L2-harmonic 1-forms on Complete Manifolds

ZHU PENG AND ZHOU JIU-RU

(1.School of Mathematics and Physics,Jiangsu University of Technology, Changzhou,Jiangsu,213001)

(2.School of Mathematical Sciences,Yangzhou University,Yangzhou,Jiangsu,225002)

L2-harmonic 1-forms on Complete Manifolds

ZHU PENG AND ZHOU JIU-RU2

(1.School of Mathematics and Physics,Jiangsu University of Technology, Changzhou,Jiangsu,213001)

(2.School of Mathematical Sciences,Yangzhou University,Yangzhou,Jiangsu,225002)

Communicated by Rong Xiao-chun

minimal hypersurface,end,quaternionic manifold,weighted Poincar´e inequality

1 Introduction

In this paper,frstly,we can obtain the following result:

Remark 1.1Theorem 1.1 generalizes Corollary 2.5 in[6]without the restriction of the second fundamental forms.

Secondly,Lam[7]showed that if M4nis a 4n-dimensional complete noncompact quaternionic K¨ahler and the Ricci curvature of M satisfes

for a positive constant δ,where λ1(M)is the lower bound of the spectrum of the Laplacian on M,then

Suppose that M is a 4n-dimensional complete noncompact quaternionic manifold satisfying the weighted Poincar´e inequality with a non-negative weight function ρ(x)and the Ricci curvature satisfes

for a nonnegative continuous function σ(σ0).If ρ(x)=O(),where rp(x)is the distance function from x to some fxed point p and 0＜α＜2,then H1(L2(M))={0}(see [6]).It is interesting to see if a similar theorem holds without the restriction of growth rate of the weight function.The following theorems had been established:

Theorem 1.2Suppose that M is a 4n-dimensional complete noncompact quaternionic manifold satisfying the weighted Poincar´e inequality with a non-negative continuous weight function ρ(x)(ρ(x)is not identically zero).Assume that the Ricci curvature satisfes

Theorem 1.3Suppose that M is a 4n-dimensional complete noncompact quaternionic manifold satisfying the weighted Poincar´e inequality with a non-negative continuous weight function ρ(x).Assume that the Ricci curvature satisfes

then

2 One End Theorem on Hypersurfaces in Rn+1

In this section,we give the proof of Theorem 1.1.

Proof of Theorem 1.1First,a complete minimal hypersurface inRn+1is noncompact. For any point p∈M and any unit tangent vector v belonging to tangent space at p,we can choose an orthonormal frame{e1,e2,···,en}on M at p such that e1=v.Since M is a minimal hypersurface,there has the following inequality:

The Gauss equation implies that

By(2.1)and(2.2),we have

Let ω∈H1(L2(M)).Then h=|ω|satisfes a formula(see[8]):

Integration by parts implies that

for each positive constantϵ1.That is,

By the defnition of minimal δ-stable hypersurfaces,we have that

for each positive constantϵ2.Combining(2.6)with(2.7),we have

where

Thus,(2.8)implies that

Note that

Letting r→+∞,we obtain that h is a constant on M.Since M is a complete noncompact minimal hypersurface inRn+1,it implies that M has infnite volume(see[9]).Thus by (2.11),we have h=0.That is,

Since Mnis a minimal hypersurface ofRn+1(n≥3),each end of M is non-parabolic(see [2])and the number of non-parabolic end of M is bounded from above by dimH1(L2(M))+1 (see[10]).Therefore,M has only one end.

3 Vanishing Theorems on Quaternionic Manifolds

In this section,we give the proofs of Theorems 1.2 and 1.3,respectively.

If M is a quaternionic manifold and ω∈H1(L2(M)),then h=|ω|satisfes a Bochner type formula(see[11]):

Proof of Theorem 1.2Note that RicM(x)≥−αρ(x).Combining with(3.1),we have

That is,

Note that

holds for each positive constantϵ1.Since ρ is weight function,we have

for each positive constantϵ2.By(3.3),(3.4)and(3.5),we get

where

Note that(2.11)holds.Letting r→+∞,we have h is a constant on M.If h is not identically zero,then,by(2.11),the volume of the M is fnite.The weighted Poincar´e inequality implies that

Proof of Theorem 1.3Combining the fact RicM(x)≥−αρ(x)−β with(3.1),we have

That is,

Note that

for each positive constantϵ1.Since ρ is a weight function,we obtain

for each positive constantϵ1.By(3.7),(3.8)and(3.9),we have

where

Note that(2.11)holds.Letting r→+∞,we obtain that

Choosingϵ1,ϵ2→0,we get

It is well known that

for each positive constantϵ3.Substituting(2.9)into(3.14),we get

Letting r→+∞,we have

Letϵ3→0.Then we obtain that

Suppose that there exists ω∈H1(L2(M))such that h is not identically constant.Combining (3.13)and(3.17),we have

which is contradiction with the restriction of λ1(M).Thus,h is constant.By(3.17),we obtain that h is identically zero.Therefore,

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A

1674-5647(2017)01-0001-07

10.13447/j.1674-5647.2017.01.01

Received date:Dec.15,2014.

Foundation item:The NSF(11471145,11371309)of China and Qing Lan Project.

E-mail address:Zhupeng2004@126.com(Zhu P).

2010 MR subject classifcation:53C21,54C42