Machine-translated from Chinese.
子流形
令$M$为$n$维光滑流形, $f:M\rightarrow X$为浸入.
那么可以取$f^\ast g$为$M$上的诱导度量. 记$p=N-n$为余维数.
我们将$f^\ast (TX)$分解为$TM\oplus(TM)^\perp,$ 分别称为切丛和法丛.
沿$M$取标架场$e_A(m),$ 使得$e_i(m),e_\alpha(m)$分别是$m$处的切向量,
法向量.
记$\theta_A=f^\ast \omega_A,$ $\theta_{AB}=f^\ast \omega_{AB},$
那么$\theta_\alpha=0.$ 这是因为$\theta_\alpha(v)=\omega_\alpha(f_\ast v)=0.$
法向的对偶基作用在切向上取零. 由于$\theta_A$为拉回,
适用于$\omega_A,\omega_{BA}$的方程对$\theta_A,\theta_{BA}$仍满足.
回忆$d\omega_A=\omega_B\wedge\omega_{BA},$
我们进一步有$\theta_i\wedge\theta_{i\alpha}=0.$ 由Cartan引理, 我们有
$$
\theta_{i\alpha}=h_{i\alpha j}\theta_j,\quad h_{i\alpha j}=h_{j\alpha i}.
$$
微分形式
$$
\Theta=\Theta_\alpha\otimes e_\alpha,\quad \Theta_\alpha=h_{i\alpha j}\theta_i\theta_j,
$$
称为$M$在$X$中的第二基本型.
它描述了$M$作为$X$中子流形的最简单的度量信息.
定义平均曲率向量
$$
H=\frac{1}{n}h_{i\alpha i}e_\alpha,
$$
它是$M$上的法向量. $M$称为极小子流形若$H=0.$ 它是全测地的,
若$\Theta=0.$ $1$维全测地子流形就是测地线.
第一变分
设$M$紧致, 可能带边界, 它的总体积由
$$
V=\int_M \theta_1\wedge \cdots\wedge \theta_n
$$
给出. 考虑变分.
取$F:M\times (-\varepsilon,\varepsilon)\rightarrow X,$
满足$F$限制在$M\times \{t\}$上为浸入, $F|_{t=0}=f.$
考虑$M\times I$上的标架场$\{e_A\},$ 条件同前(切向, 法向). 此时记
$$
F^\ast \omega_i=\theta_i+a_idt,\quad F^\ast \omega_\alpha=a_\alpha dt,
$$
$$
F^\ast \omega_{i\alpha}=\theta_{i\alpha}+a_{i\alpha} dt.
$$
限制在$\{t=0\}$上, 这与原先记号相同. $a_Ae_A$称为形变向量.
在$M\times I$上, 将外微分分解为
$$
d=d_M+d_t\frac{\partial {} }{\partial {}t}.
$$
在$X$上, 我们有
$$
d(\omega_1\wedge\cdots\wedge \omega_n)=\omega_\alpha\wedge\Omega_\alpha,
$$
$$
\Omega_\alpha=-\sum_i \omega_1\wedge\cdots\wedge \omega_{i-1}\wedge\omega_{i\alpha}\wedge \omega_{i+1}\wedge\cdots\wedge \omega_n.
$$
拉回到$M$上, 我们得到
$$
LHS=d\{(\theta_1\wedge\cdots\wedge \theta_n)+dt\wedge\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n\},
$$
$$
RHS=-dt\wedge a_\alpha \widetilde\Theta_\alpha,
$$
$$
\widetilde\Theta_\alpha=\sum_i\theta_1\wedge\cdots\wedge \theta_{i-1}\wedge \theta_{i\alpha}\wedge\theta_{i+1}\wedge\cdots\wedge \theta_{n}=h_{i\alpha i}\mathrm{vol}_M.
$$
将带$dt$项的式子取等, 我们得到
$$
\frac{\partial {}\mathrm{vol}_M}{\partial {}t}=d_M\left(\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n\right)-a_\alpha\widetilde\Theta_\alpha
$$
积分就给出了
$$
V'(0)=\int_M -a_\alpha\widetilde\Theta_\alpha+\int_{\partial M}\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n.
$$
当$a_i=0$时, 右端第二项消失, 即形变向量垂直于$\partial M.$
这更是在$\partial M$固定时成立.
而对任意$\{a_\alpha\}$首项为零当且仅当$H=0,$ 即$M$为极小子流形.
故我们有定理
定理 1. 黎曼流形的极小子流形固定边界时体积变分为临界值.
文章最后更新于 2023-03-02 19:47:46
submanifold
Let $M$ be the $n$-dimensional smooth manifold, and $f:M\rightarrow X$ be the immersion.
Then $f^\ast g$ can be taken as the induced metric on $M$. Let $p=N-n$ be the codimension.
We decompose $f^\ast (TX)$ into $TM\oplus(TM)^\perp,$, which are called tangent bundles and normal bundles respectively.
Take the frame field $e_A(m),$ along $M$ so that $e_i(m),e_\alpha(m)$ is the tangent vector at $m$ respectively,
normal vector.
Note $\theta_A=f^\ast \omega_A,$ $\theta_{AB}=f^\ast \omega_{AB},$
So $\theta_\alpha=0.$ this is because $\theta_\alpha(v)=\omega_\alpha(f_\ast v)=0.$
The dual basis action in the normal direction takes zero in the tangential direction. Since $\theta_A$ is pullback,
The equations that apply to $\omega_A,\omega_{BA}$ still satisfy $\theta_A,\theta_{BA}$.
Memories$d\omega_A=\omega_B\wedge\omega_{BA},$
We further have $\theta_i\wedge\theta_{i\alpha}=0.$ By Cartan's lemma, we have
$$
\theta_{i\alpha}=h_{i\alpha j}\theta_j,\quad h_{i\alpha j}=h_{j\alpha i}.
$$
differential form
$$
\Theta=\Theta_\alpha\otimes e_\alpha,\quad \Theta_\alpha=h_{i\alpha j}\theta_i\theta_j,
$$
called $M$ in $X$ second basic type.
It describes $M$ as the simplest metric information of the $X$ neutron manifold.
definition mean curvature vector
$$
H=\frac{1}{n}h_{i\alpha i}e_\alpha,
$$
It is the normal vector on $M$. $M$ is called minimal submanifold if $H=0.$ it is Fully geodesic,
If $\Theta=0.$ $1$ the dimensional complete geodesic submanifold is a geodesic.
first variation
Assume $M$ is compact and may have boundaries, and its total volume is given by
$$
V=\int_M \theta_1\wedge \cdots\wedge \theta_n
$$
Given. Consider variations.
Take $F:M\times (-\varepsilon,\varepsilon)\rightarrow X,$
Satisfying the $F$ limit is immersion on $M\times \{t\}$, $F|_{t=0}=f.$
Consider the frame field $\{e_A\},$ on $M\times I$ with the same conditions as before (tangential direction, normal direction). At this time, record
$$
F^\ast \omega_i=\theta_i+a_idt,\quad F^\ast \omega_\alpha=a_\alpha dt,
$$
$$
F^\ast \omega_{i\alpha}=\theta_{i\alpha}+a_{i\alpha} dt.
$$
is restricted to $\{t=0\}$, which is the same as the original notation. $a_Ae_A$ is called deformation vector.
On $M\times I$, decompose the exterior derivative into
$$
d=d_M+d_t\frac{\partial {} }{\partial {}t}.
$$
On $X$, we have
$$
d(\omega_1\wedge\cdots\wedge \omega_n)=\omega_\alpha\wedge\Omega_\alpha,
$$
$$
\Omega_\alpha=-\sum_i \omega_1\wedge\cdots\wedge \omega_{i-1}\wedge\omega_{i\alpha}\wedge \omega_{i+1}\wedge\cdots\wedge \omega_n.
$$
Pulling back to $M$, we get
$$
LHS=d\{(\theta_1\wedge\cdots\wedge \theta_n)+dt\wedge\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n\},
$$
$$
RHS=-dt\wedge a_\alpha \widetilde\Theta_\alpha,
$$
$$
\widetilde\Theta_\alpha=\sum_i\theta_1\wedge\cdots\wedge \theta_{i-1}\wedge \theta_{i\alpha}\wedge\theta_{i+1}\wedge\cdots\wedge \theta_{n}=h_{i\alpha i}\mathrm{vol}_M.
$$
Equality the equation with the $dt$ term, we get
$$
\frac{\partial {}\mathrm{vol}_M}{\partial {}t}=d_M\left(\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n\right)-a_\alpha\widetilde\Theta_\alpha
$$
Points are given
$$
V'(0)=\int_M -a_\alpha\widetilde\Theta_\alpha+\int_{\partial M}\sum_i (-1)^{i-1}a_i\theta_1\wedge\cdots\wedge \widehat\theta_i\wedge\cdots\wedge \theta_n.
$$
When $a_i=0$, the second term on the right disappears, that is, the deformation vector is perpendicular to $\partial M.$
This is especially true when $\partial M$ is fixed.
For any $\{a_\alpha\}$, the leading term is zero if and only if $H=0,$, that is, $M$ is a minimal submanifold.
Therefore we have theorem
Theorem 1. When the minimal submanifold of a Riemannian manifold has a fixed boundary, the volume change reaches a critical value.
The article was last updated on 2023-03-02 19:47:46
