Linear Weingarten spacelike hypersurfacein locally symmetric Lorentz space

Wang Peijun Chao Xiaoli Bai Miaomiao

(School of Mathematics, Southeast University, Nanjing 211189, China)

Abstract：The rigidity of spacelike hypersurface Mn immersed in locally symmetric space is investigated, where the (normalized) scalar curvature R and mean curvature H of Mn satisfy R=aH+b, and a,b are real constants. First, an estimate of the upper bound of the function L(nH) is given, where L is a second-order differential operator. Then, under the assumption that the square norm of the second fundamental form is bounded by a given positive constant, it is proved that Mn must be either totally umbilical or contain two distinct principle curvatures, one of which is simple. Moreover, a similar result is obtained for complete noncompact spacelike hypersurfaces in locally symmetric Einstein spacetime. Hence, some known rigidity results for hypersurface with constant scalar curvature are extended for the linear Weingarten case.

Key words：spacelike hypersurface; linear Weingarten; locally symmetric Lorentz space

On the other hand, as a natural generalization of hypersurface with constant scalar curvature or with constant mean curvature, the linear Weingarten hypersurface has been extensively studied during the past decades[6-10]. A hypersurface is said to be linear Weingarten if its (normalized) scalar curvatureRand its mean curvatureHsatisfyR=aH+b, whereaandbare real constants. Motivated by this observation, Yang[10]extended the theorems in Refs.[4-5] to the linear Weingarten case.

1 Preliminaries

1≤A,B,C,…≤n+1, 1≤i,j,k,…≤n

The structure equations ofMnare

The Gauss equations are

(1)

The Codazzi and Ricci equations are

where the covariant derivative ofhijis defined as

Similarly, the componentshijklof the second derivative2hare given as

We then choose a local frame of orthonormal vector fields {ei} such that atp∈Mn,

hij=λiδij

Then it follows, atp, that

(2)

Setφij=hij-Hδij, and it is easy to confirm thatφis traceless and

Following Cheng and Yau[12], we introduce the operatorWassociated withφacting on any smooth functionfby

(3)

Then, settingf=nHin Eq.(3), we obtain

(4)

Moreover,the equality holds if and only if at least (n-1) ofβi′sare equal.

Lemma3[6]LetXbe a smooth vector field on the complete non-compact Riemannian manifoldMn, such that divMXdoes not change sign onMn, where div represents the divergence operator. If |X|∈L1(M), then divMX=0.

wherec=2c2+c1/n.

(5)

(6)

2nc2(S-nH2)

(7)

Then the lemma can be proven easily by substituting Eqs.(5) to (7) into Eq.(2).

(8)

wherec=2c2+c1/n.

ProofFirst, we obtain from Eq.(1) that

(9)

(10)

Applying Lemmas 2 and 4 to Eq.(10), we obtain

(11)

Letμi=λi-H, we can obtain

Using Lemma 1, we obtain that

(12)

Substituting (12) into (11), we have

2 Main Results

ProofFirst, we consider the quadratic form

By using the orthogonal transformation

We obtain that

(13)

Substituting(13) into (8), we obtain

(14)

On the other hand, sinceLis self-adjoint andMnis compact,

(15)

Furthermore, whenMnis complete noncompact, we have the following extension of Theorem 1.

ProofAccording to the proof in Theorem 1, we obtain

(16)

Noting thatLis elliptic andHattains its maximum onMn, by using the maximum principle, we can obtain thatHis a constant. Consequently,

Hence,λiis constant for eachi=1,2,…,n. Furthermore,L(nH)=0 and we obtain from (16) that

(17)

By using the same argument as in Theorem 1, the proof can be completed easily.

Remark1SinceSis bounded,His bounded as well. ThenHcan attain its maximum onMnsincehijk≥0. Hence, the assumption thatHcan attain the maximum onMnin Theorem 2 can be removed. So, the main difference between Theorem 2 and Theorem 1.6 in Ref.[10] lies in the assumption that supHis attained at some points or not.

If the metric and Ricci tensors of a Lorentz space are homotetic[15], we call it Einstein spacetime. For the spacelike hypersurface in Einstein spacetime, we have the following result.

ProofAccording to Ref.[15], we have

L(nH)=divM(P(H))

|P(H)|∈L1(M)

(18)

Thus, from (16), (18) and Lemma 3, we obtain thatL(nH)=0. By using the same argument as in Theorem 2, the proof is completed easily.