# Biharmonic Hypersurfaces in Pseudo-Riemannian Space Forms with at Most Two Distinct Principal Curvatures

Authors: Chao Yang; Jiancheng Liu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we show that biharmonic hypersurfaces with at most two distinct principal curvatures in pseudo-Riemannian space formNsn+1c with constant sectional curvature c and index s have constant mean curvature. Furthermore, we find that such biharmonic hypersurfaces Mr2k−1 in even-dimensional pseudo-Euclidean space Es2k, Ms−12k−1 in even-dimensional de Sitter space Ss2kcc>0, and Ms2k−1 in even-dimensional anti-de Sitter space ℍs2kcc<0 are minimal.


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## Body

## 1. Introduction

LetNsn+1c be a (n+1)-dimensional pseudo-Riemannian space form with index 1≤s≤n+1 and constant sectional curvature c. According to c=0, c>0, and c<0, Nsn+1c is isometric to pseudo-Euclidean space Esn+1, de Sitter space Ssn+1c, and anti-de Sitter space ℍsn+1c.Suppose thatϕ:Mrn→Nsn+1c, r=s−1, or s be an isometric immersion of a pseudo-Riemannian hypersurface Mrn into Nsn+1c. The hypersurface Mrn is said to be biharmonic if its bitension field τ2ϕ vanishes identically, i.e.,
(1)τ2ϕ≔trace∇ϕ∇ϕ−∇∇ϕτϕ−traceR~dϕ,τϕdϕ=0,where τϕ≔divdϕ, R~, ∇ϕ, and ∇ are the curvature tensor of Nsn+1c, the induced connection by ϕ on the bundle ϕ−1TNsn+1c, and the connection of Mrn, respectively. If the mean curvature of the hypersurface Mrn is zero, then we call Mrn 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 spaceEsn+1is minimal. Up to now, this conjecture has been examined for many biharmonic hypersurfaces, such as Mr2 of Es3 (cf. [2, 3]), M13 of E14 (cf. [1]), M23 of E24 (cf. [10]), and Mrn in Esn+1 with at most three distinct principal curvatures and diagonalizable shape operator (cf. [5, 7]).When the ambient space is de Sitter spaceSsn+1c, there are also some papers that studied the above problem. Sasahara in [11] considered biharmonic hypersurfaces Mr2 of S13c and proved that it must be minimal when r=0, but may not when r=1. Investigators studied biharmonic hypersurfaces with at most two distinct principal curvatures in Ssn+1c whose shape operator is diagonalizable in [6, 8] and showed that such hypersurface Ms−1n is minimal, but the hypersurface Msn may not. Naturally, there is a question as to whether any biharmonic hypersurface Ms−1n in de Sitter space Ssn+1c is minimal.The situation is quite different when the ambient space is anti-de Sitter spaceℍsn+1c. For biharmonic hypersurface Mr2 of ℍ13c, it must be minimal when r=1 and may not when r=0 (cf. [11]). For biharmonic hypersurface Mrn with at most two distinct principal curvatures in ℍsn+1c whose shape operator is diagonalizable, it is minimal when r=s and may not when r=s−1 (cf. [6, 8]). A natural question is whether any biharmonic hypersurface Msn in anti-de Sitter space ℍsn+1c is minimal.In this paper, we study biharmonic hypersurfaces with at most two distinct principal curvatures in pseudo-Riemannian space formsNsn+1c, 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 Mr2k−1 in even-dimensional pseudo-Euclidean space Es2k, Ms−12k−1 in even-dimensional de Sitter space Ss2kc, and Ms2k−1 in even-dimensional anti-de Sitter space ℍs2kc are minimal.


## 2. Preliminaries

### 2.1. Notions and Formulas of Hypersurfaces inNsn+1c

LetNsn+1c be a pseudo-Riemannian space form with index s and constant sectional curvature c. A nonzero vector X in Nsn+1c is called time-like, space-like, or light-like, according to whether X,X is negative, positive, or zero, respectively. Let Mrn be a nondegenerate hypersurface in Nsn+1c. ξ→ denotes a unit normal vector field to Mrn, then ε=ξ→,ξ→=±1. Denote by ∇ and ∇~ the Levi-Civita connections of Mrn and Nsn+1c, respectively. For any vector fields X,Y tangent to Mrn, the Gauss formula and Weingarten formula are given by
(2)∇~XY=∇XY+hX,Yξ→,∇~Xξ→=−AX,where h is the scalar-valued second fundamental form and A is the shape operator of Mrn associated to ξ→. The mean curvature vector field H can be expressed as H→=Hξ→, with mean curvature H=1/nεtrA. For any vector fields X,Y,Z tangent to Mrn, the Codazzi and Gauss equations are (cf. [9])
(3)∇XAY=∇YAX,(4)RX,YZ=cY,ZX−X,ZY+εAY,ZAX−εAX,ZAY,here RX,YZ=∇X∇YZ−∇Y∇XZ−∇X,YZ.A hypersurfaceMrn of Nsn+1c is biharmonic if and only if its mean curvature H satisfies the following two equations (cf. [4]):
(5)A∇H=−n2εH∇H,(6)ΔH+εHtrA2=ncH,where
(7)ΔH=−∑i=1neieiH−∇eieiH,here e1,e2,⋯,en is a local orthonormal frame of TxMrn.


### 2.2. The Shape Operator ofMrn in Nsn+1c

According to [9] (exercise 18, pp. 260-261), the tangent space TxMrn at x∈Mrn can be expressed as a direct sum of subspaces Vk, 1≤k≤m, that are mutually orthogonal and invariant under the shape operator A, and each AVk (the restriction of A on Vk) has form (a) or (b) as follows.
(a)
AVk has the form(8)AVk=λk1λk⋱⋱1λk1λk,with respect to a basis Bk=uk1,uk2,⋯,ukαk of Vk. The inner products of the basis elements in Bk are all zero except
(9)ukb,ukd=εk=±1,b+d=αk+1.(b)
AVk has the form(10)AVk=γkτk−τkγk10γkτk01−τkγk10γkτk01−τkγk⋱10γkτk01−τkγk,τk≠0,with respect to a basis B¯k=u¯11,v¯11,⋯,u¯kβk,v¯kβk of Vk. The inner products of the basis elements in B¯k are all zero except
(11)u¯kb,u¯kd=1=−v¯kb,v¯kd,b+d=βk+1.We denote byt the number of terms AVk having form (a). We adjust the order of Vk, 1≤k≤m, such that AVk have form (a) for 1≤k≤t and AVk have form (b) for t+1≤k≤m. Denote Ai=AVi, 1≤i≤t, and A¯j=AVj, t+1≤j≤m. Collecting all the vectors in B1,⋯,Bt,B¯t+1,⋯,B¯m in order, we get a basis B of TxMrn. With respect to this basis B, the shape operator A of the hypersurface Mrn in Nsn+1c can be expressed as an almost diagonal matrix:
(12)A=diagA1,⋯,At,A¯t+1,⋯,A¯m,and the inner products of the elements in B are all zero except
(13)uia,uib=εi=±1,a+b=αi+1,1≤i≤t,u¯jc,u¯jd=1=−v¯jc,v¯jd,c+d=βj+1,t+1≤j≤m,where
(14)α1+α2+⋯+αt+2βt+1+βt+2+⋯+βm=n.Observe the forms (a) and (b); we see thatAi,1≤i≤t, has only a simple eigenvalue λi and A¯j,t+1≤j≤m, has eigenvalues γj+τj−1, γj−τj−1. It follows from the form of the shape operator A that Mrn has principal curvatures
(15)λ1,⋯,λt,γt+1±τt+1−1,⋯,γm±τm−1.So, under the assumption thatMrn has at most two distinct principal curvatures, the shape operator A has the following two possible forms:
(I)
t=m, i.e., A=diagA1,A2,⋯,Am, and there are at most two distinct values among λ1,λ2,⋯,λm(II)
t=0, i.e., A=diagA¯1,A¯2,⋯,A¯m and γ1=γ2=⋯=γm=γ, τ1=τ2=⋯=τm=τ, τ≠0For the form (I), we have(16)Auia=λiuia+uia+1,Auiαi=λiuiαi,1≤i≤m,1≤a≤αi−1.And for the form (II), we have(17)Au¯jb=νu¯jb−τv¯jb+u¯jb+1,Au¯jβj=νu¯jβj−τv¯jβj,Av¯jb=τu¯jb+νv¯jb+v¯jb+1,Av¯jβj=τu¯jβj+νv¯jβj,with t+1≤j≤m,1≤b≤βj.


## 2.1. Notions and Formulas of Hypersurfaces inNsn+1c

LetNsn+1c be a pseudo-Riemannian space form with index s and constant sectional curvature c. A nonzero vector X in Nsn+1c is called time-like, space-like, or light-like, according to whether X,X is negative, positive, or zero, respectively. Let Mrn be a nondegenerate hypersurface in Nsn+1c. ξ→ denotes a unit normal vector field to Mrn, then ε=ξ→,ξ→=±1. Denote by ∇ and ∇~ the Levi-Civita connections of Mrn and Nsn+1c, respectively. For any vector fields X,Y tangent to Mrn, the Gauss formula and Weingarten formula are given by
(2)∇~XY=∇XY+hX,Yξ→,∇~Xξ→=−AX,where h is the scalar-valued second fundamental form and A is the shape operator of Mrn associated to ξ→. The mean curvature vector field H can be expressed as H→=Hξ→, with mean curvature H=1/nεtrA. For any vector fields X,Y,Z tangent to Mrn, the Codazzi and Gauss equations are (cf. [9])
(3)∇XAY=∇YAX,(4)RX,YZ=cY,ZX−X,ZY+εAY,ZAX−εAX,ZAY,here RX,YZ=∇X∇YZ−∇Y∇XZ−∇X,YZ.A hypersurfaceMrn of Nsn+1c is biharmonic if and only if its mean curvature H satisfies the following two equations (cf. [4]):
(5)A∇H=−n2εH∇H,(6)ΔH+εHtrA2=ncH,where
(7)ΔH=−∑i=1neieiH−∇eieiH,here e1,e2,⋯,en is a local orthonormal frame of TxMrn.


## 2.2. The Shape Operator ofMrn in Nsn+1c

According to [9] (exercise 18, pp. 260-261), the tangent space TxMrn at x∈Mrn can be expressed as a direct sum of subspaces Vk, 1≤k≤m, that are mutually orthogonal and invariant under the shape operator A, and each AVk (the restriction of A on Vk) has form (a) or (b) as follows.
(a)
AVk has the form(8)AVk=λk1λk⋱⋱1λk1λk,with respect to a basis Bk=uk1,uk2,⋯,ukαk of Vk. The inner products of the basis elements in Bk are all zero except
(9)ukb,ukd=εk=±1,b+d=αk+1.(b)
AVk has the form(10)AVk=γkτk−τkγk10γkτk01−τkγk10γkτk01−τkγk⋱10γkτk01−τkγk,τk≠0,with respect to a basis B¯k=u¯11,v¯11,⋯,u¯kβk,v¯kβk of Vk. The inner products of the basis elements in B¯k are all zero except
(11)u¯kb,u¯kd=1=−v¯kb,v¯kd,b+d=βk+1.We denote byt the number of terms AVk having form (a). We adjust the order of Vk, 1≤k≤m, such that AVk have form (a) for 1≤k≤t and AVk have form (b) for t+1≤k≤m. Denote Ai=AVi, 1≤i≤t, and A¯j=AVj, t+1≤j≤m. Collecting all the vectors in B1,⋯,Bt,B¯t+1,⋯,B¯m in order, we get a basis B of TxMrn. With respect to this basis B, the shape operator A of the hypersurface Mrn in Nsn+1c can be expressed as an almost diagonal matrix:
(12)A=diagA1,⋯,At,A¯t+1,⋯,A¯m,and the inner products of the elements in B are all zero except
(13)uia,uib=εi=±1,a+b=αi+1,1≤i≤t,u¯jc,u¯jd=1=−v¯jc,v¯jd,c+d=βj+1,t+1≤j≤m,where
(14)α1+α2+⋯+αt+2βt+1+βt+2+⋯+βm=n.Observe the forms (a) and (b); we see thatAi,1≤i≤t, has only a simple eigenvalue λi and A¯j,t+1≤j≤m, has eigenvalues γj+τj−1, γj−τj−1. It follows from the form of the shape operator A that Mrn has principal curvatures
(15)λ1,⋯,λt,γt+1±τt+1−1,⋯,γm±τm−1.So, under the assumption thatMrn has at most two distinct principal curvatures, the shape operator A has the following two possible forms:
(I)
t=m, i.e., A=diagA1,A2,⋯,Am, and there are at most two distinct values among λ1,λ2,⋯,λm(II)
t=0, i.e., A=diagA¯1,A¯2,⋯,A¯m and γ1=γ2=⋯=γm=γ, τ1=τ2=⋯=τm=τ, τ≠0For the form (I), we have(16)Auia=λiuia+uia+1,Auiαi=λiuiαi,1≤i≤m,1≤a≤αi−1.And for the form (II), we have(17)Au¯jb=νu¯jb−τv¯jb+u¯jb+1,Au¯jβj=νu¯jβj−τv¯jβj,Av¯jb=τu¯jb+νv¯jb+v¯jb+1,Av¯jβj=τu¯jβj+νv¯jβj,with t+1≤j≤m,1≤b≤βj.


## 3. Theorems

Theorem 1.
LetNsn+1cbe an+1-dimensional pseudo-Riemannian space form with index s and constant sectional curvature c and Mrn be a nondegenerate biharmonic hypersurface of Nsn+1c with at most two distinct principal curvatures, then Mrn has constant mean curvature.Proof.
From Section2, the shape operator A has the form (I) or (II). If A has the form (II), then its eigenvalues are not real and −n/2εH is not an eigenvalue. It follows from (5) that ∇H=0, which tells us H is a constant.For the form (I), if we assume thatH is not a constant, then (5) implies that −n/2εH is an eigenvalue of the shape operator A. When λ1=⋯=λm, then trA=−n2/2εH. On the other hand, trA=nεH. These two expressions imply H=0, a contradiction.So, in the following, we need only to discuss the situation where there are two distinct values amongλ1,⋯,λm. Expression (5) also informs us that ∇H is an eigenvector of A with corresponding eigenvalue −n/2εH. In view of (16), ∇H is one of the directions uiαi,1≤i≤m. Without loss of generality, we suppose ∇H is in the direction of u1α1; 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 basisB given in Section 2, we can express
(18)∇H=∑i=1m∑a=1αiεiuiαi−a+1Huia.Since∇H is in the direction of u1α1, the above equation implies that
(19)u11H≠0,uiaH=0,ia≠11.Let∇uiaujb=∑k=1m∑d=1αkΓiajbkdukd. Applying compatibility condition to calculate
(20)∇uDuia,uia,∇uDuia,uib,∇uDuia,ujd,we conclude
(21)ΓDiaiαi−a+1=0,(22)ΓDiaiαi−b+1=−ΓDibiαi−a+1,ΓDiajαj−d+1=−εiεjΓDjdiαi−a+1,for D∈ke,1≤k≤m,1≤e≤αk, 1≤i,j≤m, 1≤a,b≤αi, and 1≤d≤αj.From the expression∇uBuC−∇uCuBH=uB,uCH, B,C∈kd,1≤k≤m,1≤d≤αk, and (19), we easily get
(23)ΓBC11=ΓCB11,B,C≠11.We state that in the proof, if not otherwise specified, then foria in uia and the connection’s coefficients, the ranges of i and a are as follows: 1≤i≤m and 1≤a≤αi. For the equations about the connection’s coefficients, when a=αi (or a=1), then the terms about ia+1 (or ia−1) disappear. And when b=βj (or b=1), then the terms about j¯b+1 (or j¯b−1) and j~b+1 (or j~b−1) disappear.It follows from the Codazzi equation (3) that for any vector fields X,Y,Z tangent to Mrn,
(24)∇XAY,Z=∇YAX,Z.Start with this equation, we can get a series of equations about the coefficients of connection.(i)
ForX,Y,Z=u1a,u1b,u1α1 in (24), then combining (19), we obtain(25)Γ1a1b+111=Γ1b1a+111.Applying (21), (22), and (23), we get from the above equation that
(26)Γ1α1−11α11α1−1=⋯=Γ121α112=Γ111α111=0.Note that ifα1=1, then (26) tells us nothing.
(ii)
ForX,Y,Z=u1a,uib,u1α1, 2≤i≤m, in (24), then combining (19), we have(27)λi+n2εHΓ1aib11+Γ1aib+111=Γib1a+111,2≤i≤m,which together with (23) implies that if λi=−n/2εH, then
(28)Γ1a+1iαi11=0,and if λi≠−n/2εH, then
(29)Γ1aib11=0.(iii)
ForX,Y,Z=uia,u1α1,u1α1−b+1, 2≤i≤m, in (24), then(30)−Γia1α11b−1=λi+nεH2Γ1α1ia1b+Γ1α1ia+11b−Γ1α1ia1b−1,which together with (22), (23), (28), and (29) implies that if λi≠−n/2εH, then
(31)Γ1α1ia1b=0,2≤i≤m.(iv)
ForX,Y,Z=uia,uib,u1α1,2≤i≤m, in (24), then combining (23), we know(32)Γiaib+111=Γibia+111.From (22), (23), and the above relation, we conclude that
(33)Γiαiib+111=0,2≤i≤m,(34)Γia1α1ib=Γia+11α1ib+1,2≤i≤m,1≤a,b≤αi−1.It follows from (33) and (34) that if a>b, then
(35)Γia1α1ib=0.(v)
ForX,Y,Z=uia,u1α1,uiαi−a+1,2≤i≤m, in (24), then(36)−n2εH−λiΓia1α1ia−Γia1α1ia−1=u1α1λi+Γ1α1ia+1ia−Γ1α1iaia−1,which together with (22) and (34) gives that
(37)Γ1α1i2i1=Γ1α1i3i2=⋯=Γ1α1iαiiαi−1=0,2≤i≤m,(38)u1α1λi=−n2εH−λiΓiαi1α1iαi,2≤i≤m.(vi)
ForX,Y,Z=uia,u1α1,uiαi,2≤i≤m,2≤a≤αi, in (24), then(39)−n2εH−λiΓia1α1i1=Γ1α1ia+1i1,which combining (35) and (37), implies that
(40)Γ1α1ibi1=0,2≤i≤m,2≤b≤αi.(vii)
ForX,Y,Z=uia,ujb,u1α1, 2≤i,j≤m, in (24), then combining (23), we have(41)λj−λiΓiajb11+Γiajb+111=Γjbia+111,which together with (23) tells us that when λi≠λj, then
(42)Γiajb11=0,2≤i,j≤m,and when λi=λj, then
(43)Γiαijb+111=0,(44)Γia1α1jb=Γia+11α1jb+1,with 2≤i,j≤m,1≤a≤αi,1≤b≤αj−1. (43) and (44) give that if a>b, then
(45)Γia1α1jb=0.(viii)
ForX,Y,Z=uia,u1α1,ujαj−b+1 with 2≤i,j≤m, and i≠j in (24), then(46)−nεH2−λjΓia1α1jb−Γia1α1jb−1=λi−λjΓ1α1iajb+Γ1α1ia+1jb−Γ1α1iajb−1.(ix)
ForX,Y,Z=uia,ujb,uiαi−d+1, a≠d, in (24), with λj=−n/2εH and λi≠−n/2εH, then(47)−n2εH−λiΓiajbid+Γiajb+1id−Γiajbid−1=Γjbia+1id−Γjbiaid−1.Let(48)Ai,1,jb=Γi2jbi1+Γi3jbi2+⋯+Γiαijbiαi−1,Ai,2,jb=Γi3jbi1+Γi4jbi2+⋯+Γiαijbiαi−2,⋮Ai,αi−1,jb=Γiαijbi1,with 1≤i,j≤m,1≤b≤αj and λj=−n/2εH, λi≠−n/2εH. Since (47), we find
(49)−n2εH−λiAi,e,jb+Ai,e,jb+1−Ai,e+1,jb=0,1≤b≤αj,1≤e≤αi−2,−n2εH−λiAi,αi−1,jb+Ai,αi−1,jb+1=0,1≤b≤αj,which implies that
(50)Ai,e,jb=0,1≤b≤αj,1≤e≤αi−1,with λj=−n/2εH, λi≠−n/2εH, and 1≤i,j≤m.
(x)
ForX,Y,Z=uia,ujb,uiαi−a+1 in (24), with λj=−n/2εH and λi≠−n/2εH, then(51)−n2εH−λiΓiajbia+Γiajb+1ia−Γiajbia−1=ujbλi+Γjbia+1ia−Γjbiaia−1,which combining (50) gives that
(52)−n2εH−λi∑a=1αiΓiajbia+∑a=1αiΓiajb+1ia=αiujbλi.(xi)
ForX,Y,Z=uia,ujb,ukαk−d+1 in (24), with λj=−n/2εH, λi=λk≠−n/2εH, and i≠k, then(53)−n2εH−λkΓiajbkd+Γiajb+1kd−Γiajbkd−1=Γjbia+1kd−Γjbiakd−1.Let(54)Bi,k,0,jb=Γi1jbk1+Γi2jbk2+⋯+Γiy0jbky0,y0=minαi,αk,Bi,k,1,jb=Γi2jbk1+Γi3jbk2+⋯+Γiy1jbky1−1,y1=minαi,αk+1,Bi,k,0,jb=Γi3jbk1+Γi4jbk2+⋯+Γiy2jbky2−2,y2=minαi,αk+2,⋮Bi,k,αi−1,jb=Γiαijbk1,with λj=−n/2εH, λi=λk≠−n/2εH, and i≠k. It follows from (53) that for 1≤b≤αj and 0≤e≤αi−2,
(55)−n2εH−λkBi,k,e,jb+Bi,k,e,jb+1−Bi,k,e+1,jb=0,−n2εH−λkBi,k,αi−1,jb+Bi,k,αi−1,jb+1=0,which implies that for λj=−n/2εH, λi=λk≠−n/2εH, and i≠k,
(56)Bi,k,e,jb=0,1≤b≤αj,0≤e≤αi−1.Now, we treat the two cases that∇H is not light-like or light-like separately and get contradictions.Case 1.
∇H is not light-like.
As∇H is in the direction of u1α1, ∇H is not light-like means that α1=1. Observe equation (38); we find if λi=−n/2εH, then u11H=0, which contradicts with (19). So, we conclude that λi≠−n/2εH, for 2≤i≤m. Since there are two distinct values among λ1,⋯,λm and trA=nεH, we have λ2=⋯=λm=3nεH/2n−1. It follows from (34) and (38) that
(57)Γia11ia=−3u11Hn+2H,2≤i≤m,1≤a≤αi.
DenoteW=Γia11ia, we have the following lemmas.Lemma 2.
We have(58)u11W+W2=3ε1n2H24n−1−cε1.Proof.
Sinceλi=3nεH/2n−1 with 2≤i≤m, it follows from (46) that
(59)−n2εH−λjΓia11j1=Γ11ia+1j1,2≤i,j≤m,i≠j,combining (43), we have
(60)Γ11ia+1j1=0,with 2≤i,j≤m, i≠j, and 2≤a≤αi−1.
Using Gauss equation forRu11,uiαiu11,ui1, with 2≤i≤m, combining (29), (31), (35), (40), (42), (43) and (60), we have
(61)u11Γiαi11iαi=−Γiαi11iαi2−∑2≤j≤m,j≠iΓiαi11jαjΓ11jαjiαi−Γjαj11iαiΓ11iαijαj−Γjαj−111iαiΓ11iαijαj−1+Γiαi11jαjΓjαj11iαi+3ε1n2H24n−1−cε1.
From (22) and (23), we know Γiαi11jαj=εiεjΓj111i1, Γjαj−111iαi=εiεjΓi111j2, with 2≤i,j≤m. So, we can rewrite the above equation as
(62)u11Γiαi11iαi=−Γiαi11iαi2−∑2≤j≤m,j≠iΓi111j1Γ11j1i1+Γi111j2Γ11j2i1−Γj111i1Γ11i1j1+Γi111j1Γj111i1+3ε1n2H24n−1−cε1,2≤i≤m.
For2≤i,j≤m and i≠j, if αj<αi, then we have Γj111i1=Γjαj11iαj=0 from (22), (43), and (44). And if αj=αi, from (56), we know
(63)Γj111i1+Γj211i2+⋯+Γjαj11iαi=0,which together with (44), tells us that
(64)Γj111i1=0.
So, for2≤i,j≤m and i≠j, if αj≤αi, then
(65)Γj111i1=0.
For2≤i,j≤m and i≠j, if αj>αi, combining (65), we get from (46) that
(66)−nn+2εH2n−1Γi111j2=Γ11i2j2−Γ11i1j1,−nn+2εH2n−1Γi211j3=Γ11i3j3−Γ11i2j2,⋮−nn+2εH2n−1Γiαi−111jαi=Γ11iαijαi−Γ11iαi−1jαi−1,−nn+2εH2n−1Γiαi11jαi+1=−Γ11iαijαi.
Ifαj>αi+1, then (43) and (44) give us that
(67)Γi111j2=⋯=Γiαi11jαi+1=0.
As (67), it follows from (66) that for 2≤i,j≤m and i≠j, if αj>αi+1, then
(68)Γ11i1j1=0.
For2≤i,j≤m and i≠j, if αj=αi+1, then combining (22) and (23), we get from (44) and (66) that
(69)Γi11α1j2=Γiαi1α1jαj=εiεjΓj11α1i1,Γ1α1i1j1=αiΓ1α1iαijαi=εiεjnn+2εH2n−1αiΓj11α1i1.
Therefore, from (59), (65), (67), (68), and (69), (62) can be simplified to
(70)nn+2εH2n−1εi∑j=2,αj=αi−1mεjαjΓi111j12−∑j=2,αj=αi+1mεjαi+1Γj111i12+u11Γi111i1+Γi111i12=3ε1n2H24n−1−cε1,2≤i≤m.
Choosingα11,α12,⋯,α1ξ1,α21,α22,⋯,α2ξ2,⋯,αh1,αh2,⋯,αhξh in α2,⋯,αm, such that
(71)α~i≔αi1=αi2=⋯=αiξi,α~i+1=α~i+1,1≤i≤h,and α~1−1,α~h+1∉α2,⋯,αm.
Taking sum in (70) for i=k1,k2,⋯,kξk, with 1≤k≤h, we have
(72)−α~1+1Q1=ξ1U,α~1Q1−α~2+1Q2=ξ2U,α~2Q2−α~3+1Q3=ξ3U,⋮α~h−2Qh−2−α~h−1+1Qh−1=ξh−1U,α~h−1Qh−1=ξhU,where
(73)U=−u11W−W2+3ε1n2H24n−1−cε1,Qk=nn+2εH2n−1∑1≤p≤ξk1≤q≤ξk+1ερεσΓσ11α1ρ12,ρ=kp,σ=k+1q.
From (72), we can get Qk=0, 1≤k≤h−1, and U=0, i.e., (58) holds.Lemma 3.
We have(74)∑i=2m∑a=1αiΓia11iau11H+u11u11H−ε1εn+8n2H34n−1+ncε1H=0.Proof.
For1≤k≤m, if αk is an even number, we put
(75)eka=uka−ukαk−a+12,kαk−a+1=uka+ukαk−a+12,with 1≤a≤αk/2. And if αk is an odd number, we take
(76)eka=uka−ukαk−a+12,kαk−a+1=uka+ukαk−a+12,1≤a<αk+12,and ekαk+1/2=ukαk+1/2. We easily find Ek=ek1,ek2,⋯,ekαk is an orthonormal basis of Vk with 1≤k≤m. Thus, E=eka,1≤k≤m,1≤a≤αk is an orthonormal basis of TxMrn.
From (19), (75), and (76), we have when αk is an even number,
(77)∇ekaekaH=12Γkaka11+Γkαk−a+1kαk−a+111−Γkakαk−a+111−Γkαk−a+1ka11u11H,∇ekαk−a+1ekαk−a+1H=12Γkaka11+Γkαk−a+1kαk−a+111+Γkakαk−a+111+Γkαk−a+1ka11u11H,with 1≤α≤αk/2, and when αk is an odd number,
(78)∇ekaekaH=12Γkaka11+Γkαk−a+1kαk−a+111−Γkakαk−a+111−Γkαk−a+1ka11u11H,∇ekαk−a+1ekαk−a+1H=12Γkaka11+Γkαk−a+1kαk−a+111+Γkakαk−a+111+Γkαk−a+1ka11u11H,∇ekαk+1/2ekαk+1/2H=Γkαk+1/2kαk+1/211u11H,with 1≤α<αk+1/2
Note thate11=u11; it follows from (19) that
(79)e11H≠0,eiaH=0,2≤i≤m,1≤a≤αi.
By calculating, we get(80)trA2=n+8n2H2410n−1.
Since the above, we obtain from (6) that
(81)∑i=2m∑a=1αiΓia11iau11H+u11u11H−ε1εn+8n2H34n−1+ncε1H=0.Now, we continue the proof of Theorem1 for Case 1.Combining (57) and (58), we get
(82)u11u11H−n+2n+5H9W2−n+2n2ε1H34n−1+13n+2cε1H.Substitute the above equation into (74), considering α2+⋯+αm=nα1=1 and (57), we have
(83)2n−4n+29W2+n+2+εn+8n2ε1H24n−1−232n+1cε1H=0.Act on (83) with u11, and using (57) and (58), we obtain
(84)2n−4n+29W2+n+2+εn+8n2ε1H24n−1−232n+1cε1u11H−2n+22n−49W2+n−10−εn+8n2ε1H212n−1+2n−49cε1HW=0.As (83) and W≠0, the above equation implies that
(85)2n−49W2+n−10−εn+8n2ε1H212n−1+2n−49cε1H=0,which together with (83) gives that
(86)−n+2n−13+n+8n+5εn2ε1H312n−1−29n−1n+5cε1H=0,which implies that H is a constant, a contradiction.Case 2.
∇H is light-like.
Notice that∇H is light-like means α1≥2. Since trA=nεH, we can suppose λ1=⋯=λp=−n/2εH and λp+1=⋯=λm=2+α1+⋯+αpnεH/2+αp+1+⋯+αm. It follows from (19) and (38) that
(87)Γi11α1i1=0,p+1≤i≤m.CalculateRu12,uiau1α1,uiαi−a+1 and Ru11,uiau1α1,uiαi−a+1 with p+1≤i≤m by Gauss equation, and letting i and a for sum with p+1≤i≤m and 1≤a≤αi, combining (21), (22), (23) (26), (28), (29), (35), (42), (45), and (87), we get
(88)∑i=p+1m∑a=1αi∑d=3α1−1Γ121α11dΓia1dia+∑k=2p∑d=2αkΓ121α1kdΓiakdia−∑i=p+1m∑1≤b<a≤αiΓia12ibΓib1α1ia−∑p+1≤k≤mk≠i∑a=1αi∑1≤b≤αkb<aΓia12kbΓkb1α1ia=0,(89)∑i=p+1m∑a=1αi∑d=2α1−1Γ111α11dΓia1dia+∑k=2p∑d=1αkΓ121α1kdΓiakdia−∑i=p+1m∑1≤b<a≤αiΓia11ibΓib1α1ia−∑p+1≤k≤mk≠i∑a=1αi∑1≤b≤αkb<aΓia11kbΓkb1α1ia=αp+1+⋯+αmcε1−12nε1Hλm.Since (50) and (56), we find
(90)∑i=p+1m∑1≤b<a≤αiΓia1dibΓib1α1ia=∑p+1≤k≤mk≠i∑a=1αi∑1≤b≤αkb<aΓia1dkbΓkb1α1ia=0.So, (88) and (89) can be simplified to
(91)∑i=p+1m∑a=1αi∑d=3α1−1Γ121α11dΓia1dia+∑k=2p∑d=2αkΓ121α1kdΓiakdia=0,(92)∑i=p+1m∑a=1αi∑d=2α1−1Γ111α11dΓia1dia+∑k=2p∑d=1αkΓ111α1kdΓiakdia=αp+1+⋯+αmcε1−12nε1Hλm.It follows from (52) that
(93)∑i=p+1m∑a=1αi−n2εH−λmΓiakdia+Γiakd+1ia=0,with 1≤k≤p and kd≠11.Multiply both sides of (92) with −n/2εH−λm, we get
(94)∑d=2α1−1Γ111α11d∑i=p+1m∑a=1αi−n2εH−λmΓia1dia+∑k=2p∑d=1αkΓ1a1α1kd∑i=p+1m∑a=1αi−n2εH−λmΓiakdia=ε1αp+1+⋯+αm−n2εH−λmc−12nHλm.Combining (87) and (93), it gives us
(95)∑d=2α1−2Γ111α11d∑i=p+1m∑a=1αiΓia1d+1ia−∑k=2p∑d=1αk−1Γ1a1α1kd∑i=p+1m∑a=1αiΓ1akd+1kd=ε1αp+1+⋯+αm−n2εH−λmc−12nHλm.Since (91), we get from (95) that
(96)−n2εH−λmc−12nHλm=0,which combining λm≠−n/2εH and λm=2+α1+⋯+αpnεH/2αp+1+⋯+αm, tells us that c−2+α1+⋯+αpn2εH2/4αp+1+⋯+αm=0.So,H is a constant, a contradiction.In view of the two cases, we complete the proof of Theorem1.Remark 4.
With the assumption that the shape operator is diagonalizable, the result of Theorem1 was proved in [6, 8, 12] when the hypersurface Mrn has at most three distinct principal curvatures.
Applying Theorem1, we can get the following theorem.Theorem 5.
LetMrn be a nondegenerate biharmonic hypersurface of pseudo-Riemannian space form Nsn+1c, with cε≤0. Suppose that Mrn has at most two distinct principal curvatures, which are all real numbers, then Mrn is minimal.Proof.
We know from Theorem1 that H is a constant; it follows from (6) that
(97)HtrA2=ncεH.
Whencε≤0, it is easy to see from (97) that H=0. When cε≤0, then (97) implies H=0, or trA2=0. Since trA2 is equal to the sum of the squares of all principal curvatures and trA2 is equal to the sum of all principal curvatures, so trA2=0 tells us trA2=0. Combining trA=nεH, we have H=0, i.e., Mrn 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 Theorem5, the following theorems are true.Theorem 6.
LetMr2k−1 be a nondegenerate biharmonic hypersurface of pseudo-Euclidean space Es2k. Suppose that Mr2k−1 has at most two distinct principal curvatures, then Mr2k−1 is minimal.Theorem 7.
LetMs−12k−1 be a nondegenerate biharmonic hypersurface of de Sitter space Ss2kc. Suppose that Ms−12k−1 has at most two distinct principal curvatures, then Ms−12k−1is minimal.Theorem 8.
LetMs2k−1 be a nondegenerate biharmonic hypersurface of anti-de Sitter space ℍs2kc. Suppose that Ms2k−1 has at most two distinct principal curvatures, then Ms2k−1 is minimal.Remark 9.
Under the assumption that the shape operator is diagonalizable, the results of Theorems6, 7, and 8 were proven not only for odd-dimensional hypersurfaces but also for even-dimensional hypersurfaces (cf. [6, 8]).



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