Let $CP^{n+p}$ be a $2(n+p)$-dimensional complex projective space endowed with the Fubini-Study metric of constant holomorphic sectional curvature 4. Let $M^n$ be an $n$-dimensional submanifold in $CP^{n+p}$. $M^n$ is called totally real if each tangent space of $M^n$ is mapped into the normal space by the complex structure $J$ of $CP^{n+p}$. It plays an important role in geometry of submanifolds to investigate rigidity of totally real submanifolds in complex projective space. Totally real submanifolds in complex projective space were extensively studied and many rigidity theorems were proved, see, for example [1-6], etc.

Recently, Cao [7] and Gu, Xu [8] proved the following rigidity theorems, respectively, by using the DDVV inequality verified by Ge and Tang [9], Lu [10].

Theorem A (see [7])   Let $M^n$ be an $n$-dimensional oriented closed minimal submanifold in an $n$-dimensional simply connected and locally symmetric Riemannian manifold $N^{n+p}$. Suppose the sectional curvature $K_{N}$ of $N$ satisfies $\delta\leq K_N\leq1$. If the sectional curvature $K_{M}$ of $M^n$ satisfies

$ K_M\geq\frac{4}{3n(p+1)}(n-1)^{\frac{1}{2}}(p-1)(p+2)(1-\delta) +(\frac{p+2}{2(p+1)}-\frac{\delta}{p+1}){\rm sgn}(p-1), $

then either $M$ is totally geodesic, or $N^{n+p}=S^{n+p}$ and $M$ is isometric to the standard immersion of the product of two spheres or the Veronese surface in $S^4$.

Theorem B (see [8])   Let $M^n$ be an n-dimensional oriented compact submanifold with parallel mean curvature $H\neq0$ in $F^{n+p}(c)$. If $c+H^2>0$ and

$ K_M\geq\frac{{\rm sgn}(p-2)(p-1)}{2p}(c+H^2), $

then $M$ is either a totally umbilical sphere $S^n(\frac{1}{\sqrt{c+H^2}})$ in $F^{n+p}(c)$, the standard immersion of the product of two spheres or the Veronese surface in $S^4(\frac{1}{\sqrt{c+H^2}})$.

In this paper, we study totally real submanifolds in $CP^{n+p}$, obtain two rigidity theorems and an integral inequality, by using moving-frame method and the DDVV inequality.

Theorem 1.1   Let $M^n$ be an $n$$(n\geq2)$-dimensional compact totally real submanifolds with parallel mean curvature vector $\xi \neq 0$ in $CP^{n+p}$ $(p \geq1)$. If the sectional curvature $R_M$ of $M^n$ satisfies

$ R_{M}\geq \frac{n+2p-1}{2(n+2p)}(1+H^2), $

then $M^n$ is a totally umbilical sphere $S^n(\frac{1}{\sqrt{1+H^2}})$, where $H$ is the mean curvature of $M^n$.

Theorem 1.2   Let $M^n$ be an $n$$(n\geq2)$-dimensional complete totally real pseudo-umbilical submanifold in $CP^{n+p}$ $(p \geq1)$. If $J\xi$ is normal to $M^n$, then either $M^n$ is totally umbilical or inf $\rho \leq n(1+H^2)(n-\frac{5}{3})$, where $\xi$, $\rho$, $H$ are the mean curvature vector, the scalar curvature, the mean curvature of $M^n$.

Compared with the result in [5], we do not need the submanifolds to have parallel mean curvature vector condition in Theorem 1.2.

Theorem 1.3   Let $M^n$ be an $n$$(n\geq2)$-dimensional compact totally real pseudo-umbilical submanifold in $CP^{n+p}$ $(p >0)$.If $J\xi$ is tangent to $M^n$, then

$ \int_{M^n}[2(1+4H^2)nS-3S^2-5n^2H^4-4n^2H^2+2nH^2]dV\leq 0, $

where $S$, $H$ are the length square of the second fundamental form, the mean curvature of $M^n$.

Let $M^n$ be an $n$$(n\geq2)$-dimensional totally real submanifold in $CP^{n+p}$. Choose a local field of orthonormal frames

$ 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} $

in $CP^{n+p}$, in such a way that, restricted to $M^n$, $e_1, \cdots, e_{n}$ are tangent to $M^n$ and

$ e_{n+1}, \cdots, e_{n+p}, e_{1^*}, \cdots, e_{n^*}, e_{(n+1)^*}, \cdots, e_{(n+p)^*} $

are normal to $M^n$. We shall make use of the following convention on the range of indices:

$ 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^*. $

Let $\omega^A$ and $\omega_{B}^{A}$ be the dual frame field and the connection 1-forms of $CP^{n+p}$, respectively, then the stucture equations of $CP^{n+p}$ are given by

$ d{\omega _A} = \sum\limits_B {{\omega _{AB}}} \wedge {\omega _B},{\omega _{AB}} + {\omega _{BA}} = 0, $

(2.1)

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

(2.2)

where

$ \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} $

(2.3)

Restricting these forms to $M^n$, we have

$ \begin{array}{l} {\omega _\alpha } = 0,\;\;{\omega _{\alpha i}} = \sum\limits_j {h_{ij}^\alpha } {\omega _j},\;\;h = \sum\limits_{ij\alpha } {h_{ij}^\alpha } {\omega _i} \otimes {\omega _j} \otimes {e_\alpha },\\ h_{jk}^{{i^*}} = h_{ik}^{{j^*}} = h_{ij}^{{k^*}},\;\;\;\xi = \frac{1}{n}\sum\limits_\alpha {(\sum\limits_i {h_{ii}^\alpha } )} {e_\alpha }, \end{array} $

(2.4)

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

(2.5)

$ h^{\alpha}_{ijk}-h^{\alpha}_{ikj}\;\;=\;\;-K_{\alpha ijk}, $

(2.6)

$ 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}, $

(2.7)

$ R_{\alpha\beta kl} \;\;=\;\;K_{\alpha\beta kl}+\sum\limits_{m}(h^{\alpha}_{km}h^{\beta}_{ml}-h^{\alpha}_{lm}h^{\beta}_{km}), $

(2.8)

where $h$, $\xi$, $R_{ijkl}$, $R_{\alpha\beta kl}$ are the second fundamental form, the mean curvature vector, the curvature tensor, the normal curvature tensor of $M^n$ and $h^\alpha_{ijk}$ is the covariant of $h^\alpha_{ij}$. We define

$ \begin{eqnarray}S=|h|^2, H=|\xi|, H_\alpha=(h^\alpha_{ij})_{n\times n}.\end{eqnarray} $

(2.9)

The scalar curvature $\rho$ of $M^n$ is given by

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

(2.10)

Denoting the first and second covariant derivatives of $h_{ij}^\alpha$ by $h_{ijk}^\alpha$ and $h_{ijkl}^\alpha$, respectively, we have

$ \begin{eqnarray*} \sum\limits_{k}h_{ijk}^{\alpha}\omega_k&=&dh_{ij}^{\alpha}-\sum\limits_{k}h_{kj}^{\alpha}\omega_{ki} -\sum\limits_{k}h_{ik}^{\alpha}\omega_{kj}-\sum\limits_{\beta}h_{ij}^{\beta}\omega_{\beta\alpha}, \\ \sum\limits_{l}h_{ijkl}^{\alpha}\omega_l&=&dh_{ijk}^{\alpha}-\sum\limits_{l}h_{ljk}^{\alpha}\omega_{li} -\sum\limits_{l}h_{ilk}^{\alpha}\omega_{lj}-\sum\limits_{l}h_{ijl}^{\alpha}\omega_{lk} +\sum\limits_{\beta}h_{ijk}^{\beta}\omega_{\beta\alpha}. \end{eqnarray*} $

Then the Laplacian of $h_{ij}^{\alpha}$ is

$ \begin{eqnarray}\triangle h_{ij}^{\alpha}=\sum\limits_{k}h_{ijkk}^{\alpha}=\sum\limits_{k}h_{kkij}^{\alpha}+ \sum\limits_{km}(h^{\alpha}_{km}R_{mijk}+h^{\alpha}_{mi}R_{mkjk})-\sum\limits_{\beta k}h^{\beta}_{ki}R_{\alpha\beta jk}.\end{eqnarray} $

(2.11)

Lemma2.1(see [9, 10])   Let $B_1, \ldots, B_m$ be symmetric $(n\times n)$-matrices, then

$ \sum\limits_{r, s=1}^{m}\|[B_r, B_s]\|^2\leq(\sum\limits_{r=1}^{m}\|B_r\|^2)^2, $

where the equality holds if and only if under rotation all $B_r$'s are zero except two matrices which can be written as

$ \tilde{B_r}=P \left( \begin{array}{ccccc} 0&\mu&0&\cdots&0 \\ \mu&0 &0& \cdots& 0\\ 0& 0&0&\cdots &0 \\ \vdots&\vdots&\vdots&\ddots &\vdots \\ 0&0&0&\cdots &0 \end{array} \right)P^t, \ \ \tilde{B_s}=P \left( \begin{array}{ccccc} \mu&0&0&\cdots&0 \\ 0&-\mu &0& \cdots& 0\\ 0& 0&0&\cdots &0 \\ \vdots&\vdots&\vdots&\ddots &\vdots \\ 0&0&0&\cdots &0 \end{array} \right)P^t, $

where $P$ is an orthogonal $( n\times n )$-matrix, $[B_r, B_s]=B_r B_s -B_s B_r$ is the commutator of the matrices $B_r, B_s$.

Lemma2.2(see [11])   Let $A_1, A_2, \cdots, A_m$ ($m\geq 2$) be symmetric $(n\times n)$-matrices. Then

$ -2\sum\limits_{\alpha\beta=1}^m[{\rm tr}(A_\alpha^2 B_\beta^2)-{\rm tr}(A_\alpha A_\beta)^2] -\sum\limits_{\alpha\beta=1}^m[{\rm tr}(A_\alpha A_\beta)]^2\geq -\frac{3}{2}(\sum\limits_{\alpha=1}^m{\rm tr}(A^2_\alpha))^2. $

Proof of Theorem 1.1   $M^n$ is a submanifold with parallel mean curvature vector $\xi$. Choose $e_{n+1}$ such that it is parallel to $\xi$, and

$ \begin{eqnarray}{\rm tr} H_{n+1}=nH, {\rm tr} H_\alpha=0, \alpha\neq n+1.\end{eqnarray} $

(3.1)

The mean curvature vector $\xi$ is parallel, so we have

$ \begin{eqnarray}D^\bot\xi=dH e_{n+1}+H D^\bot e_{n+1}=dH e_{n+1}+H \sum\limits_\beta \omega_{n+1 \beta} e_{\beta}=0.\end{eqnarray} $

(3.2)

(3.2) and (2.7) imply

$ \begin{eqnarray}d\omega_{n+1\beta}=-\sum\limits_{\gamma}\omega_{n+1\gamma}\wedge\omega_{\gamma\beta}+\frac{1}{2}\sum\limits_{kl} R_{n+1\beta kl}\omega_{k}\wedge\omega_{l}=\frac{1}{2}\sum\limits_{kl} R_{n+1\beta kl}\omega_{k}\wedge\omega_{l}=0.\end{eqnarray} $

(3.3)

From (3.3), we know $R_{n+1\beta kl}=0$. Set $S_H={\rm tr} H_{n+1}^2$, $\tau=S-{\rm tr} H_{n+1}^2$. From (2.11), noting that $M^n$ has parallel mean curvature vector and $\sum\limits_{k}h^\alpha_{kkij}=0, $ one gets

$ \begin{eqnarray}\frac{1}{2}\vartriangle S_H &=&\sum\limits_{ijk} (h^{n+1}_{ijk})^2+ \sum\limits_{ij }h^{n+1}_{ij}\vartriangle h^{n+1}_{ij}\nonumber\\ &=&\sum\limits_{ijk} (h^{n+1}_{ijk})^2+ \sum\limits_{ijkm}h^{n+1}_{ij}(h^{n+1}_{km}R_{mijk}+h^{n+1}_{mi}R_{mkjk}). \end{eqnarray} $

(3.4)

Denote $R_{M}(p, \pi)$ the sectional curvature of $M^n$ for 2-plane $\pi\subset T_pM$ at point $p\in M^n$. Set $R_{\min}(p)=\min_{\pi\subset T_pM}R_M(p, \pi)$. We choose the orthonormal fields $\{e_i\}$ such that $h^{n+1}_{ij}=\lambda_{i}\delta_{ij}$, hence, we get

$ \begin{eqnarray}\sum\limits_{ijkm}h^{n+1}_{ij}(h^{n+1}_{km}R_{mijk}+h^{n+1}_{mi}R_{mkjk})= \frac{1}{2}\sum\limits_{ij}(\lambda_{i}-\lambda_{j})^2R_{ijij}\geq \frac{1}{2}\sum\limits_{ij}(\lambda_{i}-\lambda_{j})^2R_{\min}.\end{eqnarray} $

(3.5)

It follows from (3.4) and (3.5) that

$ \frac{1}{2}\vartriangle S_H\geq \sum\limits_{ijk} (h^{n+1}_{ijk})^2+ \frac{1}{2}\sum\limits_{ij}(\lambda_{i}-\lambda_{j})^2R_{\min}. $

It follows from $R_{M}\geq \frac{n+2p-1}{2(n+2p)}(1+H^2)$ and lemma of Hopf that $S_H$ is a constant, and

$ \begin{eqnarray}\frac{1}{2}\sum\limits_{ij}(\lambda_{i}-\lambda_{j})^2R_{\min}=0.\end{eqnarray} $

(3.6)

(3.6) implies that $\lambda_i=\lambda_j$, $\forall i, j$ and $M^n$ is pseudo-umbilical. From (2.11), nothing that $M^n$ has parallel mean curvature vector and $\sum\limits_{k}h^\alpha_{kkij}=0, $ one gets

$ \begin{eqnarray}\frac{1}{2}\vartriangle \tau&=&\sum\limits_{\alpha\neq n+1}\sum\limits_{ijk} (h^{\alpha}_{ijk})^2+ \sum\limits_{\alpha\neq n+1}\sum\limits_{ijkm}h^{\alpha}_{ij}(h^{\alpha}_{km}R_{mijk}+h^{\alpha}_{mi}R_{mkjk})\nonumber\\ & &-\sum\limits_{\alpha\neq n+1}\sum\limits_{\beta ijk}h^{\alpha}_{ij}h^{\beta}_{ki}R_{\alpha\beta jk}.\end{eqnarray} $

(3.7)

By using (2.3), (2.5), (3.1) and the fact that $M^n$ is pseudo-umbilical, we can get

$ \begin{eqnarray}&&\sum\limits_{\alpha\neq n+1}\sum\limits_{ijkm}h^{\alpha}_{ij}(h^{\alpha}_{km}R_{mijk}+h^{\alpha}_{mi}R_{mkjk}) \nonumber\\ &=&n(1+H^2)\tau+\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)] -\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})]^2.\end{eqnarray} $

(3.8)

Combining (2.3), (2.8) and (3.1), we obtain

$ \begin{eqnarray}\sum\limits_{\alpha\neq n+1}\sum\limits_{\beta ijk}h^{\alpha}_{ij}h^{\beta}_{ki}R_{\alpha\beta jk} =-\sum\limits_i{\rm tr}H_{i^*}^2-\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)].\end{eqnarray} $

(3.9)

Substituting (3.8) and (3.9) into (3.7), for any real number $a$, we have

$ \begin{eqnarray} \frac{1}{2}\vartriangle \tau&=&\sum\limits_{\alpha\neq n+1}\sum\limits_{ijk} (h^{\alpha}_{ijk})^2+\sum\limits_itrH_{i^*}^2 -an(1+H^2)\tau\nonumber\\ & &+(1+a)\sum\limits_{\alpha\neq n+1}\sum\limits_{ijkm}h^{\alpha}_{ij}(h^{\alpha}_{km}R_{mijk}+h^{\alpha}_{mi}R_{mkjk}) +a\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})]^2\nonumber\\ & &+(1-a)\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)].\end{eqnarray} $

(3.10)

For fixed $\alpha$, we choose the orthonormal frame field $\{e_i\}$ such that $h_{ij}^\alpha=\lambda_i^\alpha\delta_{ij}$. From (3.1), we get

$ \begin{eqnarray}\sum\limits_{ijkm}h^{\alpha}_{ij}(h^{\alpha}_{km}R_{mijk}+h^{\alpha}_{mi}R_{mkjk}) &=&\frac{1}{2}\sum\limits_{ij}(\lambda_{i}^\alpha-\lambda_{j}^\alpha)^2R_{ijij} \geq \frac{1}{2}\sum\limits_{ij}(\lambda_{i}^\alpha-\lambda_{j}^\alpha)^2R_{\min}\nonumber\\ &=&n{\rm tr}H_{\alpha}^2 R_{\min}.\end{eqnarray} $

(3.11)

(3.11) implies

$ \begin{eqnarray}\sum\limits_{\alpha\neq n+1}\sum\limits_{ijkm}h^{\alpha}_{ij}(h^{\alpha}_{km}R_{mijk}+ h^{\alpha}_{mi}R_{mkjk})\geq n\tau R_{\min}.\end{eqnarray} $

(3.12)

By a direct computation and the DDVV inequality, we obtain

$ \begin{eqnarray}\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}^2H_{\beta}^2)-{\rm tr}(H_{\alpha}H_{\beta})^2] &=&\frac{1}{2}\sum\limits_{\alpha\beta\neq n+1}{\rm tr}(H_\alpha H_\beta-H_\beta H_\alpha)^2\nonumber\\ &\leq& \frac{1}{2}(\sum\limits_{\alpha\neq n+1}{\rm tr}H^2_\alpha)^2=\frac{1}{2}\tau^2.\end{eqnarray} $

(3.13)

We also have

$ \begin{eqnarray}\sum\limits_{\alpha\beta\neq n+1}[{\rm tr}(H_{\alpha}H_{\beta})]^2\geq \frac{1}{n+2p-1}\tau^2.\end{eqnarray} $

(3.14)

Taking $a=\frac{n+2p-1}{n+2p+1}$ in (3.10), it follows from (3.12), (3.13) and (3.14) that

$ \begin{eqnarray}\frac{1}{2}\vartriangle \tau\geq [-\frac{n+2p-1}{n+2p+1}(1+H^2)+\frac{2n+4p}{n+2p+1}R_{\min}]n\tau. \end{eqnarray} $

(3.15)

Hence, if $R_{M}\geq \frac{n+2p-1}{2(n+2p)}(1+H^2)$, then $\frac{1}{2}\vartriangle \tau\geq0$. Thus, by a well-known lemma of Hopf, we have $\frac{1}{2}\vartriangle \tau=0$, consequently we have either

(ⅰ) $\tau=0$, or (ⅱ) $ R_{M}=\frac{n+2p-1}{2(n+2p)}(1+H^2)$.

(ⅰ) If $\tau=0$, then $M^n$ is totally umbilical. From (2.3) and (2.5), we obtain

$ R_{ijij}=1+H^2 $

therefore, $M^n$ is a totally umbilical sphere $S^n(\frac{1}{\sqrt{1+H^2}})$.

(ⅱ) If $R_{M}= \frac{n+2p-1}{2n+4p}(1+H^2)$, then inequality signs in (3.12), (3.13), (3.14) and (3.15) become equalities. Now, we will prove that case (ⅱ) can not occur. The equality of (3.13) implies that either all $H_{\alpha}$'s are zero or two of the $H_{\alpha}$'s are nonzero $ (\alpha\neq n+1)$. When inequality signs in (3.14) and (3.15) become equality, respectively, we get that

$ {\rm tr} H_\alpha^2={\rm tr}H_\beta^2 (\alpha, \beta\neq n+1), $

and $\sum\limits_{i}{\rm tr}H^2_{i^*}=0$. Hence, we have that all $H_{\alpha}$'s are zero $ (\alpha\neq n+1)$. Thus, $M^n$ is totally umbilical, $R_{ijij}=1+H^2$. This leads to a contradiction.

Proof of Theorem 1.2   $J\xi$ is normal to $M^n$. Without loss of generality, we can choose $e_{n+1}$ such that it is parallel to $\xi$, and

$ \begin{eqnarray}{\rm tr}H_{n+1}=nH, {\rm tr} H_\alpha=0, \alpha\neq n+1.\end{eqnarray} $

(3.16)

From (2.11), we have

$ \begin{array}{l} \frac{1}{2}\vartriangle S = \sum\limits_{\alpha ijk} {{{(h_{ijk}^\alpha )}^2}} + \sum\limits_{\alpha ijk} {h_{ij}^\alpha } h_{kkij}^\alpha \\ \;\;\;\;\;\;\;\;\;\; + \sum\limits_{\alpha ijkm} {h_{ij}^\alpha } (h_{km}^\alpha {R_{mijk}} + h_{mi}^\alpha {R_{mkjk}}) - \sum\limits_{\alpha \beta ijk} {h_{ij}^\alpha } h_{ki}^\beta {R_{\alpha \beta jk}}. \end{array} $

(3.17)

Combining (2.3), (2.5), (2.8), (3.16) and the fact that $M^n$ is pseudo-umbilical, we can get

$ \begin{array}{l} \sum\limits_{\alpha ijkm} {h_{ij}^\alpha } (h_{km}^\alpha {R_{mijk}} + h_{mi}^\alpha {R_{mkjk}})\\ = n\left( {1 + {H^2}} \right)S - {n^2}{H^2} + \sum\limits_{\alpha \beta } {\left[ {{\rm{tr}}{{\left( {{H_\alpha }{H_\beta }} \right)}^2} - {\rm{tr}}\left( {H_\alpha ^2H_\beta ^2} \right)} \right] - } {\sum\limits_{\alpha \beta } {\left[ {{\rm{tr}}\left( {{H_\alpha }{H_\beta }} \right)} \right]} ^2}, \end{array} $

(3.18)

$ \sum\limits_{\alpha\beta ijk}h^{\alpha}_{ij}h^{\beta}_{ki}R_{\alpha\beta jk} =-\sum\limits_i{\rm tr}H_{i^*}^2-\sum\limits_{\alpha\beta}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)]. $

(3.19)

Using (3.16) and pseudo-umbilical condition $h^{n+1}_{ij}=H\delta_{ij}$, we have

$ \sum\limits_{\alpha ijk}h^{\alpha}_{ij}h^{\alpha}_{kkij}=nH\triangle H, $

(3.20)

$ \sum\limits_{\alpha ijk} (h^{\alpha}_{ijk})^2\geq \sum\limits_{ik}(h^{n+1}_{iik})^2=n\sum\limits_{i}(\nabla_iH)^2, $

(3.21)

$ \frac{1}{2}\triangle H^2=H\triangle H+\sum\limits_{i}(\nabla_iH)^2. $

(3.22)

By Lemma 2.2 and pseudo-umbilical condition $h^{n+1}_{ij}=H\delta_{ij}$, we have

$ \begin{array}{l} \;\;\;\;\;2\sum\limits_{\alpha \beta } {\left[ {{\rm{tr}}{{\left( {{H_\alpha }{H_\beta }} \right)}^2} - {\rm{tr}}\left( {H_\alpha ^2H_\beta ^2} \right)} \right]} {\rm{ - }}\sum\limits_{\alpha \beta } {{\rm{tr}}{{\left( {{H_\alpha }{H_\beta }} \right)}^2}} \\ {\rm{ = 2}}\sum\limits_{\alpha \beta \ne n + 1} {\left[ {{\rm{tr}}{{\left( {{H_\alpha }{H_\beta }} \right)}^2} - {\rm{tr}}\left( {H_\alpha ^2H_\beta ^2} \right)} \right]} - \sum\limits_{\alpha \beta \ne n + 1} {\left[ {{\rm{tr}}{{\left( {{H_\alpha }{H_\beta }} \right)}^2}} \right]} - {\left( {{\rm{tr}}\mathit{H}_{n + 1}^2} \right)^2}\\ \ge - \frac{3}{2}{\tau ^2} - {n^2}{H^4} = - \frac{3}{2}{(S - n{H^2})^2} - {n^2}{H^4}. \end{array} $

(3.23)

Substituting (3.18)--(3.23) into (3.17), we have

$ \begin{eqnarray} \frac{1}{2}\vartriangle S&\geq & \frac{1}{2}n\triangle H^2+n(1+H^2)S-\frac{3}{2}(S-nH^2)^2-n^2H^4-n^2H^2\nonumber\\ &=& \frac{1}{2}n\triangle H^2+(S-nH^2)[n(1+H^2)-\frac{3}{2}(S-nH^2)]\nonumber\\ &=& \frac{1}{2}n\triangle H^2+\tau[n(1+H^2)-\frac{3}{2}\tau] .\end{eqnarray} $

(3.24)

By the same argument as in [5], we conclude that either $M^n$ is totally umbilical or

$ {\rm inf}\rho \leq n(1+H^2)(n-\frac{5}{3}). $

Proof of Theorem 1.3   $J\xi$ is tangent to $M^n$. Without loss of generality, we can choose $e_{1^*}$ such that it is parallel to $\xi$, and tr$H_{1^*}=nH, {\rm tr}H_\alpha=0, \alpha\neq 1^*.$ This, together with (2.3) and (2.8), implies

$ \begin{eqnarray} \sum\limits_{\alpha\beta ijk}h^{\alpha}_{ij}h^{\beta}_{ki}R_{\alpha\beta jk} &=&n^2H^2-\sum\limits_i{\rm tr}H_{i^*}^2-\sum\limits_{\alpha\beta}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)]\nonumber\\ &\leq& n^2H^2-{\rm tr}H_{1^*}^2-\sum\limits_{\alpha\beta}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)]\nonumber\\ &=& n^2H^2-nH^2-\sum\limits_{\alpha\beta}[{\rm tr}(H_{\alpha}H_{\beta})^2-{\rm tr}(H_{\alpha}^2H_{\beta}^2)] .\end{eqnarray} $

(3.25)

By the same argument as in Theorem 1.2, we conclude that

$ \frac{1}{2}\vartriangle S\geq \frac{1}{2}n\triangle H^2+n(1+H^2)S-\frac{3}{2}(S-nH^2)^2-n^2H^4-2n^2H^2+nH^2. $

As this and $M^n$ is compact, we obtain

$ \int_{M^n}[2(1+4H^2)nS-3S^2-5n^2H^4-4n^2H^2+2nH^2]dV\leq 0. $