Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 3 Issue 3(2011), Pages 97-108.
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b IN DE
SITTER SPACE FORM Spn+p (c)
(COMMUNICATED BY UDAY CHAND DE)
YINGBO HAN, SHUXIANG FENG
Abstract. In this paper, we investigate n-dimensional complete spacelike
sumbmanifolds M n (n ≥ 3) with R = aH + b in de Sitter space form Spn+p (c).
Some rigidity theorems are obtained for these spacelike submanifolds.
1. Introduction
Spn+p (c)
A de Sitter space form
is an (n + p)-dimensional connected pseudoRiemannian manifold of index p with constant sectional curvature c > 0. A submanifold immersed in Spn+p (c) is said to be spacelike if the induced metric in M n
from the metric of the ambient space Spn+p (c) is positive definite. Since Goddard’s
conjecture (see [6]), several papers about spacelike hypersurfaces with constant
mean curvature in de Sitter space have been published. For the study of spacelike hypersurface with constant scalar curvature in de Sitter space, there are also
many results such as [2, 9, 15, 16]. There are some results about space like sumanifolds with constant scalar curvature and higher codimension in de Sitter space
form Spn+p (c) such as [17]. Recently, F.E.C.Camargo, et al.[5] and Chao X.L. [4]
obtained some interesting characters for space like submanifolds with parallel normalized mean vector(which is much weaker than the condition to have parallel
mean curvature vector) in Spn+p (c). In this note, we consider complete space like
submanifolds with R = aH + b in de Sitter space and we get the following results:
Theorem 1.1. Let M n (n ≥ 3) be a complete space like submanifold
with R =
√
aH + b, (n − 1)a2 + 4nc − 4nb ≥ 0 and a ≥ 0 in Spn+p (c). If S < 2 (n − 1)c, then
M n is totally umbilical.
Theorem 1.2. Let M n (n ≥ 3) be a complete space like submanifold with R =
aH + b, (n − 1)a2 + 4nc − 4nb ≥ 0 and a ≥ 0 in de Sitter space form Spn+p (c).
Suppose that M n has bounded mean curvature H :
2000 Mathematics Subject Classification. 53C42,53B30.
Key words and phrases. space like submanifolds, de Sitter space.
c
⃝2011
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted June 1, 2011. Published June 20, 2011.
Y.H. was supported by NSFC No. 10971029, NSFC-TianYuan Fund No.11026062, Project of
Henan Provincial Department of Education No.2011A110015 and Talent youth teacher fund of
Xinyang Normal University.
97
98
YINGBO HAN, SHUXIANG FENG
4(n−1)
2
n
(n−2)2 p+4(n−1) c, then S = nH and M is totally umbilical.
2
(2) If sup(H)2 = (n−2)4(n−1)
and M n is totally
2 p+4(n−1) c, then either S = nH
2 2
p +4(n−1)
umbilical, or sup(S) = nc (n−2)
(n−2)2 p+4(n−1) .
2
(3-a) If sup(H)2 > c > (n−2)4(n−1)
and M n is totally
2 p+4(n−1) c, then either S = nH
2
+
(1) If sup(H)2 <
umbilical, or n sup H < sup S ≤ S .
2
(3-b) If sup(H)2 = c > (n−2)4(n−1)
and M n is totally
2 p+4(n−1) c, then either S = nH
umbilical, or n sup H 2 < sup S ≤ S + .
2
(3-c) If c > sup(H)2 > (n−2)4(n−1)
and M n is totally
2 p+4(n−1) c, then either S = nH
umbilical, or S − ≤ sup S ≤ S + .
√
2
2 −4(n−1)c
2(n−2)2 p2 +4(n−1)(p+1)
Where S + = n(n−2)
sup H 2 +sup |H|p2 [(n−2) p+4(n−1)]supH
]−
2(n−1) [
2(n−2)
p
√
2 2
2
2 −4(n−1)c
2(n−2) p +4(n−1)(p+1)
sup H 2 −sup |H|p2 [(n−2) p+4(n−1)]supH
]−
npc, and S − = n(n−2)
2(n−1) [
2(n−2)
p
npc.
Remark. (i) We take the parallel normalized mean curvature vector off from our
theorems. (ii)From the conclusion (1) in theorem 1.2, we obtain the theorem 1.2 in
[4] .
Recently, the first author in [7] obtained an intrinsic inequality for space like
submanifolds in Spn+p (c),
Theorem 1.3. [7] If M n (n > 1) is a complete space-like submanifold of indefinite
space form Mpn+p (c)(c > 0) (p ≥ 1), Ric and R are Ricci curvature tensor and the
normalized scalar curvature of M n , respectively, then
|Ric|2 ≥ 2cRn(n − 1)2 − c2 n(n − 1)2 .
(1.1)
Moreover, |Ric|2 = 2cRn(n − 1)2 − c2 n(n − 1)2 if and only if M n is a spacelike
Einstein submanifolds with Ricij = c(n − 1)gij , where g is the Riemannian metric
of M n .
In this note, we also obtain the following result:
Theorem 1.4. Let M n (n ≥ 3) be a complete space-like submanifold of de Sitter
space form Spn+p (c) (p > 1). If the mean curvature satisfies the following inequality:
H2 <
c
,
(n − 1)(1 − p1 )
(1.2)
then |Ric|2 = 2cRn(n − 1)2 − c2 n(n − 1)2 if and only if M n is totally geodesic,
where Ric and R are Ricci curvature tensor and the normalized scalar curvature of
M n , respectively.
2. Preliminaries
We choose a local field of semi-Riemannian orthonormal frames {e1 , · · · , en , en+1 ,
· · · , en+p } in Spn+p (c) such that, restricted to M n , e1 , · · · , en are tangent to M n .
Let ω1 , · · · , ωn+p be its dual frame field such that the semi-Riemannian metric of
∑n+p
Spn+p (c) is given by ds2 = A=1 ϵA (ωA )2 , where ϵi = 1, i = 1, · · · , n and ϵα = −1,
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b
99
α = n + 1, · · · , n + p. Then the structure equations of Spn+p (c) are given by
∑
dωA = −
ϵB ωAB ∧ ωB , ωAB + ωBA = 0,
(2.1)
B
dωAB = −
∑
ϵC ωAC ∧ ωCB +
C
1∑
KABCD ωC ∧ ωD ,
2
(2.2)
CD
KABCD = cϵA ϵB (δAC δBD − δAD δBC ).
(2.3)
n
We restrict these forms to M , then
α = n + 1, · · · , n + p,
ωα = 0,
n
2
and the Riemannian metric of M is written as ds =
∑
0 = dωα = −
ωα,i ∧ ωi ,
(2.4)
∑
2
i ωi .
Since
(2.5)
i
by Cartan’s lemma we may write
∑
ωα,i =
hα
ij ωj ,
α
hα
ij = hji .
(2.6)
j
From these formulas, we obtain the structure equations of M n :
∑
dωi = −
ωij ∧ ωj , ωij + ωji = 0,
j
dωij = −
∑
ωik ∧ ωkj +
k
1∑
Rijkl ωk ∧ ωl ,
2
Rijkl = c(δik δjl − δil δjk ) −
k,l
α
(hα
ik hjl
α
− hα
il hjk ),
(2.7)
n
where Rijkl are the components of curvature tensor of M . We call
∑
B=
hα
ij ωi ⊗ ωj ⊗ eα
(2.8)
i,j,α
∑
the second fundamental form of M n . The mean curvature vector is h = n1 i,α hα
ii eα =
∑ α
∑
∑ α
1
α
α 2
2
2
α σ eα , where σ = n
i,j,α (hij ) , and H = |h| . We
i hii . We denote S =
call that M n is maximal if its mean curvature field vanishes, i.e. h = 0.
α
Let hα
ij,k and hij,kl denote the covariant derivative and the second covariant
α
α
derivative of hij . Then we have hα
ij,k = hik,j and
∑
∑
∑ β
α
hij Rαβkl ,
(2.9)
hα
hα
hα
ij,kl − hij,lk = −
im Rmjkl −
mj Rmikl −
m
m
β
where Rαβkl are the components of the normal curvature tensor of M n .
∑
β
β α
Rαβkl = −
(hα
ik hil − hik hil ),
(2.10)
i
Ricik = (n − 1)cδik −
∑
∑∑
α
α
(
hα
hα
ll )hik +
il hlk ,
α
l
n(n − 1)R = n(n − 1)c + S − n2 H 2 ,
where R is the normalized scalar curvature.
(2.11)
l
(2.12)
100
YINGBO HAN, SHUXIANG FENG
∑ α
α
α
The Laplacian △hα
ij of hij is defined by △hij =
k hijkk . From (2.9) we have
∑
∑
∑
∑ β
△hα
hα
hα
hα
(2.13)
hki Rαβjk .
ij =
kkij −
km Rmijk −
mi Rmkjk −
k
Now, we assume H > 0. We choose en+1 =
tr(H n+1 ) = nH,
α
where H denote the matrix
h
H.
α ̸= n + 1,
tr(H α ) = 0,
(hα
ij ).
Hence
(2.14)
Let us define
Φn+1
= hn+1
− Hδij ,
ij
ij
α
Φα
ij = hij ,
α ̸= n + 1.
(2.15)
Φn+1 = H n+1 − HI,
Φα = H α ,
α ̸= n + 1.
(2.16)
Therefore
α
where Φ denotes the matrix
(Φα
ij ).
Then
∑
|Φn+1 |2 = tr(H n+1 )2 − nH 2 ,
|Φα |2 =
α̸=n+1
∑
2
(hα
ij ) ,
tr(Φβ ) = 0, (2.17)
α̸=n+1
for ∀β ∈ {n + 1, · · · , n + p}. Thus
n+p
∑
|Φ|2 =
|Φα |2 = S − nH 2 .
(2.18)
α=n+1
∑
2
Set S1 = tr(H n+1 )2 and S2 = α̸=n+1 (hα
ij ) , so S = S1 + S2 , where S1 , S2 are
well defined on M .
By replacing (2.7) (2.10) and (2.14) into (2.13), we get the following equations:
∑
∑
β
β
β
β
△hn+1
= nchn+1
− nHcδij + nHij +
hn+1
hn+1
ij
ij
km hmk hij − 2
km hmj hik
∑
+
β,k,m
β
β
hn+1
mi hmk hkj
∑
− nH
and
= nchα
ij − nHcδij + nHij +
β
β
hα
mi hmk hkj
− nH
β
β
hα
km hmk hij − 2
n+1
hα
mi hmj
∑
+
m
β,k,m
Since
∑
∑
β,k,m
∑
β
β
hn+1
jm hmk hki (2.19)
β,k,m
β,k,m
∑
+
+
m
β,k,m
△hα
ij
n+1
hn+1
mi hmj
∑
β
β
hα
km hmj hik
β,k,m
β
β
hα
jm hmk hki .
(2.20)
β,k,m
∑
∑
1
2
α
(hα
hα
△S =
ijk ) +
ij △hij ,
2
α,ij
(2.21)
α,ijk
from (2.19) and (2.20), we have that
1
△S
2
=
∑
n+1
2
2
2
(hα
ijk ) + hij (nH)ij + ncS − n cH − nH
α,ijk
+
∑
α,β
α
β
2
[tr(H H )] +
∑
∑
tr(H n+1 (H α )2 )
α
N (H H − H H ),
α
α,β
where N (A) = trAA′ for all matrix A = (aij ).
β
β
α
(2.22)
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b
101
In this note we consider the spacelike submanifolds with R = aH + b in de Sitter
space from Spn+p (c), where a, b are real constants. Following Cheng-Yau [3], Chao
X.L. in [4] introduced a modified operator acting on any C 2 -function f by
∑
n−1
L(f ) =
(nHδij − hn+1
a∆f.
(2.23)
ij )fij +
2
ij
We need the following Lemmas.
Lemma 2.1. [4]Let M n be a spacelike submanifolds in Spn+p (c) with R = aH + b,
and (n − 1)a2 + 4nc − 4nb ≥ 0. We have the following:
(1)
∑
2
2
2
(hα
(2.24)
ijk ) ≥ n |∇H| .
(2)If the mean curvature H of M n is bounded and a ≥ 0, then there is a sequence
of points {xk } ∈ M n such that
lim nH(xk ) = sup(nH),
k→∞
lim |∇nH(xk )| = 0,
k→∞
lim sup(L(nH)(xk )) ≤ 0.
(2.25)
k→∞
Lemma
2.2. [1, 11] Let µ1 , · · · , µn be real numbers such that
∑ 2
2
=
β
, where β ≥ 0 is constant. Then
µ
i i
∑
n−2
|
µ3i | ≤ √
β3,
n(n
−
1)
i
∑
i
µi = 0 and
(2.26)
and equality holds if and only if at least n − 1 of µ′i s are equal.
∑
Lemma
∑ 2.3. [13] Let xi , yi i = 1, · · · , n, be the real numbers such that i xi =
0 = i yi . Then
∑
∑
∑
1
n−2
|
(2.27)
x2i yi | ≤ √
(
x2i )(
yi2 ) 2 .
n(n
−
1)
i
i
i
∑
Lemma 2.4. [7] Let a1 , · · · , an and b1 , · · · , bn be real numbers such that i bi = 0.
Then
∑
∑
∑
n
ai aj (bi − bj )2 ≤ √
(
a2i )(
b2i ).
(2.28)
n
−
1
ij
i
i
3. Proof of the theorems
First, we prove the following algebraic lemmas,
Lemma 3.1. Let A, B be two real symmetric matrixes such that trA = trB = 0.
Then
1
n−2
|tr(A2 B)| ≤ √
[trA2 ][tr(B 2 )] 2 .
(3.1)
n(n − 1)
Proof. We can find an orthogonal matrix Q such that QAQ−1 = (af
f
ij ) where a
ij =
−1
f
af
δ
.
Since
Q
is
an
orthogonal
matrix,
we
have
trB
=
trQBQ
=
tr(
b
).
ii ij
ij
So we have
∑ 2
2 f
f
tr(A2 B) = tr(QAQ−1 QAQ−1 QBQ−1 ) = tr((af
af
(3.2)
ij ) (bij )) =
ii bii
i
102
YINGBO HAN, SHUXIANG FENG
Since trA = trB = 0, we have
|tr(A2 B)| = |
∑
∑
i
af
ii =
∑ f
i bii = 0. By using Lemma 2.3, we have
2f
af
ii bii | ≤
i
≤
=
∑ 2 ∑ 2 1
n−2
√
(
af
bf
ii )(
ii ) 2
n(n − 1) i
i
∑ 2 ∑ 2 1
n−2
√
(
af
bf
ii )(
ij ) 2
n(n − 1) i
ij
1
n−2
√
[trA2 ][trB 2 ] 2 .
n(n − 1)
This proves this Lemma.
Lemma 3.2. Let A and B be real symmetric matrixes satisfying tr(A) = 0. Then
n
tr(B)tr(A2 B) − (tr(AB))2 ≤ √
tr(A2 )tr(B 2 ).
2 n−1
(3.3)
Proof. We can find an orthogonal matrix Q such that QAQ−1 = (af
f
ij ) where a
ij =
−1
f
af
= tr(bij ). By
ii δij . Since Q is an orthogonal matrix, we have trB = trQBQ
using trA = 0 and Lemma 2.4, we have
tr(B)tr(A2 B) − (tr(AB))2
=
∑
i
=
≤
≤
=
This proves this Lemma.
∑ 2
∑
f
f2
bf
af
af
ii [
ii bii ] − [
ii bii ]
i
i
1 ∑ ff
2
bii bjj (f
aii − af
jj )
2 ij
∑ 2 ∑ 2
n
√
(
af
bf
ii )(
ii )
2 n−1 i
i
∑ 2 ∑ 2
n
√
(
af
bf
ii )(
ij )
2 n−1 i
ij
n
√
tr(A2 )tr(B 2 ).
2 n−1
Proof of Theorem 1.1: Choose a local orthonormal frame field {e1 , · · · , en }
such that hn+1
= λn+1
δij and Φn+1
= λn+1
δij − Hδij . Let µi = λn+1
− H and
ij
i
ij
i
i
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b
denote Φ21 =
L(nH) =
∑
i
∑
µ2i . From (2.12) and (2.22) and the relation R = aH + b, we have
(nHδij − hn+1
ij )(nH)ij +
ij
=
nH△(nH) −
∑
ij
=
=
=
=
≥
103
(n − 1)a
△(nH)
2
1
hn+1
ij (nH)ij + △(n(n − 1)R − n(n − 1)b)
2
∑
1
△[(nH)2 + n(n − 1)R] − n2 |∇H|2 −
hn+1
ij (nH)ij
2
ij
∑
1
△[n(n − 1)c + S] − n2 |∇H|2 −
hn+1
ij (nH)ij
2
ij
∑
1
△S − n2 |∇H|2 −
hn+1
ij (nH)ij
2
ij
(3.4)
1
△S − n2 |∇H|2 − hn+1
ij (nHij )
2
∑
2
2
2
2
2
n+1 n+1 2
(hα
H
)) − nHtr(H n+1 )3
ijk ) − n |∇H| + ncS1 − n cH + (tr(H
|
{z
}
ijkα
ncS2 − nH
∑
tr(H
α̸=n+1
|
n+1
α 2
(H ) ) +
∑
I
[tr(H
n+1
H β )]2 +
β̸=n+1
{z
∑
[tr(H α H β )]2 .
α,β̸=n+1
II
Firstly, we estimate (I):
− nHtr(H n+1 )3
∑
∑
(λn+1
)3 = −nH[ (µ3i ) + 3S1 H − 3nH 3 + nH 3 ]
i
∑
= −3nS1 H 2 + 2n2 H 4 − nH
(µ3i ).
(3.5)
= −nH
By applying Lemma 2.2 to real numbers µ1 , · · · , µn , we get
n(n − 2)
|H||Φ1 |3 .
− nHtr(H n+1 )3 ≥ −3nS1 H 2 + 2n2 H 4 − √
n(n − 1)
(3.6)
So
I
n(n − 2)
|H||Φ1 |).
≥ |Φ1 |2 (nc − nH 2 + |Φ1 |2 − √
n(n − 1)
(3.7)
xy + y 2 . By the orthogonal
Consider the quadratic form P (x, y) = −x2 − √n−2
n−1
transformation
√
√
1
u = √ ((1 + n − 1)y + (1 − n − 1)x)
2n
√
√
1
v = √ ((−1 + n − 1)y + (1 + n − 1)x)
2n
√
P (x, y) = 2√nn−1 (u2 −v 2 ). Take x = nH and y = |Φ1 |; we obtain u2 +v 2 = x2 +y 2 ,
and by (3.7), we have
n
n
n
I ≥ |Φ1 |2 (nc + √
(u2 − v 2 )) ≥ |Φ1 |2 (nc − √
(u2 + v 2 ) + √
2u2 )
2 n−1
2 n−1
2 n−1
n
n
≥ |Φ1 |2 (nc − √
(u2 + v 2 )) ≥ |Φ1 |2 (nc − √
S1 ).
(3.8)
2 n−1
2 n−1
}
104
YINGBO HAN, SHUXIANG FENG
Finally, we estimate (II):
Since trH α = 0, we can use Lemma 3.2 and get
n
− nHtr(H n+1 (H α )2 ) + (trH n+1 H α )2 ≥ − √
S1 tr(H α )2 .
2 (n − 1)
(3.9)
so we have
n
II ≥ ncS2 − nHtr(H n+1 (H α )2 ) + (trH n+1 H α )2 ≥ S2 (nc − √
S1 ).(3.10)
2 (n − 1)
From the inequalities (2.24) (3.4) (3.8)and (3.10), we get
n
n
S1 ) + S2 (nc − √
S1 )
L(nH) ≥ |Φ1 |2 (nc − √
2 n−1
2 (n − 1)
n
n
≥ |Φ|2 (nc − √
S1 ) ≥ |Φ|2 (nc − √
S),
2 (n − 1)
2 (n − 1)
that is,
n
S).
L(nH) ≥ |Φ|2 (nc − √
2 (n − 1)
√
From the assumption S < 2 n − 1c and Eq. (2.12), we have
√
nH 2 + n(n − 1)(aH + b) − n(n − 1)c = S < 2 n − 1c,
(3.11)
(3.12)
So we know that H is bounded. According to Lemma 2.1 (2), there exists a sequence
of points {xk } in M n such that
lim nH(xk ) = sup(nH),
k→∞
lim sup(L(nH)(xk )) ≤ 0.
k→∞
(3.13)
From Eq.(2.12) and (2.18), we have
|Φ|2 = n(n − 1)(H 2 − c + R) = n(n − 1)(H 2 − c + aH + b).
(3.14)
Notice that limk→∞ nH(xk ) = sup(nH) and R is constant, so we have
lim |Φ|2 (xk ) = sup |Φ|2 ,
k→∞
lim S(xk ) = sup S
(3.15)
k→∞
Evaluating (3.11) at the points xk of the sequence, taking the limit and using (3.13),
we obtain that
n
0 ≥ lim sup(L(nH)(xk )) ≥ sup |Φ|2 (nc − √
sup S)
(3.16)
k→∞
2 (n − 1)
√
If S < 2(n − 1)c, we have sup |Φ| = 0, that is, Φ = 0. Thus, we infer that
S = nH 2 and M n is totally umbilical. This proves Theorem 1.1.
Proof of Theorem 1.2:
∑
−nH
tr(H n+1 (H α )2 )
α̸=n+1
= −nH
∑
tr[(H n+1 − HI)(H α )2 ] − nH 2
α̸=n+1
= −nH
∑
∑
tr(H α )2
α̸=n+1
tr(Φ
n+1
(Φ ) ) − nH S2
α 2
2
(3.17)
α̸=n+1
By applying Lemma 3.1 to the matrixes Φn+1 , · · · , Φn+p , we get
∑
n−2
− nH
tr(H n+1 (H α )2 ) ≥ −n|H| √
|Φ1 |S2 − nH 2 S2 .
n(n
−
1)
α̸=n+1
(3.18)
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b
So
II
≥ ncS2 − nH
∑
tr(H n+1 (H α )2 ) +
α̸=n+1
∑
105
[tr(H α H β )]2
α,β̸=n+1
∑
≥ ncS2 − n|H| √
|Φ1 |S2 − nH 2 S2 +
[tr(Φα )2 ]2 . (3.19)
n(n − 1)
α̸=n+1
n−2
From the inequalities (2.24) (3.4) (3.7)and (3.19), we get
L(nH) ≥ |Φ|2 (nc − nH 2 +
n(n − 2)
|Φ|2
−√
|H||Φ|).
p
n(n − 1)
(3.20)
According to Lemma 2.1 (2), there exists a sequence of points {xk } in M n such
that
lim nH(xk ) = sup(nH),
k→∞
lim sup(L(nH)(xk )) ≤ 0.
k→∞
(3.21)
Evaluating (3.20) at the points xk of the sequence, taking the limit and using
(3.21), we obtain that
0 ≥
≥
lim sup(L(nH)(xk ))
k→∞
sup |Φ|2 (nc − n sup H 2 +
sup |Φ|2
n(n − 2)
−√
sup |H| sup |Φ|).(3.22)
p
n(n − 1)
Consider the following polynomial given by
Lsup H (x) =
x2
n(n − 2)
sup |H|x + nc − n sup H 2 .
−√
p
n(n − 1)
(3.23)
(1) If sup(H)2 < (n−2)4(n−1)
2 p+4(n−1) c, it is easy to check that the discriminant of
Lsup H (x) is negative. Hence, for any x, Lsup H (x) > 0, so does Lsup H (sup |Φ|) > 0.
From (3.22), we know that sup |Φ| = 0, that is |Φ| = 0. Thus, we infer that
S = nH 2 and M n is totally umbilical.
(2) If sup(H)2 = (n−2)4(n−1)
2 p+4(n−1) c, we have
√
n(n − 2)p
c
√
Lsup H (x) = (sup |Φ| −
)2 ≥ 0.
(3.24)
p(n − 2)2 + 4(n − 1)
n
√
c
2
√
If (sup |Φ|− n(n−2)p
p(n−2)2 +4(n−1) ) > 0, from (3.22) we have sup |Φ| = 0, that is
n
|Φ| = 0.√Thus, we infer that sup(S) = nH 2 and M n is totally umbilical. If sup |Φ| =
n(n−2)p
(n−2)2 p2 +4(n−1)
c
√
p(n−2)2 +4(n−1)c , from (2.18) we have that sup(S) = nc (n−2)2 p+4(n−1) .
n
(3) If sup(H)2 > (n−2)4(n−1)
2 p+4(n−1) c, we know that Lsup H (x) has two real roots
−
+
xsup H and xsup H given by
√
√
n
[(n − 2)2 p + 4(n − 1)] sup H 2 − 4(n − 1)c
{(n − 2) sup |H| −
}
x−
sup H = p
4(n − 1)
p
√
√
n
[(n − 2)2 p + 4(n − 1)] sup H 2 − 4(n − 1)c
+
xsup H = p
{(n − 2) sup |H| +
}
4(n − 1)
p
106
YINGBO HAN, SHUXIANG FENG
−
it is easy to say that x+
sup H is always positive; on the other hand, xsup H < 0 if and
−
−
only if sup H 2 > c, xsup H = 0 if and only if sup H 2 = c, and xsup H > 0 if and only
if sup H 2 < c.
In this case, we also have that
Lsup H (x) =
1
+
(sup |Φ| − x−
sup H )(sup |Φ| − xsup H )
p
(3.25)
From (3.22) and (3.25), we have that
1
+
0 ≥ sup |Φ|2 (sup |Φ| − x−
sup H )(sup |Φ| − xsup H )
p
(3.26)
−
(3-a) If sup(H)2 > c > (n−2)4(n−1)
2 p+4(n−1) c, then we have xsup H < 0. Therefore,
from (3.26), we have sup |Φ|2 = 0, i.e. M n is totally umbilical or 0 < sup |Φ| ≤
x+
sup H , i.e.
n sup H 2 < sup S ≤ S + ,
√
2(n−2)2 p2 +4(n−1)(p+1)
[(n−2)2 p+4(n−1)] sup H 2 −4(n−1)c
2
2
where S + = n(n−2)
[
sup
H
+sup
|H|p
]−
2(n−1)
2(n−2)
p
npc.
−
(3-b) If sup(H)2 = c > (n−2)4(n−1)
2 p+4(n−1) c, then we have xsup H = 0. Therefore,
from (3.26), we have sup |Φ|2 = 0, i.e. M n is totally umbilical or 0 < sup |Φ| ≤
x+
sup H , i.e.
n sup H 2 < sup S ≤ S + .
−
(3-c) If c > sup(H)2 > (n−2)4(n−1)
2 p+4(n−1) c, then we have xsup H > 0. Therefore,
from (3.26), we have sup |Φ|2 = 0, i.e. M n is totally umbilical or x−
sup H ≤ sup |Φ| ≤
+
xsup H , i.e.
S − ≤ sup S ≤ S + ,
√
[(n−2)2 p+4(n−1)] sup H 2 −4(n−1)c
2(n−2)2 p2 +4(n−1)(p+1)
2
2
where S − = n(n−2)
[
sup
H
−sup
|H|p
]−
2(n−1)
2(n−2)
p
npc.
This proves theorem 1.2.
Proof of Theorem 1.4: From theorem 1.3, we only need to prove that M n
is a spacelike Einstein submanifolds with Ricij = c(n − 1)gij if and only if M n is
totally geodesic.
If M n is totally geodesic in Spn+p (c), we have Ricij = c(n − 1)gij by the equation
(2.11).
Conversely, if M n is a spacelike Einstein submanifolds with Ricij = c(n − 1)gij ,
we have R = c = 0H + c and S = n2 H 2 . From inequality (3.20), we have the
following:
L(nH) ≥ |Φ|2 (nc − nH 2 +
n(n − 2)
|Φ|2
−√
|H||Φ|).
p
n(n − 1)
Because Ricij = c(n − 1)δij , we see by the Bonnet-Myers theorem that M n is
bounded and hence compact. Since L is self-adjoint, we have
∫
|Φ|2
n(n − 2)
|Φ|2 (nc − nH 2 +
−√
|H||Φ|).
(3.27)
0≥
p
n(n − 1)
Mn
ON SPACE LIKE SUBMANIFOLDS WITH R = aH + b
107
Since n2 |H|2 = S and |Φ|2 = S − nH 2 = n(n − 1)H 2 , we have
nc − nH 2 +
|Φ|2
n(n − 2)
−√
|H||Φ|
p
n(n − 1)
n(n − 1)H 2
− n(n − 2)H 2
p
1
= nc − n(n − 1)(1 − )H 2 .
p
= nc − nH 2 +
If H 2 <
c
1 ,
(n−1)(1− p
)
√n(n−2) |H||Φ|) > 0, which together
we have (nc−nH 2 + |Φ|
p −
2
n(n−1)
with (3.27) yields |Φ|2 = 0. That is, S = nH 2 , so we know that n2 H 2 = nH 2 , so
we have H = 0, i.e. S = nH 2 = 0, so M n is totally geodesic. This proves Theorem
1.4.
√n(n−2) |H||Φ| =
Remark. When p = 1, since n2 H 2 = S, we have nc−nH 2 + |Φ|
p −
2
n(n−1)
nc > 0, so we know that |Φ| = 0, i.e. S = nH 2 , so M n is totally geodesic.
References
[1] H.Alencar, M.do Carmo, Hypersurfacs with constant mean curvature in spheres. Pro.
Amer.Math. Soc. 120 (1994), 1223-1229.
[2] Q.M.Cheng, S.Ishikawa, Space-like hypersurfaces with constant scalar curvature, Manu.
Math.95 (1998), 499-505.
[3] S.Y. Cheng, S.T. Yau, Hypersurfaces with constant scalar curvature, Math. Ann. 225 (1977),
195-204.
[4] X.L. Chao, On complete spacelike submanifolds in semi-Riemannian space form with parallel
normalized mean curvature vector, Kodai Math.J. 34(2011), 42-54.
[5] F.E.C.Camargo,R.M.B.Chaves and L.A.M.Sousa Jr, New characterizations of complete
spacelike submanifolds in semi-Riemannian space forms, Kodai Math.J. 32 (2009), 209-230.
[6] A.J.Goddard, Some remarks on the existence of space-like hypersurfaces of constant mean
curvature, Math. Proc.Camgridge Philos.Soc. 82 (1977), 489-495.
[7] Y.B. Han, Spacelike submanifolds in indefinite space form Mpn+p (c), Archivum Math.(Brno)
46 (2010), 79-86.
[8] X.M. Liu, Spacelike submanifolds with constant scalar curvature, C.R.Acad.Sci.Paris,t. 332
(2001), 729-734.
[9] X.M. Liu, Complete space-like hypersurfaces with constant scalar curvature, Manu. Math.
105 (2001), 367-377.
[10] J.C. Liu, Z.Y. Sun, On spacelike hypersurfaces with constant scalar curvature in locally
symmetric Lorentz spaces, J.Math.Anal. Appl. 364 (2010), 195-203.
[11] M.Okumura, Hypersurfaces and a pinching problem on the second fundamental tensor,
Amer.J.Math. 96 (1974), 207-213.
[12] H. Omori, Isometric immersions of Riemmanian manifolds, J.Math. Soc. Japan 19 (1967),
205-214.
[13] W.Santos, Submanifolds with parallel mean curvature vector, Tohoku Math. J. 46 (1994),
403-415.
[14] Rosa M.B.Chves, L.A.M. Sousa Jr., Some applications of a Simons’ type formula for complete
spacelike submanifolds in a semi-Riemannian space form, Diff. Geom. Appl. 25 (2007), 419432.
[15] Y. Zheng, On spacelike hypersurfaces in the de Sitter spaces, Ann. of Global Anal. and Geom.
13 (1995), 317-321.
[16] Y. Zheng, Space-like hypersurfaces with constant scalar curvature in de Sitter space, Diff.
Geom. Appl. 6 (1996), 51-54.
[17] J.F.Zhang, Spacelike submanifolds in the de Sitter space Spn+p (c) with constant scalar curvature, Appl. Math. J.Chin.Unvi. ser.B 20 (2005), 183-196.
108
YINGBO HAN, SHUXIANG FENG
Yingbo Han
College of Mathematics and information Science, Xinyang Normal University, Xinyang
464000, Henan, P. R. China
E-mail address: yingbhan@yahoo.com.cn
Shuxiang Feng
College of Mathematics and information Science, Xinyang Normal University, Xinyang
464000, Henan, P. R. China
E-mail address: fsxhyb@yahoo.com.cn