设$CP^{n+p}$是具有Fubini-Study度量的复$n+p$维复射影空间, 全纯截面曲率为常数4.设$J$为$CP^{n+p}$的复结构, $M^{n}$为$CP^{n+p}$的实$n$维子流形.如果$M^{n}$上每点切空间被$J$变换到自身, 则称$M^{n}$是$CP^{n+p}$的全纯子流形.与此相反, 若$M^{n}$上每点的切空间被$J$变换到该点法空间, 则称$M^{n}$为$CP^{n+p}$的全实子流形.

关于具有常数量曲率子流形, Cheng和Yau最早研究了空间形式中的常数量曲率超曲面(见文献[1]), 引入了一个自共轭的二阶椭圆算子, 这个算子现在仍是我们研究具有常数量曲率子流形的重要工具.本文研究复射影空间中常数量曲率的完备全实子流形, 得到了

定理 设$M^{n}$是$CP^{n+p}$中具有常数量曲率($R\geq1$)的完备全实子流形则

$1)$当$p=1$或$p\leq2, n\geq3$或$p\geq3, n>7$时, 若$\sup S\leq2\sqrt{n-1}$, 则$M^{n}$是$CP^{n+p}$中的全脐子流形.

$2)$当$p\geq3, 3<n\leq7$时, 若$\sup S\leq\frac{2}{3}n$, 则$M^{n}$是$CP^{n+p}$中的全脐子流形.其中$S$为第二基本形式模长平方.

设$M^{n}$是$CP^{n+p}$中实$n$维全实子流形, $J$为$CP^{n+p}$的复结构.在$CP^{n+p}$上选取局部规范正交标架场

$ e_1, \cdots, e_n, e_{n+1}, \cdots, e_{n+p}, e_{1*}=Je_1, \cdots, e_{n^*}=Je_n, e_{(n+1)^{*}}=Je_{n+1}, \cdots, e_{(n+p)^{*}}=Je_{n+p} $

使得限制于$M^{n}, \{e_1, \cdots, e_{n}\}$与$M^{n}$相切, 约定各类指标的取值范围

$ A, B, C, \cdots=1, \cdots, n+p, 1^{*}, \cdots, (n+p)^*; i, j, k, \cdots=1, \cdots, n;\\ \alpha, \beta, \gamma, \cdots=n+1, \cdots, n+p, 1^{*}, \cdots, (n+p)^*;\\ \lambda, \mu\cdots=n+1, \cdots, n+p.$

以$\{\omega_A\}$表示$\{e_A\}$的对偶标架场, 则$CP^{n+p}$的结构方程为

$\begin{eqnarray} d\omega_A=-\sum\limits_B\omega_{AB}\wedge \omega_B, \end{eqnarray}$

(2.1)

$\begin{eqnarray} d\omega_{AB}=-\sum\limits_C\omega_{AC}\wedge \omega_{CB}+\frac{1}{2}\sum\limits_{CD}K_{ABCD}\omega_C\wedge \omega_D, \end{eqnarray}$

(2.2)

其中

$\begin{eqnarray} \omega_{ij}=\omega_{i^*j^*}, \omega_{i^*j}=\omega_{j^*i}, \omega_{\lambda\mu}=\omega_{\lambda^*\mu^*};\nonumber\\ \omega_{\lambda^*\mu}=\omega_{\mu^*\lambda}, \omega_{i\mu}=\omega_{i^*\mu^*}, \omega_{i^*\lambda}=\omega_{\lambda^*i};\end{eqnarray}$

(2.3)

$\begin{eqnarray} K_{ABCD}=\delta_{AC}\delta_{BD}-\delta_{AD}\delta_{BC}+J_{AC}J_{BD}-J_{AD}J_{BC}+2J_{AB}J_{CD}, \end{eqnarray}$

(2.4)

这里$(J_{AB})$为线性变换$J$关于$\{e_A\}$的变换矩阵, 即

$\begin{equation}J_{AB}=\left( \begin{array}{cc} 0 I_{n+p} \\ -I_{n+p} 0 \\ \end{array} \right), \end{equation}$

(2.5)

其中$I_{n+p}$为$n+p$阶单位矩阵.

将上述形式限制在$M$上, 则有

$\begin{eqnarray} \omega_\alpha=0, \omega_{\alpha i}=\sum\limits_j h^{\alpha}_{ij}\omega_j, \end{eqnarray}$

(2.6)

$\begin{eqnarray} h=\sum\limits_{\alpha, i, j}h_{ij}^\alpha \omega_i\otimes \omega_j\otimes e_\alpha, h_{jk}^{i^*}=h_{ik}^{j^*}=h_{ij}^{k^*}, \end{eqnarray}$

(2.7)

$\begin{eqnarray} \begin{cases} d\omega_i=-\sum\limits_j\omega_{ij}\wedge \omega_j, \\ d\omega_{ij}=-\sum\limits_k\omega_{ik}\wedge \omega_{kj}+\frac{1}{2}\sum\limits_{kl}R_{ijkl}\omega_k\wedge \omega_l, \end{cases}\end{eqnarray}$

(2.8)

$\begin{eqnarray} R_{ijkl}=K_{ijkl}+\sum\limits_\alpha(h_{ik}^\alpha h_{jl}^\alpha-h_{il}^\alpha h_{jk}^\alpha), \end{eqnarray}$

(2.9)

$\begin{eqnarray} d\omega_{\alpha\beta}=-\sum\limits_\gamma \omega_{\alpha \gamma}\wedge \omega_{\gamma\beta}+\frac{1}{2}\sum\limits_{kl}R_{\alpha \beta kl}\omega_k\wedge \omega_l, \end{eqnarray}$

(2.10)

$\begin{eqnarray} R_{\alpha \beta ij}=K_{\alpha \beta ij}+\sum\limits_k(h_{ik}^\alpha h_{kj}^\beta-h_{jk}^\alpha h_{ki}^\beta), \end{eqnarray}$

(2.11)

其中$R_{ijkl}, R_{\alpha\beta ij}$分别是$M^n$的Riemann曲率张量场和法曲率张量场关于$\{e_A\}$的分量.进一步, $M^n$的平均曲率向量场$\xi$, 平均曲率$H, $第二基本形式模长平方$S$及标准化数量曲率$R$可表示为

$\begin{eqnarray} \xi=\frac{1}{n}\sum\limits_\alpha(\sum\limits_ih_{ii}^\alpha)e_\alpha, H=\parallel \xi\parallel^2, S=\parallel h \parallel^{2}, \end{eqnarray}$

(2.12)

$\begin{eqnarray} n(n-1)R =n(n-1)+n^2H^2-S.\end{eqnarray}$

(2.13)

用$h_{ijk}^\alpha $及$h_{ijkl}^\alpha $表示$h_{ij}^\alpha $的共变导数, 则

$\begin{eqnarray} h_{ijk}^\alpha-h_{ikj}^\alpha=-K_{\alpha ijk}=0, \end{eqnarray}$

(2.14)

$\begin{eqnarray} h_{ijkl}^\alpha-h_{ijlk}^\alpha=\sum\limits_m(h_{im}^\alpha R_{mjkl}+h_{mj}^\alpha R_{mikl})-\sum\limits_\beta h_{ij}^\beta R_{\alpha\beta k l}, \end{eqnarray}$

(2.15)

由(2.14), (2.15) 式, 且记$h_{ij}^\alpha$的Laplacian为$\bigtriangleup h_{ij}^\alpha$, 则

$\begin{eqnarray} \frac{1}{2}\Delta S =\sum\limits_{i, j, k, \alpha}(h_{ijk}^{\alpha})^2+\sum\limits_{i, j, k, \alpha}h_{ij}^{\alpha} h_{kkij}^{\alpha}+nS-n^2H^2+2\sum\limits_{\alpha, \beta >n+1}[{\hbox{tr}}(A_{\alpha} A_{\beta})^2-{\hbox{tr}}(A^{2}_{\alpha} A^{2}_{\beta})]\nonumber\\ -\sum\limits_{\alpha, \beta >n+1}[{\hbox{tr}}A_{\alpha} A_{\beta}]^{2} -\sum\limits_{\alpha >n+1}[{\hbox{tr}}A{_\alpha} A_{n+1}]^{2}-\sum\limits_{\beta}[\hbox{tr}A_{n+1}A_{\beta}]^2+2\sum\limits_\beta[{\hbox{tr}}(A_{n+1}A{_\beta})^2\nonumber\\-A^{2}_{n+1} A^{2}_{\beta})]+\sum\limits_{\alpha, \beta}({\hbox{tr}}A{_\alpha}^{2} A_{\beta})\cdot {\hbox{tr}}A_{\beta}+\sum\limits_{m}{\hbox{tr}}A_{m^*}^2. \end{eqnarray}$

(2.16)

可选取$e_{n+1}=\frac{\xi}{H}^{[2]}$, 此时(2.16) 式变为

$\begin{eqnarray} \frac{1}{2}\Delta S =\sum\limits_{i, j, k, \alpha}(h_{ijk}^{\alpha})^2+\sum\limits_{i, j}h_{ij}^{n+1} (nH)_{ij}+nS-n^2H^2\\ +2\sum\limits_{\alpha, \beta >n+1}[{\hbox{tr}}(A_{\alpha} A_{\beta})^2-{\hbox{tr}}(A^{2}_{\alpha} A^{2}_{\beta})]\nonumber\\ -\sum\limits_{\alpha, \beta >n+1}[\hbox{tr}A_{\alpha} A_{\beta}]^{2} -\sum\limits_{\alpha >n+1}[{\hbox{tr}}A{_\alpha} A_{n+1}]^{2}-\sum\limits_{\beta}[\hbox{tr}A_{n+1}A_{\beta}]^2\\ +2\sum\limits_\beta[\hbox{tr}(A_{n+1}A{_\beta})^2\nonumber\\-{\hbox{tr}}A^{2}_{n+1} A^{2}_{\beta})]+nH\sum\limits_{\alpha>n+1}({\hbox{tr}}A{_\alpha}^{2} A_{n+1})+nH{\hbox{tr}}(A_{n+1}^{3})+\sum\limits_{m}{\hbox{tr}}A_{m^*}^2. \end{eqnarray}$

(2.17)

另外, 我们需要下面的引理

引理1 ^[3]  设$a_1, a_2, \cdots, a_n, b_1, \cdots, b_n(n\geq2)$是实数, 且$\sum\limits_ia_i=0$, 则

$\sum\limits_{i, j}b_ib_j(a_i-a_j)^2\geq-\frac{n}{\sqrt{n-1}}(\sum\limits_ia_i^2)(\sum\limits_ib_i^2).$

引理2 ^[4] 设$a_1, a_2, \cdots, a_n(n\geq 2)$是实数, 且满足$\sum\limits_ia_i=0$.则有

$|\sum\limits_ia_i^3|\leq\frac{n-2}{\sqrt{n(n-1)}}(\sum\limits_ia_i^2)^{\frac{3}{2}}$

且等号成立当且仅当至少有$n-1$个$a_i$相等.

引理3 ^[5-6]  设$M$是Ricci曲率有下界的完备黎曼流形, 若$F$是$M$上$C^{\infty}$有上界的函数, 则$\forall\varepsilon>0, \exists x\in M$使得

$\sup F-\varepsilon＜F(x), \|{\hbox{grad}}F\|＜\varepsilon, \Delta F(x)＜\varepsilon.$

(2.18)

定理证明还需要下面的引理.

引理4  设$M$是$CP^{n+p}$中具有常数量曲率的完备子流形, 若$S$有上界, 且$F=-(f^{2}+a)^{-\frac{1}{2}}$, 这里$a\in R, f=\sqrt{nH}$是$M$上的$C^2$非负函数, 则对任意收敛序列$\{\varepsilon_{m}\}$满足$\varepsilon_m>0, \varepsilon_m\rightarrow 0(m\rightarrow 0)$, 存在点列{$x_m$}使得

$\lim\limits_{m\rightarrow\infty}\sup\triangle(nH)(x_m)\leq 0, F_0:=\lim\limits_{m\rightarrow\infty}F(x_m)=\sup F, f_0:=\lim\limits_{m\rightarrow\infty}f(x_m)=\sup f.$

证  通过直接计算

$\begin{equation}F\triangle F=3\|{\hbox{grad}}F\|^2-\frac{1}{2}F^4\Delta f^2.\end{equation}$

(3.1)

由$S$有上界及$R$为常数知F有上界.由Gauss方程知$M$的Ricci曲率有下界.由引理3及(3.1) 式我们有

$ \begin{equation}\frac{1}{2}F^4(x)\Delta f^2(x)=3\|{\hbox{grad}}F\|^2(x)-F(x)\Delta F(x)＜3\varepsilon^2-\varepsilon F(x).\end{equation}$

(3.2)

这样对任意收敛序列$\{\varepsilon_m\}$满足$\varepsilon_m>0, \varepsilon_m\rightarrow 0(m\rightarrow 0), $存在点列$\{\tilde{x_m}\}$使得(2.18) 式成立.因此$\varepsilon_{m} (3 \varepsilon -F(\tilde{x_m}))\rightarrow 0(m\rightarrow 0)$.

另一方面, 由(2.18) 式知$F(\tilde{x_m})>\sup F-\varepsilon_m.$又$F$有上界, 可选$\{\tilde{x_m}\}$的子列记作$\{x_m\}$使得$F(x_m)\rightarrow F_0(m\rightarrow\infty)$并且(2.18) 式成立.由上确界的定义及(2.18) 式知$F_0=\sup F$且由F的定义知$f(x_m)\rightarrow f_0=\sup f(m\rightarrow\infty)$, 由(3.2) 式有

$3\varepsilon_m^2-F(x_m)\varepsilon_m>\frac{1}{2}F^4(x_m)\Delta f^2(x_m)=\frac{1}{2}F^4(x_m)\Delta(nH)(x_m), $

即$\lim\limits_{m\rightarrow\infty}\sup\triangle(nH)(x_m)\leq 0.$

现在证明定理.

令$u_i = h_{ii}^{n+1} - H$, 于是$\sum\limits_iu_i = 0, \sum\limits_iu_i^2 = {\hbox{tr}}A_{n+1}^2 - nH^2 \triangleq \big| z\big|^2.$

由引理2, 有

$\begin{eqnarray} nH{\hbox{tr}}(A_{n+1}^3) = nH\sum\limits_i(h_{ii}^{n+1})^3\nonumber \\ = 3nH^2 \cdot | z|^2 + n^2H^4 + nH \cdot \sum\limits_iu_i^3\nonumber\\ \geq 3nH^2 \cdot |z|^2 + n^2H^4 - \frac{n(n-2)}{\sqrt{n(n-1)}}H\cdot |z|^3 \nonumber\\ = |z|^2( nH^2 - \frac{n(n-2)}{\sqrt{n(n-1)}}H\cdot |z| - |z|^2 ) + ({\hbox{tr}}A_{n+1}^2)^2, \end{eqnarray}$

(3.3)

考虑以$\pm \frac{n}{2\sqrt{n-1}}$为特征值的二次型

$ F(x, y) = x^2 - \frac{n-2}{\sqrt{n-1}}xy - y^2 $

作正交变换

$\begin{eqnarray} \begin{cases} u = \frac{1}{\sqrt{2n}}\left[(1+\sqrt{n-1})x + (1-\sqrt{n-1})y \right], \\ v = \frac{1}{\sqrt{2n}}\left[(\sqrt{n-1}-1)x + (\sqrt{n-1}+1)y \right], \end{cases}\\ F(x, y) = \frac{n}{2\sqrt{n-1}}(u^2 - v^2). \nonumber\end{eqnarray}$

(3.4)

令$x = \sqrt{n} H, y = \big|z\big|$, 则$x^2+y^2 = \hbox{tr}A_{n+1}^2$且(3.4) 式是正交变换, 于是从$x^2+y^2 = u^2 + v^2 = {\hbox{tr}}A_{n+1}^2$得

$\begin{eqnarray} nH^2 - \frac{n(n-2)}{\sqrt{n(n-1)}}H\cdot \big|z\big| - \big|z\big|^2 = x^2 - \frac{n-2}{\sqrt{n-1}}xy - y^2 \nonumber\\ =\frac{n}{2\sqrt{n-1}}(u^2 - v^2) =\frac{n}{2\sqrt{n-1}}(2u^2 -{\hbox{ tr}}A_{n+1}^2)\nonumber\\ \geq -\frac{n}{2\sqrt{n-1}}{\hbox{tr}}A_{n+1}^2, \end{eqnarray}$

(3.5)

又

$\begin{equation} \sum\limits_{\beta}[\hbox{tr}(A_{n+1}A_\beta)]^2 = \sum\limits_{\beta > n+1}( \sum\limits_{i, j}(h_{ij}^{n+1} - H\delta_{ij})h_{ij}^\beta )^2 \leq \big|z\big|^2 \cdot S, \end{equation}$

(3.6)

由(2.13) 式及引理1知

$\begin{aligned} nH{\hbox{tr}}(A_\alpha^2 A_{n+1}) - \left[{\hbox{ tr}}(A_\alpha A_{n+1}) \right]^2 = \frac{1}{2}\sum\limits_{i, j}h_{ii}^{n+1}h_{jj}^{n+1}(h_{ii}^\alpha - h_{jj}^\alpha)^2 \\ \geq -\frac{n}{2\sqrt{n-1}}{\hbox{ tr}}A_\alpha^2 {\hbox{ tr}}A_{n+1}^2 \quad (\alpha > n+1). \end{aligned}$

故

$\begin{equation} nH\sum\limits_{\alpha > n+1}{\hbox{ tr}}(A_\alpha^2 A_{n+1}) - \sum\limits_{\alpha > n+1} \left[{\hbox{ tr}} (A_\alpha A_{n+1}) \right]^2 \geq -\frac{n}{2\sqrt{n-1}}tr A_{n+1}^2 \sum\limits_{\alpha > n+1}{\hbox{ tr}} A_\alpha^2, \end{equation}$

(3.7)

再由文献[7]及简单计算有

$\begin{eqnarray} -2\sum\limits_{\alpha, \beta > n+1}\left[{\hbox{ tr}}(A_\alpha^2 A_\beta^2)-{\hbox{ tr}} (A_\alpha A_\beta)^2 \right] - \sum\limits_{\alpha, \beta > n+1}\left[{\hbox{ tr}} (A_\alpha A_\beta) \right]^2\nonumber\\ \geq -\left[1+ \frac{1}{2}\textrm{sgn}(p-2) \right]\left( \sum\limits_{\alpha > n+1}{\hbox{ tr}} A_\alpha^2 \right)^2, \end{eqnarray}$

(3.8)

对$\omega_{n+1\alpha}=0$外微分, 注意到$K_{n+1\alpha ij}=0$,

$0= d\omega_{n+1\alpha}=\sum\limits_{i}\omega_{n+1i}\wedge \omega_{i\alpha}+\sum\limits_{\beta}\omega_{n+1\beta}\wedge \omega_{\beta\alpha}\\ = -\sum\limits_{i, j, k}h_{ij}^{n+1}h_{ik}^\alpha \omega_j\wedge \omega_k =\sum\limits_{j<k}\sum\limits_i(h_{ij}^{n+1}h_{ik}^\alpha-h_{ik}^{n+1}h_{ij}^{\alpha})\omega_j\wedge \omega_k, $

从而

$\begin{equation}H_{n+1}H_\alpha=H_\alpha H_{n+1}, \end{equation}$

(3.9)

于是对于任意$\alpha, H_\alpha, H_{n+1}$可同时对角化, 由(2.17), (3.3)-(3.9) 式有

$\begin{equation}\frac{1}{2}\Delta S\geq\sum\limits_{i, j, k, \alpha}(h_{ijk}^{\alpha})^2+\sum\limits_{i, j}h_{ij}^{n+1}(nH)_{ij}+nS-n^2H^2-|z|^2 \frac{n}{2\sqrt{n-1}}S-MS\sum\limits_{\alpha>n+1}\hbox{tr}A^2_{\alpha}.\end{equation}$

(3.10)

其中$M = \max\left\{ 1+ \frac{1}{2}\textrm{sgn}(p-2), \frac{n}{2\sqrt{n-1}} \right\}.$

定义$M^n$上的对称张量场$T=\sum\limits_{i, j}T_{ij}w_i\otimes w_j$如下

$T_{ij}=nH\delta_{ij}-h_{ij}^{n+1}.$

我们引入一个与$T$有关的算子$\Box$, 它作用在函数$f\in C^2(M^n)$上为

$\square f=\sum\limits_{i, j}T_{ij}f_{ij}=nH\triangle f-\sum\limits_{i, j}h_{ij}^{n+1}f_{ij}.$

由于R是常数, 应用自伴算子$\Box$到$nH$并利用(2.13) 式有

$\begin{eqnarray} \Box(nH) =\sum\limits_{i, j}(nH\delta_{ij}-h^{n+1}_{ij})(nH)_{ij}=\frac{1}{2}\Delta(nH)^2-\|{\hbox{grad}}(nH)\|^2-\sum\limits_{i, j}h^{n+1}_{ij}(nH)_{ij}\nonumber\\ =\frac{1}{2}\Delta S-\|{\hbox{grad}}(nH)\|^2-\sum\limits_{i, j}h^{n+1}_{ij}(nH)_{ij}. \end{eqnarray}$

(3.11)

由R(R$\geq 1$)是常数, 利用(2.13) 式有$nH\nabla_k(nH)=\sum\limits_{i, j, \alpha}h^{\alpha}_{ij}h^{\alpha}_{ijk}$, 利用施瓦兹不等式有

$\begin{eqnarray} n^2H^2\|{\hbox{grad}}(nH)\|^2 =\sum\limits_k(\sum\limits_{i, j, \alpha}h^{\alpha}_{ij}h^{\alpha}_{ijk})^2\leq(\sum\limits_{i, j, \alpha}(h^\alpha_{ij})^2) (\sum\limits_k\sum\limits_{i, j, \alpha}(h^\alpha_{ijk})^2)\nonumber\\ = S\sum\limits_k\sum\limits_{i, j, \alpha}(h^\alpha_{ijk})^2\leq n^2H^2\sum\limits_{i, j, k, \alpha}(h^\alpha_{ijk})^2, \end{eqnarray}$

(3.12)

将(3.10), (3.12) 式代入(3.11) 式得

$\Box(nH)\geq n(S-nH^2)-|z|^2\frac{n}{2\sqrt{n-1}}S-MS\sum\limits_{\alpha>n+1}{\hbox{tr}}A^2_{\alpha}.$

(3.13)

另一方面

$\Box(nH)=\sum\limits_{i}(nH-h^{n+1}_{ii})(nH)_{ii}\leq(nH_{\max}-\lambda)\Delta(nH), $

(3.14)

其中$H_{\max}$是平均曲率H的最大值，$\lambda$是主曲率$\{\lambda_i\}^{n+p}_{i=1}$的最小值.

由(2.13) 式和引理3, 取一收敛序列$\{\varepsilon_m\}$满足$\varepsilon_m>0, \varepsilon_m\rightarrow 0(m\rightarrow\infty), $存在点列$\{x_m\}$使得$H(x_m)\rightarrow \sup H(m\rightarrow\infty)$, 由(2.13) 式知$S(x_m)\rightarrow \sup S(m\rightarrow\infty).$

下面分两种情况讨论

(1) 当$p=1$或$p\leq 2, n\geq 3$或$p\geq 3, n>7$时, 此时$M=\frac{n}{2\sqrt{n-1}}$, 则由引理4, (3.13), (3.14) 式得

$\sup(S-nH^2)(n-\frac{n}{2\sqrt{n-1}}\sup S)\leq 0.$

(3.15)

若$\sup S< 2\sqrt{n-1}$, 则由(3.15) 式知$S=nH^2$, 即$M$是$CP^{n+p}$中全脐子流形.

(2) 当$p\geq 3, n<3\leq 7$时, $M=\frac{3}{2}$, 此时由引理4, (3.13), (3.14) 式得

$\sup(S-nH^2)(n-\frac{3}{2}\sup S)\leq 0.$

(3.16)

若$\sup S<\frac{2}{3}n$, 则由(3.16) 式知$S=nH^2$, 即$M$是$CP^{n+p}$中全脐子流形.