Machine-translated from Chinese.
记号
记空间形式$R^{n+p}(c)$为$n+p$维黎曼流形, 具常截面曲率$c$. 我们知道,
$$
R^{n+p}(c)=\begin{cases}
E^{n+p},&c=0\\
S^{n+p},&c=1,\quad x_1^2+\cdots+x_{n+p+1}^2=1\\
H^{n+p},&c=-1,\quad x_1^2+\cdots+x_{n+p}^2-x_{n+p+1}^2=-1
\end{cases}
$$
考虑$x:M^n\rightarrow S^{n+1}\hookrightarrow E^{n+p+1},$
$\left<{}x,x\right>=1.$ 类似地考虑$\{e_i\},\{e_\alpha\}.$
那么$\{e_i,e_\alpha,x\}$构成$E^{n+p+1}$中的正交标架. 我们有
$$
dx=\theta_ie_i,\quad de_i=\theta_{ij}e_j+\theta_{i\alpha}e_\alpha-\theta_ix.
$$
$$
de_\alpha=\theta_{\alpha i}e_i+\theta_{\alpha\beta}e_\beta,\quad \theta_{i\alpha}=h_{i\alpha j}\theta_j.
$$
$x$项系数是通过内积得到的, 如
$$
\left<{}de_i,x\right>=-\left<{}e_i,dx\right>=-\theta_i.
$$
极小子流形
设有浸入
$$
x:M\hookrightarrow S^{n+p}\hookrightarrow \mathbb{R}^{n+p+1},
$$
$\,\forall\,a\in \mathbb{R}^{n+p+1},$ 我们来计算
$$
\Delta\left<{}x,a\right>=n\left<{}a,H\right>-n\left<{}x,a\right>.
$$
记$f=\left<{}x,a\right>,$ 通过类似的计算,
我们有$f_i=\left<{}e_i,a\right>,$
$$
f_{ij}\theta_j=df_i+f_j\theta_{ji}=\left<{}de_i,a\right>+\left<{}e_j,a\right>\theta_{ji}=\left<{}\theta_{ij}e_j+h_{i\alpha j}\theta_je_\alpha-\theta_ix,a\right>+\left<{}e_j,a\right>\theta_ji=\left<{}h_{i\alpha j}\theta_je_\alpha-\theta_ix,a\right>,
$$
$$
\Delta\left<{}x,a\right>=f_{ii}=n\left<{}a,H\right>-n\left<{}x,a\right>.
$$
从而
定理 1. $x:M\rightarrow S^{n+p}$极小当且仅当$\Delta\left<{}x,a\right>=-n\left<{}x,a\right>,$ 当且仅当$\Delta x=-nx.$
例子
大圆
$n$-大圆都是全测地的. 取固定的$a\neq 0,$ 大圆可表示为
$$
M_0:=\{x\in S^{n+1}|\left<{}x,a\right>=0\},
$$
$$
dx=\theta_i e_i,\quad de_i=\theta_{ij}e_j+h_{ij}e_je_{n+1}-\theta_ix,
$$
$$
de_{n+1}=-h_{ij}\theta_je_i.
$$
对$\left<{}x,a\right>=0$求导,
我们有$\left<{}dx,a\right>=0,$ 因此$a\perp T_pM.$ 又由$a\perp x,$
$$
a=ke_{n+1}.
$$
由$\left<{}e_i,a\right>=0,$ $\left<{}de_i,a\right>=0,$
$$
\left<{}\theta_{ij}e_j+h_{ij}\theta_je_{n+1}-\theta_ix,a\right>=0,\quad h_{ij}\theta_j=0,
$$
于是$h_{ij}=0.$
对于$n$-小圆
$$
M_c:=\{x\in S^{n+1}|\left<{}x,a\right>=c,0<c<1\},
$$
第二基本型特征值均为固定非零常数,
即$\lambda_1=\cdots=\lambda_n=const\neq 0,$ 称为totally umbilical的.
Clifford tori
考虑极小超曲面$x:S^r(a_1)\times S^{n-r}(a_2)\hookrightarrow S^{n+1}\hookrightarrow \mathbb{R}^{n+2},$
$1\le r\le n-1.$ 我们通过
$$
x:S^r(a_1)\times S^{n-r}(a_2)\hookrightarrow E^{r+1}\times E^{n-r+1}=E^{n+2}
$$
来实现这样的嵌入,
那么设$x=(a_1\xi_1,a_2\xi_2),$ $|\xi_1|=|\xi_2|=1,$ $\xi_1\in E^{r+1},$
$\xi_2\in E^{n-r+1},$
$$
\left<{}x,x\right>=a_1^2\left<{}\xi_1,\xi_1\right>+a_2^2\left<{}\xi_2,\xi_2\right>=a_1^2+a_2^2=1.
$$
$$
dx=(a_1d\xi_1,a_2d\xi_2),\quad \mathrm{I}=\left<{}dx,dx\right>=a_1^2\left<{}d\xi_1,d\xi_1\right>+a_2^2\left<{}d\xi_2,d\xi_2\right>.
$$
选取$e_{n+1}=(-a_2\xi_1,a_1\xi_2),$ 那么我们选取到了正交基. 第二基本型
$$
\mathrm{II}=-\left<{}dx,de_{n+1}\right>=-\left<{}(a_1d\xi_1,a_2d\xi_2),(-a_2d\xi_1,a_1d\xi_2)\right>=a_1a_2\left(\left<{}d\xi_1,d\xi_1\right>-\left<{}d\xi_2,d\xi_2\right>\right),
$$
主曲率$\lambda_1=\cdots=\lambda_r=\frac{a_2}{a_1},$
$\lambda_{r+1}=\cdots=\lambda_n=-\frac{a_1}{a_2}.$ 于是
$$
n H=\lambda_1+\cdots+\lambda_n=r\frac{a_2}{a_1}-(n-r)\frac{a_1}{a_2}=0\Leftrightarrow r\frac{a_2}{a_1}=(n-r)\frac{a_1}{a_2},
$$
回忆$a_1^2+a_2^2=1,$ 从而这当且仅当$a_1=\sqrt{\frac{r}{n} },$
$a_2=\sqrt{\frac{n-r}{n} },$ 称为Clifford tori, $1\le r\le n-1.$
Veronese曲面
考虑$S^2(\sqrt{3})\hookrightarrow S^4(1),$
$(x,y,z)\mapsto (u_1,\cdots,u_5),$
$$
u_1=\frac{1}{\sqrt{3} }yz,\quad u_2=\frac{1}{\sqrt{3} }zx,\quad u_3=\frac{1}{\sqrt{3} }xy,
$$
$$
u_4=\frac{1}{2\sqrt{3} }(x^2-y^2),\quad u_5=\frac{1}{6}(x^2+y^2-2z^2).
$$
希望它是极小曲面, 因此要有$\Delta u_i=-2u_i.$ 显然$\Delta_E u_i=0.$
我们可以通过研究两者的联系证明, 也可以直接进行计算.
该嵌入可以拉到$\mathbb{R}\mathrm{P}^2$上, 称为Veronese曲面.
Gauss映射
设$x:M\rightarrow S^{n+1}$为极小曲面. 类似于第三章, 我们有
$$
\Delta \left<{}a,e_{n+1}\right>=-\sigma\left<{}a,e_{n+1}\right>,
$$
取$f=\left<{}a,e_{n+1}\right>,$ 那么
$$
df=-\left<{}a,h_{ji}\theta_ie_j\right>=f_i\theta_i,
$$
$$
f_{ij}\theta_j=df_i+f_j\theta_{ji}=-d(h_{ij}\left<{}a,e_j\right>)-h_{jk}\left<{}a,e_k\right>\theta_{ji},
$$
$$
f_{ij}=-h_{ikj}\left<{}a,e_k\right>-h_{ik}\left<{}a,h_{kj}e_{n+1}-x\delta_{kj}\right>,
$$
$$
\Delta f=f_{ii}=-h_{ij}^2f,\quad \Delta\left<{}a,e_{n+1}\right>=-\sigma\left<{}a,e_{n+1}\right>.
$$
$M$定向时, 有良定义的$e_{n+1}:M\hookrightarrow S^{n+1}$给出了Gauss映射.
定理 2. 令$M^n\rightarrow S^{n+1}$为定向闭极小超曲面. 若Gauss映射像在某个开半球内, 则$M^n$全测地.
即存在$a,$ 使得$\left<{}a,e_{n+1}\right>>0.$ 注意到
$$
0=\int_M\Delta \left<{}a,e_{n+1}\right>dA=-\int_M\sigma \left<{}a,e_{n+1}\right>dA.
$$
因此$\sigma=0,$ 即$M$全测地.
定理 3. 设$x:M^2\hookrightarrow S^3$为极小曲面, 那么$x(S^2)$为$S^3$上的大圆.
思路是, 首先构造一个全纯二次型$Q(dz)^2,$ 由$g(S^2)=0,$ 推出$Q\equiv 0$.
这样就有$h_{ij}=0,$ $M=S^2(1).$
存在等温坐标系
$$
ds^2=\lambda^2(dx^2+dy^2)=\lambda^2dzd\overline z,\quad z=x+iy,\quad \lambda\in C^\infty(M).
$$
记
$$
\alpha=\theta_1+i\theta_2=\lambda dz, \quad h=h_{11}+ih_{12},
$$
$$
\beta=\theta_{13}+i\theta_{23}=h_{11}\theta_1+h_{12}\theta_2+i(h_{12}\theta_1-h_{11}\theta_2)=h_{11}(\theta_1-i\theta_2) +ih_{12}(\theta_1-i\theta_2)=h\overline\alpha.
$$
那么
$$
d\alpha=\theta_{12}\wedge \theta_2+i\theta_{21}\wedge \theta_1=\theta_{12}\wedge (\theta_2-i\theta_1)=-i\theta_{12}\wedge\alpha,
$$
$$
(d\lambda + i\lambda \theta_{12})\wedge dz=0, \quad d\lambda=-i\lambda \theta_{12}+\Lambda_1dz.
$$
$$
d\beta=d\theta_{13}+id\theta_{23}=\theta_{12}\wedge\theta_{23}+i\theta_{21}\wedge\theta_{13}=-i\theta_{12}\wedge \beta.
$$
那么,
$$
dh\overline\alpha=dh\wedge\overline\alpha+hd\overline\alpha=dh\wedge\overline\alpha+h(i\theta_{12}\wedge \overline\alpha)=-hi\theta_{12}\wedge \overline\alpha,
$$
于是,
$$
\lambda(dh+2ih\theta_{12})\wedge\overline z=0,\quad d\overline h-2i\overline h\theta_{12}=\Lambda_2 d z.
$$
取$q=\lambda^2\overline h,$ 那么
$$
\begin{aligned}
d(\lambda^2\overline h)=&2\lambda d\lambda \overline h+\lambda^2 d\overline h\\
=&2\lambda (-i\lambda \theta_{12}+\Lambda_1dz)\overline h+\lambda^2(2i\overline h\theta_{12}+\Lambda_2dz)\\
=&\lambda (2\overline h\Lambda_1+\lambda\Lambda_2)dz.
\end{aligned}
$$
于是$q$是全纯形式,
$\frac{\partial {} }{\partial {}\overline z}(\lambda^2\overline h)=0.$
文章最后更新于 2023-03-16 15:51:56
mark
Let the space form $R^{n+p}(c)$ be the $n+p$ Veliemannian manifold with constant cross-section curvature $c$. We know that,
$$
R^{n+p}(c)=\begin{cases}
E^{n+p},&c=0\\
S^{n+p},&c=1,\quad x_1^2+\cdots+x_{n+p+1}^2=1\\
H^{n+p},&c=-1,\quad x_1^2+\cdots+x_{n+p}^2-x_{n+p+1}^2=-1
\end{cases}
$$
Consider $x:M^n\rightarrow S^{n+1}\hookrightarrow E^{n+p+1},$
$\left<{}x,x\right>=1.$ Consider $\{e_i\},\{e_\alpha\}.$ similarly
Then $\{e_i,e_\alpha,x\}$ constitutes the orthogonal frame in $E^{n+p+1}$. We have
$$
dx=\theta_ie_i,\quad de_i=\theta_{ij}e_j+\theta_{i\alpha}e_\alpha-\theta_ix.
$$
$$
de_\alpha=\theta_{\alpha i}e_i+\theta_{\alpha\beta}e_\beta,\quad \theta_{i\alpha}=h_{i\alpha j}\theta_j.
$$
$x$ term coefficient is obtained through inner product, such as
$$
\left<{}de_i,x\right>=-\left<{}e_i,dx\right>=-\theta_i.
$$
minimal submanifold
with immersion
$$
x:M\hookrightarrow S^{n+p}\hookrightarrow \mathbb{R}^{n+p+1},
$$
$\,\forall\,a\in \mathbb{R}^{n+p+1},$ Let’s calculate
$$
\Delta\left<{}x,a\right>=n\left<{}a,H\right>-n\left<{}x,a\right>.
$$
Note $f=\left<{}x,a\right>,$ through similar calculations,
We have $f_i=\left<{}e_i,a\right>,$
$$
f_{ij}\theta_j=df_i+f_j\theta_{ji}=\left<{}de_i,a\right>+\left<{}e_j,a\right>\theta_{ji}=\left<{}\theta_{ij}e_j+h_{i\alpha j}\theta_je_\alpha-\theta_ix,a\right>+\left<{}e_j,a\right>\theta_ji=\left<{}h_{i\alpha j}\theta_je_\alpha-\theta_ix,a\right>,
$$
$$
\Delta\left<{}x,a\right>=f_{ii}=n\left<{}a,H\right>-n\left<{}x,a\right>.
$$
thereby
Theorem 1. $x:M\rightarrow S^{n+p}$ is minimal if and only if $\Delta\left<{}x,a\right>=-n\left<{}x,a\right>,$ if and only if $\Delta x=-nx.$
Example
big circle
$n$-Great circles are all completely geodesic. Taking a fixed $a\neq 0,$ great circle can be expressed as
$$
M_0:=\{x\in S^{n+1}|\left<{}x,a\right>=0\},
$$
$$
dx=\theta_i e_i,\quad de_i=\theta_{ij}e_j+h_{ij}e_je_{n+1}-\theta_ix,
$$
$$
de_{n+1}=-h_{ij}\theta_je_i.
$$
Take the derivative of $\left<{}x,a\right>=0$,
We have $\left<{}dx,a\right>=0,$ and therefore $a\perp T_pM.$ and thus $a\perp x,$
$$
a=ke_{n+1}.
$$
From $\left<{}e_i,a\right>=0,$ $\left<{}de_i,a\right>=0,$
$$
\left<{}\theta_{ij}e_j+h_{ij}\theta_je_{n+1}-\theta_ix,a\right>=0,\quad h_{ij}\theta_j=0,
$$
So$h_{ij}=0.$
For $n$-small circle
$$
M_c:=\{x\in S^{n+1}|\left<{}x,a\right>=c,0<c<1\},
$$
The second basic type eigenvalues are all fixed non-zero constants,
That is, $\lambda_1=\cdots=\lambda_n=const\neq 0,$ is called totally umbilical.
Clifford tori
Consider the minimal hypersurface $x:S^r(a_1)\times S^{n-r}(a_2)\hookrightarrow S^{n+1}\hookrightarrow \mathbb{R}^{n+2},$
$1\le r\le n-1.$ We pass
$$
x:S^r(a_1)\times S^{n-r}(a_2)\hookrightarrow E^{r+1}\times E^{n-r+1}=E^{n+2}
$$
To achieve such embedding,
Then let $x=(a_1\xi_1,a_2\xi_2),$ $|\xi_1|=|\xi_2|=1,$ $\xi_1\in E^{r+1},$
$\xi_2\in E^{n-r+1},$
$$
\left<{}x,x\right>=a_1^2\left<{}\xi_1,\xi_1\right>+a_2^2\left<{}\xi_2,\xi_2\right>=a_1^2+a_2^2=1.
$$
$$
dx=(a_1d\xi_1,a_2d\xi_2),\quad \mathrm{I}=\left<{}dx,dx\right>=a_1^2\left<{}d\xi_1,d\xi_1\right>+a_2^2\left<{}d\xi_2,d\xi_2\right>.
$$
Select $e_{n+1}=(-a_2\xi_1,a_1\xi_2),$, then we select the orthonormal basis. The second basic form
$$
\mathrm{II}=-\left<{}dx,de_{n+1}\right>=-\left<{}(a_1d\xi_1,a_2d\xi_2),(-a_2d\xi_1,a_1d\xi_2)\right>=a_1a_2\left(\left<{}d\xi_1,d\xi_1\right>-\left<{}d\xi_2,d\xi_2\right>\right),
$$
Principal curvature$\lambda_1=\cdots=\lambda_r=\frac{a_2}{a_1},$
$\lambda_{r+1}=\cdots=\lambda_n=-\frac{a_1}{a_2}.$ So
$$
n H=\lambda_1+\cdots+\lambda_n=r\frac{a_2}{a_1}-(n-r)\frac{a_1}{a_2}=0\Leftrightarrow r\frac{a_2}{a_1}=(n-r)\frac{a_1}{a_2},
$$
Recall $a_1^2+a_2^2=1,$ thus this if and only if $a_1=\sqrt{\frac{r}{n} },$
$a_2=\sqrt{\frac{n-r}{n} },$ called Clifford tori, $1\le r\le n-1.$
Veronese surface
Consider $S^2(\sqrt{3})\hookrightarrow S^4(1),$
$(x,y,z)\mapsto (u_1,\cdots,u_5),$
$$
u_1=\frac{1}{\sqrt{3} }yz,\quad u_2=\frac{1}{\sqrt{3} }zx,\quad u_3=\frac{1}{\sqrt{3} }xy,
$$
$$
u_4=\frac{1}{2\sqrt{3} }(x^2-y^2),\quad u_5=\frac{1}{6}(x^2+y^2-2z^2).
$$
I hope it is a minimal surface, so there must be $\Delta u_i=-2u_i.$ and obviously $\Delta_E u_i=0.$
We can prove it by studying the connection between the two, or we can calculate it directly.
This embedding can be pulled onto $\mathbb{R}\mathrm{P}^2$ and is called a Veronese surface.
Gauss mapping
Let $x:M\rightarrow S^{n+1}$ be a minimal surface. Similar to Chapter 3, we have
$$
\Delta \left<{}a,e_{n+1}\right>=-\sigma\left<{}a,e_{n+1}\right>,
$$
Take $f=\left<{}a,e_{n+1}\right>,$ then
$$
df=-\left<{}a,h_{ji}\theta_ie_j\right>=f_i\theta_i,
$$
$$
f_{ij}\theta_j=df_i+f_j\theta_{ji}=-d(h_{ij}\left<{}a,e_j\right>)-h_{jk}\left<{}a,e_k\right>\theta_{ji},
$$
$$
f_{ij}=-h_{ikj}\left<{}a,e_k\right>-h_{ik}\left<{}a,h_{kj}e_{n+1}-x\delta_{kj}\right>,
$$
$$
\Delta f=f_{ii}=-h_{ij}^2f,\quad \Delta\left<{}a,e_{n+1}\right>=-\sigma\left<{}a,e_{n+1}\right>.
$$
When $M$ is oriented, the well-defined $e_{n+1}:M\hookrightarrow S^{n+1}$ gives a Gauss map.
Theorem 2. Let $M^n\rightarrow S^{n+1}$ be a directed closed minimal hypersurface. If the Gauss map image is in an open hemisphere, then $M^n$ is completely geodesic.
That is, there is $a,$ such that $\left<{}a,e_{n+1}\right>>0.$ notices
$$
0=\int_M\Delta \left<{}a,e_{n+1}\right>dA=-\int_M\sigma \left<{}a,e_{n+1}\right>dA.
$$
Therefore $\sigma=0,$ is the total geodesic of $M$.
Theorem 3. Assume $x:M^2\hookrightarrow S^3$ is a minimal surface, then $x(S^2)$ is a great circle on $S^3$.
The idea is to first construct a holomorphic quadratic form $Q(dz)^2,$ and derive $Q\equiv 0$ from $g(S^2)=0,$.
This way we have $h_{ij}=0,$ $M=S^2(1).$
There is an isothermal coordinate system
$$
ds^2=\lambda^2(dx^2+dy^2)=\lambda^2dzd\overline z,\quad z=x+iy,\quad \lambda\in C^\infty(M).
$$
remember
$$
\alpha=\theta_1+i\theta_2=\lambda dz, \quad h=h_{11}+ih_{12},
$$
$$
\beta=\theta_{13}+i\theta_{23}=h_{11}\theta_1+h_{12}\theta_2+i(h_{12}\theta_1-h_{11}\theta_2)=h_{11}(\theta_1-i\theta_2) +ih_{12}(\theta_1-i\theta_2)=h\overline\alpha.
$$
Then
$$
d\alpha=\theta_{12}\wedge \theta_2+i\theta_{21}\wedge \theta_1=\theta_{12}\wedge (\theta_2-i\theta_1)=-i\theta_{12}\wedge\alpha,
$$
$$
(d\lambda + i\lambda \theta_{12})\wedge dz=0, \quad d\lambda=-i\lambda \theta_{12}+\Lambda_1dz.
$$
$$
d\beta=d\theta_{13}+id\theta_{23}=\theta_{12}\wedge\theta_{23}+i\theta_{21}\wedge\theta_{13}=-i\theta_{12}\wedge \beta.
$$
Then,
$$
dh\overline\alpha=dh\wedge\overline\alpha+hd\overline\alpha=dh\wedge\overline\alpha+h(i\theta_{12}\wedge \overline\alpha)=-hi\theta_{12}\wedge \overline\alpha,
$$
So,
$$
\lambda(dh+2ih\theta_{12})\wedge\overline z=0,\quad d\overline h-2i\overline h\theta_{12}=\Lambda_2 d z.
$$
Take $q=\lambda^2\overline h,$ then
$$
\begin{aligned}
d(\lambda^2\overline h)=&2\lambda d\lambda \overline h+\lambda^2 d\overline h\\
=&2\lambda (-i\lambda \theta_{12}+\Lambda_1dz)\overline h+\lambda^2(2i\overline h\theta_{12}+\Lambda_2dz)\\
=&\lambda (2\overline h\Lambda_1+\lambda\Lambda_2)dz.
\end{aligned}
$$
So $q$ is the holomorphic form,
$\frac{\partial {} }{\partial {}\overline z}(\lambda^2\overline h)=0.$
The article was last updated on 2023-03-16 15:51:56
