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Revista de la Unión Matemática Argentina

versión On-line ISSN 1669-9637

Rev. Unión Mat. Argent. v.47 n.1 Bahía Blanca ene./jun. 2006

 

On Complete Spacelike Submanifolds in the De Sitter Space With Parallel Mean Curvature Vector

Rosa Maria S. Barreiro Chaves and Luiz Amancio M. Sousa Jr.

 

Abstract: The text surveys some results concerning submanifolds with parallel mean curvature vector immersed in the De Sitter space. We also propose a semi-Riemannian version of an important inequality obtained by Simons in the Riemannian case and apply it in order to obtain some results characterizing umbilical submanifolds and a product of submanifolds in the (n + p)-dimensional De Sitter space 𝕊n+pp .

2000 Mathematics Subject Classification. Primary 53C42, 53A10

Key words and phrases.De Sitter space, Simons type formula, complete spacelike submanifolds, parallel mean curvature vector

 

1. Introduction

Let  n+p+1 ℝ p be an (n + p + 1) -dimensional real vector space endowed with an inner product of index p given by

 p n+p+1 ∑ ∑ < x,y >= - xiyi + xjyj, i=1 j=p+1

where x = (x1,x2,⋅⋅⋅ ,xn+p+1) is the natural coordinate of ℝnp+p+1 .

We also define the semi-Riemannian manifold 𝕊n+p p , by

 p n+p+1 n+p n+p+1 ∑ 2 ∑ 2 𝕊p = {(x1,x2,⋅⋅⋅xn+p+1) ∈ ℝ p ∕ - xi + xj = 1} i=1 j=p+1 .

 n+p 𝕊p is called (n + p) -dimensional De Sitter space of index p .

Let M n be an n -dimensional semi-Riemannian manifold immersed in 𝕊np+p . M n is said to be spacelike if the induced metric on M n from the metric of 𝕊n+p p is positive definite.

From now on, we will consider spacelike submanifolds  n M of  n+p 𝕊 p with parallel mean curvature vector h . Let H = ∣h∣ be the mean curvature of  n M . If h is parallel it is easy to verify that H is constant and, when p = 1 , these two conditions are equivalent. We say that M n is a maximal submanifold if h vanishes identically.

It was proved by E. Calabi [6] (for n ≤ 4 ) and by S.Y. Cheng and S.T. Yau [8] (for all n) that a complete maximal spacelike hypersurface in  n+1 ℝ1 is totally geodesic. In [17], S. Nishikawa obtained similar results for others Lorentzian manifolds. In particular, he proved that a complete maximal spacelike hypersurface in 𝕊n+11 is totally geodesic. We recall that a submanifold M n is said totally geodesic if its second fundamental form B vanishes identically.

A. Goddard [11] conjectured that the complete spacelike hypersurfaces of  n+1 𝕊1 with H constant must be totally umbilical. The totally umbilical hypersurfaces of  n+1 𝕊1 are obtained by intersecting  n+1 𝕊 1 with linear hyperplanes through the origin of  n+2 ℝ1 , where  n+1 𝕊1 can be viewed as hypersphere of ℝn+12 .

J. Ramanathan [19] proved Goddard's conjecture for 𝕊31 and 0 ≤ H ≤ 1 . Moreover, if H > 1 he showed that the conjecture is false as can be seen from an example due to Dajczer-Nomizu [10]. In his proof, Ramanthan used the complex structure of  3 𝕊 1 . K. Akutagawa [2] proved that Goddard's conjecture is true when n = 2 and  2 H ≤ 1 or when n ≥ 3 and H2 < 4(nn-21) . He also constructed complete spacelike rotation surfaces in 𝕊31 with constant H satisfying H > 1 and which are not totally umbilical.

In [15], S. Montiel proved that Goddard's conjecture is true provided that  n M is compact. Furthermore, he exhibited examples of complete spacelike hypersurfaces with constant H satisfying H2 ≥ 4(nn-21) and being not totally umbilical - the so called hyperbolic cylinders (cf. [2] and [13]), which are isometric to the Riemannian product ℍ1(sinh r) × 𝕊n-1(cosh r) of a hyperbolic line and an (n - 1) -dimensional sphere of constant seccional curvatures  2 1 - coth r and  2 1 - tanh r , respectively. Later, Montiel [16] studied complete spacelike hypersurfaces with constant mean curvature  4(n-1) H2 = -n2--- and proved the following result.

Theorem 1.1. Let M n be a complete spacelike hypersurfaces in 𝕊n+1 1 with constant mean curvature  2 4(n-1) H = --n2-- . If  n M is not connected at infinity, that is, if  n M has at least two ends, then M n is, up to isometry, a hyperbolic cylinder.

Concerning to submanifolds M n of 𝕊np+p with parallel mean curvature vector we may cite the following remarkable results. In [12], T. Ishihara proved the following theorem that generalizes for higher codimension the result of Cheng-Yau [8]

Theorem 1.2. Let  n M be an n-dimensional complete Riemannian manifold isometrically immersed in ℝn+pp or 𝕊np+p . If M n is maximal, then the immersion is totally geodesic and M n is a Riemannian space of constant curvature.

In [7], Q.M. Cheng showed that Akutagawa's result [2] is valid for higher codimensional complete spacelike submanifolds in 𝕊np+p with parallel mean curvature vector. More precisely, he proved the following result.

Theorem 1.3. Let M n be an n-dimensional complete spacelike submanifold in 𝕊np+p with parallel mean curvature vector. If H2 ≤ 1 , when n=2 or n2H2 < 4(n - 1) , when n ≥ 3, then M n is totally umbilical.

In [14], H. Li obtained the following extension of Theorem 1.1.

Theorem 1.4. Let M n be an n-dimensional complete spacelike submanifold in 𝕊np+p with parallel mean curvature vector. If  2 4(n-1)- H = n2 and  n M is not connected at infinity, that is, if  n M has at least two ends, then  n M is, up to isometry, a hyperbolic cylinder in  n+1 𝕊1 .

R. Aiyama [1] studied compact spacelike submanifold in 𝕊n+pp with parallel mean curvature vector and proved the following results:

Theorem 1.5. Let  n M be an n-dimensional compact spacelike submanifold in  n+p 𝕊p with parallel mean curvature vector. If the normal connection of M n is flat, then M n is totally umbilical.

Theorem 1.6. Let  n M be an n-dimensional compact spacelike submanifold in  n+p 𝕊p with parallel mean curvature vector. If the sectional curvature of  n M is non-negative, then M n is totally umbilical.

We point out that L. Alias and A. Romero [3] also obtained results related to complete spacelike submanifolds in 𝕊np+p with parallel mean curvature vector.

Let 𝕊n(r) be an n-dimensional sphere in ℝn+1 with radius r and let M n be an n -dimensional submanifold minimally immersed in  n+p 𝕊 (1) . Denote by B the second fundamental form of this immersion and by S the square of the length of B . In his pioneering work, J. Simons [20] proved the following inequality for ΔS

1 ( ( 1) ) -ΔS ≥ S n - 2 - -- S . 2 p
(1.1)

As an application of formula (1.1), Simons [20] obtained the following result.

Theorem 1.7. Let M n be a closed minimal submanifold of 𝕊n+p(1) . Then either M n is totally geodesic, or S = -n-1 2- p , or sup S > -n1- 2- p .

Two years later, S.S. Chern, M. do Carmo and S. Kobayashi [9], determined all the minimal submanifolds of 𝕊n+p(1) satisfying S = --n1 2- p . More precisely, they proved:

Theorem 1.8. Let M n be a closed minimal submanifold of 𝕊n+p(1) . Assume that S ≤ -n-- 2- 1p . Then:
(i) Either S = 0 (and M n is totally geodesic) or S = 2n- 1 p .
(ii)  -n-- S = 2- 1p if and only if:
a) p = 1 and  n M is locally a Clifford torus  ∘ -- ∘ ---- k ( k) n- k ( n-k) 𝕊 n × 𝕊 n .
b) p = n = 2 and M 2 is locally a Veronese surface in 𝕊4(1) .

In the case of a submanifold M n of 𝕊n+p(1) with non-zero parallel mean curvature vector h , it is convenient to modify slightly the second fundamental form B and to introduce the tracelless tensor Φ = B - Hg , where H =∣ h ∣ is the mean curvature and g stands for the induced metric on  n M . W. Santos [21] established the following inequality for the Laplacian of  2 ∣ Φ ∣

 ( ( ) ) 1- 2 2 2 -n(n---2)-- 2p---3- 2 2Δ ∣ Φ ∣ ≥∣ Φ ∣ n(1 + H ) - ∘ ---------∣ g(Φ,h) ∣ - p - 1 ∣ Φ ∣ . n(n - 1)

Let  n M be a complete spacelike maximal submanifold of  n+p 𝕊p . In [12], T. Ishihara derived the following inequality for ΔS

 ( ) 1-ΔS ≥ S n + S- . 2 p
(1.2)

As an important application of (1.2), Ishihara proved Theorem 1.2.

If M n is a spacelike hypersurface of 𝕊n+11 with constant mean curvature H , as in the Riemannian case, it is convenient to consider the tensor Φ . U.H. Ki, H.J. Kim and H. Nakagawa [13], established the following inequality for Δ ∣ Φ ∣2

 ( ) 1 n(n - 2) -Δ ∣ Φ ∣2≥ ∣ Φ ∣2 ∣ Φ ∣2 - ∘---------H ∣ Φ ∣ +n(1 - H2) . 2 n(n - 1)
(1.3)

By applying (1.3) they obtained a constant S+ that depends on n and H and such that S ≤ S+ . They also characterized the hyperbolic cylinders as the only complete spacelike hypersurfaces of 𝕊n1+1 with non-zero constant H and S = S+ . Moreover, they proved that a complete spacelike hypersurface of 𝕊n+1 1 with non-zero constant H and non-negative sectional curvature is totally umbilical, provided that S < S+ .

A. Brasil, G. Colares and O. Palmas [5] obtained the following gap theorem.

Theorem 1.9. Let  n M , n ≥ 3 , be a complete spacelike hypersurface in  n+1 𝕊1 with constant mean curvature H > 0 . Then  2 sup ∣ Φ ∣< ∞ and
a)either sup ∣ Φ ∣= 0 and M n is totally umbilical or
b) ∘ ---------- B- ≤ sup ∣ Φ ∣2 ≤ B+ H H , where B - ≤ B+ H H are the roots of the polynomial

P (x) = x2 - ∘n(n---2)-Hx + n(1 - H2). H n(n - 1)

Recently, A. Brasil, R.M.B. Chaves and G. Colares [4] extended the above result for complete spacelike submanifolds in 𝕊np+p with parallel mean curvature vector.
Let M n be a spacelike submanifold of Qn+p(c) p with non-zero parallel mean curvature vector h and let H = ∣ h ∣ . Define the second fundamental form with respect to the normal direction  h- ξ = H by  ξ h . If  ξ 2 ∣ h ∣ denotes the squared norm of  ξ h , set ∣ μ ∣2=∣ hξ ∣2 - nH2 . In [7], Q. M. Cheng proved that

 ( ) 1- 2 2 2 -n(n---2)-- 2 2 Δ ∣ μ ∣≥ ∣ μ ∣ ∣ μ ∣ - ∘n(n----1)H ∣ μ ∣ +n(1 - H ) .
(1.4)

Now we are going to state our main results. Theorem 1.10 is a Simons' type inequality for submanifolds in De Sitter space 𝕊n+p p .

Theorem 1.10. Let M n be a spacelike submanifold immersed in 𝕊np+p with parallel mean curvature. Then the following inequality holds

 ( 2 ) 1Δ ∣ Φ ∣2 ≥ ∣ Φ ∣2 ∣ Φ-∣-- ∘n(n----2)-H ∣ Φ ∣ +n(1 - H2) . 2 p n(n - 1)
(1.5)

Next Theorem is a Lorentzian version of results obtained by K. Yano and S. Ishihara [22] and also by S.T. Yau [23] for Riemannian submanifolds.

Theorem 1.11. Let M n be a complete spacelike submanifold in 𝕊np+p with parallel mean curvature vector and non-negative sectional curvature. If M n has constant scalar curvature R, then M n is totally umbilical or a product M × M × ⋅⋅⋅ × M 1 2 k , where each Mi is a totally umbilical submanifold of  n+p 𝕊 p and the  ′ M is are mutually perpendicular along their intersections.

As we saw in the Theorem 1.6, compact spacelike submanifolds in  n+p 𝕊p with parallel mean curvature vector and non-negative sectional curvature are totally umbilic.

The following result is an application of formula (1.5).

Theorem 1.12. Let  n M be a complete spacelike submanifold in  n+p 𝕊p with parallel mean curvature vector. If sup K denotes the function that assigns to each point of M n the supremum of the sectional curvatures at that point, there exists a constant β(n, p,H) such that if sup K ≤ β(n, p,H) , then either:
(i) n = 2 and M 2 is totally umbilical or
(ii) n ≥ 3 and M n is totally geodesic.

2. Preliminaries

In this section we will introduce some basic facts and notations that will appear on the paper. Let  n M be an n -dimensional Riemannian manifold immersed in  n+p 𝕊p . As the indefinite Riemannian metric of  n+p 𝕊 p induces the Riemannian metric of M n , the immersion is called spacelike. We choose a local field of semi-Riemannian orthonormal frames e1,...,en+p in 𝕊np+p such that, at each point of M n , e1,...,en span the tangent space of M n . We make the following standard convention of indices

1 ≤ A, B, C,⋅⋅⋅ ≤ n + p, 1 ≤ i,j,k,⋅⋅⋅ ≤ n, n + 1 ≤ α,β,γ, ⋅⋅⋅ ≤ n + p.

Take the correspondent dual coframe {ω1, ...,ωn+p} such that the semi-Riemannian metric of 𝕊np+p is given by  2 ∑ 2 ∑ 2 ∑ 2 ds¯ = ωi - ωα = ɛAωA ,ɛi = 1,ɛα = - 1, 1 ≤ i ≤ n,n + 1 ≤ α ≤ n + p. i α A Then the structure equations of  n+p 𝕊p are given by

 ∑ dωA = ɛB ωAB ∧ ωB, ωAB + ωBA = 0. B
(2.1)
dω = ∑ ɛ ω ∧ ω - 1-∑ ɛ ɛ K ω ∧ ω . AB C AC CB 2 C D ABCD C D C C,D
(2.2)
KABCD = ɛA ɛB(δACδBD - δAD δBC).
(2.3)

Next, we restrict those forms to M n . First of all we get

ωα = 0, n + 1 ≤ α ≤ n + p.
(2.4)

So the Riemannian metric of M n is written as  ∑ ds2 = ωi2 i .

Since  ∑ 0 = dωα = ωαi ∧ ωi, i from Cartan's lemma, we can write

 ∑ ωαi = h αijωj, hαij = h αji. j
(2.5)

Set  ∑ α B = hijωiωjeα α,i,j ,  ( ) 1-∑ ∑ α h = n hii eα α i and  ∘ ------------- 1- ∑ ∑ α 2 H = ∣h∣ = n ( hii) α i the second fundamental form, the mean curvature vector and the mean curvature of M n , respectively.

Using the structure equations we obtain the Gauss equation

 ∑ ( α α α α ) Rijkl = (δikδjl - δilδjk) - hikhjl - h ilhjk . α
(2.6)

The scalar curvature R is given by

 2 2 R = n(n - 1) - n H + S,
(2.7)

where  ∑ S = (h αij)2 α,i,j is the squared norm of the second fundamental form of M n .

We also have the structure equations of the normal bundle of M n

dω = ∑ ω ∧ ω , ω + ω = 0. α αβ β αβ βα β
(2.8)
 ∑ 1 ∑ dω αβ = ωαγ ∧ ωγβ - -- R αβijωi ∧ ωj, γ 2 i,j
(2.9)

where

 ∑ ( α β α β) R αβij = hilhlj - hjlhli . l
(2.10)

The covariant derivatives  α hijk of  α h ij satisfy

∑ α α ∑ α ∑ α ∑ β hijkωk = dhij + hikωkj + hjkωki - hijω βα. k k k β
(2.11)

Then, by exterior differentiation of (2.5), we obtain the Codazzi equation

 α α α h ijk = hjik = hikj.
(2.12)

Similarly, we have the second covariant derivatives hα ijkl of hα ij so that

∑ α α ∑ α ∑ α hijklωl = dhijk + hljkωli + h ilkωlj+ ∑l ∑ l l hαijlωlk - hβijkωβα. l β
(2.13)

By exterior differentiation of (2.11), we can get the following Ricci formula

 ∑ ∑ ∑ hαijkl - hαijlk = hαimRmjkl + hαjmRmikl + hβijRαβkl. m m β
(2.14)

The Laplacian △h αij of hαij is defined by  ∑ △h αij = k hαijkk . From (2.12) and (2.14), we have

 α ∑ α ∑ α ∑ α ∑ β △h ij = h kkij + h kmRmijk + hmiRmkjk + h ikR αβjk. k m,k m,k k,β
(2.15)

If H ⁄= 0 , we choose  h en+1 = H-. Thus

 1 1 Hn+1 = --trhn+1 = H and H α = --trh α = 0,α ≥ n + 2, n n
(2.16)

where hα denotes the matrix [h α]. ij

From (2.6), (2.10), (2.15) and (2.16) it is straightforward to verify that

 1 ∑ ∑ -△S = (hαijk)2 + n h αijH αij+ 2 α,i,j,k α,i,j 2 2 ∑ n+1 α 2 (nS - n H ) - nH tr(h (h ))+ α ∑ ∑ [tr(hαh β)]2 + N (h αhβ - hβh α), α,β α,β
(2.17)

where  t N (A) = tr(AA ) , for all matrix A = [aij].

Recall that  n M is a submanifold with parallel mean curvature vector h if  ⊥ ∇ h ≡ 0, where ∇ ⊥ is the normal connection of M n in 𝕊np+p. Note that this condition implies that H = ∣h∣ is constant and

∑ α hkki = 0, ∀i,α. k
(2.18)

We will need the following generalized Maximum Principle due to Omori and Yau (cf. [18] and [23]).

Lemma 2.1. Let M n be a complete Riemannian manifold with Ricci curvature bounded from below and let F : M n → ℝ be a C2 -function which is bounded from below on M n . Then there is a sequence of points {p } k in M n such that

lki→m∞ F (pk) = inf(F), lkim→∞ ∣ ∇F (pk) ∣= 0 and limk→ in∞f ΔF (pk) ≥ 0.

We also will need the following algebraic Lemma (for a proof see [21]).

Lemma 2.2. Let A, B : ℝn → ℝn be symmetric linear maps such that AB - BA = 0 and trA = trB = 0. Then

 2 ---n --2--- ∘ ------ ∣ trA B ∣≤ ∘n--(n --1)N (A) N (B)
(2.19)

and the equality holds if and only if n - 1 of the eigenvalues xi of A and the corresponding eigenvalues yi of B satisfy

 ∘ --------- ∣ x ∣= -N-(A)--, x x ≥ 0, i n(n - 1) i j ∘ --------- ( ∘ ---------) -N-(B)--- -N-(B)--- yi = n(n - 1) resp.yi = - n(n - 1) .
(2.20)

3. Proof of Simons' type Inequality

Proof of Theorem 1.10. If H ⁄= 0 , set  α α α Φ ij = hij - H δij and consider the following symmetric tensor

 ∑ Φ = Φ αijωiωje α. α,i,j
(3.1)

It is easy to check that Φ is traceless and

 α α α 2 N (Φ ) =∑ N (h ) - n(H ) ; ∣ Φ ∣2= N (Φ α) = S - nH2, α
(3.2)

where Φα denotes the matrix [Φα ] ij .

Because h is parallel, we have H constant. Moreover, as H ⁄= 0 , we can choose a local field of orthonormal frames {e1, e2,⋅ ⋅⋅ ,en+p} such that  h en+1 = --- H . With this choice (2.16) implies that

hn+1h α = hαhn+1, Φn+1 = hn+1 - H δij, Nij(Φn+1) ij= tr(hn+1)2 - nH2 = N (hn+1) - nH2, n+1 3 n+1 3 n+1 3 tr(h ) = tr(Φ ) + 3HN (Φ ) + nH .
(3.3)
Φ α = hα , N (Φ α) = N (hα), α ≥ n + 2. ij ij
(3.4)

Since h is parallel, from (2.17), (3.2), (3.3) and (3.4) we have

1- 2 1- 2 2 ∑ n+1 α 2 ∑ α β 2 2 Δ ∣ Φ ∣= 2ΔS ≥ n(1 - H ) ∣ Φ ∣ - nH tr(Φ (Φ ) ) + (tr Φ Φ ) . α α,β
(3.5)

As the matrices  α Φ and  n+1 Φ are traceless and the matrix  n+1 Φ comutes with all the matrices  α Φ , we can apply Lemma 2.2 in order to obtain

∑ n - 2 ∘ --------- tr(Φn+1( Φα)2) ≤ ∘----------- N (Φn+1) ∣ Φ ∣2 α n(n - 1) ---n---2--- 3 ≤ ∘ ---------∣ Φ ∣ . n(n - 1)
(3.6)

Due to Cauchy-Schwarz inequality we can write

 ∑ ∑ ∣ Φ ∣4≤ p N 2(Φ α) ≤ p (tr ΦαΦ β)2. α α,β
(3.7)

It follows from (3.5), (3.6) and (3.7) that formula (1.5) holds.

If H ≡ 0 , M n is said to be maximal. In this case, from (1.2) we have

1 (S ) --ΔS ≥ S --+ n . 2 p
(3.8)

□

4. Proofs of Theorems 1.11 and 1.12

Proof of Theorem 1.11. Since the mean curvature vector h is parallel and  ∑ β 1 ∑ hαijhkiR αβjk = -- N (h αhβ - hβh α) α,β,i,j,k 2 α,β , from (2.15) we have

1- 1-∑ α 2 ∑ α 2 ∑ α α 2ΔS = 2 Δ(h ij) = (hijk) + hijΔh ij ∑ α,i,j ∑α,i,j,k α,i,j = (hα )2 + 1- N (hαhβ - h βhα) α,i,j,k ijk 2 α,β ∑ α α ∑ α α + hijhkmRmijk + h ijhmiRmkjk. α,i,j,k,m α,i,j,k,m
(4.1)

Next, we will obtain a pointwise estimate for the last two terms. For each fixed α, let λα i be an eigenvalue of  α h , i.e.  α α h ij = λi δij, and denote by inf K the infimum of the sectional curvatures at a point p of  n M . Then

 ( ∑ ∑ ) 2 h αhα R + hα hα R = ij km mijk ij mi mkjk ∑ i,j,k,m α α ∑ i,j,(k,mα 2 α 2) (- 2λi λ k)Rikik + (λi ) + (λ k) Rikik = i∑,k i,k ∑ (λα - λα )2Rikik ≥ (inf K) (λα- λα)2 = i,k i k i,k i k α 2 α 2 α (inf K)(2nN (h ) - 2n (H ) ) = 2n(inf K)N (Φ ).
(4.2)

It implies that

 ∑ α α ∑ α α hijh kmRmijk + hijhmiRmkjk ≥ α,i,j,k,m ∑ α,i,j,k,m n(inf K) N (Φ α) = n(inf K) ∣ Φ ∣2 . α
(4.3)

As h parallel implies H constant, by (2.7) we see that  2 2 S = R + n H - n(n - 1) is also constant, thus ΔS = 0 .

Since Rijij ≥ 0 , from (4.1) and (4.3), we get

 1- ∑ α 2 2 0 = 2ΔS ≥ (hijk) + n(inf K) ∣ Φ ∣ ∑ α,i,j,k + 1- N (hαh β - hβhα) ≥ 0. 2 α,β
(4.4)

It turns out that:
i) h αhβ = h βhα , for all α and β and so the normal bundle of M n is flat. Hence, all the matrices h α can be diagonalized simultaneously;
ii) hα = 0, ∀i,j,k,α ijk and so the second fundamental form B is parallel. In particular, it implies that  α λi is constant for all i,α .

From i), ii), (4.1) and (4.2) we can write  ∑ 0 = (λαi - λαj )2Rijij α,i,j and, since Rijij ≥ 0 , we obtain (λαi - λ αj)Rijij = 0 .

Consequentely, we may apply the same methods used by Ishihara (see [12], Lemmas 5.1, 5.2 and Theorem 1.3) to conclude that  n M is totally umbilical or a product M1 × M2 ⋅⋅⋅ × Mk, where Mi is a totally umbilical submanifold in  n+p 𝕊 p and the  ′ M is are mutually perpendicular along their intersections. □

Remark: Let M n be a complete spacelike submanifold in 𝕊n+p (c) p with parallel mean curvature vector and non-negative sectional curvature. In (4.4), we got the inequality ΔS ≥ 0 , which shows that S is a subharmonic smooth function. Therefore, if the supremum of S is attained on  n M , it follows from the Maximum Principle that S is constant and we have the same conclusions as in Theorem 1.11.

Proof of Theorem 1.12. In the proof of Theorem 1.10 we used the following inequality

 ∑ hα hα R + ∑ h αhα R = ij km mijk ij mi mkjk α,i,j,k,m ∑ ( α,i,j,k,m ) ∑ n ∣ Φ ∣2 - nH tr hn+1(h α)2 + (tr(hαhβ))2+ α α,β 1∑ α β β α 2 N (h h - h h ) ≥ α,β ( ) 2 ∣ Φ ∣2 n(n - 2) 2 ∣ Φ ∣ ------- ∘----------H ∣ Φ ∣ +n(1 - H ) . p n(n - 1)
(4.5)

Applying the same arguments as in the proof of the inequality (4.3), we obtain

 ∑ α α ∑ α α hijh kmRmijk + hijhmiRmkjk ≤ α,i,j,k,m ∑ α,i,j,k,m n supK N (Φ α) = n sup K ∣ Φ ∣2 . α
(4.6)

For technical reasons, we will write the expression (4.1) for the Laplacian of S as

 ( ∑ ∑ ) 1-Δ ∣ Φ ∣2≥ (1 - a) h αhα R + hαh α R 2 ij km mijk ij mi mkjk ( α,i,j,k,m α,i,j,k),m ∑ α α ∑ α α +a hijhkmRmijk + hijhmiRmkjk . α,i,j,k,m α,i,j,k,m
(4.7)

Thus, from (4.5), (4.6) and (4.7), if a ≥ 1 , we have

 ( 1 ∣ Φ ∣2 n(n - 2) -Δ ∣ Φ ∣2 ≥ a ∣ Φ ∣2 ------ ∘----------H ∣ Φ ∣ 2 (p ) n(n - 1)) 2 1 - a +n[1 - H + ------ sup K] . a
(4.8)

Using similar arguments as in [14], it is possible to show that ∣ Φ ∣2< ∞ . Therefore, we can apply Lemma 2.1 to the function ∣ Φ ∣2 and obtain a sequence of points {pk} in M n such that

lim ∣ Φ ∣2 (p ) = sup ∣ Φ ∣2= (sup ∣ Φ ∣)2, k→∞ k lim ∣ ∇ ∣ Φ ∣2 (pk) ∣= 0 and lim sup Δ ∣ Φ ∣2 (pk) ≤ 0. k→∞ k→∞
(4.9)

By applying inequality (4.8) at pk , taking the limit, and using (4.9) we get

 ( -1- 2 2 sup-∣ Φ-∣2 0 ≥ 2a limk→su∞p Δ ∣ Φ ∣ ≥ (sup ∣ Φ ∣) p ( ) ) - ∘n(n---2)--H sup ∣ Φ ∣ + n[1 - H2 + 1---a- supK] . n (n - 1) a
(4.10)

If  a ( 2 2) sup K ≤ β(n, p,H) = --------------- 4(n - 1) - [p(n - 2) + 4(n - 1)]H 4(a - 1)(n - 1) , it can be easily checked that

( 2 ( ) ) (sup-∣ Φ-∣) - ∘n(n----2)-H sup∣ Φ ∣ + n[1 - H2 + 1 --a- sup K] ≥ 0, p n (n - 1) a

and the equality holds if and only if sup K = β(n, p,H) and  --pn(n---2)-- sup ∣ Φ ∣= 2 ∘n-(n----1) .

Thus, if sup K < β(n, p,H) , from (4.10) and the last inequality we conclude that sup ∣ Φ ∣= 0 and  n M is totally umbilical.

If sup K = β(n, p,H) , we will suppose that M n is not totally umbilical and derive a contradiction. First, let us prove that p = 1 . Notice that

 ( ) (sup ∣ Φ ∣)2 n(n - 2) (1 - a) (sup ∣ Φ ∣)2 ------------ ∘----------H sup ∣ Φ ∣ + n[1 - H2 + ------ sup K] = 0. p n (n - 1) a

It shows that all the estimates used to obtain inequality (4.10) turn into equalities. More precisely, (3.6) and (3.7) can now be written as

∘ --------- N (Φn+1) ∣ Φ ∣2= ∣ Φ ∣3 .
(4.11)
 ∑ ∣ Φ ∣4= p N 2(Φ α). α
(4.12)

As mentioned before, taking subsequences if necessary, we can arrive to a sequence {pk} in  n M , which satisfies (4.9) and such that

lim N (Φ α)(p ) = Cα, α ≥ n + 1. k→∞ k
(4.13)

By evaluating (4.11) at pk , taking the limit for k → ∞ and using (4.13) it gives

√ ----- Cn+1 (sup ∣ Φ ∣)2 = sup ∣ Φ ∣3= (sup ∣ Φ ∣)3 ,
(4.14)

Since sup ∣ Φ ∣> 0, we have

 n+1 2 2 ∑ α C = (sup ∣ Φ ∣) = sup(∣ Φ ∣) = C . α
(4.15)

Hence, C α = 0, ∀α ≥ n + 2. By evaluating (4.12) at pk and taking the limit for k → ∞ , from (4.13) and (4.15), we get

 ∑ (sup ∣ Φ ∣)4 = p (C α)2 = p(Cn+1)2 = p(sup ∣ Φ ∣)4, α

which implies p = 1 .

Next, let us prove that sup K = 0. Since h is parallel and the equality holds in (4.6) and (4.7), we arrive to

0 = lim sup 1-Δ ∣ Φ ∣2 (pk) = lim sup 1-ΔS(pk) = n(sup K) sup ∣ Φ ∣2= n(sup K)(sup ∣ Φ ∣)2. k→ ∞ 2 k→ ∞ 2

Therefore, sup K = 0 .

Now we are in position to prove that  n M is totally umbilical. Observe that sup K = 0 and p = 1 yield

 ( ) 0 = sup K = β(n,1, H) = -------a------- 4(n - 1) - n2H2 . 4(a - 1)(n - 1)

Hence  4(n-1) H2 = --n2-- . In this case, according to Montiel (cf. [16], Proposition 2), either M n is a totally umbilical hypersurface or n > 2 and the supremum of the scalar curvature of M n is equal to (n - 2)2 .

As M n is not totally umbilical, we conclude that the supremum of the scalar curvature of  n M is equal to  2 (n - 2) , which contradicts the fact that sup K = 0 . Therefore,  n M is totally umbilical.

Because a is arbitrary, taking the limit for a → ∞ in  ( ) supK ≤ β(n, p,H) = -------a------- 4(n - 1) - [p(n - 2)2 + 4(n - 1)]H2 4(a - 1)(n - 1) , we get  ----1----( 2 2) supK ≤ β(n, p,H) = 4(n - 1) 4(n - 1) - [p(n - 2) + 4(n - 1)]H .

Moreover, since M n is totally umbilical, if n ≥ 3 we obtain
 1 ( ) 1 - H2 = sup K ≤ --------- 4(n - 1) - [p(n - 2)2 + 4(n - 1)]H2 , 4(n - 1) thus
 2 p(n - 2)H ≤ 0 , which implies H = 0 and shows that  n M is totally geodesic. □

Acknowledgements. The authors would like to express their thanks to Fernanda Ester C. Camargo for valuable comments and suggestions about this paper, as well as to the referee for his careful reading of the original manuscript. This work was carried out while the second author was visiting the Institute of Mathematics and Statistics at the University of São Paulo (Brazil). He would like to thank Professor Claudio Gorodski and Professor Paolo Piccione for the warm hospitality and financial support, during his visit.

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Rosa Maria S. Barreiro Chaves
Instituto de Matemática e Estatística
Universidade de São Paulo, Rua do Matão, 1010,
São Paulo - SP, Brazil, CEP 05508-090
rosab@ime.usp.br

Luiz Amancio M. Sousa Jr.
Departamento de Matemática e Estatística
Universidade Federal do Estado do Rio de Janeiro, Avenida Pasteur, 458,
Urca, Rio de Janeiro - RJ, Brazil, CEP 22290-240
amancio@impa.br

Recibido: 17 de noviembre de 2005
Aceptado: 22 de septiembre de 2006