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Rigidity theorems for closed hypersurfaces in a unit sphere
Journal of Geometry and Physics 55 (2005) 227–240
Rigidity theorems for closed hypersurfaces in a unit sphere夽 Qiaoling Wang∗ , Changyu Xia Departamento de Matem´atica-IE, Universidade de Bras´ılia, Campus Universit´ario, 70910-900 Bras´ılia, DF, Brasil Received 15 July 2004; received in revised form 6 December 2004; accepted 13 December 2004
Abstract In this paper, we prove some rigidity theorems for closed hypersurfaces in a unit sphere. © 2004 Elsevier B.V. All rights reserved. MSC: 53C20; 53C42 PACS: 02.40.H, M; 02.40.K
JGP SC: Riemannian geometry Keywords: Rigidity; Hypersurfaces; Sphere
1. Introduction Let M n be an n-dimensional closed hypersurface in a unit sphere S n+1 (1) of dimension n + 1 and denote by S the squared norm of the second fundamental form of M n . Many metric and topological rigidity theorems about M n have been obtained. Simons [19], Chern 夽 Both authors were partially supported by CNPq. ∗
Corresponding author. Tel.: +55 612733356257; fax: +55 612732737. E-mail addresses: [email protected] (Q. Wang), [email protected] (Changyu Xia).
0393-0440/$ – see front matter © 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.geomphys.2004.12.005
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Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
et al. [10] and Lawson [11] proved that if M n is minimal and if S ≤ n then M n is either totally geodesic or a Clifford minimal hypersurface. Further results about rigidity of minimal hypersurfaces in S n+1 (1) can be found, e.g., in [15–17], etc. As a natural generalization, the rigidity phenomenon for hypersurfaces in S n+1 (1) with constant mean curvature has also been studied. It has been proven by Nomizu and Smyth [13] that if M n has constant mean curvature and non-negative sectional curvature then M n is either totally umbilic or a Riemannian product S k (c1 ) × S n−k (c2 ), 1 ≤ k ≤ n − 1, where S k (c) denotes the sphere of radius c. Alencar and do Carmo [2] have shown that if M n has constant mean curvature H and if S ≤ nH 2 + C(H, n), where C(H, n) is a constant that√depends only on H and n, then M n is either totally umbilic or a Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1/n). On the other hand, some rigidity theorems for hypersurfaces with constant scalar curvature have been proven. It has been shown by Cheng and Yau that if M n has nonnegative sectional curvature and constant scalar curvature n(n − 1)r with r ≥ 1, then M n is isometric to either a totally umbilic hypersurface or a Riemannian product S k (c1 ) × S n−k (c2 ), 1 ≤ k ≤ n − 1 [9]. Li [12] proved that if M n has constant scalar curvature n(n − 1)r with r ≥ 1 and if S ≤ (n − 1)(n(r − 1) + 2)/(n − 2) + (n − 2)/(n(r − 1) + 2), n is isometric to either a totally umbilic hypersurface or the Riemannian prodthen M√ 1 uct S ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 2)/(nr). Recently, Alencar et al. [3] obtained a gap theorem for closed hypersurfaces with constant scalar curvature n(n − 1) in a unit sphere. √ In this paper, we prove a rigidity theorem for the Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1/n) without the constancy condition on the mean curvature or the scalar curvature. Namely, we have the following theorem. Theorem 1.1. Let M n be an n-dimensional closed hypersurface in S n+1 (1). Denote by S and H the squared norm of the second fundamental form and the mean curvature of M n , respectively. Assume that the fundamental group π1 (M n ) of M n is infinite and that S ≤ S(n, H), where S(n, H) = n +
n(n − 2)|H| 2 2 n3 H 2 − n H + 4(n − 1). 2(n − 1) 2(n − 1)
(1.1)
√ Then S is constant, S = S(n, H) and M is isometric to a Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1)/n. Montiel and Ros [14] have proven that an embedded closed hypersurface in a half-sphere with constant r-mean curvature for some r ∈ {1, . . . , n}, is a totally umbilical hypersphere. In the second part of this paper, we give another characterization for the totally umbilical hypersphere. Theorem 1.2. Let M n be an n-dimensional closed orientable hypersurface in S n+1 (1). Assume that the squared norm S of the second fundamental form of M n is constant and that one of the following conditions holds: (a) The mean curvature H of M n satisfies H 2 ≥ {(n − 1)/n2 }S on M n .
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229
(b) The Ricci curvatures of M n are bounded from below by n − 1. Then M n is a totally umbilical hypersphere.
2. Preliminaries Let M n be an n-dimensional Riemannian manifold. We choose a local frame of orthonormal vector fields {e1 , . . . , en } on M n and let {ω1 , . . . , ωn } be the dual coframe. The connection forms {ωij } of M n are characterized by the structure equations: ωij ∧ ωj , ωij + ωji = 0, (2.1) dωi = j
dωij =
ωik ∧ ωkj −
k
1 Rijkl ωk ∧ ωl , 2
(2.2)
k,l
where Rijkl are the components of the curvature tensor of M n . For any C2 -function f defined on M n , we define its gradient and Hessian by the following formulas df = fi ω i , (2.3)
i
fij ωj = dfi +
j
fj ωji .
(2.4)
j
The Laplacian of f is given by f = i fii . Let φ = i,j φij ωi ⊗ ωj be a symmetric tensor defined on M n . The covariant derivative of φ is defined by φijk ωk = dφij + φkj ωki + φik ωkj . (2.5) k
Let |φ|2
k
2 i,j φij
k
= and tr φ = i φii . Choose a frame field {ω1 , . . . , ωn } so that φij = λi δij . If φ satisfies the “Codazzi equation” φijk = φikj ,
then we have [9] 1 1 2 |φ|2 = φijk + λi (tr φ)ii + Rijij (λi − λj )2 . 2 2 i,j,k
i
(2.6)
i,j
The Weyl curvature tensor W = (Wijkl ) of M n is defined by 1 (Rik δjl − Ril δjk + Rjl δik − Rjk δil ) n−2 R + (δik δjl − δil δjk ), (n − 1)(n − 2)
Wijkl = Rijkl −
(2.7)
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where Rij and R are the components of Ricci curvature tensor and the scalar curvature of M n , respectively. The Beck tensor B = (Bijk ) of M n is given by 1 1 (Rijk − Rikj ) − (δij Rk − δik Rj ), n−2 2(n − 1)(n − 2)
Bijk =
(2.8)
where Rijk are the components of the covariant derivative of the Ricci curvature tensor of M n and Rk = ek R. The following fact is needed in the proof of Theorem 1.1. Lemma 2.1. [1] If the Ricci curvature of a compact Riemannina manifold is non-negative and positive at a point, then the manifold carries a metric of positive Ricci curvature. M n is said to be locally conformally flat if, for each x ∈ M n , there exists a conformal diffeomorphism of a neighborhood of x onto an open set of the Euclidean n-space Rn . When n ≥ 4, M n is locally conformally flat if and only if Wijkl = 0, and that Bijkl = 0 on M n in this case. When n = 3, we always have Wijkl = 0 on M 3 and M 3 is a locally conformally flat Riemannian manifold if and only if Bijk = 0. Thus, if M n is a locally conformally flat Riemannian manifold, then Rijkl =
1 R (Rik δjl − Ril δjk + Rjl δik − Rjk δil ) − (δik δjl − δil δjk ), n−2 (n − 1)(n − 2) (2.9)
and Rijk −
1 1 δij Rk = Rikj − δik Rj . 2(n − 1) 2(n − 1)
(2.10)
n sphere S n+1 (1) and denote by Let M be a hypersurface in a unit (n + 1)-dimensional n h = i,j hij ωi ⊗ ωj the second fundamental form of M . The squared norm S of h and the mean curvature H of M n are given by
S=
i,j
h2ij ,
nH =
hii ,
i
respectively. The Gauss equation of M n can be written as Rijkl = δik δjl − δil δjk + (hik hjl − hil hjk ). The Ricci tensor and the scalar curvature R of M n are then given by hik hkj , Rij = (n − 1)δij + nHhij −
(2.11)
(2.12)
k
R = n(n − 1) + n2 H 2 − S,
(2.13)
respectively. The covariant derivative hijk of h satisfies hijk = hikj . The eigenvalues λ1 , . . . , λn of (hij ) are the principal curvatures of M n .
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231
3. Proofs of the results Proof of Theorem 1.1. Assume that {e1 , . . . , en } diagonalizes the second fundamental form of M n so that i,j hij ωi ⊗ ωj = i λi ωi ⊗ ωi . In this case, we have Rijkl = (1 + λi λj )(δik δjl − δil δjk ),
(3.1)
Rij = 0, i = j,
(3.2)
Rii = n − 1 + nHλi − λ2i , i = 1, . . . , n.
(3.3)
and For any fixed j ∈ {1, . . . , n}, since 2
(nH − λj )2 =
λk ≤ (n − 1)
j=k
k=j
λ2k = (n − 1)(S − λ2j ),
we have n2 H 2 − (n − 1)S + nλ2j − 2nHλj ≤ 0.
(3.4)
It is easy to see from
(λi − H) = 0,
i
(λi − H)2 = S − nH 2 ,
i
that (λj − H)2 ≤
n−1 (S − nH 2 ), n
which, combining with (3.4), implies that 0 ≥ n(λ2j − nHλj ) + (n − 2)n(λj − H)H + 2(n − 1)nH 2 − (n − 1)S n−1 2 ≥ n(λj − nHλj ) − (n − 2)n|H| (S − nH 2 ) + 2(n − 1)nH 2 − (n − 1)S. n (3.5) It then follows from (3.3) that
n−1 (n − 1) (S − nH 2 ) + 2(n − 1)H 2 − S n n
n−1 n = n + 2nH 2 − S − (n − 2)|H| · · S − nH 2 n n−1
n−1 n 1 = S − nH 2 + (n − 2)|H| + n2 H 2 + 4(n − 1) n 2 n−1
Rjj ≥ (n − 1) − (n − 2)|H|
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×
n 1 −(n − 2)|H| + n2 H 2 + 4(n − 1) . − S − nH 2 + 2 n−1 (3.6)
Observe that our condition S ≤ S(n, H) is equivalent to 2 n 1 2 2 S − nH ≤ , −(n − 2)|H| + n H + 4(n − 1) 2 n−1 2
that is n 1 2 −(n − 2)|H| + n2 H 2 + 4(n − 1) ≥ 0. − S − nH + 2 n−1 Thus Rjj ≥ 0, ∀j, which, combining with (3.2), implies that M n has non-negative Ricci curvature. Since our M n has infinite fundamental group, we conclude from Bonnet–Myer’s theorem [7] and Lemma 2.1 that ∀p ∈ M n , there exists a unit vector u ∈ Tp M n , such that the Ricci curvature Ric of M n satisfies Ric(u, u) = 0. Note that Ric attains its maximum and minimum in the principal directions. We can assume without loss of generality that at any fixed point x ∈ M n , Rnn = 0. Thus, when n = j, at the point x, the inequality (3.5) should be an equality and S(x) = S(n, H)(x), which in turn implies that when n = j, the above inequalities should take equality sign at x. Consequently, we know that λ1 (x) = · · · = λn−1 (x). Since x ∈ M n is arbitrary, one then deduces that our M n has at most two distinct principal curvatures and, when M n has exactly two distinct principal curvatures, one of these two distinct principal curvatures should be simple. Let us assume that λ1 = · · · = λn−1 = λ and λn = µ on M n . Since Rii ≥ 0 and one of R11 , . . . , Rnn is zero, we deduce from (3.3) that 1 + λµ = 0. Therefore, we conclude that M n has exactly two distinct principal curvatures one of which is simple and S = S(n, H) holds on M. We want to show that λ and µ are constant functions on M. Notice that our M n is a closed manifold with non-negative Ricci curvature. It is well-known that [18, p. 220] the Riemannian universal covering space ˜ n of M n can be decomposed as M ¯ n−s × Rs for some s ∈ {0, 1, . . . , n}, where M ¯ n−s is M a closed simply connected (n − s)-dimensional Riemannina manifold with non-negative Ricci curvature and Rs is the s-dimensional Euclidean space with its standard flat metric. ¯ n−s × Rs is non-compact and so we have s ≥ 1. The infinity of π1 (M n ) implies that M 1 + λµ = 0 on M n , we know from (3.1) that if Sinceλ1 = · · · = λn−1 n= λ, λn = µ and n n u = i=1 ai ei , v = j=1 bj ej ∈ Tp M with |u| = |v| = 1, u, v = 0, then the sectional curvature K(u ∧ v) of the plane spanned by u and v is given by
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
K(u ∧ v) =
i,j,k,l
= 1+
ai bj ak bl Rijkl =
233
ai bj ak bl (1 + λi λj )(δik δjl − δil δjk )
i,j,k,l
i
2 λi ai2 λj bj2 − λi a i b i j
= 1 + λ(1 − an2 ) + µan2
= (1 + λ2 )(1 − an2 − bn2 ).
i
λ(1 − bn2 ) + µbn2 − (λ(−an bn ) + µan bn )2
(3.7)
Lemma 3.1. A vector u ∈ Tp M n with |u| = 1 satisfies the following condition (∗) if v ∈ Tp M n with u, v = 0, |v| = 1, then K(u ∧ v) = 0, if and only if u = ±en (p). Proof of Lemma 3.1. If u = ±en (p) and v satisfies |v| = 1, u, v = 0, then we can write v=
n−1
ai ei (p),
i=1
n−1 i=1
Thus K(u ∧ v) =
ai aj Rninj =
i,j
=
i
ai2 = 1.
ai aj (1 + λn λi )(δnn δij − δni δnj )
i,j
(1 + µλi )ai2 − an2 (1 + µ2 ) = 0.
(3.8)
On the other hand, if (∗) is satisfied and suppose by contradiction that u = aen (p) + w,
n−1
n−1
(3.9)
where w, en (p) = 0, w = 0. Let w = i=1 ci ei (p) and take a vector z = i=1 di ei (p) ∈ Tp M n satisfying |z| = 1, z, w = 0. Then z, u = 0 and so we have from (3.7) that K(u ∧ z) = (1 + λ2 )(1 − a2 ) = (1 + λ2 )|w|2 > 0, which contradicts to (∗). Thus u is parallel to en (p), since |u| = 1, we know that u = ±en (p). Lemma 3.1 is proved. ¯ n−s × Rs are locally isometric, Let us go on the proof of Theorem 1.1. Since M and M ¯ n−1 has constant sectional it follows from Lemma 3.1 that s = 1. We claim that M curvature and so is isometric to a Euclidean (n − 1)-sphere. In order to see this, let us ¯ n−1 × R → M be the natural projection; then π is a assume first that n ≥ 4. Let π : M n−1 ¯ ¯ n−1 . local isometry. For any x ∈ M , take an orthonormal base {f1 , . . . , fn−1 } of Tx M n−1 n−1 ¯ ¯ Since T(x,0) (M × R)=Tx M × T0 R, we know that {(f1 , 0), . . . , (fn−1 , 0), (0, 1)}
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¯ n−1 . ¯ n−1 × R), where 0 is the zero-vector of Tx M is an orthonormal base of T(x,0) (M n−1 ¯ × R) with v, (0, 1) = 0, |v| = 1, it holds Observe that for any v ∈ T(x,0) (M ˜ ˜ denotes the sectional curvature of M ¯ n−1 × R. It follows that K((0, 1) ∧ v) = 0, where K n K(dπ(x,0) ((0, 1)) ∧ z) = 0, ∀z ∈ Tπ(x,0) M with |z| = 1 and dπ(x,0) ((0, 1)), z = 0. Thus we know from Lemma 3.1 that dπ(x,0) ((0, 1)) = ±en (y) and so for any j = 1, . . . , n − 1, ˜ i , 0) ∧ dπ(x,0) ((fj , 0)) ∈ span{e1 (y), . . . , en−1 (y)}, where y = π(x, 0). Hence, K((f 2 (fj , 0)) = K(dπ(x,0) ((fi , 0)) ∧ dπ(x,0) ((fj , 0))) = 1 + λ (y), i = j ∈ {1, . . . , n − 1}, ¯ n−1 and any two-dimensional plane P ⊂ Tx M ¯ n−1 , the which shows that for any x ∈ M n−1 ¯ ˜ ¯ ¯ on P must satisfy K(P) = K(P) = g(x) > 0, where sectional curvature K(P) of M ¯ n−1 . Thus, since dim(M ¯ n−1 ) ≥ 3, we know from g(x) = 1 + λ2 (π(x, 0)) is a function on M n−1 ¯ the well-known Schur Lemma [4, p. 106] that M has constant sectional curvature. Consider now the case that n = 3. Let us first show that (2.10) is satisfied and so M 3 is a locally conformally flat manifold. In fact, since n = 3, we get by taking the covariant derivatives of (2.12) that Rijk = 3Hk hij + 3Hhijk −
(hilk hlj + hil hljk ).
(3.10)
l
Also, one has from (2.13) that Rk = 18HHk − Sk .
(3.11)
Hence 1 Bijk = Rijk − Rikj − (δij Rk − δik Rj ) 4 = 3Hk hij − 3Hj hik + (hilj hlk − hilk hlj ) l
−
1 4
δij (18HHk − Sk ) − δik (18HHj − Sj ) .
From hij = λi δij , we have Sk = 2
l,m
hlm hlmk = 2
λl hllk .
l
Thus
3 3 1 Bijk = 3Hk λi − H δij − 3Hj λi − H δik + hikj (λk − λj ) + (δij Sk − δik Sj ) 2 2 4
3 1 = 3(Hk δij − Hj δik ) λi − H + hikj (λk − λj ) + (δij hllk − δik hllj )λl . 2 2 l
(3.12)
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
Taking hlm = λl δlm in the equality dhij + (hlj ωli + hil ωlj ) = hijl ωl , l
235
(3.13)
l
one easily gets hijk = δij ek λi + (λi − λj )ωij (ek ),
(3.14)
and so we have hiik = ek λi .
(3.15)
Since hijk = hikj , we know from (3.14) that if i, j, k are all distinct, then (λi − λj )ωij (ek ) = (λi − λk )ωik (ej ).
(3.16)
Since λ1 = λ2 = λ, λ3 = µ, we conclude that when i, j, k are all distinct Bijk = (λk − λj )hikj = (λk − λj )(λi − λk )ωik (ej ) = (λk − λj )(λi − λj )ωij (ek ) = 0. (3.17) It is trivial to see from (3.12) that Biii = 0. Let A be the Weingarten operator defined by the second fundamental form, that is, for any p ∈ M and all X, Y ∈ Tp M, A : Tp M → Tp M, AX, Y = h(X, Y ). Set Dp (λ) = {X ∈ Tp M : AX = λp X} and let D(λ) be the assignment of Dp (λ) to each point p ∈ M. Since the multiplicity of the principal curvature λ is greater than one, it follows from [15] that D(λ) is a completely integrable distribution on M and that λ is constant on each leaf of D(λ). Thus we have e1 λ = e2 λ = 0. If i = k, we have from (3.12) and (3.15) that
3 1 Biik = 3Hk λi − H + hiik (λk − λi ) + hllk λl 2 2 l
1 1 = (2ek λ + ek µ) λi − λ − µ + (ek λi )(λk − λi ) + hllk λl . (3.18) 2 2 l
Thus
1 1 B113 = B223 = (2e3 λ + e3 µ) − µ + (e3 λ)(µ − λ) + (e3 λ)λ + (e3 µ)µ = 0, 2 2 B331 = e1 µ
µ
1 − λ + (e1 µ)(λ − µ) + (e1 µ)µ = 0. 2 2
Similarly, we have B332 = 0, Bikk = Biki = 0, i = k.
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˜ 3 is also locally conformally flat. Recall Therefore, M 3 is locally conformally flat and so M 3 2 2 ¯ ¯ ˜ that M = M × R, where M has non-negative Gaussian curvature κ. We shall use the same notations Rijkl and Rij , etc. to denote the components of the curvature tensor and the ˜ 3 , respectively. Take an orthonormal local frame {v1 , v2 , v3 } Ricci curvature tensor, etc. of M 3 ¯ 2 . Since M ˜ 3 is a product, one can see easily that ˜ of M such that v1 and v2 are tangent to M R11 = R22 = R1212 + R1313 = R1212 = κ, R12 = R13 = R23 = R33 = 0,
R = 2κ.
Taking i = j = 1, k = 2 in the equality (2.10), we get R112 − R121 =
1 R2 . 4
From the definition, we have R121 = (dR12 )(v1 ) +
3
Rl2 ωl1 (v1 ) +
l=1
R112 = (dR11 )(v2 ) +
3
3
R1l ωl2 (v1 ) = 0,
l=1
Rl1 ωl1 (v2 ) +
l=1
3
Rl2 ωl2 (v2 ) = v2 R11 = v2 κ.
l=1
Therefore, v2 κ =
1 v2 κ, 2
and so v2 κ = 0. ¯ n−1 Similarly, we have v1 κ = 0. Thus κ is a constant function. Hence, for any n ≥ 3, M is a Euclidean sphere. Consequently our M n has constant scalar curvature. But the scalar curvature of M n is given by r=
Rii = (n − 1)(n − 2)(1 + λ2 ).
i
Hence λ is a constant and so µ = −1/λ is also constant. That is, M n is an isoparametric hypersurface in S n+1 (1) with two distinct principal curvatures √ one of which is simple. n = S 1 ( 1 − c2 ) × S n−1 (c) for some From the work of Cartan [6], we conclude that M √ c ∈ (0, 1). Since the principal curvatures of S 1 ( 1 − c2 ) × S n−1 (c) are λ1 = · · · = λn−1 =
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
237
1/(c2 − 1), λn = − c2 /(1 − c2 ) and S = S(n, H), we know that c2 ≤ (n − 1/n). This completes the proof of Theorem 1.1. Before proving Theorem 1.2, we recall an algebraic fact. Lemma 3.2. [8] Let x1 , . . . , xn be n(≥)2 real numbers satisfying the inequality ( i xi )2 ≥ (n − 1) i xi2 . Then for any distinct i, j, 1 ≤ i < j ≤ n, we have xi xj ≥ 0. Proof of Theorem 1.2. Let h = ni,j=1 hij ωi ⊗ ωj be the second fundamental form of M n and assume that hij = λi δij , where λi , i = 1, . . . , n are the principal curvatures of M n . It follows from the Gauss equation that for any i = j (3.19) Rijij = 1 + λi λj , Let |h|2 = ni,j=1 h2ij = i=1 λ2i and tr h = ni=1 λi . Since |h|2 is constant, we have from (2.6) that 0=
n i,j,k=1
h2ijk +
n
λi (tr h)ii +
i=1
n 1 Rijij (λi − λj )2 . 2
(3.20)
i,j=1
Consider first the case that H 2 ≥ {(n − 1)/n2 }S. In this case, we know from Lemma 2.1 that λi λj ≥ 0, ∀i = j. Thus one finds from (3.19) that the sectional curvatures of M n are bounded from below by 1. It then follows from Theorem 1.1 in [5] that either M n is totally geodesic or M n is the boundary of a convex body in an open half-sphere, which implies that the second fundamental form of M n is always semi-positive definite if we choose the unit normal vector field of M n properly. Therefore, we can assume that λi ≥ 0 on M n , ∀i = 1, . . . , n. Take a point p ∈ M n such that (tr h)(p) = minn (tr h)(x). x∈M
Then we have from the maximal principle that (trh)ii (p) ≥ 0, i = 1, . . . , n,
(3.21)
which, combining with (3.20), gives 0≥
1 (λi − λj )2 (p), 2 i,j
and so λ1 (p) = · · · = λn (p). Now for any q ∈ M n , we have from S(q) = S(p) and (3.22) that
(3.22)
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(λi (q) − λj (q)) = 2n 2
i,j
(λi (q)) − 2 2
n
i
≤ 2n
n
2 λi (q)
i
(λi (p))2 − 2
n
i
2 λi (p)
= 0.
(3.23)
i
Thus M n is totally umbilical. Assume now that the Ricci curvature of M n is bounded from below by n − 1. Then we have from Gauss equation that
n λi λj − λ2j ≥ 0, j = 1, . . . , n, on M n . (3.24) i=1
Suppose without loss of generality that λ1 ≥ λ2 ≥ · · · ≥ λn . We claim that λi λj ≥ 0, ∀1 ≤ i, j ≤ n, on M n ,
(3.25)
which, combining with the above arguments above, will imply that M n is totally umbilical. Let us verify that for any p ∈ M n , there holds either λi (p) ≥ 0, i = 1, . . . , n, or λi (p) ≤ 0, i = 1, . . . , n, and so (3.25) holds. We shall prove this fact by contradiction. Thus suppose that there exists a q ∈ M n such that λ1 (q) > 0 If
n
i=1 λi (q) n−1
and
λn (q) < 0.
≥ 0, then we have
λi (q) ≥ −λn (q) > 0.
i=1
Thus
n i=1
λi (q)
λn (q) − λ2n (q)
=
n−1 i=1
λi (q) λn (q) < 0,
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
which contradicts to (3.24). On the other hand, if n
n
i=1 λi (q)
239
< 0, then
λi (q) < −λ1 (q) < 0,
i=2
which gives
n
λi (q)
λ1 (q) − λ21 (q)
i=1
=
n
λi (q) λ1 (q) < 0,
i=2
contradicting to (3.24) again. This completes the proof of Theorem 1.2.
Acknowledgement The authors would like to thank the referee for the encouragements and the careful reading of the manuscript.
References [1] T. Aubin, Some nonlinear problems in Riemannian geometry, Springer Monographs in Mathematics, Springer-Verlag, Berlin, 1998. [2] H. Alencar, M. do Carmo, Hypersurfaces with constant mean curvature in spheres, Proc. Am. Math. Soc. 120 (1994) 1223–1229. [3] H. Alencar, M. do Carmo, W. Santos, A gap theorem for hypersurfaces of the sphere with constant scalar curvature one, Comment. Math. Helv. 77 (2002) 549–562. [4] M. do Carmo, Riemannian Geometry, Birkh¨auser, Boston, MA, 1992. [5] M. do Carmo, F.W. Warner, Rigidity and convexity of hypersurfaces in spheres, J. Diff. Geom. 4 (1970) 133–144. [6] E. Cartan, Familles de surfaces isoparametriques dans les espaces a curvatura constante, Annali di Mat. 17 (1938) 177–191. [7] J. Cheeger, D. Ebin, Comparison Theorems in Riemannian Geometry, North-Holland, 1975. [8] B.Y. Chen, M. Okumura, Scalar curvature, inequalities and submanifolds, Proc. Am. Math. Soc. 38 (1973) 605–608. [9] S.Y. Cheng, S.T. Yau, Hypersurfaces with constant scalar curvature, Math. Ann. 225 (1977) 195–204. [10] S.S. Chern, M. do Carmo, S. Kobayashi, Minimal submanifolds of a sphere with second fundamental form of constant length, in: F. Browder (Ed.), Functional Analysis and Related Fields, Springer-Verlag, Berlin, 1970, pp. 59–75. [11] B. Lawson, Local rigidity theorems for minimal hypersurfaces, Ann. Math. 89 (1969) 187–197. [12] H.Z. Li, Hypersurfaces with constant scalar curvature in space forms, Math. Ann. 305 (1996) 665–672. [13] K. Nomizu, B. Smyth, A formula of Simons’ type and hypersurfaces with constant mean curvature, J. Diff. Geom. 3 (1969) 367–377. [14] S. Montiel, A. Ros, Compact hypersurfaces: the Alexandrov theorem for higher order mean curvatures, Differential Geometry, Pitman Monogr. Surveys Pure Appl. Math., vol. 52, Longman Sci. Tech., Harlow, 1991, pp. 279–296. [15] T. Otsuki, Minimal hypersurfaces in a Riemannian manifold of constant curvature, Am. J. Math. 92 (1970) 145–173.
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[16] C.K. Peng, C.L. Terng, Mimimal hypersurfaces of spheres with constant scalar curvature, Ann. Math. Stud. 103 (1983) 177–198. [17] C.K. Peng, C.L. Terng, The scalar curvature of minimal hypersurfaces in spheres, Math. Ann. 266 (1983) 105–113. [18] T. Sakai, Riemannian Geometry, Translated from the 1992 Japanese original by the author, Translations of Mathematical Monographs, vol. 149, American Mathematical Society, Providence, RI, 1996, xiv + 358 pp. [19] J. Simons, Minimal varieties in Riemannian manifolds, Ann. Math. 88 (1968) 62–105.
Rigidity theorems for closed hypersurfaces in a unit sphere夽 Qiaoling Wang∗ , Changyu Xia Departamento de Matem´atica-IE, Universidade de Bras´ılia, Campus Universit´ario, 70910-900 Bras´ılia, DF, Brasil Received 15 July 2004; received in revised form 6 December 2004; accepted 13 December 2004
Abstract In this paper, we prove some rigidity theorems for closed hypersurfaces in a unit sphere. © 2004 Elsevier B.V. All rights reserved. MSC: 53C20; 53C42 PACS: 02.40.H, M; 02.40.K
JGP SC: Riemannian geometry Keywords: Rigidity; Hypersurfaces; Sphere
1. Introduction Let M n be an n-dimensional closed hypersurface in a unit sphere S n+1 (1) of dimension n + 1 and denote by S the squared norm of the second fundamental form of M n . Many metric and topological rigidity theorems about M n have been obtained. Simons [19], Chern 夽 Both authors were partially supported by CNPq. ∗
Corresponding author. Tel.: +55 612733356257; fax: +55 612732737. E-mail addresses: [email protected] (Q. Wang), [email protected] (Changyu Xia).
0393-0440/$ – see front matter © 2004 Elsevier B.V. All rights reserved. doi:10.1016/j.geomphys.2004.12.005
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et al. [10] and Lawson [11] proved that if M n is minimal and if S ≤ n then M n is either totally geodesic or a Clifford minimal hypersurface. Further results about rigidity of minimal hypersurfaces in S n+1 (1) can be found, e.g., in [15–17], etc. As a natural generalization, the rigidity phenomenon for hypersurfaces in S n+1 (1) with constant mean curvature has also been studied. It has been proven by Nomizu and Smyth [13] that if M n has constant mean curvature and non-negative sectional curvature then M n is either totally umbilic or a Riemannian product S k (c1 ) × S n−k (c2 ), 1 ≤ k ≤ n − 1, where S k (c) denotes the sphere of radius c. Alencar and do Carmo [2] have shown that if M n has constant mean curvature H and if S ≤ nH 2 + C(H, n), where C(H, n) is a constant that√depends only on H and n, then M n is either totally umbilic or a Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1/n). On the other hand, some rigidity theorems for hypersurfaces with constant scalar curvature have been proven. It has been shown by Cheng and Yau that if M n has nonnegative sectional curvature and constant scalar curvature n(n − 1)r with r ≥ 1, then M n is isometric to either a totally umbilic hypersurface or a Riemannian product S k (c1 ) × S n−k (c2 ), 1 ≤ k ≤ n − 1 [9]. Li [12] proved that if M n has constant scalar curvature n(n − 1)r with r ≥ 1 and if S ≤ (n − 1)(n(r − 1) + 2)/(n − 2) + (n − 2)/(n(r − 1) + 2), n is isometric to either a totally umbilic hypersurface or the Riemannian prodthen M√ 1 uct S ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 2)/(nr). Recently, Alencar et al. [3] obtained a gap theorem for closed hypersurfaces with constant scalar curvature n(n − 1) in a unit sphere. √ In this paper, we prove a rigidity theorem for the Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1/n) without the constancy condition on the mean curvature or the scalar curvature. Namely, we have the following theorem. Theorem 1.1. Let M n be an n-dimensional closed hypersurface in S n+1 (1). Denote by S and H the squared norm of the second fundamental form and the mean curvature of M n , respectively. Assume that the fundamental group π1 (M n ) of M n is infinite and that S ≤ S(n, H), where S(n, H) = n +
n(n − 2)|H| 2 2 n3 H 2 − n H + 4(n − 1). 2(n − 1) 2(n − 1)
(1.1)
√ Then S is constant, S = S(n, H) and M is isometric to a Riemannian product S 1 ( 1 − c2 ) × S n−1 (c) with c2 ≤ (n − 1)/n. Montiel and Ros [14] have proven that an embedded closed hypersurface in a half-sphere with constant r-mean curvature for some r ∈ {1, . . . , n}, is a totally umbilical hypersphere. In the second part of this paper, we give another characterization for the totally umbilical hypersphere. Theorem 1.2. Let M n be an n-dimensional closed orientable hypersurface in S n+1 (1). Assume that the squared norm S of the second fundamental form of M n is constant and that one of the following conditions holds: (a) The mean curvature H of M n satisfies H 2 ≥ {(n − 1)/n2 }S on M n .
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229
(b) The Ricci curvatures of M n are bounded from below by n − 1. Then M n is a totally umbilical hypersphere.
2. Preliminaries Let M n be an n-dimensional Riemannian manifold. We choose a local frame of orthonormal vector fields {e1 , . . . , en } on M n and let {ω1 , . . . , ωn } be the dual coframe. The connection forms {ωij } of M n are characterized by the structure equations: ωij ∧ ωj , ωij + ωji = 0, (2.1) dωi = j
dωij =
ωik ∧ ωkj −
k
1 Rijkl ωk ∧ ωl , 2
(2.2)
k,l
where Rijkl are the components of the curvature tensor of M n . For any C2 -function f defined on M n , we define its gradient and Hessian by the following formulas df = fi ω i , (2.3)
i
fij ωj = dfi +
j
fj ωji .
(2.4)
j
The Laplacian of f is given by f = i fii . Let φ = i,j φij ωi ⊗ ωj be a symmetric tensor defined on M n . The covariant derivative of φ is defined by φijk ωk = dφij + φkj ωki + φik ωkj . (2.5) k
Let |φ|2
k
2 i,j φij
k
= and tr φ = i φii . Choose a frame field {ω1 , . . . , ωn } so that φij = λi δij . If φ satisfies the “Codazzi equation” φijk = φikj ,
then we have [9] 1 1 2 |φ|2 = φijk + λi (tr φ)ii + Rijij (λi − λj )2 . 2 2 i,j,k
i
(2.6)
i,j
The Weyl curvature tensor W = (Wijkl ) of M n is defined by 1 (Rik δjl − Ril δjk + Rjl δik − Rjk δil ) n−2 R + (δik δjl − δil δjk ), (n − 1)(n − 2)
Wijkl = Rijkl −
(2.7)
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where Rij and R are the components of Ricci curvature tensor and the scalar curvature of M n , respectively. The Beck tensor B = (Bijk ) of M n is given by 1 1 (Rijk − Rikj ) − (δij Rk − δik Rj ), n−2 2(n − 1)(n − 2)
Bijk =
(2.8)
where Rijk are the components of the covariant derivative of the Ricci curvature tensor of M n and Rk = ek R. The following fact is needed in the proof of Theorem 1.1. Lemma 2.1. [1] If the Ricci curvature of a compact Riemannina manifold is non-negative and positive at a point, then the manifold carries a metric of positive Ricci curvature. M n is said to be locally conformally flat if, for each x ∈ M n , there exists a conformal diffeomorphism of a neighborhood of x onto an open set of the Euclidean n-space Rn . When n ≥ 4, M n is locally conformally flat if and only if Wijkl = 0, and that Bijkl = 0 on M n in this case. When n = 3, we always have Wijkl = 0 on M 3 and M 3 is a locally conformally flat Riemannian manifold if and only if Bijk = 0. Thus, if M n is a locally conformally flat Riemannian manifold, then Rijkl =
1 R (Rik δjl − Ril δjk + Rjl δik − Rjk δil ) − (δik δjl − δil δjk ), n−2 (n − 1)(n − 2) (2.9)
and Rijk −
1 1 δij Rk = Rikj − δik Rj . 2(n − 1) 2(n − 1)
(2.10)
n sphere S n+1 (1) and denote by Let M be a hypersurface in a unit (n + 1)-dimensional n h = i,j hij ωi ⊗ ωj the second fundamental form of M . The squared norm S of h and the mean curvature H of M n are given by
S=
i,j
h2ij ,
nH =
hii ,
i
respectively. The Gauss equation of M n can be written as Rijkl = δik δjl − δil δjk + (hik hjl − hil hjk ). The Ricci tensor and the scalar curvature R of M n are then given by hik hkj , Rij = (n − 1)δij + nHhij −
(2.11)
(2.12)
k
R = n(n − 1) + n2 H 2 − S,
(2.13)
respectively. The covariant derivative hijk of h satisfies hijk = hikj . The eigenvalues λ1 , . . . , λn of (hij ) are the principal curvatures of M n .
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3. Proofs of the results Proof of Theorem 1.1. Assume that {e1 , . . . , en } diagonalizes the second fundamental form of M n so that i,j hij ωi ⊗ ωj = i λi ωi ⊗ ωi . In this case, we have Rijkl = (1 + λi λj )(δik δjl − δil δjk ),
(3.1)
Rij = 0, i = j,
(3.2)
Rii = n − 1 + nHλi − λ2i , i = 1, . . . , n.
(3.3)
and For any fixed j ∈ {1, . . . , n}, since 2
(nH − λj )2 =
λk ≤ (n − 1)
j=k
k=j
λ2k = (n − 1)(S − λ2j ),
we have n2 H 2 − (n − 1)S + nλ2j − 2nHλj ≤ 0.
(3.4)
It is easy to see from
(λi − H) = 0,
i
(λi − H)2 = S − nH 2 ,
i
that (λj − H)2 ≤
n−1 (S − nH 2 ), n
which, combining with (3.4), implies that 0 ≥ n(λ2j − nHλj ) + (n − 2)n(λj − H)H + 2(n − 1)nH 2 − (n − 1)S n−1 2 ≥ n(λj − nHλj ) − (n − 2)n|H| (S − nH 2 ) + 2(n − 1)nH 2 − (n − 1)S. n (3.5) It then follows from (3.3) that
n−1 (n − 1) (S − nH 2 ) + 2(n − 1)H 2 − S n n
n−1 n = n + 2nH 2 − S − (n − 2)|H| · · S − nH 2 n n−1
n−1 n 1 = S − nH 2 + (n − 2)|H| + n2 H 2 + 4(n − 1) n 2 n−1
Rjj ≥ (n − 1) − (n − 2)|H|
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×
n 1 −(n − 2)|H| + n2 H 2 + 4(n − 1) . − S − nH 2 + 2 n−1 (3.6)
Observe that our condition S ≤ S(n, H) is equivalent to 2 n 1 2 2 S − nH ≤ , −(n − 2)|H| + n H + 4(n − 1) 2 n−1 2
that is n 1 2 −(n − 2)|H| + n2 H 2 + 4(n − 1) ≥ 0. − S − nH + 2 n−1 Thus Rjj ≥ 0, ∀j, which, combining with (3.2), implies that M n has non-negative Ricci curvature. Since our M n has infinite fundamental group, we conclude from Bonnet–Myer’s theorem [7] and Lemma 2.1 that ∀p ∈ M n , there exists a unit vector u ∈ Tp M n , such that the Ricci curvature Ric of M n satisfies Ric(u, u) = 0. Note that Ric attains its maximum and minimum in the principal directions. We can assume without loss of generality that at any fixed point x ∈ M n , Rnn = 0. Thus, when n = j, at the point x, the inequality (3.5) should be an equality and S(x) = S(n, H)(x), which in turn implies that when n = j, the above inequalities should take equality sign at x. Consequently, we know that λ1 (x) = · · · = λn−1 (x). Since x ∈ M n is arbitrary, one then deduces that our M n has at most two distinct principal curvatures and, when M n has exactly two distinct principal curvatures, one of these two distinct principal curvatures should be simple. Let us assume that λ1 = · · · = λn−1 = λ and λn = µ on M n . Since Rii ≥ 0 and one of R11 , . . . , Rnn is zero, we deduce from (3.3) that 1 + λµ = 0. Therefore, we conclude that M n has exactly two distinct principal curvatures one of which is simple and S = S(n, H) holds on M. We want to show that λ and µ are constant functions on M. Notice that our M n is a closed manifold with non-negative Ricci curvature. It is well-known that [18, p. 220] the Riemannian universal covering space ˜ n of M n can be decomposed as M ¯ n−s × Rs for some s ∈ {0, 1, . . . , n}, where M ¯ n−s is M a closed simply connected (n − s)-dimensional Riemannina manifold with non-negative Ricci curvature and Rs is the s-dimensional Euclidean space with its standard flat metric. ¯ n−s × Rs is non-compact and so we have s ≥ 1. The infinity of π1 (M n ) implies that M 1 + λµ = 0 on M n , we know from (3.1) that if Sinceλ1 = · · · = λn−1 n= λ, λn = µ and n n u = i=1 ai ei , v = j=1 bj ej ∈ Tp M with |u| = |v| = 1, u, v = 0, then the sectional curvature K(u ∧ v) of the plane spanned by u and v is given by
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
K(u ∧ v) =
i,j,k,l
= 1+
ai bj ak bl Rijkl =
233
ai bj ak bl (1 + λi λj )(δik δjl − δil δjk )
i,j,k,l
i
2 λi ai2 λj bj2 − λi a i b i j
= 1 + λ(1 − an2 ) + µan2
= (1 + λ2 )(1 − an2 − bn2 ).
i
λ(1 − bn2 ) + µbn2 − (λ(−an bn ) + µan bn )2
(3.7)
Lemma 3.1. A vector u ∈ Tp M n with |u| = 1 satisfies the following condition (∗) if v ∈ Tp M n with u, v = 0, |v| = 1, then K(u ∧ v) = 0, if and only if u = ±en (p). Proof of Lemma 3.1. If u = ±en (p) and v satisfies |v| = 1, u, v = 0, then we can write v=
n−1
ai ei (p),
i=1
n−1 i=1
Thus K(u ∧ v) =
ai aj Rninj =
i,j
=
i
ai2 = 1.
ai aj (1 + λn λi )(δnn δij − δni δnj )
i,j
(1 + µλi )ai2 − an2 (1 + µ2 ) = 0.
(3.8)
On the other hand, if (∗) is satisfied and suppose by contradiction that u = aen (p) + w,
n−1
n−1
(3.9)
where w, en (p) = 0, w = 0. Let w = i=1 ci ei (p) and take a vector z = i=1 di ei (p) ∈ Tp M n satisfying |z| = 1, z, w = 0. Then z, u = 0 and so we have from (3.7) that K(u ∧ z) = (1 + λ2 )(1 − a2 ) = (1 + λ2 )|w|2 > 0, which contradicts to (∗). Thus u is parallel to en (p), since |u| = 1, we know that u = ±en (p). Lemma 3.1 is proved. ¯ n−s × Rs are locally isometric, Let us go on the proof of Theorem 1.1. Since M and M ¯ n−1 has constant sectional it follows from Lemma 3.1 that s = 1. We claim that M curvature and so is isometric to a Euclidean (n − 1)-sphere. In order to see this, let us ¯ n−1 × R → M be the natural projection; then π is a assume first that n ≥ 4. Let π : M n−1 ¯ ¯ n−1 . local isometry. For any x ∈ M , take an orthonormal base {f1 , . . . , fn−1 } of Tx M n−1 n−1 ¯ ¯ Since T(x,0) (M × R)=Tx M × T0 R, we know that {(f1 , 0), . . . , (fn−1 , 0), (0, 1)}
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¯ n−1 . ¯ n−1 × R), where 0 is the zero-vector of Tx M is an orthonormal base of T(x,0) (M n−1 ¯ × R) with v, (0, 1) = 0, |v| = 1, it holds Observe that for any v ∈ T(x,0) (M ˜ ˜ denotes the sectional curvature of M ¯ n−1 × R. It follows that K((0, 1) ∧ v) = 0, where K n K(dπ(x,0) ((0, 1)) ∧ z) = 0, ∀z ∈ Tπ(x,0) M with |z| = 1 and dπ(x,0) ((0, 1)), z = 0. Thus we know from Lemma 3.1 that dπ(x,0) ((0, 1)) = ±en (y) and so for any j = 1, . . . , n − 1, ˜ i , 0) ∧ dπ(x,0) ((fj , 0)) ∈ span{e1 (y), . . . , en−1 (y)}, where y = π(x, 0). Hence, K((f 2 (fj , 0)) = K(dπ(x,0) ((fi , 0)) ∧ dπ(x,0) ((fj , 0))) = 1 + λ (y), i = j ∈ {1, . . . , n − 1}, ¯ n−1 and any two-dimensional plane P ⊂ Tx M ¯ n−1 , the which shows that for any x ∈ M n−1 ¯ ˜ ¯ ¯ on P must satisfy K(P) = K(P) = g(x) > 0, where sectional curvature K(P) of M ¯ n−1 . Thus, since dim(M ¯ n−1 ) ≥ 3, we know from g(x) = 1 + λ2 (π(x, 0)) is a function on M n−1 ¯ the well-known Schur Lemma [4, p. 106] that M has constant sectional curvature. Consider now the case that n = 3. Let us first show that (2.10) is satisfied and so M 3 is a locally conformally flat manifold. In fact, since n = 3, we get by taking the covariant derivatives of (2.12) that Rijk = 3Hk hij + 3Hhijk −
(hilk hlj + hil hljk ).
(3.10)
l
Also, one has from (2.13) that Rk = 18HHk − Sk .
(3.11)
Hence 1 Bijk = Rijk − Rikj − (δij Rk − δik Rj ) 4 = 3Hk hij − 3Hj hik + (hilj hlk − hilk hlj ) l
−
1 4
δij (18HHk − Sk ) − δik (18HHj − Sj ) .
From hij = λi δij , we have Sk = 2
l,m
hlm hlmk = 2
λl hllk .
l
Thus
3 3 1 Bijk = 3Hk λi − H δij − 3Hj λi − H δik + hikj (λk − λj ) + (δij Sk − δik Sj ) 2 2 4
3 1 = 3(Hk δij − Hj δik ) λi − H + hikj (λk − λj ) + (δij hllk − δik hllj )λl . 2 2 l
(3.12)
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
Taking hlm = λl δlm in the equality dhij + (hlj ωli + hil ωlj ) = hijl ωl , l
235
(3.13)
l
one easily gets hijk = δij ek λi + (λi − λj )ωij (ek ),
(3.14)
and so we have hiik = ek λi .
(3.15)
Since hijk = hikj , we know from (3.14) that if i, j, k are all distinct, then (λi − λj )ωij (ek ) = (λi − λk )ωik (ej ).
(3.16)
Since λ1 = λ2 = λ, λ3 = µ, we conclude that when i, j, k are all distinct Bijk = (λk − λj )hikj = (λk − λj )(λi − λk )ωik (ej ) = (λk − λj )(λi − λj )ωij (ek ) = 0. (3.17) It is trivial to see from (3.12) that Biii = 0. Let A be the Weingarten operator defined by the second fundamental form, that is, for any p ∈ M and all X, Y ∈ Tp M, A : Tp M → Tp M, AX, Y = h(X, Y ). Set Dp (λ) = {X ∈ Tp M : AX = λp X} and let D(λ) be the assignment of Dp (λ) to each point p ∈ M. Since the multiplicity of the principal curvature λ is greater than one, it follows from [15] that D(λ) is a completely integrable distribution on M and that λ is constant on each leaf of D(λ). Thus we have e1 λ = e2 λ = 0. If i = k, we have from (3.12) and (3.15) that
3 1 Biik = 3Hk λi − H + hiik (λk − λi ) + hllk λl 2 2 l
1 1 = (2ek λ + ek µ) λi − λ − µ + (ek λi )(λk − λi ) + hllk λl . (3.18) 2 2 l
Thus
1 1 B113 = B223 = (2e3 λ + e3 µ) − µ + (e3 λ)(µ − λ) + (e3 λ)λ + (e3 µ)µ = 0, 2 2 B331 = e1 µ
µ
1 − λ + (e1 µ)(λ − µ) + (e1 µ)µ = 0. 2 2
Similarly, we have B332 = 0, Bikk = Biki = 0, i = k.
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˜ 3 is also locally conformally flat. Recall Therefore, M 3 is locally conformally flat and so M 3 2 2 ¯ ¯ ˜ that M = M × R, where M has non-negative Gaussian curvature κ. We shall use the same notations Rijkl and Rij , etc. to denote the components of the curvature tensor and the ˜ 3 , respectively. Take an orthonormal local frame {v1 , v2 , v3 } Ricci curvature tensor, etc. of M 3 ¯ 2 . Since M ˜ 3 is a product, one can see easily that ˜ of M such that v1 and v2 are tangent to M R11 = R22 = R1212 + R1313 = R1212 = κ, R12 = R13 = R23 = R33 = 0,
R = 2κ.
Taking i = j = 1, k = 2 in the equality (2.10), we get R112 − R121 =
1 R2 . 4
From the definition, we have R121 = (dR12 )(v1 ) +
3
Rl2 ωl1 (v1 ) +
l=1
R112 = (dR11 )(v2 ) +
3
3
R1l ωl2 (v1 ) = 0,
l=1
Rl1 ωl1 (v2 ) +
l=1
3
Rl2 ωl2 (v2 ) = v2 R11 = v2 κ.
l=1
Therefore, v2 κ =
1 v2 κ, 2
and so v2 κ = 0. ¯ n−1 Similarly, we have v1 κ = 0. Thus κ is a constant function. Hence, for any n ≥ 3, M is a Euclidean sphere. Consequently our M n has constant scalar curvature. But the scalar curvature of M n is given by r=
Rii = (n − 1)(n − 2)(1 + λ2 ).
i
Hence λ is a constant and so µ = −1/λ is also constant. That is, M n is an isoparametric hypersurface in S n+1 (1) with two distinct principal curvatures √ one of which is simple. n = S 1 ( 1 − c2 ) × S n−1 (c) for some From the work of Cartan [6], we conclude that M √ c ∈ (0, 1). Since the principal curvatures of S 1 ( 1 − c2 ) × S n−1 (c) are λ1 = · · · = λn−1 =
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
237
1/(c2 − 1), λn = − c2 /(1 − c2 ) and S = S(n, H), we know that c2 ≤ (n − 1/n). This completes the proof of Theorem 1.1. Before proving Theorem 1.2, we recall an algebraic fact. Lemma 3.2. [8] Let x1 , . . . , xn be n(≥)2 real numbers satisfying the inequality ( i xi )2 ≥ (n − 1) i xi2 . Then for any distinct i, j, 1 ≤ i < j ≤ n, we have xi xj ≥ 0. Proof of Theorem 1.2. Let h = ni,j=1 hij ωi ⊗ ωj be the second fundamental form of M n and assume that hij = λi δij , where λi , i = 1, . . . , n are the principal curvatures of M n . It follows from the Gauss equation that for any i = j (3.19) Rijij = 1 + λi λj , Let |h|2 = ni,j=1 h2ij = i=1 λ2i and tr h = ni=1 λi . Since |h|2 is constant, we have from (2.6) that 0=
n i,j,k=1
h2ijk +
n
λi (tr h)ii +
i=1
n 1 Rijij (λi − λj )2 . 2
(3.20)
i,j=1
Consider first the case that H 2 ≥ {(n − 1)/n2 }S. In this case, we know from Lemma 2.1 that λi λj ≥ 0, ∀i = j. Thus one finds from (3.19) that the sectional curvatures of M n are bounded from below by 1. It then follows from Theorem 1.1 in [5] that either M n is totally geodesic or M n is the boundary of a convex body in an open half-sphere, which implies that the second fundamental form of M n is always semi-positive definite if we choose the unit normal vector field of M n properly. Therefore, we can assume that λi ≥ 0 on M n , ∀i = 1, . . . , n. Take a point p ∈ M n such that (tr h)(p) = minn (tr h)(x). x∈M
Then we have from the maximal principle that (trh)ii (p) ≥ 0, i = 1, . . . , n,
(3.21)
which, combining with (3.20), gives 0≥
1 (λi − λj )2 (p), 2 i,j
and so λ1 (p) = · · · = λn (p). Now for any q ∈ M n , we have from S(q) = S(p) and (3.22) that
(3.22)
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Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
(λi (q) − λj (q)) = 2n 2
i,j
(λi (q)) − 2 2
n
i
≤ 2n
n
2 λi (q)
i
(λi (p))2 − 2
n
i
2 λi (p)
= 0.
(3.23)
i
Thus M n is totally umbilical. Assume now that the Ricci curvature of M n is bounded from below by n − 1. Then we have from Gauss equation that
n λi λj − λ2j ≥ 0, j = 1, . . . , n, on M n . (3.24) i=1
Suppose without loss of generality that λ1 ≥ λ2 ≥ · · · ≥ λn . We claim that λi λj ≥ 0, ∀1 ≤ i, j ≤ n, on M n ,
(3.25)
which, combining with the above arguments above, will imply that M n is totally umbilical. Let us verify that for any p ∈ M n , there holds either λi (p) ≥ 0, i = 1, . . . , n, or λi (p) ≤ 0, i = 1, . . . , n, and so (3.25) holds. We shall prove this fact by contradiction. Thus suppose that there exists a q ∈ M n such that λ1 (q) > 0 If
n
i=1 λi (q) n−1
and
λn (q) < 0.
≥ 0, then we have
λi (q) ≥ −λn (q) > 0.
i=1
Thus
n i=1
λi (q)
λn (q) − λ2n (q)
=
n−1 i=1
λi (q) λn (q) < 0,
Q. Wang, C. Xia / Journal of Geometry and Physics 55 (2005) 227–240
which contradicts to (3.24). On the other hand, if n
n
i=1 λi (q)
239
< 0, then
λi (q) < −λ1 (q) < 0,
i=2
which gives
n
λi (q)
λ1 (q) − λ21 (q)
i=1
=
n
λi (q) λ1 (q) < 0,
i=2
contradicting to (3.24) again. This completes the proof of Theorem 1.2.
Acknowledgement The authors would like to thank the referee for the encouragements and the careful reading of the manuscript.
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