\Theta$ is the determinant of $\phi_{ij}$.
$$
\mathcal{H}^{n-1}(\partial^\ast B_r^g(p))=\int_{(\partial B_r^n(0)\cap \widehat{S}_p)/r}\Theta(r,\theta)dS^{n-1},
$$
$$
\mathcal{H}^{n-1}(\partial B_r^g(p)\cap N_p)=\frac{1}{2}\int_{(\partial B_r^n(0)\cap \widehat{N}_p)/r}\Theta(r,\theta)dS^{n-1}.
$$
For almost everywhere $r,$ $\partial B_r^g(p)$ there is only the first formula, but for the bad $r,$
To the measure, add the expression in the second line.
Application
Mark $\operatorname{Cut}_p=N_p\cup R_p,$ for $\dim M=2,$
$R_p:=L_p\cup Q_p$ has zero dimension.
$$
E:=\{r\in (0,+\infty)|\partial B_r^g(p)\cap R_p\}\neq \varnothing.
$$
Proposition 1. $\dim E=0.$
Since $\dim R_p=0,$ is fixed $\tau>0,$ $\widehat{B}_\infty^\tau(R_p)=0.$
Thus $\,\exists\,x_i,r_i,$ makes $x_i\in R_p,$
$R_p\subset \cup \overline{B}_{r_i}(x_i),$
$\sum_{i=1}^\infty r_i^\tau<\varepsilon.$
$\,\forall\,r\in E,$ $\,\exists\,x\in R_p\cap \partial B_r^g(p),$
$\,\exists\,i,$ makes $x\in \overline{B}_{r_i}(x_i).$ remember $a_i:=d(p,x_i),$
$$
|r-a_i|=|d(p,x)-a_i|\le d(x,x_i)\le r_i.
$$
Thus $r\in \cup [a_i-r_i,a_i+r_i]\supset E.$
$$
\widehat{B}^\tau_{\infty}(E)\le \sum \omega(\tau)r_i^\tau\le C\varepsilon.
$$
Let $\varepsilon\rightarrow 0$ get $\widehat{B}^\tau_{\infty}(E)=0.$
There is a conclusion from the arbitrariness of $\tau$.
$r\in \mathbb{R}_+$ is said to be regular if $r\notin E,$
$\partial B_r^g(p)$ intersects $N_p$ transversely. At this time $\partial B_r^g(p)$ is piecewise smooth.
Regular $r$ is almost everywhere, and so is this property.
The article was last updated on 2022-09-22 23:12:19
