Yaqi Guo
School of Science, Chongqing University of Technology, Chongqing, China.
DOI: 10.4236/ojapps.2026.165102   PDF    HTML   XML   20 Downloads   100 Views   Citations

Abstract

This paper focuses on the geometric rigidity of 3-dimensional proper biconservative submanifolds $M_r^3$ with a parallel normalized mean curvature vector field in the 5-dimensional pseudo-Euclidean space $E_s^5$ . Grounded in the theories of harmonic maps, biharmonic maps, and stress-energy tensors, we employ fundamental tools from pseudo-Riemannian geometry including the Gauss equation, Weingarten formula, and Codazzi equation to derive the equivalent conditions characterizing biconservative submanifolds. We further establish that if the shape operator of such a submanifold is diagonalizable and admits at most two distinct principal curvatures in the direction of the mean curvature vector field $H$ , its scalar curvature takes the form of a fractional-power polynomial in the mean curvature $f$ with non-vanishing coefficients. In particular, any such submanifold with constant scalar curvature necessarily possesses constant mean curvature, revealing a striking rigidity property.

Keywords

Pseudo-Euclidean Space Form, Principal Curvatures, Proper Biconservative, Scalar Curvature

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1. Introduction

Let $M_r^m$ and $N_q^{m+p}$ be pseudo-Riemannian manifolds with indexs $r$ and $q$ $\left(q\ge r\ge 0\right)$ , respectively, and we consider a smooth map $\varphi :M_r^m\to N_q^{m+p}$ . The energy functional is defined by

$E\left(\varphi \right)={\displaystyle {\int }_{M_r^m}e\left(\varphi \right)v_g},$

where $e\left(\varphi \right)=\frac{1}{2}{\left|d\left(\varphi \right)\right|}^2$ is called the energy density of $\varphi$ . Critical points of

$E\left(\varphi \right)$ are called harmonic maps. The theory of harmonic maps has been applied to various fields in differential geometry, we refer to [1] [2] for a review.

The study of biharmonic submanifolds began in the mid-1980s, proposed by Chen in his research on finite type submanifolds in Euclidean and pseudo-Euclidean spaces [3]. Meanwhile, In [4] and [5], Jiang introduced the concept of $k$ -harmonic mapping proposed by Eells and Sampson in [6] to study the double harmonic isometric immersion between Riemannian manifolds. Harmonic mappings are characterized by the vanishing of the tension field as the criterion, corresponding to the critical points of the energy functional, and describe the smoothest mapping relationship between manifolds [6], while biharmonic mappings are defined by the vanishing of the biharmonic field and are the critical points of the bienergy functional [7].

In order to understand the geometric characteristics of biharmonic systems, some geometers have begun to focus on studying doubly conservative submanifolds [8]-[11]. For example, the general notion of biconservative submanifolds was introduced in [8]. Aslo, the complete classification of biconservative hypersurfaces in Euclidean spaces with three distinct principal curvatures is obtained by the second named author in [10].

The stress-energy tensor was intitated by G. Y. Jiang in [12] and afterwards developed by E. Loubeau, S. Montaldo and C. Oniciuc in [13], defining the stress-energy tensor $S_2$ with

$div{S}_2=-\langle {\tau }_2\left(\phi \right),d\phi \rangle$

Moreover, Jiang [12] demonstrated that $\phi$ is biharmonic if and only if it satisfies the Euler-Lagrange equation related to the dual energy functional, i.e.,${\tau }_2\left(\phi \right)=0$ . where ${\tau }_2\left(\phi \right)$ is the bitension field of $\phi$ defined by

${\tau }_2\left(\phi \right)=\text{Δ}\tau \left(\phi \right)-tr\tilde{R}\left(d\phi ,\tau \left(\phi \right)\right)d\phi ,$

where Δ is the Rough-Laplacian. If the condition $\langle {\tau }_2\left(\phi \right),d\phi \rangle =0$ is satisfied, then $\phi$ is called a biconservative mapping. Pseudo-Riemannian manifolds are generalizations of Riemannian manifolds. The condition for a Riemannian submanifold to be biconservative can be used to obtain the condition for a pseudo-Riemannian submanifold to be bi-conservative as follows (cf [8] [14])

$trace{A}_{D(.)H(.)}+\frac{n}{4}grad\langle H,H\rangle =0,$

where $D$ is the normal connection, $A_{\xi }$ is the shape operator with respect to the normal vector field $\xi$ and $H$ is the mean curvature vector filed of $M_r^m$ . In [11], authors studied geometrical properties of PNMCV surfaces of $E^4$ and proved that a biharmonic PNMCV surface in $E^4$ is minimal. Motivated by above paragraphs, in this paper, we will study 3-dimensional biconservative of $E_s^5$ with parallel normal mean curvture vector (PNMCV). we prove

Theorem 1.1. If a proper biconservative submanifold $M_r^3$ $\left(\Vert gradf\ne 0\Vert \right)$ in $E_s^5$ has a diagonalizable shape operator and at most two distinct principal curvatures in the direction of $H$ , then its scalar curvature is a power-law polynomial in the mean curvature $f$ with non-zero coefficients. Furthermore, the expression for the scalar curvature is derived from the Gauss equation and the results concerning the shape operator.

Theorem 1.2. Let $M_r^3$ be a biconservative submanifold with constant scalar curvature in the pseudo-Euclidean space $E_s^5$ , possessing a parallel normaliazed mean curvature vector field. If, in the direction of $f$ , it has at most two distinct principal curvatures and its shape operator is diagonalizable, then $M_r^3$ necessarily has constant mean curvature.

Remark 1.3. It is easy to see that a PNMC submanifold is CMC if and only if it is PMC. As a corollary of Theorem 1.3, let $M_r^3$ be a biconservative pnmc submanifold with constant scalar curvature in $E_s^5$ . Then $M_r^3$ is PMC provided that it has diagonalizable shape operator with two distinct principal curvatures in the direction of $f$ .

Remark 1.4. Let $M_r^3$ be a biconservative pnmc submanifold with constant scalar curvature in the pseudo-Euclidean space from $E_s^5$ . Assume that $M_r^3$ has two distinct principal curvatures in the direction of $f$ . As an immediate consequence of Theorem 1.3, we know that $M_r^3$ has constant mean curvature.

2. Preliminaries

Let $E_s^n$ denote the pseudo-Euclidean n-space with the metric tensor $\tilde{g}$ given by

$\tilde{g}=\langle \dots \rangle =-{\displaystyle \sum }_{i=1}^{s}d{x}_i\otimes d{x}_i+{\displaystyle \sum }_{j=s+1}^{n}d{x}_j\otimes d{x}_j,$ (1)

Let $\phi :M_r^3\to E_s^5$ be an isometric immersion of an 3-dimensional pseudo-Riemannian manifold $M_r^3$ into a pseudo-Euclidean 5-space. Denote the Levi-Civita connections of $M_r^3$ and $E_s^3$ by $\nabla$ and $\tilde{\nabla }$ , respectively. Then the Gauss and Weingarten formulae are given by (cf [15] [16])

${\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+h\left(X,Y\right),$ (2)

and

${\tilde{\nabla }}_X\xi =-A_{\xi }X+{\nabla }_X^{\perp }\xi ,$ (3)

respectively, for any vectors $X$ , $Y$ tangent to $M_r^3$ and $\xi$ normal to $M_r^3$ , where $h$ and $A_{\xi }$ are the second fundamental form and the shape of $M_r^3$ along the normal direction $\xi$ , respectively and ${\nabla }^{\perp }$ is the normal connection, It is well known that $h$ and $A_{\xi }$ are related by

$\langle h\left(X,Y\right),\xi \rangle =\langle A_{\xi }X,Y\rangle .$ (4)

If $R$ and $\tilde{R}$ stand for the curvature tensor of $M_r^3$ and $E_s^3$ respectively,

then, the Codazzi equation ${\left(\tilde{R}\left(X,Y\right)Z\right)}^{\perp }=0$ and the Gauss equation

${\left(\tilde{R}\left(X,Y\right)Z\right)}^{\top }=0$ become

$\left({\tilde{\nabla }}_{X}h\right)\left(Y,Z\right)=\left({\tilde{\nabla }}_{X}h\right)\left(Y,Z\right).$ (5)

where $\tilde{\nabla }h$ is defined by

$\left({\tilde{\nabla }}_{X}h\right)\left(Y,Z\right)={\nabla }_X^{\perp }h\left(Y,Z\right)-h\left({\nabla }_{X}Y,Z\right)-h\left(Y,{\nabla }_{X}Z\right).$ (6)

$R\left(X,Y\right)Z,W)=R\left(X,Y,Z,W\right)=\langle h\left(Y,Z\right),h\left(X,W\right)\rangle -\langle h\left(X,Z\right),h\left(Y,W\right)\rangle .$ (7)

where

$R\left(X,Y\right)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{\left[X,Y\right]}Z.$ (8)

Let $e_1,e_2,e_3,e_4,e_5$ be a pseudo-Euclidean orthonormal field on $E_s^3$ such that $e_1,e_2,e_3$ are tangent to $M_r^3$ and $e_4,e_5$ are normal to $M_r^3$ , and we denote the connection forms corresponding to this fram field by ${\omega }_{ij}$ . Then, we have

${\nabla }_{e_i}{e}_j={\displaystyle \sum }_{k=1}^5{\epsilon }_k{\omega }_{ij}\left(e_k\right),1\le i,j\le 5,{\omega }_{ii}=0,{\omega }_{ij}+{\omega }_{ji}=0,\left(i\ne j\right).$ (9)

The mean curvature vector $f$ of $M_r^3$ is defined by

$H=\frac{1}{3}\text{tr}h=\frac{1}{3}{\displaystyle \sum }_{i=1}^3{\epsilon }_{i}h\left(e_i,e_i\right)=\frac{1}{3}{\displaystyle \sum }_4^5{\epsilon }_4\text{tr}{A}_4{e}_4,$ (10)

where $A_4=A_{e_4}$ . The mean curvature $f$ of $M_r^3$ in $E_s^5$ is expressed as $f={\langle H,H\rangle }^{1/2}$ .

At a point $p\in M$ , a 2-dimensional linear subspae $\pi$ of the tangent space $T_{p}M$ is called a plane section. For a given basis $\nu$ , $\omega$ of the palne section $\pi$ , we define a real number by

$Q\left(\nu ,\omega \right)=\langle \nu ,\nu \rangle \langle \omega ,\omega \rangle -{\langle \nu ,\omega \rangle }^2.$ (11)

The plane section $\pi$ is called nondegenerate if and only if $Q\left(\nu ,\omega \right)\ne 0$ .

$Q\left(\nu ,\omega \right)$ is positive when ${g|}_{\pi }$ is definite, and is negative when ${g|}_{\pi }$ is indefinte.

The absolute value $\Vert Q\left(\mu ,\nu \right)\Vert$ is the square of the area of the parallelogram with

sides $\mu$ and $\nu$ .

For a nondegenerate plane setion $\pi$ at $p$ , the number

$K\left(\mu ,\nu \right)=\frac{\langle R\left(\mu ,\nu \right)\nu ,\mu \rangle }{Q\left(\mu ,\nu \right)}.$ (12)

is independent of the choice of basis $\nu$ , $\omega$ for $\pi$ , which is called the setional curvature $K\left(\pi \right)$ of $\pi$ .

3. Some Key Lemmas

According to [17]-[19]:

Lemma 2.1 Let $\phi :M_r^3\to E_s^5$ be an isometric immersion of an 3-dimensional pseudo-Riemannian manifold $M_r^3$ into a pseudo-Euclidean space. $\phi$ is bicon-servative if and only if the equation

$m\nabla \langle H,H\rangle +4{\displaystyle \sum }_{i=1}^m{\epsilon }_i{A}_{{\nabla }_{e_i}^{\perp }H}\left(e_i\right)=0.$ (13)

is satisfied, where $m$ is the imension of $M$ . By Lemma 2.1, we can obtain

Lemma 2.2 (cf [18]) Let $M_r^3$ be a submanifolds with parallel normalized mean curvature vector field in $E_s^5$ . Then $M_r^3$ is biconservative if and only if the equation holds:

$A_{e_4}\left(\nabla f\right)=\frac{-3{\epsilon }_{4}f}{2}\left(\nabla f\right),$ (14)

where

$\nabla f={\displaystyle \sum }_i^3{\epsilon }_i{e}_i\left(f\right)e_i.$ (15)

Proof. Essentially this lemma is a special case of Lemma 2.2 in [18] We can choose a pesudo-Riemannian orthonormal frame field $e_1,e_2,e_3,e_4,e_5$ , such that

$e_1$ is parallel to $\nabla f$ , $e_2,e_3$ are tangent to $M_r^3$ , $e_4=\frac{H}{f}$ , $e_5$ are normal to

$M_r^3$ , then

$H=f{e}_4.$ (16)

Note that ${\nabla }_{e_i}^{\perp }{e}_4={\nabla }_{e_i}^{\perp }\left(\frac{H}{f}\right)=0$ , $i=1,2,3$ , then it follows form (16) we have

${\nabla }_{e_i}^{\perp }H={\nabla }_{e_i}^{\perp }\left(f{e}_4\right)=e_i\left(f\right)e_4,$ (17)

which together with (15) and (16) we have

${\displaystyle \sum }_{i=1}^3{\epsilon }_i{A}_{{\nabla }_{e_i}^{\perp }H}\left(e_i\right)=A_4\left(\nabla f\right).$ (18)

using (16), we can obtain

$\nabla \langle H,H\rangle =\nabla \langle f{e}_4,f{e}_4\rangle ={\epsilon }_4\nabla f^2=2{\epsilon }_{4}f\nabla f$ (19)

Combining with (13), (18) and (19) we can obtain $A_{e_4}\left(\nabla f\right)=\frac{-3{\epsilon }_{4}f}{2}\left(\nabla f\right)$ .

According to [17]-[20]:

Lemma 2.3 Assum that $\phi$ has parallel normalized mean curvature vector $e_4$ .

In this case, the Ricci equation ${\left(\tilde{R}\left(X,Y\right)\xi \right)}^{\top }=0$ yields that all the shape

operators of $\phi$ can be diagonalized simulataneously (see [21]). Therefore, by abusung the terminology, we are going to call $X$ as principal direction of $\phi$ , if $A_{e_4}X=\lambda X$ , where the smooth function $\lambda$ is going to be called as the corrrs-ponding principal curvature. Note that there exist an orthonormal fram field $e_1,e_2,e_3,e_4,e_5$ such that

$A_{e_4}=diag\left({\lambda }_1,{\lambda }_2,{\lambda }_3\right),A_{e_5}=diag\left({\mu }_1,{\mu }_2,{\mu }_3\right).$ (20)

for some smooth fuctions ${\lambda }_i$ , ${\mu }_i$ satisfying ${\lambda }_1+{\lambda }_2+{\lambda }_3=3{\epsilon }_{4}f$ and

${\mu }_1+{\mu }_2+{\mu }_3=0$ . We are going to call a biconservative PNMCV immersion as proper if $\Vert \nabla f\Vert$ does not vanish. Assum that $\phi$ is proper biconservative

PNMCV immersion. By calculating ${\nabla }_{e_i}^{\perp }\langle e_4,e_4\rangle =0$ and ${\nabla }_{e_i}^{\perp }\langle e_4,e_5\rangle =0$ , where

$i=1,2,3$ . we obtain

${\nabla }_{e_i}^{\perp }{e}_4={\nabla }_{e_i}^{\perp }{e}_5=0$ (21)

where $e_5$ is a unit normal vector field orthogonal to $e_4$ . If $e_1$ is chosen to be proportional to $\Vert \nabla f\Vert$ , then (14) implies

$e_1\left(f\right)\ne 0,e_2\left(f\right)=e_3\left(f\right)=0.$ (22)

and ${\lambda }_1=\frac{-3{\epsilon }_{4}f}{2}$ .

Lemma 2.4 Let $\phi :M_r^3\to E_s^5$ be an isometric immersion with two distinct principal curvatures, if $\phi$ is proper biconservative PNMCV then there exists an orthonormal frame field $e_1,e_2,e_3,e_4,e_5$ such that the shape operators $A_{e_4}$ has the form

$A_{e_4}=\left(\begin{array}{ccc}\frac{-3{\epsilon }_{4}f}{2}& & \\ & \frac{9{\epsilon }_{4}f}{4}& \\ & & \frac{9{\epsilon }_{4}f}{4}\end{array}\right)$ (23)

and $A_{e_5}$ has the form

(I) $A_{e_5}=0$ , or (24)

(II) $A_{e_5}=\left(\begin{array}{ccc}0& & \\ & g_2{f}^{\frac{3}{5}}& \\ & & -g_3{f}^{\frac{3}{5}}\end{array}\right)$ (25)

where $g_2$ and $g_3$ are nonzero constants with $g_2+g_3=0$ , or.

(III) $A_{e_5}=\left(\begin{array}{ccc}{c}_1{f}^{\frac{9}{5}}& & \\ & -\frac{c_1}{2}{f}^{\frac{9}{5}}+g_2{f}^{\frac{3}{5}}& \\ & & -\frac{c_1}{2}{f}^{\frac{9}{5}}-g_3{f}^{\frac{3}{5}}\end{array}\right)$ (26)

for $c_1$ and $g_2$ , $g_3$ are nonzero constants with $g_2+g_3=0$ .

Proof. We have from (10) and (16) that

$f=\frac{{\epsilon }_4}{3}trace{A}_{e_4},trae{A}_{e_5}=0.$ (27)

When $M_r^3$ has the same principal curvatures in the direction of $H$ ,

${\lambda }_1={\lambda }_2={\lambda }_3=\frac{-3{\epsilon }_{4}f}{2}$ , from (27), it follows that

$\left(2+3\right)f=0,$ (28)

which shows $f=0$ , it is contradition. There for we are going to consider the case ${\lambda }_1\ne {\lambda }_2={\lambda }_3$ . from (20) and (22) we have ${\lambda }_2$ , ${\lambda }_3$ satisfy

${\lambda }_2={\lambda }_3=\frac{9{\epsilon }_{4}f}{4}.$ (29)

Now, we start to derive the explicit expressions of the shape operator $A_{e_5}$ of $M_r^3$ Combing with (4) and (20) yields, for any $i,j=1,2,3$

$h\left(e_i,e_j\right)={\displaystyle \sum }_{\beta =4}^5{\epsilon }_{\beta }\langle h\left(e_i,e_j\right),e_{\beta }\rangle e_{\beta }={\epsilon }_4\langle e_i,e_j\rangle {\lambda }_i{e}_4+{\epsilon }_5\langle e_i,e_j\rangle {\mu }_i{e}_5$ (30)

which means that

$\begin{array}{l}h\left(e_1,e_2\right)=0,h\left(e_1,e_3\right)=0,h\left(e_1,e_1\right)={\epsilon }_4{\epsilon }_1{\lambda }_1{e}_4+{\epsilon }_5{\epsilon }_1{\mu }_1{e}_5,\\ h\left(e_2,e_2\right)={\epsilon }_4{\epsilon }_2{\lambda }_2{e}_4+{\epsilon }_5{\epsilon }_1{\mu }_2{e}_5,\\ h\left(e_3,e_3\right)={\epsilon }_4{\epsilon }_1{\lambda }_3{e}_4+{\epsilon }_5{\epsilon }_1{\mu }_3{e}_5.\end{array}$

Calculating $\left(\tilde{\nabla }{e}_{1}h\right)\left(e_A,e_1\right)=\left(\tilde{\nabla }{e}_{A}h\right)\left(e_1,e_1\right)$ , for $A=2,3$ . Uising (6), (9), (22), (29), (30), we get

${\epsilon }_5{\epsilon }_1{e}_A\left({\mu }_1\right)e_5={\omega }_{1A}\left(e_1\right){\epsilon }_4\left({\lambda }_1-{\lambda }_A\right)e_4+{\epsilon }_5\left({\mu }_1-{\mu }_A\right)e_5.$ (31)

from (22) and (29) we have

${\epsilon }_5{\epsilon }_1{e}_A\left({\mu }_1\right)e_5={\omega }_{1A}\left(e_1\right){\epsilon }_4\left(-\frac{15{\epsilon }_{4}f}{4}\right)e_4+{\epsilon }_5\left({\mu }_1-{\mu }_A\right)e_5.$ (32)

Furthermore, we have

${\epsilon }_5\left({\epsilon }_1{e}_A\left({\mu }_1\right)-{\omega }_{1A}\left(e_1\right)\left({\mu }_1-{\mu }_A\right)\right)e_5=-\frac{15{\epsilon }_4}{4}{\omega }_{1A}\left(e_1\right)f{e}_4.$ (33)

whih implies that

${\omega }_{1A}\left(e_1\right)=0,e_A\left({\mu }_1\right)=0,A=2,3$ (34)

Similary, calculating $\left(\tilde{\nabla }{e}_{A}h\right)\left(e_1,e_A\right)=\left(\tilde{\nabla }{e}_{A}h\right)\left(e_A,e_A\right)$ , from (6), (9), (22), (29), (30), we find

$e_1\left({\mu }_A\right)={\epsilon }_A{\omega }_{1A}\left(e_A\right)\left({\mu }_1-{\mu }_A\right)$ (35)

and

$e_1\left({\lambda }_A\right)={\epsilon }_A{\omega }_{1A}\left(e_A\right)\left({\lambda }_1-{\lambda }_A\right)={\epsilon }_A{\omega }_{1A}\left(e_A\right)\left(-\frac{15{\epsilon }_4}{4}\right).$ (36)

This together with (29) and (36) deduces to

${\omega }_{1A}\left(e_A\right)=-\frac{3{\epsilon }_A}{5}\frac{e_1\left(f\right)}{f}.$ (37)

Substituting (37) into (35), we have

$e_1\left({\mu }_A\right)=\frac{3}{5}\frac{e_1\left(f\right)}{f}\left({\mu }_A-{\mu }_1\right).$ (38)

Furthermore we have

$e_1\left({\displaystyle \sum }_{A=2}^3{\mu }_A\right)=\frac{3}{5}\frac{e_1\left(f\right)}{f}\left({\displaystyle \sum }_{A=2}^3{\mu }_A-2{\mu }_1\right)$ (39)

using (21), (16) and (39), we get

$e_1\left({\mu }_1\right)=\frac{9}{5}\frac{e_1\left(f\right)}{f}{\mu }_1.$ (40)

case (1) when ${\mu }_1=0$ , then (38) becomes

$e_1\left({\mu }_A\right)=\frac{3}{5}\frac{e_1\left(f\right)}{f}{\mu }_A,A=2,3.$ (41)

if ${\mu }_2={\mu }_3=0$ , then $A_{e_5}$ has the form ($I$ ).

If ${\mu }_2,{\mu }_3$ are not equal to 0, Then it follows from (36) that

$\frac{e_1\left({\mu }_A\right)}{{\mu }_A}=\frac{3}{5}\frac{e_1\left(f\right)}{f},A=2,3.$ (42)

Integrating (42), we have

${\mu }_A=g_A{f}^{\frac{3}{5}},A=2,3.$ (43)

where $g_A$ are nonzero constents. Substituting (43) into (20) we get

$g_2+g_3=0,A=2,3.$ (44)

Thus $A_{e_5}$ takes the form

$A_{e_5}=\left(\begin{array}{ccc}0& & \\ & g_2{f}^{\frac{3}{5}}& \\ & & -g_3{f}^{\frac{3}{5}}\end{array}\right)$

Case (2) If ${\mu }_1\ne 0$ for $A=2,3$ , we have form (36) that

${\mu }_1=c_1{f}^{\frac{3}{5}}.$ (45)

where $c_1$ is a nonzero constant, substituting (45) into (38) we have

$e_1\left({\mu }_A\right)=\frac{3}{5}\frac{e_1\left(f\right)}{f}\left({\mu }_A-c_1{f}^{\frac{3}{5}}\right).$ (46)

By integrating (46). It fllows that

${\mu }_A=-\frac{c_1}{2}{f}^{\frac{9}{5}}+g_A{f}^{\frac{3}{5}}.$ (47)

where $e_1\left(g_A\right)=0$ and $g_A$ are smooth function. Also $g_2+g_3=0$ . Thus

$A_{e_5}=\left(\begin{array}{ccc}{c}_1{f}^{\frac{9}{5}}& & \\ & -\frac{c_1}{2}{f}^{\frac{9}{5}}+g_2{f}^{\frac{3}{5}}& \\ & & -\frac{c_1}{2}{f}^{\frac{9}{5}}-g_3{f}^{\frac{3}{5}}\end{array}\right)$ (48)

In the following, we will prove that $g_A$ ($A=2,3$ ) are constants.

If $g_2,g_3$ are not constants, then we know from (47) that $e_A\left(g_2\right)$ and $e_A\left(g_3\right)$ are not equal to 0 for $A=2,3$ .

Calculating the equation $\left(\tilde{\nabla }{e}_{1}h\right)\left(e_2,e_3\right)=\left(\tilde{\nabla }{e}_{2}h\right)\left(e_1,e_3\right)$ and

$\left(\tilde{\nabla }{e}_{1}h\right)\left(e_3,e_2\right)=\left(\tilde{\nabla }{e}_{2}h\right)\left(e_1,e_2\right)$ , combing (6), (9), we have

${\omega }_{23}\left(e_1\right)\left({\epsilon }_{2}h\left(e_2,e_2\right)-{\epsilon }_{3}h\left(e_3,e_3\right)\right)={\omega }_{31}\left(e_2\right)\left({\epsilon }_{3}h\left(e_3,e_3\right)-{\epsilon }_{1}h\left(e_1,e_1\right)\right).$

${\omega }_{32}\left(e_1\right)\left({\epsilon }_{3}h\left(e_3,e_3\right)-{\epsilon }_{2}h\left(e_2,e_2\right)\right)={\omega }_{21}\left(e_3\right)\left({\epsilon }_{2}h\left(e_2,e_2\right)-{\epsilon }_{1}h\left(e_1,e_1\right)\right).$

Combing (30) we obtain

$\begin{array}{l}{\epsilon }_4{\omega }_{23}\left(e_1\right)\left({\lambda }_2-{\lambda }_3\right)e_4+{\epsilon }_5{\omega }_{23}\left(e_1\right)\left({\mu }_2-{\mu }_3\right)e_5\\ ={\epsilon }_4{\omega }_{31}\left(e_2\right)\left({\lambda }_3-{\lambda }_1\right)e_4+{\epsilon }_5{\omega }_{31}\left(e_1\right)\left({\mu }_3-{\mu }_1\right)e_5.\end{array}$ (49)

$\begin{array}{l}{\epsilon }_4{\omega }_{32}\left(e_1\right)\left({\lambda }_3-{\lambda }_2\right)e_4+{\epsilon }_5{\omega }_{32}\left(e_1\right)\left({\mu }_3-{\mu }_2\right)e_5\\ ={\epsilon }_4{\omega }_{21}\left(e_3\right)\left({\lambda }_2-{\lambda }_1\right)e_4+{\epsilon }_5{\omega }_{21}\left(e_1\right)\left({\mu }_2-{\mu }_1\right)e_5.\end{array}$

Note that ${\lambda }_2={\lambda }_3$ , then it follows from (46) that

${\omega }_{13}\left(e_2\right)=0,{\omega }_{12}\left(e_3\right)=0,{\omega }_{23}\left(e_1\right)=0.$ (50)

A short calulation from (9) shows that

$\begin{array}{l}{\nabla }_{e_1}{\nabla }_{e_2}{e}_1=e_1\left({\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_2\right)\right)e_k+{\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_2\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{kt}\left(e_1\right)e_t\\ {\nabla }_{e_1}{\nabla }_{e_3}{e}_1=e_1\left({\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_3\right)\right)e_k+{\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_3\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{kt}\left(e_1\right)e_t\\ {\nabla }_{e_2}{\nabla }_{e_1}{e}_1=e_2\left({\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_1\right)\right)e_k+{\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_1\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{kt}\left(e_2\right)e_t\\ {\nabla }_{e_3}{\nabla }_{e_1}{e}_1=e_3\left({\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_1\right)\right)e_k+{\displaystyle \sum }_{k=1}^3{\epsilon }_k{\omega }_{1k}\left(e_1\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{kt}\left(e_3\right)e_t\\ {\nabla }_{\left[e_1,e_2\right]}{e}_1={\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{2k}\left(e_1\right)-{\omega }_{1k}\left(e_2\right)\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{1t}\left(e_k\right)e_t\\ {\nabla }_{\left[e_1,e_3\right]}{e}_1={\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{3k}\left(e_1\right)-{\omega }_{1k}\left(e_3\right)\right){\displaystyle \sum }_{t=1}^3{\epsilon }_t{\omega }_{1t}\left(e_k\right)e_t\end{array}$ (51)

Then it follows from (8), (9), (34), (51)

$\begin{array}{c}R\left(e_1,e_2\right)e_1,e_2)=e_1\left({\omega }_{12}\left(e_2\right)\right)-e_2\left({\omega }_{12}\left(e_1\right)\right)-{\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{2k}\left(e_1\right)-{\omega }_{1k}\left(e_2\right)\right){\omega }_{12}\left(e_k\right)\\ +{\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{1k}\left(e_2\right){\omega }_{k2}\left(e_1\right)-{\omega }_{1k}\left(e_1\right){\omega }_{k2}\left(e_2\right)\right)\\ =e_1\left({\omega }_{12}\left(e_2\right)\right)+{\epsilon }_2{\left({\omega }_{12}\left(e_2\right)\right)}^2\\ R\left(e_1,e_3\right)e_1,e_3)=e_1\left({\omega }_{12}\left(e_3\right)\right)-e_3\left({\omega }_{12}\left(e_1\right)\right)-{\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{3k}\left(e_1\right)-{\omega }_{1k}\left(e_3\right)\right){\omega }_{13}\left(e_k\right)\\ +{\displaystyle \sum }_{k=1}^3{\epsilon }_k\left({\omega }_{1k}\left(e_3\right){\omega }_{k2}\left(e_1\right)-{\omega }_{1k}\left(e_1\right){\omega }_{k3}\left(e_3\right)\right)\\ =e_1\left({\omega }_{13}\left(e_3\right)\right)+{\epsilon }_3{\left({\omega }_{13}\left(e_3\right)\right)}^2\end{array}$ (52)

Also using Guauss equation we from (30) that

$\begin{array}{l}R\left(e_1,e_2\right)e_1,e_2)=-{\epsilon }_1{\epsilon }_2\left({\epsilon }_4{\lambda }_1{\lambda }_2+{\epsilon }_5{\mu }_1{\mu }_2\right)\\ R\left(e_1,e_3\right)e_1,e_3)=-{\epsilon }_1{\epsilon }_3\left({\epsilon }_4{\lambda }_1{\lambda }_3+{\epsilon }_5{\mu }_1{\mu }_3\right)\end{array}$ (53)

Those two facts shows that

$\begin{array}{c}{e}_1\left({\omega }_{12}\left(e_2\right)\right)+{\epsilon }_2{\left({\omega }_{12}\left(e_2\right)\right)}^2=-{\epsilon }_1{\epsilon }_2\left({\epsilon }_4{\lambda }_1{\lambda }_2+{\epsilon }_5{\mu }_1{\mu }_2\right)\\ =-{\epsilon }_1{\epsilon }_2\left(-\frac{27{\epsilon }_4{f}^2}{8}+{\epsilon }_5\left(-\frac{c_1^2}{2}{f}^{\frac{12}{5}}+c_1{g}_2{f}^{\frac{6}{5}}\right)\right)\\ e_1\left({\omega }_{13}\left(e_3\right)\right)+{\epsilon }_3{\left({\omega }_{13}\left(e_3\right)\right)}^2=-{\epsilon }_1{\epsilon }_3\left({\epsilon }_4{\lambda }_1{\lambda }_3+{\epsilon }_5{\mu }_1{\mu }_3\right)\\ =-{\epsilon }_1{\epsilon }_2\left(-\frac{27{\epsilon }_4{f}^2}{8}+{\epsilon }_5\left(-\frac{c_1^2}{2}{f}^{\frac{12}{5}}+c_1{g}_3{f}^{\frac{6}{5}}\right)\right)\end{array}$ (54)

Calculating $\left({\nabla }_{e_1}{e}_2-{\nabla }_{e_2}{e}_1\right)f=\left[e_1,e_2\right]\left(f\right)=e_1{e}_2\left(f\right)-e_2{e}_1\left(f\right)$ and

$\left({\nabla }_{e_1}{e}_3-{\nabla }_{e_3}{e}_1\right)f=\left[e_1,e_3\right]\left(f\right)=e_1{e}_3\left(f\right)-e_3{e}_1\left(f\right)$ , yeid

$e_2{e}_1\left(f\right)=0,e_3{e}_1\left(f\right)=0.$ (55)

Calculating $\left({\nabla }_{e_1}{e}_2-{\nabla }_{e_2}{e}_1\right)\left(e_{1}f\right)=\left[e_1,e_2\right]\left(e_{1}f\right)=e_1{e}_2{e}_1\left(f\right)-e_2{e}_1{e}_1\left(f\right)$ and $\left({\nabla }_{e_1}{e}_3-{\nabla }_{e_3}{e}_1\right)\left(e_{1}f\right)=\left[e_1,e_3\right]\left(e_{1}f\right)=e_1{e}_3{e}_1\left(f\right)-e_3{e}_1{e}_1\left(f\right)$ , yeid

$e_2{e}_1{e}_1\left(f\right)=0,e_3{e}_1{e}_1\left(f\right)=0.$ (56)

Differentiating (54) along $e_A$ , for $A=2,3$ we have

${\mu }_1{e}_A{\mu }_A=0$ (57)

Combing (45), (47), (57), we obtain

$\frac{-c_1^2}{2}{e}_A\left(f^{\frac{9}{5}}\right)f^{\frac{9}{5}}+c_1{e}_A\left(g_A\right)f^{\frac{12}{5}}+c_1{g}_A{e}_A\left(f^{\frac{3}{5}}\right)f^{\frac{9}{5}}=0.$ (58)

Noting that $e_A\left(g_A\right)\ne 0$ and $f\ne 0$ , then it follows from (22) and (56) that $c_1=0$ , which together with (45) proves ${\mu }_1=0$ , a contradiction. There for $g_2$ and $g_3$ are constant function.

4. Proofs of Main Theorems

Proof of Theorem 1.1. we choose a pseudo-Riemannian orthonormal frame field $e_1,e_2,\cdots ,e_n$ . For $i<j$ , we have from (11) that

${\Vert e_i\wedge e_j\Vert }^2=\langle e_i,e_i\rangle \langle e_j,e_j\rangle -{\langle e_i,e_j\rangle }^2={\epsilon }_i{\epsilon }_j.$ (59)

Using the Gauss equation, it follows from (30) that

$\begin{array}{c}R\left(e_i,e_j,e_i,e_j\right)=\langle h\left(e_j,e_i\right),h\left(e_i,e_j\right)\rangle -\langle h\left(e_i,e_i\right),h\left(e_j,e_j\right)\rangle \\ =-{\epsilon }_i{\epsilon }_j\left({\epsilon }_5{\lambda }_i{\lambda }_j+{\epsilon }_5{\mu }_i{\mu }_j\right).\end{array}$ (60)

Combing (11), (12), (20) we have

$\begin{array}{c}R={\displaystyle \sum }_{i,j=1}^{3}K\left(e_i,e_j\right)\\ ={\displaystyle \sum }_{i,j=1}^3\frac{{\epsilon }_i{\epsilon }_j({\epsilon }_4{\lambda }_i{\lambda }_j+{\epsilon }_5{\mu }_i{\mu }_j}{{\epsilon }_i{\epsilon }_j}\\ ={\displaystyle \sum }_{i,j=1}^3\left({\epsilon }_4{\lambda }_i{\lambda }_j+{\epsilon }_5{\mu }_i{\mu }_j\right)\\ ={\epsilon }_4{\left({\lambda }_1+{\lambda }_2+{\lambda }_3\right)}^2-\left({\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2\right)+{\epsilon }_5{\left({\mu }_1+{\mu }_2+{\mu }_3\right)}^2-\left({\mu }_1^2+{\mu }_2^2+{\mu }_3^2\right)\\ ={\epsilon }_4{\left({\lambda }_1+{\lambda }_2+{\lambda }_3\right)}^2-\left({\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2\right)-{\epsilon }_5\left({\mu }_1^2+{\mu }_2^2+{\mu }_3^2\right).\end{array}$ (61)

When $A_{e_5}$ has the form (I), Combining with (24) and (61) we obtain

$\begin{array}{c}R={\epsilon }_4{\left({\lambda }_1+{\lambda }_2+{\lambda }_3\right)}^2-\left({\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^2\right)\\ ={\epsilon }_4{\left(\frac{-3{\epsilon }_{4}f}{2}+\frac{9{\epsilon }_{4}f}{4}+\frac{9{\epsilon }_{4}f}{4}\right)}^2-{\epsilon }_4\left({\left(\frac{-3{\epsilon }_{4}f}{2}\right)}^2+{\left(\frac{9{\epsilon }_{4}f}{4}\right)}^2+{\left(\frac{9{\epsilon }_{4}f}{4}\right)}^2\right)\\ =-\frac{27{\epsilon }_4{f}^2}{8}.\end{array}$ (62)

When $A_{e_5}$ has the form (II), Combining with (25) and (61) we obtain

$R=-\frac{27{\epsilon }_4{f}^2}{8}-{\epsilon }_5{g}_2^2{f}^{\frac{6}{5}}-{\epsilon }_5{g}_3^2{f}^{\frac{6}{5}}.$

where $g_2$ and $g_3$ are constants with $g_2+g_3=0$ .

When $A_{e_5}$ has the form (III), Combining with (26) and (61) we obtain

$R=-\frac{27{\epsilon }_4{f}^2}{8}-{\epsilon }_5{g}_2^2{f}^{\frac{6}{5}}-{\epsilon }_5{g}_3^2{f}^{\frac{6}{5}}-\frac{3{\epsilon }_5{c}_1{}^2}{2}{f}^{\frac{18}{5}}.$

where $g_2$ and $g_3$ are constants satisfying $g_2+g_3=0$ . From the above, it follows that $R$ is expressed as a fractional power polynomial in the mean curvature with non-zero coefficients, which completes the proof of Theorem 1.1.

Assuming that $f$ is non-constant, a contradiction can be derived by applying Theorem 1.2, thereby concluding the proof of Theorem 1.2.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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