Shaoxiang ZHANGJu TAN

Abstract Boek (1980) has introduced a class of solvable Lie groups Gn with arbitrary odd dimension to construct irreducible generalized symmetric Riemannian space such that the identity component of its full isometry group is solvable.In this article, the authors provide the set of all left-invariant minimal unit vector fields on the solvable Lie group Gn,and give the relationships between the minimal unit vector fields and the geodesic vector fields, the strongly normal unit vectors respectively.

Keywords Solvable Lie groups,Lagrangian multiplier method,Minimal unit vector fields, Geodesic vector fields, Strongly normal unit vectors

1 Introduction

Let (M,g) be a Riemannian manifold and (T1M,gs) be the unit tangent sphere bundle equipped with the Sasaki metric.Every smooth unit vector field determines a mapping between(M,g) and (T1M,gs), embeddingMinto its tangent unit sphere bundleT1M.Every smooth unit vector fieldXonMcan be viewed as a submanifold ofT1M, then if the manifoldMis compact and orientable, we can define the volume ofXas the volume of the immersion.

Gluck and Ziller firstly considered the problem of determining unit vector fields which have minimal volume in [8].They proved that the unit vector fields of minimum volume on the unit sphereS3are precisely the Hopf vector fields.However, this is no longer true for the higher dimensional sphereS2n+1,n≥2 (see [10–11, 13–14]).In [7], the authors proved that a unit vector fieldVis a critical point of the volume functional restricted to the set of unit vector fields(M) if and only ifV:M→T1Mis a minimal immersion.So such unit vector fields are called minimal even though the manifold is not compact.

Some examples of Lie groups equipped with minimal unit vector fields are provided in [4, 7,9, 15–18].For three dimensional Lie groups, Tsukada and Vanhecke gave all the left invariant minimal unit vector fields in [16].Yi in [18] obtained all the left-invariant minimal unit vector fields on the semi-direct product Rn?PR, wherePis a nonsingular diagonal matrix.But for most of examples of Lie groups, it is difficult to determine all the left invariant minimal unit vector fields, there are just some special minimal unit vector fields.

Bo?ek introduced a class of important solvable Lie groups Gnwith arbitrary odd dimension to construct irreducible generalized symmetric Riemannian space such that the identity component of its full isometry group is solvable in [5].In rencent years, a great deal of mathematical effort has been devoted to the study of the solvable Lie group Gn.In [6], Calvaruso, Kowalski and Marinosci studied homogeneous geodesic of solvable Lie groups Gn.Aghasi and Nasehi in [3] generalized this study to the Randers setting of Douglas type, they proved that these homogeneous Randers spaces are locally projectively flat Finsler spaces.In [1], the authors studied some other geometrical properties on these spaces with dimension five, and extended those geometrical properties for an arbitrary odd dimension in both Riemannian and Lorentzian cases in [2].

Thus, it is an interesting question to determine the left invariant minimal unit vector fields on Lie groups Gn.The study of this problem will deepen our understanding of this kind of Lie groups undoubtedly.In this paper, the aim is to provide the set of all left invariant minimal unit vector fields on these Lie groups Gnby Lagrange multiplier method.For an integern≥2,a unimodular solvable Lie group Gnis defined as follows:

where (x0,x1,···,xn,u1,···,un) ∈R2n+1andConsidering the following left invariant vector fields

and

By [2], we can equip Gnwith a left invariant Riemannian metric as follow

The inner product which is induced by this metric shows that the set{X0,X1,···,Xn,U1,···,Un} is an orthonormal frame field for the Lie algebra gnof Gn.

The main result can be shown as follows.

Theorem 1.1For Lie groupGn, n≥2, the set of left invariant minimal unit vector fields is

This paper is organized as follows.We recall some basic notions and facts about minimal unit vector fields in Section 2.In Section 3, we give all left invariant minimal unit vector fields on Gn, i.e., Theorem 1.1.Finally we devote Section 4 to discuss the relationships between the minimal unit vector and the geodesic vector, the strongly normal vector in Theorems 4.1 and 4.2, respectively.

2 Preliminaries

Let (M,g) be an-dimensional smooth Riemannian manifold and(M) be the set of all vector fields onM.Furthermore, ▽denotes the Levi-Civita connection.

whereI, (▽V)?denote the identity map and the adjoint operator of (▽V), respectively.And letthen for a compact closed oriented manifoldM, the volume functional of vector fields Vol:X1(M)→R is given by

where dvis the volume form on (M,g).

Now we give a (1,1)-tensor fieldKVand a 1-formωVassociated toV.They are defined as

We can easily get

Let HVdenote the distribution consisting of the tangent vectors orthogonal toV.In [7] it is proved thatVis a critical point for the volume functional Vol on(M) if and only ifωVvanishes on HV.

For an orthonormal basis {E1,E2,···,En} of the tangent space,ωV(X) can be written as

Besides, it is shown thatVis critical if and only if the submanifold of(T1M,gs)determined byVis minimal (see [7]), wheregsis the Sasaki metric.

Definition 2.1(see [18])A unit vector field V on a Riemannian manifold(M,g)is called minimal if ωV(X)=0for all X∈HV.

Now we consider left invariant unit vector fields on Lie groups.LetGbe an-dimensional connected Lie group equipped with a left invariant metric,and g be the Lie algebra ofG.Then the left invariant metric onGdetermines the inner product 〈·,·〉 on g.Furthermore, let S be the unit sphere of g.By the left invariance, the functionfcan be considered as a function on S.The distribution HVis invariant with respect to the left translation and can be equal to orthogonal complementV⊥ofVin g.SoV⊥is identified with the tangent spaceTVS of the unit sphere S atV.Therefore we have the following lemma.

Lemma 2.1(see [16])A left invariant unit vector field V on a Lie group G is minimal if and only if the linear map ωVongvanishes on V⊥TVS.

Now we compute the differential dfof the functionfon S atV.And we have the following proposition:

Proposition 2.1(see [17])For X∈TVS, we have

and V is minimal if and only if

for all X∈TVS.

If the Lie groupGis a unimodular Lie group, that is tradX=0 for allX∈g (see [12]), we can easily get the following corollary.

Corollary 2.1A left invariant unit vector field V on a unimodular Lie group G is minimal if and only if V is a critical point of the function f onS.

3 Left-Invariant Minimal Unit Vector Fields on Gn

For anyn≥1, the unimodular solvable Lie group Gnis as follows:

where (x0,x1,···,xn,u1,···,un) ∈R2n+1andConsidering the following left invariant vector fields

and

By [2], we can equip Gnwith a left invariant Riemannian metric as following

The inner product which is induced by this metric shows that the set{X0,X1,···,Xn,U1,···,Un}is an orthonormal frame field for the Lie algebra gnof Gnand the Lie bracket is introduced as follows:

By Koszul’s formula in [12],

wherei,j,α,β=1,···,n.

and by (2.1), we can get

So we can easily get

Proof of Theorem 1.1By Corollary 2.1, for a unimodular Lie groupG, a left invariant unit vector fieldVis minimal if and only ifVis a critical point of the functionfon S.

LetH(k0,k1,···,k2n) ?detLV, we will find the set of critical points of the functionH(k0,k1,···,k2n) with the constraint

Using the Lagrange multiplier method, we need to solve the following system of equations:

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wherei=1,2,···,n.

For the casen= 1, the result is already known in [16] by Tsukada and Vanhecke.So we will restrict ourselves ton≥2.Firstly, we have the following assertion.

Claim 3.1λ≠0.

ProofIt is easy to find that

Ifλ=0, by (3.3)–(3.4) we have

Then we can simplify (3.5) to

ifkn+i= 0, we obtainthen we getkn+1= ··· =k2n= 0.This contradicts to(3.2).

So allkn+j(j=1,···,n) are not equal to 0, by (3.6), we have

Case 3.1k0≠0.

Now, by (3.3) we can get

Ifk1≠0, according to (3.4), we have

Then with the help of Mathematica, we solve the system of (3.2), (3.7)–(3.8), the solution set is empty.So we havek1=0.In the same way, we can getk2=···=kn=0.

Ifkn+1≠0, from (3.5) we obtain

Sincek1=···=kn=0, (3.7) can be reduced to

So we have

Solving the system of(3.2),(3.10)by Mathematica,the solution set is empty.We havekn+1=0.Similarly,kn+2=···=k2n=0.

In conciusion, we havek0=±1, so ±X0are minimal unit vector fields.

Case 3.2k0=0.

Case 3.2.1?l(1 ≤l≤n) such thatki1≠ 0,···,kil≠ 0,kn+1= ··· =k2n= 0, where 1 ≤i1<···

Thus, the set of left invariant minimal vector fields is

Case 3.2.2?l(1 ≤l≤n) such thatki1≠0,···,kil≠0, where 1 ≤i1<···

Assumeki≠0, (3.4) can be written as

so if ?β≠μ,kβ≠0,kμ≠0, we have

Then forkn+i, according to (3.5), we have

Ifkn+i= 0, we can obtainifkn+i≠ 0, we also haveThen we can easily get at least twokn+j≠0.

Thus for ?kn+α(1 ≤α≤n),(3.5) can be reduced to

Ifkn+α≠0, solving the system of (3.12), (3.14), we can get

According to (3.15), (3.2) andwe obtain

wherekn+iβ≠0 orkiβ≠0,β=1,···,l.

Thus, the set of left invariant minimal vector field is

Case 3.2.3k1=···=kn=0.

In this case, (3.5) can be reduced to

Ifkn+α≠0, the equation above can be written as

? If ?kn+i= 0, by (3.18), we haveThen we can simplify the equation above to

Therefore ifkn+α,kn+β≠0,α,β=1,···,n, we havekn+α2=kn+β2.

According to

we obtain that the set of minimal unit vector fields is

? Ifkn+1≠0,···,k2n≠0, according to (3.1), we have

Then taking natural logarithm, we have

Applying Lagrange multiplier method, we get

Forα≠β, we can obtain

Ifkn+α?kn+β≠0, we have

With the help of Mathematica,whenkn+α+kn+β≠0 andthere are no solutions.

Therefore, we have

According to the constraint conditionwe can get the following results.

Whennis odd,

Whennis even,

So we obtain the set of minimal vector fields is

In conclusion, combining with (3.11), (3.17), (3.21), (3.26) and case 3.1, we obtain all left invariant minimal unit vector fields on the solvable Lie group Gn(n≥2) as follows:

This completes the proof of Theorem 1.1.

4 Geodesic Vector Fields and Strongly Normal Unit Vectors on Gn

In this section, firstly we determine all geodesic vector fields on Gnand obtain the relationship between geodesic vector fields and minimal unit vector fields.Then we study the strongly normal unit vectors on Gn.

Definition 4.1A unit vector field V on a Riemannian manifold(M,g)is called a geodesic vector field if▽VV=0.

The set of all geodesic vector fields on Gnis given as follows.

Theorem 4.1For n≥2, the set of all left invariant geodesic vector fields on the solvable Lie groupGnis

ProofLetsince the Lie brackets of gnare as follows:

and the non-vanishing Riemannian connection components are given by

wherei,j,α,β=1,···,n.

Then we have

IfVis a geodesic vector field, by ▽VV=0, we have

Ifk0= 0, thenki= 0,i= 1,···,n, so the geodesic vector fieldwhere; Ifk0≠ 0, thenki≠ 0 andkn+i= 0,i= 1,···,n, so the geodesic vector fieldwhere

Therefore the set of all left invariant geodesic vector fields on the solvable Lie group Gnis

By Theorem 1.1 and Theorem 4.1, we can easily obtain the sets of vector fields which are both the minimal unit vector fields and the geodesic vector fields on the Lie group Gn,n≥2 as follows:

Definition 4.2A unit vector field V on a Riemannian manifold(M,g)is called strongly normal if g(▽XAVY,Z)=0,?X,Y,Z∈HV, where AV=?▽V.

It is difficult to calculate all strongly normal unit vectors on the solvable Lie group Gn.So we study the set of vector fields which are both the minimal unit vector fields and the strongly normal vector fields as follows.

Theorem 4.2For n≥2, The set of vector fields which are both the minimal unit vector fields and the strongly normal vector fields on the solvable Lie groupGnis

ProofLetandAssume thatX,Y,Z,Vsatisfy the conditions

According to the non-vanishing Riemannian connection components

wherei,j,α,β=1,···,n.Then

▽XAVY=?▽X(▽YV)+▽▽XYV

IfVis a strongly normal unit vector, then

IfV= ±X0, that isk0= ±1, thena0=b0=c0= 0, according to the above equation, we can easily get ▽XAVY=0, soV=±X0are the strongly normal unit vectors.Similarly, ±Xi,i=1,···,nalso are the strongly normal unit vectors.

IfV= {±Ui}, without loss of generality, letV=U1, that iskn+1= 1,k0= ··· =kn=kn+2=···=k2n=0, thenan+1=bn+1=cn+1=0, we have

We can easily give a counter-example thatVis not a strongly normal unit vector.LetX=X0,Y=X0,Z=U2, theng(▽XAVY,Z) = 1 ≠ 0.In the same way, we can prove thatare not the strongly normal unit vectors.

Therefore, The set of vector fields which are both the minimal unit vector fields and the strongly normal vector fields on the Lie group Gn,n≥2 is