STUDY ON 2-DIMENSIONAL SUBMANIFOLDS WITH CONSTANT DETERMINANT OF BLASCHKE TENSOR

YU Ying-jia,GUO Zhen

(Yunnan Normal University,Kunming 650500,China)

Abstract:In this paper,we study the rigidity of 2-dimensional submanifolds in S2+p.Let M2 be a 2-dimensional submanifold in the(2+p)-dimensional unit sphere S2+pwithout umbilic points.Four basic invariants of M2under the Moebius transformation group of S2+pare Moebius metric g,Blaschke tensor A,Moebius form Φ and Moebius second fundamental form B.In this paper,by using inequality estimation,we proved the following rigidity theorem:Let x:M2→S2+pbe a 2-dimensional compact submanifold in the(2+p)-dimensional unit sphere S2+pwith vanishing Moebius form Φ and Det A=c(const)＞ 0,if trA ≥ ,then either x(M2)is Moebius equivalent to a minimal submanifold with constant scalar curvature in S2+p,orin.Our results complement the case 2-dimensional submanifolds in document[3].

Keywords:2-dimensional submanifolds;Moebius metric;Moebius form;Moebius second fundamental form;Blaschke tensor

1 Introduction

where Hessijand?are the Hessian-matrix and the gradient with respect to the induced metric I=dx·dx.Let?⊥be normal connection,and theis defined byMoreover,we introduce the trace-free Blaschke tensor

Hu and Li studied the dimension of submanifold is m ≥ 3,the Moebius form Φ =0,and the constant scalar curvature.In this paper,we proved the dimension of submanifold is m=2,the Moebius form Φ =0,and Det A=c(const)＞ 0,we get the following theorem.

Theorem 1.1Let x:M2→S2+pbe a 2-dimensional compact submanifold in the(2+p)-dimensional unit sphere S2+pwith vanishing Moebius form Φ and 0 ＜ Det A=c(const)＜,then

and x(M2)is Moebius equivalent to a minimal submanifold with constant curvature in S2+p;or

2 Preliminaries

In this section,we give the Moebius invariants and review its structural equations for surfaces in S2+p,for details we refer to[2].

Then we have the following.

Theorem 2.1(see[2])Two submanifolds x,:M→Snare Moebius equivalent if and only if there exists T in the Lorentz group O(n+1,1)insuch that

From Theorem 2.1,we know that the 2-form

is a Moebius invariant(see[1]).Let Δ be the Laplace operator with respect to g.Then we have〈ΔY,ΔY〉=1+4K,where K is the sectional curvature of g([1]).By defining

then we have([1])

Let{E1,E2}be a local orthonormal basis for(M2,g)with dual basis{ω1,ω2},write Yi=Ei(Y),then

Let{Eα}be an orthonormal basis of V,where

where the coefficients ωijbelong to the connection form of the Moebius metric g,and we have the symmetries Aij=Aji,Bij=Bji.It is clear that

are Moebius invariants,and

De fine the covariant derivatives of A,B and Φ by([1])

The integrability conditions for the structure equations(2.10)-(2.14)are given by([1])

where R1212and Rαβ12denote the sectional curvature of g and the normal curvature of the normal connection.Set K=R1212.The second covariant derivative of Aijandare defined by([1])

3 Integral Inequality

Let x:M2→S2+pbe a submanifold in S2+pwithout umbilic points,the Moebius metric is g= ρ2dx·dx,and so the canonical lift of x is given by Y= ρ(k,x)([1]).Then along with M2,we can choose a moving frame{Y,N,Y1,Y2,E3,···,E2+p}in R4+p1,and we replace Eαin(2.9)by Eα=(Hαk,eα+Hαx).For the Moebius invariants A,B and Φ appearing in the structure equation(2.11)-(2.14),by calculation,we can get the expression(1.1)for Φ,(1.3)for B and(1.2)should be changed to([3])

Lemma 3.1For any positive constants k＞a＞0,the torus xa,k:Ma,k=S1(a)×S1(b)→S3(k),a2+b2=k2,choose unit frame field{e1}and{e2}in S1(a)and S1(b)respectively,the Moebius invariants components of the torus are as follows:

From(3.7)we see that

The Moebius metric g is given by

and

From(3.1),(3.8),(3.9)and(3.10),we have

where

Then we have

So the conclusion holds.

then we have

Choose a basis{Ei}such that(Aij)is diagonalized,i.e.,

Then we have

Proof

then

From Φ=0,we obtain

Lemma 3.3Let x:M2→S2+pbe a compact submanifold in the unit sphere S2+pwith vanishing Moebius form Φ,we have

ProofLet L:C∞(M)→C∞(M)be L operator([5]),and L(f)is defined by L(f):=(Aij-trAδij)f,ij,from(3.18)we have

then we have

Integrating both sides of the above equation,according to the properties of Δ and L,we get

From Cauchy-Schwarz Inequality,we see

Lemma 3.4Let x:M2→S2+pbe a compact submanifold in the unit sphere S2+pwith vanishing Moebius form Φ,Det A=c ＞ 0,we have

where equality holds if and only if?A=0.

Proof

i.e.,

Take the derivative of the left hand side of the equation,we have

Take the derivative of the right hand side of the equation,we have

thus

Square both ends of the above equation,we have

On the left hand side,we use the Cauchy-Schwarz inequality to get

taking(3.47)in(3.46),we have

Then we have

where the equality holds if and only if?A=0.

From the above lemma,we can get the following theorem:

Lemma 3.5Let x:M2→S2+pbe a compact submanifold in the unit sphere S2+pwith vanishing Moebius form Φ,Det A=c(const)＞ 0,we have the following inequality

4 Integral Inequality

Theorem 4.1Let x:M2→S2+pbe a 2-dimensional compact submanifold in the(2+p)-dimensional unit sphere S2+pwith vanishing Moebius form Φ,and 0 ＜ Det A=c(const)＜,then

and x(M2)is Moebius equivalent to a minimal submanifold with constant curvature in S2+p;or

Which implies

then we get

From Lemma 3.5,we have

i.e,

by use of(2.18),we see

We claim that we can choose the normal frame field{Eα},such that

In fact,we can choose a new orthonormal frame{}in the normal bundle N(M2)such that,and then define a new orthonormal frame{}in the Moebius normal bundle by,whereis the mean curvature vector of M2,thenand with respect to.If{eα},{}are two orthonormal frames in the normal bundle withP,where(σαβ)is an orthogonal matrix,then we have,when α ≥ 4,we have

that means that span{e4,e5,···,e2+p}is totally umbilical in the normal bundle N(M2).From(4.10)we have

where θαβis the Euclidean normal connection of N(M2)([1]).

From(2.19)and?A=0,we have

thus,we have

Since a Riemannian universal coverage is locally equidistant and not general,we can assume that M2is simply connected.From above,TM2has the following decomposition TM2=V1⊕V2,where V1and V2are the 1-dimensional eigenspace of the Blaschke tensor A with eigenvalues a1and a2.

Form(4.17),we can get that the eigenspaces V1and V2are both integrable.We can write M=M1×M2,where M1and M2are both 1-dimensional submanifold.We can define g1=and g2=,then we have

From(2.12)and(4.16),we have

Since M2is compact submanifold,M1and M2are also compact submanifold.

From(4.21)-(4.23),we have a 1-dimensional manifold with the same curvature as(S1(r1),)and(M1,g1),(s1(r2),)and(M2,g2)are also 1-dimensional manifolds with the same curvature.Thus there exist isometries ψ1:(M1,g1) → (S1(r1),)and ψ2:(M2,g2) →(S1(r2),),let ψ =(ψ1,ψ2),then ψ :M2→ S1(r1)× S1(r2)holds the Moebius metric and the Moebius shape operator.(M2,g)and(S1(r1)×S1(r2),)have the same Moebius metric,Moebius second fundamental form,Moebius shape operator,Blaschke tensor,so the x(M2)is Moebius equivalent to

This completes the proof of the main theorem.