The minimality of biharmonic hypersurfaces in pseudo-Euclidean spaces

1.   Introduction

Let $\phi:M^n_r\rightarrow \mathbb{E}^{n+p}_s$ be an isometric immersion of a submanifold $M^n_r$ into the pseudo-Euclidean space $\mathbb{E}^{n+p}_s$. Denoted by $\Delta$ and $\vec{H}$ the Laplace-Beltrami operator and the mean curvature vector field of $M^n_r$, respectively. It is well known that the position vector of $M^n_r$ satisfies

A submanifold $M^n_r$ is called biharmonic if and only if

It is easy to see that minimal submanifolds (i.e., $\vec{H} = 0$) are automatically biharmonic. Naturally, we consider the problem to determine if there exist biharmonic submanifolds of $\mathbb{E}^{n+p}_s$, other than the minimal ones. Concerning this problem, B. Y. Chen in 1991 proposed the following

Chen's conjecture: any biharmonic submanifold in a Euclidean space $\mathbb{E}^{n+p}$ is minimal.

The conjecture was proved to be true in the low dimension. We refer to ^[1,2] for $n = 3$, ^[3,4] for $n = 4$ and ^[5] for $n = 5$. For higher dimensional cases, the conjecture is still true under additional geometric conditions, and we refer the reader to ^[6,7,8] for a review, ^[9] with references therein for recent progress. We need to point out that Chen's conjecture is still open widely.

When the ambient space is pseudo-Euclidean, Chen's conjecture is not necessarily true. B. Y. Chen and S. Ishikawa gave some nonminimal biharmonic space-like surfaces in $\mathbb{E}^4_s (s = 1, 2)$ (cf. ^[10]), and nonminimal biharmonic pseudo-Riemannian surfaces in $\mathbb{E}^4_s (s = 1, 2, 3)$ (cf. ^[1]).

However, the minimality of biharmonic hypersurfaces in pseudo-Euclidean spaces is still one of the central topics in this area, and many important signs of progress have been made during the last four decades. For example, B. Y. Chen and S. IshiKawa (cf. ^[1,10]) proved that any biharmonic surface in $\mathbb{E}^3_s\ (s = 1, 2)$ is minimal. Later, F. Defever, G. Kaimakamis, V. Papantoniou in ^[11] proved that the hypersurface $M^3_r$ with diagonalizable shape operator in $\mathbb{E}^4_s$ is minimal. Y. Fu showed in ^[12] that such hypersurfaces $M^4_r$ in $\mathbb{E}^5_s$ are minimal. More generally, it is shown in ^[13] that hypersurfaces $M^n_r$ with diagonalizable shape operator and at most three distinct principal curvatures in $\mathbb{E}^{n+1}_s$ are minimal. The same conclusion holds for the hypersurfaces in $\mathbb{E}^{n+1}_s$ provided that the number of distinct principal curvatures $\leq6$. We refer to ^[14] for details.

In this paper, we will continue to focus on the minimality of biharmonic hypersurfaces in $\mathbb{E}^{n+1}_s$ with no restriction for the number of distinct principal curvatures and prove the following.

Main theorem Let $M^n_r$ be a biharmonic hypersurface with a diagonalizable shape operator in a pseudo-Euclidean space. If one among the Ricci operator, the curvature operator, the Jacobi operator or shape operator of $M^n_r$ is recurrent, then $M^n_r$ must be minimal.

2.   Preliminaries

Let $\mathbb{E}_s^{n+1}$, $0 < s < n+1$, be a pseudo-Euclidean space with metric given by

where $(y_1, y_2, \ldots, y_{n+1})$ is the natural coordinate system of $\mathbb{E}_s^{n+1}$.

Let $M_r^n$ be an $n$-dimensional hypersurface in $\mathbb{E}_s^{n+1}$. Denote by $\nabla$ and $\overline{\nabla}$ the Levi-Civita connections of $M_r^n$ and $\mathbb{E}_s^{n+1}$, respectively. Let $\xi$ be a local unit normal vector field to $M_r^n$ in $\mathbb{E}_s^{n+1}$, and we denote $\varepsilon = \langle\xi, \xi\rangle = \pm1$, then the Gauss and Weingarten formulas are given, respectively, by (cf. ^[15])

and

where $h$ denotes the second fundamental form, and $A_\xi$ denotes the shape operator with respect to $\xi$. As it is well known, $h$ and $A_\xi$ are related by

The mean curvature vector is given by $\vec{H} = H\xi$, with $H = \frac{\langle\xi, \xi\rangle}{n}\text{trace} A_{\xi}$ the mean curvature of $M^n_r$ in $\mathbb{E}_s^{n+1}$.

Then the Guass and Codazzi equations are given by

where

Let $T$ be a tensor on the pseudo-Riemannian manifold $M_r^n$. Then $T$ is said to be recurrent if there exists a certain 1-form $\eta$ on $M_r^n$ satisfying $\nabla_X T = \eta(X)T$ for any $X\in TM_r^n$. Thus, the recurrent (1, 1)-tensor can be considered as an extension of the parallel one.

A hypersurface is said to be biharmonic if

The condition is equivalent to

By comparing the vertical and horizontal parts, the above equation is equivalent to the following two equations:

where $A$ and $\Delta^\perp$ denote by the Weingarten operator and the Laplace operator in the normal bundle of $M_r^n$ in $\mathbb{E}_s^{n+1}$, respectively.

Finally, we give a useful lemma to complete our main theorem.

Lemma 2.1. Let $M_r^n$ be a biharmonic hypersurface with a diagonalizable shape operator in $\mathbb{E}_s^{n+1}$. Assume that the mean curvature $H$ is not constant, then

where $\{e_i\}_{i = 1}^n$ is a local orthonormal frame with $\varepsilon_i = \langle e_i, e_i \rangle = \pm 1$, and $\omega_{ij}^k$, $i, j, k = 1, 2, \ldots, n$, are called connection forms.

Proof. Since $H$ is not a constant, there exists a point $p\in U$, where $U$ is an open subset of $M_r^n$ such that $\text{grad}H\neq0$ on $U$. We know from (2.3) that $\text{grad}H$ is a principal direction corresponding to the principal curvature $\frac{-n\varepsilon}{2}H$. Without loss of generality, we denote by

We choose a local orthonormal frame field $\{e_1, e_2, \ldots, e_n\}$ such that $e_1$ is parallel to $\text{grad}H$, where $\text{grad}H = \sum_{i = 1}^n\varepsilon_ie_i(H)e_i$, and $A$ takes the following form

which means that $e_i$ is a principal direction of $A$ with the principal curvature $\lambda_i$, $i = 1, 2, \ldots, n$. Then

Meanwhile, it follows from (2.1) that, for $i, j = 1, 2, \ldots, n$,

We write

Then, using the Codazzi equation $(\nabla_{e_i}A)e_k = (\nabla_{e_k}A)e_i$, we obtain

which implies, for distinct $i, j, k = 1, 2, \ldots, n$,

Note that (2.5) and (2.7) lead to

Then it is easy to compute that

which means

Taking $j = 1$ and $2\leq i, k\leq n$, (2.11) becomes

together with (2.12), we see

Applying compatibility condition to calculate $\nabla_{e_k}\langle e_i, e_j\rangle = 0$, we conclude

It follows from the first equation in (2.14) that $\omega_{k1}^1 = 0$. Also, it follows from (2.10) and (2.14) that $\omega_{1i}^1 = 0$ and $\omega_{11}^i = 0$, which together with (2.13), we get

Putting this all together, we obtain the claim.

3.   Proof of main theorem

Proof of main theorem The idea of the proof is the following. First, we prove that the mean curvature $H$ of $M^n_r$ is a constant by using a proof by contradiction. Then, combining with (2.3), we show that $H = 0$, i.e., $M^n_r$ is minimal.

Now we start to prove that $H$ is a constant.

Suppose on the contrary that $\text{grad}H\neq0$. Recalled, from (2.3) that $\text{grad}H$ is a principal direction corresponding to the principal curvature $\frac{-n\varepsilon}{2}H$. We choose a local orthonormal frame field $\{e_1, e_2, \ldots, e_n\}$, such that $e_1$ is parallel to $\text{grad}H$, and (2.6) holds.

Case 1: When the Ricci operator is recurrent.

According to ^[15], we know

and so

Thus, a short calculation together with (2.6) and (2.8) shows

where

Because the Ricci operator is recurrent, i.e.,

for any $X, Y\in TM_r^n$, we have from (2.9) that

Using (2.9) again, it follows from (3.1) that

When $\alpha_j = 0$, for some $j = 1, 2, \ldots, n$, then, (3.3) and (3.4) show

which tells us that $\alpha_k = 0$, for $k = 2, 3, \ldots, n$. Then it follows from (3.2) that $\lambda_k = 0$ or $\lambda_k = \varepsilon nH$. Thus, $M_r^n$ has three distinct principal curvatures $0$, $\varepsilon nH$ and $\frac{-n\varepsilon}{2}H$.

When $\alpha_j \neq 0$, for all $j = 1, 2, \ldots, n$, then (2.15), (3.3) and (3.4) give, for fix indexes $i, j$,

which implies

It follows the first equation of (3.6) that

Taking $j = 1$ in the above equation, we have

The above facts show that $e_i\ln{|\alpha_j|} = e_i\ln{|\alpha_1|}$, which implies that $\ln |\frac{\alpha_j}{\alpha_1}| =$ constant. Further, we get that $\alpha_j = c \alpha_1$ ($c$ is a constant), for any $j \neq 1$. Thus, we get that $M^n_r$ has at most three principal curvatures.

In conclusion, $M^n_r$ has at most three distinct principal curvatures. We know from ^[13] or ^[14] that $H$ is a constant, then $\text{grad}H = 0$. It is a contradiction.

Case 2: When the curvature operator is recurrent.

Since the curvature operator $R$ is recurrent, there exists a 1-form $\eta$ such that

for any $\ X, Y, Z, W\in TM_r^n$. Note that $R(e_i, e_j)e_k = 0$, for distinct $i, j, k$, and so

Meanwhile, according to the Guass equation, we have from (2.6) and (2.9) that

Since the linear independent of $\{e_i\}_{i = 1}^n$, it follows from (2.15) that, for distinct $i, j, k, l = 2, 3, \ldots, n$,

which means that $\omega_{il}^k = 0$ (otherwise, $M_r^n$ has two distinct principal curvatures), then it follows from $(2.11)$ that

Then $\lambda_i = \lambda_k$ or $\omega_{li}^k = 0$, for $i \neq k$. In particular, if $\omega_{li}^k = 0$, then we have from (2.11) that $\lambda_l = \lambda_k$. Thus, $M_r^n$ has at most two distinct principal curvatures. The same situation happens as above, and it should be modified.

Case 3: When the Jacobi operator is recurrent.

Because the Jacobi operator is recurrent, i.e.,

where $R_X(Z) = R(Z, X)X$, for any $\ X, Y, Z\in TM_r^n$. Then using the Guass equation, and combining with (2.6) and (2.9), we get

Similarly, it follows that

By comparing the above two equations, we obtain

When $\lambda_j = 0$, then it is obvious to see that $M_r^n$ has two distinct principal curvatures.

When $\lambda_j \neq 0$, then

● If $\lambda_l = \lambda_k$, then it follows from (2.11) that

which means $\lambda_i = \lambda_j$. So $M_r^n$ has two distinct principal curvatures.

● If $\lambda_l \neq \lambda_k$, we will make the inner product of the two sides of (3.7) with $e_j$ and

Then, we have

Since $\lambda_k \neq 0$, then it follows that $\omega_{ik}^j = 0$. This together with (2.11), we have

Hence $\lambda_i = \lambda_j$, $i \neq j$. To sum up, we know that $M^n_r$ has at most two distinct principal curvatures. Applying the same arguments as in Case 2, we obtain that $H$ is a constant.

Case 4: When the shape operator is recurrent.

Since the shape operator is recurrent, i.e., $(\nabla_X A)Y = \eta(X)A(Y)$, we have from (2.6) that

Then, we have from (2.2) that

which shows that $\lambda_j-\lambda_k = 0$ or $\omega_{ij}^k = 0$.

We claim that $\lambda_j = \lambda_k$.

Suppose on contrary that $\lambda_j\neq\lambda_k$, then $\omega_{ij}^k = 0$. Together with (2.11), it reduces to

which means that $\lambda_i = \lambda_k$, it is impossible. So, $M_r^n$ has two distinct principal curvatures. By making use of the similar methods in Case 3, we know that $H$ is a constant.

Summarizing the above four cases, we obtain that $H$ is a constant.

Next, we prove that $M^n_r$ is minimal.

Since $H$ is a constant, it follows from (2.3) and (2.6) that

which implies that $H = 0$ or $\lambda_i = 0$, $i = 1, 2, \ldots, n$. Note that $H = \frac{\varepsilon}{n}\sum_{i = 1}^n\lambda_i$, then, $M^n_r$ must be minimal.

We complete the proof of main theorem.

Acknowledgments

This work is supported by the Natural Science Foundation of Chongqing (No. cstc2021jcyj-msxmX0388); the Scientific Research Starting Foundation of Chongqing University of Technology (No. 2017ZD52)

Conflict of interest

The authors declare there is no conflicts of interest.