ARCHIVUM MATHEMATICUM (BRNO)
Tomus 47 (2011), 151–161
ON COMPLETE SPACELIKE HYPERSURFACES WITH
R = aH + b IN LOCALLY SYMMETRIC LORENTZ SPACES
Yingbo Han† , Shuxiang Feng, and Liju Yu‡
Abstract. In this note, we investigate n-dimensional spacelike hypersurfaces
M n with R = aH +b in locally symmetric Lorentz space. Two rigidity theorems
are obtained for these spacelike hypersurfaces.
1. Introduction
Let M1n+1 be an (n + 1)-dimensional Lorentz space, i.e. a pseudo-Riemannian
manifold of index 1. When the Lorentz space M1n+1 is of constant curvature c, we
call it a Lorentz space form, denoted by M1n+1 (c). A hypersurface M n of a Lorentz
space is said to be spacelike if the induced metric on M n from that of the Lorentz
space is positive definite. Since Goddard’s conjecture (see [7]), several papers about
spacelike hypersurfaces with constant mean curvature in de Sitter space S1n+1 (1)
have been published. For a more complete study of spacelike hypersurfaces in
general Lorentzian space with constant mean curvature, we refer to [2]. For the
study of spacelike hypersurface with constant scalar curvature in de Sitter space
S1n+1 (1), there are also many results such as [4, 9, 14, 15]. There are some results
about spacelike hypersurfaces with constant scalar curvature in general Lorentzian
space, such as [8] and [13].
It is natural to study complete spacelike hypersurfaces in the more general
Lorentz spaces, satisfying the assumptions R = aH + b, where R is the normalized
scalar curvature at a point of space-like hypersurface, H is the mean curvature
and a, b ∈ R are constants. First of all, we recall that Choi et al. [6, 12] introduced
the class of (n + 1)-dimensional Lorentz spaces M1n+1 of index 1 which satisfy the
following two conditions for some fixed constants c1 and c2 :
(i) for any spacelike vector u and any timelike vector v,
c1
K(u, v) = − ,
n
(ii) for any spacelike vectors u and v,
K(u, v) ≥ c2 .
2010 Mathematics Subject Classification: primary 53C42; secondary 53B30.
Key words and phrases: spacelike submanifolds, locally symmetric Lorentz spaces.
† Supported by NSFC No. 10971029 and NSFC-TianYuan Fund No.11026062.
‡ Supported by Project of Zhejiang Provincial Department of Education No.Y200909563.
Received October 31, 2010. Editor O. Kowalski.
152
Y. HAN, S. FENG AND L. YU
(Here, and in the sequel, K denotes the sectional curvature of M1n+1 .)
Convention. When M1n+1 satisfies conditions (i) and (ii), we shall say that M1n+1
satisfies condition (∗).
We compute the scalar curvature at a point of Lorentz space M1n+1 ,
(1) R̄ =
X
A
A R̄AA = −2
n
X
i=1
R̄n+1iin+1 +
X
R̄ijji = −2c1 +
ij
X
R̄ijji ,
ij
where R̄n+1iin+1 = −K(ei , en+1 ) = cn1 , for i = 1, . . . , n.
n+1
It
is locally symmetric,
Pis known that R̄ is constant when the Lorentz space M1
so ij R̄ijji is constant. In this note, we shall prove the following main results:
Theorem 1.1. Let M n be a complete spacelike hypersurface with bounded mean
curvature in locally symmetric P
Lorentz space M1n+1 satisfying the condition (∗).
If R = aH + b, (n − 1)2 a2 + 4 ij R̄ijji − 4n(n − 1)b ≥ 0, and a ≥ 0, then the
following properties hold.
c1
2
n
(1) If sup H 2 < 4(n−1)
n2 c, where c = n + 2c2 , then c > 0, S = nH and M is
totally umbilical.
2
n
(2) If sup H 2 = 4(n−1)
is totally
n2 c, then c ≥ 0 and either S = nH and M
umbilical, or sup S = nc.
(3-a) If c < 0, then either S = nH 2 and M n is totally umbilical, or n sup H 2 <
sup S ≤ S + .
2
n
(3-b) If c ≥ 0 and sup H 2 ≥ c > 4(n−1)
is
n2 c, then either S = nH and M
totally umbilical, or n sup H 2 < sup S ≤ S + .
2
n
(3-c) If c ≥ 0 and c > sup H 2 > 4(n−1)
n2 c, then either S = nH and M is totally
−
+
umbilical, or S ≤ sup S ≤ S .
(4)
p
n
(2) S ≡
[n2 sup H 2 + (n − 2) sup |H| n2 sup H 2 − 4(n − 1)c] − nc ,
2(n − 1)
if and only if M is an isoparametric hypersurface with two distinct principal
curvatures one of which is simple.
p
n
Here S + = 2(n−1)
[n2 sup H 2 + (n − 2) sup |H| n2 sup H 2 − 4(n − 1)c] − nc, and
p
n
S − = 2(n−1)
[n2 sup H 2 − (n − 2) sup |H| n2 sup H 2 − 4(n − 1)c] − nc.
Theorem 1.2. Let M n (n > 1) be a complete spacelike hypersurface in locally
symmetric Lorentz space M1n+1 satisfying the condition (∗). If c = cn1 + c2 > 0,
c2 > 0 and
(3)
W 2 = tr(W )W ,
where W is the shape operator with respect to en+1 , then M n must be totally
geodesic.
Remark 1.3. The Lorentz space form M1n+1 (c) satisfies the condition (∗), where
− cn1 = c2 = const.
ON COMPLETE SPACELIKE HYPERSURFACES WITH R = aH + b
153
2. Preliminaries
Let M n be a spacelike hypersurface of Lorentz space M1n+1 . We choose a local
field of semi-Riemannian orthonormal frames {e1 , . . . , en , en+1 } in M1n+1 such
that, restricted to M n , e1 , . . . , en are tangent to M n and en+1 is the unit timelike
normal vector. Denote by {ωA } the corresponding dual coframe and by {ωAB } the
connection forms of M1n+1 . Then the structure equations of M1n+1 are given by
(4)
dωA = −
X
B ωAB ∧ ωB ,
ωAB + ωBA = 0 ,
i = 1 ,
n+1 = −1 ,
B
(5)
dωAB = −
X
C ωAC ∧ ωCB −
C
1X
C D R̄ABCD ωC ∧ ωD ,
2
CD
where A, B, C, · · · = 1, . . . , n + 1 and i, j, l, · · · = 1, . . . , n. The components R̄CD of
the Ricci tensor and the scalar curvature R̄ of M1n+1 are given by
R̄CD =
(6)
X
B R̄BCDB ,
R̄ =
B
X
A R̄AA .
A
The components R̄ABCD;E of the covariant derivative of the Riemannian curvature
tensor R̄ are defined by
X
E R̄ABCD;E ωE = dR̄ABCD −
X
E
E (R̄EBCD ωEA
E
+ R̄AECD ωEB + R̄ABED ωEC + R̄ABCE ωED ) .
(7)
n
We restrict these forms
P to2 M , then ωn+1 = 0 and the Riemannian metric of
2
M is written as ds = i ωi . Since
n
(8)
0 = dωn+1 = −
X
ωn+1,i ∧ ωi ,
i
by Cartan’s lemma we may write
(9)
ωn+1,i =
X
hij ωj ,
hij = hji .
j
From these formulas, we obtain the structure equations of M n :
dωi = −
X
ωij ∧ ωj ,
ωij + ωji = 0 ,
ωik ∧ ωkj −
1X
Rijkl ωk ∧ ωl ,
2
j
dωij = −
X
k
(10)
k,l
Rijkl = R̄ijkl − (hil hjk − hik hjl ) ,
154
Y. HAN, S. FENG AND L. YU
where Rijkl are the components of curvature tensor of M n . Components Rij of
Ricci tensor and scalar curvature R of M n are given by
X
X
X
(11)
Rij =
R̄kijk −
hkk hij +
hik hjk ,
k
(12)
n(n − 1)R =
k
X
k
2
2
R̄ijji + S − n H .
ij
We call
B=
(13)
X
hij ωi ⊗ ωj ⊗ en+1
i,j,α
P
the second fundamental form of M n . The mean curvature vector is h = n1 i hii en+1 .
P
We denote S = i,j (hij )2 , H 2 = |h|2 and W = (hij )ni,j=1 . We call that M n is
maximal if its mean curvature vector vanishes, i.e. h = 0.
Let hijk and hijkl denote the covariant derivative and the second covariant
derivative of hα
ij . Then we have hijk = hikj + R̄(n+1)ijk and
X
X
(14)
hijkl − hijlk = −
him Rmjkl −
hmj Rmikl .
m
m
Restricting the covariant derivative R̄ABCD;E on M n , then R̄(n+1)ijk;l is given by
(15)
R̄(n+1)ijk;l = R̄(n+1)ijkl + R̄(n+1)i(n+1)k hjl
X
+ R̄(n+1)ij(n+1) hkl +
R̄mijk hml ,
m
where R̄(n+1)ijkl denotes the covariant derivative of R̄(n+1)ijk as a tensor on M n
so that
X
X
R̄(n+1)ijkl = g R̄(n+1)ijk −
R̄(n+1)ljk ωli −
R̄(n+1)ilk ωlj
l
−
(16)
X
l
R̄(n+1)ijl ωlk .
l
P
The Laplacian 4hij is defined by 4hij = k hijkk . Using Gauss equation, Codazzi
equation Ricci identity and (2), a straightforward calculation will give
X
X
1
4S =
h2ijk +
hij 4hij
2
ij
ijk
X
X
X
=
h2ijk +
(nH)ij hij +
(R̄(n+1)ijk;k + R̄(n+1)kik;j )hij
ij
ijk
ijk
X
X
−(
nHhij R̄(n+1)ij(n+1) + S
R̄(n+1)k(n+1)k )
ij
(17)
−2
X
ijkl
k
(hkl hij R̄lijk + hil hij R̄lkjk ) − nH
X
hil hlj hij + S 2 .
ijl
Set Φij = hij − Hδij , it is easy to check that Φ is traceless and |Φ|2 = S − nH 2 .
In this note we consider the spacelike hypersurface with R = aH + b in locally
ON COMPLETE SPACELIKE HYPERSURFACES WITH R = aH + b
155
symmetric Lorentz space M1n+1 , where a, b are real constants. Following Cheng-Yau
[5], we introduce a modified operator acting on any C 2 -function f by
X
n−1
(18)
L(f ) =
(nHδij − hij )fij +
a∆f .
2
ij
We need the following algebraic Lemmas.
Lemma 2.1 ([11]). Let M n be an n-dimensional complete Riemannian manifold
whose sectional curvature is bounded from below and F : M n → R be a smooth
function which is bounded above on M n . Then there exists a sequence of points
xk ∈ M n such that
lim F (xk ) = sup(F ) ,
k→∞
lim |∇F (xk )| = 0 ,
k→∞
lim sup max{(∇2 (F )(xk ))(X, X) : |X| = 1} ≤ 0 .
k→∞
Lemma
2.2 ([1, 10]). Let µ1 , . . . , µn be real numbers such that
P 2
2
i µi = β , where β ≥ 0 is constant. Then
X n−2
µ3i ≤ p
(19)
β3 ,
n(n
−
1)
i
P
i
µi = 0 and
and equality holds if and only if at least n − 1 of µi ’s are equal.
3. Proof of the theorems
First, we give the following lemma.
Lemma 3.1. Let M n be a complete spacelike hypersurface in locally symmetric
Lorentz space P
M1n+1 satisfying the condition (∗). If R = aH + b, a, b ∈ R and
2 2
(n − 1) a + 4 ij R̄ijji − 4n(n − 1)b ≥ 0.
(1) We have the following inequality,
n(n − 2)
(20)
|H||Φ| + nc − nH 2 .
L(nH) ≥ |Φ|2 |Φ|2 − p
n(n − 1)
where c = 2c2 + cn1 .
(2) If the mean curvature H is bounded, then there is a sequence of points
{xk } ∈ M such that
lim nH(xk ) = sup(nH) ,
lim |∇nH(xk )| = 0 ,
k→∞
lim sup L(nH)(xk ) ≤ 0 .
k→∞
(21)
k→∞
Proof. (1) Choose a local orthonormal frame field {e1 , . . . , en }Psuch that hij =
λi δij and Φij = λi δij − Hδij . Let µi = λi − H and denote Φ2 = i µ2i . From (12),
156
Y. HAN, S. FENG AND L. YU
(18) and the relation R = aH + b, we have
X
(n − 1)a
L(nH) =
(nHδij − hij )(nH)ij +
4(nH)
2
ij
X
1
= nH4(nH) −
hij (nH)ij + 4(n(n − 1)R − n(n − 1)b)
2
ij
X
1
hij (nH)ij
= 4[(nH)2 + n(n − 1)R] − n2 |∇H|2 −
2
ij
i
X
1 hX
R̄ijji + S − n2 |∇H|2 −
hij (nH)ij
= 4
2
ij
ij
X
1
= 4S − n2 |∇H|2 −
hij (nH)ij .
2
ij
From (17) and M1n is locally symmetric, we have
X
X
λ3i + S 2
L(nH) =
h2ijk − n2 |∇H|2 −nH
i
ijk
−
X
nHλi R̄(n+1)ii(n+1) + S
ij
}|
{z
|
I
X
{z
}
II
X
(λk λi R̄kiik + λ2i R̄ikik ) .
R̄(n+1)k(n+1)k − 2
ijkl
k
|
{z
}
III
Firstly, we estimate (I):
From Gauss equation, we have
X
(22)
R̄ijji + S − n2 H 2 = n(n − 1)R = n(n − 1)(aH + b) ,
ijji
Taking the covariant derivative of the above equation, we have
X
(23)
2
hij hijk = 2n2 HHk + n(n − 1)aHk .
ijk
Therefore
2
X
XX
(24) 4S
h2ijk ≥ 4
hij hijk = [2n2 H + n(n − 1)a]2 |∇H|2 .
ijk
k
ij
Since we know
[2n2 H + n(n − 1)a]2 − 4n2 S = 4n4 H 2 + n2 (n − 1)2 a2 + 4n3 (n − 1)aH
h
i
X
− 4n2 n2 H 2 + n(n − 1)R −
R̄ijji
ij
2
h
2 2
= n (n − 1) a + 4
X
ijji
i
R̄ijji − 4n(n − 1)b ≥ 0 .
ON COMPLETE SPACELIKE HYPERSURFACES WITH R = aH + b
157
if follows that
X
(25)
h2ijk ≥ n2 |∇H|2 .
ijk
Secondly, we estimate (II):
It is easy to know that
X
X
X
(26)
λ3i = nH 3 + 3H
µ2i +
µ3i .
i
i
i
By applying Lemma 2.2 to real numbers µ1 , . . . , µn , we get
X
X
S 2 − nH
λ3i = (|Φ|2 + nH 2 )2 − n2 H 4 − 3nH 2 |Φ|2 − nH
µ3i
i
i
n(n − 2)
|H||Φ|3 .
≥ |Φ| − nH |Φ| − p
n(n − 1)
4
(27)
2
2
Finally, we estimate (III):
Using curvature condition (∗), we get
X
X
nHλi R̄(n+1)ii(n+1) + S
(28) −
R̄(n+1)k(n+1)k = c1 (S − nH 2 ) .
ij
k
1
2n
2
Notice that S − nH 2 =
ij (λi − λj ) , we also have
X
X
−2
(λk λi R̄kiik + λ2i R̄ikik ) = −2
(λi λk − λ2i )Rikki
P
ik
ik
≥ c2
(29)
X
(λi − λk )2 = 2nc2 (S − nH 2 ) .
ik
From (25), ??, (28), (29) and set c = 2c2 + cn1 , we have
n(n − 2)
L(nH) ≥ |Φ|2 |Φ|2 − p
|H| |Φ| + nc − nH 2 .
n(n − 1)
(2) Choose a local orthonormal frame field {e1 , . . . , en } such that hij = λi δij . By
P
P
definition, L(nH) = i (nH − λi )(nH)ii + (n−1)a
i (nH)ii . If H ≡ 0 the result is
2
obvious. Let suppose that H is not identically zero. By changing the orientation of
M n if necessary, we may assume that sup H > 0. From
X
(λi )2 ≤ S = n2 H 2 + n(n − 1)R −
R̄ijji
ij
2
2
= n H + n(n − 1)(aH + b) −
X
R̄ijji
ij
(30)
= (nH +
X
(n − 1)a 2 1
) − (n − 1)2 a2 −
R̄ijji + n(n − 1)b
2
4
ij
≤ (nH +
(n − 1)a 2
) ,
2
158
Y. HAN, S. FENG AND L. YU
we have
(n − 1)a
|.
2
Since H is bounded and Eq. (30), we know that S is also bounded. From the
Eq. (10),
|λi | ≤ |nH +
(31)
Rijji = R̄ijji − hii hjj + (hij )2 ≥ c2 − hii hjj
= c2 − λi λj ≥ c2 − S .
(32)
This shows that the sectional curvatures of M n are bounded from below because
S is bounded. Therefore we may apply Lemma 2.1 to the function nH, and obtain
a sequence of points {xk } ∈ M n such that
lim nH(xk ) = sup(nH) ,
lim sup nHii (xk ) ≤ 0 .
k→∞
(33)
lim
k→∞
|∇(nH)(xk )| = 0 ,
k→∞
Since H is bounded, taking subsequences if necessary, we can arrive to a sequence
{xk } ∈ M n which satisfies (33) and such that H(xk ) ≥ 0 (by changing the
orientation of M n if necessary). Thus from (31) we get
(n − 1)a
(n − 1)a
− |λi (xk )| ≤ nH(xk ) +
− λi (xk )
2
2
(n − 1)a
(n − 1)a
(34)
≤ nH(xk ) +
+ |λi (xk )| ≤ 2(nH(xk ) +
).
2
2
Using once the fact that H is bounded, from (34) we infer that {nH(xk ) −
λn+1
(xk )} is non-negative and bounded. By applying L(nH) at xk , taking the
i
limit and using (33) and (34) we have
0 ≤ nH(xk ) +
(35)
lim sup(L(nH))(xk )
k→∞
≤
X
i
lim sup(nH +
k→∞
(n − 1)a
− λi )(xk )nHii (xk ) ≤ 0 .
2
Remark 3.2. When a = 0, then R = b is constant, the inequality (20) appeared
in [3, 8, 13].
Proof of Theorem 1.1. According to Lemma 3.1 (2), there exists a sequence of
points {xk } in M n such that
(36)
lim nH(xk ) = sup(nH) ,
k→∞
lim sup L(nH)(xk ) ≤ 0 .
k→∞
From Gauss equation, we have that
(37)
|Φ|2 = S − nH 2 = n(n − 1)H 2 + n(n − 1)(aH + b) −
X
ij
R̄ijji .
ON COMPLETE SPACELIKE HYPERSURFACES WITH R = aH + b
Notice that limk→∞ (nH)(xk ) = sup(nH), a ≥ 0 and
have
(38)
P
ij
159
R̄ijji is constant, we
lim |Φ|2 (xk ) = sup |Φ|2 .
k→∞
Evaluating (20) at the points xk of the sequence, taking the limit and using (36),
we obtain that
0 ≥ lim sup L(nH)(xk )
k→∞
n(n − 2)
(39)
≥ sup |Φ|2 sup |Φ|2 − p
sup |H| sup |Φ| + nc − n sup H 2 .
n(n − 1)
Consider the following polynomial given by
(40)
n(n − 2)
Psup H (x) = x2 − p
sup |H|x + nc − n sup H 2 .
n(n − 1)
(1) If sup H 2 < 4(n−1)
n2 c holds, then we have c > 0 and P (sup |Φ|) > 0. From
(39), 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 = 4(n−1)
then we have c ≥ 0 and P (|Φ|) = |Φ| −
n2 c holds,
√ 2
√ 2
n−2
n−2
√
√
c
0.
If
|Φ|−
c
0,
from (39) we have, sup |Φ| = 0, that is |Φ| = 0.
≥
>
n
n
√
2
n
√
Thus, we infer that S = nH and M is totally umbilical. If sup |Φ| = n−2
c, we
n
have that sup S = nc.
−
+
(3) If sup H 2 > 4(n−1)
n2 c, we know that P (x) has two real roots xsup H and xsup H
given by
r
p
n
x−
=
(n − 2) sup |H| − n2 sup H 2 − 4(n − 1)c
sup H
4(n − 1)
r
p
n
x+
(n − 2) sup |H| + n2 sup H 2 − 4(n − 1)c
sup H =
4(n − 1)
It is easy to know that x+
sup H is always positive. In this case, we also have that
(41)
+
Psup H (x) = (sup |Φ| − x−
sup H )(sup |Φ| − xsup H ) .
From (39) and (41), we have that
(42)
+
0 ≥ sup |Φ|2 (sup |Φ| − x−
sup H )(sup |Φ| − xsup H ) .
(3-a) If c < 0, we know that x−
sup H < 0. Therefore, from (42), we have, sup |Φ| =
0, in this case M n is totally umbilical, or 0 < sup |Φ| ≤ x+
sup H , i.e.
n sup H 2 < sup S ≤ S + .
−
(3-b) If c ≥ 0 and sup(H)2 ≥ c > 4(n−1)
n2 c, we know that xsup H < 0. Therefore,
n
from (42), we have, sup |Φ| = 0, in this case M is totally umbilical, or 0 <
sup |Φ| ≤ x+
sup H , i.e.
n sup H 2 < sup S ≤ S + .
160
Y. HAN, S. FENG AND L. YU
−
(3-c) If c ≥ 0 and c > sup(H)2 > 4(n−1)
n2 c, then we have xsup H > 0. Therefore,
from (39), we have that sup |Φ| = 0, in this case M n is totally umbilical or
+
x−
sup H ≤ sup |Φ| ≤ xsup H , i.e.
S − ≤ sup S ≤ S + .
p
n
(4) If S ≡ 2(n−1)
[n2 sup H 2 + (n − 2) sup |H| n2 sup H 2 − 4(n − 1)c] − nc holds,
P
from Gauss equation, we have S = nH 2 + n(n − 1)(aH + b) − ij R̄ijji . Since S is
constant, then H is also constant. We know that these inequalities in the proof of
Lemma 2.2, and (27) are equalities and S > nH 2 . Hence, we have H 2 ≥ 4(n−1)
n2 c
from (1) in Theorem 1.1. Thus, we can infer that n − 1 of the principal curvatures
λi are equal. Since S and H is constant, we know that principal curvatures are
constant on M n . Thus, M n is an isoparametric hypersurface with two distinct
principal curvatures one of which is simple. This proves Theorem 1.1.
Proof of Theorem 1.2. From (3), we have that
X
(43)
hik hjk = nHhij , for i, j ∈ {1, . . . , n} ,
k
and
(44)
X
h2ij = n2 H 2 ,
i.e. S = n2 H 2 .
ij
Choose a local orthonormal frame field {e1 , . . . , en } such that Rij = υi δij . From
P
(11) and (43), we have Rii = k R̄kiik ≥ (n − 1)c2 > 0, that is, υi ≥ (n − 1)c2 > 0,
so we know that Ric = (Rij ) ≥ (n − 1)c2 I, we see by the Bonnet-Myers theorem
that M n is bounded and hence compact.
P
From (12) and (44), we have that n(n − 1)R = ij R̄ijji is constant, then from
Lemma 3.1 for a = 0, we have the following inequality
n(n − 2)
L(nH) ≥ |Φ|2 |Φ|2 − p
(45)
H|Φ| + nc − nH 2 .
n(n − 1)
Since L is self-adjoint and M n is compact, we have
Z
n(n − 2)
(46)
0≥
|Φ|2 |Φ|2 − p
|H||Φ| + nc − nH 2 .
n(n − 1)
Mn
Since n2 |H|2 = S and |Φ|2 = S − nH 2 = n(n − 1)H 2 , we have
n(n − 2)
nc − nH 2 + |Φ|2 − p
|H| |Φ|
n(n − 1)
= nc − nH 2 + n(n − 1)H 2 − n(n − 2)H 2 = nc > 0 .
so we know that |Φ|2 = 0, that is, S = nH 2 . From Eq. (44), 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 1.2.
Acknowledgement. The authors would like to thank the referee whose valuable
suggestions make this paper more perfect.
ON COMPLETE SPACELIKE HYPERSURFACES WITH R = aH + b
161
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College of Mathematics and information Science,
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College of Mathematics Physics and Information Engineering,
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