Novi Sad J. Math.
Vol. XX, No. Y, 20ZZ, ??-??
Draft version November 26, 2015
⃗ = αH
⃗ with complex
Lorentz Hypersurfaces satisfying △H
eigen values
Deepika1 and Ram Shankar Gupta2
Abstract. In this paper, we study Lorentz hypersurface M1n in E1n+1
⃗ = αH
⃗ with minimal polynomial [(y − λ)2 + µ2 ](y − λ1 )(y −
satisfying △H
λn ) having shape operator (2.11). We prove that every such Lorentz hypersurface in E1n+1 having at most four distinct principal curvatures has
constant mean curvature.
AMS Mathematics Subject Classification (2000): 53D12, 53C40, 53C42
Key words and phrases: Pseudo-Euclidean space, Biharmonic submanifolds, Mean curvature vector.
1.
Introduction
The study of submanifolds with harmonic mean curvature vector field was
initiated by B. Y. Chen in 1985 and arose in the context of his theory of
submanifolds of finite type. For a survey on submanifolds of finite type and
various related topics was presented in [8, 9]. Let Mrn be an n-dimensional,
⃗
connected submanifold of the pseudo-Euclidean space Esm . Denote by ⃗x, H,
and △ respectively the position vector field, mean curvature vector field of Mrn ,
and the Laplace operator on Mrn , with respect to the induced metric g on Mrn ,
from the indefinite metric on the ambient space Esm . As it is well known that
[7]
(1.1)
⃗
△⃗x = −nH.
A submanifold Mrn of Esm satisfying the condition
(1.2)
⃗ = 0,
△H
is called biharmonic submanifold. In view of (1.1), condition (1.2) is equivalent
to △2 ⃗x = 0. Equation (1.2) is the special case of the equation
(1.3)
⃗ = αH.
⃗
△H
1 University School of Basic and Applied Sciences, Guru Gobind Singh Indraprastha University, Sector-16C, Dwarka, New Delhi-110078, India,
e-mail: sdeep2007@gmail.com
2 Assistant Professor, University School of Basic and Applied Sciences, Guru Gobind Singh
Indraprastha University, Sector-16C, Dwarka, New Delhi-110078, India,
e-mail: ramshankar.gupta@gmail.com
Deepika, Ram Shankar Gupta
As remarked, minimal submanifolds are immediately seen to be biharmonic.
Conversely, the question arises whether the class of submanifolds with harmonic
mean curvature vector field is essentially larger than the class of minimal submanifolds. Concerning this problem B. Y. Chen conjectured the following:
Conjecture: The only biharmonic submanifolds of Euclidean spaces are the
minimal ones.
In Euclidean spaces, we have the following results, which indeed support the
above mentioned conjecture. B. Y. Chen proved in 1985 that every biharmonic
surface in E 3 is minimal. Thereafter, I. Dimitric generalized this result [13].
In [17], it was proved that every biharmonic hypersurface in E 4 is minimal.
Recently, it was proved that every biharmonic hypersurface with three distinct
principal curvatures in E n+1 with arbitrary dimension is minimal [16].
The study of equation (1.3) for submanifolds in pseudo-Euclidean spaces
was originated by Ferrandez et al. in [4, 5]. They showed that if the minimal
polynomial of the shape operator of a hypersurface Mrn−1 (r = 0, 1) in E1n
is at most of degree two, then Mrn−1 has constant mean curvature. Also, in
[8] various classification theorems for submanifolds in a Minkowski spacetime
were obtained. In [1], it was proved that every hypersurface Mr3 (r = 0, 1,
2, 3) of Es4 satisfying equation (1.3) whose shape operator is diagonal, has
constant mean curvature. Also, in [3] the same conclusion was obtained for
every hypersurface M13 in E14 . Recently, it was proved that every hypersurface
having at most three distinct principal curvatures in Esn+1 satisfying (1.3) with
diagonal shape operator has constant mean curvature [14].
In contrast to the submanifolds of Euclidean spaces, Chen’s conjecture is not
true always for the submanifolds of the pseudo-Euclidean spaces. For example,
B. Y. Chen et al. [11, 12] obtained some examples of proper biharmonic surfaces
in 4-dimensional pseudo-Euclidean spaces Es4 for s = 1, 2, 3 (see also [10]).
But for hypersurfaces in pseudo-Euclidean spaces, it is reasonable that Chen’s
conjecture is also right. This is supported by the following facts: B. Y. Chen et
al. proved in [11, 12] that biharmonic surfaces in pseudo-Euclidean 3-spaces are
minimal, and A. Arvanitoyeorgos et al. [2] proved that biharmonic Lorentzian
hypersurfaces in Minkowski 4-spaces are minimal.
In this paper, we study Lorentz hypersurfaces M1n in E1n+1 satisfying (1.3)
and having shape operator (2.11).
2.
Preliminaries
Let (M1n , g) be a n-dimensional Lorentz hypersurface isometrically immersed
in a n + 1-dimensional pseudo-Euclidean space (E1n+1 , g) and g = g |M1n . We
denote by ξ unit normal vector to M1n with g(ξ, ξ) = 1.
Let ∇ and ∇ denote linear connections on E1n+1 and M1n , respectively.
Then, the Gauss and Weingarten formulae are given by
(2.1)
∇X Y = ∇X Y + h(X, Y ), ∀ X, Y ∈ Γ(T M1n ),
(2.2)
∇X ξ = −Sξ X, ∀ ξ ∈ Γ(T M1n )⊥ ,
Draft version November 26, 2015
2
⃗ = αH
⃗ with complex eigen values
Lorentz Hypersurfaces satisfying △H
3
where h is the second fundamental form and S is the shape operator. It is well
known that the second fundamental form h and shape operator S are related
by
Draft version November 26, 2015
(2.3)
g(h(X, Y ), ξ) = g(Sξ X, Y ).
The mean curvature vector is given by
⃗ = 1 traceh.
H
n
(2.4)
The Gauss and Codazzi equations are given by
(2.5)
R(X, Y )Z = g(SY, Z)SX − g(SX, Z)SY,
(2.6)
(∇X S)Y = (∇Y S)X,
respectively, where R is the curvature tensor, S = Sξ for some unit normal
vector field ξ and
(2.7)
(∇X S)Y = ∇X (SY ) − S(∇X Y ),
for all X, Y, Z ∈ Γ(T M1n ).
The necessary and sufficient conditions for M1n to have proper mean curvature in E1n+1 [1] are
(2.8)
(2.9)
△H + HtraceS 2 = αH,
S(gradH) +
n
HgradH = 0,
2
where H denotes the mean curvature. Also, the Laplace operator △ of a scalar
valued function f is given by [11]
(2.10)
△f = −
n
∑
ϵi (ei ei f − ∇ei ei f ),
i=1
where {e1 , e2 , ..., en } is an orthonormal local tangent frame on M1n with ϵi =
±1.
A vector X in Esn+1 is called spacelike, timelike or lightlike according as
g(X, X) > 0, g(X, X) < 0 or g(X, X) = 0, respectively. A non-degenerate
hypersurface Mrn of Esn+1 is called Riemannian or pseudo-Riemannian according as the induced metric on Mrn+1 from the indefinite metric on Esn+1 is
definite or indefinite. A shape operator of pseudo-Riemannian hypersurfaces is
not diagonalizable always unlike the Riemannian hypersurfaces.
4
Deepika, Ram Shankar Gupta
The matrix representation of shape operator of M1n in E1n+1 having minimal
polynomial [(y−λ)2 +µ2 ](y−λ1 )(y−λn ) with respect to a suitable orthonormal
base field of the tangent bundle takes the form [6, 15]
(2.11)
λ −µ
 µ λ



S=




λ1
···




,




···
λ1
λn
for some smooth functions λ, λ1 , λn and µ.
⃗ = αH
⃗
Lorentz Hypersurfaces in E1n+1 satisfying △H
3.
We assume that H is not constant and gradH ̸= 0. Assuming non constant
mean curvature implies the existence of an open connected subset U of M1n ,
with gradp H ̸= 0 for all p ∈ U . From (2.9), it is easy to see that gradH is an
eigenvector of the shape operator S with the corresponding principal curvature
− n2 H. In view of (2.11), the shape operator S of hypersurfaces will take the
following form
(3.1)
S(e1 ) = λe1 + µe2 , S(e2 ) = −µe1 + λe2 , S(eb ) = λ1 eb , S(en ) = λn en ,
for b = 3, 4, . . . , n−1, with respect to orthonormal basis {e1 , e2 , ..., en } of Tp M1n ,
which satisfies
(3.2)
g(e1 , e1 ) = −1,
g(ei , ei ) = 1, i = 2, 3, ..., n,
and
g(ei , ej ) = 0, f or i ̸= j.
(3.3)
We write
(3.4)
∇ei ej =
n
∑
k
ωij
ek ,
i, j = 1, 2, ..., n.
k=1
Using (3.4) and taking covariant derivatives of (3.2) and (3.3) with ek , we
find
(3.5)
i
ωki
= 0,
for i ̸= j and i, j, k = 1, 2, ..., n.
j
i
ωkj
= −ωki
,
Draft version November 26, 2015


⃗ = αH
⃗ with complex eigen values
Lorentz Hypersurfaces satisfying △H
5
Draft version November 26, 2015
Now, we consider following two cases:
Case A: If λ1 ̸= λn . We can choose gradH in the direction of en or ek for
k = 3, 4, .∑
. . , n − 1. In both the cases, gradH is space like. We can express
n
gradH = i=1 ei (H)ei . Assuming gradH in the direction of en , we get gradH=
en (H)en . Also, from (2.9), we obtain the corresponding eigen value λn = − nH
2 .
3nH
2λ
Using (2.4) and (2.11), we find λ1 = 2(n−3)
− n−3
. Also, we have
(3.6)
en (H) ̸= 0,
e1 (H) = e2 (H) = ... = en−1 (H) = 0.
Using (3.4), (3.6) and the fact that [ei ej ](H) = 0 = ∇ei ej (H) − ∇ej ei (H),
for i ̸= j and i, j ̸= n, we find
n
n
ωij
= ωji
.
(3.7)
Taking inner product of (2.6) with Z ∈ T M1n , we get
(3.8)
g((∇X S)Y, Z) = g((∇Y S)X, Z).
Using (3.1), (3.4), (3.6), (3.8) and the value of λ1 and λn , we get following
equations (3.9)∼(3.26), where j ̸= k and j, k = 3, 4, ..., n − 1:
For X = e1 , Y = e2 , Z = en
[λ +
(3.9)
nH
n
n
n
n
).
+ ω11
) = µ(ω22
− ω21
](ω12
2
For X = e1 , Y = ek , Z = ek
(3.10)
e1 (−
2λ
(n − 1)λ
3nH
k
)=[
−
]ω k + µωk2
.
n−3
n−3
2(n − 3) k1
For X = e1 , Y = ek , Z = ej
(3.11)
[
3nH
(n − 1)λ
j
−
]ω j + µωk2
= 0.
n−3
2(n − 3) k1
For X = e1 , Y = ek , Z = en
(3.12)
[
n2 H
2λ
nH n
n
n
−
]ω1k
= [λ +
]ωk1 + µωk2
.
2(n − 3) n − 3
2
For X = e2 , Y = ek , Z = ek
(3.13)
e2 (−
2λ
(n − 1)λ
3nH
k
.
)=[
−
]ω k − µωk1
n−3
n−3
2(n − 3) k2
For X = e2 , Y = ek , Z = ej
(3.14)
[
3nH
(n − 1)λ
j
−
]ω j − µωk1
= 0.
n−3
2(n − 3) k2
6
Deepika, Ram Shankar Gupta
For X = e2 , Y = ek , Z = en
(3.15)
[
n2 H
2λ
nH n
n
n
−
]ω2k
= [λ +
]ωk2 − µωk1
.
2(n − 3) n − 3
2
− (λ +
(3.16)
nH 1
2
)ω1n + µω1n
= en (λ).
2
For X = e1 , Y = en , Z = e2
− (λ +
(3.17)
nH 2
1
)ω1n − µω1n
= en (µ).
2
For X = e1 , Y = en , Z = en
(3.18)
(λ +
nH n
n
)ωn1 + µωn2
= 0.
2
For X = e2 , Y = en , Z = e1
− (λ +
(3.19)
nH 1
2
)ω2n + µω2n
= −en (µ).
2
For X = e2 , Y = en , Z = e2
− (λ +
(3.20)
nH 2
1
)ω2n − µω2n
= en (λ).
2
For X = e2 , Y = en , Z = en
(λ +
(3.21)
nH n
n
= 0.
)ωn2 − µωn1
2
For X = ek , Y = en , Z = e1
(3.22)
− (λ +
nH 1
(n − 1)λ
3nH
2
2
)ωkn + µωkn
= −[
−
]ω 1 + µωnk
.
2
n−3
2(n − 3) nk
For X = ek , Y = en , Z = e2
(3.23)
− (λ +
nH 2
(n − 1)λ
3nH
1
1
)ωkn − µωkn
= −[
−
]ω 2 − µωnk
.
2
n−3
2(n − 3) nk
For X = ek , Y = en , Z = ek
(3.24)
en (
2λ
n2 H
2λ
3nH
−
) = −[
−
]ω k .
2(n − 3) n − 3
2(n − 3) n − 3 kn
For X = ek , Y = en , Z = ej
(3.25)
j
ωkn
= 0.
Draft version November 26, 2015
For X = e1 , Y = en , Z = e1
⃗ = αH
⃗ with complex eigen values
Lorentz Hypersurfaces satisfying △H
7
For X = ek , Y = en , Z = en
n
ωnk
= 0.
(3.26)
From (3.25), (3.26) and (3.5), we have
j
n
n
k
ωkn
= ωnk
= ωkj
= ωnn
= 0,
Draft version November 26, 2015
(3.27)
for j ̸= k and j, k = 3, 4, ..., n − 1.
Using (3.12), (3.15), (3.7) and (3.5), we find
(3.28)
n
k
n
2
n
n
1
k
ω2k
= ω2n
= ωk2
= ωkn
= ωk1
= ω1k
= ωkn
= ω1n
= 0,
for k = 3, 4, ..., n − 1.
Also, from (3.16), (3.17), (3.19), (3.20), (3.9), (3.7) and (3.5), we obtain
(3.29)
1
n
n
2
n
n
2
1
ω2n
= ω21
= ω12
= ω1n
= ω22
= ω11
= ω2n
= ω1n
= 0.
Using (3.11), (3.14) and (3.5), we get
j
j
1
2
ωk1
= ωk2
= ωkj
= ωkj
= 0,
(3.30)
for j ̸= k and j, k = 3, 4, ..., n − 1.
Similarly, from (3.18), (3.21) and (3.5), we find
n
2
n
1
ωn2
= ωnn
= ωn1
= ωnn
= 0.
(3.31)
Also, from (3.22), (3.23), (3.28) and (3.5), we obtain
1
2
k
k
ωnk
= ωnk
= ωn1
= ωn2
= 0,
(3.32)
for k = 3, 4, ..., n − 1.
Now, we have the following:
Lemma 3.1. Let M1n be a Lorentz hypersurface in E1n+1 , having the shape
operator (2.11) with respect to suitable orthonormal basis {e1 , e2 , ..., en }. If
gradH is space like and in the direction of en , then
∇e1 e1 =
∑
p
ω11
e p , ∇ e1 e 2 =
p̸=1,n
∇e2 en = 0, ∇ek ek =
∑
p
ω12
e p , ∇ e2 e 1 =
p̸=2,n
∑
p
ωkk
ep , ∇ek ej =
p̸=k
∑
p̸=1,n
∑
p
ωkj
ep , ∇ek en =
p̸=1,2,j,n
1
2
∇en e1 = ωn1
e2 , ∇en e2 = ωn2
e1 , ∇en ek =
∑
p̸=1,2,k,n
for j ̸= k and j, k = 3, 4, ..., n − 1.
∑
p
ω21
ep , ∇e2 e2 =
p
ω22
ep ,
p̸=2,n
∑
p̸=1,2,j,n
p
ωnk
ep , ∇en en = 0,
p
ωkn
ep ,
8
Deepika, Ram Shankar Gupta
Using Lemma 3.1 and (2.5) to evaluate g(R(e2 , en )e2 , en ), we obtain
nH
λ=0
2
(3.33)
or λ = 0,
k
2
k
1
ωk2
= ωkk
= ωk1
= ωkk
= 0,
(3.34)
for k = 3, 4, ..., n − 1.
2
(n+6)H 2
Also, using traceS 2 = n 4(n−3)
−2µ2 , (2.10) and Lemma 3.1, the equation
(2.8) with respect to the basis {e1 , e2 , ..., en } reduces
(3.35)
− en en (H) +
n−1
∑
n
ωkk
en (H) + H(
k=3
n2 (n + 6)H 2
− 2µ2 ) = αH.
4(n − 3)
Using (2.5) and Lemma 3.1 to evaluate g(R(ek , en )en , ek ), we get
(3.36)
n
n 2
en (ωkk
) − (ωkk
) =−
3n2 H 2
,
4(n − 3)
for k = 3, 4, ..., n − 1.
Using (3.33) in (3.24), we have
n
ωkk
=
(3.37)
3en (H)
,
nH
for k = 3, 4, ..., n − 1.
From (3.37), we find
(3.38)
n
n
n
ω33
= ω44
= · · · = ω(n−1)(n−1)
.
Differentiating (3.37) along en and using (3.36), we get
(3.39)
en en (H) =
n(n + 3)H n 2
n3 H 3
(ωkk ) −
,
9
4(n − 3)
for k = 3, 4, ..., n − 1.
Using (3.38) and (3.39) in (3.35), we obtain
(3.40)
2
n2 (n + 3) 2
n 2
n(n − 6)(ωkk
) +
H − 2µ2 = α,
9
2(n − 3)
for k = 3, 4, ..., n − 1.
Differentiating (3.40) with respect to en and using (3.17), (3.29), (3.36) and
(3.37), we find
(3.41)
4
3n2 2
n 2
(n − 6)(ωkk
) +
H = 0,
9
n−3
Draft version November 26, 2015
as H ̸= 0.
Using (3.33) and (3.5) in (3.10) and (3.13), we find
⃗ = αH
⃗ with complex eigen values
Lorentz Hypersurfaces satisfying △H
9
for k = 3, 4, ..., n − 1.
Again, acting along en on (3.41) and using (3.36) and (3.37), we get
(3.42)
8(n − 6) n 2 4n2 (n + 3) 2
(ωkk ) +
H = 0,
9
3(n − 3)
Draft version November 26, 2015
for k = 3, 4, ..., n − 1.
Hence, from (3.41) and (3.42), we obtain that H must be zero.
Case B: If λ1 = λn , then we get λ1 = λn = −nH
2 . Using (2.11) and (2.4),
n2 H
we find λ = 4 . Then, from (3.24), we get en (H) = 0, which is a contradiction of (3.6), and therefore H must be constant.
Combining case A and case B, we have
⃗ = αH,
⃗
Theorem 3.2. Every Lorentz hypersurface M1n in E1n+1 satisfying △H
having shape operator given by (2.11) with at most four distinct principal curvatures has constant mean curvature.
Acknowledgement
The authors are thankful to the referee for useful suggestions to improve the
article. The first author is also grateful to Guru Gobind Singh Indraprastha
University for providing IPRF fellowship to pursue research.
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