Abstract
In this paper, we show that biharmonic hypersurfaces with at most two distinct principal curvatures in pseudo-Riemannian space form with constant sectional curvature and index have constant mean curvature. Furthermore, we find that such biharmonic hypersurfaces in even-dimensional pseudo-Euclidean space , in even-dimensional de Sitter space , and in even-dimensional anti-de Sitter space are minimal.
1. Introduction
Let be a ()-dimensional pseudo-Riemannian space form with index and constant sectional curvature . According to , , and , is isometric to pseudo-Euclidean space , de Sitter space , and anti-de Sitter space .
Suppose that , , or be an isometric immersion of a pseudo-Riemannian hypersurface into . The hypersurface is said to be biharmonic if its bitension field vanishes identically, i.e., where , , , and are the curvature tensor of , the induced connection by on the bundle , and the connection of , respectively. If the mean curvature of the hypersurface is zero, then we call as minimal. It is generally known that minimal hypersurfaces are biharmonic ones. Conversely, the natural question is whether any biharmonic hypersurface is minimal.
For biharmonic hypersurfaces in pseudo-Euclidean spaces, there is a conjecture in [2] that every biharmonic hypersurface of pseudo-Euclidean spaceis minimal. Up to now, this conjecture has been examined for many biharmonic hypersurfaces, such as of (cf. [2, 3]), of (cf. [1]), of (cf. [10]), and in with at most three distinct principal curvatures and diagonalizable shape operator (cf. [5, 7]).
When the ambient space is de Sitter space , there are also some papers that studied the above problem. Sasahara in [11] considered biharmonic hypersurfaces of and proved that it must be minimal when , but may not when . Investigators studied biharmonic hypersurfaces with at most two distinct principal curvatures in whose shape operator is diagonalizable in [6, 8] and showed that such hypersurface is minimal, but the hypersurface may not. Naturally, there is a question as to whether any biharmonic hypersurface in de Sitter space is minimal.
The situation is quite different when the ambient space is anti-de Sitter space . For biharmonic hypersurface of , it must be minimal when and may not when (cf. [11]). For biharmonic hypersurface with at most two distinct principal curvatures in whose shape operator is diagonalizable, it is minimal when and may not when (cf. [6, 8]). A natural question is whether any biharmonic hypersurface in anti-de Sitter space is minimal.
In this paper, we study biharmonic hypersurfaces with at most two distinct principal curvatures in pseudo-Riemannian space forms , without the restriction that the shape operator is diagonalizable. We proved such biharmonic hypersurfaces have constant mean curvature. Furthermore, we find that such biharmonic hypersurfaces in even-dimensional pseudo-Euclidean space , in even-dimensional de Sitter space , and in even-dimensional anti-de Sitter space are minimal.
2. Preliminaries
2.1. Notions and Formulas of Hypersurfaces in
Let be a pseudo-Riemannian space form with index and constant sectional curvature . A nonzero vector in is called time-like, space-like, or light-like, according to whether is negative, positive, or zero, respectively. Let be a nondegenerate hypersurface in . denotes a unit normal vector field to , then . Denote by and the Levi-Civita connections of and , respectively. For any vector fields tangent to , the Gauss formula and Weingarten formula are given by where is the scalar-valued second fundamental form and is the shape operator of associated to . The mean curvature vector field can be expressed as , with mean curvature . For any vector fields tangent to , the Codazzi and Gauss equations are (cf. [9]) here .
A hypersurface of is biharmonic if and only if its mean curvature satisfies the following two equations (cf. [4]): where here is a local orthonormal frame of .
2.2. The Shape Operator of in
According to [9] (exercise 18, pp. 260-261), the tangent space at can be expressed as a direct sum of subspaces , , that are mutually orthogonal and invariant under the shape operator , and each (the restriction of on ) has form (a) or (b) as follows. (a) has the formwith respect to a basis of . The inner products of the basis elements in are all zero except (b) has the formwith respect to a basis of . The inner products of the basis elements in are all zero except
We denote by the number of terms having form (a). We adjust the order of , , such that have form (a) for and have form (b) for . Denote , , and , . Collecting all the vectors in in order, we get a basis of . With respect to this basis , the shape operator of the hypersurface in can be expressed as an almost diagonal matrix: and the inner products of the elements in are all zero except where
Observe the forms (a) and (b); we see that , has only a simple eigenvalue and , has eigenvalues , . It follows from the form of the shape operator that has principal curvatures
So, under the assumption that has at most two distinct principal curvatures, the shape operator has the following two possible forms: (I), i.e., , and there are at most two distinct values among (II), i.e., and , ,
For the form (I), we have
And for the form (II), we have with .
3. Theorems
Theorem 1. Let be a-dimensional pseudo-Riemannian space form with index and constant sectional curvature and be a nondegenerate biharmonic hypersurface of with at most two distinct principal curvatures, then has constant mean curvature.
Proof. From Section 2, the shape operator has the form (I) or (II). If has the form (II), then its eigenvalues are not real and is not an eigenvalue. It follows from (5) that , which tells us is a constant.
For the form (I), if we assume that is not a constant, then (5) implies that is an eigenvalue of the shape operator . When , then . On the other hand, . These two expressions imply , a contradiction.
So, in the following, we need only to discuss the situation where there are two distinct values among . Expression (5) also informs us that is an eigenvector of with corresponding eigenvalue . In view of (16), is one of the directions . Without loss of generality, we suppose is in the direction of ; it may be a light-like vector or not. We will follow different processes to lead contradictions for these two cases.
First of all, we give a lot of equations deduced from compatibility and symmetry of the connection, as well as the Codazzi equation.
Observe the inner products of the elements in basis given in Section 2, we can express
Since is in the direction of , the above equation implies that
Let . Applying compatibility condition to calculate we conclude for , , , and .
From the expression , , and (19), we easily get
We state that in the proof, if not otherwise specified, then for in and the connection’s coefficients, the ranges of and are as follows: and . For the equations about the connection’s coefficients, when (or ), then the terms about (or ) disappear. And when (or ), then the terms about (or ) and (or ) disappear.
It follows from the Codazzi equation (3) that for any vector fields tangent to ,
Start with this equation, we can get a series of equations about the coefficients of connection. (i)For in (24), then combining (19), we obtain
Applying (21), (22), and (23), we get from the above equation that
Note that if , then (26) tells us nothing. (ii)For , , in (24), then combining (19), we havewhich together with (23) implies that if , then and if , then (iii)For , , in (24), thenwhich together with (22), (23), (28), and (29) implies that if , then (iv)For , in (24), then combining (23), we know
From (22), (23), and the above relation, we conclude that
It follows from (33) and (34) that if , then (v)For , in (24), thenwhich together with (22) and (34) gives that (vi)For , in (24), thenwhich combining (35) and (37), implies that (vii)For , , in (24), then combining (23), we havewhich together with (23) tells us that when , then and when , then with . (43) and (44) give that if , then (viii)For with , and in (24), then(ix)For , , in (24), with and , then
Let with and , . Since (47), we find which implies that with , , and . (x)For in (24), with and , thenwhich combining (50) gives that (xi)For in (24), with , , and , then
Let with , , and . It follows from (53) that for and , which implies that for , , and ,
Now, we treat the two cases that is not light-like or light-like separately and get contradictions.
Case 1. is not light-like.
As is in the direction of , is not light-like means that . Observe equation (38); we find if , then , which contradicts with (19). So, we conclude that , for . Since there are two distinct values among and , we have . It follows from (34) and (38) that
Denote , we have the following lemmas.
Lemma 2. We have
Proof. Since with , it follows from (46) that
combining (43), we have
with , , and .
Using Gauss equation for , with , combining (29), (31), (35), (40), (42), (43) and (60), we have
From (22) and (23), we know , , with . So, we can rewrite the above equation as
For and , if , then we have from (22), (43), and (44). And if , from (56), we know
which together with (44), tells us that
So, for and , if , then
For and , if , combining (65), we get from (46) that
If , then (43) and (44) give us that
As (67), it follows from (66) that for and , if , then
For and , if , then combining (22) and (23), we get from (44) and (66) that
Therefore, from (59), (65), (67), (68), and (69), (62) can be simplified to
Choosing in , such that
and .
Taking sum in (70) for , with , we have
where
From (72), we can get , , and , i.e., (58) holds.
Lemma 3. We have
Proof. For , if is an even number, we put
with . And if is an odd number, we take
and . We easily find is an orthonormal basis of with . Thus, is an orthonormal basis of .
From (19), (75), and (76), we have when is an even number,
with , and when is an odd number,
with
Note that ; it follows from (19) that
By calculating, we get
Since the above, we obtain from (6) that
Now, we continue the proof of Theorem 1 for Case 1.
Combining (57) and (58), we get
Substitute the above equation into (74), considering and (57), we have
Act on (83) with , and using (57) and (58), we obtain
As (83) and , the above equation implies that which together with (83) gives that which implies that is a constant, a contradiction.
Case 2. is light-like.
Notice that is light-like means . Since , we can suppose and . It follows from (19) and (38) that
Calculate and with by Gauss equation, and letting and for sum with and , combining (21), (22), (23) (26), (28), (29), (35), (42), (45), and (87), we get
So, (88) and (89) can be simplified to
It follows from (52) that with and .
Multiply both sides of (92) with , we get
Combining (87) and (93), it gives us
Since (91), we get from (95) that which combining and , tells us that .
So, is a constant, a contradiction.
In view of the two cases, we complete the proof of Theorem 1.
Remark 4. With the assumption that the shape operator is diagonalizable, the result of Theorem 1 was proved in [6, 8, 12] when the hypersurface has at most three distinct principal curvatures.
Applying Theorem 1, we can get the following theorem.
Theorem 5. Let be a nondegenerate biharmonic hypersurface of pseudo-Riemannian space form , with . Suppose that has at most two distinct principal curvatures, which are all real numbers, then is minimal.
Proof. We know from Theorem 1 that is a constant; it follows from (6) that
When , it is easy to see from (97) that . When , then (97) implies , or . Since is equal to the sum of the squares of all principal curvatures and is equal to the sum of all principal curvatures, so tells us . Combining , we have , i.e., is minimal.
For odd-dimensional hypersurfaces, under the assumption that hypersurfaces have at most two distinct principal curvatures, the principal curvatures are all real. So, from Theorem 5, the following theorems are true.
Theorem 6. Let be a nondegenerate biharmonic hypersurface of pseudo-Euclidean space . Suppose that has at most two distinct principal curvatures, then is minimal.
Theorem 7. Let be a nondegenerate biharmonic hypersurface of de Sitter space . Suppose that has at most two distinct principal curvatures, then is minimal.
Theorem 8. Let be a nondegenerate biharmonic hypersurface of anti-de Sitter space . Suppose that has at most two distinct principal curvatures, then is minimal.
Remark 9. Under the assumption that the shape operator is diagonalizable, the results of Theorems 6, 7, and 8 were proven not only for odd-dimensional hypersurfaces but also for even-dimensional hypersurfaces (cf. [6, 8]).
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was supported by the NSF of China (11761061), the University Project of Gansu Province (2019B-045), and the Project of Northwest Normal University (NWNU-LKQN2019-23).