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\markboth{Shichang Shu and Junfeng Chen}{Willmore spacelike submanifolds in a Lorentzian space form $N^{n+p}_p(c)$}

\title{Willmore spacelike submanifolds in a Lorentzian space form $N^{n+p}_p(c)$\thanks{Project supported by the NSF of
Shaanxi Education Department (11JK0479).}}

\author[S.\,Shu and J.\,Chen]{Shichang Shu\affil{1}\comma\corrauth and Junfeng Chen\affil{1}}

\address{\affilnum{1}\ School of Mathematics and Information Science, Xianyang Normal University, Xianyang, 712\,000 Shaanxi, P.\,R.\,China}

\emails{{\tt shushichang@126.com} (S.\,Shu), {\tt mailjunfeng@163.com} (J.\,Chen)}


\begin{abstract}
Let $N^{n+p}_p(c)$ be an $(n+p)$-dimensional connected Lorentzian
space form of constant sectional curvature $c$ and $\varphi: M
\rightarrow N^{n+p}_p(c)$ an $n$-dimensional spacelike
submanifold in $N^{n+p}_p(c)$. The immersion $\varphi: M
\rightarrow N^{n+p}_p(c)$ is called a Willmore spacelike
submanifold in $N^{n+p}_p(c)$ if it is a critical submanifold to
the Willmore functional
\[
W(\varphi)=\int_M\rho^ndv=\int_M(S-nH^2)^{\frac{n}{2}}dv,
\]
where
$S$, $H$ and $\rho^2$ denote the norm square of the second
fundamental form, the mean curvature and the non-negative function
$\rho^2=S-nH^2$ of $M$. In this article, by calculating the first
variation of $W(\varphi)$, we obtain the Euler-Lagrange equation
of $W(\varphi)$ and prove some rigidity theorems for
$n$-dimensional Willmore spacelike submanifolds in $N^{n+p}_p(c)$.
\end{abstract}

\keywords{Willmore spacelike submanifold, Lorentzian space form, Euler-Lagrange equation, totally umbilical}

\ams{53C42, 53C40}

\maketitle


%%%% Start %%%%%%
\section{Introduction}
Let $N^{n+p}_p(c)$ be an $(n+p)$-dimensional connected Lorentzian
space form of constant sectional curvature $c$. If $c>0$, $c=0$ or
$c<0$, we call $N^{n+p}_p(c)$ a Minkowski space $R^{n+p}_p$, a de
Sitter space $S^{n+p}_p(c)$ or an anti-de Sitter space
$H^{n+p}_p(c)$. A submanifold in $N^{n+p}_p(c)$ is said to be
spacelike if the induced metric on the submanifold is positive
definite. Let
\[
\varphi: M \rightarrow N^{n+p}_p(c)
\]
be an
$n$-dimensional spacelike submanifold in $N^{n+p}_p(c)$. Denote by
$h^{\alpha}_{ij}, S, \vec{H}$ and $H$ the second fundamental form,
the norm square of the second fundamental form, the mean curvature
vector and the mean curvature of $M$ and denote by $\rho^2$ the
non-negative function $ \rho^2=S-nH^2$. We define the Willmore
functional (see \cite{s4, s9,s16}):
\be\label{(1.1)}
W(\varphi)=\int_M\rho^ndv=\int_M(S-nH^2)^{\frac{n}{2}}dv, %\eqno{(1.1)}
\ee
which vanishes if and only if $M$ is a totally umbilical
submanifold, so the functional $W(\varphi)$ measures how far
$\varphi(M)$ is from being a totally umbilical submanifold. If the
critical points of the Willmore functional $ W(\varphi)$ are
submanifolds in $N^{n+p}_p(c)$, we call them Willmore spacelike
submanifolds.

Due to their backgrounds in mathematics, we know that Willmore
submanifolds in a unit sphere were extensively studied by many
mathematicians. For example, the well-known Willmore conjecture,
which says that $W(\varphi)\geq 4\pi^2$ holds for all immersed
tori $\varphi: M \rightarrow S^{3}$, was investigated by Willmore
\cite{s20,s21}, Li and Yau \cite{s10} and many others; it was recently
proved by Marques and Neves \cite{s12}. We should notice that the topic
of Willmore submanifolds and their rigidity problem was also
studied by Wang \cite{s19} (using conformal invariance), Li \cite{s8,s9} and
the first author \cite{s18} (using metric invariants) and by
Mondino-Riviere \cite{s13} (who established a divergence form of the
Willmore equation in manifolds and exploited it to get rigidity
results). On the other hand, we should see that the parallel
problem in Lorentzian conformal geometry is also an important and
interesting topic. As far as the authors know, the earliest work
in this direction is L. Alias and B. Palmer's paper \cite{s2}, in which
they essentially used the conformal invariance. We notice that one
of Alias and Palmer's main contributions is the generalization of
Willmore surfaces to Lorentz geometry and a Bernstein type theorem
for them, which implies that compact Willmore surfaces in
$3$-dimensional Lorentz space forms must be totally umbilic
spheres. This research was motivated by Barros et al. \cite{s3}, Li and
Nie \cite{s11}, Nie et al. \cite{s15,s14} and others. In this article, we
consider the Willmore functional on spacelike submanifolds in
Lorentzian space forms. By using the metric invariants, we compute
the first variation of the Willmore functional $W(\varphi)$ and
obtain the Euler-Lagrange equation and some rigidity results of
$n$-dimensional Willmore spacelike submanifolds in
$N^{n+p}_p(c)$.
\begin{theorem}\label{tm1.1}
Let $\varphi: M \rightarrow
N^{n+p}_p(c)$ be an $n$-dimensional spacelike submanifold in
$N^{n+p}_p(c)$. Then $M$ is an $n$-dimensional Willmore spacelike
submanifold if and only if for $n+1\leq \alpha, \beta \leq n+p$
\be\label{1.2}
\rho^{n-2}\Big\{SH^\alpha&+&\sum\limits_{i,j,\beta}H^\beta
h^\beta_{ij}h^\alpha_{ij}
-\sum\limits_{i,j,k,\beta}h^\alpha_{ij}h^\beta_{ik}h^\beta_{kj}
-nH^2H^\alpha\Big\}\nonumber\\ %%\tag {1.2}\\
&+&(n-1)\rho^{n-2}\Delta^\bot
H^\alpha+2(n-1)\sum\limits_i(\rho^{n-2})_i
H^\alpha_{,i}\\
&+&(n-1)H^\alpha\Delta(\rho^{n-2})-\Box^\alpha(\rho^{n-2})=0,\nonumber
\ee
where
\bee
\Delta(\rho^{n-2})&=&\sum\limits_i(\rho^{n-2})_{,ii},\\
\Delta^\bot H^\alpha&=&\sum\limits_iH^\alpha_{,ii},\\
\Box^\alpha(\rho^{n-2})&=&\sum\limits_{i,j}(\rho^{n-2})_{,ij}(nH^\alpha\delta_{ij}-
h^\alpha_{ij}),
\eee
and $(\rho^{n-2})_{,ij}$ is the Hessian of
$\rho^{n-2}$ with respect to the induced metric, $H^\alpha_{,i}$
and $H^\alpha_{,ij}$ are defined by \eqref{2.14} and \eqref{2.15}.
\end{theorem}

\begin{remark} \label{Remark 1.1.}
We should notice that in Theorem~\ref{tm1.1} (also
in Proposition~\ref{Proposition 4.1.} - \ref{Proposition 4.3.} and Corollary~\ref{Corollary 4.1.}), when $n=3$ and $n=5$,
we need to assume that $M$ has no umbilical points to guarantee
$(\rho^{n-2})_{,ij}$ is continuous on $M$. In fact, if we denote
$x:=S-nH^2$, then $\rho^{n-2}=x^{\frac{n-2}{2}}$. Thus, we have
$(\rho^{n-2})_{,i}=\frac{n-2}{2}x^{\frac{n-4}{2}}x_{,i}$ and
\[
(\rho^{n-2})_{,ij}=\frac{n-2}{2}\Big(\frac{n-4}{2}x^{\frac{n-6}{2}}x_{,j}x_{,i}+x^{\frac{n-4}{2}}x_{,ij}\Big).
\]
From the above equation, we see that when $n=2, 4$ or $n\geq 6$,
$(\rho^{n-2})_{,ij}$ is continuous on $M$ and when $n=3$ or $5$,
$(\rho^{n-2})_{,ij}$ is not continuous on the umbilical points of
$M$. Therefore, the assumption $n\neq3, 5$ is needed in Theorem~\ref{Theorem 1.2.} -Theorem~\ref{Theorem 1.4.} .
\end{remark}

\begin{remark} \label{Remark 1.2.}  We also notice that in conformal geometry
of conformal spacelike submanifolds Nie and Wu \cite{s14} obtain the
Willmore equation (Euler-Lagrange equation) in terms of
conformal invariants.

When $n=2$, since $R_{ij}=\frac{R}{2}\delta_{ij}$ and
$S=R-2c+4H^2$, from the Gauss equation \eqref{2.6} we see that
\bee
-\sum\limits_{i,j,k,\beta}h^\alpha_{ij}h^\beta_{ik}h^\beta_{kj}&=&-\sum\limits_{i,j}h^\alpha_{ij}\big(R_{ij}-c\delta_{ij}+2\sum\limits_{\beta}H^\beta
h^\beta_{ij}\big)\\
&=&-SH^\alpha+4H^2H^\alpha-2\sum\limits_{i,j,\beta}H^\beta
h^\beta_{ij}h^\alpha_{ij}.
\eee
Thus, \eqref{1.2} reduces to
\be\label{1.3}
\Delta^\bot H^\alpha+2H^2H^\alpha-\sum\limits_{i,j,\beta}H^\beta
h^\beta_{ij}h^\alpha_{ij}=0, %%\tag {1.3}
\ee
where $3\leq \alpha, \beta \leq 2+p$.
From \eqref{1.3}, we easily see
\end{remark}

\begin{proposition}\label{Proposition 1.1.}
Every maximal spacelike surface $\varphi: M
\rightarrow N^{2+p}_p(c)$ in a Lorentzian space form
$N^{2+p}_p(c)$
is a Willmore spacelike surface.
\end{proposition}

\begin{proposition}\label{Proposition 1.2.}
Every $n (n\geq3)$-dimensional maximal and
Einstein spacelike submanifold $\varphi: M \rightarrow
N^{n+p}_p(c)$ in a Lorentzian space form $N^{n+p}_p(c)$ is a
Willmore spacelike submanifold.
\end{proposition}

In fact, since $M$ is maximal and Einstein, we have $H^{\alpha}=0$
for all $\alpha$ and $R_{ij}=\frac{R}{n}\delta_{ij}=constant$.
Thus, from \eqref{2.7}, we see that $\rho^2=S=R-n(n-1)c=constant$. From \eqref{1.2}, we only need to prove $
\sum\limits_{i,j,k,\beta}h^\alpha_{ij}h^\beta_{ik}h^\beta_{kj}=0$.
From the Gauss equation \eqref{2.6}, we have
\bee
\sum\limits_{i,j,k,\beta}h^{\alpha}_{ij}h^\beta_{ik}h^{\beta}_{kj}
&=&\sum\limits_{i,j}h^{\alpha}_{ij}\Big(\sum\limits_{k,\beta}h^{\beta}_{ik}h^{\beta}_{kj}\Big)
=\sum\limits_{i,j}\big[R_{ij}-(n-1)c\delta_{ij}\big]h^{\alpha}_{ij}\\
&=&\sum\limits_{i,j}\Big[\frac{R}{n}\delta_{ij}-(n-1)c\delta_{ij}\Big]h^\alpha_{ij}
=\Big[\frac{R}{n}-c(n-1)\Big]nH^{\alpha}=0.
\eee
We also have the example of Willmore spacelike hypersurfaces of
$H^{n+1}_1(-1)$.

\begin{example} \label{Example 1.1.}
The hyperbolic cylinders
\[
H^k\big(\sqrt{\frac{n-k}{n}}\big)\times
H^{n-k}\big(\sqrt{\frac{k}{n}}\big)\subset H^{n+1}_1(-1), 1\leq
k\leq n-1,
\]
have two distinct principal curvatures
$\sqrt{k/(n-k)}$ and $-\sqrt{(n-k)/(k)}$ with
multiplicities $k$ and $n-k$, respectively. We may easily check
that they are Willmore spacelike hypersurfaces in
$H^{n+1}_1(-1)$ (see \cite{s15}) and $\rho^2=S-nH^2=n$.
\end{example}

\begin{remark}\label{Remark 1.3.}
It is unknown whether there exist non-trivial
examples of closed Willmore spacelike submanifolds whose normal
bundle is timelike.
\end{remark}

Denote by $K$ and $Q$ the functions which assign to each point of
$M$ the infimum of the sectional curvature and the Ricci curvature
at the point. We obtain the following integral inequalities of
Simons'
type and rigidity theorems in terms of $\rho^2$, $K$, $Q$ and~$H$.
\begin{theorem}\label{Theorem 1.2.}
Let $\varphi: M \rightarrow
N^{n+p}_p(c)$ be an $n(n\geq 2)$-dimensional compact Willmore
spacelike submanifold in a Lorentzian space form
$N^{n+p}_p(c)(c=1, 0, -1)$. If $n\neq 3, 5$, then
\begin{enumerate}
\item[$($1$)$] for $p= 1$, we have
\begin{itemize}
\item[$($i$)$] if $c=1, 0$, then $M$ is totally umbilical;

\item[$($ii$)$] if $c=-1$ and $\rho^2\geq n$, then $M$ is totally
umbilical;
\end{itemize}
\item[$($2$)$] for $p\geq 2$, we have
\be\label{1.4}
\int_M\rho^{n}\left\{\frac{1}{p}\rho^2+nc-nH^2\right\}dv\leq0. %%\tag{1.4}
\ee
\end{enumerate}
In particular, if
\begin{align*}
\rho^2\geq np(H^2-c),
\end{align*}
then $M$ is totally umbilical.
\end{theorem}

\begin{theorem}\label{Theorem 1.3.}
Let $\varphi: M \rightarrow
N^{n+p}_p(c)$ be an $n(n\geq 2)$-dimensional compact Willmore
spacelike submanifold in a Lorentzian space form
$N^{n+p}_p(c)(c=1, 0, -1)$. If $n\neq 3, 5$, then the following
integral inequality holds
\be \label{1.5}
\int_M\rho^{n}\left\{K-\frac{n-2}{\sqrt{n(n-1)}}H\rho\right\}dv\leq0. %%\tag{1.5}
\ee
In particular, if
\bee
K\geq \frac{n-2}{\sqrt{n(n-1)}}H\rho,
\eee
then $M$ is totally umbilical.
\end{theorem}

\begin{theorem}\label{Theorem 1.4.}
{\it Let $\varphi: M \rightarrow
N^{n+p}_p(c)$ be an $n(n\geq 2)$-dimensional compact Willmore
spacelike submanifold in a Lorentzian space form
$N^{n+p}_p(c)(c=1, 0, -1)$. If $n\neq 3, 5$, then the following
integral inequality holds
\be \label{1.5}
\int_M\rho^{n}\left\{Q-(n-p-1)(c-H^2)\right \}dv\leq0. %%\tag{1.5}
\ee
In particular, if
\begin{align*}
Q\geq (n-p-1)\left(c-H^2\right),
\end{align*}
then $M$ is totally umbilical.}
\end{theorem}

\begin{remark}\label{Remark 1.4.}
If $p= 1$ and $c=1, 0$, from Theorem~\ref{Theorem 1.2.} we
know that $M$ is totally umbilical. Thus, the conditions
\[
K\geq
\frac{n-2}{\sqrt{n(n-1)}}H\rho \quad and \quad Q\geq
(n-p-1)\left(c-H^2\right)
\]
can be omitted from Theorem~\ref{Theorem 1.3.} and
Theorem~\ref{Theorem 1.4.}  if $p= 1$ and $c=1, 0$.
\end{remark}

\begin{remark}\label{Remark 1.5.}
For the Willmore spacelike surfaces, L. Alias
and B. Palmer \cite{s2} proved that compact Willmore spacelike surfaces
in $3$-dimensional Lorentz space forms must be totally umbilical
spheres. Thus, we notice that our results above generalize Alias
and Palmer's uniqueness result to high dimension and high
co-dimension Willmore spacelike submanifolds.
\end{remark}

If $\varphi: M \rightarrow N^{2+p}_p(c)$ is a maximal spacelike
surface in a Lorentzian space form $N^{2+p}_p(c)$, from
Proposition \ref{Proposition 1.1.} and Theorem~\ref{Theorem 1.2.} - Theorem~\ref{Theorem 1.4.}, we easily have
the following result:

\begin{corollary}\label{Corollary 1.1.}
Let $\varphi: M \rightarrow
N^{2+p}_p(c)$ be a compact maximal spacelike surface in a
Lorentzian space form $N^{2+p}_p(c)(c=1, 0, -1)$. Then
\begin{enumerate}
\item[$($1$)$] if $c=1, 0$, $M$ is totally geodesic;

\item[$($2$)$] if $c=-1$, $S\geq 2p$ or $K\geq 0$ or $Q\geq p-1$,
$M$ is totally geodesic.
\end{enumerate}
\end{corollary}

\begin{remark}\label{Remark 1.6.}
 We notice that the result $(1)$ of Corollary~\ref{Corollary 1.1.} was obtained by \cite{s7}.
\end{remark}


\section{ Preliminaries}

Let $N^{n+p}_p(c)$ be an $(n+p)$-dimensional Lorentzian space form
with index $p$. Let $M$ be an $n$-dimensional connected spacelike
submanifold immersed in $N^{n+p}_p(c)$. We choose a local field of
semi-Riemannian orthonormal frames $e_1, \ldots, e_{n+p}$ in
$N^{n+p}_p(c)$ so that at each point of $M$, $e_1, \ldots, e_n$
span the tangent space of $M$ and form an orthonormal frame there.
We use the following convention on the range of indices:
\[
1\leq A, B, C, \ldots \leq n+p,  \ \ 1\leq i, j, k, \ldots \leq n,
\ \ n+1\leq \alpha, \beta, \gamma, \ldots \leq n+p.
\]
Let $\omega_1, \ldots, \omega_{n+p}$ be its dual frame field so
that the semi-Riemannian metric of $N^{n+p}_p(c)$ is given by
\[
d\overline{s}^2=\sum\limits_i\omega^2_i-\sum\limits_\alpha\omega^2_\alpha=
\sum\limits_A\varepsilon_A\omega^2_A,
\]
where $\varepsilon_i=1$ and
$\varepsilon_\alpha=-1$. Then the structure equations of
$N^{n+p}_p(c)$ are given by
\be
d\omega_A&=&\sum\limits_B\varepsilon_B\omega_{AB}\wedge\omega_B, \
\
\omega_{AB}+\omega_{BA}=0,\label{2.1}\\ %%\tag {2.1}
d\omega_{AB}&=&\sum\limits_C\varepsilon_C\omega_{AC}\wedge\omega_{CB}-\frac{1}{2}\sum\limits_{C,D}
\varepsilon_C\varepsilon_DK_{ABCD}\omega_C\wedge\omega_D,
\label{2.2}\\ %%\tag {2.2}
K_{ABCD}&=&c\varepsilon_A\varepsilon_B(\delta_{AC}\delta_{BD}-\delta_{AD}\delta_{BC}). \label{2.3} %%\tag {2.3}
\ee
If we restrict these forms to $M$, then $ \omega_\alpha=0$,
$n+1\leq \alpha \leq n+p$ and
\be\label{2.4}
\omega_{\alpha i}=\sum\limits_jh^\alpha_{ij}\omega_j,\quad
h^\alpha_{ij}=h^\alpha_{ji}. %% \eqno{(2.4)}
\ee
The Gauss equations are
\be
R_{ijkl}&=&c(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})-
\sum\limits_\alpha(h^\alpha_{ik}h^\alpha_{jl}-h^\alpha_{il}h^\alpha_{jk}),
\label{2.5} \\  %%\eqno{(2.5)}
R_{ik}&=&(n-1)c\delta_{ik}-\sum\limits_\alpha(\sum\limits_l
h^\alpha_{ll})h^\alpha_{ik}+\sum\limits_{\alpha,j}h^\alpha_{ij}
h^\alpha_{jk}, \label{2.6}\\  %%\eqno{(2.6)}
R&=&n(n-1)c+S-n^2H^2, \label{2.7}  %%\eqno{(2.7)}
\ee
where
\[
S=\sum\limits_{i,j,\alpha}(h^{\alpha}_{ij})^2,\quad
\vec{H}=\sum\limits_{\alpha}H^{\alpha}e_{\alpha},\quad
H^{\alpha}=\frac{1}{n}\sum\limits_{k}h^{\alpha}_{kk},\quad
H=|\vec{H}|
\]
and $R$ is the scalar curvature of $M$.

Define the first and the second covariant derivatives of
$h^\alpha_{ij}$, say $h^\alpha_{ijk}$ and $h^\alpha_{ijkl}$, by
\be
\sum\limits_kh^\alpha_{ijk}\omega_k\!&=&\!dh^\alpha_{ij}+\sum\limits_kh^\alpha_{ik}\omega_{kj}+
\sum\limits_k h^\alpha_{jk}\omega_{ki}-\sum\limits_\beta
h^\beta_{ij}\omega_{\beta\alpha},\label{2.8}\\ %%\tag {2.8}\\
\sum\limits_lh^\alpha_{ijkl}\omega_l\!&=&\!dh^\alpha_{ijk}\!+\!\!\sum\limits_m
h^\alpha_{mjk}\omega_{mi}\!+\!\!\sum\limits_mh^\alpha_{imk}\omega_{mj}\!+\!\!
\sum\limits_mh^\alpha_{ijm}\omega_{mk}\!-\!\!\sum\limits_\beta
h^\beta_{ijk}\omega_{\beta\alpha}. \label{2.9} %% \tag {2.9}
\ee
The Codazzi equations and the Ricci identities are
\be
h^\alpha_{ijk}&=&h^\alpha_{ikj}, \label{2.10}\\  %%\eqno{(2.10)}
h^\alpha_{ijkl}-h^\alpha_{ijlk}&=&\sum\limits_mh^\alpha_{im}R_{mjkl}+
\sum\limits_mh^\alpha_{jm}R_{mikl}+\sum\limits_\beta
h^\beta_{ij}R_{\alpha\beta kl}. \label{2.11}  %%\eqno{(2.11)}
\ee
The Ricci equations are
\be
R_{\alpha\beta
kl}=\sum\limits_m(h^\alpha_{km}h^\beta_{ml}-h^\alpha_{lm}h^\beta_{mk}). \label{2.12} %%\eqno{(2.12)}
\ee
The Laplacian of $h^\alpha_{ij}$ is defined by $\Delta
h^\alpha_{ij}=\sum\limits_k h^\alpha_{ijkk}$. From \eqref{2.11}, for any $\alpha, n+1 \leq \alpha \leq n+p$, we
obtain
\be\label{2.13}
\Delta h^\alpha_{ij}=\sum\limits_k
h^\alpha_{kkij}+\sum\limits_{k,m}h^\alpha_{km}R_{mijk}+\sum\limits_{k,m}h^\alpha_{im}
R_{mkjk}+\sum\limits_{k,\beta}h^\beta_{ik}R_{\alpha\beta jk}.
%% \eqno{(2.13)}
\ee
Define the first, second covariant derivatives and Laplacian of
the mean curvature vector field
$\vec{H}=\sum\limits_{\alpha}H^{\alpha}e_{\alpha}$ in the normal
bundle $N(M)$ as follows
\be\label{2.14}
\sum\limits_iH^{\alpha}_{,i}\theta_i&=&dH^{\alpha}+\sum\limits_{\beta}
H^{\beta}\theta_{\beta\alpha}, \\ %%\eqno{(2.14)}
\sum\limits_j H^{\alpha}_{,ij}\theta_j&=&dH^{\alpha}_{,i}+
\sum\limits_jH^{\alpha}_{,j}\theta_{ji}+\sum\limits_{\beta}H^{\beta}_{,i}\theta_{\beta\alpha},
\label{2.15}\\ %% \eqno{(2.15)}
\Delta^{\bot}H^{\alpha}&=&\sum\limits_iH^{\alpha}_{,ii},\quad
H^{\alpha}=\frac{1}{n}\sum\limits_kh^{\alpha}_{kk}. \label{2.16} %%\eqno{(2.16)}
\ee
Let $f$ be a smooth function on $M$. The first, second covariant
derivatives $f_i,f_{,ij}$ and Laplacian of $f$ are defined by
\be
df=\sum\limits_if_i\theta_i,\quad
\sum\limits_jf_{,ij}\theta_j=df_i+\sum\limits_jf_j\theta_{ji},\quad
\Delta f=\sum\limits_if_{,ii}. \label{2.17} %% \eqno{(2.17)}
\ee
For the fix index $\alpha(n+1\leq\alpha\leq n+p)$, we introduce an
operator $\Box^{\alpha}$ due to Cheng-Yau \cite{s5} by
\be
\Box^{\alpha}f=\sum\limits_{i,j}(nH^{\alpha}\delta_{ij}-h^{\alpha}_{ij})f_{,ij}.
\label{2.18} %% \eqno{(2.18)}
\ee
Since $M$ is compact, the operator $\Box^{\alpha}$ is self-adjoint
 (see \cite{s5}) if and only if
\be
\int_M(\Box^{\alpha}f)gdv=\int_Mf(\Box^{\alpha}g)dv, \label{2.19}  %% \eqno{(2.19)}
\ee
where $f$ and $g$ are smooth functions on $M$.
We need the following:
\begin{lemma}[See \cite{s17}] \label{Lemma 2.1}
Let $A, B$ be symmetric $n\times
n$ matrices satisfying $AB=BA$ and $\mathrm{tr} A=\mathrm{tr}
B=0$. Then
\be\label{2.20}
|\mathrm{tr} A^2B|\leq\frac{n-2}{\sqrt{n(n-1)}}(\mathrm{tr}
A^2)(\mathrm{tr} B^2)^{1/2}, %% \eqno{(2.20)}
\ee
and the equality holds if and only if $(n-1)$ of the eigenvalues
$x_i$ of $B$ and the corresponding eigenvalues $y_i$ of $A$
satisfy
\bee
|x_i|&=&(\mathrm{tr}B^2)^{1/2}/\sqrt{n(n-1)}, \quad x_i x_j\geq 0, \\
y_i&=&(\mathrm{tr}A^2)^{1/2}/\sqrt{n(n-1)}.
\eee
\end{lemma}

By the same method as in the proof of Lemma 4.2 in \cite{s9}, we also
have the following:
\begin{lemma}\label{Lemma 2.2.}
Let $\varphi: M\rightarrow
N^{n+p}_p(c)$ be an $n$-dimensional $(n\geq 2)$ spacelike
submanifold in $N^{n+p}_p(c)$. Then we have
\be\label{2.21}
|\nabla h|^2\geq\frac{3n^2}{n+2}|\nabla^{\bot}\vec{H}|^2,
%% \eqno{(2.21)}
\ee
where $|\nabla
h|^2=\sum\limits_{i,j,k,\alpha}(h^{\alpha}_{ijk})^2,\quad
|\nabla^{\bot}\vec{H}|^2=\sum\limits_{i,\alpha}(H^{\alpha}_{,i})^2.$
\end{lemma}


\section{ First variation and Euler-Lagrange equation}

In this section, we shall calculate the first variation of the
Willmore functional $W(\varphi_0)$ and obtain the Euler-Lagrange
equation \eqref{1.2}.

Let $\varphi_0 : M \rightarrow N^{n+p}_p(c)$ be an $n$-dimensional
compact spacelike submanifold in $N^{n+p}_p(c)$ with (possibly
empty) boundary $\partial M$. If otherwise, we will consider the
variation with compact support. Let $\varphi : M \times R
\rightarrow N^{n+p}_p(c)$ be a smooth variation of $\varphi_0$
such that $\varphi(\cdot , t) = \varphi_0$ on the boundary. Along
$\varphi : M \times R \rightarrow N^{n+p}_p(c)$, we choose a local
orthonormal basis $\{e_A\}$ for $TN^{n+p}_p(c)$ with dual basis
$\{\omega_A\}$, so that $\{e_i(\cdot , t)\}$ forms a local
orthonormal basis for $\varphi_t : M \times \{t\}\rightarrow
N^{n+p}_p(c)$. Since $T^{\ast}(M \times R) = T^{\ast}M \oplus
T^{\ast}R$, the pullback of $\{\omega_A\}$ and $\{\omega_{AB}\}$
on $N^{n+p}_p(c)$ through $\varphi : M \times R \rightarrow
N^{n+p}_p(c)$ have the decomposition
\be
\varphi^{\ast}\omega_{\alpha}&=&V_{\alpha} dt,\quad
\varphi^{\ast}\omega_i = \theta_i + V_i dt,\label{3.1}\\  %%\eqno{(3.1)}
\varphi^{\ast}\omega_{ij} &=& \theta_{ij} + L_{ij} dt,\quad
\varphi^{\ast}\omega_{i\alpha} = \theta_{i\alpha} +M_{i\alpha}dt,\quad
\varphi^{\ast}\omega_{\alpha \beta}=\theta_{\alpha
\beta}+N_{\alpha \beta}dt,\label{3.2}  %%\eqno{(3.2)}
\ee
where $\{V_i , V_{\alpha} , L_{ij} , M_{i \alpha} , N_{\alpha
\beta}\}$ are local functions on $M\times R$ with $L_{ij}
=-L_{ji}$, $N_{\alpha \beta} =-N_{\beta \alpha}$ and
\be
V = \frac{d}{dt}|_{t=0} \varphi_t
=\sum\limits_{i}V_id\varphi_0(e_i)+\sum\limits_{\alpha}V_{\alpha}
e_{\alpha},\label{3.3} %% \eqno{(3.3)}
\ee
is the variation vector field of $\varphi_t : M \rightarrow
N^{n+p}_p(c)$. We note that forms $\{\theta_i, \theta_{ij}
, \theta_{i\alpha}$, $\theta_{\alpha \beta}\}$ are defined on $M
\times \{t\}$, for $t=0$, they reduce to the forms with the same
notation on $M$. We denote by $d_M$ the differential operator on
$T^{\ast}M$; then $d=d_M+dt\frac{\partial}{\partial t}$ on
$T^{\ast}(M \times R$).

Let $K_{ABCD}$ be the components of the Riemannian curvature
tensor of $N^{n+p}_p(c)$. On $M \times \{t\}$, if we assume that
$h^{\alpha}_{ij}$ and the covariant derivatives $V_{i,j},
V_{\alpha,i}$ and $M_{i\alpha,j}$ are defined similarly to \cite{s6}
(see (3.7) - (3.10) in \cite{s6}), by the proof similar to Lemma~3.1 and
Lemma~3.2 in \cite{s6}, we have the
following lemmas:

\begin{lemma} \label{Lemma 3.1.}
Under the above notations, we have
\be
\frac{\partial\theta_i}{\partial t}&=&\sum\limits_j(V_{i,j}
+L_{ij})\theta_j+\sum\limits_{j,\alpha}h^{\alpha}_{ij}V_{\alpha}
\theta_j,\label{3.4}\\ %% \tag {3.4}\\
M_{i\alpha}&=&V_{\alpha,i}+\sum\limits_{j}h^{\alpha}_{ij}V_j,\label{3.5} %% \tag {3.5}\\
\\
&&\hspace*{-1.29cm}\begin{array}{ll}\label{3.6}
\displaystyle\frac{\partial\theta_{i\alpha}}{\partial
t}=&\displaystyle\sum\limits_{j}\Big(M_{i\alpha,j}+\sum\limits_{k}L_{ik}h^{\alpha}_{
jk}-\sum\limits_{\beta}N_{\beta
\alpha}h^{\beta}_{ij} \\
&\displaystyle-\sum\limits_{k}K_{i \alpha kj}V_k- \sum\limits_{\beta}K_{i
\alpha j\beta}V_{\beta}\Big)\theta_j.
\end{array}
\ee
\end{lemma}

\begin{lemma} \label{Lemma 3.2.}
\be \label{3.7}
\begin{array}{rl}
\displaystyle\frac{\partial h^{\alpha}_{ij}}{\partial
t}=&\displaystyle V_{\alpha,ij}+\sum\limits_{k}(L_{ik}h^{\alpha}_{kj}+L_{jk}h^{\alpha}_{ki}
+h^{\alpha}_{ijk}V_k) \\[3ex]
&\displaystyle+\sum\limits_{\beta}(N_{\alpha \beta}h^{\beta}_{ij}-K_{\alpha i
\beta
j}V_{\beta})-\sum\limits_{k,\beta}h^{\alpha}_{ik}h^{\beta}_{kj}
V_{\beta}.
\end{array}
\ee
\end{lemma}

\proof[{\bf Proof of Theorem~\ref{tm1.1}}]

By reasoning as in \cite{s6}, setting
$i=j$ in \eqref{3.7} and summing over $i$ by using
$\sum\limits_{i,k}L_{ik}h^{\alpha}_{ki}=0$, we have
\be \label{3.8}
\begin{array}{rl}
\displaystyle\frac{\partial H^{\alpha}}{\partial t}=&\displaystyle\frac{1}{n}\Delta^\bot
V_{\alpha}+\sum\limits_{k}H^{\alpha}_{,k}V_{k}+\sum\limits_{\beta}N_{\alpha
\beta}H^{\beta} \\[3ex]%% \tag {3.8}\\
&\displaystyle-\frac{1}{n}\sum\limits_{i,k,\beta}h^{\alpha}_{ik}h^{\beta}_{ki}
V_{\beta}-\frac{1}{n}\sum\limits_{i,\beta}K_{\alpha i \beta
i}V_{\beta}.
\end{array}
\ee
Since $\displaystyle\sum\limits_{i,j,\alpha,\beta}N_{\alpha
\beta}h^{\alpha}_{ij}h^{\beta}_{ij}=0$ and
$\displaystyle\sum\limits_{i,j,k,\alpha}L_{jk}h^{\alpha}_{ki}h^{\alpha}_{ij}=0$,
from \eqref{3.7} we have
\be \label{3.9}
\begin{array}{rl}
\displaystyle\frac{1}{2} \frac{\partial S}{\partial
t}=&\displaystyle\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}V_{\alpha,ij}+\frac{1}{2}\sum\limits_{k}S_{,k}V_{k} \\[3ex] %% \tag {3.9}\\
&\displaystyle-\sum\limits_{i,j,\alpha,\beta}K_{\alpha i \beta
j}h^{\alpha}_{ij}V_{\beta}-\sum\limits_{i,j,k,\alpha,\beta}h^{\alpha}_{ij}h^{\alpha}_{ik}h^{\beta}_{kj}V_{\beta}.
\end{array}
\ee
From \eqref{3.8} and $\displaystyle\sum\limits_{\alpha,\beta}N_{\alpha
\beta}H^{\alpha}H^{\beta}=0$, we have
\be \label{3.10}
\begin{array}{rl}
\displaystyle\frac{1}{2}\frac{\partial(nH^2)}{\partial
t}=&\displaystyle\sum\limits_{\alpha}H^{\alpha}\Delta^\bot
V_{\alpha}+\frac{n}{2}\sum\limits_{k}(H^2)_{,k}V_{k}  \\[3ex]  %% \tag{3.10}\\
&\displaystyle-\sum\limits_{i,j,\alpha,\beta}H^{\alpha}h^{\alpha}_{ij}h^{\beta}_{ij}V_{\beta}
-\sum\limits_{i,\alpha,\beta}H^{\alpha}K_{\alpha i \beta
i}V_{\beta}.
\end{array}
\ee
For $\varphi_t: M\rightarrow N^{n+p}_p(c)$, we consider the
non-negative functional
\be\label{3.11}
W(\varphi_t)=\int_M\rho^ndv=\int_M(S-nH^2)^{\frac{n}{2}}\theta_1\wedge
\cdots \wedge \theta_n. %% \eqno{(3.11)}
\ee
From \eqref{3.4}, we have
\be \label{3.12}
\begin{array}{rl}
\frac{\partial}{\partial t}(\theta_1\wedge \cdots \wedge
\theta_n)=&\sum\limits_{i}\theta_1 \wedge \cdots \wedge
\frac{\partial \theta_i }{\partial t}\wedge \cdots \wedge
\theta_n \\[3ex] %% \tag {3.12}\\
=&\Big(\sum\limits_{i}V_{i,i}+n\sum\limits_{\alpha}H^{\alpha}V_{\alpha}\Big)\theta_1\wedge
\cdots \wedge \theta_n.
\end{array}
\ee
From \eqref{3.9} and \eqref{3.10}, we see that
\be
\frac{\partial \rho^n}{\partial
t}&=&n\rho^{n-2}\Big\{\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}V_{\alpha,ij}
+\frac{1}{2}\sum\limits_{k}(\rho^2)_{,k}V_{k}
-\sum\limits_{i,j,\alpha,\beta}K_{\alpha i \beta
j}h^{\alpha}_{ij}V_{\beta}\nonumber\\ %% \tag {3.13}\\
&&-\sum\limits_{\alpha}H^{\alpha}\Delta^\bot V_{\alpha}
-\sum\limits_{i,j,k,\alpha,\beta}h^{\alpha}_{ij}h^{\alpha}_{ik}h^{\beta}_{kj}V_{\beta} \label{3.13}\\
&&+\sum\limits_{i,j,\alpha,\beta}H^{\alpha}h^{\alpha}_{ij}h^{\beta}_{ij}V_{\beta}
+\sum\limits_{i,\alpha,\beta}H^{\alpha}K_{\alpha i \beta
i}V_{\beta}\Big\}.\nonumber
\ee
From \eqref{3.11} - \eqref{3.13}, we have
\be
\frac{\partial w(\varphi_t)}{\partial
t}&=&\int_M\rho^{n-2}\Big\{\big[n\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}V_{\alpha,ij}-n\sum\limits_{\alpha}H^{\alpha}\Delta^\bot
V_{\alpha}+\frac{n}{2}\sum\limits_{k}(\rho^2)_{,k}V_{k}\nonumber\\ %% \tag{3.14}\\
&&+\rho^2\sum\limits_{k}V_{k,k}\big]
+n\sum\limits_{\alpha}\big[-\sum\limits_{i,j,\beta}K_{ \beta i
\alpha
j}h^{\beta}_{ij}-\sum\limits_{i,j,k,\beta}h^{\beta}_{ij}h^{\beta}_{ik}h^{\alpha}_{kj} \label{3.14}\\
&&+\sum\limits_{i,j,\beta}H^{\beta}h^{\beta}_{ij}h^{\alpha}_{ij}+\sum\limits_{i,\beta}H^{\beta}K_{\beta
i \alpha i}+\rho^2H^{\alpha}\big]V_{\alpha}\Big\}dv.\nonumber
\ee
By the same reason as in \cite{s6}, we see that
\be
\frac{\partial w(\varphi_t)}{\partial
t}&=&n\int_M\sum\limits_{\alpha}\Big\{\rho^{n-2}\big[
-\sum\limits_{i,j,k,\beta}h^{\beta}_{ij}h^{\beta}_{ik}h^{\alpha}_{kj}
-\sum\limits_{i,j,\beta}K_{ \beta i \alpha
j}h^{\beta}_{ij}\nonumber\\ %% \tag {3.15}\\
&&+\sum\limits_{i,j,\beta}H^{\beta}h^{\beta}_{ij}h^{\alpha}_{ij}+\sum\limits_{i,\beta}H^{\beta}K_{\beta
i \alpha
i}+\rho^2H^{\alpha}\big] \label{3.15}\\
&&+\sum\limits_{i,j}(\rho^{n-2}h^{\alpha}_{ij})_{ij}-\Delta^\bot
\rho^{n-2}H^{\alpha}\Big\}V_{\alpha}dv.\nonumber
\ee
From \eqref{2.3}, we see that
\[
-\sum\limits_{i,j,\beta}K_{ \beta i
\alpha j}h^{\beta}_{ij}+\sum\limits_{i,\beta}H^{\beta}K_{\beta i
\alpha i}=0.
\]
Thus, by \eqref{3.3} and \eqref{3.15} with restriction to $t=0$,
we obtain the Euler-Lagrange equation \eqref{1.2}. This completes the
proof of Theorem~\ref{tm1.1}.
\endproof


\section{Integral equalities of Willmore spacelike
submanifolds}

Define tensors
\be\label{4.1}
\tilde{h}^{\alpha}_{ij}&=&h^{\alpha}_{ij}-H^{\alpha}\delta_{ij},
%% \eqno{(4.1)}
\\
\tilde{\sigma}_{\alpha\beta}&=&\sum\limits_{i,j}\tilde{h}^{\alpha}_{ij}
\tilde{h}^{\beta}_{ij},\quad
\sigma_{\alpha\beta}=\sum\limits_{i,j}h^{\alpha}_{ij}h^{\beta}_{ij}.\label{4.2}
%% \eqno{(4.2)}
\ee
Then the $(p\times p)$-matrix $(\tilde{\sigma}_{\alpha\beta})$ is
symmetric and can be assumed to be diagonalized for a suitable
choice of $e_{n+1},\cdots,e_{n+p}$. We set
\be\label{4.3}
\tilde{\sigma}_{\alpha\beta}=\tilde{\sigma}_{\alpha}\delta_{\alpha\beta}.
%% \eqno{(4.3)}
\ee
By a direct calculation, we have
\be\label{4.4}
\sum\limits_k\tilde{h}^{\alpha}_{kk}&=&0,\quad
\tilde{\sigma}_{\alpha\beta}=\sigma_{\alpha\beta}-nH^{\alpha}H^{\beta},\quad
\rho^2=\sum\limits_{\alpha}\tilde{\sigma}_{\alpha}=S-nH^2,
%% \eqno{(4.4)}
\\
\sum\limits_{i,j,k,\alpha}h^{\beta}_{kj}h^{\alpha}_{ij}h^{\alpha}_{ik}&=&
\sum\limits_{i,j,k,\alpha}\tilde{h}^{\beta}_{kj}\tilde{h}^{\alpha}_{ij}
\tilde{h}^{\alpha}_{ik}+2\sum\limits_{i,j,\alpha}H^{\alpha}
\tilde{h}^{\alpha}_{ij}\tilde{h}^{\beta}_{ij}+H^{\beta}\rho^2+
nH^2H^{\beta}.\label{4.5} %% \eqno{(4.5)}
\ee
From \eqref{4.1}, \eqref{4.4} and \eqref{4.5}, the new Euler-Lagrange equation \eqref{1.2}
can be rewritten as
\begin{proposition} \label{Proposition 4.1.}
Let $M$ be an $n$-dimensional spacelike
submanifold in $N^{n+p}_p(c)$. Then $M$ is a Willmore spacelike
submanifold if and only if for $n+1\leq\alpha\leq n+p$
\be \label{4.6}
\begin{array}{rl}
\displaystyle\Box^{\alpha}(\rho^{n-2})=&\displaystyle(n-1)\rho^{n-2}\Delta^{\bot}H^{\alpha}+2(n-1)
\sum\limits_i(\rho^{n-2})_iH^{\alpha}_{,i} \\[3ex] %% \tag {4.6}\\
&\displaystyle+(n-1)H^{\alpha}\Delta(\rho^{n-2})-\rho^{n-2}\Big(\sum\limits_{\beta}
H^{\beta}\tilde{\sigma}_{\alpha\beta}+\sum\limits_{i,j,k,\beta}
\tilde{h}^{\alpha}_{ij}\tilde{h}^{\beta}_{ik}
\tilde{h}^{\beta}_{kj}\Big).
\end{array}
\ee
\end{proposition}

Setting $f=nH^{\alpha}$ in \eqref{2.18}, we have
\be \label{4.7}
\begin{array}{rl}
\displaystyle\Box^{\alpha}(nH^{\alpha})=&\displaystyle\sum\limits_{i,j}(nH^{\alpha}\delta_{ij}-h^{\alpha}_{ij})(nH^{\alpha})_{,ij}\\[3ex] %%\tag {4.7}\\
=&\displaystyle\sum\limits_i(nH^{\alpha})(nH^{\alpha})_{,ii}-\sum\limits_{i,j}
h^{\alpha}_{ij}(nH^{\alpha})_{,ij}.
\end{array}
\ee
We also have
\be
\frac{1}{2}\Delta(nH)^2&=&\frac{1}{2}\Delta\sum\limits_{\alpha}(nH^{\alpha})^2=
\frac{1}{2}\sum\limits_{\alpha}\Delta(nH^{\alpha})^2\nonumber\\ %% \tag {4.8}\\
&=&\frac{1}{2}\sum\limits_{\alpha,i}[(nH^{\alpha})^2]_{,ii}=
\sum\limits_{\alpha,i}[(nH^{\alpha})_{,i}]^2+
\sum\limits_{\alpha,i}(nH^{\alpha})(nH^{\alpha})_{,ii}\label{4.8}\\
&=&n^2|\nabla^{\bot}\vec{H}|^2+\sum\limits_{\alpha,i}(nH^{\alpha})
(nH^{\alpha})_{,ii}.\nonumber
\ee
Therefore, from \eqref{4.7} and \eqref{4.8}, we get
\be
\sum\limits_{\alpha}\Box^{\alpha}(nH^{\alpha})&=&
\frac{1}{2}\Delta(nH)^2-n^2|\nabla^{\bot}\vec{H}|^2-
\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}(nH^{\alpha})_{,ij}\nonumber\\ %% \tag {4.9}\\
&=&\frac{1}{2}\Delta[n(n-1)H^2-\rho^2+S]-n^2|\nabla^{\bot}\vec{H}|^2
-\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}(nH^{\alpha})_{,ij}\label{4.9}\\
&=&\frac{1}{2}\Delta S+\frac{1}{2}n(n-1)\Delta
H^2-\frac{1}{2}\Delta\rho^2-n^2|\nabla^{\bot}\vec{H}|^2
-\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}(nH^{\alpha})_{,ij}.\nonumber
\ee
On the other hand, we have
\be
\frac{1}{2}\Delta S &=&\sum\limits_{i,j,k,\alpha}(h^{\alpha}_{ijk})^2+
\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}\Delta h^{\alpha}_{ij}\nonumber\\ %% \tag {4.10}\\
&=& |\nabla h|^2+\sum\limits_{i,j,\alpha}h^{\alpha}_{ij}(nH^{\alpha})_{,ij}+
\sum\limits_{\alpha}\sum\limits_{i,j,k,l}h^{\alpha}_{ij}(h^{\alpha}_{kl}
R_{lijk}+h^{\alpha}_{li}R_{lkjk})\label{4.10}\\
 && +\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}h^{\alpha}_{ij}
 h^{\beta}_{ki} R_{\alpha \beta jk}.\nonumber
\ee
Putting \eqref{4.10} into \eqref{4.9}, we have
\be \label{4.11}
\begin{array}{rl}
\displaystyle\sum\limits_{\alpha}\Box^{\alpha}(nH^{\alpha})=&\displaystyle|\nabla h|^2-
n^2|\nabla^{\bot}\vec{H}|^2+\frac{1}{2}n(n-1)\Delta
H^2-\frac{1}{2}\Delta\rho^2 \\[3ex] %% \tag{4.11}\\
&\displaystyle+\sum\limits_{\alpha}\sum\limits_{i,j,k,l}h^{\alpha}_{ij}
(h^{\alpha}_{kl}R_{lijk}+h^{\alpha}_{li}R_{lkjk})+
\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}
h^{\alpha}_{ij}h^{\beta}_{ki}R_{\alpha\beta jk}.
\end{array}
\ee
Multiplying \eqref{4.11} by $\rho^{n-2}$ and taking the integral, using \eqref{2.19}, we have
\be \label{4.12}
\begin{array}{rl}
\displaystyle\sum\limits_{\alpha}\int_M(nH^{\alpha})\Box^{\alpha}(\rho^{n-2})dv
=&\displaystyle\int_M\rho^{n-2}(|\nabla h|^2-n^2|\nabla^{\bot}\vec{H}|^2)dv \\[3ex] %% \tag{4.12}\\
&\displaystyle+\frac{1}{2}n(n-1)\int_M\rho^{n-2}\Delta
H^2dv-\frac{1}{2}\int_M\rho^{n-2}\Delta\rho^2 dv\\[3ex]
&\displaystyle+\int_M\rho^{n-2}\sum\limits_{\alpha}\sum\limits_{i,j,k,l}
h^{\alpha}_{ij}(h^{\alpha}_{kl}R_{lijk}+h^{\alpha}_{li}R_{lkjk})dv\\[3ex]
&\displaystyle+\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}h^{\alpha}_{ij}
h^{\beta}_{ki}R_{\alpha \beta jk}dv.
\end{array}
\ee
Taking the Willmore equation \eqref{4.6} into \eqref{4.12} and making use of
the following:
\bee
\int_M\rho^{n-2}\sum\limits_{\alpha}H^{\alpha}\triangle^{\bot}H^{\alpha}
dv &=&
\frac{1}{2}\int_M\rho^{n-2}\sum\limits_{\alpha}\Delta^{\bot}(H^{\alpha})^2dv
-\int_M\rho^{n-2}\sum\limits_{i,\alpha}(H^{\alpha}_{,i})^2dv \\
&=&\frac{1}{2}\int_M\rho^{n-2}\Delta
H^2dv-\int_M\rho^{n-2}|\nabla\vec{H}|^2 dv,
\\
\int_MH^2\Delta(\rho^{n-2})dv&=&\int_M\sum\limits_{\alpha}(H^{\alpha})^2
\sum\limits_i(\rho^{n-2})_{,ii}dv\\
&=&\sum\limits_{\alpha,i}\int_M(H^{\alpha})^2(\rho^{n-2})_{,ii}dv
=-\!\sum\limits_{\alpha,i}\int_M(\rho^{n-2})_i((H^{\alpha})^2)_{,i}dv\\
&=&-2\int_M\!\sum\limits_{\alpha}H^{\alpha}\sum\limits_i(\rho^{n-2})_iH^{\alpha}_{,i}dv,\\
-\frac{1}{2}\int_M\rho^{n-2}\Delta\rho^2dv
&=&-\frac{1}{2}\sum\limits_i
\int_M\rho^{n-2}(\rho^2)_{,ii}dv\\
&=&\frac{1}{2}\sum\limits_i\int_M(\rho^2)_i(\rho^{n-2})_idv =
(n-2)\int_M\rho^{n-2}|\nabla\rho|^2dv,
\eee
by a direct calculation, we have the following:

\begin{proposition} \label{Proposition 4.2.}
Let $M$ be an $n$-dimensional compact
Willmore spacelike submanifold in $N^{n+p}_p(c)$. Then
\be \label{4.13}
\begin{array}{l}
\displaystyle
\int_M\rho^{n-2}(|\nabla h|^2-n|\nabla^{\bot}\vec{H}|^2)dv+(n-2)
\int_M\rho^{n-2}|\nabla\rho|^2dv \\[3ex] %% \tag{4.13}\\
\displaystyle\qquad+\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}nH^{\alpha}(H^{\beta}
\tilde{\sigma}_{\alpha\beta}+\sum\limits_{i,j,k}\tilde{h}^{\alpha}_{ij}
\tilde{h}^{\beta}_{ik}\tilde{h}^{\beta}_{kj})dv\\[3ex]
\displaystyle\qquad+\int_M\rho^{n-2}\sum\limits_{\alpha}\sum\limits_{i,j,k,l}h^{\alpha}_{ij}
(h^{\alpha}_{kl}R_{lijk}+h^{\alpha}_{li}R_{lkjk})dv\\[3ex]
\displaystyle\qquad+\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}h^{\alpha}_{ij}
h^{\beta}_{ki}R_{\alpha\beta jk}dv=0.
\end{array}
\ee
\end{proposition}
In general, for a matrix $A=(a_{ij})$ we denote by $N(A)$ the
square of the norm of $A$, that is,
\[
N(A)=\mathrm{trace}(A\cdot A^t)=\sum\limits_{i,j}(a_{ij})^2.
\]
Clearly, $N(A)=N(T^tAT)$ for any orthogonal matrix $T$. From \eqref{2.12}, we have
\be \label{4.14}
\begin{array}{rl}
\displaystyle\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}h^{\alpha}_{ij}
h^{\beta}_{ki}R_{\alpha\beta jk}&\displaystyle=
\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k,l}h^{\alpha}_{ij}
h^{\beta}_{ki}(h^{\beta}_{kl}h^{\alpha}_{lj}-h^{\beta}_{jl}h^{\alpha}_{lk})\\[3ex] %% \tag{4.14}\\
&\displaystyle=\frac{1}{2}\sum\limits_{\alpha,\beta,j,k}\Big(\sum\limits_lh^{\alpha}_{jl}
h^{\beta}_{lk}-\sum\limits_l
h^{\beta}_{jl}h^{\alpha}_{lk}\Big)^2\\[3ex]
&\displaystyle=\frac{1}{2}\sum\limits_{\alpha,\beta,j,k}\Big(\sum\limits_l\tilde{h}^{\alpha}_{jl}
\tilde{h}^{\beta}_{lk} -\sum\limits_l
\tilde{h}^{\beta}_{jl}\tilde{h}^{\alpha}_{lk}\Big)^2\\[3ex]
&\displaystyle=\frac{1}{2}\sum\limits_{\alpha,\beta}N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha}),
\end{array}
\ee
where $\tilde{A}_{\alpha}:=(\tilde{h}^{\alpha}_{ij})=
(h^{\alpha}_{ij}-H^{\alpha}\delta_{ij})$.

By using \eqref{2.6}, \eqref{2.12}, \eqref{4.2}, \eqref{4.4}, \eqref{4.5} and \eqref{4.14}, we
conclude that
\be \label{4.15}
\begin{array}{l}
\displaystyle\sum\limits_{\alpha}\sum\limits_{i,j,k,l}
h^{\alpha}_{ij}(h^{\alpha}_{kl}R_{lijk}+h^{\alpha}_{li}R_{lkjk})\\[3ex] %% \tag{4.15}\\
\displaystyle\qquad=nc\rho^2+\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k,l}
h^{\alpha}_{ij}h^{\beta}_{ij}h^{\alpha}_{lk}h^{\beta}_{lk}-
n\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}H^{\beta}h^{\beta}_{kj}
h^{\alpha}_{ij}h^{\alpha}_{ik}-\!\!\!\!\sum\limits_{\alpha,\beta,i,j,k}
h^{\alpha}_{ji}h^{\beta}_{ik}R_{\beta\alpha jk}\\[3ex]
\displaystyle\qquad=nc\rho^2+\sum\limits_{\alpha,\beta}\sigma^2_{\alpha\beta}
-n\sum\limits_{\alpha,\beta}
\sum\limits_{i,j,k}H^{\beta}\tilde{h}^{\beta}_{kj}
\tilde{h}^{\alpha}_{ij}\tilde{h}^{\alpha}_{ik}
-2n\sum\limits_{\alpha,\beta}\sum\limits_{i,j}H^{\alpha}H^{\beta}\tilde{h}^{\alpha}_{ij}
\tilde{h}^{\beta}_{ij}\\[3ex]
\displaystyle\qquad\quad-n\sum\limits_{\beta}(H^{\beta})^2\rho^2-
n^2H^2\sum\limits_{\beta}(H^{\beta})^2+\frac{1}{2}
\sum\limits_{\alpha,\beta}N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha})\\[3ex]
\displaystyle\qquad=nc\rho^2+\sum\limits_{\alpha,\beta}\tilde{\sigma}^2_{\alpha\beta}-nH^2\rho^2-
n\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}H^{\beta}\tilde{h}^{\beta}_{kj}
\tilde{h}^{\alpha}_{ij}\tilde{h}^{\alpha}_{ik}\\[3ex]
\displaystyle\qquad\quad+\frac{1}{2}\sum\limits_{\alpha,\beta}N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha}).
\end{array}
\ee
Putting \eqref{4.14} and \eqref{4.15} into \eqref{4.13}, we have the following:
\begin{proposition}\label{Proposition 4.3.}
Let $M$ be an $n$-dimensional compact
Willmore spacelike submanifold in $N^{n+p}_p(c)$. Then
\be \label{4.16}
\begin{array}{l}
\displaystyle\int_M\rho^{n-2}(|\nabla
h|^2-n|\nabla^{\bot}\vec{H}|^2)dv+(n-2)\int_M\rho^{n-2}|\nabla
\rho|^2dv \\[3ex] %% \tag{4.16}\\
\displaystyle\qquad +n\int_M\rho^{n-2}\Big(\sum\limits_{\alpha,\beta}H^{\alpha}H^{\beta}\tilde{\sigma}_{\alpha\beta}-H^2\rho^2\Big)
dv+nc\int_M\rho^ndv\\[3ex]
\displaystyle\qquad +\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}(N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha})+\tilde{\sigma}_{\alpha\beta}^2)dv=0.
\end{array}
\ee
\end{proposition}

\begin{corollary}\label{Corollary 4.1.}
Let $M$ be an $n$-dimensional compact
Willmore spacelike hypersurface in $N^{n+p}_p(c)$. Then
\be \label{4.17}
\int_M\rho^{n\!-\!2}(|\nabla h|^2\!-\!n|\nabla
H|^2)dv\!+\!(n\!-\!2)\!\!\int_M\rho^{n\!-\!2}|\nabla\rho|^2dv   %%\tag{4.17}\\
\!+\!\!\int_{M}\rho^n(nc\!+\!\rho^2)dv=0.
\ee
\end{corollary}

\section{Proofs of Theorems}

From Remark~\ref{Remark 1.1.}, in the proofs of Theorem~\ref{Theorem 1.2.} - Theorem~\ref{Theorem 1.4.}, we should
assume that $n\neq 3, 5$.

\proof[{\bf Proof of Theorem~\ref{Theorem 1.2.}}] $(1)$ For $p= 1$, from Lemma~\ref{Lemma 2.2.}
and \eqref{4.17}, we have
\be \label{5.1}
\begin{array}{rl}
0=&\displaystyle\int_M\rho^{n-2}(|\nabla
h|^2-\frac{3n^2}{n+2}|\nabla^{\bot}\vec{H}|^2)dv+\int_M\rho^{n-2}(\frac{3n^2}{n+2}-n)|\nabla^{\bot}\vec{H}|^2dv\\[3ex] %% \tag{5.1}\\
&\displaystyle+(n-2)\int_M\rho^{n-2}|\nabla\rho|^2dv +\int_{M}\rho^n(nc+\rho^2)dv\geq \int_{M}\rho^n(nc+\rho^2)dv.
\end{array}
\ee

$(i)$ If $c=1$, since $nc+\rho^2>0$, from \eqref{5.1}, it follows that
$\rho^2=0$ and $M$ is totally umbilical. If $c=0$, since
$nc+\rho^2=\rho^2$, from \eqref{5.1}, we easily see that $\rho^2=0$,
thus $M$ is totally umbilical.

$(ii)$ If $c=-1$ and $\rho^2\geq n$, since
$nc+\rho^2=-n+\rho^2\geq0$, from \eqref{5.1}, we have $\rho^2=0$ and $M$
is totally umbilical or $\rho^2=n$. In the latter case, since
$\rho^2=n>0$, from \eqref{5.1} we have  that
\[
\int_M\rho^{n-2}(\frac{3n^2}{n+2}-n)|\nabla^{\bot}\vec{H}|^2dv=0
\quad \mbox{and} \quad
\int_M\rho^{n-2}(|\nabla
h|^2-\frac{3n^2}{n+2}|\nabla^{\bot}\vec{H}|^2)dv=0.
\]
Thus
$\nabla^{\bot}\vec{H}=0$ and $\nabla h=0$, that is, $H= constant$
and the second fundamental form of $M$ is parallel. It easily
follows that $M$ is an isoparametric spacelike hypersurface with
two distinct constant principal curvatures. By the congruence
Theorem of Abe, Koike and Yamaguchi (see Theorem 5.1 of \cite{s1}), we
know that $M$ is isometric to Example~\ref{Example 1.1.}. This is impossible
since $M$ is compact.

$(2)$ For $p\geq 2$, from \eqref{4.3}, we get
\be\label{5.2}
\sum\limits_{\alpha,\beta}\tilde{\sigma}^2_{\alpha\beta}=\sum\limits_{\alpha}
\tilde{\sigma}^2_{\alpha}\geq\frac{1}{p}\Big(\sum\limits_{\alpha}
\tilde{\sigma}_{\alpha}\Big)^2=\frac{1}{p}\rho^4. %% \eqno{(5.2)}
\ee
From \eqref{4.16} and \eqref{5.2} and
\be \label{5.3}
\sum\limits_{\alpha,\beta}N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha})&\geq& 0, %% \tag{5.3}
\\
\sum\limits_{\alpha,\beta}H^{\alpha}H^{\beta}\tilde{\sigma}_{\alpha\beta}=\sum\limits_{\alpha}(H^{\alpha})^2\tilde{\sigma}_{\alpha}&\geq&0,\label{5.4}
%% \tag{5.4}
\ee
we have
\be \label{5.5}
\begin{array}{rl}
0=&\displaystyle\int_M\rho^{n-2}(|\nabla
h|^2-n|\nabla^{\bot}\vec{H}|^2)dv+(n-2)\int_M\rho^{n-2}|\nabla
\rho|^2dv \\[3ex] %% \tag{5.5}\\
&\displaystyle +\,n\int_M\rho^{n-2}\Big(\sum\limits_{\alpha,\beta}H^{\alpha}H^{\beta}\tilde{\sigma}_{\alpha\beta}-H^2\rho^2\Big)
dv+nc\int_M\rho^ndv\\[3ex]
&\displaystyle +\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}(N(\tilde{A}_{\alpha}\tilde{A}_{\beta}-
\tilde{A}_{\beta}\tilde{A}_{\alpha})+\tilde{\sigma}_{\alpha\beta}^2)dv\\[3ex]
\displaystyle\geq &\displaystyle\int_M\rho^{n}\left\{\frac{1}{p}\rho^2+nc-nH^2\right\}dv.
\end{array}
\ee
In particular, if
\[
\rho^2\geq np(H^2-c),
\]
from \eqref{5.5}, we see that $\rho^2=0$ and $M$ is totally umbilical or
$\rho^2= np(H^2-c)$. In the latter case, from \eqref{5.5} we have that
\[
\int_M\rho^{n-2}\sum\limits_{\alpha,\beta}H^{\alpha}H^{\beta}\tilde{\sigma}_{\alpha\beta}dv=0,
\]
that is
\be \label{5.6}
\int_M\rho^{n-2}\sum\limits_{\alpha}(H^{\alpha})^2\tilde{\sigma}_{\alpha}dv=0. %% \tag{5.6}
\ee
If $\rho^{2}=0$, that is, $M$ is totally umbilical; if
$\rho^{2}\neq 0$, from \eqref{5.6} it follows that
\[
\sum\limits_{\alpha}(H^{\alpha})^2\tilde{\sigma}_{\alpha}=0.
\]
Thus, we see that $H^{\alpha}=0$ and $H=0$. If $c=1$, we have
$\rho^2= -np<0$, a contradiction; if $c=0$, we have $\rho^2=0$,
also a contradiction since we assume that $\rho^{2}\neq 0$; if
$c=-1$, we have $\rho^2= np$. Since $H=0$ and $M$ is maximal, it
follows that $S= np$. From a result of T Ishihara \cite{s7} (see Theorem
1.3 of \cite{s7}), $M$ is isometric to
\[
H^{n_1}(\sqrt{\frac{n_1}{n}})\times \cdots \times
H^{n_{p+1}}(\sqrt{\frac{n_{p+1}}{n}}),
\]
where
$n_1+\cdots+n_{p+1}=n$. This is impossible
since $M$ is compact. This completes the proof of Theorem~\ref{Theorem 1.2.}.
\endproof

\proof[{\bf Proof of Theorem~\ref{Theorem 1.3.}}]
For a fixed $\alpha,
n+1\leq\alpha\leq n+p$, we can take a local orthonormal frame
field $\{e_1, \ldots, e_n\}$ such that
$h^{\alpha}_{ij}=\lambda^{\alpha}_i\delta_{ij}$, then
$\tilde{h}^{\alpha}_{ij}=\mu^{\alpha}_i\delta_{ij}$ with
$\mu^{\alpha}_i=\lambda^{\alpha}_i-H^{\alpha}, \
\sum\limits_i\mu^{\alpha}_i=0$. Thus
\be \label{5.7}
\sum\limits_{\alpha,i,j,k,l}h^{\alpha}_{ij}(h^{\alpha}_{kl}R_{lijk}+
h^{\alpha}_{li}R_{lkjk})&=&\frac{1}{2}\sum\limits_{\alpha,i,j}
(\lambda^{\alpha}_i-\lambda^{\alpha}_j)^2R_{ijij}\\ %% \tag{5.7}\\
&=&\frac{1}{2}\sum\limits_{\alpha,i,j}(\mu^{\alpha}_i-\mu^{\alpha}_j)^2R_{ijij}
\geq nK\rho^2,\nonumber
\ee
and the equality in \eqref{5.7} holds if and only if $R_{ijij}=K$ for
any $i\neq j$.

Let $\displaystyle\sum\limits_i(\tilde{h}^{\beta}_{ii})^2=\tau_{\beta}$. Then
$\displaystyle\tau_{\beta}\leq\sum\limits_{i,j}(\tilde{h}^{\beta}_{ij})^2=\tilde{\sigma}_{\beta}$.
Since $\displaystyle\sum\limits_i\tilde{h}^{\beta}_{ii}=0, \sum\limits_i
\mu^{\alpha}_i=0$ and
$\displaystyle\sum\limits_i(\mu^{\alpha}_i)^2=\tilde{\sigma}_{\alpha}$,
from Lemma~\ref{Lemma 2.1} we have that
\be \label{5.8}
\sum\limits_{\alpha,\beta}\sum\limits_{i,j,k}
H^{\alpha}\tilde{h}^{\alpha}_{ij}\tilde{h}^{\beta}_{kj}\tilde{h}^{\beta}_{ik}
&=&\sum\limits_{\beta,\alpha}\sum\limits_{i,j,k} H^{\beta}
\tilde{h}^{\beta}_{ij}\tilde{h}^{\alpha}_{kj}\tilde{h}^{\alpha}_{ik}=\sum\limits_{\alpha,\beta}H^{\beta}\sum\limits_i
\tilde{h}^{\beta}_{ii}(\mu^{\alpha}_i)^2\\ %% \tag{5.8}\\
&\geq&-\frac{n-2}{\sqrt{n(n-1)}}\sum\limits_{\alpha,\beta}|H^{\beta}|
\tilde{\sigma}_{\alpha}\sqrt{\tau_{\beta}}\nonumber\\
&\geq&-\frac{n-2}{\sqrt{n(n-1)}}\sum\limits_{\alpha}
\tilde{\sigma}_{\alpha}\sum\limits_{\beta}|H^{\beta}|\sqrt{\tilde{\sigma}_{\beta}}\nonumber\\
&\geq&-\frac{n-2}{\sqrt{n(n-1)}}\rho^2\sqrt{\sum\limits_{\beta}(H^{\beta})^2\sum\limits_{\beta}
\tilde{\sigma}_{\beta}}=-\frac{n-2}{\sqrt{n(n-1)}}H\rho^3.\nonumber
\ee
From \eqref{4.13}, \eqref{4.14}, \eqref{5.3}, \eqref{5.4}, \eqref{5.7} and \eqref{5.8}, we have
\be \label{5.9}
\begin{array}{rl}
0\geq&\displaystyle
\int_M\rho^{n-2}\sum\limits_{\alpha}n(H^{\alpha})^2\tilde{\sigma}_{\alpha}
-\int_M\rho^{n-2}
\frac{n(n-2)}{\sqrt{n(n-1)}}H\rho^3dv\\[3ex] %% \tag{5.9}\\
&\displaystyle+\int_M\rho^{n-2}nK\rho^2dv \geq
\int_Mn\rho^{n}\left\{K-\frac{n-2}{\sqrt{n(n-1)}}H\rho\right\}dv.
\end{array}
\ee
In particular, if
\[
K\geq\frac{n-2}{\sqrt{n(n-1)}}H\rho,
\]
from \eqref{5.9} we see that $\rho^2=0$ and $M$ is totally umbilical or
\[
K=\frac{n-2}{\sqrt{n(n-1)}}H\rho.
\]
In the latter case, from
\eqref{5.9}, we know that \eqref{5.6} holds. If $\rho^{2}=0$, that is, $M$ is
totally umbilical; if $\rho^{2}\neq 0$, it follows from \eqref{5.6} that
$\sum_{\alpha}(H^{\alpha})^2\tilde{\sigma}_{\alpha}=0$. Thus, we
see that $H^{\alpha}=0$ and $H=0$. It also follows from \eqref{5.7} that
$R_{ijij}=K$ for any $i\neq j$. Since
\[
K=\frac{n-2}{\sqrt{n(n-1)}}H\rho=0,
\]
we have $R_{ijij}=0$ for any
$i\neq j$. From the Gauss equation \eqref{2.5}, we have $n(n-1)c+S=0$.
If $c=1$, we have $n(n-1)+S=0$, a contradiction; if $c=0$, we have
$S=0$, thus $\rho^2=0$, also a contradiction since we assume that
$\rho^{2}\neq 0$; if $c=-1$, we have $S=n(n-1)$. Since $M$ is
maximal and $S= np$, where $p=n-1$, from a result of T Ishihara
\cite{s7}, $M$ is isometric to
\[
H^{n_1}(\sqrt{\frac{n_1}{n}})\times
\cdots \times H^{n_{n}}(\sqrt{\frac{n_{n}}{n}}),
\]
where
$n_1+\cdots+n_{n}=n$. This is impossible
since $M$ is compact. This completes the proof of Theorem~\ref{Theorem 1.3.}.
\endproof

\proof[{\bf Proof of Theorem~\ref{Theorem 1.4.}}]
From \eqref{2.6} and \eqref{4.1}, we have
\[
Q\leq
R_{ii}=(n-1)c-(n-2)\sum\limits_{\alpha}H^{\alpha}\tilde{h}^{\alpha}_{ii}
-(n-1)H^2+\sum\limits_{\alpha,j} (\tilde{h}^{\alpha}_{ij})^2.
\]
Thus
\be \label{5.10}
\rho^2=\sum\limits_{\alpha,i, j} (\tilde{h}^{\alpha}_{ij})^2\geq
nQ-n(n-1)(c-H^2), %% \tag{5.10}
\ee
and
\be \label{5.11}
\sum\limits_{\alpha,\beta}\tilde{\sigma}^2_{\alpha\beta}\geq\frac{1}{p}\rho^4\geq\frac{1}{p}\rho^2[nQ-n(n-1)(c-H^2)].
%% \eqno{(5.11)}
\ee
From \eqref{4.16}, \eqref{5.3}, \eqref{5.4} and \eqref{5.11}, we have
\be  \label{5.12}
0&\geq&
\int_M\rho^{n-2}\sum\limits_{\alpha}n(H^{\alpha})^2\tilde{\sigma}_{\alpha}
-n\int_M\rho^{n-2} H^2\rho^2dv\nonumber\\
&&+nc\int_M\rho^{n}dv+\int_M\rho^{n-2}\frac{1}{p}\rho^{2}[nQ-n(n-1)(c-H^2)]\\ %% \tag{5.12}\\
&\geq& \frac{n}{p}\int_M\rho^{n}\left\{Q-(n-p-1)(c-H^2)\right\}dv.\nonumber
\ee
In particular, if
$$
Q\geq (n-p-1)(c-H^2),
$$
from \eqref{5.12}, we see that $\rho^2=0$ and $M$ is totally umbilical
or $Q= (n-p-1)(c-H^2)$. In the latter case, from \eqref{5.12}, we know
that \eqref{5.6} holds. If $\rho^{2}=0$, that is, $M$ is totally
umbilical; if $\rho^{2}\neq 0$, it follows from \eqref{5.6} that
\[
\sum_{\alpha}(H^{\alpha})^2\tilde{\sigma}_{\alpha}=0.
\]
Thus, we
see that $H^{\alpha}=0$ and $H=0$. It also follows from \eqref{5.12}
that the equality in \eqref{5.10} holds, that is, $\rho^2=
nQ-n(n-1)(c-H^2)=nQ-n(n-1)c$. Since we also know that $Q=
(n-p-1)c$, we see that $\rho^2=-npc$, by reasoning as in the proof
of Theorem~\ref{Theorem 1.2.}, we know that this is impossible. This completes
the proof of Theorem~\ref{Theorem 1.4.}.
\endproof

\section*{Acknowledgement}
The authors would like to thank the
referees for their many valuable comments and suggestions that
have really improved the paper.


%%%% Acknowledgment %%%%%%%%
%\section*{Acknowledgement}



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