Closed Geodesics and Volume Growth of Open Manifolds with Sectional Curvature Bounded from Below?

Yi SHIGuanghan LIChuanxi WU

1 Introduction

Let(M,g)be ann-dimensional complete noncompact Riemannian manifold with sectional curvature satisfyingKM≥c,wherec≤0 is a constant.Denote byαn(r,c)the volume of a geodesic ball of radiusrin ann-dimensional space form of constant curvaturec.The relative volume comparison theorem(see[1])implies that the function

is monotone decreasing,whereB(p,r)is the open metric ball with centerpand radiusrinM.It is well known that

where

andωmis the volume of Sm(1).

For anyp∈M,we set

and define

One always has

Notice that 0≤νc(M)≤1,andMis isometric to ann-dimensional space form of constant curvaturecif and only ifνc(M)=1.Moreover,forc=0,ν0(p)is independent of the choice of the base pointp,i.e.,ν0(M)=ν0(p).

Riemannian manifolds with large volume growth,i.e.,νc(M)>0,have been studied extensively in the last two decades,see for examples[3,6,9–10]and the references therein.In this paper,we shall study the relationship between the existence of closed geodesics and volume growth.

It is well known that any compact Riemannian manifold contains at least one closed geodesic(see[5]),but this is not true for an open Riemannian manifold,since there is the following theorem(see[4,9]).

Theorem 1.1If N is a closed minimal k-submanifold of a nonnegatively curved n-manifold M,then

where B(N,r)={x∈M:d(x,N)

As an application of Theorem 1.1,we shall prove the following result which improves Marenich and Toponogov’s theorem(see[7,9]).

Theorem 1.2Let M be an n-dimensional complete noncompact Riemannian manifold with sectional curvature KM≥0.If

for some p∈M and s>n?1,then M is diffeomorphic ton.

Note that ifs=n,then the above theorem is just Marenich and Toponogov’s theorem.

By Theorem 1.2,it is natural to considerThus we set

and define

Note that,the limitμ(p)may be infinity.

Using Wu’s method in[9],we shall obtain the following rigidity theorem.

Theorem 1.3Let M be an n-dimensional complete noncompact Riemannian manifold with sectional curvature KM≥0.If M contains a closed geodesic σ with length L(σ),thenμ(M)≤L(σ),and the equality holds if and only if M is isometric toS1×n?1with flat metric.

Next we discuss a Riemannian manifold with sectional curvatureKM≥?κ2andν?κ2(M)>0,whereκ>0.For the topology of this kind of manifolds,the reader can refer to[10]for more details.Here we prove the following theorem about the nonexistence of closed geodesics for this kind of manifolds.

Theorem 1.4Given ν∈(0,1)and κ>0,let M be a complete Riemannian n(≥2)-manifold with KM≥?κ2,and ν?κ2(M)>ν.Assume thatis thesolution to

Iftanh?1(cosθ0),then M does not contain any closed geodesic γ with length L(γ)<2r0.

In Section 2,we shall recall some fundamental facts and several important lemmas.The main results are proved in Section 3.

2 Lemmas

In this section,we recall some fundamental facts for later use.Throughout this paper,all geodesics are assumed to have a unit speed.

By the first variation formula of arc length,it is easy to get the following lemma.

Lemma 2.1Let N be a smooth compact submanifold of a complete Riemannian manifoldM.Assume thatand that γ:[0,a]→M is a minimal geodesic from N to q.Thenfor anyv∈Tγ(0)N.

Lemma 2.2Let M be an n-dimensional complete noncompact Riemannian manifold.Givens>0,iffor some p∈M,then for any q∈M,

ProofLetd=d(p,q)be the distance betweenpandq.For anyr>d,one checks that

Then we have

Lettingr→∞,we get the conclusion.

Lemma 2.3(cf.[2])Let M be a complete Riemannian manifold with KM≥c.Denote by M2(c)the complete simply connected surface of constant curvature c.Given l1,l2>0,let γ1:[0,l1]→M,γ2:[0,l2]→M be two geodesic segments in M such that γ1(l1)=γ2(0)and∠(?γ′1(l1),γ′2(0))=α.We call such a configuration a hinge and denote it by(γ1,γ2,α).Letbe two geodesic segments such thatand

Let γ1be minimal,and ifthen the following holds

where dcdenotes the distance function in M2(c).

3 Proofs of the Main Results

Proof of Theorem 1.2Here we use a similar method as that of Theorem 3.1 in[8].First by Theorem 1.1,if a nonnegatively curved open Riemanniann-manifoldMcontains a closed geodesicσ,then

for anyp∈σands>n?1.Now by Lemma 2.2,the above limit holds for anyp∈M.

IfMis not diffeomorphic ton,by Cheeger-Gromoll’s soul theorem,the soul ofMis not a point.Then the soul must contain a closed geodesicσ(since any compact Riemannian manifold contains at least one closed geodesic).Because the soul is a totally geodesic submanifold,we have thatσis also a closed geodesic ofM,whichis a contradiction to the assumption.This finishes the proof of Theorem 1.2.

Before proving Theorem 1.3,letσ:S1→M,u→σ(u)be a closed geodesic ofM.The normal space ofσ(u)inMis given by

where?X,Y?=g(X,Y)is the inner product of vectorsXandY,andgis the Riemannian metric ofM.The corresponding normal bundle is

We consider the following map

It is easy to show that,whenis sufficiently small,the tangent mapF?|(σ(u),ξ)is a linear map offull rank.

The closed geodesicσis of course a smooth compact submanifold ofM,so by Lemma 2.1,we know thatFis a surjective map.

Proof of Theorem 1.3Assume thatσ=σ(u),andu∈[0,L(σ)]is the unit speed closed geodesic.Let

and we have

whered(σ,q)is the distance fromqtoσ.For anyu0∈u,we haveB(σ(u0),r)?F(Drσ),and then

We consider the followingn?1 Jacobi fieldsalongγ(t)

and

Then we derive

and

Now

where

anddenotes the distance to the cut points ofσ(u0)along the geodesic

Notice that

so by(3.1)–(3.2)and the Rauch comparison theorem,we get

SinceF?(t)is a positive definite symmetric matrix,we then have

From(3.4)we obtain

and by(3.3)we know

It is clear by(3.5)thatμ(σ(u0))≤L(σ),soμ(M)≤L(σ).

Ifμ(M)=L(σ),then

and

By the Gaussian lemma,we have

Therefore the metric onMis of the following form

which implies thatMis isometric to S1×n?1with flat metric.This completes the proof of Theorem 1.3.

We use a similar method as that of Theorem 2 in[3]to prove Theorem 1.4.

Proof of Theorem 1.4Assume that there is a normal closed geodesicγwith lengthL(γ)=2r1<2r0onM,and letp=γ(0)andq=γ(r1).Denote byγ1the part ofγfromptoqwithandγ2the part ofγfromptoqwithand thenLetbe the set of two unit vectors inTpM.For anylet

So

whereV(θ0)denotes the volume of a geodesic ball of radiusθ0in an(n?1)-unit sphere andωmis again the volume of Sm(1).

For eachu∈SpM,letc(u)denote the distance to the cut points ofpalong the geodesic expp(tu).We claim that for anyu∈Γ(θ0),

In fact,letu∈Γ(θ0)and setr=c(u).Thenσ1(t)=expp(tu),andt∈[0,r]is a minimal geodesic.Without loss of generality,we can takeLetz=expp(ru)andt=d(q,z).Lemma 2.3 applies to the hinge(σ1,γ1,β)to give

Take a minimal geodesicσjoiningqwithz.Sinceand

we can assume thatLemma 2.3 applies to the hinge(σ,γ1,β1)to give

Thus,

Simplifying the above inequality,we get

This proves our claim.

Let

be the volume form in the geodesic spherical coordinates aroundp,where dμp(ξ)is the Riemannian measure onSpMinduced by the Euclidean Lebesgue measure onTpM.SinceKM≥ ?κ2,we get from the Bishop-Gromov comparison theorem(cf.[1])that(see Section 1 for the definition ofThus,for anyr≥R0,we have from(3.6)that

whereB(R0)denotes theR0-ball in the space form of constant curvature?κ2.

Dividing both sides of the above inequality byαn(r,?κ2)and lettingr→∞,we obtain

This is a contradiction,completing the proof of Theorem 1.4.

AcknowledgementThe authors would like to thank the anonymous referees for the helpful comments and valuable suggestions.

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