设$M$是一光滑无边的流形, $X_{0}: {M}\rightarrow{R^{n+m}}$是一个浸入子流形, 考虑欧式空间中一簇单参数光滑超曲面浸入映射$X(0, t):{M}\times{[0, T)}\rightarrow{R^{n+m}}$, 它满足如下发展方程:

$ \begin{equation}\left\{\begin{array}{l} \frac{d}{dt}X(x, t)=H(x, t), \\ X(x, 0)=X_{0}(x), \end{array}\right. {x}\in{M}. \end{equation} $

(1.1)

其中$H(x, t)$是$X(x, t)$平均曲率向量.如果$H=-\frac{X^{N}}{2}$且$X^{N}=\langle X, e_{\alpha}\rangle$, 其中$e_{\alpha}$是流形$M$的法向量场且$\alpha=n+1, n+2, \cdots, n+m$, 则$M$称为self-shrinkers.当$m=1$时, $M$则为超曲面, 这时$X^{N}=\langle X, v\rangle$, 这里$v$是$M$的单位法向量.

对于光滑紧致定向的$n$维黎曼流形$M$, 令$\wedge^{p}(M)$表示$p$阶光滑微分形式构成的空间, $\Delta$是作用在$\wedge^{p}(M)$上的拉普拉斯算子.再令spec$^{p}(M)$表示算子$\Delta$在$\wedge^{p}(M)$上的谱.关于谱与流形之间有一个著名的问题:黎曼流形$M$的spec$^{\ast}(M)$是否决定$M$的几何结构?一般情况下是不成立的, Milnor ^[1], Vigneras^[2]和Ikeda ^[3]分别给除了一些反例.但是, 对于一些特殊流形还是有肯定的答案的. Berger, Patodi以及Tanno等分别在文献[4-6]得到了一些有意义的结果; 李光汉和吴传喜在文献[8]中得到了关于球面中具有常平均曲率超曲面的谱特征; 李振和和王伟在文献[7]中得到了关于球面上Willmore超曲面的谱特征, 即若此超曲面与Willmore torus具有相同的平均曲率, 且二者的谱相等, 则此超曲面为Willmore torus.

本文主要研究了欧氏空间中self-shrinkers谱特征的问题.

主要定理  设$M$是$R^{n+1}$上的$n(n\geq2)$维闭self-shrinkers, 且$M$和$S^{n}(\sqrt{2n})$有相同的平均曲率, 如果spec$^{p}(M) =$spec$^{p}(S^{n}(\sqrt{2n}))~(p=0, 1, 2)$, 则$M$是$S^{n}(\sqrt{2n})$.

设$M$是$R^{n+1}$中$n$维紧致self-shrinker.令$R$, Ric和$\rho$分别表示$M$的黎曼曲率张量, Ricci曲率张量和数量曲率张量, 并用$R_{ijkl}$和$R_{ij}$分别表示$R$和Ric的分量, 则Guass方程表示为

$ \begin{equation} R_{ijkl}=h_{ik}h_{jl}-h_{il}h_{jk}, \end{equation} $

(2.1)

这里这里$h_{ij}$表示$M$在$R^{n+1}$上的第二基本形式$B$的分量.

对于任意固定的${x}\in{M}$, 都可以选取幺正标架场$e_{1}, e_{2}, \cdots, e_{n}$, 使得$(h_{ij})$在$x$点可以对角化, 即

$ \begin{equation} h_{ij}=\kappa_{i}\delta_{ij}. \end{equation} $

(2.2)

令$H$和$S$分别表示$M$的平均曲率和第二基本形式模长的平方, 则

$ \begin{equation} H=\sum\limits_{i=1}^{n}\kappa_{i}, S=\sum\limits_{i=1}^{n}\kappa_{i}^{2}. \end{equation} $

(2.3)

由(2.2) 式知

$ \begin{equation} R_{ijkl}=\kappa_{i}\kappa_{j}(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}), \end{equation} $

(2.4)

则有

$ \begin{eqnarray} && R_{ij}=(H\kappa_{i}-\kappa_{i}\kappa_{j})\delta_{ij}, \end{eqnarray} $

(2.5)

$ \begin{eqnarray} && \rho=H^{2}-S. \end{eqnarray} $

(2.6)

故由模长的定义可知

$ \begin{equation} |R|^{2}=2S^{2}-2\sum\limits_{i=1}^{n}\kappa_{i}^{4}, |{\hbox{Ric}}|^{2}=H^{2}S+\sum\limits_{i=1}^{n}\kappa_{i}^{4}-2H\sum\limits_{i=1}^{n}\kappa_{i}^{3}. \end{equation} $

(2.7)

另外, 对于$p=0, 1, 2, \cdots, n$有

$ \begin{equation} {\rm spec}^{p}(M)=\{0\leq\kappa_{0, p}\leq\kappa_{1, p}\leq\kappa_{2, p}\leq\cdots\uparrow\infty\}. \end{equation} $

(2.8)

而对于这些特征值, 有Minakshisundaram-Pleijel渐进展开式:

$ \begin{equation} \sum\limits_{i=0}^{\infty}e^{-\lambda_{i, p}t}\sim(4{\pi}t)^{-\frac{n}{2}}(a_{0, p}+a_{1, p}t+a_{2, p}t^{2}+\cdots)~~ (t\rightarrow{0^{+}}), \end{equation} $

(2.9)

这里系数${a_{k, p}}(k=0, 1, 2)$ (参看文献[5]), 有以下计算式:

$ \begin{eqnarray} && {a}_{0, p}=\left( \begin{array}{c} n \\ k \\ \end{array}\right){\hbox{vol}}(M), \end{eqnarray} $

(2.10)

$ \begin{eqnarray} && {a}_{1, p}=\left( \frac{1}{6}\left(\begin{array}{c} n \\ p \\ \end{array}\right)-\left(\begin{array}{c} n-2 \\ p-1\\ \end{array}\right)\right)\int_{M}\rho dv, \end{eqnarray} $

(2.11)

$ \begin{eqnarray} a_{2, p}=\int_{M}(c_{1}(n, p)\rho^{2}+c_{2}(n, p)|{\hbox{Ric}}|^{2}+c_{3}(n, p)|R|^{2})dv, \end{eqnarray} $

(2.12)

这里$dv$表示$M$的体积元, 且

$ \begin{eqnarray} && {c_{1}}(n, p)=\frac{1}{72}\left(\begin{array}{c} n \\ p \\ \end{array}\right)-\frac{1}{6}\left(\begin{array}{c} n-2 \\ p-1 \\ \end{array}\right)+\frac{1}{2}\left(\begin{array}{c} n-4 \\ p-2 \\ \end{array}\right), \end{eqnarray} $

(2.13)

$ \begin{eqnarray} && {c_{2}}(n, p)=-\frac{1}{180}\left(\begin{array}{c} n \\ p \\ \end{array}\right)+\frac{1}{2}\left(\begin{array}{c} n-2 \\ p-1 \\ \end{array}\right)-2\left(\begin{array}{c} n-4 \\ p-2 \\ \end{array}\right), \end{eqnarray} $

(2.14)

$ \begin{eqnarray} && {c_{3}}(n, p)=\frac{1}{180}\left(\begin{array}{c} n \\ p \\ \end{array}\right)-\frac{1}{12}\left(\begin{array}{c} n-2 \\ p-1 \\ \end{array}\right)+\frac{1}{4}\left(\begin{array}{c} n-4 \\ p-2 \\ \end{array}\right). \end{eqnarray} $

(2.15)

这里当$l<0$或$p<0$或$l<p$时,

$ \begin{equation} \left( \begin{array}{c}\nonumber l \\ p \\ \end{array}\right)=0. \end{equation} $

设$S^{n}(\sqrt{2n})$是$R^{n+1}$中的超球面, 因为$\lambda_{1}=\lambda_{2}=\cdots=\lambda_{n}=\sqrt{\frac{1}{2n}}$, 则$S^{n}(\sqrt{2n})$的平均曲率$H_{0}=\sum\limits_{i=1}^{n}\lambda_{i}=n\sqrt{\frac{1}{2n}}=\sqrt{\frac{n}{2}}$, 并且$S_{0}= |B_{0}|^{2}=\sum\limits_{i=1}^{n}\lambda_{i}^{2}=n(\sqrt{\frac{1}{2n}})^{2}=\frac{1}{2}$.

对于self-shrinkers, 在文献[10]中有以下定理:

定理 1  设$M$是$R^{n+m}$中完备的$n$维self-shrinker, 且$|B|^{2}\leq\frac{1}{2}$, 则要么$|B|^{2}\equiv0$, 此时$M$是一个平面; 要么$|B|^{2}\equiv\frac{1}{2}$, 此时$M$是${S^{k}(\sqrt{2k})}\times{R^{n-k}}~~({1}\leq{k}\leq{n})$.

当$n=2$时, 由文献[10]中的定理4.2即得结论.现假设$n\geq4$, 令$a_{k, p}$和$a_{k, p}^{0}$分别表示$M$和$M_{0}=S^{n}$ $(\sqrt{2n})$的Minakshisundaram-Pleijel渐进展开式的系数, 若spec$^{p}(M)=$spec$^{p}(M_{0})~(p=0, 1, 2)$, 则有$a_{k, p}=a_{k, p}^{0}$, 结合(2.10)-(2.15) 式以及

$ \begin{equation} \frac{1}{6}\left(\begin{array}{c} n \\ p \\ \end{array}\right)-\left(\begin{array}{c} n-2 \\ p-1\\ \end{array}\right)\neq0\nonumber \end{equation} $

(对一些$p$), 得

$ \begin{eqnarray}&& {\hbox{vol}}(M)={\hbox{vol}}(M_{0}), \end{eqnarray} $

(3.1)

$ \begin{eqnarray} && \int_{M}dv=\int_{{M}_{0}}dv_{0}, \end{eqnarray} $

(3.2)

$ \begin{eqnarray} &&\int_{M}(c_{1}(n, p)\rho^{2}+c_{2}(n, p)|R'|^{2}+c_{3}(n, p)|R|^{2})dv\nonumber\\ &=&\int_{M_{0}}(c_{1}(n, p)\rho_{0}^{2}+c_{2}(n, p)|R_{0}'|^{2}+c_{3}(n, p)|R_{0}|^{2})dv_{0}. \end{eqnarray} $

(3.3)

由(3.1)-(3.3) 式以及$H=H_{0}$知

$ \begin{equation} \int_{M}Sdv=\int_{M_{0}}S_{0}dv_{0}, \end{equation} $

(3.4)

$ \;\;\;\;\int_{M}((c_{1}(n, p)+2c_{3}(n, p))S^{2}+ (c_{2}(n, p)-2c_{3}(n, p))\sum\limits_{i=1}^{n}\lambda_{i}^{4}-(2c_{2}(n, p))H\sum\limits_{i=1}^{n}\lambda_{i}^{3})dv\nonumber\\ =\int_{M_0}((c_1({n, p})+2c_{3}(n, p))S_{0}^{2}+ (c_{2}(n, p)-2c_{3}(n, p))\sum\limits_{i=1}^{n}(\lambda_{i}^{0})^{4}\\-2c_{2}(n, p)H_{0}\sum\limits_{i=1}^{n}(\lambda_{i}^{0})^{3})dv_{0}.\nonumber\\ $

(3.5)

若将上式看作关于$\int_{M}S^{2}dv-\int_{M_0}S^{2}_0dv_0$, $\int_{M}\sum\limits_{i=1}^{n}\lambda_{i}^{4}dv-\int_{M_0}\sum\limits_{i=1}^{n}\lambda{_{i}^0}^{4}dv_0$和$H\int_{M}\sum\limits_{i=1}^{n}\lambda_{i}^{3}dv-\int_{M_0}H_0\sum\limits_{i=1}^{n}\lambda{_{i}^0}^{3}dv_0$的线性方程组, 并且有

$ \begin{equation} \left|\begin{array}{cccc} c_{1}(n, o)+2c_{3}(n, 0)& c_{2}(n, 0)-2c_{3}(n, 0) &-2c_{2}(n, 0) \\ c_{1}(n, 1)+2c_{3}(n, 1)& c_{2}(n, 1)-2c_{3}(n, 1) &-2c_{2}(n, 1) \\ c_{1}(n, 2)+2c_{3}(n, 2)& c_{2}(n, 2)-2c_{3}(n, 2) &-2c_{2}(n, 2) \end{array}\right| =-\frac{1}{180}\neq{0}, \nonumber \end{equation} $

则(3.5) 式有唯一的解, 得

$ \begin{equation} \int_{M}S^{2}dv=\int_{M_{0}}S_{0}^{2}dv_{0}.\nonumber \end{equation} $

由柯西-斯瓦兹不等式及(3.4) 式得

$ \begin{equation} S_{0}{\hbox{vol}}(M_{0})=\int_{M}Sdv\leq(\int_{M}S^{2}dv)^{\frac{1}{2}}(\int_{M}dv)^{\frac{1}{2}} =(\int_{M_{0}}S_{0}^{2})^{\frac{1}{2}}({\hbox{vol}}(M))^{\frac{1}{2}}=S_{0}{\hbox{vol}}(M_{0}), \nonumber \end{equation} $

故有$ S=S_{0}, $所以$ |B|^{2}=S=S_{0}=|B_{0}|^{2}=\frac{1}{2}. $再由第二部分中的定理1即得$M$是超球面$S^{n}(\sqrt{2n})~(n\geq4)$.

当$n=3$时, 仍有(3.1)-(3.5) 式成立, 并且, 此时有

$ \begin{equation} 3S^{2}+8H\sum\limits_{i=1}^{3}\lambda_{i}^{3}-6\sum\limits_{i=1}^{3}\lambda_{i}^{4}= 5H^{4}-12H^{2}\sum\limits_{i>j=1}^{3}\lambda_{i}\lambda_{j}. \end{equation} $

(3.6)

将(3.6) 式代入(3.1)-(3.3) 式中得到

$ \begin{eqnarray*} &&\int_{M}((c_{1}(3, p)+\frac{3}{4}c_{2}(3, p)+2c_{3}(3, p))S^{2}- (\frac{1}{2}c_{2}(3, p)+2c_{3}(3, p))\sum\limits_{i=1}^{3}\lambda_{i}^{4})dv\\ &=&\int_{M_{0}}((c_{1}(3, p)+\frac{3}{4}c_{2}(3, p)+2c_{3}(3, p))S_{0}^{2}- (\frac{1}{2}c_{2}(3, p)+2c_{3}(3, p))\sum\limits_{i=1}^{3}(\lambda_{i}^{0})^{4})dv_{0}, \nonumber \end{eqnarray*} $

其中$p=0, 1$.

同时将上式看作$\int_{M}S^{2}dv-\int_{M_0}S^{2}_0dv_0$和$\int_{M}\sum\limits_{i=1}^{n}\lambda_{i}^{4}dv-\int_{M_0}\sum\limits_{i=1}^{n}\lambda{_{i}^0}^{4}dv_0$的线性方程组, 且

$ \begin{equation} \left|\begin{array}{ccc} c_{1}(3, 0)+\frac{3}{4}c_{2}(3, 0)+2c_{3}(3, 0)& \frac{1}{2}c_{2}(3, 0)-2c_{3}(3, 0) \\ c_{1}(3, 1)+\frac{3}{4}c_{2}(3, 1)+2c_{3}(3, 1) & \frac{1}{2}c_{2}(3, 1)-2c_{3}(3, 1) \end{array}\right| \neq{0}, \nonumber \end{equation} $

则以上方程有唯一的解, 因此$ \int_{M}S^{2}dv=\int_{M_{0}}S_{0}^{2}dv_{0}. $同$n\geq4$情形的方法可得$M=S^{3}(\sqrt{6})$.

综上, 定理得证.