On the Tangent Bundle of a Hypersurface in a Riemannian Manifold?

Zhonghua HOU Lei SUN

1 Introduction

Let(M,g)be a Riemannian manifold,and TM be the tangent bundle of M.Let Gsbe the Sasaki metric on TM introduced by Sasaki[13]in terms of g.

The geometry of(TM,Gs)and the unit tangent sphere bundle(S(TM),Gs)have attracted many mathematicians in the last decades.Kowalski[5]showed that if(TM,Gs)is locally symmetric,then the base metric is flat and so does Gs.Musso and Tricerri[9]proved that(TM,Gs)has constant scalar curvature if and only if(M,g)is flat.Nagano[10],Tachibana and Okumura[15]studied the almost complex structure on(TM,Gs).Nagy[11]studied the geometry of the unit tangent sphere bundle of a surface.Klingenberg and Sasaki[2]showed that(S(TS2(1)),Gs)is isometric to the elliptic space of curvatureNagy[12],Sasaki[14],Konno and Tanno[3–4]studied the geodesics and Killing vector fields on(S(TM),Gs).Tashiro[16–17]studied the contact structure on S(TM).

Deshmukh,Al-Odan and Shaman[1]considered an orientable hypersurface Mnof the Euclidean space Rn+1and observed that the tangent bundle TM of M is an immersed submanifold of the Euclidean space R2n+2.They obtained expressions for the horizontal and vertical lifts of the vector fields on M and showed that the induced metric on TM is not a natural metric in general.In the special case that the induced metric on TM becomes a natural metric,they proved that the tangent bundle TM is trivial.

In this paper,we suppose that Mnis a hypersurface of a Riemannian manifold(Nn+1,G).We use the moving frame method to study the geometry of the tangent bundle TMnwith the induced metric from(TN,Gs).

In Section 3,we study the extrinsic geometry of TMnin(TNn+1(c),Gs)where Nn+1(c)is a space form of constant curvature c.In Section 4,we study the integrability of the almost complex structure J on TM and the K¨ahlerian form ω on TM induced by J.

2 Preliminaries

Suppose that(N,G)is an(n+1)-dimensional Riemannian manifold.Let D be the Levi-Civita connection in N,and π :TN →N be the natural projection.Through out this paper,we use the Einstein convention and the following ranges of indices:

Let({yA})and({yA},{vB})be local coordinate systems in N and TN,respectively.Denote

At first,we introduce the following lemmas.

Lemma 2.1(cf.[5])Let(N,G)be a Riemannian manifold and Y ∈ Γ(TN)be a vector field on N which is locally represented byThen the vertical and horizontal lifts YVand YHof Y over TN are given by

respectively,whereare the Christo ff el symbols of D.

Lemma 2.2(cf.[5])The Lie bracket of vector fields over TN is completely determined by

where X,Y ∈ Γ(TN)andis the Riemannian curvature operator of N.

The Sasaki metric Gson TN can be described as follows.

Definition 2.1(cf.[13])Let(N,G)be a Riemannian manifold.The Sasaki metric Gson TN is defined by

for any point(y,v)∈TN and vectors X,Y∈TyN.

By direct computation,we have the following lemma.

Lemma 2.3(cf.[7])Let(N,G)be a Riemannian manifold,and TN be the tangent bundle with the Sasaki metric Gs.Then the Levi-Civita connectionon(TN,Gs)is determined by

for any point(y,v)∈TN and vectors X,Y∈TyN.

3 Geometry of the Tangent Bundle of a Hypersurface

Suppose that f:Mn→Nn+1is a smooth immersion from Mninto Nn+1.Let

df(x):TxM→Tf(x)N

be the differential map of f at any x∈M.We define the smooth immersion F:TM→TN to be

for any point(x,u)∈TM.

Lemma 3.1Suppose that h and ν are the second fundamental form and the unit normal vector field of M in N,respectively.Let Xhand Xvbe the horizontal and the vertical lifts of X ∈ Γ(TM)onto TM,respectively,with respect to g=f?G.Then the differential map dF of F is defined by

for any point(x,u)∈TM and the vector field X ∈ Γ(TM).

ProofFor any(x,u)∈TM,we have F(x,u)=(f(x),df(x)u)∈TN.Let()andbe the local coordinates around(x,u)and F(x,u).Letbe the Levi-Civita connection of the induced metric g.Then the local representation of f(x) is of the form

(f1(x1,···,xn),···,fn+1(x1,···,xn)).

Moreover,we have

Therefore,

It follows from(3.3)–(3.4)that

On the other hand,from Lemma 2.1,we have

Therefore,

It follows from(3.3)that

Lemma 3.1 follows immediately from(3.6)and(3.8).

From Lemma 3.1,we have the following lemma.

Lemma 3.2Denote=F?Gs.Then at any point(x,u)∈ TM,we have

for any X,Y ∈ Γ(TM).

Remark 3.1From(3.9)of Lemma 3.2,we can see that(TM,g)is not a natural metric.

Suppose that(Nn+1(c),G)is a space form of constant section curvature c.The Riemannian curvature operator of(Nn+1(c),G)is given by

for any X,Y,Z∈Γ(TN).

Suppose that f:(M,g)→(Nn+1(c),G)is an isometrical immersion of Mninto Nn+1(c).Let A be the shape operator of M in Nn+1(c).Consider F:(TM,g)→(TNn+1(c),Gs)defined by(3.1).Then TM is a submanifold of TN with codimension 2.

In the sequel,we proceed to study the extrinsic geometry of TM in(TNn+1(c),Gs).

The two local orthonormal normal vector fields ν1,ν2of TM in TNn+1(c)are given by

where τ2=1+g(A(u),A(u))with τ>0 at any point(x,u) ∈ TM,so that the normal bundle T⊥(M)of TM in TN is locally spanned by ν1and ν2.

From now on,we denote brie fly X:=[df(x)X]f(x)and

for any X∈TxM.

Let{e1,···,en}be a local orthonormal frame field on(M,g),and{θ1,···,θn}be its dual frame field.We denote

for any u=Γ(TM).It follows from(3.9)that

Moreover,(3.2)and(3.12)turn into

respectively.From(3.17),we obtain

Using(3.11),from Lemma 2.3,we have

Moreover,we also have

From(3.19)–(3.20),we can see that

From(3.19)to(3.21),we immediately obtain the following proposition.

Proposition 3.1Let?⊥be the normal connection of(TM,)in TNn+1(c).Then

where for any 1≤i≤n,

ProofSince{ν1,ν2}is a local orthonormal frame field of T⊥M,we have

for any 1 ≤ i≤ n and α =1,2.Substituting(3.19)–(3.21)into(3.25),we obtain(3.23).The proof of Proposition 3.1 is completed.

From(3.19)–(3.21),we can immediately obtain the following lemma.

Lemma 3.3Denote byαthe shape operator of TM with respect to ναfor α =1,2.Then

Using Lemma 3.3,we can prove the following proposition.

Proposition 3.2The second fundamental form σ of TM is determined by

for any 1≤i,j≤n.

ProofIt is well-known that σ = σανα,where for any,∈Γ(TF(x,u)TM).Substituting(3.26)into(3.28),we obtain.

This completes the proof of Proposition 3.1.

Theorem 3.1Let Mnbe an immersed hypersurface of a space form Nn+1(c).Denote bythe mean curvature vector field of TM in TNn+1(c).Suppose that the length ofis invariant along every fibre of TM.Then we have that

(1)If c≥0,M is totally geodesic in Nn+1(c).

(2)If c<0,M is an isoparametric hypersurface with at most three distinct principal curvatures{?√?c,0,√?c}with multiples{m?,m0,m+},whose second fundamental form is parallel.

ProofLetbe the mean curvature vector field of TM.Choose{ei}such that hij= λiδij.Then it follows from Proposition 3.2 that

where H is the mean curvature of M,and{Hk}are the coefficients of the covariant derivative of H.Taking the squared length on both sides of the above equation,we obtain

Since the length ofis invariant along the fibres,by the above equation,we have

It follows that

So we have

where

From(3.31),we have F2=F4=F6=0 for any(u1,···,un),which is equivalent to

From the first equality of(3.32),we have Hk=0 for any 1≤k≤n,which means that M is of the constant mean curvature.

The second equality of(3.32)implies that λi(+c)=0 for any 1 ≤ i≤ n.It follows that every principal curvature λiis constant for all 1 ≤ i≤ n.

When c≥ 0,λi=0,for all 1≤ i≤ n.In this case,Mnis totally geodesic in Nn+1(c).

When c<0,λi=0,?√?c or√?c.We suppose that λi=0 for 1≤ i≤ m0,λi= ?√?c for m0+1≤i≤m0+m?and λi=?√?c for m0+m?+1≤i≤m0+m?+m+=n.

The third equality of(3.32)turns into

It follows from the above equality and the assumptions that

for any 1≤i

Suppose that 1≤i≤j≤k≤n.It is seen from(3.34)that hijk=0,and now that i,j or j,k lie in the same range of indices.For i,j,k lying in the di ff erent ranges of indices,from(3.33),we have hijk=0.It follows that hijk=0 for any 1≤i,j,k≤n,which means that Mnhas a parallel second fundamental form in Nn+1(c).This completes the proof of Theorem 3.1.

Remark 3.2Miyaoka[6]studied the geometries of isoparametric hypersurfaces with at most three distinct principal curvatures in a space form Nn+1(c)of the constant curvature c with c≥0.Our result gives a geometrical description of this kind of hypersurfaces with c<0.

By Theorem 3.1,we immediately obtain the following corollary.

Corollary 3.1Let Mnbe a smooth hypersurface of a space form Nn+1(c)with c≥0.Then the following statements are equivalent:

(1)The length of the mean curvature fieldof TM is invariant along the fibres of TM;

(2)TM is totally geodesic in(TNn+1(c),Gs);

(3)TM is minimal in(TNn+1(c),Gs);

(4)M is totally geodesic in Nn+1(c).

Munteanu[8]computed the Riemannian curvature tensor of TN endowed with the general metric Ga,b.For(TNn+1(c),Gs),we have the following lemma.

Lemma 3.4(cf.[8])Suppose that(Nn+1(c),G)is a space form of the constant sectional curvature c.Then the Riemannian curvature tensorof(TNn+1(c),Gs)is given by

for any X,Y,Z∈Γ(TN)at point(y,v)∈TN.

By direct computation,we obtain the following lemma.

Lemma 3.5Let Mnbe a hypersurface of(Nn+1(c),G).Let{e1,···,en}be a local orthonormal frame field on M,and ν be the unit normal vector field of M.Then

where V=and

Lemma 3.6Under the assumptions as in Lemma 3.5,suppose in addition thatand{ν1,ν2}are chosen as in(3.17).Then we have

Denote bythe Riemannian curvature tensor and by⊥the normal curvature tensor of TM in TNn+1(c).Then we have the following Gauss-Codazzi equations:

for any X,Y,Z ∈ Γ(T(TM))and ξ∈ Γ(T⊥(TM)).

By Proposition 3.1,we have the following theorem.

Theorem 3.2Let Mnbe a smooth hypersurface of a space form Nn+1(c).If the normal bundle of TM in TN is flat,then Mnis flat and totally geodesic in Nn+1(c),and vise versa.

ProofLet{e1,···,en}be a local orthonormal frame field on M such that hij= λiδij.Then(3.16)and(3.26)–(3.27)turn into

where

Since the normal curvatures of TM are determined by

it follows that the normal bundle of TM is flat in TN if and only if

for any(x,u)∈TM and 1≤i,j≤n.From the Wiengarten formula(3.43),we have

Using(3.41)and(3.44),we havefrom which we get

It follows that

By a direct but not difficulty computation,we can see that

Substituting(3.49)–(3.50)into(3.47)and sorting it,we get

for all 1≤i,j≤n.From(3.48)and(3.51),we obtain

for all 1 ≤ i,j≤ n.Suppose in(3.52)that u=ukek=0,and fromat any point x∈M for all 1≤i,j≤n,we have+c=0 at any x∈M for all 1≤i≤n,which implies that Nn+1(c)is of the non-positive curvature.Suppose in(3.52)that δij=1 and c<0,i.e.,λi≠0 at any x∈M for all 1≤i≤n,and we have that

which implies that λi=0 at any x ∈ M for all 1 ≤ i≤ n.It is contradictory to our assumption,and we have that=0 at any point x∈M for all 1≤i,j≤n if and only if M is flat and totally geodesic in Nn+1(c).It is seen from(3.53)that,in this case,=0 for all 1≤i,j≤n.

The reverse is trivial.This completes the proof of Theorem 3.2.

4 The Almost Complex Structure on TM

In this section,we study the almost complex structure J on TM,which is compatible withand the K¨ahlerian form ω on TM induced by J.

4.1 The induced almost complex structure on TM

Let{e1,e2,···,en}be a local orthonormal frame field and{}be the associated connection forms on M.We describe the almost complex structure J on TM as follows:

where α,β,γ and ρ are the smooth functions on TM to be determined.Since J is compatible with,we have

Substituting(4.1)into(4.2),we get

Thus,J is determined by

for any 1≤i≤n.

The Nijenhuis tensor of J is defined to be

NJ(X,Y)=[X,Y]+J[JX,Y]+J[X,JY]?[JX,JY]

for any X,Y ∈ Γ(T(x,u)TM).It is easy to see that

NJ(Y,X)=?NJ(X,Y),NJ(X,Y)=?NJ(JX,JY),NJ(X,Y)=JNJ(JX,Y).

Therefore,we have

NJ(Xv,Yv)=?NJ(JXv,JYv),NJ(Xv,Yh)=JNJ(JXv,Yh)

for any point(x,u)∈TM and X,Y∈TxM.Since J is an isomorphism from H(x,u)to V(x,u),it follows that J is integrable if and only if NJ(Xh,Yh)=0 for any X,Y∈Γ(TM).

Let us computefor any 1≤i≤n.It is known that

from which we have

At first,we have

From the Definition of,we can see that

Substituting(4.7)into(4.6)and using(4.5),we obtain

Using(4.8)–(4.9),we have

where we have used

Therefore,

It follows that

Since=0,we have

It follows that for any differentiable function ? on TM,

By using(4.12),we have

and

Substituting(4.13)–(4.15)into(4.11),we obtain

It follows from(4.5),(4.10)and(4.16)that

where

Theorem 4.1Let Mnbe a smooth hypersurface of Nn+1,and J be the almost complex structure on(TM,).Then J is integrable if and only if M is flat and is locally a product of a part of the principal curvature line and a piece of the(n?1)-dimensional totally geodesic submanifold of Nn+1.

ProofFrom(4.17),one can see that J is integrable if and only ifat any point(x,u)∈TM for any 1≤i,j,k≤n.

Let{ei}be a local orthonormal frame field on M such that hij= λiδij.ThenIt follows from(4.12)that

for any 1≤l,p≤n.Therefore,we have for any 1≤l≤n,that

Suppose thatfor any 1≤i,j,k≤n at any point(x,u)∈TM.Then()=0 for any 1≤i,j,k,l≤n at any point(x,u)∈TM.

Computingand putting u=0,we get

for any 1≤i,j,k,l≤n at any point x ∈ M.Substituting(4.23)into(4.19),we obtain

At the point x∈M where(4.24)is trivial.

Suppose thatat the given point x∈ M.For u ∈ TxM with τ≠1,we have

Multiplying λkukon both sides of(4.25)and taking sum with respect to k,we obtain

It follows that

Substituting(4.26)into(4.25),we get

for any i≠j.Suppose that λ1≠0.Then from(4.27),we have

for any j>1.The condition that τ≠1 implies that It follows from(4.28)thatλ1u1≠0.Taking j=1 in(4.27),we obtain that λi=0 for any i>1.

So there is at most one nonzero principal curvature of the shape operator A at any point of M.It follows that

for any i≠j and k≠l,which together with(4.23)implies that=0 for any i≠j and k≠l,which implies that M is flat.

Suppose that=0 for any 1≤i,j,k≤n at any point(x,u)∈TM.Then(4.18)turns into

or equivalently

for any fixed 1≤ i,j,r≤ n.Denote Λij=hpijup.Then(4.29)turns into

for any fixed 1≤i,j,r≤n with i≠j.

At the point x∈M where(4.30)is trivial.At the point x ∈ M where λk≠for some 1 ≤ k ≤ n,we can suppose that λ1≠0 and λk=0 for 2≤ k ≤ n.

Suppose that i=1 in(4.30).Then

for any fixed 1

for any fixed 1

for all 1 ≤ i,j,k ≤ n where λj,k=ek(λj).It follows from(4.33)that

for any 1

which implies that

for all 1

for all 1≤ j,k ≤ n.It follows from(4.36)–(4.37)that

for any 1

Taking the partial derivative on both sides of(4.39)with respect to u1and using(4.20),we have

It follows from(4.36)and(4.39)–(4.40)that

for all 1

for all 1

for all 1

Let γ be a part of the curvature line with respect to the principal curvature λ1and Un?1be the maximal integral submanifold of Ln?1through every point of γ.Then it follows from(4.36)that Un?1is totally geodesic in Nn+1.The second equality of(4.41)implies that Mnis locally a Cartesian product of γ and Un?1.

The proof of sufficiency is trivial.This completes the proof of Theorem 4.1.

4.2 The induced Khlerian form on TM

It is known that the Khlerian 2-form ω of TM is defined to be

for all vector fields,∈(TM),where J is determined by(4.4).Let{e1,e2,···,en}be an orthonormal frame on M such that hij= λiδij.By direct computation,we have

It is known that the exterior differential of ω is defined by

for all vector fields,,∈(TM).

By using(4.5),we have

Theorem 4.2Suppose that Mnis a smooth hypersurface of Nn+1and ωis the K¨ahlerian2-form on(TM,g,J).If(TM,g,J,ω)is almost K¨ahlerian,then M is locally a product of a part of the principal curvature line and a piece of the(n?1)-dimensional totally geodesic submanifold of Nn+1.

ProofBy Definition,(TM,g,J,ω)is almost K¨ahlerian if dω=0 onTM,which implies that the right-hand sides of(4.50)–(4.53)are zero.

We suppose that dω=0 onTM.From(4.52),we have

for any fixed j≠k and all(x,u)∈TM.

Choose uk=1 and uj=0 in(4.54)for any other j≠k.Then it follows that

which implies that there is at most one nonzero principal curvature at any x∈M,whose multiple is 1.When λ1= ···= λn=0 at x ∈ M,the right-hand sides of(4.50)–(4.53)are identically zero.Without loss of generality,we suppose that λ1≠0 and λk=0 for all 2 ≤ k ≤ n in an open subset W of M.From(4.51),we have

for all 2≤j,k≤n at any(x,u)∈TM where x∈W.It follows immediately from(4.55)that

for all 1≤i≤n and 2≤j,k≤n.Substituting(4.56)with i=1 into(4.56),we obtain

from which we have

for all 2≤j,p≤n at any(x,u)∈TM where x∈M and u∈TxM with u1≠0.Let u1tend to 0 in(4.59)and(4.60).It follows that

Substituting(4.61)into(4.59)and(4.60),and using(4.61)once more,we obtain

for all 1≤i≤n and 2≤j≤n at any x∈W.It follows from(4.57)and(4.62)that all of hijk’s are zero except for h111.Note that for hij= λiδij,

for all 1 ≤ i,j,k ≤ n where λj,k=ek(λj).It follows that

for any 2≤i,j≤n and 1≤k≤n.From(4.64),we can see that

By applying the similar discussion as in the proof of Theorem 4.1,we can see that M is locally a product of a part of the principal curvature line and a piece of the(n?1)-dimensional totally geodesic submanifold of Nn+1.This completes the proof of Theorem 4.2.

AcknowledgementThe authors wish to make a grateful acknowledgement to the referees for their advice on the original manuscript.

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