定义第$r$阶平均曲率

$\begin{eqnarray*}\displaystyle &&{S_{r}=\frac{1}{r!} \varepsilon ^{i_{1}\cdots i_{r}} _{j_{1}\cdots j_{r}}h_{i_{1}j_{1}}\cdots h_{i_{r}j_{r}}},\\ &&H_{r}=(\begin{array}{c} n \\ r \end{array})^{-1}S_{r}, \end{eqnarray*}$

$S_{r}$与基底的选取无关, $S_{r}$不变.事实上, 对基底作正交变换

$\det (\lambda I-\bar{h})=\det (\lambda I-T^{-1}hT)=\det (T^{-1}(\lambda I-h)T)=\det (\lambda I-h).$

若$H_{r+1}=0$, 则称$x:M^{n}\rightarrow S^{n+1}$是$r$ -极小的.

第$r$个牛顿算子$\displaystyle T_{rij}=\frac{1}{r!} \varepsilon ^{i_{1}\cdots i_{r}i} _{j_{1}\cdots j_{r}j}h_{i_{1}j_{1}}\cdots h_{i_{r}j_{r}}.$定义

$\begin{eqnarray*}&&T_{r}:T_{p}M\rightarrow T_{p}M,\\ &&T_{r}(e_{i})=T_{rij}e_{j}.\end{eqnarray*}$

记$B=(h_{ij}),$易有

$\begin{aligned} \varepsilon ^{i_{1}\cdots i_{r}i} _{j_{1}\cdots j_{r}j}&=\left( \begin{array}{cccc} \delta _{i_{1}j_{1}}&\cdots& \delta _{i_{1}j_{r}}& \delta _{i_{1}j} \\ \vdots&\vdots&\vdots&\vdots \\ \delta _{i_{r}j_{1}}&\cdots& \delta _{i_{r}j_{r}}& \delta _{i_{r}j} \\ \delta _{ij_{1}}&\cdots& \delta _{ij_{r}}& \delta _{ij} \\ \end{array} \right)\\ &=\displaystyle{\delta _{ij}\varepsilon ^{i_{1}\cdots i_{r}} _{j_{1}\cdots j_{r}}-\delta _{i_{1}j}\varepsilon ^{ii_{2}\cdots i_{r}} _{j_{1}j_{2}\cdots j_{r}} -\delta _{i_{2}j}\varepsilon ^{i_{1}ii_{3}\cdots i_{r}} _{j_{1}j_{2}j_{3}\cdots j_{r}}-\cdots-\delta _{i_{r}j}\varepsilon ^{i_{1}\cdots i_{r-1}i} _{j_{1}\cdots j_{r-1}j_{r}}}, \end{aligned}$

$\begin{aligned} \frac{1}{r!} \varepsilon ^{i_{1}\cdots i_{r}i} _{j_{1}\cdots j_{r}j}h_{i_{1}j_{1}}\cdots h_{i_{r}j_{r}} =&\delta _{ij}S_{r}-\frac{1}{r!} \varepsilon ^{ii_{2}\cdots i_{r}} _{j_{1}j_{2}\cdots j_{r}}h_{jj_{1}}h_{i_{2}j_{2}}\cdots h_{i_{r}j_{r}}-\frac{1}{r!} \varepsilon ^{i_{1}ii_{3}\cdots i_{r}} _{j_{1}j_{2}j_{3}\cdots j_{r}}h_{i_{1}j_{1}}h_{jj_{2}}\cdots h_{i_{r}j_{r}}\\&-\cdots-\frac{1}{r!} \varepsilon ^{i_{1}\cdots i_{r-1}i} _{j_{1}\cdots j_{r-1}j_{r}}h_{i_{1}j_{1}}h_{i_{2}j_{2}}\cdots h_{jj_{r}}\\ =&\delta _{ij}S_{r}-\frac{(r-1)!}{r!}T_{(r-1)ij_{1}}h_{j_{1}j}-\frac{(r-1)!}{r!}T_{(r-1)ij_{2}}h_{j_{2}j}\\&-\cdots-\frac{(r-1)!}{r!}T_{(r-1)ij_{r}}h_{j_{r}j}\\ =&\delta _{ij}S_{r}-T_{(r-1)ik}h_{kj}, \end{aligned}$

因此有$T_{r}=S_{r}I-T_{r-1}B.$此外

$\begin{aligned} {\rm tr}(T_{r-1}B)&=T_{(r-1)ij}h_{ij} =\frac{1}{(r-1)!}\varepsilon ^{i_{1}\cdots i_{r-1}i} _{j_{1}\cdots j_{r-1}j}h_{i_{1}j_{1}}\cdots h_{i_{r-1}j_{r-1}}h_{ij}\\ &=\frac{r!}{(r-1)!}\cdot \frac{1}{r!}\varepsilon ^{i_{1}\cdots i_{r}} _{j_{1}\cdots j_{r}}h_{i_{1}j_{1}}\cdots h_{i_{r}j_{r}} =rS_{r}. \end{aligned}$

即$rS_{r}={\rm tr}(T_{r-1}B).$当$r=1$时, 记$T_{1ij}=T_{ij}.$

定义算子

$\begin{eqnarray*}&&L_{r}:C^{\infty}(M)\rightarrow C^{\infty}(M),\\ &&L_{r}(f):=\sum _{i,j}T_{rij}f_{i,j},\forall f\in C^{\infty}(M).\end{eqnarray*}$

特别地,

$L_{0}(f)=\sum\limits _{i,j}T_{0ij}f_{i,j}=\sum\limits _{i,j}\delta _{ij}f_{i,j}=\triangle f.$

即$L_{0}=\triangle$.

易知此算子为$\triangle$算子的推广.当$r=1$时, 记$L_{1}=L$.

经计算有$L_{r}x=(r+1)S_{r+1}e_{n+1}-c(n-r)S_{r}x,$当$r=0$时, 就是Takahashi定理$\Delta x=nHe_{n+1}-ncx.$

最近, 有作者用高阶平均曲率给出了球面的一些特征(如文献[1, 2]), 本文研究高阶极小超曲面及其高斯映射的进一步性质.

设$x:M^{n}\rightarrow S^{n+1}$为紧致的极小(0 -极小)超曲面, $\Delta x=-nx (\text {此时} H=0,c=1).$

若$x(M^{n})$包含在$S^{n+1}$的闭半球中, 则有常向量$\vec{a}$, 使$\langle x,\vec{a}\rangle \geq 0.$故

$\langle \Delta x,\vec{a}\rangle=\langle -nx,\vec{a}\rangle=-n\langle x,\vec{a}\rangle\leq 0.$

令

$\begin{eqnarray*}&&E_{A}=(0,\cdots ,1_{A},0,\cdots ,0),1\leq A\leq n+2,\\ &&\displaystyle{x=x^{A}E_{A}=(x^{1},\cdots ,x^{n+2})\in R^{n+2}},\end{eqnarray*}$

$x^{A}$是定义在上$M^{n}$的函数, 且

$\begin{eqnarray*} \| x\| ^{2}&=&\sum _{A=1}^{n+2}(x^{A})^{2}=1,\\ \Delta x=(\Delta x^{1},\Delta x^{2},\cdots \Delta x^{n+2})&=&\sum _{A=1}^{n+2}\Delta x^{A}E_{A}, \vec{a}=\sum _{A=1}^{n+2}a^{A}E_{A},\\ \langle \Delta x,\vec{a}\rangle&=&\langle \sum _{A=1}^{n+2}\Delta x^{A}E_{A},\vec{a}\rangle=\sum _{A=1}^{n+2}\Delta x^{A}\langle E_{A},\vec{a}\rangle\\ &=&\sum _{A=1}^{n+2}\Delta x^{A}a^{A}=\sum _{A=1}^{n+2}\Delta (x^{A}a^{A})\\ &=&\Delta \sum _{A=1}^{n+2}(x^{A}a^{A})=\Delta \langle x,\vec{a}\rangle. \end{eqnarray*}$

故$\Delta \langle x,\vec{a}\rangle=\langle \Delta x,\vec{a}\rangle \leq 0,$即$\langle x,\vec{a}\rangle $为下调和函数.由Hopf引理, 及$\Delta \langle x,\vec{a}\rangle=0$知$\langle x,\vec{a}\rangle=0,$则$x(M^{n})=S^{n}\subset S^{n+1}$.而$S^{n}$在$S^{n+1}$中是全测地的, 也就下述经典定理:

定理A $S^{n+1}$中紧致极小超曲面不全含在$S^{n+1}$中的一个开半球中.

推广上面定理有

引理1 对$S^{n+1}$中紧致$r$ -极小超曲面, 若$T_{r}$有定, 则其不全含在$S^{n+1}$的一个开半球中.

证 假设若能全含在$S^{n+1}$的一个开半球中, 则存在常向量$\vec{a}$, 使

$\begin{eqnarray*}&& \langle x, \vec{a}\rangle >0, \\ && L_{r}x=(r+1) S_{r+1}e_{n+1}-c(n-r)S_{r}x =-(n-r)S_{r}x, \end{eqnarray*}$

此时$S_{r+1}=(\begin{array}{c} n \\ r+1 \end{array})H_{r+1}=0,c=1.$令$\varphi (x)=\langle x,\vec{a}\rangle,$类似$\Delta $算子的证明有

$L_{r}(\varphi )=\langle L_{r}x,\vec{a}\rangle =-(n-r)S_{r}\langle x,\vec{a}\rangle =-(n-r)S_{r}\varphi .$

由于$T_{r}=S_{r}I-T_{r-1}B,rS_{r}={\rm tr}(T_{r-1}B)$, 则

${\rm tr}(T_{r})={\rm tr}(S_{r}I-T_{r-1}B)=nS_{r}-rS_{r}=(n-r)S_{r}.$

若$(T_{rij})$正定的, 则$S_{r}=\frac{1}{n-r}{\rm tr}(T_{r})>0.$由于易有$L_{r}$关于$L^{2}$ -内积是自伴的, 则有$\displaystyle\int _{M}L_{r}(\varphi )d_{M}=0$.故

$\int _{M}-(n-r)S_{r}\varphi d_{M}=0\Rightarrow \int _{M}S_{r}\varphi d_{M}=0({S_{r}>0)\Rightarrow \varphi =0},$

即$\langle x,\vec{a}\rangle=0$, 与假设矛盾.

引理2 ^[2] 对于球中$r$ -极小超曲面, $L_{r}$是椭圆算子的充要条件为rank$(h_{ij})>r$.

从此引理及引理1就有

定理1 对于$S^{n+1}$中紧致$r$ -极小超曲面, 如果rank$(h_{ij})>r$, 则其不全含在$S^{n+1}$的一个开半球中.

有类似Hopf引理如下

定理* 若$(T_{ij})$有定, $L(f)=0$(或$L(f)\geq 0$或$L(f)\leq 0$), 则$f=$const.

证 

$\begin{aligned} L(\frac{1}{2}f^{2})&=\frac{1}{2}T_{ij}(f^{2})_{i,j}=\frac{1}{2}(2T_{ij}(ff_{i})_{j})\\ &=T_{ij}f_{i}f_{j}+fT_{ij}f_{i,j}=T_{ij}f_{i}f_{j}+fL(f). \end{aligned}$

由于$\displaystyle\int _{M}gL(f)d_{M}=\int _{M}fL(g)d_{M}$, 则有$\displaystyle\int _{M}L(f)d_{M}=0.$若$L(f)=0$, $(T_{ij})$正定, 则

$0=\int _{M}L(\frac{1}{2}f^{2})d_{M}=\int _{M}T_{ij}f_{i}f_{j}d_{M}\geq 0.$

故$\displaystyle \int _{M}T_{ij}f_{i}f_{j}d_{M}=0.$又因$T_{ij}f_{i}f_{j}\geq 0$ ($(T_{ij})$正定), 则$T_{ij}f_{i}f_{j}=0.$故$f_{i}=0$即$f=$const.

若$L(f)\geq 0$, 且$\displaystyle\int _{M}L(f)d_{M}=0$, 则$L(f)=0$.故$f=$const, 对于$L(f)\leq 0$, 亦成立.

同理, 当$(T_{ij})$负定时, 类似证明.

定义$f:=\langle x,A\rangle$, A为常向量, 则

$df=\langle dx,A\rangle=\langle w_{i}e_{i},A\rangle=\langle e_{i},A\rangle w_{i}, df=f_{i}w_{i},$

故有$f_{i}=\langle e_{i},A\rangle$.由于

$de_{i}=w_{ij}e_{j}+h_{ij}w_{j}e_{n+1}-cw_{i}x=w_{ij}e_{j}+h_{ij}w_{j}e_{n+1}-c\delta _{ij}xw_{j},$

则

$\begin{aligned} f_{i,j}w_{j}&=df_{i}+f_{j}w_{ji}\\ &=d\langle e_{i},A\rangle +\langle e_{j},A\rangle w_{ji}=\langle de_{i}+e_{j}w_{ji},A\rangle \\ &=h_{ij}\langle e_{n+1},A\rangle w_{j}+\langle -c\delta _{ij}xw_{j},A\rangle \\ &=h_{ij}\langle e_{n+1},A\rangle w_{j}-c\langle x,A\rangle \delta _{ij}w_{j}. \end{aligned}$

故有

$ f_{i,j}=h_{ij}\langle e_{n+1},A\rangle -c\langle x,A\rangle \delta _{ij}, x^{A}=\langle x,E_{A}\rangle \in C^{\infty }(M^{n}),$

则有

$\begin{eqnarray*}x^{A}_{,ij}&=&h_{ij}\langle e_{n+1},E_{A}\rangle -c\langle x,E_{A}\rangle \delta _{ij},\\ L(x^{A})&=&T_{ij}x^{A}_{,ij} =T_{ij}h_{ij}\langle e_{n+1},E_{A}\rangle -cT_{ij}\langle x,E_{A}\rangle \delta _{ij}\\ &=&T_{ij}h_{ij}\langle e_{n+1},E_{A}\rangle -cT_{ii}\langle x,E_{A}\rangle,\\ L(x)&=&L(x^{A})E_{A}=T_{ij}h_{ij}\langle e_{n+1},E_{A}\rangle E_{A} -cT_{ii}\langle x,E_{A}\rangle E_{A}\\ &=&\langle T_{ij}h_{ij}e_{n+1},E_{A}\rangle E_{A} -cT_{ii}\langle x,E_{A}\rangle E_{A}\\ &=&T_{ij}h_{ij}e_{n+1}-cT_{ii}x, \end{eqnarray*}$

即

$\begin{eqnarray*}&&L(x)=(\sum _{i,j}h_{ij}T_{ij})e_{n+1}-c\sum _{i}T_{ii}x,\\ &&L\langle x,\vec{a}\rangle=\langle L(x),\vec{a}\rangle=(\sum _{i,j}h_{ij}T_{ij})\langle e_{n+1},\vec{a}\rangle-c\sum _{i}T_{ii}\langle x,\vec{a}\rangle .\end{eqnarray*}$

由于$T_{ij}=nH\delta _{ij}-h_{ij}\Rightarrow T_{ii}=nH-h_{ii},$则有

$\begin{eqnarray*}&&\sum _{i}T_{ii}=\sum _{i}(nH-h_{ii})=n^{2}H-nH=n(n-1)H,\\ &&\sum _{i,j}h_{ij}T_{ij}=\sum _{i,j}h_{ij}(nH\delta _{ij}-h_{ij})=n^{2}H^{2}-S,\end{eqnarray*}$

在$S^{n+1}(c)$中, 纯量曲率$R=n^{2}H^{2}-S+n(n-1)c,$其中$S=\sum _{i,j}(h_{ij})^{2},$则

$L\langle x,\vec{a}\rangle=(R-n(n-1)c)\langle e_{n+1},\vec{a}\rangle -cn(n-1)H\langle x,\vec{a}\rangle. (**)$

若$H=0$即极小时, $R\neq n(n-1)c\Leftrightarrow n^{2}H^{2}\neq S\Leftrightarrow S_{2}\neq 0$, $\langle e_{n+1},\vec{a}\rangle \geq 0$即$x$的Gauss映射像包含在一个闭半球中, 则当$(T_{ij})$有定时, 利用定理*有$\langle e_{n+1},\vec{a}\rangle =0.$

故得到

定理2 对$S^{n+1}(c)$中的紧致极小超曲面, 若$(T_{ij})$有定且$S_{2}\neq 0$, 则其Gauss映射像不全含在$S^{n+1}(c)$的一个开半球中.

设$x:M^{n}\rightarrow R^{n+p} (p\geq 2)$为紧致常平均曲率($H$=const)子流形, $e_{n+1}$为法化平均曲率, 则$\Delta x=nHe_{n+1}-ncx=nHe_{n+1} (\text {此时} c=0)$.若$\langle e_{n+1},\vec{a}\rangle \geq 0$, 即$x$的法化平均曲率的Gauss映射($e_{n+1}:M^{n}\rightarrow S^{n+p-1}$)像包含在的一个闭半球中.因为

$\Delta \langle x,\vec{a}\rangle =\langle \Delta x,\vec{a}\rangle =nH\langle e_{n+1},\vec{a}\rangle \geq 0,$

所以$\langle x,\vec{a}\rangle$为上调和函数.故$\langle e_{n+1},\vec{a}\rangle =0.$ $x$的法化平均曲率的Gauss映射像落在$S^{n+p-2}(\subset S^{n+p-1})$中, 进而$\langle x,\vec{a}\rangle =0,x(M^{n})\subset R^{n+p-1}.$

当$p=1$时, $e_{n+1}:M^{n}\rightarrow S^{n}.$易证$e_{n+1}$为一满射, 即此时覆盖了球至少一次.

事实上, 设$n_{0}\in S^{n}$是一个固定点, 考虑函数$f=\langle x,n_{0}\rangle: M^{n}\rightarrow R.$由于$M^{n}$是紧致的, 则$f$在$M^{n}$上有最大值, 即存在$x_{0}\in M^{n}$, 使得

$df|_{x_{0}}=0, d^{2}f|_{x_{0}}\leq 0,$

即

$\langle dx|_{x_{0}}, n_{0}\rangle =0, \langle d^{2}x|_{x_{0}}, n_{0}\rangle \leq 0,$

亦即$n_{0}$是$M^{n}$在$x_{0}$的法向量, 则存在$x_{0}$, 使$e_{n+1}(x_{0})=n_{0}$, 即$e_{n+1}$是满射.故对于$p=1$时相关的假设不成立.

故得到

定理3 $R^{n+p}$中常平均曲率子流形$M^{n}$的法化平均曲率的Gauss映射像若含在$S^{n+p-1} (p\geq 2)$的一个开半球中, 则$M^{n}$必为$R^{n+p-1}$中子流形.