Biharmonic Hypersurfaces in Pseudo-Riemannian Space Forms with at Most Two Distinct Principal Curvatures
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.
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  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. ), of (cf. ), 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  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. ). 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.
2.1. 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. ) here .
A hypersurface of is biharmonic if and only if its mean curvature satisfies the following two equations (cf. ): where here is a local orthonormal frame of .
2.2. The Shape Operator of in
According to  (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 .
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 ,
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
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 . 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