# Four dimensional hypersurfaces with proper mean curvature vector field in
pseudo-Riemannian space forms

Chao Yang<EMAIL_ADDRESS>Jiancheng Liu<EMAIL_ADDRESS>Li Du
<EMAIL_ADDRESS>College of Mathematics and Statistics, Northwest Normal
University, Lanzhou, China  School of Science, Chongqing University of
Technology, Chongqing, China

###### Abstract

In this paper, we study four dimensional hypersurface $M^{4}_{r}$ with proper
mean curvature vector field (i.e. $\Delta\vec{H}$ is proportional to
$\vec{H}$) in pseudo-Riemannian space form $N^{5}_{s}(c)$, and show that it
has constant mean curvature, and give the range of this constant. As an
application, we get that biharmonic hypersurfaces in $N^{5}_{s}(c)$ are
minimal in some specific cases, which partially confirms B.-Y. Chen’s
conjecture.

###### keywords:

proper mean curvature vector field, pseudo-Riemannian space form, four
dimensional hypersurfaces, mean curvature

###### MSC:

53C50

##  1 Introduction

Let $N^{n+1}_{s}(c)$ be a ($n+1$)-dimensional pseudo-Riemannian space form
with index $0\leq s\leq n+1$ and constant sectional curvature $c$. Specially,
the pseudo-Riemannian space form $N^{n+1}_{s}(0)$ is isometric to
($n+1$)-dimensional pseudo-Euclidean space $\mathbb{E}^{n+1}_{s}$ with index
$s$. Let $x:M^{n}_{r}\to N^{n+1}_{s}(c)$ be an isometric immersion of a
pseudo-Riemannian hypersurface $M^{n}_{r}$ into $N^{n+1}_{s}(c)$. Denote by
$\vec{H}$ and $\Delta$ the mean curvature vector field and the Laplace
operator of $M^{n}_{r}$. The hypersurface $M^{n}_{r}$ is said to have proper
mean curvature vector field, if it satisfies the equation

$\Delta\vec{H}=\lambda\vec{H},$

for some real constant $\lambda$. Specially, when $\lambda=nc$, the
hypersurface $M^{n}_{r}$ is biharmonic.

In 1988, B.-Y. Chen initiated the study of hypersurface $M^{n}_{r}$ with
proper mean curvature vector field in $\mathbb{E}^{n+1}_{s}$ in [4], and
proved that when $n=2,s=0$, surface $M^{2}$ is minimal, or an open part of a
circular cylinder. And then, A. Ferrández and P. Lucus [10] classified such
non-minimal surfaces for $n=2,s=1$. For $n=3$, the result of classification
has not been gotten, but it was proved that hypersurface $M^{3}_{r}$ of
$\mathbb{E}^{4}_{s}$ ($s=0,1,2$) has constant mean curvature ($s=0$ by F.
Defever in [6] and T. Hasanis etc. in [18], $s=1,2$ by A. Arvanitoyeorgos etc.
in [1, 2, 3]).

Naturally, there is a conjecture in [3] that: _any hypersurface having proper
mean curvature vector field in pseudo-Euclidean space $\mathbb{E}^{n+1}_{s}$
has constant mean curvature,_ which is closely related with the well known
B.-Y. Chen’s conjecture about biharmonic hypersurfaces in [5]: _Any biharmonic
hypersurface in $\mathbb{E}^{n+1}_{s}$ is minimal._ In [11], Y. Fu, M.-C. Hong
and X. Zhan illustrated the significance of solving Chen’s conjecture for
$n=4$, by analogy with the famous Bernstein problem. In contrast to the Chen’s
conjecture, we realized that to solve the conjecture about hypersurfaces with
proper mean curvature vector field for $n=4$ is necessary.

In 2021, Y. Fu and X. Zhan give an affirmative answer to this conjecture for
$n=4,s=0$ in [12]. However, when $s>0$, the related research is difficult to
carry out, since the shape operator of the hypersurface $M^{4}_{r}$ is not
necessarily diagonalizable, and the principal curvatures may not all be real.
In 2022, under the assumption that the hypersurface has constant scalar
curvature and diagonalizable shape operator, L. Du and J. Ren obtained the
same conclusion for $n=4,s>0$ in [8]. There are also some papers ([7, 9, 14,
15, 16]) studied the conjecture for $n>4$, but some strong restrictions are
attached.

In this paper, we overcome the difficulties caused by the non-diagonalizable
shape operator and the imaginary principal curvatures, and show that the
hypersurface $M^{4}_{r}$ with proper mean curvature vector field in pseudo-
Riemannian space form ${N}^{5}_{s}(c)$ has constant mean curvature in section
3 (cf. Theorem 3.1). Thus, the above conjecture about hypersurfaces with
proper mean curvature vector field is true for $n=4$.

Once we know the mean curvature of the hypersurface $M^{4}_{r}$ is a constant,
we continue in section 4 to estimate that constant. When $M^{4}_{r}$ has at
most two distinct principal curvatures, we get the value of the mean curvature
$H$ (cf. Theorems 4.1, 4.6 and 4.8), and find the value depends on
$\varepsilon c$, $\varepsilon\lambda$ and the multiplies of the principal
curvatures, where $\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle$, $\vec{\xi}$
denote a unit normal vector field to $M^{4}_{r}$. Its a pity that when the
number of distinct principal curvatures of $M^{4}_{r}$ is larger than 2, we do
not obtain the value of the mean curvature. However, we give a value range of
$H$ when the principal curvatures of $M^{4}_{r}$ are all real (cf. Theorem
4.12). As an application of these results, we can partially answer B.-Y.
Chen’s conjecture, details are as follows:

  * 1.

Biharmonic hypersurface $M^{4}_{r}$ in $N^{5}_{s}(c)$ with two distinct
principal curvatures, which are imaginary, is minimal (cf. Corollary 4.11);

  * 2.

Biharmonic hypersurface $M^{4}_{r}$ in $N^{5}_{s}(c)$ satisfying
$c\varepsilon\leq 0$ and without imaginary principal curvatures, is minimal
(cf. Corollary 4.13).

##  2 Preliminaries

###  2.1 The formulas of hypersurface $M^{n}_{r}$ in $N^{n+1}_{s}(c)$

Let $N^{n+1}_{s}(c)$ be a pseudo-Riemannian space form with index $s$ and
constant sectional curvature $c$. A non-zero vector $X$ in $N^{n+1}_{s}(c)$ is
called _time-like_ , _space-like_ or _light-like_ , according to whether
$\langle X,X\rangle$ is negative, positive or zero.

Let $M^{n}_{r}$ be a nondegenerate hypersurface in $N^{n+1}_{s}(c)$.
$\vec{\xi}$ denote a unit normal vector field to $M^{n}_{r}$, then
$\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$. $\nabla$ and
$\tilde{\nabla}$ denote the Levi-Civita connections of $M^{n}_{r}$ and
$N^{n+1}_{s}(c)$, respectively. For any vector fields $X,Y$ tangent to
$M^{n}_{r}$, the Gauss formula is given by

$\widetilde{\nabla}_{X}Y=\nabla_{X}Y+h(X,Y)\vec{\xi},$

where $h$ is the scalar-valued second fundamental form. Denote by $A$,
$\vec{H}$, and $H$ the shape operator of $M^{n}_{r}$ associated to
$\vec{\xi}$, the mean curvature vector field and the mean curvature, then
$\vec{H}=H\vec{\xi}$ and $H=\frac{1}{n}\varepsilon\text{tr}A$.

For any vector fields $X,Y,Z$ tangent to $M^{n}_{r}$, the Codazzi and Gauss
equations are given by

$\langle(\nabla_{X}A)Y,Z\rangle=\langle(\nabla_{Y}A)X,Z\rangle,$ (2.1)

and

$R(X,Y)Z=c(\langle Y,Z\rangle X-\langle X,Z\rangle Y)+\varepsilon\langle
A(Y),Z\rangle A(X)-\varepsilon\langle A(X),Z\rangle A(Y),$

where $R(X,Y)Z=\nabla_{X}\nabla_{Y}Z-\nabla_{Y}\nabla_{X}Z-\nabla_{[X,Y]}Z$.

###  2.2 The equivalent equations of $\Delta\vec{H}=\lambda\vec{H}$

The equation $\Delta\vec{H}=\lambda\vec{H}$ can be rewritten as

$-\Delta\vec{H}-\text{trace}\,\tilde{R}(\text{d}x,\vec{H})\text{d}x=(\lambda-
nc)\vec{H},$

which is equivalent to

$\tau_{2}(x)=(\lambda-nc)\tau(x),$

where $\tau(x)$ and $\tau_{2}(x)$ are the tension and bitension fields of $x$,
respectively, and $\tilde{R}$ is the curvature tensor of $N^{n+1}_{s}(c)$.

Thus, according to [7], the hypersurface $M^{n}_{r}$ satisfies
$\Delta\vec{H}=\lambda\vec{H}$ if and only if

$A(\nabla H)=-\frac{n}{2}\varepsilon H(\nabla H),$ (2.2)

and

$\Delta H+\varepsilon H\text{tr}A^{2}=\lambda H,$ (2.3)

where the Laplace operator $\Delta$ acting on scalar-valued function $f$ is
given by

$\Delta f=-\sum^{n}_{i=1}\varepsilon_{i}(e_{i}e_{i}-\nabla_{e_{i}}e_{i})f,$

with $\\{e_{i}\\}_{i=1}^{n}$ be a local orthonormal frame such that $\langle
e_{i},e_{i}\rangle=\varepsilon_{i}=\pm 1$.

###  2.3 The shape operator of hypersurface $M^{n}_{r}$

The tangent space $T_{p}M^{n}_{r}$ at $p\in M^{n}_{r}$ can be expressed as a
direct sum of subspaces $V_{k}$, $1\leq k\leq m$, that are mutually orthogonal
and invariant under the shape operator $A$. According to [17, exercise 18, pp.
260-261], there exists an integer $t$, with $0\leq t\leq m$, such that
$A|_{V_{i}}$ (the restriction of $A$ on $V_{i}$), $1\leq i\leq t$, has the
form

$\displaystyle A_{i}=\left(\begin{array}[]{ccccc}\lambda_{i}&&&&\\\
1&\lambda_{i}&&&\\\ &\ddots&\ddots&&\\\ &&1&\lambda_{i}&\\\
&&&1&\lambda_{i}\\\ \end{array}\right),$

with respect to a basis
$\mathfrak{B}_{i}=\\{u_{i_{1}},u_{i_{2}},\cdots,u_{i_{\alpha_{i}}}\\}$ of
$V_{i}$, and $A|_{V_{j}}$, $t+1\leq j\leq m$, has the form

$\displaystyle\overline{A}_{j}=\left(\begin{array}[]{cccccccccc}\gamma_{j}&\tau_{j}&&&&&&&&\\\
-\tau_{j}&\gamma_{j}&0&&&&&&&\\\ 1&0&\gamma_{j}&\tau_{j}&&&&&&\\\
&1&-\tau_{j}&\gamma_{j}&0&&&&&\\\ &&1&0&\gamma_{j}&\tau_{j}&&&&\\\
&&&1&-\tau_{j}&\gamma_{j}&0&&&\\\ &&&&\ddots&\ddots&\ddots&\ddots&&\\\
&&&&&1&0&\gamma_{j}&\tau_{j}\\\ &&&&&&1&-\tau_{j}&\gamma_{j}\\\
\end{array}\right),\tau_{j}\neq 0,$

with respect to a basis
$\overline{\mathfrak{B}}_{j}=\\{u_{\bar{j}_{1}},u_{\tilde{j}_{1}},u_{\bar{j}_{2}},u_{\tilde{j}_{2}},\cdots,u_{\bar{j}_{\beta_{j}}},u_{\tilde{j}_{\beta_{j}}}\\}$
of $V_{j}$. The inner products of the basis elements in $\mathfrak{B}_{i}$,
$1\leq i\leq t$, and $\overline{\mathfrak{B}}_{j}$, $t+1\leq j\leq n$, are all
zero except

$\langle u_{i_{a}},u_{i_{b}}\rangle=\varepsilon_{i}=\pm 1,\ \
a+b=\alpha_{i}+1,\ \ 1\leq i\leq t,$

and

$\langle u_{\bar{j}_{c}},u_{\bar{j}_{d}}\rangle=1=-\langle
u_{\tilde{j}_{c}},u_{\tilde{j}_{d}}\rangle,\ \ c+d=\beta_{j}+1,\ \ t+1\leq
j\leq m.$

Certainly, the sum of the dimensions of $V_{k}$, $1\leq k\leq m$ is equal to
$n$, i.e.

$\sum_{i=1}^{t}\alpha_{i}+2\sum_{j=t+1}^{m}\beta_{j}=n.$

Collecting all vectors in
$\mathfrak{B}_{1},\cdots,\mathfrak{B}_{t},\overline{\mathfrak{B}}_{t+1},\cdots,\overline{\mathfrak{B}}_{m}$
in order, we get a basis

$\mathfrak{B}\\!=\\!\\{u_{i_{1}},u_{i_{2}},\cdots,u_{i_{\alpha_{i}}},u_{\bar{j}_{1}},u_{\tilde{j}_{1}},u_{\bar{j}_{2}},u_{\tilde{j}_{2}},\cdots,u_{\bar{j}_{\beta_{j}}},u_{\tilde{j}_{\beta_{j}}}|1\leq
i\leq t,t+1\leq j\leq m\\}$

of $T_{x}M^{n}_{r}$. With respect to this basis $\mathfrak{B}$, the shape
operator $A$ of the hypersurface $M^{n}_{r}$ in $N^{n+1}_{s}(c)$ can be
expressed as an almost diagonal matrix

$A=\text{diag}\\{A_{1},\cdots,A_{t},\overline{A}_{t+1},\cdots,\overline{A}_{m}\\}.$

Observe the form of $A_{i}$ and $\overline{A}_{j}$, with $1\leq i\leq t$,
$t+1\leq j\leq m$, we find $A_{i}$ has only a simple eigenvalue $\lambda_{i}$,
and $\overline{A}_{j}$ has eigenvalues $\gamma_{j}+\tau_{j}\sqrt{-1}$,
$\gamma_{j}-\tau_{j}\sqrt{-1}$. So, the shape operator $A$ has real
eigenvalues

$\lambda_{1},\lambda_{2},\cdots,\lambda_{t},$

and imaginary eigenvalues

$\gamma_{t+1}+\tau_{t+1}\sqrt{-1},\gamma_{t+1}-\tau_{t+1}\sqrt{-1},\cdots,\gamma_{m}+\tau_{m}\sqrt{-1},\gamma_{m}-\tau_{m}\sqrt{-1}.$

According to the above explanations for the form of the shape operator, we
conclude that for four dimensional hypersurface $M^{4}_{r}$, if there exists
real principal curvatures, then the shape operator $A$ and corresponding
metric matrix $G$ have possible forms:

$\begin{array}[]{|c|c|c|c|}\hline\cr\textbf{Form (I)}&\textbf{Form
(I\\!I)}&\textbf{Form (I\\!I\\!I)}\\\ \hline\cr\begin{array}[]{c}t=m=4,\\\
\alpha_{i}=1,i=1,\cdots,4.\end{array}&\begin{array}[]{c}t=m=3,\\\
\alpha_{1}=2,\alpha_{2}=\alpha_{3}=1.\end{array}&\begin{array}[]{c}t=m=2,\\\
\alpha_{1}=\alpha_{2}=2.\end{array}\\\
G=\left(\begin{array}[]{cccc}\varepsilon_{1}&0&0&0\\\ 0&\varepsilon_{2}&0&0\\\
0&0&\varepsilon_{3}&0\\\
0&0&0&\varepsilon_{4}\end{array}\right)&G=\left(\begin{array}[]{cccc}0&\varepsilon_{1}&0&0\\\
\varepsilon_{1}&0&0&0\\\ 0&0&\varepsilon_{2}&0\\\
0&0&0&\varepsilon_{3}\end{array}\right)&G=\left(\begin{array}[]{cccc}0&\varepsilon_{1}&0&0\\\
\varepsilon_{1}&0&0&0\\\ 0&0&0&\varepsilon_{2}\\\
0&0&\varepsilon_{2}&0\end{array}\right)\\\
\mathfrak{B}=\\{u_{1_{1}},u_{2_{1}},u_{3_{1}},u_{4_{1}}\\}&\mathfrak{B}=\\{u_{1_{1}},u_{1_{2}},u_{2_{1}},u_{3_{1}}\\}&\mathfrak{B}=\\{u_{1_{1}},u_{1_{2}},u_{2_{1}},u_{2_{2}}\\}\\\
A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\ 0&\lambda_{2}&0&0\\\
0&0&\lambda_{3}&0\\\
0&0&0&\lambda_{4}\end{array}\right)&A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\
1&\lambda_{1}&0&0\\\ 0&0&\lambda_{2}&0\\\
0&0&0&\lambda_{3}\end{array}\right)&A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\
1&\lambda_{1}&0&0\\\ 0&0&\lambda_{2}&0\\\
0&0&1&\lambda_{2}\end{array}\right)\\\ \hline\cr\textbf{Form
(I\\!V)}&\textbf{Form (V)}&\textbf{Form (V\\!I)}\\\
\hline\cr\begin{array}[]{c}t=m=2,\\\
\alpha_{1}=3,\alpha_{2}=1.\end{array}&\begin{array}[]{c}m=3,t=2,\\\
\alpha_{1}=\alpha_{2}=1,\beta_{3}=1.\end{array}&\begin{array}[]{c}m=2,t=1,\\\
\alpha_{1}=2,\beta_{2}=1.\end{array}\\\
G=\left(\begin{array}[]{cccc}0&0&\varepsilon_{1}&0\\\ 0&\varepsilon_{1}&0&0\\\
\varepsilon_{1}&0&0&0\\\
0&0&0&\varepsilon_{2}\end{array}\right)&G=\left(\begin{array}[]{cccc}\varepsilon_{1}&0&0&0\\\
0&\varepsilon_{2}&0&0\\\ 0&0&1&0\\\
0&0&0&-1\end{array}\right)&G=\left(\begin{array}[]{cccc}0&\varepsilon_{1}&0&0\\\
\varepsilon_{1}&0&0&0\\\ 0&0&1&0\\\ 0&0&0&-1\end{array}\right)\\\
\mathfrak{B}=\\{u_{1_{1}},u_{1_{2}},u_{1_{3}},u_{2_{1}}\\}&\mathfrak{B}=\\{u_{1_{1}},u_{2_{1}},u_{\bar{3}_{1}},u_{\tilde{3}_{1}}\\}&\mathfrak{B}=\\{u_{1_{1}},u_{1_{2}},u_{\bar{2}_{1}},u_{\tilde{2}_{1}}\\}\\\
A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\ 1&\lambda_{1}&0&0\\\
0&1&\lambda_{1}&0\\\
0&0&0&\lambda_{2}\end{array}\right)&A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\
0&\lambda_{2}&0&0\\\ 0&0&\gamma_{3}&\tau_{3}\\\
0&0&-\tau_{3}&\gamma_{3}\end{array}\right)&A=\left(\begin{array}[]{cccc}\lambda_{1}&0&0&0\\\
1&\lambda_{1}&0&0\\\ 0&0&\gamma_{2}&\tau_{2}\\\
0&0&-\tau_{2}&\gamma_{2}\end{array}\right)\\\ \hline\cr\end{array}.$

And if the principal curvatures of $M^{4}_{r}$ are all imaginary, then $A$ and
$G$ have possible forms:

$\begin{array}[]{|c|c|}\hline\cr\textbf{Form (V\\!I\\!I)}&\textbf{Form
(V\\!I\\!I\\!I)}\\!\\\ \hline\cr\begin{array}[]{c}m=1,t=0,\\\
\beta_{1}=2.\end{array}&\begin{array}[]{c}m=2,t=0\\!\\\
\beta_{1}=\beta_{2}=1.\end{array}\\!\\\
G=\\!\left[\\!\begin{array}[]{cccc}\\!0\\!&\\!0\\!&\\!1\\!&\\!0\\!\\\
\\!0\\!&\\!0\\!&\\!0\\!&\\!-1\\!\\\ \\!1\\!&\\!0\\!&\\!0\\!&\\!0\\!\\\
\\!0\\!&\\!-1\\!&\\!0\\!&\\!0\\!\\\
\end{array}\\!\right]\\!&\\!G=\\!\left[\\!\begin{array}[]{cccc}\\!1\\!&\\!0\\!&\\!0\\!&\\!0\\!\\\
\\!0\\!&\\!-1\\!&\\!0\\!&\\!0\\!\\\ \\!0\\!&\\!0\\!&\\!1\\!&\\!0\\!\\\
\\!0\\!&\\!0\\!&\\!0\\!&\\!-1\\!\\\ \end{array}\\!\right]\\!\\\
\mathfrak{B}\\!=\\!\\{u_{\bar{1}_{1}},u_{\tilde{1}_{1}},u_{\bar{1}_{2}},u_{\tilde{1}_{2}}\\}\\!&\\!\mathfrak{B}\\!=\\!\\{u_{\bar{1}},u_{\tilde{1}},u_{\bar{2}},u_{\tilde{2}}\\}\\!\\\
A\\!=\\!\left[\\!\begin{array}[]{cccc}\\!\gamma_{1}\\!&\\!\tau_{1}\\!&\\!0\\!&\\!0\\!\\\
\\!-\tau_{1}\\!&\\!\gamma_{1}\\!&\\!0\\!&\\!0\\!\\\
\\!1\\!&\\!0\\!&\\!\gamma_{1}\\!&\\!\tau_{1}\\!\\\
\\!0\\!&\\!1\\!&\\!-\tau_{1}\\!&\\!\gamma_{1}\\!\\\
\end{array}\\!\right]\\!&\\!A\\!=\\!\left[\\!\begin{array}[]{cccc}\\!\gamma_{1}\\!&\\!\tau_{1}\\!&\\!0\\!&\\!0\\!\\\
\\!-\tau_{1}\\!&\\!\gamma_{1}\\!&\\!0\\!&\\!0\\!\\\
\\!0\\!&\\!0\\!&\\!\gamma_{2}\\!&\\!\tau_{2}\\!\\\
\\!0\\!&\\!0\\!&\\!-\tau_{2}\\!&\\!\gamma_{2}\\!\\\
\end{array}\\!\right]\\!\\\ \hline\cr\end{array}.$

From these forms of $A$, we can calculate $\text{tr}A$ and $\text{tr}A^{2}$,
combining $\text{tr}A=4\varepsilon H$, we find

$\begin{cases}\text{tr}A=\sum_{i=1}^{t}\alpha_{i}\lambda_{i}+2\sum_{j=t+1}^{m}\beta_{j}\gamma_{j}=4\varepsilon
H,\\\
\text{tr}A^{2}=\sum_{i=1}^{t}\alpha_{i}\lambda_{i}^{2}+2\sum_{j=t+1}^{m}\beta_{j}(\gamma_{j}^{2}-\tau_{j}^{2}).\end{cases}$
(2.4)

##  3 The result that $M^{4}_{r}$ has constant mean curvature

Theorem 3.1  _Let $N^{5}_{s}(c)$ be a $5$-dimensional pseudo-Riemannian space
form with constant sectional curvature $c$, and $M^{4}_{r}$ be a nondegenerate
hypersurface of $N^{5}_{s}(c)$ with proper mean curvature vector field, then
$M^{4}_{r}$ has constant mean curvature._

To prove Theorem 3.1, we adopt contradiction method. Assume that $H$ is not a
constant, we need to deduce a contradiction. When the principal curvatures of
$M^{4}_{r}$ are all imaginary, (2.2) implies $-2\varepsilon H$ (real) is an
eigenvalue of $A$, a contradiction. When $M^{4}_{r}$ has real principal
curvatures, there are six possible forms of $A$ (see section 2). These
different forms of $A$ will generate different expressions about connection
coefficients.To derive contradictions, we need to use these expressions. So,
we cannot deal with these six forms uniformly.

###  3.1 The shape operator has the form (I)

Proposition 3.2  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (I), then $M^{4}_{r}$ has constant
mean curvature._

When $M^{4}_{r}$ has at most three distinct principal curvatures, the proof
has been provided by L. Du etc. in [7]. So, we suppose the principal
curvatures $\lambda_{1},\lambda_{2},\lambda_{3}$ and $\lambda_{4}$ of
$M^{4}_{r}$ are distinct with each other. At the beginning, we give some
equations and Lemmas, under the assumption that the mean curvature $H$ of
$M^{n}_{r}$ is not a constant, and $\nabla H$ is in the direction $u_{1}$.

####  3.1.1 Some equations and Lemmas

Denote $u_{i}=u_{i_{1}}$, with $1\leq i\leq 4$. Let
$\nabla_{u_{i}}u_{j}=\Gamma^{k}_{ij}u_{k}$, $i,j=1,2,3,4$. we get from
compatibility of the connection $\nabla$ that

$\Gamma_{ki}^{j}=-\varepsilon_{i}\varepsilon_{j}\Gamma_{kj}^{i},$ (3.1)

with $i,j,k=1,2,3,4$. Assume that $H$ is not a constant, and $\nabla H$ is in
the direction of $u_{1}$, then (2.2) implies $\lambda_{1}=-2\varepsilon H$,
and we have

$u_{1}(H)\neq 0,\ u_{2}(H)=u_{3}(H)=u_{4}(H)=0.$ (3.2)

As the equation (3.2), we obtain from
$(\nabla_{u_{i}}u_{j}-\nabla_{u_{j}}u_{i})(H)=[u_{i},u_{j}](H)$ that

$\Gamma^{1}_{ij}=\Gamma^{1}_{ji},\quad i,j=2,3,4.$ (3.3)

Combining the equations (3.1), (3.2) and (3.3), we deduce from Codazzi
equation that

$\Gamma_{1i}^{1}=0,\ \ \Gamma_{ij}^{1}=0,\ \text{with}\ i\neq j,\ i,j=2,3,4,$
(3.4)

and

$\begin{cases}u_{i}(\lambda_{j})=(\lambda_{i}-\lambda_{j})\Gamma_{ji}^{j},\
i\neq j,\\\
(\lambda_{i}-\lambda_{j})\Gamma_{ki}^{j}=(\lambda_{k}-\lambda_{j})\Gamma_{ik}^{j},\
i,k\neq j.\end{cases}$ (3.5)

Using Gauss equation for $\langle R(u_{1},u_{i})u_{1},u_{i}\rangle$,
$i=2,3,4$, combining (3.1), (3.3) and (3.4), we have

$\displaystyle
u_{1}(\Gamma_{i1}^{i})=-(\Gamma_{i1}^{i})^{2}+(2H\lambda_{i}-c)\varepsilon_{1},\
i=2,3,4.$ (3.6)

Considering the expressions (3.1) and (3.2), the equation (2.3) can be written
as

$u_{1}u_{1}(H)+\sum_{i=2}^{4}\Gamma_{i1}^{i}u_{1}(H)-\varepsilon\varepsilon_{1}H\text{tr}A^{2}+\varepsilon_{1}\lambda
H=0.$ (3.7)

In the following, we discuss $f_{k}:=\sum_{i=2}^{4}(\Gamma_{i1}^{i})^{k}$,
with $k=1,2,\cdots,5$, and get the following Lemmas 3.3 and 3.4. And then,
applying these two Lemmas, we prove
$u_{i}(\Gamma_{j1}^{j})=u_{i}(\lambda_{j})=0$, with $i,j=2,3,4$, i.e. Lemma
3.5, which will play an important role in the proof of Proposition 3.2.

Lemma 3.3  _We have_

$\displaystyle f_{2}=$ $\displaystyle-
u_{1}(f_{1})+12H^{2}\varepsilon\varepsilon_{1}-3c\varepsilon_{1};$ (3.8)
$\displaystyle f_{3}=$
$\displaystyle\frac{1}{2}u_{1}^{(2)}(f_{1})-(4\varepsilon
H^{2}+c)\varepsilon_{1}f_{1}-24\varepsilon\varepsilon_{1}Hu_{1}(H);$
$\displaystyle f_{4}=$
$\displaystyle-\frac{1}{6}u_{1}^{(3)}(f_{1})+\frac{4}{3}(4\varepsilon
H^{2}+c)\varepsilon_{1}u_{1}(f_{1})+\frac{20}{3}\varepsilon\varepsilon_{1}Hu_{1}(H)f_{1}+8\varepsilon\varepsilon_{1}(u_{1}(H))^{2}$
$\displaystyle+16\varepsilon\varepsilon_{1}Hu_{1}^{(2)}(H)-32H^{4}+2\varepsilon
H^{2}\lambda-12\varepsilon H^{2}c+3c^{2};$ $\displaystyle f_{5}=$
$\displaystyle\frac{1}{24}u_{1}^{(4)}(f_{1})-\frac{5}{6}(4\varepsilon
H^{2}+c)\varepsilon_{1}u_{1}^{(2)}(f_{1})-\frac{25}{3}\varepsilon\varepsilon_{1}Hu_{1}(H)u_{1}(f_{1})-[16H^{4}$
$\displaystyle-\frac{13}{3}Hu_{1}^{(2)}(H)-\frac{1}{3}\varepsilon\varepsilon_{1}(u_{1}(H))^{2}+8\varepsilon
H^{2}c+c^{2}]f_{1}-8\varepsilon\varepsilon_{1}Hu_{1}^{(3)}(H)$
$\displaystyle-\frac{20}{3}\varepsilon\varepsilon_{1}u_{1}(H)u_{1}^{(2)}(H)+[16(8+\frac{2}{3}\varepsilon\varepsilon_{1})H^{3}+38\varepsilon
Hc-\frac{5}{3}H\varepsilon\lambda]u_{1}(H),$

_where $u_{1}^{(k)}(f_{1})$ and $u_{1}^{(k)}(H)$ with $k=2,3,4$ denote the
$k$-order derivatives of $f_{1}$ and $H$ along $u_{1}$, respectively. _

Proof Since $\sum_{i=1}^{4}\lambda_{i}=\text{tr}A=4\varepsilon H$ and
$\lambda_{1}=-2\varepsilon H$, we know

$\sum_{i=2}^{4}\lambda_{i}=6\varepsilon H.$

Taking sum for $j$ from $2$ to $4$ in the first equation of (3.5) with $i=1$,
we find

$\displaystyle\sum_{i=2}^{4}\lambda_{i}\Gamma_{i1}^{i}=-2\varepsilon
Hf_{1}-6\varepsilon u_{1}(H).$ (3.9)

Multiply $\lambda_{j}$ on both sides of the first equation in (3.5) with
$i=1$, and then we know

$\displaystyle\sum_{i=2}^{4}\lambda_{i}^{2}\Gamma_{i1}^{i}=-2\varepsilon
H\sum_{i=2}^{4}\lambda_{i}\Gamma_{i1}^{i}-\frac{1}{2}u_{1}(\text{tr}A^{2})+4Hu_{1}(H).$

Differentiate $\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k}$ along $u_{1}$,
combining (3.5) and (3.6), we get

$\displaystyle(k+1)\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k+1}=$
$\displaystyle-
u_{1}(\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k})+2k\varepsilon_{1}H\sum_{i=2}^{4}\lambda_{i}^{2}(\Gamma_{i1}^{i})^{k-1}$
$\displaystyle-
kc\varepsilon_{1}\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k-1}-2\varepsilon
Hf_{k+1},\ \ k=1,2,$

which together with (3.9) and the above equation gives the expressions of
$\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k}$, with $k=2,3$. Take the sum
of $i$ from 2 to 4 in (3.6), we can express $f_{2}$ as

$f_{2}=-u_{1}(f_{1})+12H^{2}\varepsilon\varepsilon_{1}-3c\varepsilon_{1}.$

Multiplying $(\Gamma_{i1}^{i})^{k}$ on both sides of (3.6), and then taking
sum for $i$, we obtain

$\displaystyle
f_{k+2}=-\frac{1}{k+1}u_{1}(f_{k+1})+2\varepsilon_{1}H\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k}-c\varepsilon_{1}f_{k},\
k=1,2,3.$

As the expressions of $\sum_{i=2}^{4}\lambda_{i}(\Gamma_{i1}^{i})^{k}$, with
$k=1,2,3$, we conclude from the above equation that (3.8) holds.
$\hfill\square$

Lemma 3.4  _For $i=2,3,4$, we have_

$u_{i}(f_{1})=0.$

Proof With the notions $f_{k}$, $k=1,2,\cdots,5$, we find the equations

$\begin{cases}f_{1}^{4}-6f_{1}^{2}f_{2}+3f_{2}^{2}+8f_{1}f_{3}-6f_{4}=0,\\\
f_{1}^{5}-5f_{1}^{3}f_{2}+5f_{1}^{2}f_{3}+5f_{2}f_{3}-6f_{5}=0\end{cases}$
(3.10)

(cf. [11]) also hold. Substitute the expressions of $f_{k}$, $k=1,2,\cdots,5$
in Lemma 3.3 into (3.10), we have

$\displaystyle
F_{1}:=u_{1}^{(3)}(f_{1})+4f_{1}u_{1}^{(2)}(f_{1})+3(u_{1}(f_{1}))^{2}+[6f_{1}^{2}-104\varepsilon\varepsilon_{1}H^{2}+10\varepsilon_{1}c]u_{1}(f_{1})$
$\displaystyle+f_{1}^{4}-104\varepsilon\varepsilon_{1}H^{2}f_{1}^{2}+10\varepsilon_{1}cf_{1}^{2}-232\varepsilon\varepsilon_{1}Hu_{1}(H)f_{1}-48\varepsilon\varepsilon_{1}(u_{1}(H))^{2}$
$\displaystyle-96\varepsilon\varepsilon_{1}Hu_{1}^{(2)}(H)+624H^{4}-144\varepsilon
H^{2}c-12\varepsilon H^{2}\lambda+9c^{2}=0,$

and

$\displaystyle F_{2}:=$ $\displaystyle-
u_{1}^{(4)}(f_{1})+(-10u_{1}(f_{1})+10f_{1}^{2}+200\varepsilon\varepsilon_{1}H^{2}-10c\varepsilon_{1})u_{1}^{(2)}(f_{1})+4f_{1}^{5}$
$\displaystyle+[20f_{1}^{3}+20(4\varepsilon
H^{2}+c)\varepsilon_{1}f_{1}+680\varepsilon\varepsilon_{1}Hu_{1}(H)]u_{1}(f_{1})-320\varepsilon\varepsilon_{1}H^{2}f_{1}^{3}$
$\displaystyle+40\varepsilon_{1}f_{1}^{3}c-480\varepsilon\varepsilon_{1}Hu_{1}(H)f_{1}^{2}+104Hu_{1}^{(2)}(H)f_{1}+192\varepsilon\varepsilon_{1}Hu_{1}^{(3)}(H)$
$\displaystyle-192\varepsilon
H^{2}cf_{1}-1344H^{4}f_{1}+8\varepsilon\varepsilon_{1}(u_{1}(H))^{2}f_{1}+160\varepsilon\varepsilon_{1}u_{1}(H)u_{1}^{(2)}(H)$
$\displaystyle+36c^{2}f_{1}-128(69+2\varepsilon\varepsilon_{1})H^{3}u_{1}(H)+(528c+40\lambda)\varepsilon
Hu_{1}(H)=0.$

Subsequently, we will eliminate $u_{1}^{(k)}(f_{1})$, $k=1,2,3,4$ from the
equations $F_{1}=0$ and $F_{2}=0$. Let
$F_{3}=\frac{1}{4}(u_{1}(F_{1})+F_{2})-f_{1}F_{1}$, then $F_{3}=0$, i.e.

$\displaystyle
12\varepsilon\varepsilon_{1}H^{2}u_{1}^{(2)}(f_{1})+[36\varepsilon\varepsilon_{1}H^{2}f_{1}+30\varepsilon\varepsilon_{1}Hu_{1}(H)]u_{1}(f_{1})+12\varepsilon\varepsilon_{1}H^{2}f_{1}^{3}$
(3.11)
$\displaystyle+30\varepsilon\varepsilon_{1}Hu_{1}(H)f_{1}^{2}+(19\varepsilon\varepsilon_{1}+13)Hu_{1}^{(2)}(H)f_{1}-4\varepsilon\varepsilon_{1}(u_{1}(H))^{2}f_{1}$
$\displaystyle+6(8c+\lambda)\varepsilon
H^{2}f_{1}+12\varepsilon\varepsilon_{1}Hu_{1}^{(3)}(H)-4\varepsilon\varepsilon_{1}u_{1}(H)u_{1}^{(2)}(H)$
$\displaystyle-480H^{4}f_{1}-(792+32\varepsilon\varepsilon_{1})H^{3}u_{1}(H)+(30c+2\lambda)\varepsilon
Hu_{1}(H)=0.$

Take
$F_{4}=2H(u_{1}(F_{3})-12\varepsilon\varepsilon_{1}H^{2}F_{1}+f_{1}F_{3})-9u_{1}(H)F_{3}$,
then $F_{4}=0$, i.e.

$\displaystyle a_{1}u_{1}(f_{1})+a_{1}f_{1}^{2}+a_{2}f_{1}+a_{3}=0,$ (3.12)

where

$\displaystyle a_{1}=$
$\displaystyle(294\varepsilon\varepsilon_{1}+78)Hu_{1}^{(2)}(H)-654\varepsilon\varepsilon_{1}(u_{1}(H))^{2}+4608H^{4}+36(\lambda-12c)\varepsilon
H^{2},$ $\displaystyle a_{2}=$
$\displaystyle(186\varepsilon\varepsilon_{1}+78)H^{2}u_{1}^{(3)}(H)-(471\varepsilon\varepsilon_{1}+273)Hu_{1}(H)u_{1}^{(2)}(H)$
$\displaystyle+108\varepsilon\varepsilon_{1}(u_{1}(H))^{3}+[(13392-192\varepsilon\varepsilon_{1})H^{2}-6(90c+13\lambda)\varepsilon]H^{2}u_{1}(H),$
$\displaystyle a_{3}=$ $\displaystyle
72\varepsilon\varepsilon_{1}H^{2}u_{1}^{(4)}(H)-276\varepsilon\varepsilon_{1}Hu_{1}(H)u_{1}^{(3)}(H)+108\varepsilon\varepsilon_{1}(u_{1}(H))^{2}u_{1}^{(2)}(H)$
$\displaystyle-24\varepsilon\varepsilon_{1}H(u_{1}^{(2)}(H))^{2}+[(2160-192\varepsilon\varepsilon_{1})H^{2}+(180c+12\lambda)\varepsilon]H^{2}u_{1}^{(2)}(H)$
$\displaystyle+[72(147+4\varepsilon\varepsilon_{1})H^{2}-(630c+42\lambda)\varepsilon]H(u_{1}(H))^{2}-44928\varepsilon\varepsilon_{1}H^{7}$
$\displaystyle+(10368c+864\lambda)\varepsilon_{1}H^{5}-648\varepsilon\varepsilon_{1}H^{3}c^{2}.$

By acting on (3.12) with $u_{1}$, and then combining (3.11), we have

$(b_{1}f_{1}+b_{2})u_{1}(f_{1})+b_{1}f_{1}^{3}+b_{3}f_{1}^{2}+b_{4}f_{1}+b_{5}=0,$

where

$\displaystyle b_{1}=$
$\displaystyle-768H^{4}\varepsilon\varepsilon_{1}+6(12c-6\lambda)H^{2}\varepsilon_{1}-(13\varepsilon\varepsilon_{1}+49)Hu_{1}^{(2)}(H)+109(u_{1}(H))^{2},$
$\displaystyle b_{2}=$
$\displaystyle(8304\varepsilon\varepsilon_{1}-64)H^{4}u_{1}(H)+(52\varepsilon\varepsilon_{1}+160)u_{1}^{(3)}(H)H^{2}+363(u_{1}(H))^{3}$
$\displaystyle-(104\varepsilon\varepsilon_{1}+642)Hu_{1}(H)u_{1}^{(2)}(H)-(252c+20\lambda)H^{2}u_{1}(H)\varepsilon_{1},$
$\displaystyle b_{3}=$ $\displaystyle
3840H^{4}u_{1}(H)\varepsilon\varepsilon_{1}+(26\varepsilon\varepsilon_{1}+98)u_{1}^{(3)}(H)H^{2}+6\varepsilon_{1}(\lambda-12c)H^{2}u_{1}(H)$
$\displaystyle+327(u_{1}(H))^{3}-(13\varepsilon\varepsilon_{1}+485)Hu_{1}(H)u_{1}^{(2)}(H),$
$\displaystyle b_{4}=$ $\displaystyle
12(13\varepsilon\varepsilon_{1}+31)u_{1}^{(4)}(H)H^{3}-2(715\varepsilon\varepsilon_{1}+1021)H^{2}(u_{1}^{(2)}(H))^{2}+368640H^{8}$
$\displaystyle+96(372\varepsilon\varepsilon_{1}-43)H^{5}u_{1}^{(2)}(H)+(923\varepsilon\varepsilon_{1}+1973)H(u_{1}(H))^{2}u_{1}^{(2)}(H)$
$\displaystyle-18(13\varepsilon\varepsilon_{1}+11)u_{1}^{(3)}(H)H^{2}u_{1}(H)+6(464c+61\lambda)\varepsilon_{1}H^{2}(u_{1}(H))^{2}$
$\displaystyle+36(96c^{2}-\lambda^{2}+4c\lambda)H^{4}+96(603\varepsilon\varepsilon_{1}-16)H^{4}(u_{1}(H))^{2}-436(u_{1}(H))^{4}$
$\displaystyle-576\varepsilon(124c+3\lambda)H^{6}+12(26c\varepsilon-172c\varepsilon_{1}-13\lambda\varepsilon-47\lambda\varepsilon_{1})H^{3}u_{1}^{(2)}(H),$
$\displaystyle b_{5}=$
$\displaystyle-12(144c^{2}+\lambda^{2}+3c\lambda)H^{3}u_{1}(H)-12(13\varepsilon\varepsilon_{1}+103)u_{1}^{(3)}(H)H^{2}u_{1}^{(2)}(H)$
$\displaystyle+144u_{1}^{(5)}(H)H^{3}-2(195c\varepsilon+13\lambda\varepsilon+1779c\varepsilon_{1}+97\lambda\varepsilon_{1})H^{2}u_{1}^{(2)}(H)\varepsilon
u_{1}(H)$
$\displaystyle-8(2853\varepsilon\varepsilon_{1}+220)H^{3}(u_{1}(H))^{3}+8(12739\varepsilon\varepsilon_{1}+1435)H^{4}u_{1}^{(2)}(H)u_{1}(H)$
$\displaystyle-96(51\varepsilon\varepsilon_{1}+4)u_{1}^{(3)}(H)H^{5}+972u_{1}^{(3)}(H)H(u_{1}(H))^{2}-436(u_{1}(H))^{3}u_{1}^{(2)}(H)$
$\displaystyle+16(1476\varepsilon
c-144c\varepsilon_{1}+741\varepsilon\lambda+12\lambda\varepsilon_{1})H^{5}u_{1}(H)-264u_{1}^{(4)}(H)H^{2}u_{1}(H)$
$\displaystyle+\varepsilon_{1}(2010c+134\lambda)H(u_{1}(H))^{3}+(24576\varepsilon\varepsilon_{1}-20736)H^{7}u_{1}(H)$
$\displaystyle+(52\varepsilon\varepsilon_{1}+580)Hu_{1}(H)(u_{1}^{(2)}(H))^{2}+24\varepsilon_{1}(51c-2\lambda)u_{1}^{(3)}(H)H^{3}.$

which together with (3.12) deduce that

$c_{1}f_{1}+c_{2}=0,$

where $c_{1}$ and $c_{2}$ are polynomials about $H$ and $u_{1}^{(k)}(H)$,
$k=1,2,\cdots,5$.

From the expression $[u_{i},u_{1}]=\nabla_{u_{i}}u_{1}-\nabla_{u_{1}}u_{i}$,
as well as the relations (3.2) and $\Gamma_{i1}^{1}=\Gamma_{1i}^{1}=0$ (see
eq. (3.1) and (3.4)), we have

$\displaystyle u_{i}u_{1}^{(k)}(H)=0,\ k=0,1,2,\cdots,5.$ (3.13)

Differentiate $c_{1}f_{1}+c_{2}=0$ along $u_{i}$, with $i=2,3,4$, using
(3.13), we have

$c_{1}u_{i}(f_{1})=0.$

Suppose $u_{i}(f_{1})\neq 0$, for some $i\in\\{2,3,4\\}$, we can conclude that
$c_{1}=c_{2}=0$. And then, we can eliminate $u_{1}^{(k)}(H)$, with
$k=1,2,\cdots,5$ and obtain a polynomial equation about $H$, which implies $H$
is a constant, a contradiction. So, $u_{i}(f_{1})=0$, $i=2,3,4$.
$\hfill\square$

Lemma 3.5  _We have $u_{i}(\Gamma_{j1}^{j})=u_{i}(\lambda_{j})=0$, with
$i,j=2,3,4$._

Proof Since $u_{i}(f_{1})=0$, with $i=2,3,4$ (see Lemma 3.4), it follows

$\displaystyle u_{i}u_{1}^{(k)}(f_{1})=0,\ i=2,3,4,\ k=0,1,2,3.$ (3.14)

Differentiate the expressions of $f_{k}$, $k=2,3$ in Lemma 3.3, combining
(3.13) and (3.14), we find

$\displaystyle u_{i}(f_{k})=0,\ i=2,3,4,\ k=1,2,3.$

Note that $f_{k}=\sum_{i=2}^{4}(\Gamma_{i1}^{i})^{k}$, the above equation
tells us that

$\begin{cases}u_{i}(\Gamma_{21}^{2})+u_{i}(\Gamma_{31}^{3})+u_{i}(\Gamma_{41}^{4})=0,\\\
\Gamma_{21}^{2}u_{i}(\Gamma_{21}^{2})+\Gamma_{31}^{3}u_{i}(\Gamma_{31}^{3})+\Gamma_{41}^{4}u_{i}(\Gamma_{41}^{4})=0,\\\
(\Gamma_{21}^{2})^{2}u_{i}(\Gamma_{21}^{2})+(\Gamma_{31}^{3})^{2}u_{i}(\Gamma_{31}^{3})+(\Gamma_{41}^{4})^{2}u_{i}(\Gamma_{41}^{4})=0,\end{cases}$

with $i=2,3,4$. Observe (3.6), we know $\Gamma_{21}^{2}$, $\Gamma_{31}^{3}$
and $\Gamma_{41}^{4}$ are distinct. So, the coefficient determinant of the
above system

$\displaystyle\left|\begin{array}[]{ccc}1&1&1\\\
\Gamma_{21}^{2}&\Gamma_{31}^{3}&\Gamma_{41}^{4}\\\
(\Gamma_{21}^{2})^{2}&(\Gamma_{31}^{3})^{2}&(\Gamma_{41}^{4})^{2}\end{array}\right|=(\Gamma_{41}^{4}-\Gamma_{31}^{3})(\Gamma_{31}^{3}-\Gamma_{21}^{2})(\Gamma_{41}^{4}-\Gamma_{21}^{2})\neq
0.$

So, the above system has only zero solution, i.e. $u_{i}(\Gamma_{j1}^{j})=0$,
$i,j=2,3,4$. And then, we have

$u_{i}u_{1}(\Gamma_{j1}^{j})=0,\ i,j=2,3,4.$

Differentiate (3.6) along $u_{j}$, combining $u_{j}(\Gamma_{i1}^{i})=0$ and
$u_{j}u_{1}(\Gamma_{i1}^{i})=0$, we get $u_{j}(\lambda_{i})=0$, $i,j=2,3,4$.
$\hfill\square$

####  3.1.2 The proof of Proposition 3.2

Assume that $H$ is not a constant, then there exists a neighbourhood $U_{p}$
of $p$ such that $H\neq 0$ and $\nabla H\neq 0$. The equation (2.2) implies
$\nabla H$ is an eigenvector of $A$, with corresponding eigenvalue
$-2\varepsilon H$. Without loss of generality, we suppose $\nabla H$ is in the
direction of $u_{1}$ and $\lambda_{1}=-2\varepsilon H$, then the equations and
Lemmas in subsection 3.1.1 hold. We will use these equations and Lemma 3.5 to
deduce contradictions.

The second equation in (3.5) yields

$\displaystyle\varepsilon_{4}(\lambda_{2}-\lambda_{4})\Gamma_{32}^{4}=\varepsilon_{4}(\lambda_{3}-\lambda_{4})\Gamma_{23}^{4}=\varepsilon_{3}(\lambda_{4}-\lambda_{3})\Gamma_{24}^{3}$
(3.15) $\displaystyle=$
$\displaystyle\varepsilon_{3}(\lambda_{2}-\lambda_{3})\Gamma_{42}^{3}=\varepsilon_{2}(\lambda_{3}-\lambda_{2})\Gamma_{43}^{2}=\varepsilon_{2}(\lambda_{4}-\lambda_{2})\Gamma_{34}^{2}.$

Together (3.1) with (3.15), we have

$(\lambda_{2}-\lambda_{3})(\lambda_{3}-\lambda_{4})(\lambda_{4}-\lambda_{2})(\Gamma_{23}^{4}\Gamma_{32}^{4}+\Gamma_{24}^{3}\Gamma_{42}^{3}+\Gamma_{34}^{2}\Gamma_{43}^{2})=0,$

i.e.

$\displaystyle\Gamma_{23}^{4}\Gamma_{32}^{4}+\Gamma_{24}^{3}\Gamma_{42}^{3}+\Gamma_{34}^{2}\Gamma_{43}^{2}=0.$
(3.16)

It follows from the equation (3.5) and Lemma 3.5 that

$\Gamma_{ij}^{i}=0,\ i,j=2,3,4.$ (3.17)

Apply Gauss equation for $\langle R(u_{i},u_{j})u_{k},u_{1}\rangle$ and
$\langle R(u_{i},u_{j})u_{i},u_{j}\rangle$, with $i,j,k$ are distinct and
$i,j,k=2,3,4$, combining (3.1), (3.4) and (3.17), we have

$\displaystyle\varepsilon_{4}(\Gamma_{21}^{2}-\Gamma_{41}^{4})\Gamma_{32}^{4}=\varepsilon_{4}(\Gamma_{31}^{3}-\Gamma_{41}^{4})\Gamma_{23}^{4}=\varepsilon_{3}(\Gamma_{41}^{4}-\Gamma_{31}^{3})\Gamma_{24}^{3}$
(3.18) $\displaystyle=$
$\displaystyle\varepsilon_{3}(\Gamma_{21}^{2}-\Gamma_{31}^{3})\Gamma_{42}^{3}=\varepsilon_{2}(\Gamma_{31}^{3}-\Gamma_{21}^{2})\Gamma_{43}^{2}=\varepsilon_{2}(\Gamma_{41}^{4}-\Gamma_{21}^{2})\Gamma_{34}^{2},$

and

$\begin{cases}\varepsilon_{1}\varepsilon_{2}\varepsilon_{3}\Gamma_{21}^{2}\Gamma_{31}^{3}-2\varepsilon_{4}\Gamma_{23}^{4}\Gamma_{32}^{4}=-c\varepsilon_{2}\varepsilon_{3}-\lambda_{2}\lambda_{3}\varepsilon\varepsilon_{2}\varepsilon_{3},\\\
\varepsilon_{1}\varepsilon_{2}\varepsilon_{4}\Gamma_{21}^{2}\Gamma_{41}^{4}-2\varepsilon_{3}\Gamma_{24}^{3}\Gamma_{42}^{3}=-c\varepsilon_{2}\varepsilon_{4}-\lambda_{2}\lambda_{4}\varepsilon\varepsilon_{2}\varepsilon_{4},\\\
\varepsilon_{1}\varepsilon_{3}\varepsilon_{4}\Gamma_{31}^{3}\Gamma_{41}^{4}-2\varepsilon_{2}\Gamma_{34}^{2}\Gamma_{43}^{2}=-c\varepsilon_{3}\varepsilon_{4}-\lambda_{3}\lambda_{4}\varepsilon\varepsilon_{3}\varepsilon_{4},\end{cases}$
(3.19)

which together with (3.16) implies that

$\varepsilon_{1}(\Gamma_{21}^{2}\Gamma_{31}^{3}+\Gamma_{21}^{2}\Gamma_{41}^{4}+\Gamma_{31}^{3}\Gamma_{41}^{4})+\varepsilon(\lambda_{2}\lambda_{3}+\lambda_{3}\lambda_{4}+\lambda_{2}\lambda_{4})+3c=0.$
(3.20)

In the following, we treat the cases $\Gamma_{23}^{4}\neq 0$ at some point and
$\Gamma_{23}^{4}=0$ at any point in $U_{p}$, respectively.

_Case 1: $\Gamma_{23}^{4}\neq 0$ at some point in $U_{p}$._

Suppose $\Gamma_{23}^{4}\neq 0$ at $q\in U_{p}$, then there exists a
neighbourhood $U_{q}\subset U_{p}$, such that $\Gamma_{23}^{4}\neq 0$ on
$U_{q}$. We work on $U_{q}$ in the following discussion. The equations (3.15)
and (3.18) give

$\frac{\Gamma_{41}^{4}-\Gamma_{31}^{3}}{\lambda_{4}-\lambda_{3}}=\frac{\Gamma_{41}^{4}-\Gamma_{21}^{2}}{\lambda_{4}-\lambda_{2}}=\frac{\Gamma_{31}^{3}-\Gamma_{21}^{2}}{\lambda_{3}-\lambda_{2}}.$

As $u_{i}(\Gamma_{j1}^{j})=u_{i}(\lambda_{j})=0$, $i,j=2,3,4$, we conclude
from the above equation that there exists two smooth functions $\mu$ and
$\nu$, with $u_{i}(\mu)=u_{i}(\nu)=0$, $i=2,3,4$, such that

$\Gamma_{i1}^{i}=\mu\lambda_{i}+\nu,\ i=2,3,4.$ (3.21)

Take sum for $i$ in (3.21), then

$f_{1}=6\varepsilon H\mu+3\nu.$ (3.22)

Differentiate (3.21) along $u_{1}$, using (3.5) and (3.6), we get for
$i=2,3,4$,

$(u_{1}(\mu)-2\varepsilon
H\mu^{2}+\mu\nu-2H\varepsilon_{1})\lambda_{i}+u_{1}(\nu)+\nu^{2}-2\varepsilon
H\mu\nu+c\varepsilon_{1}=0,$

which yields

$\begin{cases}u_{1}(\mu)=2\varepsilon H\mu^{2}-\mu\nu+2H\varepsilon_{1},\\\
u_{1}(\nu)=-\nu^{2}+2\varepsilon H\mu\nu-c\varepsilon_{1}.\end{cases}$ (3.23)

Substitute (3.21) into (3.20), we obtain

$(\varepsilon_{1}\mu^{2}+\varepsilon)(\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{4})+2\varepsilon_{1}\mu\nu(\lambda_{2}+\lambda_{3}+\lambda_{4})+3\varepsilon_{1}\nu^{2}+3c=0,$

which can be rewritten as

$(\varepsilon_{1}\mu^{2}+\varepsilon)\text{tr}A^{2}=(\varepsilon_{1}\mu^{2}+\varepsilon)40H^{2}+24\varepsilon\varepsilon_{1}H\mu\nu+6\varepsilon_{1}\nu^{2}+6c.$
(3.24)

Put (3.21) and (3.22) into (3.9), combining (3.24), we have

$u_{1}(H)=\frac{-26}{3}\varepsilon
H^{2}\mu-\frac{4\varepsilon_{1}H\mu^{2}\nu+\varepsilon\varepsilon_{1}\mu\nu^{2}+c\varepsilon\mu}{\varepsilon_{1}\mu^{2}+\varepsilon}-2H\nu.$
(3.25)

Act on (3.25) by $u_{1}$, applying (3.23), we get the expression of
$u_{1}u_{1}(H)$. And then, substitute the expressions of $u_{1}(H)$ and
$u_{1}u_{1}(H)$ into (3.7), combining (3.24), it gives

$\displaystyle(36\mu^{3}\varepsilon+18\mu\varepsilon_{1})\nu^{3}+(300H\mu^{4}+156\mu^{2}H\varepsilon\varepsilon_{1}-72H)\nu^{2}+(816\mu^{5}H^{2}\varepsilon$
(3.26)
$\displaystyle+624\mu^{3}H^{2}\varepsilon_{1}+36\mu^{3}c\varepsilon\varepsilon_{1}-192\varepsilon
H^{2}\mu+18c\mu)\nu+940\mu^{4}H^{3}\varepsilon\varepsilon_{1}$
$\displaystyle+728\mu^{6}H^{3}+174\mu^{4}Hc\varepsilon_{1}+9\mu^{4}H\lambda\varepsilon_{1}-304\mu^{2}H^{3}+120\mu^{2}Hc\varepsilon$
$\displaystyle+18\mu^{2}H\lambda\varepsilon-516H^{3}\varepsilon\varepsilon_{1}-54Hc\varepsilon_{1}+9\varepsilon_{1}\lambda
H=0.$

By differentiating (3.26) along $u_{1}$, using (3.23) and (3.25), one derive

$\displaystyle d_{1}\nu^{4}+d_{2}\nu^{3}+d_{3}\nu^{2}+d_{4}\nu+d_{5}=0,$
(3.27)

where

$\displaystyle d_{1}=$
$\displaystyle-516\mu^{5}\varepsilon\varepsilon_{1}-444\mu^{3},$
$\displaystyle d_{2}=$
$\displaystyle-4800\mu^{6}H\varepsilon_{1}-4416\mu^{4}H\varepsilon+420\mu^{2}H\varepsilon_{1}+324H\varepsilon,$
$\displaystyle d_{3}=$
$\displaystyle-15872\mu^{7}H^{2}\varepsilon\varepsilon_{1}-17668\mu^{5}H^{2}-726\mu^{5}c\varepsilon-9\mu^{5}\lambda\varepsilon+1864\mu^{3}H^{2}\varepsilon\varepsilon_{1}$
$\displaystyle-618\mu^{3}c\varepsilon_{1}-18\mu^{3}\lambda\varepsilon_{1}+3660\mu
H^{2}+36c\varepsilon\mu-9\mu\lambda\varepsilon,$ $\displaystyle d_{4}=$
$\displaystyle-21824\mu^{8}H^{3}\varepsilon_{1}-31432\mu^{6}H^{3}\varepsilon-3684\mu^{6}Hc-90\mu^{6}H\lambda+5320\mu^{4}H^{3}\varepsilon_{1}$
$\displaystyle-3588\mu^{4}Hc\varepsilon\varepsilon_{1}-198\mu^{4}H\lambda\varepsilon\varepsilon_{1}+17640\mu^{2}H^{3}\varepsilon+384\mu^{2}Hc-126\mu^{2}H\lambda$
$\displaystyle+2712H^{3}\varepsilon_{1}+288Hc\varepsilon\varepsilon_{1}-18H\lambda\varepsilon\varepsilon_{1},$
$\displaystyle d_{5}=$ $\displaystyle
6024\mu^{5}H^{4}\varepsilon\varepsilon_{1}-10192\mu^{9}H^{4}\varepsilon\varepsilon_{1}-18376\mu^{7}H^{4}-3116\mu^{7}H^{2}c\varepsilon-6\mu^{7}H^{2}\lambda\varepsilon$
$\displaystyle-3544\mu^{5}H^{2}c\varepsilon_{1}-18\mu^{5}H^{2}\lambda\varepsilon_{1}-210\mu^{5}c^{2}\varepsilon\varepsilon_{1}-9\mu^{5}c\lambda\varepsilon\varepsilon_{1}+26408\mu^{3}H^{4}$
$\displaystyle+2260\mu^{3}H^{2}c\varepsilon-18\mu^{3}H^{2}\lambda\varepsilon+12200\mu
H^{4}\varepsilon\varepsilon_{1}-174\mu^{3}c^{2}-18\mu^{3}c\lambda$
$\displaystyle+2688\mu H^{2}c\varepsilon_{1}-6\mu
H^{2}\lambda\varepsilon_{1}+36\mu c^{2}\varepsilon\varepsilon_{1}-9\mu
c\lambda\varepsilon\varepsilon_{1}.$

Eliminate $\nu$ from (3.26) and (3.27), we can get a polynomial equation about
$H$ and $\mu$

$\displaystyle\sum_{k=29}^{33}\sum_{l=k-28}^{7}r_{k,l}H^{2l}\mu^{2k}+\sum_{k=4}^{28}\sum_{l=0}^{7}r_{k,l}H^{2l}\mu^{2k}+\sum_{k=0}^{3}\sum_{l=4-k}^{7}r_{k,l}H^{2l}\mu^{2k}=0,$
(3.28)

where $r_{0,l},r_{1,l},\cdots,r_{33,l}$ are all real constants.

Applying $u_{1}$ to both sides of (3.27), we obtain a polynomial equation
about $H,\mu$ and $\nu$. Then, eliminating $\nu$ from this equation and
(3.26), we derive another polynomial equation about $H$ and $\mu$

$\displaystyle(\varepsilon\varepsilon_{1}+\mu^{2})^{10}(\varepsilon\varepsilon_{1}+2\mu^{2})^{6}(g_{1}(H,\mu))^{2}g_{2}(H,\mu)=0,$

i.e.

$\displaystyle(\varepsilon\varepsilon_{1}+\mu^{2})(\varepsilon\varepsilon_{1}+2\mu^{2})g_{1}(H,\mu)g_{2}(H,\mu)=0,$
(3.29)

where

$\displaystyle g_{1}(H,\mu)=$ $\displaystyle
6160\varepsilon_{1}H^{2}\mu^{14}+16(17136c-62986\varepsilon
H^{2}+1935\lambda)\mu^{12}+69984\varepsilon H^{2}$
$\displaystyle+8\varepsilon_{1}(37935\varepsilon
c-512645H^{2}+8424\varepsilon\lambda)\mu^{10}+48(987\lambda-28138\varepsilon
H^{2}$ $\displaystyle+1884c)\mu^{8}+\varepsilon_{1}(8334\varepsilon
c+2136896H^{2}+10044\varepsilon\lambda)\mu^{6}+(6570c$
$\displaystyle+312180\varepsilon
H^{2}-1467\lambda)\mu^{4}+\varepsilon_{1}(2106\varepsilon
c-14292H^{2}-567\varepsilon\lambda)\mu^{2},$

and

$\displaystyle g_{2}(H,\mu)\\!=$
$\displaystyle\sum_{l=4}^{6}s_{17,l}H^{2l}\mu^{34}+\sum_{l=3}^{6}s_{16,l}H^{2l}\mu^{32}+\sum_{l=2}^{6}s_{15,l}H^{2l}\mu^{30}+\sum_{l=1}^{6}s_{14,l}H^{2l}\mu^{28}$
$\displaystyle+\sum_{k=2}^{13}\sum_{l=0}^{6}s_{k,l}H^{2l}\mu^{2k}+\sum_{l=1}^{6}s_{1,l}H^{2l}\mu^{2}+\sum_{l=2}^{6}s_{0,l}H^{2l},$

with $s_{0,l},s_{1,l},\cdots,s_{17,l}$ are all real constants. We can
eliminate $\mu$ from (3.28) and $\eqref{5---gHmu}$, and give a polynomial
equation of degree $696$ for $H$, which implies $H$ is a constant in $U_{q}$.
Thus, $\nabla H=0$ at $q$, a contradiction.

_Case 2: $\Gamma_{23}^{4}\equiv 0$ on $U_{p}$._

In this case, (3.19) can be simplified into

$\begin{cases}\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{31}^{3}=-c-\varepsilon\lambda_{2}\lambda_{3},\\\
\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{41}^{4}=-c-\varepsilon\lambda_{2}\lambda_{4},\\\
\varepsilon_{1}\Gamma_{31}^{3}\Gamma_{41}^{4}=-c-\varepsilon\lambda_{3}\lambda_{4}.\end{cases}$
(3.30)

As $\lambda_{2},\lambda_{3}$ and $\lambda_{4}$ are distinct, we have
$c+\varepsilon\lambda_{2}\lambda_{3}\neq 0$, or
$c+\varepsilon\lambda_{2}\lambda_{4}\neq 0$, or
$c+\varepsilon\lambda_{3}\lambda_{4}\neq 0$. So, the above equations implies
there are at least two non-zero terms in
$\\{\Gamma_{21}^{2},\Gamma_{31}^{3},\Gamma_{41}^{4}\\}$. Without loss of
generality, we suppose $\Gamma_{21}^{2},\Gamma_{31}^{3}\neq 0$ at any point in
$U_{p}$, and discuss separately the subcases $\Gamma_{41}^{4}\equiv 0$ on
$U_{p}$ and $\Gamma_{41}^{4}\neq 0$ at some point.

_Subcase (1): $\Gamma_{41}^{4}\equiv 0$ on $U_{p}$._

It follows from (3.30) that $\lambda_{4}=0$. If $c\neq 0$, then
$c+\lambda_{2}\lambda_{4}\neq 0$, which implies $\Gamma_{41}^{4}\neq 0$, a
contradiction. In the following, we suppose $c=0$. Act on both sides of
$\lambda_{2}+\lambda_{3}=6\varepsilon H$ by $u_{1}$, we get

$6\varepsilon u_{1}(H)=(-2\varepsilon
H-\lambda_{2})\Gamma_{21}^{2}+(-2\varepsilon H-\lambda_{3})\Gamma_{31}^{3}.$
(3.31)

Differentiate (3.31) along $u_{1}$, using (3.5) and (3.6), we obtain the
expression of $u_{1}u_{1}(H)$, and put it into (3.7), combining (3.30) and
(3.31), we have

$\displaystyle 2(-2\varepsilon H-\lambda_{2})(\Gamma_{21}^{2})^{2}$
$\displaystyle+2(-2\varepsilon
H-\lambda_{3})(\Gamma_{31}^{3})^{2}-34\varepsilon_{1}H\lambda_{2}\lambda_{3}$
(3.32)
$\displaystyle+468\varepsilon_{1}H^{3}-9\varepsilon\varepsilon_{1}\lambda
H=0.$

By applying (3.5), (3.6) and (3.31), differentiate (3.32) along $u_{1}$, and
combining (3.30) and (3.32), we derive

$\displaystyle K_{1}\Gamma_{21}^{2}+K_{2}\Gamma_{31}^{3}=0,$ (3.33)

where

$\displaystyle K_{1}=$
$\displaystyle-680H\lambda_{2}\lambda_{3}\varepsilon\varepsilon_{1}+6552\varepsilon
H^{3}-96H^{2}\lambda_{2}\varepsilon+264H\lambda_{2}\lambda_{3}\varepsilon$
$\displaystyle+40\lambda_{2}^{2}\lambda_{3}\varepsilon-1404H^{2}\lambda_{2}+408H^{2}\lambda_{3}-48H\lambda_{2}^{2}+80H\lambda_{2}\lambda_{3}$
$\displaystyle+34\lambda_{3}\lambda_{2}^{2}-4\lambda_{2}\lambda_{3}^{2}+9\lambda_{2}\lambda\varepsilon-162H\lambda;$
$\displaystyle K_{2}=$
$\displaystyle-680H\lambda_{2}\lambda_{3}\varepsilon\varepsilon_{1}+6552\varepsilon
H^{3}-96H^{2}\lambda_{3}\varepsilon+264H\lambda_{2}\lambda_{3}\varepsilon$
$\displaystyle+40\lambda_{3}^{2}\lambda_{2}\varepsilon-1404H^{2}\lambda_{3}+408H^{2}\lambda_{2}-48H\lambda_{3}^{2}+80H\lambda_{2}\lambda_{3}$
$\displaystyle+34\lambda_{2}\lambda_{3}^{2}-4\lambda_{3}\lambda_{2}^{2}+9\lambda_{3}\lambda\varepsilon-162H\lambda.$

Since (3.30), we can eliminate $\Gamma_{21}^{2}$ and $\Gamma_{31}^{3}$ from
(3.32) and (3.33), and give

$\displaystyle-32(\lambda_{2}^{6}\lambda_{3}^{3}+\lambda_{2}^{3}\lambda_{3}^{6})+c_{1}(\lambda_{2}^{5}\lambda_{3}^{4}+\lambda_{2}^{4}\lambda_{3}^{5})+c_{2}(\lambda_{2}^{5}\lambda_{3}^{3}+\lambda_{2}^{3}\lambda_{3}^{5})+c_{3}\lambda_{2}^{4}\lambda_{3}^{4}$
(3.34)
$\displaystyle+c_{4}(\lambda_{2}^{5}\lambda_{3}^{2}+\lambda_{2}^{2}\lambda_{3}^{5})+c_{5}(\lambda_{2}^{4}\lambda_{3}^{3}+\lambda_{2}^{3}\lambda_{3}^{4})-9216\varepsilon
H^{3}(\lambda_{2}^{5}\lambda_{3}+\lambda_{2}\lambda_{3}^{5})$
$\displaystyle+c_{6}(\lambda_{2}^{4}\lambda_{3}^{2}+\lambda_{2}^{2}\lambda_{3}^{4})+c_{7}\lambda_{2}^{3}\lambda_{3}^{3}+c_{8}(\lambda_{2}^{4}\lambda_{3}+\lambda_{2}\lambda_{3}^{4})+c_{9}(\lambda_{2}^{3}\lambda_{3}^{2}+\lambda_{2}^{2}\lambda_{3}^{3})$
$\displaystyle+c_{10}(\lambda_{2}^{3}\lambda_{3}+\lambda_{2}\lambda_{3}^{3})+c_{11}\lambda_{2}^{2}\lambda_{3}^{2}+c_{12}(\lambda_{2}^{3}+\lambda_{3}^{3})+c_{13}(\lambda_{2}^{2}\lambda_{3}+\lambda_{2}\lambda_{3}^{2})$
$\displaystyle+c_{14}(\lambda_{2}^{2}+\lambda_{3}^{2})+c_{15}\lambda_{2}\lambda_{3}+c_{16}(\lambda_{2}+\lambda_{3})+c_{17}=0,$

where $c_{i}$, $1\leq i\leq 17$ are polynomials about $H$. Act on (3.34) by
$u_{1}$, we get

$\displaystyle L_{1}\Gamma_{21}^{2}+L_{2}\Gamma_{31}^{3}=0,$ (3.35)

where $L_{1}$ and $L_{2}$ are polynomials about $H,\lambda_{2}$ and
$\lambda_{3}$. It follows from (3.33) and (3.35) that

$K_{1}L_{2}-K_{2}L_{1}=0.$

Take into account $\lambda_{2}+\lambda_{3}=6\varepsilon H$, we can eliminate
$\lambda_{2}$ and $\lambda_{3}$ from (3.34) and the above equation, and derive
a polynomial equation of degree $78$ for $H$, which implies $H$ is a constant,
a contradiction.

_Subcase (2): $\Gamma_{41}^{4}\neq 0$ at some point in $U_{p}$._

Suppose $\Gamma_{41}^{4}\neq 0$ at $q\in U_{p}$, then there exists a
neighbourhood $U_{q}\subset U_{p}$ such that $\Gamma_{41}^{4}\neq 0$ on
$U_{q}$. For this subcase, we work on $U_{q}$. Let

$\mu=\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{4}$

and

$\nu=\lambda_{2}\lambda_{3}\lambda_{4}.$

By use of (3.5), differentiate both sides of
$\lambda_{2}+\lambda_{3}+\lambda_{4}=6\varepsilon H$,
$\mu=\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{4}+\lambda_{3}\lambda_{4}$ and
$\nu=\lambda_{2}\lambda_{3}\lambda_{4}$ along $u_{1}$, and then multiply these
results by $(c+\varepsilon\lambda_{3}\lambda_{4})\Gamma_{21}^{2}$, combining
(3.30), we obtain

$\begin{cases}6\varepsilon u_{1}(H)=Q(-12\varepsilon Hc^{2}-10H\mu
c+3\varepsilon\nu c-48H^{2}\nu+2\mu\nu),\\\
u_{1}(\mu)=Q[3\nu^{2}-2(12H^{2}+\mu)c^{2}-2(\mu^{2}+6\varepsilon
H^{2}\mu-3H\nu)c-10\varepsilon H\mu\nu],\\\ u_{1}(\nu)=Q[(-2\varepsilon
H\mu-3\nu)c^{2}+(-24\varepsilon H^{2}\nu-2\varepsilon\mu\nu)c-12\varepsilon
H\nu^{2}],\end{cases}$ (3.36)

where
$Q=\frac{-\varepsilon_{1}}{(c+\varepsilon\lambda_{3}\lambda_{4})\Gamma_{21}^{2}}$.
As (3.30) and (3.36), (3.7) can be rewritten as

$\displaystyle(30\varepsilon Hc^{2}-44Hc^{2}-6\varepsilon\nu
c-2c\nu)\mu^{2}+(504H^{3}c^{2}-36Hc^{3}\varepsilon$ (3.37)
$\displaystyle-9Hc^{2}\lambda\varepsilon-12c\nu H^{2}-15\nu
c^{2}\varepsilon-78H\nu^{2})\mu+3024H^{4}\nu c$
$\displaystyle+504H^{3}\varepsilon c^{3}-108H^{2}\nu\varepsilon
c^{2}-54H^{2}\nu\lambda\varepsilon c+1080H^{3}\nu^{2}$
$\displaystyle-72H\nu^{2}\varepsilon
c-9H\nu^{2}\lambda\varepsilon-18Hc^{4}-9H\lambda
c^{3}+9\nu^{3}\varepsilon-9\nu c^{3}=0.$

Act on (3.37) by $u_{1}$, applying (3.36), it gives

$\displaystyle f_{1}\mu^{3}+f_{2}\mu^{2}+f_{3}\mu+f_{4}=0.$ (3.38)

where

$\displaystyle f_{1}=$
$\displaystyle(828\varepsilon-256)Hc^{3}+(180\varepsilon+80)\nu c^{2};$
$\displaystyle f_{2}=$ $\displaystyle
4(522\varepsilon+747)H\nu^{2}c-1728(6\varepsilon+5)H^{3}c^{3}-288(11\varepsilon-46)H^{2}\nu
c^{2}-156\nu^{3}$
$\displaystyle+4(531\varepsilon-162)Hc^{4}+18(5\varepsilon+6)H\lambda
c^{3}+6(10\varepsilon+87)\nu c^{3}-18\nu c^{2}\lambda\varepsilon;$
$\displaystyle f_{3}=$
$\displaystyle-72576H^{5}c^{3}-(22752\varepsilon+8640)H^{3}c^{4}+1296H^{3}c^{3}\lambda\varepsilon-283392H^{4}\nu
c^{2}$ $\displaystyle+2916H^{2}\nu
c^{2}\lambda\varepsilon-3312H^{3}\nu^{2}c+(17928\varepsilon+3888)H^{2}\nu
c^{3}+90H\nu^{2}c\lambda\varepsilon$
$\displaystyle+(6552\varepsilon+648)H\nu^{2}c^{2}+1152Hc^{5}+306Hc^{4}\lambda+26136H^{2}\nu^{3}$
$\displaystyle-774\nu^{3}c\varepsilon-18\nu^{3}\lambda\varepsilon+414\nu
c^{4}-45\nu c^{3}\lambda-216\nu^{3}c;$ $\displaystyle f_{4}=$ $\displaystyle
7776H^{4}\nu\lambda\varepsilon c^{2}+11664H^{3}\nu^{2}\lambda\varepsilon
c-311040H^{4}\nu^{3}-236736H^{4}\nu\varepsilon c^{3}$
$\displaystyle+17280H^{3}\nu^{2}\varepsilon c^{2}+108H\lambda\varepsilon
c^{5}+16632H^{2}\nu^{3}\varepsilon
c+1728H^{2}\nu^{3}\lambda\varepsilon-27\nu\lambda\varepsilon c^{4}$
$\displaystyle-54H\nu^{2}\lambda c^{2}+2376H^{2}\nu\lambda
c^{3}-1109376H^{5}\nu^{2}c+12096H^{2}\nu c^{4}-27\nu^{3}\lambda c$
$\displaystyle-435456H^{6}\nu c^{2}+1296H^{3}\lambda
c^{4}+2268H\nu^{2}c^{3}-72576H^{5}\varepsilon c^{4}+216H\varepsilon c^{6}$
$\displaystyle+108\nu\varepsilon
c^{5}-3348H\nu^{4}\varepsilon-12960H^{3}c^{5}-972\nu^{3}c^{2}.$

By differentiating (3.38) along $u_{1}$, using (3.36), we have another
polynomial equation about $H$, $\mu$ and $\nu$, denoted by $g(H,\mu,\nu)=0$.
Eliminating $\mu$ from (3.37), (3.38) and $g(H,\mu,\nu)=0$, we get

$\displaystyle(24336\varepsilon+18928)\nu^{15}+\sum_{i=0}^{14}h_{i}\nu^{i}=0,$
(3.39)

and

$\displaystyle
32(32436060H^{2}\varepsilon+50162468H^{2}+2184345c\varepsilon+1894023c)\nu^{19}+\sum_{i=0}^{18}l_{i}\nu^{i}=0,$
(3.40)

where $h_{i}$, with $0\leq i\leq 14$ and $l_{j}$, with $0\leq j\leq 18$ are
polynomials about $H$.

When $\varepsilon=1$, (3.39) and (3.40) can be rewritten as

$\begin{cases}(2cH-\nu)(7cH+4\nu)^{2}(c^{3}-6cH\nu-2\nu^{2})^{3}h_{1}(H,\nu)=0,\\\
(2cH-\nu)(7cH+4\nu)^{3}(c^{3}-6cH\nu-2\nu^{2})^{3}h_{2}(H,\nu)=0,\end{cases}$

where

$\begin{cases}h_{1}(H,\nu)=338\nu^{6}+\sum_{i=0}^{5}t_{i}\nu^{i},\\\
h_{2}(H,\nu)=(254898c+5162408H^{2})\nu^{9}+\sum_{i=0}^{8}s_{i}\nu^{i}.\end{cases}$

Here, $t_{i}$ and $t_{j}$ are polynomials about $H$, with $0\leq i\leq 5$ and
$0\leq j\leq 8$. If

$(2cH-\nu)(7cH+4\nu)(c^{3}-6cH\nu-2\nu^{2})\neq 0$

at some point $o$ in $U_{q}$, we can eliminate $\nu$ from $h_{1}(H,\nu)=0$ and
$h_{2}(H,\nu)=0$, and get a polynomial equation of degree $94$ for $H$, which
implies $\nabla H=0$ at $o$, a contradiction.

If

$(2cH-\nu)(7cH+4\nu)(c^{3}-6cH\nu-2\nu^{2})\equiv 0$

on $U_{q}$, then acting on it by $u_{1}$, by use of (3.36), we have

$\displaystyle
54c^{6}H\nu-336c^{7}H^{2}+3048c^{5}H^{3}\nu+12096c^{4}H^{5}\nu+2316c^{4}H^{2}\nu^{2}$
(3.41)
$\displaystyle+147c^{5}\nu^{2}+27936c^{3}H^{4}\nu^{2}-1080c^{3}H\nu^{3}-1344c^{2}H^{3}\nu^{3}-510c^{2}\nu^{4}$
$\displaystyle-10416cH^{2}\nu^{4}-2304H\nu^{5}+(98c^{4}\nu^{2}-292c^{6}H^{2}+1008c^{5}H^{4}$
$\displaystyle+130c^{5}H\nu+4344c^{4}H^{3}\nu+200c^{3}H^{2}\nu^{2}-1532c^{2}H\nu^{3}-340c\nu^{4})\mu=0.$

Furthermore, we claim that $7cH+4\nu\equiv 0$ on $U_{q}$. Because if
$7cH+4\nu\neq 0$ (i.e. $(2cH-\nu)(c^{3}-6cH\nu-2\nu^{2})=0$) at some point in
$U_{q}$, then as (3.41), we have

$(c+\lambda_{2}\lambda_{3})(c+\lambda_{2}\lambda_{4})(c+\lambda_{3}\lambda_{4})=\nu^{2}+6cH\nu+c^{2}\mu+c^{3}=0,$
(3.42)

a contradiction. We can easily derive a polynomial equation about $H$ from
$7cH+4\nu=0$ and (3.41). Then, $H$ is a constant, a contradiction.

When $\varepsilon=-1$, (3.39) and (3.40) reduces to

$\begin{cases}(37cH-2\nu)^{2}L_{1}(H,\nu)=0,\\\
(37cH-2\nu)^{3}L_{2}(H,\nu)=0,\end{cases}$

where

$\begin{cases}L_{1}(H,\nu)=-1352\nu^{13}+\sum_{i=0}^{12}r_{i}\nu^{i},\\\
L_{2}(H,\nu)=(254898c+5162408H^{2})\nu^{16}+\sum_{i=0}^{15}w_{i}\nu^{i}.\end{cases}$

Here, $r_{i}$ and $w_{j}$ are all polynomials about $H$, with $0\leq i\leq 12$
and $0\leq j\leq 8$. If $37cH-2\nu\neq 0$ at some point $o$ in $U_{q}$, we can
eliminate $\nu$ from $L_{1}(H,\nu)=0$ and $L_{2}(H,\nu)=0$, and get a
polynomial equation of degree $292$ for $H$, which implies $\nabla H=0$ at
$o$, a contradiction. If $37cH-2\nu=0$ at any point in $U_{q}$, then acting on
it by $u_{1}$, combining (3.36), we have

$-1488c\nu H^{2}-346H\mu c^{2}-444Hc^{3}+144H\nu^{2}+98\nu\mu c+147\nu
c^{2}=0,$

which together with $37cH-2\nu=0$ and (3.37) implies that $H$ is a constant in
$U_{q}$, a contradiction. $\hfill\square$

###  3.2 The shape operator has form (II)

Proposition 3.6  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (II), then $M^{4}_{r}$ has constant
mean curvature._

Proof We suppose $\lambda_{1},\lambda_{2}$ and $\lambda_{3}$ are distinct with
each other. When there are at most two distinct values in
$\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}$, we omit the proof, since it is
similar but much easier.

Denote $u_{2}=u_{2_{1}}$ and $u_{3}=u_{3_{1}}$. Let
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},2,3$, then
compatibility condition implies that

$\Gamma_{D1_{1}}^{1_{2}}=\Gamma_{D1_{2}}^{1_{1}}=\Gamma_{D2}^{2}=\Gamma_{D3}^{3}=0,$
(3.43)

and

$\displaystyle\Gamma_{D1_{1}}^{1_{1}}=-\Gamma_{D1_{2}}^{1_{2}},\Gamma_{D2}^{3}=-\varepsilon_{2}\varepsilon_{3}\Gamma_{D3}^{2},\Gamma_{Di}^{1_{1}}=-\varepsilon_{1}\varepsilon_{i}\Gamma_{D1_{2}}^{i},\Gamma_{Di}^{1_{2}}=-\varepsilon_{1}\varepsilon_{i}\Gamma_{D1_{1}}^{i},$
(3.44)

with $i=2,3$ and $D=1_{1},1_{2},2,3$. Express $\nabla H$ as

$\nabla
H=\varepsilon_{1}u_{1_{2}}(H)u_{1_{1}}+\varepsilon_{1}u_{1_{1}}(H)u_{1_{2}}+\varepsilon_{2}u_{2}(H)u_{2}+\varepsilon_{3}u_{3}(H)u_{3}.$
(3.45)

Assume that $H$ is not a constant, it follows from (2.2) that $\nabla H$ is an
eigenvector of $A$ with corresponding eigenvalue $-2\varepsilon H$, is light-
like or not.

_Case 1: $\nabla H$ is light-like._

In this case, $\nabla H$ is in the direction $u_{1_{2}}$ and
$\lambda_{1}=-2\varepsilon H$. As (3.45), we know

$u_{1_{1}}(H)\neq 0,\ u_{1_{2}}(H)=u_{2}(H)=u_{3}(H)=0.$ (3.46)

According to symmetry of the connection $\nabla$, we have

$\Gamma^{1_{1}}_{BC}=\Gamma^{1_{1}}_{CB},\quad B,C=1_{2},2,3.$ (3.47)

Investigate the equation (2.1), for
$(X,Y,Z)=(u_{1_{1}},u_{i},u_{1_{2}}),(u_{2},u_{3},u_{1_{2}}),\\\
(u_{1_{2}},u_{i},u_{i})$, with $i=2,3$, by use of (3.46) and (3.47), we obtain
that

$\Gamma_{1_{1}2}^{1_{1}}=\Gamma_{1_{1}3}^{1_{1}}=\Gamma_{23}^{1_{1}}=0,$
(3.48)

and

$u_{1_{2}}(\lambda_{i})=(-2\varepsilon H-\lambda_{i})\Gamma_{i1_{2}}^{i},\
i=2,3.$ (3.49)

Calculate $\langle R(u_{1_{2}},u_{i})u_{1_{2}},u_{i}\rangle$, $i=2,3$ by Gauss
equation, combining (3.43), (3.44), (3.47) and (3.48), we get

$u_{1_{2}}(\Gamma_{i1_{2}}^{i})=\Gamma^{1_{2}}_{1_{2}1_{2}}\Gamma_{i1_{2}}^{i}-(\Gamma_{i1_{2}}^{i})^{2},\
i=2,3.$ (3.50)

As $\lambda_{1}=-2\varepsilon H$, the equation
$2\lambda_{1}+\lambda_{2}+\lambda_{3}=4\varepsilon H$ (cf. eq. (2.4)) reduces
to $\lambda_{2}+\lambda_{3}=8\varepsilon H.$ Differentiate this equation along
$u_{1_{2}}$, applying (3.46) and (3.49), we find

$(-2\varepsilon H-\lambda_{2})\Gamma_{21_{2}}^{2}+(-2\varepsilon
H-\lambda_{3})\Gamma_{31_{2}}^{3}=0.$

By use of (3.46), (3.49) and (3.50), act $u_{1_{2}}$ on the above equation, we
have

$(-2\varepsilon H-\lambda_{2})(\Gamma_{21_{2}}^{2})^{2}+(-2\varepsilon
H-\lambda_{3})(\Gamma_{31_{2}}^{3})^{2}=0.$

Because of $\lambda_{2},\lambda_{3}\neq-2\varepsilon H$ and
$\lambda_{2}+\lambda_{3}=8\varepsilon H$, we conclude from the above two
equations that

$\Gamma_{21_{2}}^{2}=\Gamma_{31_{2}}^{3}$ (3.51)

and

$-12\varepsilon H\Gamma_{21_{2}}^{2}=0,$

which together with $H\neq 0$ implies $\Gamma_{21_{2}}^{2}=0$. Combining
(3.43), (3.44), (3.47), (3.48) and (3.51), compute $\langle
R(u_{1_{1}},u_{i})u_{1_{2}},u_{i}\rangle$ by using Gauss equation, it gives

$2H\lambda_{i}=c,\ i=2,3,$

which yields that $\lambda_{2}=\lambda_{3}$, a contradiction.

_Case 2: $\nabla H$ is not light-like._

Observe the form (II), we know $\nabla H$ is in the direction $u_{2}$ or
$u_{3}$ in this case. Without loss of generality, we suppose $\nabla H$ is in
the direction $u_{2}$, then $\lambda_{2}=-2\varepsilon H$, which together with
$2\lambda_{1}+\lambda_{2}+\lambda_{3}=4\varepsilon H$ tells us that
$\lambda_{3}=6\varepsilon H-2\lambda_{1}$. The equation (3.45) and the
symmetry of connection $\nabla$ give that

$u_{2}(H)\neq 0,\ u_{B}(H)=0,\ B=1_{1},1_{2},3.$ (3.52)

and

$\Gamma^{2}_{BC}=\Gamma^{2}_{CB},\quad B,C=1_{1},1_{2},3.$ (3.53)

By use of (3.43), (3.44), (3.52) and (3.53), we deduce from (2.1) that

$u_{1_{2}}(\lambda_{1})=\Gamma_{1_{2}3}^{1_{1}}=\Gamma_{1_{2}3}^{1_{1}}=\Gamma_{31_{1}}^{2}=\Gamma_{31_{2}}^{2}=\Gamma_{31_{2}}^{3}=\Gamma_{2B}^{2}=0,$
(3.54) $u_{1_{2}}(\lambda_{3})=(\lambda_{1}-\lambda_{3})\Gamma_{31_{2}}^{3},\
\
\Gamma_{1_{1}3}^{1_{1}}=\Gamma_{1_{2}3}^{1_{2}}=\frac{u_{3}(\lambda_{1})}{\lambda_{3}-\lambda_{1}},$
(3.55)

and

$\Gamma_{1_{1}2}^{1_{1}}=\Gamma_{1_{2}2}^{1_{2}}=\frac{u_{2}(\lambda_{1})}{-2\varepsilon
H-\lambda_{1}},\ \ \Gamma_{32}^{3}=\frac{u_{2}(\lambda_{3})}{-2\varepsilon
H-\lambda_{3}}.$ (3.56)

Let $e_{1}=\frac{u_{1_{1}}-u_{1_{2}}}{\sqrt{2}}$ and
$e_{2}=\frac{u_{1_{1}}+u_{1_{2}}}{\sqrt{2}}$, then
$\mathfrak{E}=\\{e_{1},e_{2},u_{2},u_{3}\\}$ is an orthonormal basis of
$T_{x}M^{4}_{r}$. We easily find

$\nabla_{e_{2}}e_{2}(H)-\nabla_{e_{1}}e_{1}(H)=\Gamma_{1_{1}1_{2}}^{2}+\Gamma_{1_{2}1_{1}}^{2},$

and the equation (2.3) can be reduced to

$u_{2}u_{2}(H)+(2\Gamma_{1_{1}2}^{1_{1}}+\Gamma_{32}^{3})u_{2}(H)-\varepsilon\varepsilon_{2}H\text{tr}A^{2}+\varepsilon_{2}\lambda
H=0,$ (3.57)

where

$\text{tr}A^{2}=2\lambda_{1}^{2}+\lambda_{3}^{2}+4H^{2}.$

Using Gauss equation for $\langle R(u_{2},u_{3})u_{3},u_{2}\rangle$ and
$\langle R(u_{2},u_{1_{1}})u_{1_{2}},u_{2}\rangle$, combining (3.43), (3.44),
(3.53) and (3.54), it is not difficult to check

$\begin{cases}u_{2}(\Gamma_{32}^{3})+(\Gamma_{32}^{3})^{2}=2H\varepsilon_{2}\lambda_{3}-c\varepsilon_{2},\\\
u_{2}(\Gamma_{1_{1}2}^{1_{1}})+(\Gamma_{1_{1}2}^{1_{1}})^{2}=2H\varepsilon_{2}\lambda_{1}-c\varepsilon_{2}.\end{cases}$
(3.58)

Let

$f_{k}=(\Gamma_{1_{1}2}^{1_{1}})^{k}+(\Gamma_{1_{2}2}^{1_{2}})^{k}+(\Gamma_{32}^{3})^{k},\
k=1,2,\cdots,$

then (3.10) also holds. We can write $f_{k}$, $k=2,3,4,5$, as the expressions
about $f_{1}$, $H$, and their derivatives along $u_{2}$, similarly with (3.8),
just need to replace $u_{1}$ with $u_{2}$ in (3.8). And then, follow the
process of the proof for Lemma 3.4, we have $u_{B}(f_{1})=0$, with
$B=1_{1},1_{2},3$. Furthermore, we easily get that
$u_{B}u_{2}^{(k)}(f_{1})=u_{B}u_{2}^{(k)}(H)=0$, with $B=1_{1},1_{2},3$ and
$k=1,2$. Thus, we derive

$\displaystyle u_{B}(f_{k})=0,\ B=1_{1},1_{2},3,\ \ k=1,2,$

which gives

$\begin{cases}2u_{B}(\Gamma_{1_{1}2}^{1_{1}})+u_{B}(\Gamma_{32}^{3})=0,\\\
2\Gamma_{1_{1}2}^{1_{1}}u_{B}(\Gamma_{1_{1}2}^{1_{1}})+\Gamma_{32}^{3}u_{B}(\Gamma_{32}^{3})=0.\end{cases}$

Observe (3.58), we know $\Gamma_{1_{1}2}^{1_{1}}\neq\Gamma_{32}^{3}$. So, the
above equations yields

$u_{B}(\Gamma_{1_{1}2}^{1_{1}})=u_{B}(\Gamma_{32}^{3})=0,\ B=1_{1},1_{2},3.$

Act on the equations in (3.58) by $u_{B}$, $B=1_{1},1_{2},3$, using the above
equation, we find

$u_{B}(\lambda_{1})=u_{B}(\lambda_{3})=0,\ B=1_{1},1_{2},3,$

which together with (3.55) implies

$\Gamma_{1_{1}3}^{1_{1}}=\Gamma_{31_{2}}^{3}=0.$ (3.59)

Applying Gauss equation for $\langle
R(u_{3},u_{1_{1}})u_{3},u_{1_{2}}\rangle$, combining (3.43), (3.44), (3.53),
(3.54) and (3.59), we have

$\Gamma_{32}^{3}\Gamma_{1_{1}2}^{1_{1}}=2\varepsilon\varepsilon_{2}\lambda_{1}^{2}-6\varepsilon_{2}H\lambda_{1}-c\varepsilon_{2}.$
(3.60)

Substituting $\Gamma_{32}^{3}=\frac{u_{2}(3\varepsilon
H-\lambda_{1})}{-4\varepsilon H+\lambda_{1}}$ and
$\Gamma_{1_{1}2}^{1_{1}}=\frac{u_{2}(\lambda_{1})}{-2\varepsilon
H-\lambda_{1}}$ into (3.58), and then combining (3.57) and (3.60) to eliminate
$u_{2}u_{2}(H)$ and $u_{2}u_{2}(\lambda_{1})$, we derive

$\displaystyle\big{(}8\Gamma_{1_{1}2}^{1_{1}}+4\Gamma_{32}^{3}\big{)}u_{2}(H)=$
$\displaystyle-168\varepsilon\varepsilon_{2}H^{3}+202\varepsilon_{2}H^{2}\lambda_{1}-64\varepsilon\varepsilon_{2}H\lambda_{1}^{2}+4\varepsilon_{2}\lambda_{1}^{3}$
(3.61)
$\displaystyle+26c\varepsilon_{2}H-2c\varepsilon\varepsilon_{2}\lambda_{1}+3\varepsilon_{2}\lambda
H.$

Differentiate (3.61) with $u_{2}$, applying (3.57), (3.58) and (3.60), we
obtain

$\displaystyle
f_{1}(H,\lambda_{1})\Gamma_{1_{1}2}^{1_{1}}+g_{1}(H,\lambda_{1})\Gamma_{32}^{3}=h_{1}(H,\lambda_{1})u_{2}(H),$
(3.62)

where

$\displaystyle f_{1}(H,\lambda_{1})=1228\varepsilon H^{3}+136\varepsilon
H\lambda_{1}^{2}-852H^{2}\lambda_{1}+4c\varepsilon\lambda_{1}-82cH-17\lambda
H,$ $\displaystyle g_{1}(H,\lambda_{1})=496\varepsilon H^{3}+152\varepsilon
H\lambda_{1}^{2}-500H^{2}\lambda_{1}-8\lambda_{1}^{3}+4c\varepsilon\lambda_{1}-52cH-10\lambda
H,$ $\displaystyle
h_{1}(H,\lambda_{1})=600H^{2}\varepsilon+88\varepsilon\lambda_{1}^{2}-500H\lambda_{1}-50c-3\lambda.$

By differentiating $2\lambda_{1}+\lambda_{3}=6\varepsilon H$ along $u_{2}$,
using (3.56), we know

$3\varepsilon u_{2}(H)=-(2\varepsilon
H+\lambda_{1})\Gamma_{1_{1}2}^{1_{1}}+(-4\varepsilon
H+\lambda_{1})\Gamma_{32}^{3}.$ (3.63)

Putting (3.63) into (3.62), then

$\displaystyle
f_{2}(H,\lambda_{1})\Gamma_{1_{1}2}^{1_{1}}+g_{2}(H,\lambda_{1})\Gamma_{32}^{3}=0,$
(3.64)

where

$\displaystyle f_{2}(H,\lambda_{1})=$ $\displaystyle
2484H^{3}-2156H^{2}\varepsilon\lambda_{1}-88\varepsilon\lambda_{1}^{3}+732H\lambda_{1}^{2}$
$\displaystyle-(146c+45\lambda)H\varepsilon+(62c+3\lambda)\lambda_{1},$
$\displaystyle g_{2}(H,\lambda_{1})=$ $\displaystyle
16(93-450\varepsilon)H^{3}-300(5\varepsilon-26)\lambda_{1}H^{2}+12\lambda_{1}c$
$\displaystyle-12(213\varepsilon-38)H\lambda_{1}^{2}-6(26c\varepsilon+5\varepsilon\lambda-100c$
$\displaystyle-6\lambda)H-24(\varepsilon-11)\lambda_{1}^{3}-3(50c+3\lambda)\lambda_{1}\varepsilon.$

Multiply both sides of (3.63) by $8\Gamma_{1_{1}2}^{1_{1}}+4\Gamma_{32}^{3}$,
applying (3.60) and (3.61), we have

$\displaystyle(16\varepsilon
H+8\lambda_{1})(\Gamma_{1_{1}2}^{1_{1}})^{2}+(16\varepsilon
H-4\lambda_{1})(\Gamma_{32}^{3})^{2}=h_{3}(H,\lambda_{1}),$ (3.65)

where

$\displaystyle h_{3}(H,\lambda_{1})=$
$\displaystyle-366\varepsilon\varepsilon_{2}H^{2}\lambda_{1}-4\varepsilon\varepsilon_{2}\lambda_{1}^{3}+504\varepsilon_{2}H^{3}+88\varepsilon_{2}H\lambda_{1}^{2}$
$\displaystyle-38\varepsilon
c\varepsilon_{2}H-9\varepsilon\varepsilon_{2}\lambda
H+2c\varepsilon_{2}\lambda_{1}.$

By use of (3.60), (3.64) and (3.65), we can get a polynomial equation about
$H$ and $\lambda_{1}$

$\displaystyle(109824-1053952\varepsilon)\lambda_{1}^{9}+(21186688H-2300928H\varepsilon)\lambda_{1}^{8}+\sum_{i=0}^{7}r_{i}\lambda_{1}^{i}=0,$

where $r_{i}$, $0\leq i\leq 7$ are polynomials about $H$. Acting on the above
equation by $u_{3}$ twice, and using (3.58), (3.60), (3.63) and (3.64), we
obtain another algebraic equation of $H$ and $\lambda_{1}$

$\displaystyle(-13895028523008\varepsilon$
$\displaystyle+110378156212224)\lambda_{1}^{17}-(4577376380387328\varepsilon$
$\displaystyle-672576751853568)H\lambda_{1}^{16}+\sum_{j=0}^{15}s_{j}\lambda_{1}^{j}=0,$

where $s_{j}$, $0\leq j\leq 15$ are polynomials about $H$. Eliminate
$\lambda_{1}$ from the above two equations, we derive an algebraic equation
for $H$ with constant coefficients. So, $H$ must be a constant, a
contradiction.$\hfill\square$

###  3.3 The shape operator has the form (III)

Proposition 3.7  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (III), then $M^{4}_{r}$ has
constant mean curvature._

Proof Assume that $H$ is not a constant, then $\nabla H$ is an eigenvector of
$A$ with corresponding eigenvalue $-2\varepsilon H$. Let
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},2_{1},2_{2}$,
then

$\Gamma_{Di_{a}}^{j_{b}}=-\varepsilon_{i}\varepsilon_{j}\Gamma_{Dj_{3-b}}^{i_{3-a}},\
i,j,a,b=1,2,$

with $D=1_{1},1_{2},2_{1},2_{2}$. Observe the form (III) of $A$, we know
eigenvector $\nabla H$ is in the direction $u_{1_{2}}$ or $u_{2_{2}}$. Without
loss of generality, we suppose $\nabla H$ is in the direction $u_{1_{2}}$,
then $\lambda_{1}=-2\varepsilon H$ and $\lambda_{2}=4\varepsilon H$.

It follows that

$u_{1_{1}}(H)\neq 0,\ u_{1_{2}}(H)=u_{2_{1}}(H)=u_{2_{2}}(H)=0,$

and

$\Gamma^{1_{1}}_{BC}=\Gamma^{1_{1}}_{CB},\quad B,C\neq 1_{1}.$

We deduce from Codazzi equation that

$\Gamma_{1_{1}2_{2}}^{1_{1}}=\Gamma_{2_{2}2_{2}}^{1_{1}}=\Gamma_{2_{1}1_{2}}^{2_{1}}=0.$

Using Gauss equation for $\langle
R(u_{1_{1}},u_{2_{1}})u_{1_{2}},u_{2_{2}}\rangle$, combining the above, we get

$c-8\varepsilon H^{2}=0,$

which tells us $H$ is a constant, a contradiction. $\hfill\square$

###  3.4 The shape operator has the form (IV)

Proposition 3.8  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (IV), then $M^{4}_{r}$ has constant
mean curvature._

Proof Denote $u_{2}=u_{2_{1}}$, and let
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},1_{3},2$. We
easily find

$\Gamma_{D1_{1}}^{1_{3}}=\Gamma_{D1_{3}}^{1_{1}}=\Gamma_{D1_{2}}^{1_{2}}=\Gamma_{D2}^{2}=0,$
(3.66)

and

$\displaystyle\Gamma_{D1_{a}}^{1_{b}}=-\Gamma_{D1_{4-b}}^{1_{4-a}},\
\Gamma_{D1_{a}}^{2}=-\varepsilon_{1}\varepsilon_{2}\Gamma_{D2}^{1_{4-a}},$
(3.67)

where $a,b=1,2,3$. Assume that $H$ is not a constant, then $\nabla H$ is an
eigenvector of $A$, is light-like or not.

_Case 1: $\nabla H$ is light-like._

In this case, $\nabla H$ is in the direction $u_{1_{3}}$ and
$\lambda_{1}=-2\varepsilon H$, $\lambda_{2}=10\varepsilon H$. It follows that

$u_{1_{1}}(H)\neq 0,\ u_{1_{2}}(H)=u_{1_{3}}(H)=u_{2}(H)=0.$

and

$\Gamma^{1_{1}}_{BC}=\Gamma^{1_{1}}_{CB},\quad B,C\neq 1_{1}.$

Applying Codazzi equation, we conclude

$\Gamma_{1_{1}2}^{1_{1}}=\Gamma_{1_{2}2}^{1_{1}}=\Gamma_{21_{2}}^{2}=\Gamma_{21_{3}}^{2}=0.$

Calculate $\langle R(u_{1_{1}},u_{2})u_{1_{3}},u_{2}\rangle$ by Gauss
equation, combining the above equations, we get

$c\varepsilon_{1}-20\varepsilon\varepsilon_{1}H^{2}=0,$

a contradiction.

_Case 2: $\nabla H$ is not light-like._

We know $\nabla H$ is in the direction $u_{2}$, $\lambda_{2}=-2\varepsilon H$
and $\lambda_{1}=2\varepsilon H$. So,

$u_{2}(H)\neq 0,\ u_{B}(H)=0,B\neq 2.$ (3.68)

and

$\Gamma^{2}_{BC}=\Gamma^{2}_{CB},\quad B,C\neq 2.$ (3.69)

We obtain from Codazzi equation that

$\Gamma_{21_{1}}^{2}=\Gamma_{21_{2}}^{2}=\Gamma_{21_{3}}^{2}=\Gamma^{2}_{1_{3}1_{2}}=\Gamma^{2}_{1_{3}1_{3}}=\Gamma^{1_{1}}_{21_{2}}=\Gamma^{1_{2}}_{21_{3}}=0,$
$W:=\Gamma^{1_{1}}_{1_{1}2}=\Gamma^{1_{2}}_{1_{2}2}=\Gamma^{1_{3}}_{1_{3}2},\
\ u_{2}(H)=-2HW,$

Using Gauss equation for $\langle R(u_{2},u_{1_{3}})u_{2},u_{1_{1}}\rangle$,
combining the above equations, it gives

$u_{2}(W)+W^{2}=4\varepsilon_{1}H^{2}-c\varepsilon_{1}.$ (3.70)

Let $e_{1}=\frac{u_{1_{1}}-u_{1_{3}}}{\sqrt{2}}$, $e_{2}=u_{1_{2}}$, and
$e_{3}=\frac{u_{1_{1}}+u_{1_{3}}}{\sqrt{2}}$, then
$\mathfrak{E}=\\{e_{1},e_{2},e_{3},u_{2}\\}$ is an orthonormal basis of
$T_{p}M^{4}_{r}$, and (2.3) can be rewriten as

$\displaystyle 2Wu_{2}(H)+u_{2}u_{2}(H)-16\varepsilon_{1}\varepsilon
H^{3}+\lambda\varepsilon_{1}H=0.$

Put $u_{2}(H)=-2HW$ into the above equation, combining (3.70), we have

$\displaystyle
2HW^{2}-8(1+2\varepsilon)\varepsilon_{1}H^{3}+2c\varepsilon_{1}H+\lambda\varepsilon_{1}H=0.$
(3.71)

Differentiate (3.71) along $u_{2}$, using $u_{2}(H)=-2HW$ and (3.70), we
obtain

$-2W[4HW^{2}-16(2+3\varepsilon)\varepsilon_{1}H^{3}+4c\varepsilon_{1}H+\lambda\varepsilon_{1}H]=0,$

which together with (3.71) implies

$2W[16(1+\varepsilon)\varepsilon_{1}H^{3}+\lambda\varepsilon_{1}H]=0.$

Since $u_{2}(H)\neq 0$, we know $W\neq 0$. The above equation implies $H$ is a
constant, a contradiction. $\hfill\square$

###  3.5 The shape operator has the form (V)

Proposition 3.9  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (V), then $M^{4}_{r}$ has constant
mean curvature._

For the form (V), the equations (3.74) and (3.75) deduced from Codazzi
equation and Gauss equation are very complicated, compared with the equations
for the forms (I), (II), (III) and (IV). However, by constructing
creatively the terms $b_{k}$, $c_{k}$ (see (3.76)), and letting
$f_{k}=(\Gamma_{21}^{2})^{k}+2b_{k}$, with $k=1,\cdots,5$, we provide an
opportunity to complete the proof similarly with the form (I).

Proof Denote $u_{1}=u_{1_{1}}$, $u_{2}=u_{2_{1}}$,
$u_{\bar{3}}=u_{\bar{3}_{1}}$ and $u_{\tilde{3}}=u_{\tilde{3}_{1}}$. Let
$\nabla_{u_{B}}u_{C}=\Gamma_{BC}^{D}u_{D}$, $B,C=1,2,\bar{3},\tilde{3}$, we
have

$\Gamma_{BD}^{D}=0,\ \
\Gamma_{B2}^{1}=-\varepsilon_{1}\varepsilon_{2}\Gamma_{B1}^{2},\ \
\Gamma_{Bi}^{\bar{3}}=-\varepsilon_{i}\Gamma_{B\bar{3}}^{i},\ \
\Gamma_{Bi}^{\tilde{3}}=\varepsilon_{i}\Gamma_{B\tilde{3}}^{i},\ \
\Gamma_{B\bar{3}}^{\tilde{3}}=\Gamma_{B\tilde{3}}^{\bar{3}},$

with $B,D=1,2,\bar{3},\tilde{3}$ and $i=1,2$.

Assume that $H$ is not a constant, then there exists a neighbourhood $U_{p}$
of $p$ such that $H\neq 0$ and $\nabla H\neq 0$. It follows from (2.2) that
$\nabla H$ is an eigenvector of $A$, and is in the direction $u_{1}$ or
$u_{2}$. Without loss of generality, we suppose $\nabla H$ is in the direction
$u_{1}$, then $\lambda_{1}=-2\varepsilon H$,

$u_{1}(H)\neq 0,\ u_{2}(H)=u_{\bar{3}}(H)=u_{\tilde{3}}(H)=0.$

and

$\Gamma_{BD}^{1}=\Gamma_{DB}^{1}.$

The equation (2.3) gives

$u_{1}u_{1}(H)+(\Gamma_{21}^{2}+\Gamma_{\bar{3}1}^{\bar{3}}+\Gamma_{\tilde{3}1}^{\tilde{3}})u_{1}(H)-\varepsilon\varepsilon_{1}H\text{tr}A^{2}+\varepsilon_{1}\lambda
H=0.$ (3.72)

We get from (2.1), with
$(X,Y,Z)=(u_{1},u_{B},u_{1}),(u_{B},u_{1},u_{B}),(u_{B},u_{D},u_{1})$,
$B,D=2,\bar{3},\tilde{3}$ that

$\Gamma_{1B}^{1}=\Gamma_{2\bar{3}}^{1}=\Gamma_{2\tilde{3}}^{1}=0,\ \
\Gamma_{\bar{3}1}^{\bar{3}}=\Gamma_{\tilde{3}1}^{\tilde{3}},\ \ \text{with}\
B=2,\bar{3},\tilde{3},$ (3.73)

and

$\begin{cases}u_{1}(\lambda_{2})=(\lambda_{1}-\lambda_{2})\Gamma_{21}^{2},\\\
u_{1}(\gamma_{3})=(\lambda_{1}-\gamma_{3})\Gamma_{\bar{3}1}^{\bar{3}}+\tau_{3}\Gamma_{\tilde{3}1}^{\bar{3}},\\\
u_{1}(\tau_{3})=(\lambda_{1}-\gamma_{3})\Gamma_{\tilde{3}1}^{\bar{3}}-\tau_{3}\Gamma_{\tilde{3}1}^{\tilde{3}}.\end{cases}$
(3.74)

If $\lambda_{2}=\lambda_{1}$, then the above equations imply $u_{1}(H)=0$,
which contradicts with $u_{1}(H)\neq 0$. So, $\lambda_{2}\neq\lambda_{1}$.
Using Gauss equation for $\langle R(u_{1},u_{B})u_{B},u_{1}\rangle$ and
$\langle R(u_{1},u_{\bar{3}})u_{\tilde{3}},u_{1}\rangle$, with
$B=2,\bar{3},\tilde{3}$, we have

$\begin{cases}u_{1}(\Gamma_{21}^{2})=-(\Gamma_{21}^{2})^{2}+(2H\lambda_{2}-c)\varepsilon_{1},\\\
u_{1}(\Gamma_{\bar{3}1}^{\bar{3}})=-(\Gamma_{\bar{3}1}^{\bar{3}})^{2}+(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}+(2H\gamma_{3}-c)\varepsilon_{1},\\\
u_{1}(\Gamma_{\tilde{3}1}^{\bar{3}})=-2\Gamma_{\bar{3}1}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}+2H\tau_{3}\varepsilon_{1}.\end{cases}$
(3.75)

Construct the terms $b_{k}$ and $c_{k}$, with $k=1,\cdots,5$ as follows

$\begin{cases}b_{1}=\Gamma_{\bar{3}1}^{\bar{3}},\\\
c_{1}=\Gamma_{\tilde{3}1}^{\bar{3}},\\\ b_{l}=b_{1}b_{l-1}-c_{1}c_{l-1},\
l=2,3,4,5\\\ c_{l}=b_{1}c_{l-1}+c_{1}b_{l-1},\ l=2,3,4,5.\end{cases}$ (3.76)

Set $f_{k}=(\Gamma_{21}^{2})^{k}+2b_{k}$, with $k=1,\cdots,5$, we will find
(3.8) and $u_{B}(f_{1})=0$ also hold, with $B=2,\bar{3},\tilde{3}$. And then,
we can conclude that $u_{B}(\lambda_{2})=u_{B}(\gamma_{3})=u_{B}(\tau_{3})=0$,
$B=2,\bar{3},\tilde{3}$, similarly with Lemma 3.5, which will play an
important role in the following process of the proof.

Lemma 3.10  _( 3.8) holds._

Proof As $\lambda_{1}=-2\varepsilon H$, the equation (2.4) gives
$\lambda_{2}+2\gamma_{3}=6\varepsilon H$. Differentiate
$f_{k}=(\Gamma_{21}^{2})^{k}+2b_{k}$ along $u_{1}$, with $k=1,2,3,4$, using
(3.75), we have

$f_{k+1}=-\frac{1}{k}u_{1}(f_{k})-c\varepsilon_{1}f_{k-1}+2\varepsilon_{1}HQ_{k-1},$
(3.77)

where $f_{0}=1$ and
$Q_{k-1}=\lambda_{2}(\Gamma_{21}^{2})^{k-1}+2\gamma_{3}b_{k-1}-2\tau_{3}c_{k-1}$.
Here, $b_{0}=1$ and $c_{0}=0$.

Take the $k$-th derivatives of $\lambda_{2}+2\gamma_{3}=6\varepsilon H$ along
$u_{1}$, with $k=1,2,3$, combining (3.74) and (3.75), we obain

$Q_{1}=-6\varepsilon u_{1}(H)-2\varepsilon Hf_{1},$ (3.78) $Q_{2}=3\varepsilon
u_{1}^{(2)}(H)+\varepsilon Hu_{1}(f_{1})+\varepsilon f_{1}u_{1}(H)-\varepsilon
Hf_{2}-4\varepsilon_{1}H^{3}-3\varepsilon\varepsilon_{1}Hc+\varepsilon_{1}H\text{tr}A^{2},$

and

$\displaystyle 3Q_{3}=$ $\displaystyle\varepsilon Hu_{1}(f_{2})-3\varepsilon
u_{1}^{(3)}(H)+\varepsilon f_{2}u_{1}(H)-2\varepsilon u_{1}(H)u_{1}(f_{1})$
$\displaystyle+12\varepsilon_{1}H^{2}u_{1}(H)-\varepsilon_{1}u_{1}(H)\text{tr}A^{2}-\varepsilon
f_{1}u_{1}^{(2)}(H)+15\varepsilon\varepsilon_{1}cu_{1}(H)$
$\displaystyle-\varepsilon
Hu_{1}^{(2)}(f_{1})+4\varepsilon\varepsilon_{1}cHf_{1}-2\varepsilon
Hf_{3}-\varepsilon_{1}Hu_{1}(\text{tr}A^{2})$
$\displaystyle+4\varepsilon_{1}H(\lambda_{2}^{2}\Gamma_{21}^{2}+2(\gamma_{3}^{2}-\tau_{3}^{2})\Gamma_{\bar{3}1}^{\bar{3}}-4\gamma_{3}\tau_{3}\Gamma_{\tilde{3}1}^{\bar{3}}).$

Multiply the first, second and third equation in (3.74) by $\lambda_{2}$,
$\gamma_{3}$ and $\tau_{3}$, respectively, and then we get from these results
that

$\lambda_{2}^{2}\Gamma_{21}^{2}+2(\gamma_{3}^{2}-\tau_{3}^{2})\Gamma_{\bar{3}1}^{\bar{3}}-4\gamma_{3}\tau_{3}\Gamma_{\tilde{3}1}^{\bar{3}}=-2\varepsilon
HQ_{1}-\frac{1}{2}u_{1}(\text{tr}A^{2})+4Hu_{1}(H).$

Applying (3.72) and the above equations, we deduce from (3.77) that (3.8)
holds. $\hfill\square$

By calculation, we easily find (3.10) also holds. Then, follow the process of
the proof for Lemma 3.4 (replacing the index $i\in\\{2,3,4\\}$ with
$B\in\\{2,\bar{3},\tilde{3}\\}$), we conclude that $u_{B}(f_{1})=0$, with
$B=2,\bar{3},\tilde{3}$.

Lemma 3.11  _We have_

$u_{B}(\Gamma_{D1}^{D})=u_{B}(\Gamma_{\tilde{3}1}^{\bar{3}})=u_{B}(\lambda_{2})=u_{B}(\gamma_{3})=u_{B}(\tau_{3})=0,$

_with $B,D=2,\bar{3},\tilde{3}$._

Proof Since $u_{B}(f_{1})=u_{B}(H)=0$, $B=2,\bar{3},\tilde{3}$, we conclude

$u_{B}u_{1}^{(k)}(f_{1})=u_{B}u_{1}^{(k)}(H)=0,\ B=2,\bar{3},\tilde{3},\
k=1,2,3.$

Observe the expressions of $f_{k}$, $k=2,3$ in (3.8), combining the above
equation, we have

$\displaystyle u_{B}(f_{k})=0,\ B=2,\bar{3},\tilde{3},\ k=1,2,3,$

which gives

$\displaystyle(\Gamma_{21}^{2})^{k-1}u_{B}(\Gamma_{21}^{2})+2b_{k-1}u_{B}(\Gamma_{\bar{3}1}^{\bar{3}})-2c_{k-1}u_{B}(\Gamma_{\tilde{3}1}^{\bar{3}})=0,k=1,2,3.$

Here, $b_{0}=1$ and $c_{0}=0$. The coefficient determinant of this system

$\displaystyle\left|\begin{array}[]{ccc}1&2&0\\\
\Gamma_{21}^{2}&2\Gamma_{\bar{3}1}^{\bar{3}}&-2\Gamma_{\tilde{3}1}^{\bar{3}}\\\
(\Gamma_{21}^{2})^{2}&2((\Gamma_{\bar{3}1}^{\bar{3}})^{2}-(\Gamma_{\tilde{3}1}^{\bar{3}})^{2})&-4\Gamma_{\bar{3}1}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}\end{array}\right|=-4\Gamma_{\tilde{3}1}^{\bar{3}}[(\Gamma_{21}^{2}-\Gamma_{\bar{3}1}^{\bar{3}})^{2}+(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}].$

As $\tau_{3}\neq 0$, the third equation of (3.75) implies that
$\Gamma_{\tilde{3}1}^{\bar{3}}\neq 0$. So, the above coefficient determinant
is not equal to zero, and then

$u_{B}(\Gamma_{21}^{2})=u_{B}(\Gamma_{\bar{3}1}^{\bar{3}})=u_{B}(\Gamma_{\tilde{3}1}^{\bar{3}})=0,\
B=2,\bar{3},\tilde{3}.$

Differentiate the equations in (3.75) along $u_{B}$, $B=2,\bar{3},\tilde{3}$,
combining the above equation, we get

$u_{B}(\lambda_{2})=u_{B}(\gamma_{3})=u_{B}(\tau_{3})=0,\
B=2,\bar{3},\tilde{3}.$

$\hfill\square$

Now, we continue the proof of Proposition 3.9.

Applying Lemma 3.11, we obtain from the equation (2.1) for
$(X,Y,Z)=(u_{B},u_{D},u_{B}),(u_{2},u_{\bar{3}},u_{\tilde{3}}),(u_{2},u_{\tilde{3}},u_{\bar{3}})$,
with $B,D=2,\bar{3},\tilde{3}$ that

$\begin{cases}(\gamma_{3}-\lambda_{2})\Gamma_{2\bar{3}}^{2}-\tau_{3}\Gamma_{2\tilde{3}}^{2}=0,\\\
(\gamma_{3}-\lambda_{2})\Gamma_{2\tilde{3}}^{2}+\tau_{3}\Gamma_{2\bar{3}}^{2}=0,\\\
-2\tau_{3}\Gamma_{2\tilde{3}}^{\bar{3}}=(\lambda_{2}-\gamma_{3})\Gamma_{\bar{3}2}^{\bar{3}}-\tau_{3}\Gamma_{\bar{3}2}^{\tilde{3}},\\\
2\tau_{3}\Gamma_{2\tilde{3}}^{\bar{3}}=(\lambda_{2}-\gamma_{3})\Gamma_{\tilde{3}2}^{\tilde{3}}+\tau_{3}\Gamma_{\tilde{3}2}^{\bar{3}},\\\
(\lambda_{2}-\gamma_{3})\Gamma_{\bar{3}2}^{\tilde{3}}+\tau_{3}\Gamma_{\bar{3}2}^{\bar{3}}=0,\\\
(\lambda_{2}-\gamma_{3})\Gamma_{\tilde{3}2}^{\bar{3}}-\tau_{3}\Gamma_{\tilde{3}2}^{\tilde{3}}=0,\\\
\Gamma_{\tilde{3}\bar{3}}^{\tilde{3}}=\Gamma_{\bar{3}\tilde{3}}^{\bar{3}}=0,\end{cases}$
(3.79)

which gives that

$\Gamma_{2\bar{3}}^{2}=\Gamma_{2\tilde{3}}^{2}=\Gamma_{\bar{3}\tilde{3}}^{\bar{3}}=\Gamma_{\tilde{3}\bar{3}}^{\tilde{3}}=\Gamma_{\bar{3}2}^{\bar{3}}+\Gamma_{\tilde{3}2}^{\tilde{3}}=\Gamma_{\tilde{3}2}^{\bar{3}}-\Gamma_{\bar{3}2}^{\tilde{3}}=0,$

and

$(\Gamma_{\bar{3}2}^{\bar{3}})^{2}+(\Gamma_{\bar{3}2}^{\tilde{3}})^{2}-2\Gamma_{\bar{3}2}^{\tilde{3}}\Gamma_{2\bar{3}}^{\tilde{3}}=0.$
(3.80)

Calculate $\langle R(u_{2},u_{B})u_{2},u_{B}\rangle$, $\langle
R(u_{2},u_{B})u_{2},u_{D}\rangle$, $\langle R(u_{B},u_{D})u_{B},u_{D}\rangle$
and $\langle R(u_{2},u_{B})u_{D},u_{1}\rangle$ by Gauss equation, with
$B,D=\bar{3},\tilde{3}$ and $B\neq D$, combining (3.73) and the above two
equations, we have

$\Gamma_{\bar{3}2}^{\tilde{3}}(\Gamma_{21}^{2}-\Gamma_{\bar{3}1}^{\bar{3}})+\Gamma_{\bar{3}2}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}=0,$
(3.81)

and

$\begin{cases}-\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{\tilde{3}1}^{\bar{3}}+2\varepsilon_{2}\Gamma_{\bar{3}2}^{\bar{3}}\Gamma_{2\bar{3}}^{\tilde{3}}=\varepsilon\lambda_{2}\tau_{3},\\\
\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{\bar{3}1}^{\bar{3}}+2\varepsilon_{2}\Gamma_{\bar{3}2}^{\tilde{3}}\Gamma_{2\bar{3}}^{\tilde{3}}=-c-\varepsilon\lambda_{2}\gamma_{3},\\\
\varepsilon_{1}[(\Gamma_{\bar{3}1}^{\bar{3}})^{2}+(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}]-4\varepsilon_{2}\Gamma_{\bar{3}2}^{\tilde{3}}\Gamma_{2\bar{3}}^{\tilde{3}}=-c-\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2}).\end{cases}$
(3.82)

The second and third equations of (3.82) implies

$\varepsilon_{1}[(\Gamma_{\bar{3}1}^{\bar{3}})^{2}+(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}+2\Gamma_{21}^{2}\Gamma_{\bar{3}1}^{\bar{3}}]=-3c-2\varepsilon\lambda_{2}\gamma_{3}-\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2}).$
(3.83)

_Case 1: $\Gamma_{\bar{3}2}^{\tilde{3}}\neq 0$ at some point in $U_{p}$._

We suppose $\Gamma_{\bar{3}2}^{\tilde{3}}\neq 0$ on a neighbourhood
$U_{q}\subset U_{p}$, and work on $U_{q}$. From the fifth equation in (3.79)
and (3.81), we get

$\frac{\Gamma_{21}^{2}-\Gamma_{\bar{3}1}^{\bar{3}}}{\Gamma_{\tilde{3}1}^{\bar{3}}}=\frac{\lambda_{2}-\gamma_{3}}{\tau_{3}}.$

Considering Lemma 3.11, we conclude from the above equation that there exists
two smooth functions $\mu$ and $\nu$, with $u_{B}(\mu)=u_{B}(\nu)=0$,
$B=2,\bar{3},\tilde{3}$, such that

$\Gamma_{21}^{2}=\mu\lambda_{2}+\nu,\
\Gamma_{\bar{3}1}^{\bar{3}}=\mu\gamma_{3}+\nu,\
\Gamma_{\tilde{3}1}^{\bar{3}}=\mu\tau_{3}.$ (3.84)

The equation (3.22) also holds. By use of (3.74) and (3.75), differentiating
the equations in (3.84) along $u_{1}$, it gives (3.23). Substitute (3.84) into
(3.83), we obtain

$(\varepsilon_{1}\mu^{2}+\varepsilon)(\gamma_{3}^{2}+2\lambda_{2}\gamma_{3}+\tau_{3}^{2})+2\varepsilon_{1}\mu\nu(\lambda_{2}+2\gamma_{3})+3\varepsilon_{1}\nu^{2}+3c=0.$
(3.85)

Since

$\displaystyle\gamma_{3}^{2}+2\lambda_{2}\gamma_{3}+\tau_{3}^{2}$
$\displaystyle=\frac{1}{2}[(\lambda_{2}+2\gamma_{3})^{2}-(\lambda_{2}^{2}+2\gamma_{3}^{2}-2\tau_{3}^{2})]$
$\displaystyle=20H^{2}-\frac{1}{2}\text{tr}A^{2},$

the equation (3.85) can be rewritten as (3.24). Put (3.84) into (3.78),
combining (3.22) and (3.24), we have (3.25). Follow the process for the case
$\Gamma_{23}^{4}\neq 0$ at some point in section 3.1, just correcting (3.7)
with (3.72), one derive a contradiction.

_Case 2: $\Gamma_{\bar{3}2}^{\tilde{3}}\equiv 0$ on $U_{p}$._

It follows from (3.80) that $\Gamma_{\bar{3}2}^{\bar{3}}=0$. And then, (3.82)
can be simplified into

$\begin{cases}\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{\tilde{3}1}^{\bar{3}}=-\varepsilon\lambda_{2}\tau_{3},\\\
\varepsilon_{1}\Gamma_{21}^{2}\Gamma_{\bar{3}1}^{\bar{3}}=-c-\varepsilon\lambda_{2}\gamma_{3},\\\
\varepsilon_{1}[(\Gamma_{\bar{3}1}^{\bar{3}})^{2}+(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}]=-c-\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2}).\end{cases}$
(3.86)

_Subcase (1): $\Gamma_{21}^{2}\equiv 0$ on $U_{p}$._

The first and second equations of (3.86) tells us that $\lambda_{2}=0$ and
$c+\varepsilon\lambda_{2}\gamma_{3}=0$. If $c\neq 0$, then
$c+\varepsilon\lambda_{2}\gamma_{3}\neq 0$, a contradiction. In the following,
we suppose $c=0$. Differentiate both sides of $\gamma_{3}=3\varepsilon H$
along $u_{1}$, using (3.74), we get

$3\varepsilon u_{1}(H)=(-2\varepsilon
H-\gamma_{3})\Gamma_{\bar{3}1}^{\bar{3}}+\tau_{3}\Gamma_{\tilde{3}1}^{\bar{3}}.$
(3.87)

Applying (3.86) and (3.87), we can rewrite (3.72) as

$\displaystyle 4[-5\varepsilon
H((\Gamma_{\bar{3}1}^{\bar{3}})^{2}-(\Gamma_{\tilde{3}1}^{\bar{3}})^{2})+2\tau_{3}\Gamma_{\bar{3}1}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}]-34\varepsilon_{1}H\tau_{3}^{2}+162\varepsilon_{1}H^{3}-9\varepsilon\varepsilon_{1}\lambda
H=0.$ (3.88)

By differentiating (3.88) along $u_{1}$, using (3.74) and (3.75), and
combining (3.86) and (3.88), we derive

$\displaystyle
K_{1}\Gamma_{\bar{3}1}^{\bar{3}}+K_{2}\Gamma_{\tilde{3}1}^{\bar{3}}=0,$ (3.89)

where

$\begin{cases}K_{1}=1710H^{3}-78H\tau_{3}^{2}-135\varepsilon\lambda H,\\\
K_{2}=1494H^{2}\tau_{3}\varepsilon-78\tau_{3}^{3}\varepsilon-9\tau_{3}\lambda.\end{cases}$

As (3.86), we get from (3.89) that

$\begin{cases}(K_{1}^{2}+K_{2}^{2})\Gamma_{\bar{3}1}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}-\varepsilon\varepsilon_{1}(9H^{2}+\tau_{3}^{2})K_{1}K_{2}=0,\\\
K_{1}K_{2}[(\Gamma_{\bar{3}1}^{\bar{3}})^{2}-(\Gamma_{\tilde{3}1}^{\bar{3}})^{2}]+(K_{2}^{2}-K_{1}^{2})\Gamma_{\bar{3}1}^{\bar{3}}\Gamma_{\tilde{3}1}^{\bar{3}}=0.\end{cases}$

Eliminate $\Gamma_{\bar{3}1}^{\bar{3}}$ and $\Gamma_{\tilde{3}1}^{\bar{3}}$
from (3.88) and the above equations, one derives

$\displaystyle
2741856\varepsilon\tau_{3}^{12}+c_{1}\tau_{3}^{10}+c_{2}\tau_{3}^{8}+c_{3}\tau_{3}^{6}+c_{4}\tau_{3}^{4}+c_{5}\tau_{3}^{2}+c_{6}=0,$
(3.90)

where $c_{i}$, $1\leq i\leq 6$, are polynomials about $H$. Acting on (3.90) by
$u_{1}$, we have

$\displaystyle
L_{1}\Gamma_{\bar{3}1}^{\bar{3}}+L_{2}\Gamma_{\tilde{3}1}^{\bar{3}}=0,$

where $L_{1}$ and $L_{2}$ are polynomials about $H$ and $\tau_{3}$. It follows
from (3.89) and the above equation that

$K_{1}L_{2}-K_{2}L_{1}=0.$

Eliminating $\tau_{3}$ from (3.90) and the above equation, we get a polynomial
equation of degree $92$ for $H$, which implies $H$ is a constant, a
contradiction.

_Subcase (2): $\Gamma_{21}^{2}\neq 0$ at some point in $U_{p}$._

We suppose $\Gamma_{21}^{2}\neq 0$ at $q\in U_{p}$. If
$c+\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2})=0$, then (3.86) implies
$\Gamma_{\bar{3}1}^{\bar{3}}=\Gamma_{\tilde{3}1}^{\bar{3}}=0$ and
$\lambda_{2}=c=\gamma_{3}=\tau_{3}=0$, a contradiction. So,
$c+\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2})\neq 0$. We derive from the
equations in (3.86) that

$\displaystyle(\Gamma_{21}^{2})^{2}=-\varepsilon_{1}\frac{\lambda_{2}^{2}(\gamma_{3}^{2}+\tau_{3}^{2})+c^{2}+2c\varepsilon\lambda_{2}\gamma_{3}}{c+\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2})}.$
(3.91)

Let

$\mu=2\lambda_{2}\gamma_{3}+\gamma_{3}^{2}+\tau_{3}^{2},\ \
\nu=\lambda_{2}(\gamma_{3}^{2}+\tau_{3}^{2}),$

then take derivatives of $\mu$, $\nu$ and $H$ along $u_{1}$, by use of (3.74),
(3.86) and (3.91), we get

$\begin{cases}6\varepsilon u_{1}(H)=K(12\varepsilon Hc^{2}+10H\mu
c-3\varepsilon\nu c+48H^{2}\nu-2\mu\nu),\\\
u_{1}(\mu)=K[(24H^{2}+2\mu)c^{2}+(2\mu^{2}+12\varepsilon
H^{2}\mu-6H\nu)c-3\nu^{2}+10\varepsilon H\mu\nu],\\\
u_{1}(\nu)=K[(2\varepsilon H\mu+3\nu)c^{2}+(24\varepsilon
H^{2}\nu+2\varepsilon\mu\nu)c+12\varepsilon H\nu^{2}],\end{cases}$ (3.92)

where
$K=\frac{\varepsilon_{1}}{\Gamma_{21}^{2}[c+\varepsilon(\gamma_{3}^{2}+\tau_{3}^{2})]}$.
Applying (3.86), (3.91) and (3.92), we can rewrite (3.72) as (3.37). Follow
the process for the subcase
$\Gamma_{21}^{2},\Gamma_{\bar{3}1}^{\bar{3}},\Gamma_{\tilde{3}1}^{\tilde{3}}\neq
0$ of the case $\Gamma_{2\bar{3}}^{\tilde{3}}=0$ in section 3.1, and replace
the system (3.36) with (3.92), the equation (3.42) with

$\displaystyle-\varepsilon_{1}(c+(\gamma_{3}^{2}+\tau_{3}^{2}))^{2}(\Gamma_{21}^{2})^{2}$
$\displaystyle=(c+(\gamma_{3}^{2}+\tau_{3}^{2}))(c^{2}+\lambda_{2}^{2}(\gamma_{3}^{2}+\tau_{3}^{2})+2c\lambda_{2}\gamma_{3})$
$\displaystyle=\nu^{2}+6cH\nu+c^{2}\mu+c^{3}=0,$

we deduce a contradiction. $\hfill\square$

Remark 3.12 For form (V) of the shape operator, by constructing the terms
$b_{k}$ and $c_{k}$, $k=1,2,\cdots$, we find there are many equations
similarly with the equations for the form (I) hold, by only changing some
terms for the form (I) into the terms about $b_{k}$ and $c_{k}$ for the form
(V) as follows.

$\begin{array}[]{|c|c|}\hline\cr\textbf{The terms for form (I)}&\textbf{The
terms for form (V)}\\\
\hline\cr(\Gamma_{31}^{3})^{k}+(\Gamma_{41}^{4})^{k}&2b_{k}\\\
\lambda_{3}(\Gamma_{31}^{3})^{k}+\lambda_{4}(\Gamma_{41}^{4})^{k}&2\gamma_{3}b_{k}-2\tau_{3}c_{k}\\\
\lambda_{3}^{2}(\Gamma_{31}^{3})^{k}+\lambda_{4}^{2}(\Gamma_{41}^{4})^{k}&2(\gamma_{3}^{2}-\tau_{3}^{2})b_{k}-4\gamma_{3}\tau_{3}c_{k-1}\\\
\hline\cr\end{array}$

We believe that the construction of $b_{k}$ and $c_{k}$, $k=1,2,\cdots$, will
provide a new insight for us to study hypersurfaces with imaginary principal
curvatures.

###  3.6 The shape operator has the form (VI)

Proposition 3.13  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field. Suppose that the shape
operator $A$ of $M^{4}_{r}$ has the form (VI), then $M^{4}_{r}$ has constant
mean curvature._

Proof Assume that $H$ is not a constant, then the eigenvector $\nabla H$ of
$A$ is in the direction $u_{1_{2}}$ and $\lambda_{1}=-2\varepsilon H$. As
$2\lambda_{1}+2\gamma_{2}=4\varepsilon$, we know $\gamma_{2}=4\varepsilon H$.
Let $u_{\bar{2}}=u_{\bar{2}_{1}}$, $u_{\tilde{2}}=u_{\tilde{2}_{1}}$ and
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$,
$B,C=1_{1},1_{2},\bar{2},\tilde{2}$, then

$\Gamma_{D1_{1}}^{1_{2}}=\Gamma_{D\bar{2}}^{\bar{2}}=\Gamma_{D\tilde{2}}^{\tilde{2}}=0,$

and

$\displaystyle\Gamma_{D1_{1}}^{1_{1}}=-\Gamma_{D1_{2}}^{1_{2}},\
\Gamma_{D\bar{2}}^{\tilde{2}}=\Gamma_{D\tilde{2}}^{\bar{2}},\
\Gamma_{D1_{a}}^{\bar{2}}=-\varepsilon_{1}\Gamma_{D\bar{2}}^{1_{3-a}},\
\Gamma_{D1_{a}}^{\tilde{2}}=\varepsilon_{1}\Gamma_{D\tilde{2}}^{1_{3-a}},\ \
a=1,2.$

It follows that

$u_{1_{1}}(H)\neq 0,\ u_{1_{2}}(H)=u_{\bar{2}}(H)=u_{\tilde{2}}(H)=0.$

and

$\Gamma^{1_{1}}_{BC}=\Gamma^{1_{1}}_{CB},\quad B,C\neq 1_{1}.$

From Codazzi equation (2.1), with
$(X,Y,Z)=(u_{1_{1}},u_{B},u_{1_{2}}),(u_{\bar{2}},u_{\tilde{2}},u_{1_{2}}),\\\
(u_{1_{2}},u_{B},u_{B}),(u_{1_{2}},u_{\bar{2}},u_{\tilde{2}})$,
$B=\bar{2},\tilde{2}$, we deduce

$\Gamma_{\bar{2}\bar{2}}^{1_{1}}=-\Gamma_{\tilde{2}\tilde{2}}^{1_{1}},\ \
\Gamma_{1_{1}\bar{2}}^{1_{1}}=\Gamma_{1_{1}\tilde{2}}^{1_{1}}=\Gamma_{1_{2}\bar{2}}^{\tilde{2}}=\Gamma_{1_{2}\tilde{2}}^{\bar{2}}=0,$

and

$\begin{cases}-6\varepsilon
H\Gamma_{\bar{2}1_{2}}^{\bar{2}}+\tau_{2}\Gamma_{\tilde{2}1_{2}}^{\bar{2}}=0,\\\
u_{1_{2}}(\tau_{1})=-6\varepsilon
H\Gamma_{\tilde{2}1_{2}}^{\bar{2}}-\tau_{2}\Gamma_{\bar{2}1_{2}}^{\bar{2}}.\end{cases}$
(3.93)

Using Gauss equation for $\langle
R(u_{1_{2}},u_{\bar{2}})u_{\bar{2}},u_{1_{2}}\rangle$ and $\langle
R(u_{1_{2}},u_{\bar{2}})u_{\tilde{2}},u_{1_{2}}\rangle$, combining the above
equations, it gives

$\begin{cases}u_{1_{2}}(\Gamma_{\bar{2}1_{2}}^{\bar{2}})=\Gamma^{1_{2}}_{1_{2}1_{2}}\Gamma_{\bar{2}1_{2}}^{\bar{2}}+(\Gamma_{\tilde{2}1_{2}}^{\bar{2}})^{2}-(\Gamma_{\bar{2}1_{2}}^{\bar{2}})^{2},\\\
u_{1_{2}}(\Gamma_{\tilde{2}1_{2}}^{\bar{2}})=\Gamma^{1_{2}}_{1_{2}1_{2}}\Gamma_{\tilde{2}1_{2}}^{\bar{2}}-2\Gamma_{\bar{2}1_{2}}^{\bar{2}}\Gamma_{\tilde{2}1_{2}}^{\bar{2}}.\end{cases}$
(3.94)

Applying (3.93) and (3.94), differentiate the first equation in (3.93) along
$u_{1_{2}}$, we obtain

$-6\varepsilon
H(\Gamma_{\tilde{2}1_{2}}^{\bar{2}})^{2}-\tau_{2}\Gamma_{\bar{2}1_{2}}^{\bar{2}}\Gamma_{\tilde{2}1_{2}}^{\bar{2}}=0,$

which together with the first equation of (3.93) implies that
$\Gamma_{\tilde{2}1_{2}}^{\bar{2}}=0$ and
$H\Gamma_{\bar{2}1_{2}}^{\bar{2}}=0$. Compute $\langle
R(u_{1_{1}},u_{\bar{2}})u_{1_{2}},u_{\tilde{2}}\rangle$ by Gauss equation, we
have

$\Gamma_{\bar{2}1_{2}}^{\bar{2}}\Gamma_{1_{1}\bar{2}}^{\tilde{2}}=2H\tau_{2}.$

Multiply both sides of the above equation by $H$, combining
$H\Gamma_{\bar{2}1_{2}}^{\bar{2}}=0$, we conclude that $H=0$, a contradiction.
$\hfill\square$

##  4 Estimates for the constant mean curvature

In this section, as an application of the conclusion that $M^{4}_{r}$ has
constant mean curvature, we will estimate the value of the mean curvature $H$.
When $M^{4}_{r}$ has one or two distinct principal curvatures, we will compute
the value of $H$. It’s a pity that we can not get the value of $H$ when
$M^{4}_{r}$ has three or four distinct principal curvatures. But for the case
that the principal curvatures are all real, we give a value range of the
constant mean curvature $H$.

###  4.1 When $M^{4}_{r}$ has one simple principal curvature

Theorem 4.1  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field, satisfying
$\Delta\vec{H}=\lambda\vec{H}$, $\vec{\xi}$ be a unit normal vector field to
$M^{4}_{r}$, with $\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$.
Suppose that $M^{4}_{r}$ has one simple principal curvature, then_

  * 1.

_when_ $\varepsilon\lambda\leq 0$, we have $H=0$;

  * 2.

_when_ $\varepsilon\lambda>0$, we have $H=0$ or
$H^{2}=\frac{\varepsilon\lambda}{4}$.

Proof Recall subsection 2.3, the imaginary principal curvatures of $M^{4}_{r}$
appear in conjugate pairs. So, under the assumption that $M^{4}_{r}$ has one
simple principal curvature $\mu$, $\mu$ must be real. Then, we know $t=m$ (the
signs $t,m$ see subsection 2.3) and $\lambda_{1}=\cdots=\lambda_{m}=\mu$.
Thus, equations (2.4) can be simplified as $\text{tr}A^{2}=4\mu^{2}$ and
$\mu=\varepsilon H$. According to Theorem 3.1, the mean curvature $H$ of
$M^{4}_{r}$ is a constant. Using this result and $\text{tr}A^{2}=4\mu^{2}$, we
get from (2.3) that $H=0$ or $4\mu^{2}=\varepsilon\lambda$.

When $\varepsilon\lambda\leq 0$, we assume $H\neq 0$. Then
$4\mu^{2}=\varepsilon\lambda\leq 0$, a contradiction.

When $\varepsilon\lambda>0$, suppose that $H\neq 0$, we also have
$4\mu^{2}=\varepsilon\lambda$, which together with $\mu=\varepsilon H$ implies
that

$H^{2}=\frac{\varepsilon\lambda}{4}.$

$\hfill\square$

Remark 4.2 The result of Theorem 4.1 for $c=0$ has been gotten in [16].

###  4.2 When $M^{4}_{r}$ has two distinct principal curvatures

Lemma 4.3  _Let $M^{4}_{r}$ be a non-minimal hypersurface of $N^{5}_{s}(c)$
with proper mean curvature vector field, $\vec{\xi}$ be a unit normal vector
field to $M^{4}_{r}$, with $\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm
1$. Suppose that $M^{4}_{r}$ has two distinct principal curvatures $\mu$ and
$\nu$, then_

$c+\varepsilon\mu\nu=0.$

Proof Since the number of imaginary principal curvatures is even, these two
distinct principal curvatures $\mu$ and $\nu$ are all real or all imaginary.

_Case 1: $\mu$ and $\nu$ are all real._

Let $l$ is the multiply of $\mu$, then considering the mean curvature $H$ is a
constant and $H\neq 0$, we get from (2.3) and (2.4) that

$\begin{cases}l\mu+(4-l)\nu=4\varepsilon H,\\\
l\mu^{2}+(4-l)\nu^{2}=\varepsilon\lambda,\end{cases}$ (4.1)

which implies $\mu$ and $\nu$ are all constants. In the following, by use of
that the principal curvatures are all constant, we discuss separately the
possible forms (I), (II), (III) and (IV) of $A$ for this case, and derive
the result $c+\varepsilon\mu\nu=0$.

_Subcase (1): the shape operator has the form (I)._

Without loss of generality, we suppose $\lambda_{1}=\mu$ and
$\lambda_{4}=\nu$. Let $u_{i}=u_{i_{1}}$ and
$\nabla_{u_{i}}u_{j}=\Gamma_{ij}^{k}u_{k},i,j=1,2,3,4$, then recall subsection
3.1, we know (3.1) and (3.5) also hold. Since the principal curvatures
$\lambda_{i}$, with $i=1,2,3,4$, are constants, the equations (3.1) and (3.5)
imply that $\Gamma_{ij}^{i}=0$ if $\lambda_{i}\neq\lambda_{j}$,
$\Gamma_{i4}^{1}=\Gamma_{i1}^{4}=\Gamma_{1i}^{4}=0$ if $\lambda_{i}=\mu$, and
$\Gamma_{41}^{i}=\Gamma_{i1}^{4}=0$ if $\lambda_{i}=\nu$. Using Gauss equation
for $\langle R(u_{1},u_{4})u_{1},u_{4}\rangle$, we have

$c+\varepsilon\mu\nu=0.$

_Subcase (2): the shape operator has the form (II)._

We can suppose $\lambda_{1}=\mu$ and $\lambda_{3}=\nu$. Denote
$u_{2}=u_{2_{1}}$, $u_{3}=u_{3_{1}}$, and let
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},2,3$, then (3.43)
and (3.44) hold. If $\lambda_{2}=\nu$, considering that $\mu$ and $\nu$ are
constants, we conclude from the Codazzi equation (2.1) that

$\Gamma_{1_{1}3}^{1_{1}}=\Gamma_{31_{2}}^{3}=\Gamma_{31_{1}}^{3}=\Gamma_{1_{2}3}^{1_{1}}=\Gamma_{1_{1}2}^{1_{1}}=\Gamma_{21_{2}}^{3}=\Gamma_{31_{2}}^{2}=0.$

Applying Gauss equation for $\langle
R(u_{1_{1}},u_{3})u_{1_{2}},u_{3}\rangle$, combining the above equations, we
have $c+\varepsilon\mu\nu=0.$

If $\lambda_{2}=\mu$, then we deduce from the Codazzi equation that

$\Gamma_{1_{1}3}^{1_{1}}=\Gamma_{1_{2}3}^{1_{2}}=\Gamma_{1_{2}3}^{1_{1}}=\Gamma_{31_{2}}^{3}=\Gamma_{31_{1}}^{3}=\Gamma_{32}^{3}=\Gamma_{23}^{2}=\Gamma_{23}^{1_{1}}=\Gamma_{1_{2}3}^{2}=0,$

and

$\Gamma_{21_{1}}^{3}=\Gamma_{1_{1}2}^{3}.$

Using Gauss equation for $\langle R(u_{1_{1}},u_{3})u_{1_{2}},u_{3}\rangle$
and $\langle R(u_{2},u_{3})u_{2},u_{3}\rangle$, combining the above two
equations, we have

$\begin{cases}\Gamma_{31_{2}}^{2}\Gamma_{1_{1}2}^{3}=-c\varepsilon_{1}-\varepsilon\varepsilon_{1}\mu\nu,\\\
-2\Gamma_{31_{2}}^{2}\Gamma_{1_{1}2}^{3}=-c\varepsilon_{1}-\varepsilon\varepsilon_{1}\mu\nu,\end{cases}$

which implies $\Gamma_{31_{2}}^{2}\Gamma_{1_{1}2}^{3}=0,$ and
$c+\varepsilon\mu\nu=0.$

_Subcase (3): the shape operator has the form (III)._

Suppose $\lambda_{1}=\mu$ and $\lambda_{2}=\nu$. Let
$\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},2_{1},2_{2}$. It
follows from Codazzi equation that

$\Gamma_{2_{1}1_{2}}^{2_{1}}=\Gamma_{1_{1}2_{2}}^{1_{1}}=\Gamma_{1_{2}2_{2}}^{1_{1}}=0.$

Using Gauss equation for $\langle
R(u_{1_{1}},u_{2_{1}})u_{1_{2}},u_{2_{2}}\rangle$, combining the above
equation, we have $c+\varepsilon\mu\nu=0.$

_Subcase (4): the shape operator has the form (IV)._

We can suppose $\lambda_{1}=\mu$ and $\lambda_{2}=\nu$. Let $u_{2}=u_{2_{1}}$
and $\nabla_{u_{B}}u_{C}=\Gamma^{D}_{BC}u_{D}$, $B,C=1_{1},1_{2},1_{3},2$,
then (3.66) and (3.67) hold, and we have from Codazzi equation that

$\Gamma_{1_{1}2}^{1_{1}}=\Gamma_{1_{3}2}^{1_{3}},\ \
\Gamma_{1_{1}2}^{1_{1}}+\Gamma_{1_{2}2}^{1_{2}}+\Gamma_{1_{3}2}^{1_{3}}=0,$
(4.2)
$\Gamma_{21_{1}}^{2}=\Gamma_{21_{2}}^{2}=\Gamma_{21_{3}}^{2}=\Gamma_{1_{2}2}^{1_{1}}=\Gamma_{1_{3}2}^{1_{1}}=\Gamma_{1_{3}2}^{1_{2}}=0,$
(4.3) $(\nu-\mu)\Gamma_{1_{3}2}^{1_{3}}=-\Gamma_{21_{3}}^{1_{2}},$ (4.4)

and

$(\mu-\nu)\Gamma_{1_{1}1_{2}}^{2}+\Gamma_{1_{1}1_{3}}^{2}=(\mu-\nu)\Gamma_{1_{2}1_{1}}^{2}+\Gamma_{1_{2}1_{2}}^{2}.$
(4.5)

From Gauss equation for $\langle R(u_{1_{1}},u_{2})u_{1_{3}},u_{2}\rangle$ and
$\langle R(u_{1_{3}},u_{2})u_{1_{1}},u_{2}\rangle$, using (4.3), we have

$\begin{cases}-u_{2}(\Gamma_{1_{1}3}^{2})+\Gamma_{21_{3}}^{1_{2}}\Gamma_{1_{1}1_{2}}^{2}-\Gamma_{1_{1}2}^{1_{1}}\Gamma_{1_{1}1_{3}}^{2}=-c\varepsilon_{1}-\varepsilon\varepsilon_{1}\mu\nu,\\\
-u_{2}(\Gamma_{1_{3}1_{1}}^{2})+\Gamma_{21_{3}}^{1_{2}}\Gamma_{1_{2}1_{1}}^{2}-\Gamma_{1_{3}2}^{1_{3}}\Gamma_{1_{3}1_{1}}^{2}=-c\varepsilon_{1}-\varepsilon\varepsilon_{1}\mu\nu,\end{cases}$
(4.6)

which together with (4.2) implies that

$\Gamma_{21_{3}}^{1_{2}}=0\ \ \text{or}\ \
\Gamma_{1_{1}1_{2}}^{2}=\Gamma_{1_{2}1_{1}}^{2}.$

If $\Gamma_{21_{3}}^{1_{2}}=0$, then (4.4) tells us that
$\Gamma_{1_{3}2}^{1_{3}}=0$. If
$\Gamma_{1_{1}1_{2}}^{2}=\Gamma_{1_{2}1_{1}}^{2}$, then from (4.2), (4.4) and
(4.5), we also have

$\Gamma_{1_{3}2}^{1_{3}}=\Gamma_{21_{3}}^{1_{2}}=0.$

So, we derive from (4.6) that $c+\varepsilon\mu\nu=0.$

_Case 2: $\mu$ and $\nu$ are all imaginary._

We can suppose $\mu=\gamma+\tau\sqrt{-1}$ and $\nu=\gamma-\tau\sqrt{-1}$.
Since $H$ is a nonzero constant, the equations (2.3) and (2.4) gives that

$\begin{cases}\gamma=\varepsilon H,\\\
4(\gamma^{2}-\tau^{2})=\varepsilon\lambda.\end{cases}$

So, $\gamma$ and $\tau$ are all constants. In the following, we treat the
possible forms (VII) and (VIII) of $A$, and derive the result that
$c+\varepsilon\mu\nu=0$.

_Subcase (1): the shape operator has the form (VII)._

It follows that $\nu_{1}=\nu$ and $\tau_{1}=\tau$. Let

$\displaystyle\nabla_{\bar{u}_{1_{a}}}\bar{u}_{1_{b}}=\Gamma_{\bar{1}_{a}\bar{1}_{b}}^{\bar{1}_{d}}\bar{u}_{1_{d}}+\Gamma_{\bar{1}_{a}\bar{1}_{b}}^{\tilde{1}_{d}}\bar{v}_{1_{d}},\
\nabla_{\bar{v}_{1_{a}}}\bar{v}_{1_{b}}=\Gamma_{\tilde{1}_{a}\tilde{1}_{b}}^{\bar{1}_{d}}\bar{u}_{1_{d}}+\Gamma_{\tilde{1}_{a}\tilde{1}_{b}}^{\tilde{1}_{d}}\bar{v}_{1_{d}},$
$\displaystyle\nabla_{\bar{u}_{1_{a}}}\bar{v}_{1_{b}}=\Gamma_{\bar{1}_{a}\tilde{1}_{b}}^{\bar{1}_{d}}\bar{u}_{1_{d}}+\Gamma_{\bar{1}_{a}\tilde{1}_{b}}^{\tilde{1}_{d}}\bar{v}_{1_{d}},\
\nabla_{\bar{v}_{1_{a}}}\bar{u}_{1_{b}}=\Gamma_{\tilde{1}_{a}\bar{1}_{b}}^{\bar{1}_{d}}\bar{u}_{1_{d}}+\Gamma_{\tilde{1}_{a}\bar{1}_{b}}^{\tilde{1}_{d}}\bar{v}_{1_{d}},$

with $a,b=1,2$, we get from compatibility condition that

$\Gamma_{B\bar{1}_{a}}^{\bar{1}_{b}}=-\Gamma_{B\bar{1}_{3-b}}^{\bar{1}_{3-a}},\
\Gamma_{B\tilde{1}_{a}}^{\tilde{1}_{b}}=-\Gamma_{B\tilde{1}_{3-b}}^{\tilde{1}_{3-a}},\
\Gamma_{B\bar{1}_{a}}^{\tilde{1}_{b}}=\Gamma_{B\tilde{1}_{3-b}}^{\bar{1}_{3-a}}.$

Since $\gamma$ and $\tau$ are constants, we deduce from Codazzi equation that

$\Gamma_{\bar{1}_{2}\tilde{1}_{2}}^{\bar{1}_{1}}=\Gamma_{\bar{1}_{2}\bar{1}_{2}}^{\tilde{1}_{1}}=\Gamma_{\tilde{1}_{2}\tilde{1}_{2}}^{\bar{1}_{1}}=\Gamma_{\tilde{1}_{2}\bar{1}_{2}}^{\tilde{1}_{1}}=0,$
(4.7)

and

$\begin{cases}-\Gamma_{\tilde{1}_{2}\tilde{1}_{1}}^{\tilde{1}_{1}}+\Gamma_{\tilde{1}_{2}\bar{1}_{1}}^{\bar{1}_{1}}=0,\\\
\Gamma_{\bar{1}_{2}\bar{1}_{2}}^{\tilde{1}_{2}}+\Gamma_{\bar{1}_{2}\tilde{1}_{2}}^{\bar{1}_{2}}=-\Gamma_{\tilde{1}_{2}\tilde{1}_{2}}^{\tilde{1}_{2}}+\Gamma_{\tilde{1}_{2}\bar{1}_{2}}^{\bar{1}_{2}},\\\
\Gamma_{\bar{1}_{1}\tilde{1}_{2}}^{\bar{1}_{1}}+\Gamma_{\bar{1}_{1}\bar{1}_{2}}^{\tilde{1}_{1}}=\Gamma_{\bar{1}_{2}\tilde{1}_{1}}^{\bar{1}_{1}}+\Gamma_{\bar{1}_{2}\bar{1}_{1}}^{\tilde{1}_{1}},\\\
-\Gamma_{\bar{1}_{2}\tilde{1}_{1}}^{\tilde{1}_{1}}+\Gamma_{\bar{1}_{2}\bar{1}_{1}}^{\bar{1}_{1}}=0,\\\
\Gamma_{\bar{1}_{2}\bar{1}_{2}}^{\bar{1}_{2}}-\Gamma_{\bar{1}_{2}\tilde{1}_{2}}^{\tilde{1}_{2}}=-\Gamma_{\tilde{1}_{2}\tilde{1}_{2}}^{\bar{1}_{2}}-\Gamma_{\tilde{1}_{2}\bar{1}_{2}}^{\tilde{1}_{2}},\\\
\Gamma_{\tilde{1}_{1}\bar{1}_{2}}^{\tilde{1}_{1}}+\Gamma_{\tilde{1}_{1}\tilde{1}_{2}}^{\bar{1}_{1}}=\Gamma_{\tilde{1}_{2}\bar{1}_{1}}^{\tilde{1}_{1}}+\Gamma_{\tilde{1}_{2}\tilde{1}_{1}}^{\bar{1}_{1}},\end{cases}$

which reduces to

$\Gamma_{\bar{1}_{1}\bar{1}_{2}}^{\tilde{1}_{1}}=\Gamma_{\bar{1}_{1}\tilde{1}_{2}}^{\bar{1}_{1}}=\Gamma_{\tilde{1}_{1}\bar{1}_{2}}^{\tilde{1}_{1}}=\Gamma_{\tilde{1}_{1}\tilde{1}_{2}}^{\bar{1}_{1}}=0.$
(4.8)

Using Gauss equation for $\langle
R(\bar{u}_{1_{1}},\bar{v}_{1_{1}})\bar{u}_{1_{2}},\bar{v}_{1_{2}}\rangle$,
combining (4.7) and (4.8), we have

$c+\varepsilon(\gamma^{2}+\tau^{2})=0,$

i.e. $c+\varepsilon\mu\nu=0$.

_Subcase (2): the shape operator has the form (VIII)._

In this case, we have $\gamma_{1}=\gamma_{2}=\gamma$,
$\tau_{1}=\tau_{2}=\tau$. Denote $\bar{u}_{1}=\bar{u}_{1_{1}}$,
$\bar{u}_{2}=\bar{u}_{2_{1}}$ and $\bar{v}_{1}=\bar{v}_{1_{1}}$ and
$\bar{v}_{2}=\bar{v}_{2_{1}}$. Let

$\displaystyle\nabla_{\bar{u}_{a}}\bar{u}_{b}=\Gamma_{\bar{a}\bar{b}}^{\bar{d}}\bar{u}_{d}+\Gamma_{\bar{a}\bar{b}}^{\tilde{d}}\bar{v}_{d},\
\nabla_{\bar{v}_{a}}\bar{v}_{b}=\Gamma_{\tilde{a}\tilde{b}}^{\bar{d}}\bar{u}_{d}+\Gamma_{\tilde{a}\tilde{b}}^{\tilde{d}}\bar{v}_{d},$
$\displaystyle\nabla_{\bar{u}_{a}}\bar{v}_{b}=\Gamma_{\bar{a}\tilde{b}}^{\bar{d}}\bar{u}_{d}+\Gamma_{\bar{a}\tilde{b}}^{\tilde{d}}\bar{v}_{d},\
\nabla_{\bar{v}_{a}}\bar{u}_{b}=\Gamma_{\tilde{a}\bar{b}}^{\bar{d}}\bar{u}_{d}+\Gamma_{\tilde{a}\bar{b}}^{\tilde{d}}\bar{v}_{d},$

with $a,b=1,2$, we obtain from compatibility condition that

$\Gamma_{B\bar{a}}^{\bar{b}}=-\Gamma_{B\bar{b}}^{\bar{a}},\
\Gamma_{B\tilde{a}}^{\tilde{b}}=-\Gamma_{B\tilde{b}}^{\tilde{a}},\
\Gamma_{B\bar{a}}^{\tilde{b}}=\Gamma_{B\tilde{b}}^{\bar{a}},\ \ a,b=1,2.$

The Codazzi equation gives that

$\Gamma_{\bar{1}\tilde{1}}^{\bar{1}}=\Gamma_{\tilde{1}\bar{1}}^{\tilde{1}}=\Gamma_{\bar{2}\bar{1}}^{\tilde{1}}=\Gamma_{\tilde{2}\bar{1}}^{\tilde{1}}=0,$

and

$\Gamma_{\bar{1}\tilde{2}}^{\bar{1}}+\Gamma_{\bar{1}\bar{2}}^{\tilde{1}}=\Gamma_{\bar{1}\bar{2}}^{\bar{1}}-\Gamma_{\bar{1}\tilde{2}}^{\tilde{1}}=\Gamma_{\tilde{1}\tilde{2}}^{\bar{1}}+\Gamma_{\tilde{1}\bar{2}}^{\tilde{1}}=\Gamma_{\tilde{1}{\tilde{2}}}^{\tilde{1}}-\Gamma_{\tilde{1}\bar{2}}^{\bar{1}}=0.$

Applying Gauss equation for $\langle
R(\bar{u}_{1},\bar{v}_{1})\bar{u}_{1},\bar{v}_{1}\rangle$, combining the above
two equations, we conclude $c+\varepsilon(\gamma^{2}+\tau^{2})=0,$ i.e.
$c+\varepsilon\mu\nu=0$. $\hfill\square$

Remark 4.4 The result of Lemma 4.3 that $c+\varepsilon\mu\nu=0$ coincides with
the basic identity of Cartan in [14, Theorem 2.9] for the isoparametric
hypersurface $M^{n}_{r}$ of $N^{n+1}_{s}(c)(c=0,-1,1)$ with two distinct
principal curvatures $\mu$ and $\nu$, and algebraic and geometric
multiplicities of $\mu$ or $\nu$ coincide.

By use of Lemma 4.3, we can give the value of the mean curvature $H$,
according to the principal curvatures are real or imaginary. The following
Lemma 4.5 will be used for the case that the principal curvatures are real.

Lemma 4.5  _Let $c$, $\lambda$, $l$, $\mu$ and $\nu$ are real constants, and
the equations_

$\begin{cases}\mu\nu=-c\varepsilon,\\\
l\mu^{2}+(4-l)\nu^{2}=\varepsilon\lambda\end{cases}$

_hold, where $\varepsilon=\pm 1$. Then, $\varepsilon\lambda=-4c\varepsilon\geq
0$ is the necessary condition such that $\mu=\nu$._

Proof If $\mu=\nu$, then

$-c\varepsilon=\mu^{2}\geq 0,$

and

$\varepsilon\lambda=-4c\varepsilon.$

$\hfill\square$

Theorem 4.6  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field, satisfying
$\Delta\vec{H}=\lambda\vec{H}$, $\vec{\xi}$ be a unit normal vector field to
$M^{4}_{r}$, with $\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$.
Suppose that $M^{4}_{r}$ has two distinct real principal curvatures $\mu$ and
$\nu$, and $l=1,\ 2\ \text{or}\ 3$ is the multiply of $\mu$, then_

  * 1.

_When_ $c=0$, we have

    * (a)

_if_ $\varepsilon\lambda\leq 0$, then $H=0$;

    * (b)

_if_ $\varepsilon\lambda>0$, then $H=0$, or
$H^{2}=\frac{l\varepsilon\lambda}{16}$, or
$H^{2}=\frac{(4-l)\varepsilon\lambda}{16}$;

  * 2.

_When_ $c\neq 0$, we have

    * (a)

_if_ $\varepsilon\lambda<2\sqrt{l(4-l)}|c|$, then $H=0$;

    * (b)

_if_ $\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$, then $H=0$ or

$H^{2}=\frac{1}{16}[2\varepsilon\lambda\pm(l-2)\sqrt{\lambda^{2}-4l(4-l)c^{2}}-2l(4-l)c\varepsilon].$

_Specially, if_ $\varepsilon\lambda=-4c\varepsilon>0$, then $H=0$ or
$H^{2}=-\frac{3}{4}c\varepsilon$.

Proof (i) When $c=0$, suppose $H\neq 0$, then (4.1) holds, and
$\varepsilon\lambda>0.$ It follows from Lemma 4.3 that $\mu\nu=0$, which
together with the second equation in (4.1) gives that

$\mu=0,\ \nu^{2}=\frac{\varepsilon\lambda}{4-l},\ \ \text{or}\
\mu^{2}=\frac{\varepsilon\lambda}{l},\ \nu=0.$

Since the above equation, we obtain from the first equation in (4.1) that

$H^{2}=\frac{l\varepsilon\lambda}{16},\ \ \text{or}\
H^{2}=\frac{(4-l)\varepsilon\lambda}{16}.$

(ii) For the case $c\neq 0$, suppose $H\neq 0$, then (4.1) holds, and
$\mu\nu=-c\varepsilon$, obtained from Lemma 4.3. And then, we have

$\begin{cases}\varepsilon\lambda=\sqrt{l}\mu^{2}+\sqrt{4-l}\nu^{2}>0,\\\
\varepsilon\lambda+2\sqrt{l(4-l)}c\varepsilon=(\sqrt{l}\mu-\sqrt{4-l}\nu)^{2}\geq
0,\\\
\varepsilon\lambda-2\sqrt{l(4-l)}c\varepsilon=(\sqrt{l}\mu+\sqrt{4-l}\nu)^{2}\geq
0,\end{cases}$

which implies $\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$. So, if
$\varepsilon\lambda<2\sqrt{l(4-l)}|c|$, then $H=0$.

When $\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$, if $H\neq 0$, then we
calculate $\mu^{2}$ and $\nu^{2}$ from $\mu\nu=-4c\varepsilon$ and the second
equation of (4.1) that

$\mu^{2}=\frac{\varepsilon\lambda\pm\sqrt{\lambda^{2}-4l(4-l)c^{2}}}{2l},\ \
\nu^{2}=\frac{\varepsilon\lambda\mp\sqrt{\lambda^{2}-4l(4-l)c^{2}}}{2(4-l)}.$

Together with the first equation of (4.1) and the above equation, we have

$H^{2}=\frac{2\varepsilon\lambda\pm(l-2)\sqrt{\lambda^{2}-4l(4-l)c^{2}}-2l(4-l)c\varepsilon}{16}.$

However, some values of $\mu^{2}$, $\nu^{2}$ and $H^{2}$ in the above maybe
contradict with the condition that $\mu\neq\nu$. From Lemma 4.5, considering
$\varepsilon\lambda\neq 0$, we know $\varepsilon\lambda=-4c\varepsilon>0$ is
the necessary condition of $\mu=\nu$ (a contradiction). It’s necessary for us
to discuss the values of $\mu^{2}$, $\nu^{2}$ and $H^{2}$ for the case
$\varepsilon\lambda=-4c\varepsilon>0$, which is in the range
$\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$. By calculation, when
$\varepsilon\lambda=-4c\varepsilon>0$, we find

(1) $\mu^{2}=\nu^{2}=H^{2}=-c\varepsilon$ for $l=2$;

(2) $\mu^{2}=-3c\varepsilon$, $\nu^{2}=-\frac{c\varepsilon}{3}$,
$H^{2}=-\frac{3}{4}c\varepsilon$, or $\mu^{2}=\nu^{2}=H^{2}=-c\varepsilon$,
for $l=1$;

(3) $\nu^{2}=-3c\varepsilon$, $\mu^{2}=-\frac{c\varepsilon}{3}$,
$H^{2}=-\frac{3}{4}c\varepsilon$, or $\mu^{2}=\nu^{2}=H^{2}=-c\varepsilon$,
for $l=3$.
As $\mu\nu>0$, $\mu^{2}=\nu^{2}$ is equivalent to $\mu=\nu$ (a contradiction).
Thus, we conclude that $H^{2}=-\frac{3}{4}c\varepsilon$ for the case
$\varepsilon\lambda=-4c\varepsilon>0$.

In conclusion, when $c\neq 0$ and $\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$,
we have $H=0$ or

$H^{2}=\frac{2\varepsilon\lambda\pm(l-2)\sqrt{\lambda^{2}-4l(4-l)c^{2}}-2l(4-l)c\varepsilon}{16}.$
(4.9)

Specially, if $\varepsilon\lambda=-4c\varepsilon>0$, then $H=0$ or
$H^{2}=-\frac{3}{4}c\varepsilon$.

By the way, we emphasize that the right side values of (4.9) are all greater
than or equal to zero. When $l=2$, it is obviously. When $l=1,3$, since
$\varepsilon\lambda\geq 2\sqrt{3}|c|$, it follows that

$\varepsilon\lambda-3|c|\geq(2\sqrt{3}-3)|c|\geq 0,$

which together with

$\frac{(2\varepsilon\lambda-6c\varepsilon)^{2}}{\lambda^{2}-12c^{2}}=1+\frac{3(\lambda-4c)^{2}}{\lambda^{2}-12c^{2}}\geq
1$

gives that $2\varepsilon\lambda-6c\varepsilon\geq\sqrt{\lambda^{2}-12c^{2}}$.
So, when $l=1,3$, the right side values

$\frac{2\varepsilon\lambda\pm\sqrt{\lambda^{2}-12c^{2}}-6c\varepsilon}{16}\geq
0.$

$\hfill\square$

Remark 4.7 When $c=0$ and the algebraic and geometric multiplicities of $\mu$
or $\nu$ coincide, we has gotten in [16] the values of $H^{2}$, which is agree
with the result of Theorem 4.6 for $c=0$.

Theorem 4.8  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ with proper mean curvature vector field, satisfying
$\Delta\vec{H}=\lambda\vec{H}$, $\vec{\xi}$ be a unit normal vector field to
$M^{4}_{r}$, with $\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$.
Suppose that $M^{4}_{r}$ has two imaginary principal curvatures, we have_

  * 1.

_If_ $c\varepsilon\geq 0$, then $H=0$;

  * 2.

_If_ $c\varepsilon<0$ and $|\varepsilon\lambda|\geq-4c\varepsilon$, then
$H=0$;

  * 3.

_If_ $c\varepsilon<0$ and $|\varepsilon\lambda|<-4c\varepsilon$, then $H=0$ or
$H^{2}=\frac{\varepsilon\lambda-4c\varepsilon}{8}$.

Proof Suppose that $M^{4}_{r}$ has two imaginary principal curvatures
$\gamma+\tau i$ and $\gamma-\tau i$, we get from Lemma 4.3 and its proof that
if $H\neq 0$, then $\gamma=\varepsilon H$ and

$\begin{cases}4(\gamma^{2}-\tau^{2})=\varepsilon\lambda,\\\
\gamma^{2}+\tau^{2}=-c\varepsilon.\end{cases}$ (4.10)

When $c\varepsilon\geq 0$, we assume that $H\neq 0$, then
$\gamma^{2}+\tau^{2}=-c\varepsilon\leq 0$, a contradiction.

When $c\varepsilon<0$ and $|\varepsilon\lambda|\geq-4c\varepsilon$, we suppose
$H\neq 0$, then the equations (4.10) implies that

$\begin{cases}-4c\varepsilon-\varepsilon\lambda=8\tau^{2}>0,\\\
\varepsilon\lambda-4c\varepsilon=8\nu^{2}>0,\end{cases}$

which contradicts with $|\varepsilon\lambda|\geq-4c\varepsilon$.

For the case $c\varepsilon<0$ and $|\varepsilon\lambda|<-4c\varepsilon$, if
$H\neq 0$, we also have $8\gamma^{2}=\varepsilon\lambda-4c\varepsilon$, which
combining $\gamma=\varepsilon H$ tells us that

$H^{2}=\frac{\varepsilon\lambda-4c\varepsilon}{8}.$

$\hfill\square$

Corollary 4.9  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$\mathbb{E}^{5}_{s}$ with proper mean curvature vector field. Suppose that
$M^{4}_{r}$ has two principal curvatures, which are imaginary, then
$M^{4}_{r}$ is minimal._

Remark 4.10 In [16], we also estimated the value of the mean curvature $H$ for
the hypersurface $M^{4}_{r}$ in $\mathbb{E}^{5}_{s}$ satisfying
$\Delta\vec{H}=\lambda\vec{H}$ and with two distinct imaginary principal
curvatures, but just gave a value range that $H=0$ or
$H^{2}>\frac{\varepsilon\lambda}{4}$. Clearly, the result $H=0$ of Corollary
4.9 improves the result in [16] to the greatest extent.

In non-flat space form $N^{5}_{s}(c)\ (c\neq 0)$, under the assumption that
the hypersurface $M^{4}_{r}$ has two imaginary principal curvatures, we have
from Theorem 4.8 that the mean curvature $H$ is not necessarily zero. However,
if the hypersurface $M^{4}_{r}$ is biharmonic, i.e. $\lambda=4c$, then Theorem
4.8 yields that the mean curvature is zero, which give a partial affirmative
answer to Chen’s conjecture.

Corollary 4.11  _Let $M^{4}_{r}$ be a biharmonic hypersurface of
$N^{5}_{s}(c)$. Suppose that $M^{4}_{r}$ has two principal curvatures, which
are imaginary, then $M^{4}_{r}$ is minimal._

###  4.3 When $M^{4}_{r}$ has not imaginary principal curvatures

Theorem 4.12  _Let $M^{4}_{r}$ be a nondegenerate hypersurface of
$N^{5}_{s}(c)$ satisfying $\Delta\vec{H}=\lambda\vec{H}$ ($\lambda$ a
constant), and $\vec{\xi}$ be a unit normal vector field to $M^{4}_{r}$, with
$\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$. Suppose that
$M^{4}_{r}$ has not imaginary principal curvatures, we have_

  * 1.

_If_ $\varepsilon\lambda\leq 0$, then $H=0$;

  * 2.

_If_ $\varepsilon\lambda>0$, then $H^{2}\leq\frac{\varepsilon\lambda}{4}$, the
equal sign holds if and only if the principal curvatures are all equal and
nonzero.

Proof Since $M^{4}_{r}$ has not imaginary principal curvatures, the equations
(2.4) reduce to

$\text{tr}A^{2}=\sum_{i=1}^{m}\alpha_{i}\lambda_{i}^{2},\ \ 4\varepsilon
H=\sum_{i=1}^{m}\alpha_{i}\lambda_{i}.$

Note that $\sum_{i=1}^{m}\alpha_{i}=4$. Because of that the mean curvature $H$
is a constant, combining the above equations, we have from (2.3) that $H=0$ or
$\sum_{i=1}^{m}\alpha_{i}\lambda_{i}^{2}=\varepsilon\lambda$.

When $\varepsilon\lambda\leq 0$, we assume that $H\neq 0$, then

$\sum_{i=1}^{m}\alpha_{i}\lambda_{i}^{2}=\varepsilon\lambda\leq 0,$

which implies $\lambda_{i}=0$, $i=1,\cdots,m$. And then, $4\varepsilon
H=\sum_{i=1}^{m}\alpha_{i}\lambda_{i}=0$, a contradiction.

For the case $\varepsilon\lambda>0$, if $H\neq 0$, using Cauchy inequality, we
know

$(\sum_{i=1}^{m}\alpha_{i}\lambda_{i})^{2}\leq
4\sum_{i=1}^{m}\alpha_{i}\lambda_{i}^{2},$ (4.11)

where the equal sign holds if and only if
$\lambda_{1}=\lambda_{2}=\cdots=\lambda_{m}\neq 0$. The equation (4.11) yields
that

$0<H^{2}\leq\frac{\varepsilon\lambda}{4}.$

The equal sign of the above equation is true if and only if the principal
curvatures are all equal and nonzero. $\hfill\square$

Corollary 4.13  _Let $M^{4}_{r}$ be a biharmonic hypersurface of
$N^{5}_{s}(c)$, $\vec{\xi}$ be a unit normal vector field to $M^{4}_{r}$, with
$\varepsilon=\langle\vec{\xi},\vec{\xi}\rangle=\pm 1$. If $\varepsilon c\leq
0$ and $M^{4}_{r}$ has not imaginary principal curvatures, then $M^{4}_{r}$ is
minimal._

Remark 4.14 We claim that the value
$\frac{1}{16}[2\varepsilon\lambda\pm(l-2)\sqrt{\lambda^{2}-4l(4-l)c^{2}}-2l(4-l)c\varepsilon]$
of $H^{2}$ for the case $c\neq 0$, $\varepsilon\lambda\geq 2\sqrt{l(4-l)}|c|$
and $\varepsilon\lambda\neq-4c\varepsilon$ in Theorem 4.6 is in the range
$H^{2}<\frac{\varepsilon\lambda}{4}$. In fact, in this case, we can easily
check that $\varepsilon\lambda+l(4-l)c\varepsilon\geq 0$ and

$(2\varepsilon\lambda+2l(4-l)c\varepsilon)^{2}-(l-2)^{2}(\lambda^{2}-4l(4-l)c^{2})=l(4-l)(\lambda+4c)^{2}>0.$

Thus, it follows that

$2\varepsilon\lambda+2l(4-l)c\varepsilon>\pm(l-2)\sqrt{\lambda^{2}-4l(4-l)c^{2}},$

which deduce our claim.

Remark 4.15 When $M^{4}_{r}$ has imaginary principal curvature, we have from
(2.4) that

$\text{tr}A^{2}=\sum_{i=1}^{t}\alpha_{i}\lambda_{i}^{2}+2\sum_{j=t+1}^{m}\beta_{j}(\nu_{j}^{2}-\tau_{j}^{2}),\
\ t<m.$

Since there exists negative terms in the right side of the above equation, we
can not get the range value of $H$ by using Cauchy inequality, for the case
that the number of distinct principal curvatures is larger than 2.

Acknowledgements This work was supported by the National Natural Science
Foundation of China (Nos. 12161078, 11761061), the Foundation for
Distinguished Young Scholars of Gansu Province (No. 20JR5RA515), and the
Project of Northwest Normal University (No. NWNU-LKQN2019-23).

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