设$ f:M^n\rightarrow N^{n+p} $是等距浸入, 则

1  $ M^n $是全测地子流形的充要条件是$ M^n $是全脐子流形也是极小子流形

2  $ M^n $是极小子流形的必要条件是$ M^n $是伪脐子流形

3   $ M^n $是全脐子流形的必要条件是$ M^n $是迷向子流形

因为Kähler流形的Kähler子流形必是极小子流形, 我们考虑上述结果反过来若要成立的条件.一般地, 对于一个具有度量$g_{ij} $的黎曼流形, 如果其曲率张量满足以下等式：

$ R_{ijkl} = K(g_{ik} g_{jl} - g_{il} g_{jk}), $

则称其为常曲率黎曼流形, 其中K是一个常数.

1972年, Chen和矢野[1]在研究共形平坦空间时提出了拟全脐的概念, 该文中他们得到了这类空间的一个曲率的表达式：

$ K_{kij}^h=\left( k+\alpha ^2 \right) \left( \delta _{k}^{h}g_{ij}-\delta _{j}^{h}g_{ki} \right) +\alpha \beta \left[ \left( \delta _{k}^{h}v_j-\delta _{k}^{j}v_h \right) v_i+\left( v_kg_{ji}-v_jg_{ki} \right) v^h \right]. $

1986年, 白正国[2]将这类空间命名为拟常曲率空间, 定义为曲率张量写作以下形式的黎曼空间：

$ K_{ABCD}=a\left( g_{AC}g_{BD}-g_{AD}g_{BC} \right) +b\left( g_{AC}\lambda _B\lambda _D+g_{BD}\lambda _A\lambda _C-g_{AD}\lambda _B\lambda _C-g_{BC}\lambda _A\lambda _D \right). $

这里的$ a, b $是任意光滑函数, $ \left\{ \lambda _A \right\} $是一组单位向量在标准基下的分量.

而复射影空间的曲率张量可以写作[3]：

$ K_{ABCD}=\delta _{AC}\delta _{BD}-\delta _{AD}\delta _{BC}+J_{AC}J_{BD}-J_{AD}J_{BC}+2J_{AD}J_{BC}. $

类似地, 2011年宋卫东教授在文献[4]中提出了拟复射影空间的概念, 他将拟复射影空间定义为曲率张量满足

$ \begin{equation} \begin{aligned} K_{ABCD}=&a\left( g_{AC}g_{BD}-g_{AD}g_{BC}+J_{AC}J_{BD}-J_{AD}J_{BC}+2J_{AB}J_{CD} \right) \\ &+b\left( g_{AC}\lambda _B\lambda _D+g_{BD}\lambda _A\lambda _C-g_{AD}\lambda _B\lambda _C-g_{BC}\lambda _A\lambda _D \right)\notag. \end{aligned} \end{equation} $

的复$ n+p $维黎曼流形, 做为复射影空间的推广.

一般地, 设$ p\in M, u=\sum_i^{}{u^ie_i\in T_pM} $, 其中$ \sum_i^{}{\left( u^i \right) ^2}=1 $, 称

$ \begin{equation} h_p\left( u, u \right) =\sum\limits_{\alpha , i, j}^{}{h_{ij}^{\alpha}u^iu^je_{\alpha}} \end{equation} $

(1.1)

为关于$ u $在$ p $点的法曲率张量.若$ \lVert h_p \rVert =\lambda _p $, 则称$ M^n $为$ N^{n+p} $中$ \lambda $-迷向子流形[5].

对于复射影空间的$ \lambda- $迷向子流形, 尹松庭证明了如下定理[5].

定理A  设$ M^n $为复射影空间$ CP^{n+p}\left( 4 \right) $中迷向Kähler子流形, 如果$ M^n $第二基本形式模长平方S满足

$ S<\frac{2\left( n+1 \right) \left( n+4 \right)}{4n-1}, $

则$ M^n $是全测地的.

定理B  设$ M^n $为复射影空间$ CP^{n+p}\left( 4 \right) $中迷向Kähler子流形, 如果在每一点处$ M^n $截面曲率满足

$ K\ge \frac{3\left( n+1 \right)}{4n^2+10n+3}, $

其中$ K $是$ M^n $截面曲率的下确界, 则$ M^n $是全测地的.

定理C  设$ M^n $为复射影空间$ CP^{n+p}\left( 4 \right) $中迷向Kähler子流形, 如果在每一点处$ M^n $的Ricci曲率满足

$ Q>\left( 2n+2 \right) -\frac{\left( n+1 \right) \left( n+4 \right)}{n\left( 4n-1 \right)}, $

其中$ Q $是$ M^n $的Ricci曲率的下确界, 则$ M^n $是全测地的.

本文将外围空间扩展到拟复射影空间, 得到如下结果. 首先是在局部对称的条件下, 得到了：

定理1.1  若$ M^n $是局部对称拟复射影空间$ CQ^{n+p} $的迷向Kähler子流形, 则当

$ S<\frac{\left[ a\left( 8n^2+4n-3 \right) -2\left| b \right|\left( 8n^2-6n+2 \right) \right] \left( n+1 \right)}{n\left( 4n-1 \right)} $

时, $ M^n $是全测地的.

定理1.2  若$ M^n $是局部对称$ CQ^{n+p} $的Kähler迷向子流形, 若

$ K>-\frac{a\left( 4n-1 \right)}{8n^2\left( n+1 \right)}+\frac{\text{（}2n^2+4n-1\left( \left[ 2\left| b \right|\left( 8n^2-6n+2 \right) -a\left( 8n^2+4n-3 \right) \right] \right)}{8n^3\left( n+1 \right) \left( 4n-1 \right)}, $

则$ M^n $是全测地的, 其中$ K $是$ M^n $的截面曲率下确界.

定理1.3  若$ M^n $是局部对称$ CQ^{n+p} $的Kähler迷向子流形, 若

$ Q>2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) +\frac{\left( n+1 \right) \left[ 2\left| b \right|\left( 8n^2-6n+2 \right) -a\left( 8n^2+4n-3 \right) \right]}{2n^2\left( 4n-1 \right)}, $

则$ M^n $是全测地的, 其中$ Q $是$ M^n $的Ricci曲率下确界.

推论1.4  若$ M^n $是局部对称$ CP^{n+p} $的Kähler迷向子流形, 则$ S<\frac{\left( 8n^2+4n-3 \right) \left( n+1 \right)}{n\left( 4n-1 \right)} $时, $ M^n $是全测地的.

推论1.5  若$ M^n $是局部对称$ CP^{n+p} $的Kähler迷向子流形, 则$ K>-\frac{4n-1}{8n^2\left( n+1 \right)}-\frac{\left( 2n^2+4n-1 \right) \left( 8n^2+4n-3 \right)}{8n^3\left( n+1 \right) \left( 4n-1 \right)} $时, $ M^n $是全测地的, 其中$ K $是$ M^n $的截面曲率下确界.

推论1.6  若$ M^n $是局部对称$ CP^{n+p} $的Kähler迷向子流形, 则$ Q>2\left( n+1 \right) -\frac{\left( n+1 \right) \left( 8n^2+4n-3 \right)}{2n^2\left( 4n-1 \right)} $时, $ M^n $是全测地的, 其中$ Q $是$ M^n $的Ricci曲率下确界.

另一方面, 在没有局部对称条件的情况下, 我们得到了：

定理1.7  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 记$ X=2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) $, $ Y=2n\left( 8n^2+2n-1 \right) $, 若

$ \left\{ \begin{array}{l} X>0\\ X^2>b^2Y\\ \, \, 2n\left( n+1 \right) \frac{X-\sqrt{X^2-b^2Y}}{2Y}<S<2n\left( n+1 \right) \frac{X+\sqrt{X^2-b^2Y}}{2Y}\\ \end{array} \right. $

则$ M^n $是全测地的.

定理1.8  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 记$ X=2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) $, $ Y=2n\left( 8n^2+2n-1 \right) $, 若$ K>-\frac{a\left( 4n-1 \right)}{8n^2\left( n+1 \right)}+\frac{2Y\left[ 2n^2\left( 4n-1 \right) +\frac{1}{4}b^2 \right]}{8n^2\left( n+1 \right) \left( X+\sqrt{X^2-b^2Y} \right)}, $则$ M^n $是全测地的, 其中$ K $是$ M^n $的截面曲率下确界.

定理1.9  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 记$ X=2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) $, $ Y=2n\left( 8n^2+2n-1 \right) $, 若$ Q>2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) -\frac{\left( n+1 \right) \left( X+\sqrt{X^2-b^2Y} \right)}{2Y}, $则$ M^n $是全测地的, 其中$ Q $是$ M^n $的Ricci曲率下确界.

推论1.10  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 若$ S<\frac{2n\left( n+1 \right) \left( 4n^2+2n-1 \right)}{8n^2+2n-1}, $则$ M^n $是全测地的.

推论1.11  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 若$ K>-\frac{4n-1}{8n^2\left( n+1 \right)}+\frac{n^2\left( 8n^2+2n-1 \right) \left( 4n-1 \right)}{4n\left( n+1 \right) \left( 4n^2+2n-1 \right)}, $则$ M^n $是全测地的.

推论1.12  若$ M^n $是$ CQ^{n+p} $的Kähler迷向子流形, 若$ Q>2\left( n+1 \right) -\frac{\left( n+1 \right) \left( 4n^2+2n-1 \right)}{n\left( 8n^2+2n-1 \right)}, $则$ M^n $是全测地的.

约定各类指标的取值范围如下：

$ \begin{equation} A, B, C, \cdots =1, \cdots , n+p, 1^*, \cdots , \left( n+p \right) ^* \notag \end{equation} $

$ \begin{equation} i, j, k, \cdots =1, \cdots, n, 1^*, \cdots , n^*\notag\ \end{equation} $

$ \begin{equation} \alpha , \beta , \gamma , \cdots =n+1, \cdots , n+p, \left( n+1 \right) ^*, \cdots , \left( n+p \right)^*\notag\ \end{equation} $

设$ CQ^{n+p} $是具有Kähler度量的复$ n+p $维复黎曼流形, 如果其曲率张量表示为

$ \begin{equation} \begin{aligned} K_{ABCD}&=a\left( g_{AC}g_{BD}-g_{AD}g_{BC}+J_{AC}J_{BD}-J_{AD}J_{BC}+2J_{AB}J_{CD} \right) \\&+b\left( g_{AC}\lambda _B\lambda _D+g_{BD}\lambda _A\lambda _C-g_{AD}\lambda _B\lambda _C-g_{BC}\lambda _A\lambda _D \right), \end{aligned} \end{equation} $

(2.1)

则称$ CQ^{n+p} $为拟复射影空间, 也称拟常曲率复射影空间. 其中, $ g $是$ CQ^{n+p} $上的黎曼度量, $ J $为$ CQ^{n+p} $的复结构, $ a, b $是$ CQ^{n+p} $上的光滑函数, $ \left\{ \lambda _A \right\} $是$ CQ^{n+p} $上的单位向量函数, 满足$ \sum\limits_{A, B}^{}{g^{AB}\lambda _A\lambda _B=1} $.

当$ a=\frac{c}{4}, b=0 $时, 拟复射影空间$ CQ^{n+p} $是全纯截面曲率为$ c $的复射影空间$ CP^{n+p} $. 设$ M^n $是$ CQ^{n+p} $的$ n $维全实子流形, $ J $为$ CQ^{n+p} $的复结构, 取$ CQ^{n+p} $的局部规范正交标架场$ e_1, \cdots e_{n+p}, e_{1^*}, \cdots , e_{\left( n+p \right) ^*} $满足

$ \begin{equation} Je_1=e_{1^*}, \cdots , Je_{n+p}=e_{\left( n+p \right) ^*}. \end{equation} $

(2.2)

因为$ J $是复结构, 满足$ J^2=-1 $, 对(2.2)式两边用$ J $作用, 得

$ \begin{equation} Je_{1^*}=-e_1, \cdots , Je_{\left( n+p \right) ^*}=-e_{n+p}. \end{equation} $

(2.3)

取$ \left( J_{AB} \right) $为复结构, 视为线性变换$ J $关于$ \left\{ e_A \right\} $的变换矩阵, 则由(2.2),(2.3)两式得

$ \begin{equation} J_{AB}=\left( \begin{matrix} 0& I_{n+p}\\ -I_{n+p}& 0\\ \end{matrix} \right). \end{equation} $

(2.4)

这里, $ I_{n+p} $为$ n+p $阶单位矩阵. 用$ \left\{ \omega _A \right\} $表示$ \left\{ e_A \right\} $的对偶标架场, 则$ CQ^{n+p} $的结构方程为

$ \begin{equation} d\omega _A=-\sum\limits_B^{}{\omega _{AB}\land \omega _B, \omega _{AB}}+\omega _{BA}=0\notag, \end{equation} $

$ \begin{equation} d\omega _{AB}=-\sum\limits_C^{}{\omega _{AC}\land \omega _{CB}+\frac{1}{2}}\sum\limits_{C, D}^{}{K_{ABCD}\omega _C\land \omega _D}\notag, \end{equation} $

其中$ K_{ABCD} = a (\delta_{AC} \delta_{BD} - \delta_{AD} \delta_{BC} + J_{AC} J_{BD} - J_{AD} J_{BC} + 2J_{AB} J_{CD}) + b (\delta_{AC} \lambda_{B} \lambda_{D} + \delta_{BD} \lambda_{A} \lambda_{C} - \delta_{AD} \lambda_{B} \lambda_{C} - \delta_{BC} \lambda_{A} \lambda_{D}), \quad \left( \sum\limits_{A} \lambda_{A}^{2} = 1 \right). $设$ \left\{ \omega _{AB} \right\} $是$ CQ^{n+p} $的联络1-形式, 将上述形式限制在$ M^n $上, 有[5]

$ \begin{equation} \omega_{\alpha} = 0, \quad \omega _{\alpha i}=\sum\limits_j^{}{h_{ij}^{\alpha}\omega _j}, \quad h=\sum\limits_{\alpha , i, j}^{}{h_{ij}^{\alpha}\omega _i\otimes \omega _j\otimes e_{\alpha}}, \quad d\omega_{ij} = -\sum\limits_{j} \omega_{ij} \wedge \omega_{j}.\notag \end{equation} $

$ \begin{equation} d\omega_{ij} = -\sum\limits_{k} \omega_{ik} \wedge \omega_{kj} + \frac{1}{2} \sum\limits_{k, l} R_{ijkl} \omega_{k} \wedge \omega_{l}.\notag \end{equation} $

$ \begin{equation} d\omega_{\alpha \beta} = -\sum\limits_{\gamma} \omega_{\alpha \gamma} \wedge \omega_{\gamma \beta} + \frac{1}{2} \sum\limits_{k, l} R_{\alpha \beta kl} \omega_{k} \wedge \omega_{l}.\notag \end{equation} $

$ \begin{equation} R_{ijkl} = K_{ijkl} + \sum\limits_{\alpha} \left( h_{ ik}^\alpha h_{jl}^\alpha - h_{ il}^\alpha h_{jk}^\alpha \right). \end{equation} $

(2.5)

$ \begin{equation} R_{\alpha \beta ij} = K_{\alpha \beta ij} + \sum\limits_{k} \left( h_{ik}^ \alpha h_{jk}^\beta - h_{jk}^ \alpha h_{ik}^\beta \right). \end{equation} $

(2.6)

其中, $ h, R_{ijkl}, R_{\alpha \beta ij} $分别是$ M^n $的第二基本形式, 黎曼曲率张量和法曲率张量场关于$ {e_A} $的分量, $ K_{ABCD} $是$ CQ^{n+p} $的曲率张量的分量.

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可分别表示为

$ \begin{equation} \xi = \frac{1}{n} \sum\limits_{\alpha} \left( \sum\limits_{i} h_{ii}^{\alpha} \right) e_{\alpha}, \quad H = \| \xi \|, \quad S = \| h \|^2. \end{equation} $

(2.7)

下面用$h_{ijk}^{\alpha} $及$h_{ijkl}^{\alpha} $分别表示$h_{ij}^{\alpha} $的一阶共变导数和二阶共变导数[6],

$ \begin{equation} -\sum\limits_{k} h_{ijk}^{\alpha} \omega_{k} = dh_{ij}^{\alpha} - \sum\limits_{k} h_{ik}^{\alpha} \omega_{kj} - \sum\limits_{l} h_{lj}^{\alpha} \omega_{il} - \sum\limits_{\beta} h_{ij}^{\beta} \omega_{\beta \alpha}\notag, \end{equation} $

$ \begin{equation} -\sum\limits_{l} h_{ijkl}^{\alpha} \omega_{l} = dh_{ijk}^{\alpha} - \sum\limits_{m} h_{ijm}^{\alpha} \omega_{mk} - \sum\limits_{l} h_{ljk}^{\alpha} \omega_{il} - \sum\limits_{\beta} h_{ijk}^{\beta} \omega_{\beta \alpha}\notag, \end{equation} $

则有

$ \begin{equation} h_{ijk}^{\alpha} - h_{ikj}^{\alpha} = -K_{aijk}. \end{equation} $

(2.8)

$ \begin{equation} h_{ijkl}^{\alpha} - h_{ijlk}^{\alpha} = \sum\limits_{m} \left( h_{im}^{\alpha} R_{mjkl} + h_{jm}^{\alpha} R_{mikl} \right) - \sum\limits_{\beta} h_{ij}^{\beta} R_{\beta \alpha kl}. \end{equation} $

(2.9)

$ \begin{equation} K_{\alpha \beta ij} = a \left( J_{\alpha k} J_{\beta j} - J_{\alpha j} J_{\beta k} \right). \end{equation} $

(2.10)

引理2.1  ^[7] $ f:M^n\rightarrow CQ^{n+p} $是$ \lambda $-迷向浸入当且仅当

$ \begin{equation} \sum\limits_{\alpha}^{}{\left( h_{ii}^{\alpha} \right) ^2=\lambda ^2, \sum\limits_{\alpha}^{}{h_{ii}^{\alpha}h_{ij}^{\alpha}=0}}, \end{equation} $

(2.11)

$ \begin{equation} \sum\limits_{\alpha}^{}{h_{ii}^{\alpha}h_{jj}^{\alpha}+2\sum\limits_{\alpha}^{}{\left( h_{ij}^{\alpha} \right) ^2=\lambda ^2}}, \end{equation} $

(2.12)

$ \begin{equation} \sum\limits_{\alpha}^{}{\left( h_{ii}^{\alpha}h_{jk}^{\alpha}+2h_{ij}^{\alpha}h_{ik}^{\alpha} \right) =\sum\limits_{\alpha}^{}{\left( h_{ij}^{\alpha}h_{kl}^{\alpha}+h_{il}^{\alpha}h_{jk}^{\alpha}+h_{ik}^{\alpha}h_{jl}^{\alpha} \right) =0}}, \end{equation} $

(2.13)

其中$ i, j, k, l $互不相同.

引理2.2  ^[8] 若$ M^n $是$ N^{n+p} $的极小子流形, 则$ M^n $是伪脐子流形; Kähler流形的Kähler子流形必是极小子流形.

引理2.3  ^[5] 设$ f:M^n\hookrightarrow CP^{n+p}\left( 4 \right) $为Kähler迷向浸入, 则

$ \left\{ \begin{array}{l} \sum\nolimits_{\alpha , k}^{}{h_{ik}^{\alpha}h_{kj}^{\alpha}}=\left( n+1 \right) \lambda ^2\delta _{ij}\\ S=2n\left( n+1 \right) \lambda ^2.\\ \end{array} \right. $

首先我们证明定理1.1

证  同引理2.3的证明和(2.1) 可知, 引理2.3对拟复射影空间也成立

$ \begin{equation} \begin{aligned} &\frac{1}{2}\sum\limits_i^{}{\varDelta \left[ \sum\limits_{\alpha}^{}{\left( h_{ii}^{\alpha} \right) ^2} \right]}\\ =&\sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}+\sum\limits_{\alpha , i, k}^{}{h_{ii}^{\alpha}h_{iikk}^{\alpha}}\\ =&\sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}-\sum\limits_{\alpha , i, k}^{}{h_{ii}^{\alpha}\left( K_{\alpha kiki}+K_{\alpha iikk} \right)}+\sum\limits_{\alpha .\beta , m, i, k}^{}{h_{ii}^{\alpha}\left( h_{mk}^{\alpha}R_{miik}+h_{mi}^{\alpha}R_{mkik} \right)} \\ &+h_{ii}^{\alpha}h_{ki}^{\beta}R_{\alpha \beta ki}+\sum\limits_{\alpha , i, k}^{}{h_{kkii}^{\alpha}h_{ii}^{\alpha}} . \end{aligned} \end{equation} $

(3.1)

首先对于$ \sum\limits_{\alpha , \beta , m, k}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}R_{miik}} $当$ m=i $或$ k=i $时, 由引理2.1有$ \sum\limits_{\alpha}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}}=0 $, 下面考虑$ m, k\ne i $的情况.

$ \begin{equation} \begin{aligned} &\sum\limits_{\alpha , \beta , k, i, m}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}R_{miik}}\\ =&\sum\limits_{\alpha , \beta , m, i, k}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}\left[ a\left( \delta _{mi}\delta _{ik}-\delta _{mk}\delta _{ii} \right) +b\left( \delta _{mi}\lambda _i\lambda _k+\delta _{ik}\lambda _m\lambda _i-\delta _{mk}\lambda _{i}^{2}-\delta _{ii}\lambda _m\lambda _k \right) \right]}\\ &+\sum\limits_{\alpha , \beta , m, i, k}^{}{+h_{ii}^{\alpha}h_{mk}^{\alpha}\left( h_{mi}^{\beta}h_{ik}^{\beta}-h_{mk}^{\beta}h_{ii}^{\beta} \right)}\\ =&\sum\limits_{\alpha , \beta , k, i, m}^{}{a\left( h_{ii}^{\alpha} \right) ^2-2anh_{ii}^{\alpha}h_{kk}^{\alpha}-bh_{ii}^{\alpha}h_{kk}^{\alpha}\lambda _{i}^{2}-\delta _{ii}bh_{ii}^{\alpha}h_{mk}^{\alpha}\lambda _m\lambda _k+h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mi}^{\beta}h_{ik}^{\beta}-h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mk}^{\beta}h_{ii}^{\beta}} . \end{aligned} \end{equation} $

(3.2)

首先, 由引理2.1和$ M^n $的极小性,

$ \begin{equation} \begin{aligned} \sum\limits_{\alpha , i, i\ne k}^{}{\left( h_{ii}^{\alpha} \right) ^2}=2n\lambda ^2 , \end{aligned} \end{equation} $

(3.3)

$ \begin{equation} \begin{aligned} -an\sum\limits_{\alpha , i, k, i\ne k}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}}=-an\left[ \sum\limits_{\alpha , i, k}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}}-\sum\limits_{\alpha , i}^{}{\left( h_{ii}^{\alpha} \right) ^2} \right] =4an^2\lambda ^2 . \end{aligned} \end{equation} $

(3.4)

同理, 由上式

$ \begin{equation} -b\sum\limits_{\alpha , i, k, k\ne i}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}\lambda _{i}^{2}}=2bn\lambda ^2\sum\limits_{i, i\ne k}^{}{\lambda _{i}^{2}} . \end{equation} $

(3.5)

对于第四项

$ \begin{equation} \begin{aligned} &-b\sum\limits_{\alpha , i, m, k, m\ne i, k\ne i}^{}{\delta _{ii}h_{ii}^{\alpha}h_{mk}^{\alpha}\lambda _m\lambda _k}\\ =& -b\left[ \sum\limits_{\alpha , i, m, k, m\ne i, k, k\ne i}^{}{\delta _{ii}h_{ii}^{\alpha}h_{mk}^{\alpha}\lambda _m\lambda _k}+\sum\limits_{\alpha , i, m, k, k\ne i}^{}{\delta _{ii}h_{ii}^{\alpha}h_{kk}^{\alpha}\lambda _{k}^{2}} \right] \\ =&b\left(2n-1 \right)2n\lambda ^2\sum\limits_{i, i\ne k}^{}{\lambda _{i}^{2}} . \end{aligned} \end{equation} $

(3.6)

最后两项

$ \begin{equation} \begin{aligned} &\sum\limits_{\alpha , \beta , i, m, k, i\ne k, m}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mi}^{\beta}h_{ik}^{\beta}-h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mk}^{\beta}h_{ii}^{\beta}}\\ =&\sum\limits_{\alpha , \beta , i, k, m, i\ne k, m, k\ne m}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mi}^{\beta}h_{ik}^{\beta}-h_{ii}^{\alpha}h_{mk}^{\alpha}h_{mk}^{\beta}h_{ii}^{\beta}}+\sum\limits_{\alpha , \beta , i, k, i\ne k}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}h_{ki}^{\beta}h_{ki}^{\beta}-h_{ii}^{\alpha}h_{kk}^{\alpha}h_{kk}^{\beta}h_{ii}^{\beta}}\\ =&\sum\limits_{\alpha , \beta , i, k, m, i\ne k, m, k\ne m}^{}{-2h_{mi}^{\alpha}h_{ki}^{\alpha}h_{mi}^{\beta}h_{ki}^{\beta}-4h_{mi}^{\alpha}h_{ki}^{\alpha}h_{mi}^{\beta}h_{ki}^{\beta}}+\sum\limits_{\alpha , i, k, i\ne k}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}}\sum\limits_{\beta , i, k, i\ne k}^{}{\left( h_{ki}^{\beta} \right) ^2}\\ &-\sum\limits_{\alpha , \beta , i, k, i\ne k}^{}{h_{ii}^{\alpha}h_{kk}^{\alpha}h_{kk}^{\beta}h_{ii}^{\beta}}\\ =&-4n^2\lambda ^4\left( n+1 \right) . \end{aligned} \end{equation} $

(3.7)

综上所述, 我们有

$ \begin{equation} \begin{aligned} \sum\limits_{\alpha , \beta , i, k, m}^{}{h_{ii}^{\alpha}h_{mk}^{\alpha}R_{miik}}=a\left( 4n^2+2n-1 \right) \lambda ^2+2bn\lambda ^2\sum\limits_{i, i\ne k}^{}{\lambda _{i}^{2}}-4n^3\lambda ^4-4n^2\lambda ^4 . \end{aligned} \end{equation} $

(3.8)

再来估计$ \sum\limits_{\alpha , m, k}^{}{h_{ii}^{\alpha}h_{mi}^{\alpha}R_{mkik}} $.

当$ m\ne i $时, $ \sum\limits_{\alpha , m}^{}{h_{ii}^{\alpha}h_{mi}^{\alpha}}=0 $, $ k=i $时, $ R_{mkik}=0 $, 所以接下来考虑$ m=i\ne k $的情况.

$ \begin{equation} \begin{aligned} &\sum\limits_{\alpha , k}^{}{\left( h_{ii}^{\alpha} \right) ^2R_{ikik}}\\ =&\sum\limits_{\alpha , \beta , i, k, i\ne k}^{}{a\left( h_{ii}^{\alpha} \right) ^2\delta _{ii}\delta _{kk}+b\delta _{ii}\left( h_{ii}^{\alpha} \right) ^2\lambda _{k}^{2}+b\delta _{kk}\left( h_{ii}^{\alpha} \right) ^2\lambda _{i}^{2}+\left( h_{ii}^{\alpha} \right) ^2\left( h_{ii}^{\beta} \right) \left( h_{kk}^{\beta} \right) -\left( h_{ii}^{\alpha} \right) ^2\left( h_{ik}^{\beta} \right) ^2}\\ =&2an\left( 2n-1 \right) ^2\lambda ^2+2b\left(2n-1 \right) ^2\sum\limits_{k, k\ne i}^{}{\lambda _{k}^{2}}+\left( 2n-1 \right) \lambda ^2\left\{ -2n\lambda ^2-\left[ \sum\limits_{\beta , i, k}^{}{\left( h_{ik}^{\beta} \right) ^2}-\sum\limits_{\beta , i}^{}{\left( h_{ii}^{\alpha} \right) ^2} \right] \right\}\\ =&2an\left( 2n-1 \right) \lambda ^2+2b\left( n-1 \right) ^2\lambda ^2\sum\limits_{k, k\ne i}^{}{\lambda _{k}^{2}}-\left( 4n^3+2n^2-2n \right) \lambda ^4 . \end{aligned} \end{equation} $

(3.9)

最后考虑$ \sum\limits_{\alpha , \beta , k}^{}{h_{ii}^{\alpha}h_{ki}^{\beta}R_{\alpha \beta ki}}. $由(2.1), (2.6) 得

$ R_{\alpha \beta ki}=K_{\alpha \beta ki}+\sum\limits_l^{}{\left( h_{kl}^{\alpha}h_{li}^{\beta}-h_{il}^{\alpha}h_{kl}^{\beta} \right) =}a\left( J_{\alpha k}J_{\beta i}-J_{\alpha i}J_{\beta k} \right) +\sum\limits_l^{}{\left( h_{kl}^{\alpha}h_{li}^{\beta}-h_{il}^{\alpha}h_{kl}^{\beta} \right)}, $

其中

$ \begin{equation} \begin{aligned} \sum\limits_{\alpha .k.i, k\ne i}^{}{ah_{ii}^{\alpha}h_{ki}^{\beta}\left( J_{\alpha k}J_{\beta i}-J_{\alpha i}J_{\beta k} \right)} =a\sum\limits_{k, i, k\ne i}^{}{\left( h_{ii}^{k^*}h_{ii}^{k^*}-h_{ii}^{i^*}h_{kk}^{i^*} \right)} =a\left( 4n-1 \right) \lambda ^2 , \end{aligned} \end{equation} $

(3.10)

$ \begin{equation} \begin{aligned} &\sum\limits_{\alpha , \beta , k, l, i}^{}{h_{ii}^{\alpha}h_{ki}^{\beta}h_{kl}^{\alpha}h_{li}^{\beta}-h_{ii}^{\alpha}h_{ki}^{\beta}h_{il}^{\alpha}h_{lk}^{\beta}}\\ =&\sum\limits_{\alpha , \beta , k, l, i, i\ne k, l}^{}{h_{ii}^{\alpha}h_{ki}^{\beta}h_{kl}^{\alpha}h_{li}^{\beta}-h_{ii}^{\alpha}h_{ki}^{\beta}h_{il}^{\alpha}h_{lk}^{\beta}}+\sum\limits_{\alpha , \beta , k, i, i\ne k}^{}{h_{ii}^{\alpha}h_{ki}^{\beta}h_{ki}^{\alpha}h_{ii}^{\beta}-h_{ii}^{\alpha}h_{ki}^{\beta}h_{ii}^{\alpha}h_{ik}^{\alpha}}\\ =&\sum\limits_{\alpha , \beta , k, l, i, i\ne k, l}^{}{h_{ii}^{\alpha}h_{kl}^{\alpha}h_{ki}^{\beta}h_{li}^{\beta}}+\sum\limits_{\alpha , \beta , k, i, i\ne k}^{}{-\left( h_{ii}^{\alpha} \right) ^2\left( h_{ki}^{\beta} \right) ^2}\\ =&-2n^2\left( 4n-1 \right) \lambda ^4 . \end{aligned} \end{equation} $

(3.11)

又由$ CQ^{n+p} $的局部对称性和$ M^n $的极小性, 有

$ \begin{equation} \begin{aligned} &-\sum\limits_{\alpha , i, k}^{}{h_{ii}^{\alpha}\left( K_{\alpha kiki}+K_{\alpha iikk} \right)}\\ =&-\sum\limits_{\alpha , \beta , i, k}^{}{h_{ii}^{\alpha}h_{ii}^{\beta}K_{\alpha k\beta k}}-\sum\limits_{\alpha , \beta , i, k}^{}{h_{ii}^{\alpha}h_{kk}^{\beta}K_{\alpha ii\beta}}+\sum\limits_{\alpha , i, k, m}^{}{h_{ii}^{\alpha}\left( h_{mk}^{\alpha}K_{miik}+h_{mi}^{\alpha}K_{mkik} \right)}\\ =&a\left( 4n^2-1 \right) \lambda ^2+b\left( 8n^2-6n+2 \right) \lambda ^2\sum\limits_{i, i\ne k}^{}{\lambda _{i}^{2}} . \end{aligned} \end{equation} $

(3.12)

综上所述, 我们有

$ \begin{equation} \begin{aligned} \frac{1}{2}\varDelta \left( n\lambda ^2 \right) =\lambda ^2\left[ \left( -2n^2\left( 4n-1 \right) \right) \lambda ^2+a\left( 8n^2+4n-3 \right) +2b\left( 8n^2-6n+2 \right) \sum\limits_{i, i\ne k}^{}{\lambda _{i}^{2}} \right] . \end{aligned} \end{equation} $

(3.13)

因为$ \sum_A^{}{\lambda _{A}^{2}}=1 $, 所以$ \sum_{i, i\ne k}^{}{\lambda _{i}^{2}}\le 1 $, $ n\ge 1 $时, $ -2n^2\left( 4n-1 \right) <0 $, $ 8n^2-6n+2>0 $, $ 8n^2+4n-3>0 $, 于是我们有

$ \begin{equation} \begin{aligned} \int_{M^n}{\lambda ^2\left[ -2n^2\left( 4n-1 \right) \lambda ^2+a\left( 8n^2+4n-3 \right) -2\left| b \right|\left( 8n^2-6n+2 \right) \right]}dV\le 0 . \end{aligned} \end{equation} $

(3.14)

所以当$ \lambda ^2<\frac{a\left( 8n^2+4n-3 \right) -2\left| b \right|\left( 8n^2-6n+2 \right)}{2n^2\left( 4n-1 \right)} $结合引理2.3可得

$ S<\frac{\left( n+1 \right) \left[ \left( 8n^2+4n-3 \right) -2\left| b \right|\left( 8n^2-6n+2 \right) \right]}{n\left( 4n-1 \right)} $

时, 有$ S=2n\left( n+1 \right) \lambda ^2=0. $证毕.

下面证明定理1.2.

证  记$ K $为$ M^n $截面曲率的下确界, 则

$ \begin{equation} \sum\limits_{\alpha , i, k, m}^{}{h_{ii}^{\alpha}\left( h_{mk}^{\alpha}R_{miik}+h_{mi}^{\alpha}R_{mkik} \right)}\ge 2nKS=4n^2\left( n+1 \right) K\lambda ^2 . \end{equation} $

(3.15)

将(3.10)、(3.11)、(3.12)、(3.15) 代入(3.1) 得：

$ \begin{equation} 0\ge \int_{M^n}{\lambda ^2\left[ 8n^2\left( n+1 \right) K+a\left( 4n-1 \right) +2n\lambda ^2\left( 2n^2+4n-1 \right) \right]}dV. \end{equation} $

(3.16)

于是当$ K>-\frac{a\left( 4n-1 \right) +2n\lambda ^2\left( 2n^2+4n-1 \right)}{8n^2\left( n+1 \right)} $时, $ \lambda =0 $, 进而$ M^n $是全测地的.

将定理1.1代入可得, 当

$ K>-\frac{a\left( 4n-1 \right)}{8n^2\left( n+1 \right)}+\frac{\left( 2n^2+4n-1 \right) \left[ 2\left| b \right|\left( 8n^2-6n+2 \right) -a\left( 8n^2+4n-3 \right) \right]}{8n^3\left( n+1 \right) \left( 4n-1 \right)} $

时, $ M^n $是全测地的. 证毕.

下面证明定理1.3

证  记$ Q $是$ M^n $Ricci曲率的下确界, 则由(2.1)可得：

$ \begin{equation} Q\le R_{ii}=\sum\limits_j^{}{R_{ijij}}\le 2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) +\sum\limits_{\alpha , j}^{}{\left[ h_{ii}^{\alpha}h_{jj}^{\alpha}-\left( h_{ij}^{\alpha} \right) ^2 \right]} \end{equation} $

(3.17)

其中$ \sum\limits_{\alpha , j}^{}{\left[ h_{jj}^{\alpha}h_{ii}^{\alpha}-\left( h_{ij}^{\alpha} \right) ^2 \right]}=\sum\limits_{\alpha , j\left( j\ne i \right)}^{}{\left[ h_{jj}^{\alpha}h_{ii}^{\alpha}-\left( h_{ij}^{\alpha} \right) ^2 \right]} $将引理2.1代入上式, 结合$ M^n $的极小性可得

$ \sum\limits_{\alpha , j\left( j\ne i \right)}^{}{\left[ h_{jj}^{\alpha}h_{ii}^{\alpha}-\left( h_{ij}^{\alpha} \right) ^2 \right]}=\sum\limits_{j\left( j\ne i \right)}^{}{\sum\limits_{\alpha}^{}{\left[ h_{ii}^{\alpha}h_{jj}^{\alpha}-\frac{1}{2}\left( \lambda ^2-h_{ii}^{\alpha}h_{jj}^{\alpha} \right) \right]}}=-\left( n+1 \right) \lambda ^2. $

于是有

$ Q\le 2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) -\left( n+1 \right) \lambda ^2. $

由上式, 当

$ Q>2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) +\frac{\left( n+1 \right) \left[ 2\left| b \right|\left( 8n^2-6n+2 \right) -a\left( 8n^2+4n-3 \right) \right]}{2n^2\left( 4n-1 \right)}. $

时, 由定理1可得$ M^n $是全测地的.这就证明了定理1.3. 证毕.

下面我们在没有局部对称的条件下讨论这个问题, 证明定理1.6.

证  由引理2.2可知, $ M^n $是伪脐的, 设$ \omega =\sum_{\alpha , i, j, k}^{}{\left( h_{ik}^{\alpha}K_{\alpha jij}+h_{ij}^{\alpha}K_{\alpha ijk} \right)} $, 则

$ div\omega =\sum\limits_{\alpha , i, j, k}^{}{\nabla _k\left( h_{ik}^{\alpha}K_{\alpha jij}+h_{ij}^{\alpha}K_{\alpha ijk} \right)}. $

于是得

$ \begin{equation} -\sum\limits_{\alpha , i, j, k}^{}{h_{ij}^{\alpha}\left( K_{kikj}^{\alpha}+K_{ijkk}^{\alpha} \right)} =\sum\limits_{\alpha , i, j, k}^{}{\left( h_{ikk}^{\alpha}K_{\alpha jij}+h_{ijk}^{\alpha}K_{\alpha ijk} \right)}-div\omega . \end{equation} $

(3.18)

于是$ \sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}-\sum\limits_{\alpha , i, k}^{}{h_{ii}^{\alpha}\left( K_{\alpha kiki}+K_{\alpha iikk} \right)}=\sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}+\sum\limits_{\alpha , i, k}^{}{h_{iik}^{\alpha}K_{\alpha iik}}-div\omega. $又由(2.1)得

$ \sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}+\sum\limits_{\alpha , i, k}^{}{h_{iik}^{\alpha}K_{\alpha iik}}=\sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2}+\sum\limits_{\alpha , i, k}^{}{h_{iik}^{\alpha}\left( \delta _{ik}\lambda _{i}^{2}-\delta _{ii}\lambda _{\alpha}\lambda _k \right)}. $

因为$ \sum_A^{}{\lambda _{A}^{2}}=1 $, 也就是$ \sum_i^{}{\lambda _{i}^{2}}+\sum_{\alpha}^{}{\lambda _{\alpha}^{2}}=1 $, 所以

$ \sqrt{\sum\limits_i^{}{\lambda _{i}^{2}\sum\limits_{\alpha}^{}{\lambda _{\alpha}^{2}}}}\le \frac{\sum\limits_i^{}{\lambda _{i}^{2}}+\sum\limits_{\alpha}^{}{\lambda _{\alpha}^{2}}}{2}=\frac{1}{2}. $

于是, $ 0\le \sum\limits_{i, \alpha}^{}{\left( \lambda _i\lambda _{\alpha} \right) ^2\le \frac{1}{4}}. $

因为指标$ i $有$ 2n $个取值, 指标$ \alpha $有$ 2p $个取值, 所以$ \frac{\sum\limits_{i, \alpha}^{}{\lambda _i\lambda _{\alpha}}}{4np}\le \sqrt{\frac{\sum\limits_{i, \alpha}^{}{\left( \lambda _i\lambda _{\alpha} \right) ^2}}{4np}}\le \sqrt{\frac{1}{4np}} $由柯西不等式, 有

$ \left( \sum\limits_{i, \alpha}^{}{\lambda _i\lambda _{\alpha}} \right) ^2\le \sum\limits_{i, \alpha}^{}{1^2}\sum\limits_{i, \alpha}^{}{\left( \lambda _i\lambda _{\alpha} \right) ^2\le np}, $

化简得

$ \begin{equation} -\sqrt{np}\le \sum\limits_{i, \alpha}^{}{\lambda _i\lambda _{\alpha}}\le \sqrt{np} \notag. \end{equation} $

记$ A=\sum_{\alpha , k}^{}{\lambda _{k}^{2}-n\lambda _{\alpha}\lambda _k} $, 则$ -\sqrt{np}\le A\le \sqrt{np}+1 . $由二次函数的性质, 有

$ \begin{equation} \sum\limits_{\alpha , i, k}^{}{\left( h_{iik}^{\alpha} \right) ^2+bAh_{iik}^{\alpha}}\ge \frac{b^2A\left( A-2 \right)}{4}\ge -\frac{1}{4}b^2 . \end{equation} $

(3.19)

结合(3.1)、(3.8)、(3.10)、(3.11)、(3.19), 得

$ \begin{equation} \int_{M^n}{\left\{ \lambda ^2\left[ 2a\left( 4n^2+2n-1 \right) +2n\lambda ^2\left( -8n^2-2n+1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) \right] -\frac{1}{4}b^2 \right\}}dV\le 0 . \end{equation} $

(3.20)

考虑关于$ \lambda ^2 $的二次函数

$ f\left( \lambda ^2 \right) =-2n\left( 8n^2+2n-1 \right) \lambda ^4+\left[ 2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) \right] \lambda ^2-\frac{1}{4}b^2. $

记$ \varDelta =\left[ 2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) \right] ^2-2nb^2\left( 8n^2+2n-1 \right). $

记$ X=2a\left( 4n^2+2n-1 \right) -\left| b \right|\left( 8n^2-6n+2 \right) $, $ Y=2n\left( 8n^2+2n-1 \right) $, 结合二次函数的图像性质可知, 若

$ \left\{ \begin{array}{l} X>0\\ X^2>b^2Y\\ \, \, \frac{X-\sqrt{X^2-b^2Y}}{2Y}<\lambda ^2<\frac{X+\sqrt{X^2-b^2Y}}{2Y}, \end{array} \right. $

则$ \lambda ^2=0 $, 进而$ M^n $是全测地的. 证毕.

类似定理1.2、1.3的证明过程, 我们证明定理1.8、定理1.9.

证  由(3.1)、(3.8)、(3.10)、(3.11)、(3.15) 我们可以得到

$ 0\ge \int_{M^n}{\left[ 8n^2\left( n+1 \right) K\lambda ^2+a\left( 4n-1 \right) \lambda ^2-2n^2\left( 4n-1 \right) -\frac{1}{4}b^2 \right]}dV. $

当$ K>\frac{2Y\left[ 2n^2\left( 4n-1 \right) +\frac{1}{4}b^2 \right]}{8n^2\left( n+1 \right) \left( X+\sqrt{X^2-b^2Y} \right)}-\frac{a\left( 4n-1 \right)}{8n^2\left( n+1 \right)} $时, 由定理1.7, $ M^n $是全测地的, 这样就证明了定理1.8. 证毕.

证  同定理1.3的讨论, 当$ Q>2a\left( n+1 \right) +\left| b \right|\left( 2n+1 \right) -\frac{\left( n+1 \right) \left( X+\sqrt{X^2-b^2Y} \right)}{2Y} $时, 由定理1.7, 可知$ M^n $是全测地的, 就得到了定理1.9. 证毕.

特别地, 在上述的六个定理中取$ a=1, b=0 $, 即可得对应条件下关于复射影空间的推论.