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高度函数
首先考虑外围空间$X$为$N$维欧氏空间$E^N$的情形.
此时所有切空间可以等同于$E^N.$ 考虑浸入子流形$x:M\rightarrow E^n.$ 此时
$$
dx=\theta_i e_i,\quad de_i=\theta_{ij}e_j+\theta_{i\alpha}e_\alpha=\theta_{ij}e_j+h_{i\alpha j}\theta_j e_\alpha.
$$
固定一个向量$a\in E^N,$ 那么取$M$上函数$f=\left<{}a,x\right>,$
表示沿$a$的高度函数. 那么,
$$
f_i\theta_i=df=\left<{}a,dx\right>=\left<{}a,e_i\right>\theta_i,
$$
$$
f_{ij}\theta_j=d f_i+f_j\theta_{ji}=\left<{}a,de_i\right>+\left<{}a,e_j\right>\theta_{ji}=\left<{}a,\theta_{ij}e_j+h_{i\alpha j}\theta_j e_\alpha+\theta_{ji}e_j\right>=\left<{}a,h_{i\alpha j}\theta_je_\alpha\right>.
$$
这就给出了
$$
\Delta f=f_{ii}=\left<{}a,h_{i\alpha i}e_\alpha\right>=n\left<{}a,H\right>.
$$
这在$n>0$时说明了如下定理
定理 1. 欧氏空间中的浸入子流形极小当且仅当在诱导度量下, 全部坐标函数都是调和函数. 进而欧氏空间中没有紧致无边极小子流形.
这是因为紧致无边流形上的调和函数都是常数.
曲率
回忆
$$
d\omega_{AB}-\omega_{AC}\wedge\omega_{CB}=-\frac{1}{2}\widehat R_{ABCD}\omega_C\wedge\omega_D.
$$
注意到欧氏空间中曲率为零. 因此拉回得到
$$
d\theta_{ij}-\theta_{ik}\wedge\theta_{kj}=-\theta_{i\alpha}\wedge\theta_{j\alpha}=-\frac{1}{2}(h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k})\theta_k\wedge\theta_l.
$$
记$S_{ijkl}$为$M$上度量的曲率张量, 那么
$$
S_{ijkl}=h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k}.
$$
事实上, 这是高斯方程.
若$M$是极小的, 那么它的Ricci张量为
$$
S_{ik}=S_{ijkj}=-h_{i\alpha j}h_{k\alpha j},
$$
是半负定的.
进一步它的数量曲率为
$$
S=S_{ii}=-h_{i\alpha j}^2\le 0.
$$
因此我们有如下定理
定理 2. 极小子流形$M$的Ricci张量是半负定的. 它是全测地的(为线性子空间)当且仅当它的数量曲率为零.
极小超曲面
此时余维数为$1,$ 简记$h_{i,n+1,j}=h_{ij}.$
那么$\theta_{i,n+1}=h_{ij}\theta_j.$ 我们有
$$
\begin{aligned}
0&=d\theta_{i,n+1}-\theta_{iC}\theta_{C,n+1}\\
&=dh_{ij}\wedge\theta_j+h_{ij}\theta_k\wedge\theta_{kj}-h_{jk}\theta_{ij}\wedge\theta_k\\
&=(dh_{ik}+h_{ij}\theta_{jk}+h_{jk}\theta_{ji})\wedge\theta_k\\
&=h_{ik,j}\theta_j\wedge\theta_k
\end{aligned}
$$
进而$h_{ik,j}=h_{ij,k}.$ 这就是Codazzi方程.
由于$\theta_{n+1}=0,$
$$
0=d\theta_{n+1}=\theta_i\wedge\theta_{i,n+1}=h_{ij}\theta_i\wedge\theta_j.
$$
因此$h_{ij}=h_{ji}.$ 于是$h_{ij,k}$三变元都是对称的.
于是极小超曲面任意两下标求和均为零.
记$f=\left<{}a,e_{n+1}\right>,$ 那么
$$
d e_{n+1}=\theta_{n+1,i}e_i=-h_{ij}\theta_je_i,
$$
$$
f_i=-\left<{}a,e_j\right>h_{ji},
$$
$$
f_{ij}\theta_j=-\left<{}a,\theta_{jk}e_k+h_{jk}\theta_ke_{n+1}\right>h_{ij}-\left<{}a,e_j\right>dh_{ij}-\left<{}a,e_k\right>h_{kj}\theta_{ji}.
$$
由
$$
dh_{ij}+h_{kj}\theta_{ki}+h_{ik}\theta_{kj}=h_{ij,k}\theta_k,
$$
$$
f_{ij}\theta_j=-\left<{}a,e_{n+1}\right>h_{ij}h_{jk}\theta_k-\left<{}a,e_j\right>h_{ij,k}\theta_k,
$$
$$
f_{ik}=-\left<{}a,e_{n+1}\right>h_{ij}h_{jk}-\left<{}a,e_j\right>h_{ij,k},
$$
$$
\Delta f=f_{ii}=-\left<{}a,e_{n+1}\right>h_{ij}^2=\left<{}a,e_{n+1}\right>S.
$$
考虑形如$z=z(x_1,\cdots,x_n)$的极小超曲面. 那么切向量为
$$
\frac{\partial {}r}{\partial {}x_i}=(0,\cdots,1,\cdots,0,p_i),\quad p_i=\frac{\partial {}z}{\partial {}x_i}.
$$
$$
g_{ij}=\delta_{ij}+p_ip_j.
$$
$$
e_{n+1}=\frac{1}{W}(p_1,\cdots,p_n,-1),\quad W=\sqrt{1+p_k^2}.
$$
该曲面有到超平面的一一映射. Bernstein问题为是否极小曲面总是超平面.
我们希望证明$z=a_ix_i+a_{n+1}.$ 但是这在$n\ge 8$时有反例.
现考虑$n=2$的情形.
希望利用$\Delta \left<{}a,e_{n+1}\right>=S\left<{}a,e_{n+1}\right>.$
取$a=(0,\cdots,0,1),$ 此时$\left<{}a,e_{n+1}\right>=-\frac{1}{W}.$ 因此,
$$
\Delta(\frac{1}{W})=\frac{2K}{W}\le 0.
$$
那么$\frac{1}{W}$是上调和的正值函数, 必为常数. 这就说明了结论.
或者我们有高斯方程
$$
S_{ijkl}=h_{ik}h_{jl}-h_{il}h_{jk},\quad S_{ik}=nHh_{ik}-h_{ij}h_{jk}.
$$
曲面极小时, 有
$$
S_{ik}=K\delta_{ik}=-h_{il}h_{lk}.
$$
由于
$$
\Delta\log (1+\frac{1}{W})=\frac{1}{1+\frac{1}{W} }\Delta(\frac{1}{W})-\frac{1}{(1+\frac{1}{W})^2}|\nabla (\frac{1}{W})|^2.
$$
设$f=1+\frac{1}{W},$ 那么
$$
\Delta \log f=\frac{\Delta f}{f}-\frac{f_k^2}{f^2}.
$$
回忆
$$
\left<{}a,e_3\right>_i=-\left<{}a,e_j\right>h_{ij}.
$$
由于$e_3=\frac{1}{W}(p_1,p_2,-1),$
$$
|\nabla (\frac{1}{W})|^2=|\nabla\left<{}a,e_3\right>|^2=\left<{}a,e_3\right>_i^2=\left<{}a,e_j\right>h_{ij}\left<{}a,e_k\right>h_{ik}=-K\left<{}a,e_j\right>^2.
$$
由于
$$
(0,0,1)=a=\left<{}a,e_i\right>e_i+\left<{}a,e_3\right>e_3=\left<{}a,e_i\right>e_i-\frac{1}{W}e_3,
$$
$$
1=\left<{}a,e_i\right>^2+(\frac{1}{W})^2.
$$
因此
$$
\Delta(\log (1+\frac{1}{W}))=\frac{1}{1+\frac{1}{W} }\frac{2K}{W}+\frac{1}{(1+\frac{1}{W})^2}(-K)(1-(\frac{1}{W})^2)=K.
$$
考虑一个新度量$d\sigma=(1+\frac{1}{W})ds,$ $ds$由浸入诱导.
$d\sigma\ge ds$ 是完备的. 断言在$\sigma$下, $\widetilde K\equiv 0,$
进而它是可展曲面. 那么,
$$
d\sigma^2=d\xi^2+d\eta^2.
$$
$$
\Delta_{ds^2}=\Lambda(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2}),\quad \Lambda=(1+\frac{1}{W})>0.
$$
已知$K\le 0.$ 因此
$$
0\ge K=\Delta\log(1+\frac{1}{W})= \Lambda(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2})(\log (1+\frac{1}{W})),
$$
$$
(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2})\log(1+\frac{1}{W})\le 0,
$$
上调和正函数为常数, 因此$K\equiv 0.$ 由前面的定理, 它是全测地的,
进而为超平面.
对前面的断言, 设$d\sigma^2=e^{2\lambda}ds^2,$ 那么
$$
e^{2\lambda}\widetilde K=K-\Delta \lambda,\quad e^{2\lambda}=(1+\frac{1}{W})^2,\quad \lambda=\log(1+\frac{1}{W}).
$$
于是$\widetilde K=0$当且仅当$K=\Delta \log (1+\frac{1}{W}).$
极小曲面方程
回顾$ds^2=g_{ij}dx_idx_j,$ $g_{ij}=\delta_{ij}+p_ip_j.$
$g_{ik}g^{kj}=\delta_i^k,$ $g^{ij}=\delta_{ij}-\frac{1}{W^2}p_ip_j.$
考虑第二基本型
$$
l_{ij}dx_idx_j=\Theta_{n+1}=-\left<{}dr,de_{n+1}\right>=-dx_i d(\frac{p_i}{W})+dzd(\frac{1}{W}).
$$
$$
\Theta_{n+1}=\left<{}d^2r,e_{n+1}\right>=-\frac{1}{W}\frac{\partial^2 {}z}{\partial {}x_i\partial {}x_j}dx^idx^j.
$$
因此
$$
nH=g^{ij}l_{ij}=-\frac{1}{W}g^{ij}p_{ij}=0\Leftrightarrow g^{ij}p_{ij}=0.
$$
这当且仅当
$$
\frac{\partial {} }{\partial {}x_i}(\frac{p_i}{W})=0.
$$
这是因为
$$
0=g^{ij}p_{ij}=(\delta_{ij}-\frac{p_ip_j}{W^2})p_{ij}=p_{ii}-\frac{1}{W^2}p_ip_jp_{ij}.
$$
另一方面,
$$
\frac{\partial {} }{\partial {}x_i}(\frac{p_i}{W})=\frac{1}{W}p_{ii}-\frac{1}{W^2}p_i\frac{\partial {}W}{\partial {}x_i}=\frac{1}{W}p_{ii}-\frac{1}{W^3}p_ip_kp_{ki}=\frac{1}{W}(p_{ii}-\frac{1}{W^2}p_ip_kp_{ki}).
$$
因此两者相等. 这样我们就得到了极小曲面方程.
文章最后更新于 2023-03-02 19:47:54
height function
First consider the situation where the peripheral space $X$ is the $N$-dimensional Euclidean space $E^N$.
At this time, all tangent spaces can be equal to $E^N.$. Considering the immersed submanifold $x:M\rightarrow E^n.$, at this time
$$
dx=\theta_i e_i,\quad de_i=\theta_{ij}e_j+\theta_{i\alpha}e_\alpha=\theta_{ij}e_j+h_{i\alpha j}\theta_j e_\alpha.
$$
Fix a vector $a\in E^N,$, then take the function $f=\left<{}a,x\right>,$ on $M$
represents the height function along $a$. Then,
$$
f_i\theta_i=df=\left<{}a,dx\right>=\left<{}a,e_i\right>\theta_i,
$$
$$
f_{ij}\theta_j=d f_i+f_j\theta_{ji}=\left<{}a,de_i\right>+\left<{}a,e_j\right>\theta_{ji}=\left<{}a,\theta_{ij}e_j+h_{i\alpha j}\theta_j e_\alpha+\theta_{ji}e_j\right>=\left<{}a,h_{i\alpha j}\theta_je_\alpha\right>.
$$
This gives
$$
\Delta f=f_{ii}=\left<{}a,h_{i\alpha i}e_\alpha\right>=n\left<{}a,H\right>.
$$
This explains the following theorem at $n>0$
Theorem 1. An immersed submanifold in Euclidean space is minimal if and only if under the induced metric, all coordinate functions are harmonic functions. Furthermore, there is no compact boundless minimal submanifold in Euclidean space.
This is because the harmonic functions on the compact infinite manifold are all constants.
curvature
memories
$$
d\omega_{AB}-\omega_{AC}\wedge\omega_{CB}=-\frac{1}{2}\widehat R_{ABCD}\omega_C\wedge\omega_D.
$$
Note that the curvature in Euclidean space is zero. So pulling back we get
$$
d\theta_{ij}-\theta_{ik}\wedge\theta_{kj}=-\theta_{i\alpha}\wedge\theta_{j\alpha}=-\frac{1}{2}(h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k})\theta_k\wedge\theta_l.
$$
Let $S_{ijkl}$ be the curvature tensor measured on $M$, then
$$
S_{ijkl}=h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k}.
$$
In fact, this is Gauss's equation.
If $M$ is extremely small, then its Ricci tensor is
$$
S_{ik}=S_{ijkj}=-h_{i\alpha j}h_{k\alpha j},
$$
It is negative semi-definite.
Further its quantitative curvature is
$$
S=S_{ii}=-h_{i\alpha j}^2\le 0.
$$
Therefore we have the following theorem
Theorem 2. The Ricci tensor of the minimal submanifold $M$ is negative semidefinite. It is fully geodesic (a linear subspace) if and only if its quantitative curvature is zero.
minimal hypersurface
At this time, the co-dimension is $1,$, abbreviated as $h_{i,n+1,j}=h_{ij}.$
So $\theta_{i,n+1}=h_{ij}\theta_j.$ we have
$$
\begin{aligned}
0&=d\theta_{i,n+1}-\theta_{iC}\theta_{C,n+1}\\
&=dh_{ij}\wedge\theta_j+h_{ij}\theta_k\wedge\theta_{kj}-h_{jk}\theta_{ij}\wedge\theta_k\\
&=(dh_{ik}+h_{ij}\theta_{jk}+h_{jk}\theta_{ji})\wedge\theta_k\\
&=h_{ik,j}\theta_j\wedge\theta_k
\end{aligned}
$$
Then $h_{ik,j}=h_{ij,k}.$ This is the Codazzi equation.
Due to $\theta_{n+1}=0,$
$$
0=d\theta_{n+1}=\theta_i\wedge\theta_{i,n+1}=h_{ij}\theta_i\wedge\theta_j.
$$
Therefore, $h_{ij}=h_{ji}.$ and $h_{ij,k}$ are all symmetrical.
Therefore, the sum of any two subscripts of the minimal hypersurface is zero.
Remember $f=\left<{}a,e_{n+1}\right>,$ then
$$
d e_{n+1}=\theta_{n+1,i}e_i=-h_{ij}\theta_je_i,
$$
$$
f_i=-\left<{}a,e_j\right>h_{ji},
$$
$$
f_{ij}\theta_j=-\left<{}a,\theta_{jk}e_k+h_{jk}\theta_ke_{n+1}\right>h_{ij}-\left<{}a,e_j\right>dh_{ij}-\left<{}a,e_k\right>h_{kj}\theta_{ji}.
$$
by
$$
dh_{ij}+h_{kj}\theta_{ki}+h_{ik}\theta_{kj}=h_{ij,k}\theta_k,
$$
$$
f_{ij}\theta_j=-\left<{}a,e_{n+1}\right>h_{ij}h_{jk}\theta_k-\left<{}a,e_j\right>h_{ij,k}\theta_k,
$$
$$
f_{ik}=-\left<{}a,e_{n+1}\right>h_{ij}h_{jk}-\left<{}a,e_j\right>h_{ij,k},
$$
$$
\Delta f=f_{ii}=-\left<{}a,e_{n+1}\right>h_{ij}^2=\left<{}a,e_{n+1}\right>S.
$$
Consider a minimal hypersurface of shape $z=z(x_1,\cdots,x_n)$. Then the tangent vector is
$$
\frac{\partial {}r}{\partial {}x_i}=(0,\cdots,1,\cdots,0,p_i),\quad p_i=\frac{\partial {}z}{\partial {}x_i}.
$$
$$
g_{ij}=\delta_{ij}+p_ip_j.
$$
$$
e_{n+1}=\frac{1}{W}(p_1,\cdots,p_n,-1),\quad W=\sqrt{1+p_k^2}.
$$
This surface has a one-to-one mapping to a hyperplane. Bernstein's problem is whether minimal surfaces are always hyperplanes.
We want to prove $z=a_ix_i+a_{n+1}.$ but this has a counterexample for $n\ge 8$.
Now consider the situation $n=2$.
Hope to use $\Delta \left<{}a,e_{n+1}\right>=S\left<{}a,e_{n+1}\right>.$
Take $a=(0,\cdots,0,1),$ and then $\left<{}a,e_{n+1}\right>=-\frac{1}{W}.$. Therefore,
$$
\Delta(\frac{1}{W})=\frac{2K}{W}\le 0.
$$
Then $\frac{1}{W}$ is a positive function of the upward summation and must be a constant. This illustrates the conclusion.
Or we have Gauss's equation
$$
S_{ijkl}=h_{ik}h_{jl}-h_{il}h_{jk},\quad S_{ik}=nHh_{ik}-h_{ij}h_{jk}.
$$
When the surface is extremely small, there is
$$
S_{ik}=K\delta_{ik}=-h_{il}h_{lk}.
$$
due to
$$
\Delta\log (1+\frac{1}{W})=\frac{1}{1+\frac{1}{W} }\Delta(\frac{1}{W})-\frac{1}{(1+\frac{1}{W})^2}|\nabla (\frac{1}{W})|^2.
$$
Let $f=1+\frac{1}{W},$ then
$$
\Delta \log f=\frac{\Delta f}{f}-\frac{f_k^2}{f^2}.
$$
memories
$$
\left<{}a,e_3\right>_i=-\left<{}a,e_j\right>h_{ij}.
$$
Due to $e_3=\frac{1}{W}(p_1,p_2,-1),$
$$
|\nabla (\frac{1}{W})|^2=|\nabla\left<{}a,e_3\right>|^2=\left<{}a,e_3\right>_i^2=\left<{}a,e_j\right>h_{ij}\left<{}a,e_k\right>h_{ik}=-K\left<{}a,e_j\right>^2.
$$
due to
$$
(0,0,1)=a=\left<{}a,e_i\right>e_i+\left<{}a,e_3\right>e_3=\left<{}a,e_i\right>e_i-\frac{1}{W}e_3,
$$
$$
1=\left<{}a,e_i\right>^2+(\frac{1}{W})^2.
$$
Therefore
$$
\Delta(\log (1+\frac{1}{W}))=\frac{1}{1+\frac{1}{W} }\frac{2K}{W}+\frac{1}{(1+\frac{1}{W})^2}(-K)(1-(\frac{1}{W})^2)=K.
$$
Consider a new metric $d\sigma=(1+\frac{1}{W})ds,$ $ds$ induced by immersion.
$d\sigma\ge ds$ is complete. Assert that under $\sigma$, $\widetilde K\equiv 0,$
Then it is a developable surface. Then,
$$
d\sigma^2=d\xi^2+d\eta^2.
$$
$$
\Delta_{ds^2}=\Lambda(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2}),\quad \Lambda=(1+\frac{1}{W})>0.
$$
It is known that $K\le 0.$ therefore
$$
0\ge K=\Delta\log(1+\frac{1}{W})= \Lambda(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2})(\log (1+\frac{1}{W})),
$$
$$
(\frac{\partial {} }{\partial {}\xi^2}+\frac{\partial {} }{\partial {}\eta^2})\log(1+\frac{1}{W})\le 0,
$$
The upscaling and positive functions are constants, so $K\equiv 0.$ is fully geodesic by the previous theorem,
Then it is a hyperplane.
For the previous assertion, let $d\sigma^2=e^{2\lambda}ds^2,$ then
$$
e^{2\lambda}\widetilde K=K-\Delta \lambda,\quad e^{2\lambda}=(1+\frac{1}{W})^2,\quad \lambda=\log(1+\frac{1}{W}).
$$
Then $\widetilde K=0$ iff $K=\Delta \log (1+\frac{1}{W}).$
minimal surface equation
Review$ds^2=g_{ij}dx_idx_j,$ $g_{ij}=\delta_{ij}+p_ip_j.$
$g_{ik}g^{kj}=\delta_i^k,$ $g^{ij}=\delta_{ij}-\frac{1}{W^2}p_ip_j.$
Consider the second basic form
$$
l_{ij}dx_idx_j=\Theta_{n+1}=-\left<{}dr,de_{n+1}\right>=-dx_i d(\frac{p_i}{W})+dzd(\frac{1}{W}).
$$
$$
\Theta_{n+1}=\left<{}d^2r,e_{n+1}\right>=-\frac{1}{W}\frac{\partial^2 {}z}{\partial {}x_i\partial {}x_j}dx^idx^j.
$$
Therefore
$$
nH=g^{ij}l_{ij}=-\frac{1}{W}g^{ij}p_{ij}=0\Leftrightarrow g^{ij}p_{ij}=0.
$$
This is if and only if
$$
\frac{\partial {} }{\partial {}x_i}(\frac{p_i}{W})=0.
$$
This is because
$$
0=g^{ij}p_{ij}=(\delta_{ij}-\frac{p_ip_j}{W^2})p_{ij}=p_{ii}-\frac{1}{W^2}p_ip_jp_{ij}.
$$
on the other hand,
$$
\frac{\partial {} }{\partial {}x_i}(\frac{p_i}{W})=\frac{1}{W}p_{ii}-\frac{1}{W^2}p_i\frac{\partial {}W}{\partial {}x_i}=\frac{1}{W}p_{ii}-\frac{1}{W^3}p_ip_kp_{ki}=\frac{1}{W}(p_{ii}-\frac{1}{W^2}p_ip_kp_{ki}).
$$
Therefore the two are equal. This gives us the equation of the minimal surface.
The article was last updated on 2023-03-02 19:47:54
