Semi-Simplicity of a Lie Algebra of Isometries

Anona Manelo^*
*Correspondence: Anona Manelo, Department of Mathematics and Computer Science, Faculty of Sciences, University of Antananarivo, Antananarivo 101, PB 906, Madagascar, Tel: +261 33 12 027 73, Email:

Keywords

Differentiable manifolds • Lie algebra • Spray • Nijenhuis tensor Riemannian manifolds • Infinitesimal isometry

Introduction

Frölicher-Nijenhuis and Rund H.’s works [1,2] have enabled Grifone J. [3,4]) to show a connection as a vector 1-form Γ such that JΓ=J and ΓJ=-J of which J is the natural tangent structure of the tangent bundle TM to a differentiable manifold M. Such Γ is an almost product structure (Γ²=I), I being the identity vector 1-form. This formalism allows a more algebraic approach for the study of Lie algebras defined by Γ [5].

Given a paracompact differentiable manifold M of n ≥ 2 dimension of class Γ is a connection [3]. In [5], we studied the associated Lie algebras to Γ, through their first Chevalley-Eilenberg cohomology spaces, namely, the Lie algebra of vector fields ^AΓ on TM, whose Lie derivative applied to Γ is zero, and the horizontal nullity space of the curvature R of Γ The Lie algebra ^AΓ is formed by projectable vector fields, and contains two ideals consists of projectable vector fields of the horizontal nullity space of the curvature R and an ideal of the vertical space.

In [6], Loos O. considered a spray S as a system of second order differential equations on M. If we denote by A_S the Lie algebra of vector fields on TM which commute with S, the projectable vector fields of A_S correspond to being the complete lift on the tangent bundle TM of the set of all vector fields χ(M) on M. We will denote the set such that The Lie algebra corresponds to a Lie group G_S of transformations on M of which the tangent linear mappings preserve the spray S. The group G_S acts freely on the linear frame bundle of M. The dimension of is at most equals to n²+n.

In the following, a connection Γ is linear without torsion, in this case [3], Γ is written: Γ=[J, S] and [C, S]=S, where C being the Liouville field on TM. We prove that by considering . For a linear connection without torsion such that the rank of the horizontal nullity space of the projectable vector fields of curvature R is nonzero constant, the Lie algebra . All the commutative ideals of do not always come from the horizontal nullity space of the curvature for any spray S. Let E be an energy function [3], that is to say, a mapping on TM to , null on the null section, homogeneous of degree two (L_CE=2E), L_C being the Lie derivative with respect to C, such that the 2-form Ω=dd_JE is of maximum rank. The canonical spray S is defined [5] by

i_sdd_jE=-dE, i_S being the inner product with respect to S. The 2-form Ω defines a Riemannian metric on the vertical bundle. In local coordinate system on an open set U of M, , the coordinate system on TU , the function E is written

where are positive functions such that the symmetric matrix is invertible . We denote by g such that is a commutative ideal of . We then have the following algebra inclusions: . The dimension of is at most equal to . The is also the Lie algebra of the Killing fields of projectable vectors which commute with the spray S. If the dimension of is greater than or equal to three, the Lie algebra is semi-simple if and only if the horizontal nullity space of the Nijenhuis tensor of Γ is reduced to zero. In this context, the algebra coincides with . It follows that for the flat Riemannian manifolds (R=0), the Lie algebra , the Killing algebra also, contain commutative ideals. For Riemannian manifolds with nonzero constant sectional curvature, is the isometry algebra, and it is semi-simple.

This study also makes it possible to construct numerous examples of Lie algebra containing a commutative ideal. In particular, for a semi-simple Lie algebra, the derivative ideal coincides with algebra. This examples show that this property is not sufficient to be a semi-simple Lie algebra.

This paper is part of the continuation of the studies made in [5,7-9].

Preliminaries

We will recall the bracket of two vector 1-form K and L on a manifold M [1]. If we denote χ(M) the set of vector fields on M,

(2.1)

The bracket N_L=1/2 [L, L] is called the Nijenhuis tensor of L. The Lie derivative with respect to X applied to L is written:

The exterior differentiation dL is defined by be a connection. By

h=1/2(I+ Γ) and v=1/2(I-Γ),

h is the horizontal projector, projector of the subspace corresponding to the eigenvalue +1, v the vertical projector corresponding to the eigenvalue -1. The curvature of Γ is defined by

R=1/2 [h,h];

which is equal to

R=1/8 [Γ, Γ].

The Lie algebra A_Γ is written

The nullity space of the curvature R is written:

According to the results of [5], the elements of A_Γ are projectable vectors fields. The horizontal and vertical parts of A_Γ respectively andare ideals of A_Γ . If we denote by H° the set of horizontal and projectable vector fields, then we have

(2.2)

Definition 1 [3]

We call strong torsion T of Γ the vector 1-form

where S indicates a spray, t=1/2 [J, Γ ] called weak torsion of Γ, i_S the inner product with respect to a spray S.

We recall a result of [3] on the decomposition of a connection

Γ=[J, S]+T.

The elements of are well known [5]. In the following section, we will be interested in the ideal .

Properties of the Vertical Ideal of

A vertical vector field can be written in the form JX, J being the tangent structure on TM, and X a horizontal vector field.

Proposition 1 [5]

A vertical vector field JX is an element of if and only if JX commutes with all horizontal and projectable vector fields.

Proposition 2

Let Γ be a zero weak torsion connection, X and Y two horizontal vector fields such that JX and JY are elements of . So we have

J [X, Y]=0.

In particular,

[X, Y]=R(X, Y).

Proof: The nullity of the weak torsion of Γ allows to write

v [JX, Y]+v [X, JY]=J [X, Y], (3.1)

for all X, Y horizontal vector fields. If JX and JY are two elements of , we get v [JX, Y]=0 and v [X, JY]=0, that is to say

J [X, Y]=0.

This means that the horizontal part of [X, Y] is zero, we find

[X, Y]=v [X, Y]=R (X, Y).

Proposition 3

Let Γ be a zero weak torsion connection, X a horizontal vector field. The two following conditions are equivalent:

a) X is a projectable vector field and [JX, h]=0;

b) [X, J]=0.

Proof: The nullity of the weak torsion of Γ is written:

[JX, h]Y+[X, J]Y=h[X, JY], (3.2)

for all horizontal vector fields X, Y. If X is projectable, the term h [X, JY] is zero; and if [JX, h]=0, we get [X, J]Y=0, for all horizontal vector fields Y. If Y is a vertical vector field, we have [X, J] Y=0, since X is projectable. This proves that the relation a) implies b). If [X, J]=0, the horizontal vector field X is projectable [11], so we have h [X, JY]=0; and the nullity of the weak torsion of Γ (3.2) results in [JX, h]=0. In the following section, we assume that the connection Γ has a null strong torsion, that is, Γ=[J, S] and [C, S]=S. Einstein’s convention on the summation of indices will be adopted. In natural local coordinates on an open set U of M, , the coordinates on TU, a spray S is written:

For a connection ΓΓ =[J, S], the coefficients of Γ become and the horizontal projector is

The vertical projector is written:

The curvature R=1/2 [h, h] become

With

We can associate with Γ a linear connection on TM, called Berwald connection [4] with curvature

for all X, Y and Z ∈χ(TM). The curvature R of Γ is linked to by the relation:

[JZ, R(X ,Y)]+ R([JZ, X ],Y) + R(X ,[JZ,Y])

In particular,

(X ,Y)S = −R(X ,Y).

Proposition 4

Let JX such that X be a projectable vector field, then we have

(Y, Z)X = 0,

for all Y, Z ∈χ(TM).

Proof: Let JX . This means [JX,h]=0. According the expression of R=1/2 [h, h] and the identity of Jacobi, we get [JX,R]=0. In other words, for all Y, Z ∈χ(TM), we have

[JX, R(Y, Z)]-R([JX, Y], Z)-R(Y,[JX, Z])=0.

So the expression of becomes

(Y, Z)X = J[X , R(Y, Z)].

The vector field X being projectable and R(Y, Z) vertical, the second member of the equality is null. So we find (Y, Z)X = 0.

Let be the nullity space of , Ker the kernel of :

Proposition 5

The horizontal ideal of A_Γ becomes

Proof: According to the relationfor all X, Y ∈χ(TM), we have

If X according to the previous relationship between and R, we get

if X is projectable, [JZ, X] is vertical and R ([JZ, X], Y) is therefore zero. This results in

In conformity with a result of [5], we have

Bianchi identity on is written [4]:

Proposition 6

If the module N_R ∩ H° on the ring of constant functions to the fibers is of nonzero and constant rank, there is locally a non-trivial vector field such that

Proof: We will solve the equation

(3.4)

This leads to solving, according to the proposition 3, the equations

In other words, X is a horizontal and projectable vector field of the nullity space belonging to the Lie algebra A_J. The elements of A_J are well known [11]. We can write in local coordinates:

For a linear connection, we have , the system of equations to be solved becomes

(3.5)

with i, j, l ∈{1,…,n} . The compatibility condition of the equation (3.5), according to the Frobenius theorem, is

This condition is nothing but and according to the proposition 5, we have If the rank is nonzero constant, defines a foliation. On a submanifold defined by the distribution the equation (3.5) admits a non-trivial solution.

Proposition 7

Let B be the set defined by

Then B+JB is a commutative subalgebra of and a commutative ideal of

Proof: On the one hand, let X and Y be two elements of B, according to the proposition 2, we have

[X, Y]=0

On the other hand, X and Y being projectable, we find

[JX, JY]=0.

It is obvious that B is a subalgebra of , and JB a subalgebra of . Consequently, B+JB is a commutative subalgebra of and of A_Γ .

If Y ∈ A_Γ ∩ A_J, for X ∈ , we have sinceis an ideal of A_Γ. Now let us prove , that is, [J[X, Y], h]=0. According to the proposition 3, this leads to prove that [[X, Y], J]=0 and [X, Y] is a horizontal vector field, which is immediate by the identity of Jacobi and X ∈, horizontal ideal of A_Γ .

Particular Lie Algebras from Γ and S

The Lie algebra A_J is well known [11], and it is written

in which denotes the complete lift on the tangent bundle TM of χ(M). We denote A_S ={X ∈χ(TM) such that [X, S]=0}.

Proposition 8

The vertical vectors fields of A_S are trivial.

Proof: A vertical vector field is of the form JX, where X ∈ χ(TM). JX is an element of A_S if [JX, S]=0. In conformity with the formula J [JX, S]=JX for all X ∈ χ(TM), the previous relation implies JX=0 for all JX ∈ A_S.

Proposition 9

The set of projectable vector fields of A_S coincides with and that

Proof: Let X be a projectable vectors field, the identity of Jacobi makes it possible to write

[J, [X, S]]+[X, [S, J]]+[S, [J, X]]=0.

If [X, S]=0, the above relation becomes

[[J, S], X]=[[J, X], S]. (4.1)

But we have Γ=[J, S], Γh=h, Γv=-v and h+v=I, we find

[Γ, X]Y=[ΓY, X] -Γ [Y, X]=2v[Y, X],

for all Y horizontal vector field of Γ . On the other hand, we have

[S, [J, X]] Y=[S, [J, X]Y]-[J, X] [S, Y].

As [J, X] [S, Y] and [J, X] Y are vertical since X is projectable, by applying J to the relation (4.1) we find

[J, X] Y=0, (4.2)

for all Y horizontal vectors field. If Y is vertical, it is immediate to note that

[J, X] Y=0. (4.3)

The relations (4.2) and (4.3) prove that [J, X]=0. In other words, X ∈ AJ. But , M , and taking into account the proposition 8, therefore, the set of projectable vector fields of A_S coincides with

With [J, X]=0, the relation (4.1) proves that X ∈ A_Γ . That is to say

Let us now prove that .

Let be This results in with S=hS. If we apply to S the above relation, we get

(4.4)

From the relation we find. In other words, taking into account that h is semi-basic

(4.5)

as v+h=I, the two relations (4.4) and (4.5) result in

Which proves that

Remark 1: The proposition 9 is contained in [12] in another way. In the following, we assume and the horizontal part of .

Proposition 10

The horizontal part of is such that

and is a commutative ideal of the Lie algebra .

Proof: This is to prove that the horizontal part of is only the B of the proposition 7, which is immediate according to the proposition 3.

Theorem 1

For all differentiable manifold of n dimension with a linear connection without torsion such that the rank of the nullity space of the projectable vector fields of the curvature is nonzero constant, the Lie algebra contains a nonzero commutative ideal formed by the horizontal part of .

Proof: This is the consequence of previous studies. The existence of the nonzero horizontal part of is given by the proposition 6.

Automorphisms

We denote by GS the set of diffeomorphisms φ of M such that the induced diffeomorphismson TM preserve a spray S, G_S forms a group. From the study of the equations of A_S and G_S on a local coordinate system of M, using the classical theorem of Palais, Loos O. [6] deduces the following result:

Theorem 2 [6]

a) GS acts freely on the linear frame bundle of M. In particular, the dimension of is less than or equals to n²+n.

b) There exists a unique Lie group structure on G_S such that G_S is a Lie transformation group of M with Lie algebra the set of complete vector fields in

c) If the dimension of =n2+n and M a manifold simply connected, then (M, S) is isomorphic to for an unique is given by the equations

Examples

Example 1

Let be and the connection Γ=[J, S] with

then the coefficients of Γ being the nonzero coefficients of the connection are

The horizontal space is generated by

We notice that The Lie algebra is generated by

We note that all the subalgebras of are commutative ideals of .

None of these nonzero ideals are horizontal.

Example 2

Let E be the energy function such that

The canonical spray is written:

The nonzero coefficients of the connection are

The horizontal vector fields are generated by

The horizontal nullity space of the curvature is generated by .

The elements of are generated by

The commutative ideal is generated by .

We note that for any linear connection (see the example 1), the commutative ideals do not always come from the horizontal nullity space of the curvature.

The example 2 shows that the commutative ideal is of the horizontal nullity space of the curvature. The question is to know if it is true for all canonical connection coming from an energy function in the language of [3]. This is the object of our study.

Connections on Riemannian Manifolds

In this paragraph, we are content to state the well-known results on the Riemannian manifolds which will be used.

Definition 2

We call the Riemannian manifold, the couple (M, E):

• M is a differentiable manifold;

• E a function of on TM, on the null section, and homogeneous of degree two, such that dd_JE has a maximum rank.

We notice that if E is of class on the null section, we will have a Finsler manifold. Imposing E of class on the null section implies that the function E is of class on TM.

The application E is called energy function, Ω=dd_JE its fundamental form defines a spray S [13]:

i_Sdd_JE=-dE and Γ=[J, S] is called canonical connection. (7.1)

The fundamental form Ω allows to define a metric g on the tangent bundle by

g (X, Y)=Ω (X, FY), (7.2)

X and Y being two vector fields on TM, F the almost complex structure associated with Γ, (FJ=h and Fh=-J whereThere is, [4], one and only one metric lift D of the canonical connection such that:

1.

2.

3. DJ=0,

4. DC=v,

5. DΓ=0,

6. Dg=0,

7. DF=0.

The linear connection D is called Cartan connection. For a Riemannian manifold, the Cartan connection D and the Berwald connection are identical. We have

D_JXJY=[J, JY] X, D_hXJY=[h, JY] X.

To the linear connection D, we associate a curvature

for all X, Y, Z ∈ χ(TM). In particular,

(7.4)

In natural local coordinates on an open set U of M, (xⁱ, y^j) ∈TU ∈ i, j ∈ {1,..., n} , the energy function is written [3] p.330

where are symmetric positive functions such that the matrix is invertible . the relation i_Sdd_JE=-dE gives the spray S,

with

where

By

we have

The relation (7.1) is equivalent to the following relation

d_hE=0. (7.5)

Properties of the Curvature of a Riemannian Man

From the properties stated above, we obtain the following classic results: for all X, Y, Z, T ∈ χ(TM)

Properties of the Horizontal Nullity Space of Curvatures

From the properties of the curvatures given above, we have:

Proposition 11

On a Riemannian manifold, the horizontal ideal of the Lie algebra A_Γ satisfies

Proof: According to the proposition 5, the horizontal ideal A_Γ of A_Γ is written

From the relation (8.4), we get It remains to prove that

It is immediate to notice that

Let be According to the relation (9.1), we have for all

Y ∈χ(TM),

g (R^ο (Y), JX)=0.

As Dg=0, we get

g ([J, Rο (Y)]Z, JX)=-g(Rο (Y), [J, JX]Z).

The vector field X being projectable, [J, JX]=0, we have

g ([J, R^ο (Y)]Z, JX)=0.

From the relation (8.2), namely: namely:

we conclude

According to the relation (9.6) and that JX is orthogonal to Im R^ο, we find

g (R(Y, Z ), JX ) = 0,

that is, JX⊥ ImR. Taking into account DJT g=0, and X a projectable vectors field, we have

that is,

Proposition 12

On a Riemannian manifold, the three following relations are equivalent:

i. the horizontal nullity of the curvature R is reduced to zero

ii. the horizontal nullity space of the projectable vector fields of R is reduced to zero

iii. the dimension of the image space of the curvature R is equal to n-1

Proof: If the horizontal nullity space of the projectable vector fields is not reduced to zero, there will be a nonzero horizontal and projectable vector field X, according to the proposition 11, such that , and according to the relation (9.4), JX is orthogonal to therefore orthogonal to ImR according to the relation (7.4). The image space of the curvature R is both orthogonal to JX and to the Liouville field C=JS with JX ∧ C ≠ 0. This is only possible if the dimension of the image space of R is strictly less than n-1.. So the relation (iii) implies (ii). We notice by (7.5) that d_RE=0. That is, the image space of R is in the kernel of dE. If the dimension of the image space of the curvature R is strictly less than n-1, there will be a horizontal vector field X such that JX ∈ KerdE and JX ⊥ ImR. According to the relation (9.2), we have . By developing this equality and taking into account the relation (9.1), we have R(X, Y)=R^ο[JY, X] ∀Y ∈χ(TM).. This is only possible if X=S or X∈ N_R. Since, JX ⊥C, therefore, we have X∈ N_R.

This last calculation proves at the same time that the horizontal nullity space of R is generated by projectable vectors field belonging to N_R. Consequently, we have (i)=⇒ (iii) and (ii)=⇒ (i).

Conformal Infinitesimal Transformations of a Riemannian Manifold

Definition 3

Let ω be a p-differential form defined on a manifold M. A conformal infinitesimal transformation of ω is a vector field X defined on M, such that for all elements (t, x) of the domain D(X) of the flow Φ of X, we have

where ρ is a differentiable numeric function, with positive values, defined on D(X).

We have the following result:

Proposition 13

For a vector field X to be a conformal infinitesimal transformation of p-differential form ω, it is necessary and sufficient that we have,

L_Xω=λω,

where L_X is the Lie derivative with respect to X and λ a differentiable numerical function defined on M. When this is the case, the function λ is linked to the function ρ which appears in the definition 3, by the relation

In the following, we are interested in conformal transformations of the fundamental form Ω of a Riemannian manifold.

Proposition 14

If X is a conformal infinitesimal transformation of Ω, i.e.

L_XΩ=λΩ,then λ is a constant function.

Proof: Let be X a vector field belonging to χ(TM) such that

L_XΩ=λΩ,

with the set of differentiable functions on TM. Since Ω=dd_JE, we have

λdd_JE=dL_Xd_JE.

The expression λdd_JE is an exact 2-form, this implies

(10.1)

as ddJE is of maximum rank and the dimension of the manifold M is assumed to be greater than or equal to two, the relation (10.1) results in

dλ=0.

In other words, the function λ is a constant.

In the following, we will denote A_gc the set of conformal infinitesimal transformations of Ω and

Proposition 15

if and only if

where λ is a constant given by the relation .

Proof: Letwe have

where λ is a constant according to the proposition 14. As the two derivations and d_J commute, we obtain

By derivating by i_C, the two 2-forms above, C being Liouville field, we find

The application of i_S to the above equality gives

Conversely, if with, it is immediate to note that by continuing the previous calculation, we have

Proposition 16

Proof: Let . Since i_Sdd_JE=-dE, we can write successively

From the formula we get

As dd_JE is a symplectic form, we find

This means that . For a Riemannian manifold,

coincides with . Thus,

Proposition 17

Let F be the almost complex structure associated to Γ . So by noting

we have

.

Proof: Let . According to the definition of F, namely,

FJ=h and Fh=-J,

we can write successively for all Y ∈χ(TM)

Similarly, we have

for all Y ∈χ(TM), the two relations (10.2) and (10.3) lead to

.

In other words,

Proposition 18

Let .So

is equivalent to.

Proof: a) Suppose , we have

for all Y, Z∈χ(TM). Given the relation g(Y, Z)=Ω(Y, FZ), the second member of the above equality becomes

By taking into account the propositions 16 and 17, we obtain

for all Y, Z ∈χ(TM). If , we find

.

b) Suppose . We have

(10.4)

for all Y, Z ∈χ(TM). The expression of Ω in function of g is [3]

Ω (Y, Z)=g (Y, JZ)+g (JY, Z), (10.5)

for all Y, Z ∈χ(TM).. The expression (10.4) becomes

As belongs to A_J, we get

.

If , taking into account the relation (10.5) we find

.

Automorphisms of a Riemannian Manifold

Definition 4

A vector field X on a Riemannian manifold (M, E) is said to be infinitesimal automorphism of the symplectic form Ω if

L_XΩ=0.

We notice the canonical spray S of (M, E) is an infinitesimal automorphism of the symplectic form Ω.

The set of infinitesimal automorphisms of Ω forms a Lie algebra. We will denote this Lie algebra by A_g.

Proposition 19

By supposing , we have

a) if and only if X is a projectable vector field such that X ∈A_g and L_XE=0;

b) ; the horizontal elements of form a commutative ideal of .

c) The elements of are Killing fields of projectable vectors of the metric g belonging to . The dimension of is at most equal to.

Proof: It is clear that if then L_XE=0. This is a particular case of the proposition 15 for λ=0.

Let X be a projectable vector field such that L_XΩ=0 and L_XE=0, we will prove that . The fundamental form Ω is written Ω=dd_JE, as L_XΩ=0, we have

i_SL_XΩ=0.

Besides, according to the relation (7.1) and the hypothesis L_XE=0 L_XiSΩ=-dL_XE=0. According to the formula i[_S,X]=i_SL_X-L_Xi_S, we find i[_S,X]Ω=0.

Since the fundamental form Ω is symplectic, we get [S, X]=0. The vector field X being projectable by hypothesis, according to the proposition

-The result of b) is a consequence of the proposition 16 because is a particular case of Since, we have by (7.5), the horizontal elements of belong to and form a commutative ideal of , according to the theorem 1.

- Finally, the property c) is given by the proposition 18 by taking λ=0.

In addition, g defines a metric on the vertical bundle which is of n dimension. The flow Φ of a vector field is a local isometry for g. Consequently, Φ preserves the orthogonal bases of g on the vertical bundle. With the method used in theorem 3.3 of [14] vol.1 p.238, the dimension of is at most equal to .

Theorem 3

Let (M, E) be a Riemannian manifold with a nonzero constant sectional curvature. Then coincides with .

Proof: If the sectional curvature K is constant, we have [2]

where For all , we have . If K is not zero, we get

this is only possible if , that is, according to the proposition 19

a), . In other words,

.

The proposition 19 b) gives

.

Riemannian Manifolds with Regular Curvature

Definition 5

We will say that a connection Γ has a regular curvature at a point z of TM if the vector space generated by the image of the curvature R_z is of n-1 dimension.

Proposition 20

Let (M, E) be a Riemannian manifold such that

for all horizontal vector fields X, Y linearly independent, then (M, E) has a regular curvature.

Proof: The dimension of ImR is less than or equal to n-1, since the canonical field C=JS is orthogonal to ImR.

If the dimension of ImR is less than or equal to n-2, there will be a horizontal vector field Z such that JZ ⊥ ImR and S ∧ Z ≠ 0, and according to the relation (9.2), we would have

this contradicts the hypothesis.

Remark 2: The converse of the proposition 20 is not true in general.

Theorem 4

Let (M, E) be a Riemannian manifold, Γ a canonical connection of (M, E) with regular curvature. So

.

Proof: The proposition 16 gives

.

It remains to prove that is contained in ; and for that, it is simply sufficient to prove for all , according to the proposition 15.

Let as A_Γ is identical to A_h, we have

According to the formula and by (7.5), we obtain

. (12.1)

That implies

Let H be the horizontal space of Γ, H ⊕ImR is contained in the kernel of dE. By hypothesis, the curvature R is regular, so H ⊕ ImR is of 2n-1 dimension, and the kernel of dE. Therefore, H ⊕ ImR is a completely integrable distribution. According to Frobenius theorem, at each point z of TM, it passes a maximum integral manifold which is the solution of the equation (12.1). We notice that is homogeneous of degree two and that the function is continuous and null on the null section.

This imposes the initial conditions for the solutions of the equation (12.1). Hence the result

With λ is a constant.

Remark 3: A Riemannian manifold with a constant and nonzero sectional curvature has a regular curvature, but the value of λ is zero. The theorem 3 excludes the case of λ ≠ 0.

Proposition 21

Let (M, E) be a Riemannian manifold with regular curvature, R^ο=i_SR, we have the following relation:

Proof: For all we have

Theorem 5

The Lie algebra of the Killing fields contained in of dimension greater than or equal to three is semi-simple if and only if the Riemannian manifold is with regular curvature.

Proof: If the Riemannian manifold is not with regular curvature, there is the horizontal nullity space of the projectable vector fields of the curvature R which provides the commutative ideal of , according to the propositions 19 b) and 12, which is also that of . Then is not semi- simple.

It is assumed that the Riemannian manifold has a regular curvature. Let are linearly independent. As the mapping is injective on the projectable vector fields according to the proposition 11 and the definition 5. are also linearly independent.

According to the proposition 21,

As we have the following system of equations

If we have linearly independent, we have linearly independent, and we have a system of six equations as (12.2). For , we have a system of 2p equations as (12.2). Such a system of equations do not allow us to have a commutative ideal of other than zero.

We conclude that if is of dimension superior or equals to three for a Riemannian manifold with regular curvature, is semi-simple.

Corollary 1

On a Riemannian manifold (M, E) of n ≥ 2 dimension, the Lie algebra of infinitesimal isometries contained in of dimension superior or equals to three is semi-simple if and only if the horizontal nullity space of the Nijenhuis tensor of Γ is reduced to zero. In this case, the Lie algebra coincides with .

Proof: The reasoning used in the theorem 5 is based on the existence of two linearly independent elements of . So the dimension of is assumed to be superior or equals to two. We know that there is no semisimple Lie algebra of one or two dimension. Hence, the condition of the dimension of superior or equals to three.

We notice the interlocking of the algebras . So if the dimension of is superior or equals to three, for a Riemannian manifold with regular curvature, according to the theorem 5, is semi-simple. In this case, according to the theorem =and is an ideal of . Consequently = , otherwise we have a contradiction.

Remark 4: If the Riemannian manifold is flat (R=0), the horizontal elements of constitute a nonzero commutative ideal of . If the Riemannian manifold has a nonzero constant sectional curvature, according to the theorem 3, = . It is obvious that this manifold has a regular curvature.

Remark 5: The result of the theorem 5 is not true in general if we only impose the fields of projectable vectors to be those of Killing, that is to say to belong to A_g without being elements of .

The elements of A_g even projectable form in general an algebra of infinite dimension.

Examples

Example 3

We assume The spray S is written

The nonzero coefficients of the connection are:

The horizontal vector fields are generated by

The curvature R is not zero. The elements of are

The Lie algebra coincides with . It is semi-simple.

Example 4

We assume

The spray is written:

The coefficients of the connection are:

The Γ connection has a regular curvature. We note that is generated by

According to this example, we note that the Lie algebra of Killing fields A_g even projectable is of infinite dimension.

Example 5

We take

The spray S is written

The non-zero coefficients of the connection are:

Horizontal fields are generated by

The curvature R is zero.

The elements of are generated by:

The horizontal vectors fields which form the commutative ideal are {g₃,g₇,g₁₂}. The elements in , according to the proposition 19 (Table 1), are

Table 1. Multiplication table of

\We see that the derivative ideal from coincides with .The commutative ideal is generated by {e₂, e₄, e₆}.

References

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