CHERN’S CONJECTURE ON MINIMAL HYPERSURFACES

(RUNNING TITLE: CHERN’S CONJECTURE)

Hongcang Yang* Institute of Mathematics, Academia Sinica, 100080 Beijing, China and Qing-Ming Cheng** Departmet of Mathematics, Saga University, Saga 840, Japan Fax:81-952-28-8501, e-mail: cheng@ms.saga-u.ac.jp

Abstract. In this paper we consider Chern’s conjecture on minimal hypersurfaces in the unit sphere. That is, we prove the following: Let M^{n} (n > 3) be a closed minimal hypersurface in the unit sphere S^{n+1}(1) with constant scalar curvature and S denote the squared norm of the second fundamental form of M. If S > n, then

S − n ≥ \frac{1}{3} n.

1. INTRODUCTION

The following conjecture is well known:

Chern’s Conjecture. For n-dimensional closed minimal hypersurfaces in the unit sphere S^{n+1}(1) with constant scalar curvature, the values S of the squared norm of the second fundamental forms should be discrete.

For this conjecture, Simons [4], Chern et al [1] and Lawson [2] proved that the first and the second value are 0 and n, respectively. Yau [8] (or [9]) brought out this conjecture again as an open problem in his well-known problem section. On the third value of S, Peng and Terng [3] proved that if S > n, then

(1-1)  S > n + c(n),

where

(1-2)  c(n) > 1/(12n)

is a positive constant. In addition, when n = 3, they got S ≥ 6. Hence they conjectured that the third value of S should be 2n since there exist E. Cartan’s

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 53C20, 53C42.

Key words and phrases. minimal hypersurface, second fundamental form, scalar curvature.

*Supported in part by Chinese NSF and NSF of USA

**Supported in part by NNSFC, CPSF, FECC, and SFB256 at Univ. of Bonn

isoparametric minimal hypersurfaces in the unit sphere S^{n+1}(1) satisfying S = 2n. The authors [5] and [6] obtained an important improvment and proved

(1-3)  c(n) > \frac{2}{7}n − \frac{9}{14}.

In this paper we shall give a sharper estimate of S which suggests that the conjecture S − n ≥ n when S > n is possibly true. We shall prove the following main theorem:

Theorem 1. Let M^{n} (n > 3) be a closed minimal hypersurface of the unit sphere S^{n+1}(1) with constant scalar curvature and S the squared norm of the second fundamental form. If S > n then

(1-4)  S − n ≥ \frac{1}{3}n.

In the case that ∑_{i,j,k} h_{ij}h_{jk}h_{ki} = ∑_{i} λ_{i}^{3} = const. , we shall also prove following sharper estimate:

Theorem 2. Let M^{n} (n > 3) be a closed minimal hypersurface of the unit sphere S^{n+1}(1) with constant scalar curvature and S the squared norm of the second fundamental form. Assume that ∑_{i,j,k} h_{ij}h_{jk}h_{ki} ≡ ∑_{i} λ_{i}^{3} = const., where II = ∑_{i,j} h_{ij}ω_{i}ω_{j} is the second fundamental form on M^{n}. If S > n, then

(1-5)  S − n ≥ \frac{2}{3}n.

Remark 1. When n = 3, ∑_{i} λ_{i}^{3} = 3K, where K is the Gauss-Kronecker curvature. Hence ∑_{i} λ_{i}^{3} is an interesting geometric quantity. On the other hand, when n = 3, ∑_{i} λ_{i}^{4} is constant if S is constant. For any dimension n, if we assume ∑_{i} λ_{i}^{4} or ∑_{i} λ_{i}^{3} is constant, then Theorem 2 is also true. Thus Theorem 2 is a natural generalization of case n = 3.

2. BASIC ESTIMATES

Throughout this paper, all manifolds are assumed to be smooth and connected without boundary. We work in the smooth category. Let M^{n} be an n-dimensional hypersurface in the unit sphere S^{n+1}(1). we choose a local field of orthonormal frames e_{1}, ⋯, e_{n+1} in S^{n+1}(1) and the dual coframes ω_{1}, ⋯, ω_{n+1} in such a way that e_{1}, ⋯, e_{n} are tangent to M^{n}. Then

ω_{n+1} = 0.

We see that e_{1}, ⋯, e_{n} is a local orthonormal frames on M^{n} and ω_{1}, ⋯, ω_{n} is a local field of its dual coframes on M^{n}. It follows from Cartan’s lemma that

ω_{n+1i} = ∑_{j} h_{ij}ω_{j}, h_{ij} = h_{ji}.

The second fundamental form and mean curvature H of M^{n} are defined by

II = ∑_{i,j} h_{ij}ω_{i}ω_{j}, H = \frac{1}{n} ∑_{i} h_{ii}.

We recall that M^{n} is by definition a minimal hypersurface if the mean curvature of M^{n} is identically zero.

Let F_{i} = ∇_{i}F, F_{ij} = ∇_{j}∇_{i}F, h_{ijk} = ∇_{k}h_{ij} and h_{ijkl} = ∇_{l}∇_{k}h_{ij}, respectively, where ∇_{j} is the covariant differentiation operator. From the structure equations of M^{n}, the Guass equation, the Codazzi equation and the Ricci formulas are given by

R_{ijkl} = (δ_{ik}δ_{jl} − δ_{il}δ_{jk}) + (h_{ik}h_{jl} − h_{il}h_{jk}),

h_{ijk} = h_{ikj},

h_{ijkl} − h_{ijlk} = ∑_{m} h_{im}R_{mjkl} + ∑_{m} h_{mj}R_{mikl}.

We choose a local field of orthonomal frames e_{1}, ⋯, e_{n} on M^{n} such that, at the point we consider,

h_{ij} = \{ \begin{matrix}λ_{i} & if i = j, \\ 0 & if i ≠ j. \end{matrix}

If the squared norm S of the second fundamental form is constant, we have

(2-0)  \{ \begin{matrix}∑_{i} h_{ii} = ∑_{i} λ_{i} = 0, S = ∑_{i,j} h_{ij}^{2} = ∑_{i} λ_{i}^{2}, \\ ∑_{i,j,k} h_{ijk}^{2} = S(S − n), \\ ∑_{p} h_{ijpp} = (n − S)h_{ij}, \\ h_{ijij} − h_{jiji} = (λ_{i} − λ_{j})(1 + λ_{i}λ_{j}). \end{matrix}

We define

(2-1)  u_{ijkl} := \frac{1}{4}(h_{ijkl} + h_{jkli} + h_{klij} + h_{lijk}).

By a direct computation, we get

(2-2)  \{ \begin{matrix}∑_{i,j,k,l} h_{ijkl}^{2} + S(S − n)(2n + 3 − S) + 3(2B − A) = 0, \\ A = ∑_{i,j,k} h_{ijk}^{2}λ_{i}^{2}, B = ∑_{i,j,k} h_{ijk}^{2}λ_{i}λ_{j}, \\ ∑_{i,j,k,l} h_{ijkl}^{2} ≥ ∑_{i,j,k,l} u_{ijkl}^{2} + \frac{3}{2} [S ∑_{i} λ_{i}^{4} − (∑_{i} λ_{i}^{3})^{2} − 2S^{2} + nS] . \end{matrix}

We need to estimate

(2-3)  f ≡ ∑_{i} λ_{i}^{4} − \frac{1}{S} (∑_{i} λ_{i}^{3})^{2} − \frac{S^{2}}{n} ,

or

(2-4)  f_{0} ≡ ∑_{i} λ_{i}^{4} − \frac{1}{S} (∑_{i} λ_{i}^{3})^{2} − 2S + n = f + \frac{(S − n)^{2}}{n} .

The following results in Lemma 2.1 can be found or proved in the same methods as in our early paper [6] by more careful estimates.

Lemma 2.1. For any fixed c ∈ R, there exists a point x_{c} ∈ M^{n} such that at the point x_{c},

(2-5)  \begin{matrix}(S − n) [c (∑_{i} λ_{i}^{3})^{2} − S ∑_{i} λ_{i}^{4}] = \\ = & c (∑_{i,j,k} 2λ_{i}h_{ijk}^{2}) ∑_{i} λ_{i}^{3} − (2A + B)S + 3c ∑_{j} (∑_{i} λ_{i}^{2}h_{iij})^{2} , \end{matrix}

(2-6)  f ≥ \frac{(λ_{1} − λ_{2})^{2}}{λ_{1}^{2} + λ_{2}^{2}} (\frac{S}{n} + λ_{1}λ_{2})^{2} ,

putting

(2-7)  S − n = tS,

(2-8)  A − B ≤ \frac{1}{3}(λ_{1} − λ_{2})^{2}tS^{2}(1 − α)

where the nonnegative constant α satisfies α < 1 and αtS^{2} = ∑_{i} h_{iii}^{2},

(2-9)  ∑_{k} (∑_{i} λ_{i}^{2}h_{iik})^{2} ≤ \frac{1 + 2α}{3}tS^{2}f,

where f = f_{0} − t^{2}S^{2}/n,

(2-10) (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} ≤ [\frac{4}{3}(A + 2B) − \frac{16}{3} ∑_{k} \frac{1}{S + 2λ_{k}^{2}} (∑_{i} λ_{i}^{2}h_{iik})^{2}] tS^{2}.

3. PROOF OF THEOREM 2

In the case that ∑_{i} λ_{i}^{3} = const., we can get a sharper estimate of S − n when S > n. First of all, we give an estimate of ∑_{i,j,k,l} u_{ijkl}^{2} ( see (2-1) ). Define

(3-1)  f_{ij} = ∑_{p} h_{ip}h_{pj} − xh_{ij} − \frac{S}{n} δ_{ij}, i, j = 1, …, n.

where

x = \frac{1}{S} ∑_{i} λ_{i}^{3}, δ_{ij} = \{ \begin{matrix}1, if i = j, \\ 0, if i ≠ j. \end{matrix}

It is easy to see that

(3-2)  \begin{matrix}∑_{i,j} h_{ij}f_{ij} = ∑_{i} λ_{i} (λ_{i}^{2} − xλ_{i} − \frac{S}{n} ) = 0, \\ ∑_{i,j} δ_{ij}f_{ij} = ∑_{i} (λ_{i}^{2} − xλ_{i} − \frac{S}{n} ) = 0, \\ ∑_{i,j} f_{ij}^{2} = ∑_{i} λ_{i}^{2} (λ_{i}^{2} − xλ_{i} − \frac{S}{n} ) \\ = ∑_{i} λ_{i}^{4} − \frac{1}{S} (∑_{i} λ_{i}^{3})^{2} − \frac{S^{2}}{n} = f_{0} − \frac{t^{2}S^{2}}{n} ≡ f. \end{matrix}

Since ∑_{i,j} h_{ij}^{2} = ∑_{i} λ_{i}^{2} = S, ∑_{i,j,k} h_{ijk}^{2} = tS^{2} , we have

(3-3)  ∑_{i,j} h_{ij}h_{ijkl} + ∑_{i,j} h_{ijk}h_{ijl} = 0,  ∑_{i,j,k} h_{ijk}h_{ijkl} = 0.

It follows from (3-3) that

(3-4)  \begin{matrix}∑_{i,j,k,l} h_{ijkl}h_{ij}h_{kl} = ∑_{i,j} λ_{i}h_{iijj}λ_{j} = − ∑_{i,j,k} λ_{i}h_{ijk}^{2}. \\ ∑_{i,j,k,l,p} h_{ijkl}h_{ij}h_{kp}h_{pl} = ∑_{i,j} λ_{i}h_{iijj}λ_{j}^{2} = − ∑_{i,j,k} λ_{i}^{2}h_{ijk}^{2} = −A. \end{matrix}

By (2-0), we also have

(3-5)  \begin{matrix}∑_{i,j} λ_{i}^{2}h_{iijj}λ_{j} = −A + ∑_{i,j} λ_{i}^{2}λ_{j}(λ_{i} − λ_{j})(1 + λ_{i}λ_{j}) \\ = −A + S ∑_{i} λ_{i}^{4} − (∑_{i} λ_{i}^{3})^{2} − S^{2} = −A + Sf + \frac{S^{2}t}{1 − t}. \end{matrix}

And then, by the symmetry of u_{ijkl},

(3-6)  \begin{matrix}∑_{i,j,k,l} u_{ijkl}h_{ij}f_{kl} = −A + \frac{1}{2}Sf + \frac{S^{2}t}{1 − t} + \frac{1}{S} ∑_{i,j,k} λ_{i}h_{ijk}^{2} ∑_{i} λ_{i}^{3} =: F, \\ ∑_{i,j,k} u_{ijkl}h_{ijk} = −\frac{3}{2} ∑_{i} λ_{i}^{2}h_{iil}λ_{l},  ∑_{i,j,k} f_{ij}h_{kl}h_{ijk} = ∑_{i,j} λ_{i}^{2}h_{iij}h_{jl}, \\ ∑_{i,j,k,l} u_{ijkl}h_{ij}δ_{kl} = −\frac{1}{2}S^{2}t. \end{matrix}

Since for any α_{1}, β_{i}, γ, δ ∈ R,

\begin{matrix}0 ≤ ∑_{i,j,k,l}[ & u_{ijkl} − α_{1}(h_{ij}f_{kl} + h_{kl}f_{ij}) + (β_{i}h_{jkl} + β_{j}h_{ikl} + β_{k}h_{lij} + β_{l}h_{ijk}) \\ + γh_{ij}h_{kl} + δ(h_{ij}δ_{kl} + h_{kl}δ_{ij})]^{2}, \end{matrix}

it follows from (3-4), (3-5) and (3-6) that

\begin{matrix}∑_{i,j,k,l} u_{ijkl}^{2} ≥ & 4α_{1}F − 2α_{1}^{2}Sf + 8 ∑_{i,j} α_{1}β_{j}λ_{i}^{2}h_{iij}λ_{j} + 12 ∑_{i,j} λ_{i}^{2}h_{iij}λ_{j}β_{j}− \\ − 16 ∑_{i,j,k,l} β_{l}^{2}h_{ijk}^{2} + 2γ ∑_{i,j,k} λ_{i}h_{ijk}^{2} − γ^{2}S^{2} + 2δS^{2}t − 2δ^{2}Sn. \end{matrix}

Letting

\{ \begin{matrix}γ = \frac{1}{S^{2}} ∑_{i,j,k} λ_{i}h_{ijk}^{2}, δ = \frac{tS}{2n} = \frac{t}{2(1 − t)}, β_{i} = \frac{2α_{1} + 3}{8tS^{2}} ∑_{j} λ_{j}^{2}h_{jji}λ_{i}, \\ α_{1} = \frac{1}{Sf − \frac{1}{2tS^{2}} ∑_{j} λ_{j}^{2}(∑_{i} λ_{i}^{2}h_{iij})^{2}} [F + \frac{3}{4tS^{2}} ∑_{j} λ_{j}^{2} (∑_{i} λ_{i}^{2}h_{iij})^{2}] , \end{matrix}

respectively, we have

(3-7) ∑_{i,j,k,l} u_{ijkl}^{2} ≥ \frac{9}{4tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} + \frac{t^{2}S^{2}}{2(1 − t)}

+\frac{1}{2 [Sf − \frac{1}{2tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2}]} [2F + \frac{3}{2tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2}]^{2} .

We shall use (3-7) to prove Theorem 1 in the section 4.

Now in the case ∑_{i} λ_{i}^{3} ≡ const., (3-7) is reduced to

∑_{i,j,k,l} u_{ijkl}^{2} ≥ \frac{1}{2Sf} [Sf − 2A + t (∑_{i} λ_{i}^{3})^{2} + \frac{2tS^{2}}{1 − t}]^{2} (3-8)

+ \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} + \frac{t^{2}S^{2}}{2(1 − t)}.

We also need the following estimates:

Lemma 3.1. There is a point x_{0} ∈ M such that at x_{0},

(3-9)  \{ \begin{matrix}Sf = A − B − \frac{1}{1 − t}tS^{2}, −tSf ≤ −\frac{1}{3}(2A − 5B) + \frac{1}{1 − t}tS^{2}, \\ 2tSf + t (∑_{i} λ_{i}^{3})^{2} − 2A + \frac{1}{1 − t}2S^{2}t ≥ \frac{2}{3}(A − B), \\ − (3 + t)tSf − \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} \\ ≤ (1 − t)(A − B) − 3(A − 2B) + \frac{3 + t}{1 − t}tS^{2}. \end{matrix}

Proof. Since ∑_{i} λ_{i}^{3} = const., we have t^{2}S^{2} (∑_{i} λ_{i}^{3})^{2} = (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} and there is a point x_{0} ∈ M such that tS ∑_{i} λ_{i}^{4} = 2A + B. Lemma 2.1 gives

(∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} ≤ \frac{4}{3}(A + 2B)tS^{2}.

Then one can easily see that at x_{0}

− tSf − \frac{1}{1 − t}tS^{2} = −tS ∑_{i} λ_{i}^{4} + t (∑_{i} λ_{i}^{3})^{2} ≤ −(2A + B) + \frac{4}{3}(A + 2B);

2tSf + t (∑_{i} λ_{i}^{3})^{2} − 2A + \frac{1}{1 − t}2tS^{2} = 2tS ∑_{i} λ_{i}^{4} − \frac{1}{tS^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} − 2A

≥ 2(2A + B) − 2A − \frac{4}{3}(A + 2B) = \frac{2}{3}(A − B);

− (3 + t)tSf − \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2}

= −(3 + t)tS ∑_{i} λ_{i}^{4} + \frac{3}{tS^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} + \frac{3 + t}{1 − t}tS^{2}

≤ −(3 + t)(2A + B) + 4(A + 2B) + t(A + 2B) + \frac{3 + t}{1 − t}tS^{2}

≤ (1 − t)(A − B) − 3(A − 2B) + \frac{3 + t}{1 − t}tS^{2}.

On the other hand,

A − 2B = ∑_{i,j,k,p,q} h_{ijk}(h_{ij}h_{pk}h_{pq} − h_{ip}h_{jp}h_{kp})_{q}

\begin{matrix}= & − ∑_{i,j,k,p,q}[h_{ijkq}(h_{ij}h_{pk}h_{pq} − h_{ip}h_{jp}h_{kq} − (h_{ijk}h_{ip}h_{jp}h_{kq})_{q}] \\ = & − ∑_{i,j} h_{iijj}(λ_{i}λ_{j}^{2} − λ_{i}^{2}λ_{j}) − \frac{1}{3} ∑_{p,q}(∑_{i} λ_{i}^{3})_{pq}h_{pq} \\ = & S ∑_{i} λ_{i}^{4} − (∑_{i} λ_{i}^{3})^{2} − S^{2} = Sf + \frac{1}{1 − t}tS^{2}. \end{matrix}

Hence we complete the proof of Lemma 3.1. # By Lemma 2.1 , we also have

\begin{matrix}tSf ≥ \frac{1}{3}(2A − 5B) − \frac{1}{1 − t}tS^{2} \\ = (A − 2B) − \frac{1}{3}(A − B) − \frac{t}{1 − t}S^{2} = Sf − \frac{1}{3}(A − B). \end{matrix}

Hence

(3-10)  (1 − t)Sf ≤ \frac{1}{3}(A − B).

Now we can prove Theorem 2. It follows from Lemma 3.1 that

\frac{1}{2Sf} [Sf − 2A + t (∑_{i} λ_{i}^{3})^{2} + \frac{2t}{1 − t}S^{2}]^{2}

≥ \frac{1}{2Sf} [(1 − 2t)Sf + \frac{2}{3}(A − B)]^{2}

= \frac{1}{2}(1 − 2t)^{2}Sf + \frac{2}{3}(1 − 2t)(A − B) + \frac{2}{9Sf} (A − B)^{2}

≥ \frac{1}{2} [(1 − 2t)^{2} − 4(1 − t)^{2}] Sf + (2 − \frac{8}{3}t) (A − B).

Combining it with (2-2), (2-4) and (3-8), we get

2tSf + \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} + (2 − \frac{8}{3}t)(A − B) (3-11)

≤ (2tS − S − 3 − \frac{2t}{1 − t}) tS^{2} + 3(A − 2B).

Adding up (3-10) and (3-11) to (3-9), then using Lemma 2.1, we have

(3-12) \begin{matrix}(1 − 2t − t^{2})Sf ≤ & (2tS − S + \frac{2t}{1 − t}) tS^{2} + (−\frac{2}{3} + \frac{5}{3}t) (A − B) \\ ≤ & (2tS − S + \frac{2t}{1 − t}) tS^{2} − \frac{1}{9}(λ_{1} − λ_{2})^{2}t(2 − 5t)S^{2}. \end{matrix}

We assume that S − n ≤ \frac{2}{3}n, then t ≤ \frac{2}{5}, 1 − 2t − t^{2} ≥ 0. Since (λ_{1} − λ_{2})^{2} ≤ 2S, (3-12) gives

2tS − S + \frac{2t}{1 − t} − \frac{2}{9}(2 − 5t)S ≥ 0.

Therefore,

t ≥ \frac{13}{28} − \frac{18t}{28(1 − t)S} ,

or

S − n ≥ \frac{13}{28}S − \frac{3}{7}. (since t ≤ \frac{2}{5})

Finally,

S − n ≥ \frac{13}{15}n − \frac{4}{5} ≥ \frac{2}{3}n. (when n ≥ 4)

Theorem 2 is proved.

Remark 2. In fact, we have proved that under the conditions given by Theorem 2, if n is large enough, then

(3-13)  S − n ≥ \frac{4}{5}n.

4. PROOF OF THEOREM 1

In the general case, we are not yet able to get an estimate as sharp as (1-5) because of the difficulty which we meet to estimate the term on the right side of (3-7). In this section we shall prove Theorem 1 by making an effort to deal with that term. We begin with the estimate of ∑_{i,j,k,l} u_{ijkl}^{2}.

It follows from (3-6) and (3-7) that

∑_{i,j,k,l} u_{ijkl}^{2} ≥ \frac{9}{4tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} + \frac{1}{2(1 − t)}t^{2}S^{2}

+ \frac{1}{2 [Sf − \frac{1}{2tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2}]} \{[Sf − \frac{1}{2tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2}] +

+ [−2A + \frac{2}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} + \frac{1}{1 − t}tS^{2}]\}^{2}

≥ \frac{1}{2}Sf + \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} − 2A + \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2}

+ \frac{1}{1 − t}t (2 + \frac{1}{2}t) S^{2} + \frac{1}{S^{2}} (∑_{i,j,k} λ_{i}h_{ijk}^{2})^{2} .

Then (2-2) gives

(4-1)  \begin{matrix}2Sf + \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} + \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} \\ + \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} \\ ≤ tS^{2} (2tS − S − 5 − \frac{4t}{1 − t}) + 5A − 6B. \end{matrix}

Now we apply Lemma 2.1 to the case that c = 1 + ^{\frac{22+3t+\sqrt{(22+3t)^{2}−400}}{26}} ≡ 1 + c_{0}. It is easy to see that c_{0} satisfies the following equation:

(4-2)  \frac{1}{3c_{0}} (1 + c_{0} − \frac{3}{13})^{2} = \frac{14 + t}{13}.

By Lemma 2.1, there is a point x_{c} ∈ M such that at x_{c},

− tS ∑_{i} λ_{i}^{4} + t(1 + c_{0}) (∑_{i} λ_{i}^{3})^{2}

= \frac{1 + c_{0}}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} − (2A + B)

+ \frac{3(1 + c_{0})}{S} ∑_{j} (∑_{i} λ_{i}^{2}h_{iij})^{2} .

From now on we only consider the formulas and the estimate at the point x_{c}. We define

G ≡ −\frac{13}{3}tSf − \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} (4-3)

− \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} − \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} .

Then

\begin{matrix}G = −\frac{13}{3}(2A + B) − \frac{13}{3}c_{0}t (∑_{i} λ_{i}^{3})^{2} \\ + \frac{13}{3} (\frac{10}{13} + c_{0}) \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} \end{matrix}

+ \frac{13t}{3(1 − t)}S^{2} + \frac{13(1 + c_{0})}{S} ∑_{j} (∑_{i} λ_{i}^{2}h_{iij})^{2}

− \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} − \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2}

≤ \frac{13}{3t} [\frac{1}{4c_{0}} (c_{0} + \frac{10}{13})^{2} − \frac{3t}{52}] \frac{1}{S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2}

− \frac{13}{3}(2A + B) + \frac{13t}{3(1 − t)}S^{2}

− \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{13(1 + c_{0})}{S} ∑_{i} (∑_{j} λ_{j}^{2}h_{jji})^{2}

= −\frac{13}{3}(2A + B) + \frac{14}{4} \frac{1}{tS^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} + \frac{13t}{3(1 − t)}S^{2}

− \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{13(1 + c_{0})}{S} ∑_{i} (∑_{j} λ_{j}^{2}h_{jji})^{2} .

By Lemma 2.1,

G ≤ −\frac{13}{3}(2A + B) + \frac{14}{3}(A + 2B) + \frac{13t}{3(1 − t)}S^{2}

− ∑_{i} [\frac{14}{3} \frac{4}{S + 2λ_{i}^{2}} + \frac{2(S + 2λ_{i}^{2})}{tS^{2}} − \frac{2}{tS} − \frac{13(1 + c_{0})}{S}] (∑_{j} λ_{j}^{2}h_{jji})^{2}

≤ −4A + 5B − \frac{1}{S} [8\sqrt{\frac{7}{3t}} − \frac{2}{t} − 13(1 + c_{0})] ∑_{i} (∑_{j} λ_{j}^{2}h_{jji})^{2} + \frac{13t}{3(1 − t)}S^{2}

Denote

(4-4)  \begin{matrix}g(t) ≡ \frac{1}{13t}(8\sqrt{21t} − 6) − 3(1 + c_{0}) \\ = \frac{1}{13t}(8\sqrt{21t} − 6) − 3 \{1 + \frac{1}{26} [22 + 3t − \sqrt{(22 + 3t)^{2} − 400}]\} . \end{matrix}

Since \frac{dg(t)}{dt} = 0 has only three real roots: t_{1} = 0.101786, t_{2} = 0.109566 and t_{3} = 1.61358. Hence g(t) is a decreasing function when \frac{1}{9} < t < \frac{1}{4} because \frac{dg(t)}{dt} < 0. Hence g(t) ≥ g(\frac{1}{4}) when t ∈ (\frac{1}{9}, \frac{1}{4}). S − n ≥ \frac{1}{8}n was proved in [6], i.e. t ≥ \frac{1}{9}. On the other hand, if Theorem 1 is not true, we can assume that S − n < \frac{1}{3}n, or t < \frac{1}{4}. Therefore, we can assume that t ∈ [\frac{1}{9}, \frac{1}{4}). Then, by a simple calculation, we have

(4-5)  g(t) ≥ g (\frac{1}{4}) ≥ −\frac{3}{5}.

By Lemma 2.1, combining (4-5) with (4-3) of the expression of G, we have

\begin{matrix}− \frac{13}{3} [1 + \frac{1}{5}(1 + 2α)] tSf − \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} − \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} \\ − \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} ≤ −4A + 5B + \frac{13}{3(1 − t)}tS^{2}, \end{matrix}

or

− \frac{13}{10} (1 + \frac{1}{3}α) Sf − \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2}

− \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} − \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2}

≤ −4A + 5B + \frac{13tS^{2}}{3(1 − t)}.

Then

(4-6)  \begin{matrix}− \frac{13}{10}Sf − \frac{1}{S} ∑_{i} λ_{i}^{3} ∑_{i,j,k} 2λ_{i}h_{ijk}^{2} − \frac{1}{4S^{2}} (∑_{i,j,k} 2λ_{i}h_{ijk}^{2})^{2} \\ − \frac{4}{tS^{2}} ∑_{i} λ_{i}^{2} (∑_{j} λ_{j}^{2}h_{jji})^{2} \\ ≤ \frac{1}{1 + (1/3)α} [−\frac{13}{3}(A − B) + \frac{1}{3}(A + 2B)] + \frac{13tS^{2}}{3(1 − t)}. \end{matrix}

Since A + 2B = \frac{1}{3} ∑_{i,j,k}(λ_{i} + λ_{j} + λ_{k})^{2}h_{ijk}^{2} ≥ 0, adding (4-6) up to (4-1), we get

(4-7) 0 ≤ −\frac{7}{10}Sf + \frac{1}{3} [16 − \frac{13}{1 + (1/3)α}] (A−B)+[2tS − S − \frac{2}{3} + \frac{t}{3(1 − t)}] tS^{2}.

It follows from Lemma 2.1 that

(4-8)  \begin{matrix}\frac{1}{3} [16 − \frac{13}{1 + (1/3)α}] (A − B) ≤ \\ ≤ \frac{1}{9} [16 − \frac{13}{1 + (1/3)α}] (1 − α)(λ_{1} − λ_{2})^{2}tS^{2} \\ = \frac{1}{3} (16 − \frac{13}{y}) (\frac{4}{3} − y) (λ_{1} − λ_{2})^{2}tS^{2} (where y = 1 + (1/3)α) \\ = \frac{1}{3} (\frac{64}{3} + 13 − \frac{52}{3y} − 16y) (λ_{1} − λ_{2})^{2}tS^{2} \\ ≤ \frac{7}{20}(λ_{1} − λ_{2})^{2}tS^{2}. \end{matrix}

Then by Lemma 2.1, (4-7) and (4-8), we have

0 ≤tS^{2} (2tS − S − \frac{2}{3} + \frac{1}{9}) + \frac{7}{20}(λ_{1}^{2} + λ_{2}^{2} − 2λ_{1}λ_{2})tS^{2} − \frac{7}{10} (λ_{1}λ_{2} + \frac{1}{1 − t})^{2} S

≤tS^{2} [2tS − S + \frac{7}{20}S − \frac{5}{9} + \frac{7}{10(1 − t)}] + \frac{7}{10}(zt − z^{2})S^{3}

(where zS = −λ_{1}λ_{2} − \frac{1}{1 − t})

≤tS^{2} (2tS + \frac{7}{40}tS − S + \frac{7}{20}S − \frac{5}{9} + \frac{42}{45}) = tS^{2} (\frac{87}{40}tS − \frac{13}{20}S + \frac{17}{45}) .

Therefore,

t ≥ \frac{26}{87} − \frac{16}{87S} ,

or

S − n ≥ \frac{26}{87}S − \frac{16}{87}.

Finally, we get

S − n ≥ \frac{26}{61}n − \frac{16}{61} ≥ \frac{1}{3}n, (when n ≥ 4)

which is a contradiction to the assumption that S − n < \frac{1}{3}n. Theorem 1 is proved.

Remark 3. In fact, we can prove that when n is large enough, if S − n > 0, then

S − n ≥ \frac{2}{5}n.

Acknowledgement. The first author is indebted to Professor H. Wu for his kindly support and hospitality during his visit at University of California at Berkeley, USA. The second author would like to thank Professor H. Karcher for his support and hospitality during his visit at University of Bonn, Germany.

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