follows that the Lie algebra gk is identi¬ed with the whole space J0 (Rm , Rm )0 ,

k

m

or equivalently with the space of k-jets of vector ¬elds on Rm at the origin that

vanish at the origin. Since each j0 X, X ∈ X(Rm ), has a canonical polynomial

k

representative, the elements of gk can also be viewed as polynomial vector ¬elds

m

‚

ai xµ ‚xi . Here the sum goes over i and all multi indices µ with 1 ¤

X= µ

|µ| ¤ k.

For technical reasons, we shall not use any summation convention in the rest of

this section and we shall use only subscripts for the indices of the space variables

x ∈ Rn , i.e. if (x1 , . . . , xn ) ∈ Rn , then x2 always means x1 .x1 , etc.

1

13.2. The tangent maps to the jet projections turn out to be jet projections

as well. Hence the sequence 13.1.(1) gives rise to the sequence of Lie algebra

homomorphisms

k’1

k

π2

πk’2

πk’1

gk ’ ’ gk’1 ’ ’ · · · ’ 1 g1 ’ 0

’’ ’’ ’m

’

m m

and we get the ¬ltration by ideals bl = ker πl (or bk if more suitable)

k

l

gk = b0 ⊃ b1 ⊃ · · · ⊃ bk’1 ⊃ bk = 0.

m

Let us de¬ne gp ‚ gk , 0 ¤ p ¤ k ’1, as the space of all homogeneous polynomial

m

vector ¬elds of degree p+1, i.e. gp = Lp+1 (Rm , Rm ). By de¬nition, gp is identi¬ed

sym

with the quotient bp /bp+1 and at the level of vector spaces we have

gk = g0 • g1 • · · · • gk’1 .

(1) m

For any two subsets L1 , L2 in a Lie algebra g we write [L1 , L2 ] for the linear

subspace generated by the brackets [l1 , l2 ] of elements l1 ∈ L1 , l2 ∈ L2 . A

decomposition g = g0 •g1 •. . . of a Lie algebra is called a grading if [gi , gj ] ‚ gi+j

for all 0 ¤ i, j < ∞. In our decomposition of gk we take gi = 0 for all i ≥ k.

m

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

13. Jet groups 129

Proposition. The Lie algebra gk of the Lie group Gk is the vector space

m m

{j0 X ; X ∈ X(Rm ), X(0) = 0} with the bracket

k

k k k

[j0 X, j0 Y ] = ’j0 [X, Y ]

(2)

and with the exponential mapping

exp(j0 X) = j0 FlX ,

k k

j0 X ∈ gk .

k

(3) 1 m

The decomposition (1) is a grading and for all indices 0 ¤ i, j < k we have

(4) [gi , gj ] = gi+j if m > 1, or if m = 1 and i = j.

Proof. For every vector ¬eld X ∈ X(Rm ), the map t ’ j0 FlX is a one-parameter

k

t

subgroup in Gk and the corresponding element in gk is

m m

FlX = j0 FlX = j0 X.

k k k

‚ ‚

‚t 0 j0 t t

‚t 0

Hence exp(t.j0 X) = j0 FlX , see 4.18. Now, let us consider vector ¬elds X, Y

k k

t

on Rm vanishing at the origin and let us write brie¬‚y a := j0 X, b := j0 Y .

k k

According to 3.16 and 4.18.(3) we have

‚2

FlX —¦ FlY —¦ FlX —¦ FlY

k k k

’2j0 [X, Y ] = 2j0 [Y, X] = j0 ’t ’t t t

‚t2

0

2

j0 FlX —¦j0 FlY —¦j0 FlX —¦j0 FlY

k k k k

‚

= ’t ’t t t

‚t2

0

2

‚

exp(’ta) —¦ exp(’tb) —¦ exp(ta) —¦ exp(tb)

= ‚t2

0

‚2

FlLb —¦ FlLa —¦ FlLb —¦ FlLa (e) = 2[j0 X, j0 Y ].

k k

= ’t ’t

t t

‚t2

0

So we have proved formulas (2) and (3). For all polynomial vector ¬elds a =

‚ ‚

ai x» ‚xi , b = bi xµ ‚xi ∈ gk the coordinate formula for the Lie bracket of

µ m

»

vector ¬elds, see 3.4, and formula (2) imply

‚

ci xγ

[a, b] = where

γ

‚xi

i,γ

(5)

»j bj ai ’ µj aj bi .

ci =

γ µ» »µ

1¤j¤m

µ+»’1j =γ

i

Here 1j means the multi index ± with ±i = δj and there is no implicit summation

in the brackets. This formula shows that (1) is a grading. Let us evaluate

‚ ‚ ‚

x± , xβ = (±i ’ βi )x±+β’1i

‚xi ‚xi ‚xi

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

130 Chapter IV. Jets and natural bundles

and consider two degrees p, q, 0 ¤ p + q ¤ k ’ 1. If p = q then for every γ with

|γ| = p + q + 1 and for every index 1 ¤ i ¤ m, we are able to ¬nd some ± and

β with |±| = p + 1, |β| = q + 1 and ± + β = γ + 1i , βi = ±i . Since the vector

‚

¬elds xγ ‚xi , 1 ¤ i ¤ m, |γ| = p + q + 1, form a linear base of the homogeneous

component gp+q , we get equality (4). If p = q, then the above consideration fails

only in the case γi = |γ|. But if m > 1, then we can take the bracket

[xj xp ‚xi , xq+1 ‚xj ] = xp+q+1 ‚xi ’ (q + 1)xp+q xj ‚xj

‚ ‚ ‚ ‚

j = i.

i i i i

Since the second summand belongs to [gp , gq ] this completes the proof.

13.3. Let us recall some general concepts. The commutator of elements a1 , a2

of a Lie group G is the element a1 a2 a’1 a’1 ∈ G. The closed subgroup K(S1 , S2 )

1 2

generated by all commutators of elements s1 ∈ S1 ‚ G, s2 ∈ S2 ‚ G is called

the commutator of the subsets S1 and S2 . In particular, G := K(G, G) is called

the derived group of the Lie group G. We get two sequences of closed subgroups

G(0) = G = G(0)

G(n) = (G(n’1) ) n∈N

n ∈ N.

G(n) = K(G, G(n’1) )

A Lie group G is called solvable if G(n) = {e} and nilpotent if G(n) = {e} for

some n ∈ N. Since always G(n) ⊃ G(n) , every nilpotent Lie group is solvable.

The Lie bracket determines in each Lie algebra g the following two sequences

of Lie subalgebras

g = g(0) = g(0)

g(n) = [g(n’1) , g(n’1) ] n∈N

n ∈ N.

g(n) = [g, g(n’1) ]

The sequence g(n) is called the descending central sequence of g. A Lie algebra g

is called solvable if g(n) = 0 and nilpotent if g(n) = 0 for some n ∈ N, respectively.

Every nilpotent Lie algebra is solvable. If b is an ideal in g(n) such that the factor

g(n) /b is commutative, then b ⊃ g(n+1) . Consequently Lie algebra g is solvable

if and only if there is a sequence of subalgebras g = b0 ⊃ b1 ⊃ · · · ⊃ bl = 0

where bk+1 ‚ bk is an ideal, 0 ¤ k < l, and all factors bk /bk+1 are commutative.

Proposition. [Naymark, 76, p. 516] A connected Lie group is solvable, or nilpo-

tent if and only if its Lie algebra is solvable, or nilpotent, respectively.

13.4. Let i : GL(m) ’ Gk be the map transforming every matrix A ∈ GL(m)

m

into the r-jet at zero of the linear isomorphism x ’ A(x), x ∈ Rm . This is a

splitting of the short exact sequence of Lie groups

w

w w w

k

u

π1

k

G1

(1) e B1 Gm e

m

i

so that we have the situation of 5.16.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

13. Jet groups 131

Proposition. The Lie group Gk is the semidirect product GL(m) B1 with

m

the action of GL(m) on B1 given by (1). The normal subgroup B1 is connected,

simply connected and nilpotent. The exponential map exp : b1 ’ B1 is a global

di¬eomorphism.

Proof. Since the normal subgroup B1 is di¬eomorphic to a Euclidean space,

see 13.1, it is connected and simply connected. Hence B1 is also nilpotent, for

its Lie algebra b1 is nilpotent, see 13.2.(4) and 13.3. By a general theorem, see

[Naymark, 76, p. 516], the exponential map of a connected and simply connected

solvable Lie group is a global di¬eomorphism. Since our group is even nilpotent

this also follows from the Baker-Campbell-Hausdor¬ formula, see 4.29.

13.5. We shall need some very basic concepts from representation theory. A

representation π of a Lie group G on a ¬nite dimensional vector space V is a

Lie group homomorphism π : G ’ GL(V ). Analogously, a representation of

a Lie algebra g on V is a Lie algebra homomorphism g ’ gl(V ). For every

representation π : G ’ GL(V ) of a Lie group, the tangent map at the identity

T π : g ’ gl(V ) is a representation of its Lie algebra, cf. 4.24.

Given two representations π1 on V1 and π2 on V2 of a Lie group G, or a Lie

algebra g, a linear map f : V1 ’ V2 is called a G-module or g-module homo-

morphism, if f (π1 (a)(x)) = π2 (a)(f (x)) for all a ∈ G or a ∈ g and all x ∈ V ,

respectively. We say that the representations π1 and π2 are equivalent, if there

is a G-module isomorphism or g-module isomorphism f : V1 ’ V2 , respectively.

A linear subspace W ‚ V in the representation space V is called invariant if

π(a)(W ) ‚ W for all a ∈ G (or a ∈ g) and π is called irreducible if there is no

proper invariant subspace W ‚ V . A representation π is said to be completely

reducible if V decomposes into a direct sum of irreducible invariant subspaces.

A decomposition of a completely reducible representation is unique up to the

ordering and equivalences. A classical result reads that the standard action of

GL(V ) on every invariant linear subspace of —p V ——q V — is completely reducible

for each p and q, see e.g. [Boerner, 67].

A representation π of a connected Lie group G is irreducible, or completely

reducible if and only if the induced representation T π of its Lie algebra g is

irreducible, or completely reducible, respectively, see [Naymark, 76, p. 346].

A representation π : GL(m) ’ GL(V ) is said to have homogeneous degree r if

π(t.idRm ) = tr idV for all t ∈ R \ {0}. Obviously, two irreducible representations

with di¬erent homogeneous degrees cannot be equivalent.

13.6. The GL(m)-module structure on b1 ‚ gk . Since B1 ‚ Gk is a

m m

normal subgroup, the corresponding subalgebra b1 = g1 • · · · • gk’1 is an ideal.

The (lower case) adjoint action ad of g0 = gl(m) on b1 and the adjoint action

Ad of GL(m) = G1 on b1 determine structures of a g0 -module and a GL(m)-

m

module on b1 . As we proved in 13.2, all homogeneous components gr ‚ b1 are

g0 -submodules.

Let us consider the canonical volume form ω = dx1 § · · · § dxm on Rm and

recall that for every vector ¬eld X on Rm its divergence is a function divX on

Rm de¬ned by LX ω = (divX)ω.

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

132 Chapter IV. Jets and natural bundles

In coordinates we have div( ξ i ‚/‚xi ) = ‚ξ i /‚xi and so every k-jet j0 X ∈

k

k’1

gk determines the (k ’ 1)-jet j0 (divX). Hence we can de¬ne div(j0 X) = k

m

k’1

j0 (divX) for all j0 X ∈ gk . If X is the canonical polynomial representative

k

m

k r

of j0 X of degree k, then divX is a polynomial of degree k ’ 1. Let C1 ‚ gr be

k

the subspace of all elements j0 X ∈ gr with divergence zero. By de¬nition,

div[X, Y ]ω = L[X,Y ] ω = LX LY ω ’ LY LX ω

(1)

= (X(divY ) ’ Y (divX))ω.

r

Since every linear vector ¬eld X ∈ g0 has constant divergence, C1 ‚ gr is a

gl(m)-submodule. In coordinates,

‚

ai x» r

»i ai x»’1i = 0,

∈ C1 if and only if

» »

‚xi

i,»

i.e. i (µi + 1)ai i = 0 for each µ with |µ| = r.

µ+1

‚

Further, let us notice that the Lie bracket of the ¬eld Y0 = j xj ‚xj with

r

any linear ¬eld X ∈ g0 is zero. Hence, also the subspace C2 of all vector ¬elds

Y ∈ gr of the form Y = f Y0 with an arbitrary polynomial f = f± x± of degree

r is g0 -invariant. Indeed, it holds [X, f Y0 ] = ’(Xf )Y0 .

Since div(f Y0 ) = j (±j + 1)f± x± , we see that gr = C1 • C2 . In coordinates,

r r

r

we have linear generators of C2

X± = x± ( ‚

|±| = r,

(2) xk ‚xk ),

k

r

and if m > 1 then there are linear generators of C1

|±| = r,

X±,k = x± (±k + 1)x1 ‚x1 ’ (±1 + 1)xk ‚xk ,

‚ ‚

k = 2, . . . , m

(3)

Yµ,k = xµ ‚xk ,

‚

k = 1, . . . , m, |µ| = r + 1, µk = 0.

k’1 k’1

1 2 1 2

We shall write C1 = C1 • C1 • · · · • C1 and C2 = C2 • C2 • · · · • C2 .

According to (1), C1 ‚ b1 is a Lie subalgebra. Since for smooth functions f , g on

Rm we have [f X, gX] = (g(Xf ) + f (Xg))X, C2 ‚ b1 is a Lie subalgebra as well.

So we have got a decomposition b1 = C1 • C2 . According to the general theory

this is also a decomposition into G1 + -submodules, but as all the spaces Cj are

r

m

invariant with respect to the adjoint action of any exchange of two coordinates,

the latter spaces are even GL(m)-submodules.

r r

Proposition. If m > 1, then the GL(m)-submodules C1 , C2 in gr , 1 ¤ r ¤

r

k ’ 1, are irreducible and inequivalent. For m = 1, C1 = 0, 1 ¤ r ¤ k ’ 1, and

r

all C2 are irreducible inequivalent GL(1)-modules.

Proof. Assume ¬rst m > 1. A reader familiar with linear representation the-

r

ory could verify that the modules C2 are equivalent to the irreducible modules

Electronic edition of: Natural Operations in Differential Geometry, Springer-Verlag, 1993

13. Jet groups 133

det’r C(r,r,...,r,0) , where the symbol C(r,...,r,0) corresponds to the Young™s dia-

m m

gram (r, . . . , r, 0), while C1 are equivalent to det’(r+1) C(r+2,r+1,...,r+1,0) , see e.g.

r m

[Dieudonn´, Carrell, 71]. We shall present an elementary proof of the proposi-

e

tion.

r

Let us ¬rst discuss the modules C2 . Consider one of the linear generators X±

‚

de¬ned in (2) and a linear vector ¬eld xi ‚xj ∈ gl(m). We have

[’xi ‚xj , x± ( xk ‚xk )] = ±j xi x±’1j

‚ ‚ ‚

(4) (xk ‚xk ).

k k

If j = i, we get a scalar multiplication, but in all other cases the index ±j

decreases while ±i increases by one and if ±j = 0, then the bracket is zero.

Hence an iterated action of suitable linear vector ¬elds on an arbitrary linear

combination of the base elements X± yields one of the base elements. Further,

r

formula (4) implies that the submodule generated by any X± is the whole C2 .

r

This proves the irreducibility of the GL(m)-modules C2 .

r

In a similar way we shall prove the irreducibility of C1 . Let us evaluate the

‚

action of Zi,j = xi ‚xj on the linear generators X±,k , Yµ,k .

j

[’Zi,j , X±,k ] = (±k + 1)(±j + δ1 )x±+11 +1i ’1j ‚x1 ’

‚

j

’ (±1 + 1)(±j + δk )x±+1k +1i ’1j ‚xk ’

‚

’ δ1 (±k + 1)x±+11 ‚xj + δk (±1 + 1)x±+1k ‚xj

i i

‚ ‚

[’Zi,j , Yµ,k ] = µj xµ’1j +1i ‚xk ’ δk xµ ‚xj .

i

‚ ‚