SUBMANIFOLDS WITH CONSTANT SCALAR CURVATURE*

Qing-Ming Cheng

Department of Mathematics, Faculty of Science and Engineering, Saga University, Saga 840-8502, Japan, cheng@ms.saga-u.ac.jp

Abstract. In this paper, we study n-dimensional complete submanifolds with constant scalar curvature in the Euclidean space E^{n+p} and n-dimensional compact submanifolds with constant scalar curvature in the unit sphere S^{n+p}(1). We prove that the totally umbilical sphere S^{n}(r), totally geodesic Euclidean space E^{n} and generalized cylinder S^{n−1}(c)×E^{1} are the only n-dimensional (n > 2) complete submanifolds M^{n} with constant scalar curvature n(n − 1)r in the Euclidean space E^{n+p}, which satisfy the following condition:

S ≤ \frac{n(n − 1)r}{n − 2} .

Where S denotes the squared norm of the second fundamental form of M^{n} . For compact submanifolds with constant scalar curvature in the unit sphere S^{n+p}(1), we also obtain a corresponding result (see Theorem 1.3).

1. Introduction A classical theorem of Hilbert says that a complete surface with the non-zero constant curvature in the 3-dimensional Euclidean space E^{3} must be isometric to the sphere. On the other hand, Hartman and Nirenberg [7] proved that a complete surface with zero curvature in the 3-dimensional Euclidean space E^{3} must be isometric to the plane or the cylinder. Hence, these two results give a complete classification of complete surfaces with constant curvature in the 3-dimensional Euclidean space E^{3}. As a generalization of these theorems, S.Y. Cheng and Yau [5] studied ndimensional complete hypersurfaces with constant scalar curvature in the Euclidean space E^{n+1} and in the unit sphere S^{n+1}(1). They proved the following:

Theorem CY (S. Y. Cheng and Yau [5]). An n-dimensional (n > 2) complete hypersurface M^{n} with constant scalar curvature in E^{n+1} is isometric to a sphere, a hyperplane or a generalized cylinder S^{k}(c) × E^{n−k}, 1 ≤ k ≤ n − 1, if the sectional curvature of M^{n} is nonnegative.

When the ambient space is the unit sphere S^{n+1}(1), they obtained that an n-dimensional compact hypersurface M^{n} with constant scalar curvature n(n − 1)r satisfying r ≥ 1 in the unit sphere S^{n+1}(1) is isometric to a sphere, or a Riemannian product S^{k}(c_{1}) × S^{n−k}(c_{2}), 1 ≤ k ≤ n − 1, if the sectional curvature of M^{n} is nonnegative. In [9], Li extended the results due to S.Y. Cheng and Yau [5]. He proved the following:

Key words and phrases. submanifolds, totally umblical submanifold, generalized cylinder, constant scalar curvature and second fundamental form.

*Research partially Supported by the Grant-in-Aid for Scientific Research from the Ministry of Education, Science, Sports and Culture, Japan

2000 Mathematics Subject Classification 53C42

Theorem L (Li [9]). An n-dimensional (n > 2) compact hypersurface M with constant scalar curvature n(n−1)r satisfying r ≥ 1 in the unit sphere S^{n+1}(1) is isometric to the totally umbilical sphere S^{n}(r), or the Riemannian product S^{1}(\sqrt{1 − c^{2}}) ×S^{n−1}(c) with c = \frac{n−2}{nr} if

S ≤ (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}

holds. Where S denotes the squared norm of the second fundamental form of M^{n}.

In [2] and [3], the author classified complete hypersurfaces with constant scalar curvature and with two distinct principal curvatures in the Euclidean space E^{n+1} and in the unit sphere S^{n+1}(1).

On the other hand, it is natural and very important to study n-dimensional submanifolds with constant scalar curvature and higher codimension p in the Euclidean space E^{n+p} and in the unit sphere S^{n+p}(1). But there are few results about it. In this paper, we shall propose to study complete submanifolds with constant scalar curvature in the Euclidean space E^{n+p} and in the unit sphere S^{n+p}(1). We shall prove the following:

Theorem 1.1. Let M^{n} be an n-dimensional (n > 2) complete submanifold with constant scalar curvature n(n − 1)r in the Euclidean space E^{n+p}. If

S ≤ \frac{n(n − 1)r}{n − 2}

is satisfied, then M is isometric to the totally umbilical sphere S^{n}(r), the totally geodesic Euclidean space E^{n} or the generalized cylinder S^{n−1}(c) × E^{1}. Where S denotes the squared norm of the second fundamental form of M^{n}.

Remark 1.2. Let M^{n} be an n-dimensional submanifold in a space form of constant curvature c. From Gauss equation (see the identity (2.12) in the section 2), we know that the mean curvature H and the squared norm S of the second fundamental form satisfy the following relation:

n(n − 1)r = n(n − 1)c + n^{2}H^{2} − S.

In particular, when c = 0, we know the following three conditions are equivalent with each other.

(1) S ≤ \frac{n(n−1)r}{n−2} ,

(2) S ≤ \frac{n^{2}H^{2}}{n−1} ,

(3) n^{2}H^{2} ≤ \frac{n(n−1)^{2}r}{n−2} .

Thus, if you replace the inequality in the Theorem 1.1 with one of the above inequalities (2) and (3), then the Theorem 1.1 holds also.

Theorem 1.3. Let M^{n} be an n-dimensional (n > 2) compact submanifold with constant scalar curvature n(n − 1)r satisfying r > 1 in the unit sphere S^{n+p}(1). If

S ≤ α(n, r),

then M^{n} is isometric to the totally umbilical sphere S^{n}(r) or the Riemannian product S^{1}(\sqrt{1 − c^{2}})×S^{n−1}(c) with c = \frac{n−2}{nr}, where S is the squared norm of the second fundamental form of M^{n} and

α(n, r) = \{ \begin{matrix}(n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2}, & for p ≤ 2, \\ n(r − 1) + \frac{3n−5+(n^{2}−n−1)(r−1)}{2n^{2}r^{2}} \\ −^{\frac{(n−2)\sqrt{4+2(3n−1)(r−1)+(n^{2}+2n−2)(r−1)^{2}}}{2n^{2}r^{2}}} , & for p ≥ 3. \end{matrix}

2. Preliminaries Let M^{n+p}(c) be an (n+p)-dimensional connected space form of constant curvature c and M^{n} an n-dimensional connected submanifold in M^{n+p}(c). We choose a local field of orthonormal frames {e_{1}, ⋯, e_{n+p}} adapted to the Riemannian metric of M^{n+p}(c) and the dual coframes {ω_{1}, ⋯, ω_{n+p}} in such a way that, restricted to the submanifold M^{n}, {e_{1}, ⋯, e_{n}} are tangent to M^{n}. Then the connection forms {ω_{AB}} of M^{n+p}(c) are characterized by the structure equations

(2.1)  \{ \begin{matrix}dω_{A} = − ∑_{B=1}^{n+p} ω_{AB} ∧ ω_{B}, ω_{AB} + ω_{BA} = 0, \\ dω_{AB} = − ∑_{C=1}^{n+p} ω_{AC} ∧ ω_{CB} + \frac{1}{2} ∑_{C,D=1}^{n+p} K_{ABCD}ω_{C} ∧ ω_{D}, \end{matrix}

K_{ABCD} = c(δ_{AC}δ_{BD} − δ_{AD}δ_{BC}),

where K_{ABCD} denotes the components of curvature tensor of M^{n+p}(c).

The canonic forms {ω_{A}} and connection forms {ω_{AB}} restricted to M^{n} are also denoted by the same symbols. We then have

(2.2)  ω_{α} = 0, α = n + 1, ⋯, n + p.

We see that e_{1}, ⋯, e_{n} is a local field of orthonormal frames adapted to the induced Riemannian metric on M^{n} and ω_{1}, ⋯, ω_{n} is a local field of its dual coframes on M^{n}. It follows from (2.1), (2.2) and Cartan’s Lemma that

(2.3)  ω_{αi} = ∑_{j=1}^{n} h_{ij}^{α}ω_{j}, h_{ij}^{α} = h_{ji}^{α}.

The second fundamental form II and the mean curvature vector h of M^{n} are defined by

(2.4)  II = ∑_{α=n+1}^{n+p} ∑_{i,j=1}^{n} h_{ij}^{α}ω_{i}ω_{j}e_{α},

(2.5)  h = \frac{1}{n} ∑_{α=n+1}^{n+p}(∑_{i=1}^{n} h_{ii}^{α})e_{α}.

The mean curvature H of M^{n} is defined by

(2.7)  H = \frac{1}{n}\sqrt{∑_{α=n+1}^{n+p}(∑_{i=1}^{n} h_{ii}^{α})^{2}}.

Let S = ∑_{α=n+1}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} denote the squared norm of the second fundamental form of M^{n}. The connection form of M^{n} are characterized by the structure equations

(2.8)  dω_{i} = − ∑_{j=1}^{n} ω_{ij} ∧ ω_{j}, ω_{ij} + ω_{ji} = 0,

(2.9)  dω_{ij} = − ∑_{k=1}^{n} ω_{ik} ∧ ω_{kj} + \frac{1}{2} ∑_{k,l=1}^{n} R_{ijkl}ω_{k} ∧ ω_{l},

(2.10)  R_{ijkl} = c(δ_{ik}δ_{jl} − δ_{il}δ_{jk}) + ∑_{α=n+1}^{n+p}(h_{ik}^{α}h_{jl}^{α} − h_{il}^{α}h_{jk}^{α}).

where R_{ijkl} is the components of the curvature tensor of M^{n}. Letting R_{ij} and n(n − 1)r denote the components of the Ricci curvature and the scalar curvature of M^{n}, respectively, we have from (2.10):

R_{jk} = (n − 1)cδ_{jk} + ∑_{α=n+1}^{n+p}(∑_{i=1}^{n} h_{ii}^{α}h_{jk}^{α} − ∑_{i=1}^{n} h_{ik}^{α}h_{ji}^{α}), (2.11)

n(n − 1)r = n(n − 1)c + n^{2}H^{2} − S. (2.12)

We also have

(2.13)  dω_{αβ} = − ∑_{γ=n+1}^{n+p} ω_{αγ} ∧ ω_{γβ} + \frac{1}{2} ∑_{i,j=1}^{n} R_{αβij}ω_{i} ∧ ω_{j},

(2.14)  R_{αβij} = ∑_{l=1}^{n}(h_{il}^{α}h_{lj}^{β} − h_{jl}^{α}h_{li}^{β}).

By taking the exterior differentiation of (2.3) and defining h_{ijk}^{α} by

(2.15)  ∑_{k=1}^{n} h_{ijk}^{α}ω_{k} = dh_{ij}^{α} − ∑_{k=1}^{n} h_{ik}^{α}ω_{kj} − ∑_{k=1}^{n} h_{jk}^{α}ω_{ki} − ∑_{β=n+1}^{n+p} h_{ij}^{β}ω_{βα},

we obtain the Codazzi equation by straightforward computations

(2.16)  h_{ijk}^{α} = h_{ikj}^{α} = h_{jik}^{α}.

We take the exterior differentiation of (2.15) and define h_{ijkl}^{α} by

(2.17) ∑_{l=1}^{n} h_{ijkl}^{α}ω_{l} = dh_{ijk}^{α} − ∑_{l=1}^{n} h_{ljk}^{α}ω_{li} − ∑_{l=1}^{n} h_{ilk}^{α}ω_{lj} − ∑_{l=1}^{n} h_{ijl}^{α}ω_{lk} − ∑_{β=n+1}^{n+p} h_{ijk}^{β}ω_{βα}.

Then the Ricci formula for the second fundamental form is given by

(2.18)  h_{ijkl}^{α} − h_{ijlk}^{α} = ∑_{m=1}^{n} h_{mj}^{α}R_{mikl} + ∑_{m=1}^{n} h_{im}^{α}R_{mjkl} + ∑_{β=n+1}^{n+p} h_{ij}^{β}R_{βαkl}.

The Laplacian Δh_{ij}^{α} of h_{ij}^{α} is defined by

Δh_{ij}^{α} = ∑_{k=1}^{n} h_{ijkk}^{α}.

From the Codazzi equation (2.16) and the Ricci formula (2.18), we obtain for any α, n + 1 ≤ α ≤ n + p,

Δh_{ij}^{α} = ∑_{k=1}^{n} h_{kijk}^{α} (2.19)

= ∑_{k=1}^{n} h_{kikj}^{α} + ∑_{k,m=1}^{n} h_{km}^{α}R_{mijk}

+ ∑_{k,m=1}^{n} h_{mi}^{α}R_{mkjk} + ∑_{k=1}^{n} ∑_{β=n+1}^{n+p} h_{ki}^{β}R_{βαjk}

= ∑_{k=1}^{n} h_{kkij}^{α} + ∑_{k,m=1}^{n} h_{km}^{α}R_{mijk}

+ ∑_{k,m=1}^{n} h_{mi}^{α}R_{mkjk} + ∑_{k=1}^{n} ∑_{β=n+1}^{n+p} h_{ki}^{β}R_{βαjk}.

The following Generalized Maximum Principle due to Omori [10] and Yau [12] will be used in order to prove our theorems.

Generalized Maximum Principle ( Omori [10] and Yau [12]). Let M^{n} be a complete Riemannian manifold whose Ricci curvature is bounded from below and f ∈ C^{2}(M) a function bounded from above on M^{n}. Then for any ϵ > 0, there exists a point p ∈ M^{n} such that

f(p) ≥ sup f − ϵ, ∥gradf∥(p) < ϵ, Δf(p) < ϵ.

3. Proofs of Theorems In the case of the mean curvature vector h ≠ 0 on M^{n}, we know that e_{n+1} = \frac{h}{H} is a normal vecor field defined globally on M^{n}. We define S_{1} and S_{2} by

(3.1)  S_{1} := ∑_{i,j=1}^{n}(h_{ij}^{n+1} − Hδ_{ij})^{2}, S_{2} := ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2},

respectively. Then, S_{1} and S_{2} are functions defined on M^{n} globally, which do not depend on the choice of the orthonormal frame {e_{1}, ⋯, e_{n}}. And

(3.2)  S − nH^{2} = S_{1} + S_{2}.

From the definition of the mean curvature vector h, we know nH = ∑_{i=1}^{n} h_{ii}^{n+1} and ∑_{i=1}^{n} h_{ii}^{α} = 0 for n + 2 ≤ α ≤ n + p on M^{n}.

Proposition 3.1. Let M^{n} be an n-dimensional submanifold with the mean curvature vector h ≠ 0 in a space form M^{n+p}(c). Then we have

\frac{1}{2}ΔS_{2} = ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + nc ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} (3.3)

+ nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2}) − ∑_{α=n+2}^{n+p}[trace(H_{n+1}H_{α})]^{2}

− ∑_{α,β=n+2}^{n+p} N(H_{α}H_{β} − H_{β}H_{α}) − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2}

+ ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α})^{2} − ∑_{α=n+2}^{n+p} trace(H_{n+1}^{2}H_{α}^{2}).

Proof. Since e_{n+1} = \frac{h}{H}, we have

nH = ∑_{i=1}^{n} h_{ii}^{n+1} and ∑_{i=1}^{n} h_{ii}^{α} = 0 for n + 2 ≤ α ≤ n + p

on M^{n}. From Gauss equation (2.10), we have

∑_{α=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{kl}^{α}R_{lijk} = c ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2}

+ ∑_{α,β=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{kl}^{α}(h_{lj}^{β}h_{ik}^{β} − h_{lk}^{β}h_{ij}^{β})

+ ∑_{α=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{kl}^{α}(h_{lj}^{n+1}h_{ik}^{n+1} − h_{lk}^{n+1}h_{ij}^{n+1})

= c ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} + ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α})^{2}

− ∑_{α=n+2}^{n+p}[trace(H_{n+1}H_{α})]^{2}

+ ∑_{α,β=n+2}^{n+p} trace(H_{α}H_{β})^{2} − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2},

∑_{α=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{li}^{α}R_{lkjk}

= −c ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} + nc ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2}

+ ∑_{α,β=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{li}^{α}(h_{lj}^{β}h_{kk}^{β} − h_{lk}^{β}h_{kj}^{β})

+ nH ∑_{α=n+2}^{n+p} ∑_{i,j,l=1}^{n} h_{ij}^{α}h_{li}^{α}h_{lj}^{n+1} − ∑_{α=n+2}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{li}^{α}h_{lk}^{n+1}h_{kj}^{n+1}

= (n − 1)c ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} + nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2})

− ∑_{α=n+2}^{n+p} trace(H_{n+1}^{2}H_{α}^{2}) − ∑_{α,β=n+2}^{n+p} trace(H_{α}H_{β}H_{β}H_{α}),

and

∑_{α,β=n+1}^{n+p} ∑_{i,j,k=1}^{n} h_{ij}^{α}h_{ki}^{β}R_{βαjk}

= ∑_{α,β=n+1}^{n+p} ∑_{i,j,k,l=1}^{n} h_{ij}^{α}h_{ki}^{β}(h_{lk}^{α}h_{lj}^{β} − h_{lj}^{α}h_{lk}^{β})

= ∑_{α,β=n+1}^{n+p} trace(H_{α}H_{β})^{2} − ∑_{α,β=n+1}^{n+p} trace(H_{α}H_{β}H_{β}H_{α}).

We conclude from the formula (2.19) in the section 2,

\frac{1}{2}ΔS_{2} = ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n} h_{ij}^{α}Δh_{ij}^{α} (3.4)

= ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + nc ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2}

+ nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2}) − ∑_{α=n+2}^{n+p}[trace(H_{n+1}H_{α})]^{2}

− ∑_{α,β=n+2}^{n+p} N(H_{α}H_{β} − H_{β}H_{α}) − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2}

+ ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α})^{2} − ∑_{α=n+2}^{n+p} trace(H_{n+1}^{2}H_{α}^{2}).

That is, the Proposition 3.1 is true.

In order to prove our Theorem 1.1 and Theorem 1.3, we prepare several Lemmas.

Lemma 3.2. Let a_{1}, ⋯, a_{n} and b_{ij} for i, j = 1, ⋯, n be real numbers satisfying ∑_{i=1}^{n} a_{i} = 0, ∑_{i=1}^{n} b_{ii} = 0, ∑_{i,j=1}^{n} b_{ij}^{2} = b and b_{ij} = b_{ji} for i, j = 1, ⋯, n. Then

(3.5)  −(∑_{i=1}^{n} b_{ii}a_{i})^{2} + ∑_{i,j=1}^{n} b_{ij}^{2}a_{i}a_{j} − ∑_{i,j=1}^{n} b_{ij}^{2}a_{i}^{2} ≥ − ∑_{i=1}^{n} a_{i}^{2}b.

Proof. We consider the function

(3.6)  f(x_{ij}) = −(∑_{i=1}^{n} x_{ii}a_{i})^{2} − \frac{1}{2} ∑_{i,j=1}^{n} x_{ij}^{2}(a_{j} − a_{i})^{2}

subjecting to the constraint conditions

(3.7)  ∑_{i=1}^{n} x_{ii} = 0 and ∑_{i,j=1}^{n} x_{ij}^{2} = b.

By making use of the method of Lagrange multiplier, we shall calculate the minimum of the function f(x_{ij}) with constraint condition (3.7). Let

g = f(x_{ij}) + λ ∑_{i=1}^{n} x_{ii} + μ( ∑_{i,j=1}^{n} x_{ij}^{2} − b),

where λ and μ are the Lagrange multipliers. We have

g = −(∑_{i=1}^{n} a_{i}x_{ii})^{2} − \frac{1}{2} ∑_{i,j=1}^{n} x_{ij}^{2}(a_{j} − a_{i})^{2} + λ ∑_{i=1}^{n} x_{ii} + μ( ∑_{i,j=1}^{n} x_{ij}^{2} − b).

If f attains its minimum f_{0} at some point (x_{ij}), we have

− 2 ∑_{i=1}^{n} a_{i}x_{ii}a_{j} + λ + 2μx_{jj} = 0, for j = 1, ⋯, n, (3.8)

− x_{ij}(a_{j} − a_{i})^{2} + 2μx_{ij} = 0, for i ≠ j. (3.9)

Hence,

−(∑_{i=1}^{n} a_{i}x_{ii})^{2} + μ ∑_{j=1}^{n} x_{jj}^{2} = 0,

−\frac{1}{2} ∑_{i,j=1}^{n} x_{ij}^{2}(a_{j} − a_{i})^{2} + μ ∑_{i,j=1,i≠j}^{n} x_{ij}^{2} = 0.

Thus,

f_{0} = −μb.

From (3.8) and ∑_{i=1}^{n} a_{i} = 0, we have λ = 0 and

(μ − ∑_{j=1}^{n} a_{j}^{2}) ∑_{i=1}^{n} x_{ii}a_{i} = 0,

μ ∑_{j=1}^{n} x_{jj}^{2} − (∑_{i=1}^{n} x_{ii}a_{i})^{2} = 0.

If ∑_{i=1}^{n} x_{ii}a_{i} ≠ 0, we have μ = ∑_{j=1}^{n} a_{j}^{2}. Hence,

f_{0} = −μb = − ∑_{j=1}^{n} a_{j}^{2}b.

If ∑_{i=1}^{n} x_{ii}a_{i} = 0, we have μ ∑_{j=1}^{n} x_{jj}^{2} = 0. μ = 0 yields f_{0} = 0. If μ ≠ 0, we have ∑_{j=1}^{n} x_{jj}^{2} = 0. Hence, b = 0 or there exists i ≠ j such that x_{ij} ≠ 0. From (3.9), we obtain

2μ = (a_{i} − a_{j})^{2} ≤ 2 ∑_{j=1}^{n} a_{j}^{2}.

Therefore,

f_{0} ≥ − ∑_{j=1}^{n} a_{j}^{2}b.

Since ∑_{i=1}^{n} a_{i} = 0, ∑_{i=1}^{n} b_{ii} = 0, ∑_{i,j=1}^{n} b_{ij}^{2} = b and b_{ij} = b_{ji} for i, j = 1, ⋯, n hold, we have

− (∑_{i=1}^{n} b_{ii}a_{i})^{2} + ∑_{i,j=1}^{n} b_{ij}^{2}a_{i}a_{j} − ∑_{i,j=1}^{n} b_{ij}^{2}a_{i}^{2}

= −(∑_{i=1}^{n} b_{ii}a_{i})^{2} − \frac{1}{2} ∑_{i,j=1}^{n} b_{ij}^{2}(a_{j} − a_{i})^{2} ≥ − ∑_{j=1}^{n} a_{j}^{2}b.

Thus, we complete the proof of the Lemma 3.2.

Lemma 3.3. Let b_{i} for i = 1, ⋯, n be real numbers satisfying ∑_{i=1}^{n} b_{i} = 0 and ∑_{i=1}^{n} b_{i}^{2} = B. Then

(3.10)  ∑_{i=1}^{n} b_{i}^{4} − \frac{B^{2}}{n} ≤ \frac{(n − 2)^{2}}{n(n − 1)}B^{2}.

Proof. We consider the function

f(y) = ∑_{i=1}^{n} y_{i}^{4} − \frac{B^{2}}{n}

with constraint conditions ∑_{i=1}^{n} y_{i} = 0 and ∑_{i=1}^{n} y_{i}^{2} = B.

Since ∑_{i=1}^{n} y_{i}^{2} = B, we know that at least one of y_{i}^{2}’s is not less than \frac{B}{n}. We assume y_{n}^{2} ≥ \frac{B}{n}, without loss of generality. From ∑_{i=1}^{n} y_{i} = 0, we have

y_{n}^{2} = (∑_{i=1}^{n−1} y_{i})^{2} ≤ (n − 1) ∑_{i=1}^{n−1} y_{i}^{2} = (n − 1)(B − y_{n}^{2}),

y_{n}^{2} − \frac{B}{2} = ∑_{1≤i<j≤n−1} y_{i}y_{j},

y_{n}^{2} ≤ \frac{(n − 1)B}{n} .

Hence,

f(y) = ∑_{i=1}^{n−1} y_{i}^{4} + y_{n}^{4} − \frac{B^{2}}{n}

= (∑_{i=1}^{n−1} y_{i}^{2})^{2} − 2 ∑_{1≤i<j≤n−1} y_{i}^{2}y_{j}^{2} + y_{n}^{4} − \frac{B^{2}}{n}

≤ (B − y_{n}^{2})^{2} − \frac{4}{(n − 1)(n − 2)}( ∑_{1≤i<j≤n−1} y_{i}y_{j})^{2} + y_{n}^{4} − \frac{B^{2}}{n}

= \frac{2n(n − 3)}{(n − 1)(n − 2)}(y_{n}^{4} − By_{n}^{2}) + (\frac{n − 1}{n} − \frac{1}{(n − 1)(n − 2)})B^{2}.

Since the maximum of the function t^{2} − Bt in the interval [\frac{1}{n}B, \frac{n−1}{n}B] is −\frac{n−1}{n^{2}}B^{2}, we obtain

f(y) ≤ \frac{(n − 2)^{2}}{n(n − 1)}B^{2}.

We finished the proof of the Lemma 3.3.

Lemma 3.4. Let a_{i} and b_{i} for i = 1, ⋯, n be real numbers satisfying ∑_{i=1}^{n} a_{i} = 0 and ∑_{i=1}^{n} a_{i}^{2} = a. Then

(3.11)  ∑_{i=1}^{n} a_{i}b_{i}^{2} ≥ −\sqrt{∑_{i=1}^{n} b_{i}^{4} − \frac{(∑_{i=1}^{n} b_{i}^{2})^{2}}{n}} \sqrt{a}.

Proof. By making use of the method of Lagrange multiplier, we calculate the minimum of the function g(x) = ∑_{i=1}^{n} x_{i}b_{i}^{2} with constraint conditions ∑_{i=1}^{n} x_{i} = 0 and ∑_{i=1}^{n} x_{i}^{2} = a. If the function g(x) attains its minimum g_{0} at some point x, then we have, at x,

b_{i}^{2} + λ + 2μx_{i} = 0, for i = 1, ⋯, n,

where λ and μ are the Lagrange multipliers. Hence, we have

g_{0} = −2μa, λ = −\frac{∑_{i=1}^{n} b_{i}^{2}}{n} ,

∑_{i=1}^{n} b_{i}^{4} − \frac{(∑_{i=1}^{n} b_{i}^{2})^{2}}{n} + 2μg_{0} = 0.

Thus, the Lemma 3.4 is true.

The following Lemma 3.5 can be found in [8].

Lemma 3.5 (see [8]). For symmetric matrices H_{1}, ⋯, H_{p} (p ≥ 2), put S_{αβ} = trace(H_{α}H_{β}), S = ∑_{α=1}^{p} S_{αα} and N(H_{α}) = trace(^{t}H_{α}H_{α}). Then

(3.12)  ∑_{α,β=1}^{p} N(H_{α}H_{β} − H_{β}H_{α}) + ∑_{α,β=1}^{p} S_{αβ}^{2} ≤ \frac{3}{2}S^{2},

and the equality holds if and only if one of the following conditions holds: (1) H_{1} = H_{2} = ⋯ = H_{p} = 0,

(2) only two of the matrices H_{1}, ⋯, H_{p} are different from zero. Moreover, assuming H_{1} ≠ 0, H_{2} ≠ 0 and H_{3} = ⋯ = H_{p} = 0, then S_{11} = S_{22} and there exists an orthogonal matrix T such that

tT H_{1}T = \sqrt{\frac{S_{11}}{2}} ( \begin{matrix}1 & 0 & 0 & ⋯ & 0 \\ 0 & −1 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 \end{matrix} ) ,

tT H_{2}T = \sqrt{\frac{S_{11}}{2}} ( \begin{matrix}0 & 1 & 0 & ⋯ & 0 \\ 1 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 \end{matrix} ) .

Proposition 3.6. Let M^{n} be an n-dimensional submanifold with the mean curvature vector h ≠ 0 in a space form M^{n+p}(c). Then we have

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + ncS_{2} (3.13)

+ (nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − S_{1} − \frac{3}{2}S_{2})S_{2}.

Proof. From the Proposition 3.1, we have

\frac{1}{2}ΔS_{2} = ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + nc ∑_{α=n+2}^{n+p} ∑_{i,j=1}^{n}(h_{ij}^{α})^{2} (3.14)

+ nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2}) − ∑_{α=n+2}^{n+p}[trace(H_{n+1}H_{α})]^{2}

− ∑_{α,β=n+2}^{n+p} N(H_{α}H_{β} − H_{β}H_{α}) − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2}

+ ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α})^{2} − ∑_{α=n+2}^{n+p} trace(H_{n+1}^{2}H_{α}^{2}).

According to the Lemma 3.5 and the definition of S_{2}, we obtain

(3.15)  − ∑_{α,β=n+2}^{n+p} N(H_{α}H_{β} − H_{β}H_{α}) − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2} ≥ −\frac{3}{2}S_{2}^{2}.

Since e_{n+1} = \frac{h}{H}, we have trace(H_{α}) = 0 for α = n+2, ⋯, n+p and trace(H_{n+1}) = nH.

− ∑_{α=n+2}^{n+p}{trace(H_{n+1}H_{α})}^{2}+ ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α})^{2} − ∑_{α=n+2}^{n+p} trace(H_{n+1}^{2}H_{α}^{2})

= ∑_{α=n+2}^{n+p}[−{trace(H_{n+1}H_{α})}^{2} + trace(H_{n+1}H_{α})^{2} − trace(H_{n+1}^{2}H_{α}^{2})]

= ∑_{α=n+2}^{n+p}[−{trace{(H_{n+1} − HI)H_{α}}}^{2}

+ trace{(H_{n+1} − HI)H_{α}}^{2} − trace{(H_{n+1} − HI)^{2}H_{α}^{2}}],

where I denotes the unit matrix.

For a fixed α, n + 2 ≤ α ≤ n + p, we can take a local orthonormal frame field {e_{1}, ⋯, e_{n}} such that h_{ji}^{α} = λ_{i}^{α}δ_{ij}. Thus, we have ∑_{i=1}^{n} λ_{i}^{α} = 0 and traceH_{α}^{2} = ∑_{i=1}^{n}(λ_{i}^{α})^{2}. Let B = H_{n+1} − HI = (b_{ij}). We have b_{ij} = b_{ji} for any i, j = 1, ⋯, n, ∑_{i=1}^{n} b_{ii} = 0 and ∑_{i,j=1}^{n} b_{ij}^{2} = S_{1}.

− [trace{(H_{n+1} − HI)H_{α}}]^{2}

+ trace{(H_{n+1} − HI)H_{α}}^{2} − trace{(H_{n+1} − HI)^{2}H_{α}^{2}}

= −{trace(BH_{α})}^{2} + trace(BH_{α})^{2} − trace(B^{2}H_{α}^{2})

= −(∑_{i=1}^{n} b_{ii}λ_{i}^{α})^{2} + ∑_{i=1}^{n} b_{ij}^{2}λ_{i}^{α}λ_{j}^{α} − ∑_{i=1}^{n} b_{ij}^{2}(λ_{i}^{α})^{2}.

Since λ_{i}^{α} and b_{ij} for i, j = 1, ⋯, n satisfy the conditions in the Lemma 3.2, we obtain

− [trace{(H_{n+1} − HI)H_{α}}]^{2}+trace{(H_{n+1} − HI)H_{α}}^{2}
− trace{(H_{n+1} − HI)^{2}H_{α}^{2}} ≥ −S_{1}traceH_{α}^{2}.

Therefore, we have

∑_{α=n+2}^{n+p}[−[trace{(H_{n+1} − HI)H_{α}}]^{2} (3.16)

+ trace{(H_{n+1} − HI)H_{α}}^{2} − trace{(H_{n+1} − HI)^{2}H_{α}^{2}}]

≥ −S_{1} ∑_{α=n+2}^{n+p} traceH_{α}^{2} = −S_{1}S_{2}.

nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2})

= nH ∑_{α=n+2}^{n+p} trace{(H_{n+1} − HI)H_{α}^{2}} + nH^{2} ∑_{α=n+2}^{n+p} traceH_{α}^{2}

= nH ∑_{α=n+2}^{n+p} trace{(H_{n+1} − HI)H_{α}^{2}} + nH^{2}S_{2}.

By making use of the same assertion as above, we have, for fixed α, n+2 ≤ α ≤ n+p,

trace{(H_{n+1} − HI)H_{α}^{2}} = ∑_{i=1}^{n} b_{ii}(λ_{i}^{α})^{2}.

From the Lemma 3.3 and Lemma 3.4, we obtain

trace\{(H_{n+1} − HI)H_{α}^{2}\} ≥ −\frac{n − 2}{\sqrt{n(n − 1)}}\sqrt{S_{1}}traceH_{α}^{2}.

Hence, we conclude

(3.17)  nH ∑_{α=n+2}^{n+p} trace(H_{n+1}H_{α}^{2}) ≥ nH^{2}S_{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}}S_{2}.

From (3.14), (3.15), (3.16) and (3.17), we have

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + ncS_{2}

+ (nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − S_{1} − \frac{3}{2}S_{2})S_{2}.

We finished the proof of Proposition 3.2.

Proof of Theorem 1.1. From (2.12), we have

n(n − 1)r = n^{2}H^{2} − S.

Hence, as we remarked in the Remark 1.2, these three conditions in the Remark 1.2 are equivalent with each other. Thus we can use any one of them in order to prove our Theorem 1.1. From the condition of the Theorem 1.1, we have that r ≥ 0 on M and r = 0 if and only if M^{n} is totally geodesic because r is constant. Hence, if r = 0, then M^{n} is isometric to the totally geodesic Euclidean space E^{n}. Next, we assume r > 0. From (2.12), we know H > 0 on M^{n}. Thus, the mean curvature vector h ≠ 0 on M^{n}. We know that e_{n+1} = \frac{h}{H} is a normal vector field defined on

13

M^{n} globally. From the Proposition 3.6, we have

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} (3.18)

+ (nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − S_{1} − \frac{3}{2}S_{2})S_{2}

≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2}

+ (nH^{2} − \frac{n(n − 2)}{2(n − 1)}H^{2} − \frac{n − 2}{2}S_{1} − S_{1} − \frac{3}{2}S_{2})S_{2}

= ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2}

+ (nH^{2} − \frac{n(n − 2)}{2(n − 1)}H^{2} + \frac{n^{2}H^{2}}{2} − \frac{n}{2}S + \frac{(n − 3)}{2}S_{2})S_{2}

= ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + \{\frac{n}{2}(\frac{n^{2}H^{2}}{n − 1} − S) + \frac{(n − 3)}{2}S_{2}\}S_{2}

≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + \{\frac{(n − 3)}{2}S_{2}\}S_{2} ≥ 0.

Since S ≤ \frac{n(n−1)r}{n−2} , from (2.11) and the Remark 1.2, we know that the Ricci curvature of M^{n} is bounded from below. By applying the Generalized Maximum Principle due to Omori [10] and Yau [12] in the section 2 to the function S_{2}, we have that there exists a sequence {p_{k}} ⊂ M such that

(3.19)  lim_{k→∞} S_{2}(p_{k}) = sup S_{2} and lim_{k→∞} sup ΔS_{2}(p_{k}) ≤ 0.

Since S ≤ \frac{n(n−1)r}{n−2} , we know that \{h_{ij}^{α}(p_{k})\}, for any i, j = 1, 2, ⋯, n and any α = n + 1, ⋯, n + p, is a bounded sequence. Hence, we can assume lim_{k→∞} h_{ij}^{α}(p_{k}) = h̃_{ij}^{α}, if necassary, we can take a subsequence. From (3.18) and (3.19), by taking limitation of (3.18), we know that all of inequalities is equalities. Hence, sup S_{2} = 0 for n > 3. When n = 3, if sup S_{2} ≠ 0, we know lim_{k→∞}(\frac{n^{2}H^{2}}{n−1} −

S)(p_{k}) = 0 and lim_{k→∞} \sqrt{\frac{n}{n−1}}H(p_{k}) = lim_{k→∞} \sqrt{S_{1}(p_{k})}. Let lim_{k→∞} H(p_{k}) = H̃,

lim_{k→∞} S(p_{k}) = S̃. lim_{k→∞} S_{1}(p_{k}) = S̃_{1}. We have \frac{n^{2}H̃^{2}}{n−1} = S̃, \frac{n}{n−1} H̃^{2} = S̃_{1} and S̃ = sup S_{2} + S̃_{1} + nH̃^{2} = S̃ + sup S_{2}. This is impossible. Hence, we obtain sup S_{2} = 0. That is, S_{2} = 0 on M^{n}. From (3.18), we have

(3.20)  ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} = 0

on M^{n}. From (2.15), we have, for any α ≠ n + 1,

∑_{i,k=1}^{n} h_{iik}^{α}ω_{k} = −nHω_{αn+1}.

Hence, (3.20) yields ω_{αn+1} = 0 for any α. Thus, we know that e_{n+1} is parallel in the normal bundle T^{⊥}(M^{n}) of M^{n}. Then from the Theorem 1 in [11], we know that M^{n} lies in a totally geodesic submanifold E^{n+1} of E^{n+p}. From the condition in the Theorem 1.1 and the Remark 1.2, we can prove the sectional curvature of M^{n} is nonnegative (cf. Chen [1]). Thus, M is a complete hypersurface in E^{n+1} with constant scalar curvature and its sectional curvature is nonnegative. From the Theorem CY due to S.Y. Cheng and S.T. Yau [5] in the introduction, we know that our Theorem 1.1 is true.

Proof of Theorem 1.3. Since r > 1, we have H > 0 from (2.12). Hence, the e_{n+1} = \frac{h}{H} is a normal vector field defined on M^{n} globally.

When p = 1, from the theorem L in the introduction, we know that Theorem 1.3 is true. When p = 2, since

− ∑_{α,β=n+2}^{n+p} N(H_{α}H_{β} − H_{β}H_{α}) − ∑_{α,β=n+2}^{n+p}[trace(H_{α}H_{β})]^{2} = −S_{2}^{2},

from the proof of the Proposition 3.6, we have

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + nS_{2} (3.21)

+ (nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − S_{1} − S_{2})S_{2}

= ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + {n + nH^{2}

− \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S − nH^{2} − S_{2}} − (S − nH^{2})\}S_{2}

≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + {n + nH^{2}

− \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S − nH^{2}} − (S − nH^{2})\}S_{2}.

From (2.12), we have

(3.22)  S − nH^{2} = S − \frac{n(n − 1)(r − 1) + S}{n} = \frac{n − 1}{n}\{S − n(r − 1)\},
nH = \sqrt{n(n − 1)(r − 1) + S}.

Thus, we obtain

n + nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S − nH^{2}} − (S − nH^{2}) (3.23)

= n + n(r − 1) − \frac{n − 2}{n}\{S − n(r − 1)\}

− \frac{n − 2}{n}\sqrt{\{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}}.

Since S ≤ (n − 1)\frac{n(r−1)+2}{n−2} + \frac{n−2}{n(r−1)+2} holds, we have

(3.24)  {n + n(r − 1)}^{2} ≥ (n − 2){n(r − 1) + 2}{S − n(r − 1)}.

Hence,

\{n + n(r − 1) − \frac{n − 2}{n}(S − n(r − 1))\}^{2}

≥ \frac{(n − 2)^{2}}{n^{2}} \{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}.

Therefore, from (3.24), we have

n + n(r − 1) − \frac{n − 2}{n}(S − n(r − 1)) (3.25)

≥ \frac{n − 2}{n}\sqrt{\{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}}.

Thus, (3.21), (3.23) and (3.25) yield

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2}

+ [n + n(r − 1) − \frac{n − 2}{n}\{S − n(r − 1)\}

− \frac{n − 2}{n}\sqrt{\{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}}]S_{2} ≥ 0.

From Stokes formula, we have

(3.26)  ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} = 0,

and S_{2} = 0, or S_{2} = 0 and

S = (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}.

From (2.15), we have,

∑_{i,k=1}^{n} h_{iik}^{n+2}ω_{k} = −nHω_{n+2,n+1}.

Hence, ω_{n+2,n+1} = 0 from (3.26). Thus, e_{n+1} is parallel on the normal bundle T^{⊥}(M^{n}) of M^{n}. From the Theorem 1 in Yau [11], we know that M^{n} is a hypersurface with constant scalar curvature n(n − 1)r in the totally geodesic submanifold S^{n+1}(1) of S^{n+p}(1) and satisfies

S ≤ (n − 1)\frac{n(r − 1) + 2}{n − 2} + \frac{n − 2}{n(r − 1) + 2}.

From the Theorem L in the introduction, we conclude that our Theorem is valid.

Next we consider the case of p ≥ 3, from the Proposition 3.6, we have

\frac{1}{2}ΔS_{2} ≥ ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + nS_{2} (3.27)

+ (nH^{2} − \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − S_{1} − \frac{3}{2}S_{2})S_{2}

= ∑_{α=n+2}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} + {n + nH^{2}

− \sqrt{\frac{n}{n − 1}}(n − 2)H\sqrt{S_{1}} − \frac{3}{2}(S − nH^{2}) + \frac{1}{2}S_{1}\}S_{2}.

Since e_{n+1} = \frac{h}{H} is a normal vector field defined globally on M^{n}. In this case, we consider a differential operator □ acting on function f defined by

□f = ∑_{i,j=1}^{n}(nHδ_{ij} − h_{ij}^{n+1})f_{ij},

where df = ∑_{i=1}^{n} f_{i}ω_{i}, ∑_{ij=1}^{n} f_{ij}ω_{j} = df_{i}+∑_{j=1}^{n} f_{j}ω_{ji}. If M^{n} is compact orientable Riemannian submanifold with constant scalar curvature, we can prove that □ is self-adjoint, that is,

∫_{M^{n}}(□f)gdM^{n} = ∫_{M^{n}} f(□g)dM^{n}

holds for any functions f, g ∈ C^{2}(M^{n}). From (2.12), we have

Δn^{2}H^{2} = ΔS = Δ(traceH_{n+1}^{2}) + ΔS_{2}.

Hence, we have

Δn^{2}H^{2} − ΔS_{2} = Δ(traceH_{n+1}^{2}).

We choose a local orthonormal frame field {e_{1}, ⋯, e_{n}} such that h_{ij}^{n+1} = λ_{i}δ_{ij}.

Then, traceH_{n+1} = nH, traceH_{n+1}^{2} = ∑_{i=1}^{n} λ_{i}^{2}. From (2.19), we conclude

\frac{1}{2}Δ(traceH_{n+1}^{2}) = ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} + ∑_{i,j=1}^{n} h_{ij}^{n+1}Δh_{ij}^{n+1} (3.28)

= ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} + ∑_{i,j,k=1}^{n} h_{ij}^{n+1}h_{kkij}^{n+1}

+ ∑_{i,j,k,m=1}^{n} h_{ij}^{n+1}h_{km}^{n+1}R_{mijk}

+ ∑_{i,j,k,m=1}^{n} h_{ij}^{n+1}h_{mi}^{n+1}R_{mkjk} + ∑_{i,j,k=1}^{n} ∑_{β=n+1}^{n+p} h_{ij}^{n+1}h_{ki}^{β}R_{βn+1jk}

= ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} + ∑_{i,j=1}^{n} h_{ij}^{n+1}(nH)_{ij} + n ∑_{i=1}^{n} λ_{i}^{2} − (∑_{i=1}^{n} λ_{i}^{2})^{2}

− n^{2}H^{2} + nH ∑_{i=1}^{n} λ_{i}^{3} − ∑_{β=n+2}^{n+p}{∑_{i=1}^{n}(λ_{i} − H)h_{ii}^{β}}^{2}

+ ∑_{β=n+2}^{n+p}{ ∑_{i,k=1}^{n}(λ_{i}λ_{k} − λ_{i}^{2})(h_{ik}^{β})^{2}}.

Let μ_{i} = λ_{i} − H for i = 1, ⋯, n. Then, we have ∑_{i=1}^{n} μ_{i} = 0, ∑_{i=1}^{n} μ_{i}^{2} = ∑_{i=1}^{n} λ_{i}^{2} − nH^{2} = traceH_{n+1}^{2} − nH^{2} = S_{1} and

(3.29)  ∑_{i=1}^{n} μ_{i}^{3} = −3HS_{1} − nH^{3} + ∑_{i=1}^{2} λ_{i}^{3},

− ∑_{β=n+2}^{n+p}{∑_{i=1}^{n}(λ_{i} − H)h_{ii}^{β}}^{2} + ∑_{β=n+2}^{n+p}{ ∑_{i,k=1}^{n}(λ_{i}λ_{k} − λ_{i}^{2})(h_{ik}^{β})^{2}}

= ∑_{β=n+2}^{n+p}{−(∑_{i=1}^{n} μ_{i}h_{ii}^{β})^{2} + ∑_{i,k=1}^{n}(μ_{i}μ_{k} − μ_{i}^{2})(h_{ik}^{β})^{2}}.

By applying the Lemma 3.2 to −(∑_{i=1}^{n} μ_{i}h_{ii}^{β})^{2} + ∑_{i,k=1}^{n}(μ_{i}μ_{k} − μ_{i}^{2})(h_{ik}^{β})^{2} for β = n + 2, ⋯, n + p, we have

(3.30) − ∑_{β=n+2}^{n+p}{∑_{i=1}^{n}(λ_{i} − H)h_{ii}^{β}}^{2} + ∑_{β=n+2}^{n+p}{ ∑_{i,k=1}^{n}(λ_{i}λ_{k} − λ_{i}^{2})(h_{ik}^{β})^{2}} ≥ −S_{1}S_{2}.

By making use of the following Lemma 3.7, we have

(3.31)  | ∑_{i=1}^{n} μ_{i}^{3}| ≤ \frac{n − 2}{\sqrt{n(n − 1)}}S1^{\frac{3}{2}} .

Lemma 3.7. Let a_{1}, ⋯, a_{n} be real numbers satisfying ∑_{i=1}^{n} a_{i} = 0 and ∑_{i=1}^{n} a_{i}^{2} = a. Then

| ∑_{i=1}^{n} a_{i}^{3}| ≤ \frac{n − 2}{\sqrt{n(n − 1)}}a^{\frac{3}{2}} ,

and the equality holds if and only if at least n − 1 of the a_{i}’s are equal with each other.

From (3.28), (3.29), (3.30) and (3.33), we have

\frac{1}{2}Δ(traceH_{n+1}^{2}) ≥ ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} + ∑_{i,j=1}^{n} h_{ij}^{n+1}(nH)_{ij} (3.32)

+ ntraceH_{n+1}^{2} − (traceH_{n+1}^{2})^{2} − n^{2}H^{2}

+ nH\{3HS_{1} + nH^{3} − \frac{n − 2}{\sqrt{n(n − 1)}}S1^{\frac{3}{2}} \} − S_{1}S_{2}

= ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} + ∑_{i,j=1}^{n} h_{ij}^{n+1}(nH)_{ij}

+ S_{1}\{n + 2nH^{2} − S − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}S1^{\frac{1}{2}} \}.

Since

□(nH) = ∑_{i,j=1}^{n}(nHδ_{ij} − h_{ij}^{n+1})(nH)_{ij} (3.33)

= \frac{1}{2}Δ(n^{2}H^{2}) − ∥grad(nH)∥^{2} − ∑_{i,j=1}^{n} h_{ij}^{n+1}(nH)_{ij},

we have, from (3.27), (3.32) and (3.33),

□(nH) (3.34)

≥ \frac{1}{2}ΔS_{2} + ∑_{i,j,k=1}^{n}(h_{ijk}^{n+1})^{2} − ∥grad(nH)∥^{2}

+ S_{1}\{n + 2nH^{2} − S − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}S1^{\frac{1}{2}} \}

≥ ∑_{α=n+1}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} − ∥grad(nH)∥^{2}

+ \{n + nH^{2} − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}S1^{\frac{1}{2}} − (S − nH^{2})\}(S_{1} + S_{2})

+ \{−\frac{1}{2}(S − nH^{2}) + \frac{1}{2}S_{1}\}S_{2}.

According to (2.12) and r > 1, we have H^{2} > S and

nH∇_{k}(nH) = ∑_{α=n+1}^{n+p} ∑_{i,j=1}^{n} h_{ij}^{α}h_{ijk}^{α}.

Hence, n^{2}H^{2}∥grad(nH)∥^{2}

= ∑_{k=1}^{n}{ ∑_{α=n+1}^{n+p} ∑_{i,j=1}^{n} h_{ij}^{α}h_{ijk}^{α}}^{2}

≤ S ∑_{α=n+1}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2}.

Thus we have

(3.35)  ∑_{α=n+1}^{n+p} ∑_{i,j,k=1}^{n}(h_{ijk}^{α})^{2} − ∥grad(nH)∥^{2} ≥ 0

because of n^{2}H^{2} > S > 0. From (3.34), (3.35), we have □(nH) (3.36)

≥ \{n + nH^{2} − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}S1^{\frac{1}{2}} − (S − nH^{2})\}(S_{1} + S_{2})

+ \{−\frac{1}{2}(S − nH^{2}) + \frac{1}{2}S_{1}\}S_{2}

≥ \{n + nH^{2} − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}S1^{\frac{1}{2}} − \frac{3}{2}(S − nH^{2})\}(S_{1} + S_{2})

≥ \{n + nH^{2} − \frac{(n − 2)nH}{\sqrt{n(n − 1)}}(S − nH^{2})^{\frac{1}{2}}

− \frac{3}{2}(S − nH^{2})\}(S_{1} + S_{2})

= [n + n(r − 1) − \frac{3n − 5}{2n}\{S − n(r − 1)\}

− \frac{n − 2}{n}\sqrt{\{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}}](S_{1} + S_{2}).

Since

S ≤ n(r − 1) + \frac{3n − 5 + (n^{2} − n − 1)(r − 1)}{2n^{2}r^{2}}

− \frac{(n − 2)\sqrt{4 + 2(3n − 1)(r − 1) + (n^{2} + 2n − 2)(r − 1)^{2}}}{2n^{2}r^{2}}

holds, we have n^{2}r^{2} − {3n − 5 + (n^{2} − n − 1)(r − 1)}{S − n(r − 1)}

+ \frac{(n − 1)(5n − 9)}{4n^{2}}\{S − n(r − 1)\}^{2} ≥ 0,

[nr − \frac{3n − 5}{2n}\{S − n(r − 1)\}]^{2}

≥ \frac{(n − 2)^{2}}{n^{2}} [n^{2}(r − 1)\{S − n(r − 1)\}]\{S − n(r − 1)\},

nr − \frac{3n − 5}{2n}\{S − n(r − 1)\} (3.37)

≥ \frac{n − 2}{n}\sqrt{n^{2}(r − 1)\{S − n(r − 1)\}}\sqrt{\{S − n(r − 1)\}}.

From (3.36) and (3.37), we obtain

□(nH)

≥ [n + n(r − 1) − \frac{3n − 5}{2n}\{S − n(r − 1)\}

− \frac{n − 2}{n}\sqrt{\{n(n − 1)(r − 1) + S\}\{S − n(r − 1)\}}](S_{1} + S_{2}) ≥ 0.

From Stokes formula, we have S_{1} = S_{2} = 0 on M^{n}. Thus we know that M^{n} is a totally umbilical submanifold. Therefore, M^{n} is isometric to the totally umbilical sphere S^{n}(r). We finished the proof of the Theorem 1.3.

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