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体积第一变分
本节考虑测地线的推广: $k$维极小子流形.
设在$\widetilde{M}$中给定了具有边界的浸入紧致子流形:
$$
f:M\rightarrow \widetilde{M}, \quad\dim M=n<\widetilde{n}=\dim\widetilde{M}.
$$
设有$C^\infty$映射$F:M\times [0,\varepsilon]\rightarrow\widetilde{M},$
$f_t(x):=F(x,t).$ 若$f_0=f,$ $f_t|_{\partial M}=f|_{\partial M},$
$f_t$为浸入, 则称$\{M_t\}$为$M$的正常变分, $M_t$为$f_t$的像.
沿给定浸入$f$的变分向量场为$W=dF\left(\left.\frac{d {} }{d {}t}\right|_{t=0}\right).$
体积$V(t):=V(M_t)$指$M$在$f_t^\ast \widetilde{g}$下的体积, 那么当$V'=0$时,
称$M$关于$\widetilde{M}$是稳态的.
若对$M$中每个具$C^\infty$边界$\partial D$的定向紧致子区域$D,$
$D$关于每个正常变分都是稳态的, 则称$M$为极小的.
由于$f:M\rightarrow\widetilde{M}$为浸入, 局部上$f$是一个嵌入,
因此可将$M$与$f(M)$恒同,
将$f(M)$的平均曲率法向量场$\eta$称为$M$在$\widetilde{M}$中的平均曲率法向量场,
$M$的平均曲率为$|\eta|.$
引理 1. 浸入子流形$f:M\rightarrow \widetilde{M}$是极小的, 充要条件是其平均曲率为零.
事实上它是如下更精确定理的直接推论.
命题 2 (体积第一变分). 设$M$是浸入在$\widetilde{M}$中具边界的定向紧致流形, $\{M_t\}$为其正常变分, 则
$$
V'(0)=\int_M\left<{}\eta,W\right>dv.
$$
证: 记$\Omega_t$为$M$上度量$f_t^\ast \widetilde{g}$的体积元. 我们来证明
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=(\operatorname{div}\top W)\Omega_0+\left<{}\eta,W\right>\Omega_0.
$$
一旦式子成立, 那么由Green定理, 以及$W|_{\partial M}=0,$ 即有:
$$
V'(0)=\int_M\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=\int_{\partial M}\left<{}\top W,n\right>dv+\int_M\left<{}\eta,W\right>\Omega_0=\int_M\left<{}\eta,W\right>\Omega_0.
$$
对于体积元, 我们有:
$$
\Omega_t=\sqrt{G_t}{}dx^{1}\wedge\cdots\wedge{}dx^{ {}n}.
$$
记$f_t^\ast \widetilde{g}=(g_t)_{ij}dx^idx^j=\left<{}-,-\right>_t,$
$G_t=\det[(g_t)_{ij}],$
$\Omega_{\mathbb{R} }={}dx^{1}\wedge\cdots\wedge{}dx^{ {}n},$ 那么
$$
\left<{}\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n},\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n}\right>_t=\det\left[\left<{}\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right>_t\right]=\det[(g_t)_{ij}]=G_t.
$$
于是我们有:
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=\left(\left.\frac{d {} }{d {}t}\right|_{t=0}\sqrt{G_t}\right)\Omega_{\mathbb{R} }=\frac{1}{\sqrt{G_0} }\left<{}\widetilde{D}_W\left(\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge\frac{\partial {} }{\partial {}x^n}\right),\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n}\right>_0\Omega_{\mathbb{R} }.
$$
记$\perp W=\nu,$ $\top W=\tau,$ 则:
$$
\begin{aligned}
\widetilde{D}_W\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_W\frac{\partial {} }{\partial {}x^i}\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }W\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }\tau\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}\\
&+\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }\nu\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\left(\operatorname{tr}(X\mapsto \widetilde{D}_X\tau)+\operatorname{tr}A_\nu\right)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }\\
=&(\operatorname{div}\tau+\left<{}\eta,\nu\right>)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }\\
=&(\operatorname{div}\top W+\left<{}\eta,W\right>)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }
\end{aligned}
$$
从而,
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=(\operatorname{div}\top W+\left<{}\eta,W\right>)\sqrt{G_0}\,\Omega_{\mathbb{R} }=(\operatorname{div}\top W)\Omega_0+\left<{}\eta,W\right>\Omega_0.
$$
设$M$是$\widetilde{M}$中的浸入超曲面, 则$M$的平均曲率正好是主曲率之和,
于是我们有如下推论:
推论 3. 超曲面是极小子流形的充要条件是它在每点处主曲率之和为零.
利用Gauss方程, 我们可以得到下面的结果.
推论 4. 非正截面曲率流形的极小子流形具非正Ricci曲率.
通过建立$M$上Laplace算子$\Delta$与$\widetilde{D}^2f$的关系,
我们还有如下的结论:
推论 5. 在非正截面曲率的单连通完备流形$\widetilde{M}$中不存在紧致的极小子流形.
事实上, 局部上取定坐标系后, 平均曲率$|\eta|=0$可由$f$来表示,
这将是一个拟线性的椭圆型方程组,
因此极小子流形的研究与偏微分方程关系重大. 特别地,
当$\widetilde{M}=\mathbb{R}^{n+k}$时, $M$为其中的极小图像当且仅当
$$
\sum_{i,j} g^{ij}\frac{\partial^2 {}f^A}{\partial {}y^i\partial {}y^j}=0,\quad A=n+1,\cdots,n+k,
$$
其中$g_{ij}=\delta_{ij}+\sum_A\frac{\partial {}f^A}{\partial {}y^i}\frac{\partial {}f^A}{\partial {}y^j}.$
体积第二变分
同上假设, 且我们仅讨论$V'(0)=0$基础上的情况, 因为这时第二变分才有意义.
由体积第一变分, 切向分量$\top W$不起作用,
因此不妨总假设$\{M_t\}$是法向正常变分, 即$W\perp f(M).$
设$M$是浸入在$\widetilde{M}$中的极小子流形,
若对每个具$C^\infty$边界$\partial D$的可定向紧致子域$D,$
它的每个法向变分都具非负体积第二变分, 则称$M$是稳定的.
也即稳定的极小子流形对所有具紧支集的法向正常变分,
都具非负的体积第二变分.
对任意同上条件的$D,$
若$\widetilde{M}$中任意与$D$边界相同的$C^\infty$子流形$\overline{D},$
总有$V(\overline{D})\ge V(D),$ 则称该极小子流形是(整体)极小化的.
设$\mathscr{N}(M)$为$M$的法丛, 对每个$X\in \mathscr{T}(M),$ 定义映射
$$
\nabla_X:\mathscr{N}(M)\rightarrow \mathscr{N}(M), \quad \nabla_X\nu=\perp\widetilde{D}_X\nu.
$$
容易验证$\nabla$具有联络的所有性质,
称它是在$M$法丛$\mathscr{N}(M)$上的诱导联络. 类似地,
任意取定$M$中幺正标架场$\{E_i\},$ 我们定义$\nabla$的Laplace算子:
$$
\overline{\Delta}\nu:=\operatorname{tr}\nabla^2\nu=\sum_{i=1}^n \nabla_{E_i}\nabla_{E_i}\nu-\nabla_{D_{E_i}E_i}\nu,\quad \,\forall\,\nu\in \mathscr{N}(M).
$$
对任意标架场$\{E_i\},$
我们再引入$\mathscr{N}(M)$上的Ricci张量$\overline{\operatorname{Ric} }:$
$$
\overline{\operatorname{Ric} }:\mathscr{N}(M)\rightarrow\mathscr{N}(M),\quad \overline{\operatorname{Ric} }(\nu)=\sum_{i=1}^n\widetilde{R}_{E_i\nu}E_i.
$$
对于第二基本形式$A_\nu:M_x\rightarrow M_x,$
对任意$M_x$中标准正交基$\{e_i\},$ 定义算子范数:
$$
|A_\nu|^2:=\sum_{i,j}\left<{}A_\nu(e_i),e_j\right>^2=\sum_{i,j}\left<{}S(e_i,e_j),\nu\right>^2.
$$
令$S^t$表示$S:M_x\otimes M_x\rightarrow M_x^\perp$的转置,
即$S^t:M_x^\perp\rightarrow M_x\otimes M_x,$ 使得
$$
\left<{}S^t(\mu),V\otimes W\right>=\left<{}\mu,S(V,W)\right>,\quad \,\forall\,\mu\in M_x^\perp, \: V,W\in M_x.
$$
那么由线性代数知识($\Vert A\Vert^2=\operatorname{tr}(A^TA)=\operatorname{tr}(AA^T)$),
$$
|A_\nu|^2=\left<{}(S\circ S^t)(\nu),\nu\right>.
$$
命题 6 (体积第二变分). 设$M$是极小地浸入$\widetilde{M}$中的一个具边界定向紧流形, $f:M\rightarrow\widetilde{M},$ 且$\{M_t\}$为$M$的法向正常变分, 变分向量场为$\nu\in \mathscr{N}(M),$ $V(t)=V(M_t).$ 那么,
$$
V''(0)=\int_M\left<{}-\overline{\Delta}\nu-\overline{\operatorname{Ric} }(\nu)-(S\circ S^t)(\nu),\nu\right>dv
$$
证: 首先由体积第一变分, 可以得到:
$$
V'(t)=\int_M\left<{}\eta_t,\nu_t\right>\Omega_t,
$$
其中局部上$\eta_t$是$f_t(M)\subset \widetilde{M}$的平均曲率法向量场,
$\nu_t$是$f_t(M)$上的法向变分向量场. 进一步,
$$
V''(0)=\int_M\left<{}\widetilde{D}_\nu\eta,\nu\right>\Omega+\left<{}\eta,\widetilde{D}_\nu\nu\right>\Omega+\left<{}\eta,\nu\right>\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t.
$$
由于$\eta=0,$ 通过直接的计算, 可以得到:
$$
V''(0)=\int_M\left<{}\widetilde{D}_\nu\eta,\nu\right>\Omega=\int_M\left(\left<{}-\overline{\Delta}\nu-\overline{\operatorname{Ric} }(\nu)-(S\circ S^t)(\nu),\nu\right>+f\right)\Omega,
$$
其中$f$满足$\int_Mf\Omega=0.$ 这样就得到了结论. 具体计算的过程在此略过.
当$M$为$\widetilde{M}$中的极小超曲面, $\widetilde{M}$可定向时,
取单位法向量场$N\in\mathscr{N}(M),$ 那么$\nu=uN.$ 此时$\nabla_XN=0,$
因此$\overline{\Delta} \nu=(\Delta u)N.$ 其次,
$$
\left<{}\overline{\operatorname{Ric} }(\nu),\nu\right>=u^2\sum_{i=1}^n\left<{}\widetilde{\operatorname{Ric} }_{E_iN}E_i,N\right>=u^2\widetilde{\operatorname{Ric} }(N,N).
$$
最后,
$$
\left<{}(S\circ S^t)(\nu),\nu\right>=u^2|A_N|^2=u^2\sum_i p_i^2,
$$
$\{p_i\}$为$M\subset \widetilde{M}$的主曲率. 记$A_N$为$A,$
则此时第二变分公式为:
$$
V''(0)=\int_M(-u\Delta u-u^2(\widetilde{\operatorname{Ric} }(N,N)+|A|^2))dv.
$$
由该式, 我们可推得如下命题:
命题 7. 设$M,\widetilde{M}$可定向, $M$是极小地浸入在$\widetilde{M}$中的超曲面, 则$M$稳定的充要条件是:
$$
\int_M u^2(\widetilde{\operatorname{Ric} }(N,N)+|A|^2)dv\le \int_M |du|^2 dv,\quad \,\forall\,u\in C_c^\infty(M,\mathbb{R}).
$$
定理 8 (Schoen-Yau定理). 设$\widetilde{M}$为具正数量曲率的紧致定向$3$维流形, 则$\widetilde{M}$中不存在亏格为正的, 紧致的稳定极小的浸入$2$维定向子流形.
证: 反证法, 假设$M$是这样的子流形. 由于$M$本身紧致,
可令第二变分公式中$u\equiv 1,$ 则
$$
\int_M(\widetilde{\operatorname{Ric} }(N,N)+|A|^2)dv\le 0.
$$
此外, 可以证明:
$$
\widetilde{\operatorname{Ric} }(N,N)+|A|^2=\frac{1}{2}\sigma-K+\frac{1}{2}|A|^2,\tag*{(\ast )}
$$
其中$\sigma$是$\widetilde{M}$数量曲率在$M$上的限制,
$K$是$M$上诱导度量的Gauss曲率. 积分得到:
$$
0\ge \frac{1}{2}\int_M\sigma dv-\int_MKdv+\frac{1}{2}\int_M|A|^2dv.
$$
由假设, $\int_M\sigma dv>0;$ 由Gauss-Bonnet定理,
$$
\int_MKdv=2\pi \chi(M)=4\pi(1-g)\le 0.
$$
最后结合$\int_M|A|^2dv\ge 0$即可推得矛盾.
于是我们只需证明$(\ast )$式. 取$M_x$的一组标准正交基$\{e_1,e_2\},$
使得第二基本形式$A$是对角化的, 即
$$
A(e_1)=\lambda e_1,\quad A(e_2)=\mu e_2.
$$
由于$M$极小, $\operatorname{tr}A=0,$ $\lambda=-\mu.$
因此$|A|^2=2\lambda^2.$ 令$N=e_3,$
则$\{e_1,e_2,e_3\}$构成$\widetilde{M}_x$的一组标准正交基.
记$\widetilde{K}(\operatorname{span}\{e_i,e_j\})=\widetilde{K}_{ij},$ 则
$$
\sigma=\sum_{i,j}\widetilde{K}_{ij}=2(\widetilde{\operatorname{Ric} }(N,N)+\widetilde{K}_{12}).
$$
由Gauss公式, $\widetilde{K}_{12}=K-\det(A)=K+\lambda^2,$
因此$(\ast )$式即可得证:
$$
\widetilde{\operatorname{Ric} }(N,N)+|A|^2=\left(\frac{1}{2}\sigma-\widetilde{K}_{12}\right)+2\lambda^2=\frac{1}{2}\sigma-K+\lambda^2=\frac{1}{2}\sigma-K+\frac{1}{2}|A|^2.
$$
该定理是如下结论的关键, 与正质量定理密切相关:
定理 9. 若$\mathbb{R}^3$中的黎曼度量在紧集$K$外部是平坦的, 且处处具非负数量曲率, 则它必是平坦的.
最后, 我们来看欧氏空间中极小超曲面的结果.
引理 10. 设$M$是$\mathbb{R}^{n+1}$中极小超曲面. 若$D$是$M$中充分小的区域, 具$C^\infty$边界$\partial D,$ 则对$\mathbb{R}^{n+1}$中每一个使得$\partial D'=\partial D$的超曲面$D',$ 有$V(D')\ge V(D).$
由于局部上,
每个超曲面都是某函数$f:\mathscr{U}\rightarrow \mathbb{R}(\mathscr{U}\subset \mathbb{R}^n)$的图像,
因此该引理为如下引理的直接推论:
引理 11. 设$\mathscr{U}$为$\mathbb{R}^n$中的区域, 具$C^\infty$边界$\partial \mathscr{U}.$ 有$C^\infty$函数$f:\overline{UUU}\rightarrow\mathbb{R},$ 使得$f$在$\overline{\mathscr{U} }\times\mathbb{R}$中的图像$M$是$\mathbb{R}^{n+1}$中的一个(具边界的)极小超曲面. 若$N$是$\overline{\mathscr{U} }\times \mathbb{R}$中的超曲面, 使得$\partial N=\partial M,$ 则$V(N)\ge V(M),$ 且取等当且仅当$N=M.$
证:
令$\nu=\frac{1}{W}\left(\frac{\partial {}f}{\partial {}x^1},\cdots,\frac{\partial {}f}{\partial {}x^n},-1\right)$为$M$在$(x,f(x))$处的单位外法向量,
$W=\sqrt{1+|df|^2}.$
令$\omega$为$\overline{\mathscr{U} }\times\mathbb{R}$中的$n$次微分形式:
$$
\omega:=\sum_{i=1}^{n+1}(-1)^{i-1}\nu_idx^1\wedge\cdots\wedge \widehat{dx^i}\wedge \cdots\wedge dx^{n+1}.
$$
我们来说明$d\omega=0.$ 首先,
$d\omega=\sum_i\frac{\partial {}\nu_i}{\partial {}x^i}{}dx^{1}\wedge\cdots\wedge{}dx^{ {}n+1},$
其次向量场$X_i=\frac{\partial {} }{\partial {}x^i}+\frac{\partial {}f}{\partial {}x^i}\frac{\partial {} }{\partial {}x^{n+1} }$构成了$M_y$的一组基,
同时, 计算可得
$$
A_\nu(X_i)=\frac{\partial {}\nu}{\partial {}x^i}+\frac{\partial {}f}{\partial {}x^i}\frac{\partial {}\nu}{\partial {}x^{n+1} }=\frac{\partial {}\nu}{\partial {}x^i}=\sum_j\frac{\partial {}\nu_j}{\partial {}x^i}X_j.
$$
由于$M$是极小的,
$\operatorname{tr}A_\nu=\sum_i \frac{\partial {}\nu_i}{\partial {}x^i}=0,$
从而$d\omega=0.$
设$\Sigma$为由$M,N$围成的区域, $\partial \Sigma=M-N.$ 因此由Stokes公式,
$$
\int_M\omega-\int_N\omega=\int_{\Sigma}\omega=\int_\Sigma d\omega=0,
$$
从而$\int_M \omega=\int_N\omega.$
由于单位外法向量$\nu$的对偶微分形式为$\widetilde{\nu}=\nu_idx^i,$
有$\widetilde{\nu}\wedge \omega=dx^1\wedge\cdots\wedge dx^{n+1}.$
因此$\omega|_M$就是$M$的体积元. 另一方面, 记$\Omega$为$N$的体积元, 那么
$$
V(M)=\int_N\omega=\int_N\left<{}\omega,\Omega\right>\Omega\le \int_N |\omega||\Omega| \Omega=\int_N \Omega=V(N).
$$
当$M\neq N$时, 取$p\in M,$ $q\in N$投影到$\mathscr{U}$中相同的点,
但$M_{p},N_q$并不平行.
此时在$q$点处$\left<{}\omega,\Omega\right><1$严格成立,
进而不等式在$N$上的小邻域也成立. 从而$V(M)<V(N),$ 命题得证.
推论 12. 如果$f:\mathbb{R}^n\rightarrow \mathbb{R}$的图像是$\mathbb{R}^{n+1}$中的极小超曲面, 那么它是极小化的, 进而是稳定的.
文章最后更新于 2022-04-30 14:19:16
first variation of volume
This section considers the generalization of geodesics: $k$ dimension minimal submanifold.
Let an immersed compact submanifold with boundaries be given in $\widetilde{M}$:
$$
f:M\rightarrow \widetilde{M}, \quad\dim M=n<\widetilde{n}=\dim\widetilde{M}.
$$
Let $C^\infty$ map $F:M\times [0,\varepsilon]\rightarrow\widetilde{M},$
$f_t(x):=F(x,t).$ If $f_0=f,$ $f_t|_{\partial M}=f|_{\partial M},$
$f_t$ is immersed, then $\{M_t\}$ is called $M$ normal variation, $M_t$ is the image of $f_t$.
Along the given immersion $f$ variational vector field for $W=dF\left(\left.\frac{d {} }{d {}t}\right|_{t=0}\right).$
Volume $V(t):=V(M_t)$ refers to the volume of $M$ under $f_t^\ast \widetilde{g}$, then when $V'=0$,
Say $M$ is about $\widetilde{M}$ Steady state of.
If for each directed compact subregion $D,$ with $C^\infty$ boundary $\partial D$ in $M$
$D$ is steady-state with respect to every normal variation, then $M$ is called extremely small of.
Since $f:M\rightarrow\widetilde{M}$ is an immersion, locally $f$ is an embedding.
Therefore, $M$ and $f(M)$ can be identified as
Call the mean curvature normal vector field $\eta$ of $f(M)$ in $M$ in $\widetilde{M}$ Mean curvature normal vector field,
$M$'s mean curvature for $|\eta|.$
Lemma 1. The immersed submanifold $f:M\rightarrow \widetilde{M}$ is extremely small, and the necessary and sufficient condition is that its mean curvature is zero.
In fact it is a direct corollary of the following more precise theorem.
Proposition 2 (First variation of volume). Assume $M$ is a directional compact manifold with boundaries immersed in $\widetilde{M}$, and $\{M_t\}$ is its normal variation, then
$$
V'(0)=\int_M\left<{}\eta,W\right>dv.
$$
Certificate: Let $\Omega_t$ be the volume element measuring $f_t^\ast \widetilde{g}$ on $M$. Let us prove
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=(\operatorname{div}\top W)\Omega_0+\left<{}\eta,W\right>\Omega_0.
$$
Once the formula is established, then according to Green's theorem and $W|_{\partial M}=0,$:
$$
V'(0)=\int_M\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=\int_{\partial M}\left<{}\top W,n\right>dv+\int_M\left<{}\eta,W\right>\Omega_0=\int_M\left<{}\eta,W\right>\Omega_0.
$$
For volume elements, we have:
$$
\Omega_t=\sqrt{G_t}{}dx^{1}\wedge\cdots\wedge{}dx^{ {}n}.
$$
Remember$f_t^\ast \widetilde{g}=(g_t)_{ij}dx^idx^j=\left<{}-,-\right>_t,$
$G_t=\det[(g_t)_{ij}],$
$\Omega_{\mathbb{R} }={}dx^{1}\wedge\cdots\wedge{}dx^{ {}n},$ Then
$$
\left<{}\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n},\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n}\right>_t=\det\left[\left<{}\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right>_t\right]=\det[(g_t)_{ij}]=G_t.
$$
So we have:
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=\left(\left.\frac{d {} }{d {}t}\right|_{t=0}\sqrt{G_t}\right)\Omega_{\mathbb{R} }=\frac{1}{\sqrt{G_0} }\left<{}\widetilde{D}_W\left(\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge\frac{\partial {} }{\partial {}x^n}\right),\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \frac{\partial {} }{\partial {}x^n}\right>_0\Omega_{\mathbb{R} }.
$$
Note $\perp W=\nu,$ $\top W=\tau,$ then:
$$
\begin{aligned}
\widetilde{D}_W\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_W\frac{\partial {} }{\partial {}x^i}\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }W\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }\tau\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}\\
&+\sum_i\frac{\partial {} }{\partial {}x^1}\wedge\cdots\wedge \widetilde{D}_{\frac{\partial {} }{\partial {}x^i} }\nu\wedge \cdots\wedge \frac{\partial {} }{\partial {}x^n}.\\
=&\left(\operatorname{tr}(X\mapsto \widetilde{D}_X\tau)+\operatorname{tr}A_\nu\right)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }\\
=&(\operatorname{div}\tau+\left<{}\eta,\nu\right>)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }\\
=&(\operatorname{div}\top W+\left<{}\eta,W\right>)\frac{\partial {} }{\partial {}{}x^{1} }\wedge\cdots\wedge\frac{\partial {} }{\partial {}{}x^{ {}n} }
\end{aligned}
$$
Thus,
$$
\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t=(\operatorname{div}\top W+\left<{}\eta,W\right>)\sqrt{G_0}\,\Omega_{\mathbb{R} }=(\operatorname{div}\top W)\Omega_0+\left<{}\eta,W\right>\Omega_0.
$$
Assume $M$ is the immersed hypersurface in $\widetilde{M}$, then the mean curvature of $M$ is exactly the sum of the principal curvatures,
So we have the following inference:
Corollary 3. The necessary and sufficient condition for a hypersurface to be a minimal submanifold is that the sum of its principal curvatures at every point is zero.
Using the Gauss equation, we can get the following results.
Corollary 4. The minimal submanifold of a manifold with non-positive cross-section curvature has non-positive Ricci curvature.
By establishing the relationship between Laplace operators $\Delta$ and $\widetilde{D}^2f$ on $M$,
We also have the following conclusions:
Corollary 5. There is no compact minimal submanifold in a simply connected complete manifold $\widetilde{M}$ with non-positive cross-section curvature.
In fact, after taking a fixed coordinate system locally, the mean curvature $|\eta|=0$ can be represented by $f$,
this will be a quasi-linear Elliptic equations,
Therefore, the study of minimal submanifolds is closely related to partial differential equations. In particular,
When $\widetilde{M}=\mathbb{R}^{n+k}$, $M$ is the smallest image among them if and only if
$$
\sum_{i,j} g^{ij}\frac{\partial^2 {}f^A}{\partial {}y^i\partial {}y^j}=0,\quad A=n+1,\cdots,n+k,
$$
Among them $g_{ij}=\delta_{ij}+\sum_A\frac{\partial {}f^A}{\partial {}y^i}\frac{\partial {}f^A}{\partial {}y^j}.$
second variation of volume
The same assumptions as above, and we only discuss the situation based on $V'(0)=0$, because then the second variation is meaningful.
From the first variation of volume, the tangential component $\top W$ does not play a role,
Therefore, we might as well always assume that $\{M_t\}$ is normal normal variation, that is $W\perp f(M).$
Assume $M$ is a minimal submanifold immersed in $\widetilde{M}$,
If for each orientable compact subdomain $D,$ with $C^\infty$ boundary $\partial D$
Each of its normal variations has a non-negative volume second variation, then $M$ is said to be stable.
That is to say, the stable minimal submanifold has normal variation for all normal variations with compact supports,
All have non-negative second variations of volume.
$D,$ for any of the above conditions
If any $C^\infty$ submanifold $\overline{D},$ in $\widetilde{M}$ has the same boundary as $D$
There is always $V(\overline{D})\ge V(D),$, then the minimal submanifold is said to be (overall) minimization.
Let $\mathscr{N}(M)$ be $M$ Fa Cong, define mapping for each $X\in \mathscr{T}(M),$
$$
\nabla_X:\mathscr{N}(M)\rightarrow \mathscr{N}(M), \quad \nabla_X\nu=\perp\widetilde{D}_X\nu.
$$
It is easy to verify that $\nabla$ has all the properties of contact,
Call it on the $M$ cluster $\mathscr{N}(M)$ induced contact.Similarly,
Arbitrarily determining the unitary frame field $\{E_i\},$ in $M$, we define the Laplace operator of $\nabla$:
$$
\overline{\Delta}\nu:=\operatorname{tr}\nabla^2\nu=\sum_{i=1}^n \nabla_{E_i}\nabla_{E_i}\nu-\nabla_{D_{E_i}E_i}\nu,\quad \,\forall\,\nu\in \mathscr{N}(M).
$$
For any frame field $\{E_i\},$
We then introduce $\mathscr{N}(M)$ on Ricci tensor $\overline{\operatorname{Ric} }:$
$$
\overline{\operatorname{Ric} }:\mathscr{N}(M)\rightarrow\mathscr{N}(M),\quad \overline{\operatorname{Ric} }(\nu)=\sum_{i=1}^n\widetilde{R}_{E_i\nu}E_i.
$$
For the second fundamental form $A_\nu:M_x\rightarrow M_x,$
Define the operator norm for any orthonormal basis $\{e_i\},$ in $M_x$:
$$
|A_\nu|^2:=\sum_{i,j}\left<{}A_\nu(e_i),e_j\right>^2=\sum_{i,j}\left<{}S(e_i,e_j),\nu\right>^2.
$$
Let $S^t$ represent the transpose of $S:M_x\otimes M_x\rightarrow M_x^\perp$,
That is, $S^t:M_x^\perp\rightarrow M_x\otimes M_x,$ makes
$$
\left<{}S^t(\mu),V\otimes W\right>=\left<{}\mu,S(V,W)\right>,\quad \,\forall\,\mu\in M_x^\perp, \: V,W\in M_x.
$$
Then from linear algebra knowledge ($\Vert A\Vert^2=\operatorname{tr}(A^TA)=\operatorname{tr}(AA^T)$),
$$
|A_\nu|^2=\left<{}(S\circ S^t)(\nu),\nu\right>.
$$
Proposition 6 (Second variation of volume). Assume $M$ is a boundary-oriented compact manifold that is minimally immersed in $\widetilde{M}$, $f:M\rightarrow\widetilde{M},$ and $\{M_t\}$ is the normal normal variation of $M$, and the variation vector field is $\nu\in \mathscr{N}(M),$ $V(t)=V(M_t).$. Then,
$$
V''(0)=\int_M\left<{}-\overline{\Delta}\nu-\overline{\operatorname{Ric} }(\nu)-(S\circ S^t)(\nu),\nu\right>dv
$$
Certificate: First, from the first variation of volume, we can get:
$$
V'(t)=\int_M\left<{}\eta_t,\nu_t\right>\Omega_t,
$$
Where locally $\eta_t$ is the mean curvature normal vector field of $f_t(M)\subset \widetilde{M}$,
$\nu_t$ is the normal variation vector field on $f_t(M)$. Further,
$$
V''(0)=\int_M\left<{}\widetilde{D}_\nu\eta,\nu\right>\Omega+\left<{}\eta,\widetilde{D}_\nu\nu\right>\Omega+\left<{}\eta,\nu\right>\left.\frac{d {} }{d {}t}\right|_{t=0}\Omega_t.
$$
Since $\eta=0,$ can be obtained through direct calculation:
$$
V''(0)=\int_M\left<{}\widetilde{D}_\nu\eta,\nu\right>\Omega=\int_M\left(\left<{}-\overline{\Delta}\nu-\overline{\operatorname{Ric} }(\nu)-(S\circ S^t)(\nu),\nu\right>+f\right)\Omega,
$$
Among them, $f$ satisfies $\int_Mf\Omega=0.$ and the conclusion is obtained. The specific calculation process is skipped here.
When $M$ is a minimal hypersurface in $\widetilde{M}$ and $\widetilde{M}$ can be oriented,
Take the unit normal vector field $N\in\mathscr{N}(M),$ then $\nu=uN.$ at this time $\nabla_XN=0,$
Therefore $\overline{\Delta} \nu=(\Delta u)N.$ Secondly,
$$
\left<{}\overline{\operatorname{Ric} }(\nu),\nu\right>=u^2\sum_{i=1}^n\left<{}\widetilde{\operatorname{Ric} }_{E_iN}E_i,N\right>=u^2\widetilde{\operatorname{Ric} }(N,N).
$$
Finally,
$$
\left<{}(S\circ S^t)(\nu),\nu\right>=u^2|A_N|^2=u^2\sum_i p_i^2,
$$
$\{p_i\}$ is the main curvature of $M\subset \widetilde{M}$. Let $A_N$ be $A,$
Then the second variation formula at this time is:
$$
V''(0)=\int_M(-u\Delta u-u^2(\widetilde{\operatorname{Ric} }(N,N)+|A|^2))dv.
$$
From this formula, we can derive the following proposition:
Proposition 7. Assume $M,\widetilde{M}$ can be oriented, $M$ is a hypersurface that is minimally immersed in $\widetilde{M}$, then the necessary and sufficient conditions for $M$ to be stable are:
$$
\int_M u^2(\widetilde{\operatorname{Ric} }(N,N)+|A|^2)dv\le \int_M |du|^2 dv,\quad \,\forall\,u\in C_c^\infty(M,\mathbb{R}).
$$
Theorem 8 (Schoen-Yau theorem). Assume $\widetilde{M}$ is a compact directed $3$-dimensional manifold with positive quantity curvature, then there is no compact, stable minimal immersed $2$-dimensional directed submanifold with positive genus in $\widetilde{M}$.
Certificate: Proof by contradiction, suppose $M$ is such a sub-manifold. Since $M$ itself is compact,
We can make $u\equiv 1,$ in the second variation formula then
$$
\int_M(\widetilde{\operatorname{Ric} }(N,N)+|A|^2)dv\le 0.
$$
Furthermore, it can be proven that:
$$
\widetilde{\operatorname{Ric} }(N,N)+|A|^2=\frac{1}{2}\sigma-K+\frac{1}{2}|A|^2,\tag*{(\ast )}
$$
Where $\sigma$ is the limit of the curvature of $\widetilde{M}$ on $M$,
$K$ is the Gauss curvature of the induced metric on $M$. Integration gives:
$$
0\ge \frac{1}{2}\int_M\sigma dv-\int_MKdv+\frac{1}{2}\int_M|A|^2dv.
$$
By assumption, $\int_M\sigma dv>0;$ by Gauss-Bonnet theorem,
$$
\int_MKdv=2\pi \chi(M)=4\pi(1-g)\le 0.
$$
Finally, the contradiction can be deduced by combining $\int_M|A|^2dv\ge 0$.
So we only need to prove the formula $(\ast )$. Take a set of orthonormal bases $\{e_1,e_2\},$ of $M_x$
Make the second fundamental form $A$ diagonal, that is
$$
A(e_1)=\lambda e_1,\quad A(e_2)=\mu e_2.
$$
Since $M$ is extremely small, $\operatorname{tr}A=0,$ $\lambda=-\mu.$
Therefore $|A|^2=2\lambda^2.$ lets $N=e_3,$
Then $\{e_1,e_2,e_3\}$ constitutes a set of orthonormal bases of $\widetilde{M}_x$.
Note $\widetilde{K}(\operatorname{span}\{e_i,e_j\})=\widetilde{K}_{ij},$ then
$$
\sigma=\sum_{i,j}\widetilde{K}_{ij}=2(\widetilde{\operatorname{Ric} }(N,N)+\widetilde{K}_{12}).
$$
According to Gauss formula, $\widetilde{K}_{12}=K-\det(A)=K+\lambda^2,$
Therefore, formula $(\ast )$ can be proved:
$$
\widetilde{\operatorname{Ric} }(N,N)+|A|^2=\left(\frac{1}{2}\sigma-\widetilde{K}_{12}\right)+2\lambda^2=\frac{1}{2}\sigma-K+\lambda^2=\frac{1}{2}\sigma-K+\frac{1}{2}|A|^2.
$$
This theorem is the key to the following conclusion, and positive mass theorem closely related to:
Theorem 9. If the Riemannian metric in $\mathbb{R}^3$ is flat outside the compact set $K$ and has non-negative quantity curvature everywhere, then it must be flat.
Finally, we look at the results of minimal hypersurfaces in Euclidean space.
Lemma 10. Let $M$ be the minimal hypersurface in $\mathbb{R}^{n+1}$. If $D$ is a sufficiently small region in $M$ with $C^\infty$ and boundary $\partial D,$, then for each hypersurface $D',$ in $\mathbb{R}^{n+1}$ such that $\partial D'=\partial D$ has $V(D')\ge V(D).$
Because locally,
Each hypersurface is the image of a certain function $f:\mathscr{U}\rightarrow \mathbb{R}(\mathscr{U}\subset \mathbb{R}^n)$,
Therefore this lemma is a direct corollary of the following lemma:
Lemma 11. Let $\mathscr{U}$ be the region in $\mathbb{R}^n$, with $C^\infty$ boundary $\partial \mathscr{U}.$ and $C^\infty$ function $f:\overline{UUU}\rightarrow\mathbb{R},$ such that the image $M$ of $f$ in $\overline{\mathscr{U} }\times\mathbb{R}$ is a minimal hypersurface (with boundary) in $\mathbb{R}^{n+1}$. If $N$ is a hypersurface in $\overline{\mathscr{U} }\times \mathbb{R}$, such that $\partial N=\partial M,$ then $V(N)\ge V(M),$ And take equality if and only if $N=M.$
Certificate:
Let $\nu=\frac{1}{W}\left(\frac{\partial {}f}{\partial {}x^1},\cdots,\frac{\partial {}f}{\partial {}x^n},-1\right)$ be the unit outer normal vector of $M$ at $(x,f(x))$,
$W=\sqrt{1+|df|^2}.$
Let $\omega$ be the $n$ differential form of $\overline{\mathscr{U} }\times\mathbb{R}$:
$$
\omega:=\sum_{i=1}^{n+1}(-1)^{i-1}\nu_idx^1\wedge\cdots\wedge \widehat{dx^i}\wedge \cdots\wedge dx^{n+1}.
$$
Let’s explain $d\omega=0.$ first,
$d\omega=\sum_i\frac{\partial {}\nu_i}{\partial {}x^i}{}dx^{1}\wedge\cdots\wedge{}dx^{ {}n+1},$
Secondly, the vector field $X_i=\frac{\partial {} }{\partial {}x^i}+\frac{\partial {}f}{\partial {}x^i}\frac{\partial {} }{\partial {}x^{n+1} }$ constitutes a set of bases of $M_y$,
At the same time, it can be calculated
$$
A_\nu(X_i)=\frac{\partial {}\nu}{\partial {}x^i}+\frac{\partial {}f}{\partial {}x^i}\frac{\partial {}\nu}{\partial {}x^{n+1} }=\frac{\partial {}\nu}{\partial {}x^i}=\sum_j\frac{\partial {}\nu_j}{\partial {}x^i}X_j.
$$
Since $M$ is extremely small,
$\operatorname{tr}A_\nu=\sum_i \frac{\partial {}\nu_i}{\partial {}x^i}=0,$
Thus $d\omega=0.$
Let $\Sigma$ be the area surrounded by $M,N$, $\partial \Sigma=M-N.$ therefore according to Stokes formula,
$$
\int_M\omega-\int_N\omega=\int_{\Sigma}\omega=\int_\Sigma d\omega=0,
$$
Thus $\int_M \omega=\int_N\omega.$
Since the dual differential form of the unit external normal vector $\nu$ is $\widetilde{\nu}=\nu_idx^i,$
Yes $\widetilde{\nu}\wedge \omega=dx^1\wedge\cdots\wedge dx^{n+1}.$
Therefore $\omega|_M$ is the volume element of $M$. On the other hand, let $\Omega$ be the volume element of $N$, then
$$
V(M)=\int_N\omega=\int_N\left<{}\omega,\Omega\right>\Omega\le \int_N |\omega||\Omega| \Omega=\int_N \Omega=V(N).
$$
When $M\neq N$, take $p\in M,$ $q\in N$ and project it to the same point in $\mathscr{U}$,
But $M_{p},N_q$ is not parallel.
At this time, $\left<{}\omega,\Omega\right><1$ is strictly established at point $q$,
Furthermore, the inequality also holds in the small neighborhood of $N$. Thus the proposition of $V(M)<V(N),$ is proved.
Corollary 12. If the image of $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is a minimal hypersurface in $\mathbb{R}^{n+1}$, then it is minimized and therefore stable.
The article was last updated on 2022-04-30 14:19:16
