(i) Cm = 0 and m > 1.

±+1

(1) ±+1

±

(ii) C1 = 0; Cj = 0 whenever ±j = 0 and 1 ¤ j ¤ m; C1 1 = Cj j (no

summation) for all |±| = r ’ 1, 1 ¤ j ¤ m.

(m+1) ±

(iii) Cm+n = 0, n > 1, and Cj = 0 whenever j ¤ m.

±+1j

±+1

(m+1) ±

(iv) Cm+1 = 0; Cj = 0 if j ¤ m or ±j = 0; and Cm+1m+1 = Cj (no

summation) for all |±| = r ’ 1, j ≥ m + 1.

±

Proof. Let C be a non-zero form on gr’1 with coordinates Cj in the canonical

basis of Rm+n ’ Rm . Let us consider a matrix A ∈ GL(m + n) whose ¬rst row

consists of arbitrary real parameters a1 = t1 = 0, a1 = t2 , . . . , a1 = tm , a1 = 0

m

1 2 j

i

for j > m, and let all the other elements be like in the unit matrix. Let aj be the

˜

’1

elements of the inverse matrix A . If we perform this linear transformation,

we get a new coordinate expression of C, in particular

¯ (1)

Cj = a11 · · · a1r Cs1 ...ir as .

i

(1) ˜j

i i

Hence we get

1

¯ (1)

C1 = ti1 · · · tir C11 ...ir

i

(2)

t1

tj i1 ...ir

¯ (1) Cj1 ...ir ’

i

Cj = ti1 · · · tir for 1 < j ¤ m.

(3) C

t1 1

¯ (1) ±

Formula (2) implies that either we can obtain C1 = 0 or C1 = 0 for all multi

indices ±, |±| = r. Let us assume m > 1 and try to get condition (i). According

to (3), if (i) does not hold after performing any of our transformations, then

the expression on the right hand side of (3) has to be identically zero for all

±

values of the parameters and this implies Cj = 0 whenever ±j = 0, |±| = r,

±+1

±+1

and C1 1 = Cj j for all |±| = r ’ 1, 1 ¤ j ¤ m. Hence we can summarize:

¯±

either (i) can be obtained, or (ii) holds, or Cj = 0 for all 1 ¤ j ¤ m, |±| = r, in

suitable a¬ne coordinates.

Analogously, let us take a matrix A ∈ GL(m + n) whose (m + 1)-st row

consists of real parameters t1 , . . . , tm+n , tm+1 = 0 and let the other elements be

like in the unit matrix. The new coordinates of C are obtained as above

1

¯ (m+1) i1 ...i

Cm+1 =ti1 · · · tir Cm+1r

(4)

tm+1

tj

¯ (m+1) =ti · · · ti Cj1 ...ir ’

i i1 ...i

(5) Cj Cm+1r .

1 r

tm+1

±

Now we may assume Cj = 0 whenever 1 ¤ j ¤ m, for if not then (i) or (ii) could

(m+1)

be obtained. As before, either there is a basis relative to which Cm+1 = 0 or

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196 Chapter V. Finite order theorems

±

Cm+1 = 0 for all |±| = r. Further, according to (5) either we can get (iii) or

±+1

±+1

±

Cj = 0 whenever ±j = 0, and Cm+1m+1 = Cj j , for all |±| = r ’ 1, j ≥ m + 1.

Therefore if both (iii) and (iv) do not hold after arbitrary transformations, then

±

all Cj have to be zero, but this is contradictory to the fact that C is non-zero.

21.6. Lemma. Let p, q be two degrees with p + q = r ’ 1 > 0 and p ≥ q ≥ 0.

Let m > 1 and n > 1 or n = 0, and let hp , hq be subspaces of gp , gq . Let C be a

±

non-zero linear form on gr’1 and suppose [hp , hq ] ‚ ker C. If Ci , 1 ¤ i ¤ m + n,

|±| = r, is a coordinate expression of C satisfying one of the conditions 21.5.(i)-

(iv), then

2m ’ 2, if 21.5.(i) holds

±

2m, if 21.5.(ii) holds and q > 0

m, if 21.5.(ii) holds and q = 0

codim hp + codim hq ≥

2n ’ 2, if 21.5.(iii) holds

2n, if 21.5.(iv) holds and q > 0

n, if 21.5.(iv) holds and q = 0.

Proof. De¬ne a bilinear form

f : gp — gq ’ R f (a, b) = C([a, b]) .

By our assumptions f (hp , hq ) = {0}. Hence by lemma 21.4 it su¬ces to prove

that the codimension of the f -annihilator of gq in gp has the above lower bounds.

Let h0 be this annihilator and consider elements a ∈ h0 , b ∈ gq . We get

Ci ([a, b])i = 0.

±

C([a, b]) = ±

1¤i¤m+n

|±|=r

Using formula for the bracket 13.2.(5) we obtain

µ+»’1j

»j bj ai ’ µj aj bi

0= Ci µ» »µ

1¤i,j¤m+n

|µ|=q+1

|»|=p+1

µ+»’1j µ+»’1i

µi ai bj .

»j ’ Cj

= Ci »µ

1¤i,j¤m+n

|µ|=q+1

|»|=p+1

Since b ∈ gq is arbitrary, we have got a system of linear equations for the

annihilator h0 containing one equation for each couple (j, µ), where 1 ¤ j ¤

m + n, |µ| = q + 1 and µi = 0 whenever i > m and j ¤ m. The (j, µ)-equation

reads

µ+»’1j µ+»’1i

µi ai = 0.

»j ’ Cj

(1) Ci »

1¤i¤m+n

|»|=p+1

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

21. Actions of jet groups 197

A lower bound of the codimension of h0 is given by any number of linearly

independent (j, µ)-equations and we have to discuss this separately for the cases

21.5.(i)“(iv).

(1)

Let us ¬rst assume that 21.5.(i) holds, i.e. Cm = 0, m > 1. We denote by Es

the (s, (1))-equation, 1 ¤ s < m and by Fk the (m, (1)+1k )-equation, 1 ¤ k < m

(note that if q = 0 then (1) + 1k = 1k ). We claim that this subsystem is of full

rank. In order to verify this, consider a linear combination

m’1 m’1

s

b k Fk = 0 as , bk ∈ R.

a Es +

s=1 k=1

From (1) we get

m’1

(1)+»’1s (1)+»’1i 1

δi (q + 1) as +

»s ’ Cs

(2) Ci

s=1

1¤i¤m+n

|»|=p+1

m’1

(1)+1k +»’1m

»m ’ Cm k +»’1i (δi q + δi ) bk ai = 0.

(1)+1 1 k

+ Ci »

k=1

Hence all the coe¬cients at the variables ai with 1 ¤ i ¤ m + n, » = p + 1, and

»

»j = 0 whenever j > m and i ¤ m, have to vanish. Therefore, we get equations

on reals as , bk , whenever we choose i and ». We have to show that all these

reals are zero.

(1)

First, let us substitute » = (1) and i = m. Then (2) implies Cm (p + 1)a1 = 0

and consequently a1 = 0. Now we choose » = (1) + 1v , i = m, with 1 < v < m,

(1)

and we get Cm av = 0 so that as = 0 for 1 ¤ s ¤ m ’ 1. Further, take » = (1)

(1)

and 1 < i < m to obtain ’Cm bi = 0. Finally, the choice i = 1 and » = (1)

(1)

leads to ’Cm (q + 1)b1 = 0. In this way, we have proved that the chosen 2m ’ 2

equations Es and Fk are independent and this implies the ¬rst lower bound in

21.6.

Now suppose 21.5.(ii) takes place and let us denote Es the (s, (1))-equation,

1 ¤ s ¤ m, and if q > 0, then Fk will be the (m, (1) + 1m + 1k )-equation,

m m

1 ¤ k ¤ m. As before, we assume s=1 as Es + k=1 bk Fk = 0 for some reals

as and bk and we compare the coe¬cients at ai to show that all these reals are

»

zero. But before doing this, we can simplify all (j, µ)-equations with 1 ¤ j ¤ m

using the relations from 21.5.(ii). Indeed, (1) reduces to

µ+»’1i

(»j ’ µi )ai + R = 0.

Cj »

1¤i¤m

|»|=p+1

where R involves all terms with indices i > m. Consequently Es and Fk have

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

198 Chapter V. Finite order theorems

the forms

(1)+»’1i

(»s ’ δi (q + 1))ai + R = 0

1

Cs »

1¤i¤m

|»|=p+1

Cm k +1m +»’1i (»m ’ δi (q ’ 1) ’ δi ’ δi )ai + R = 0.

(1)+1 1 k m

»

1¤i¤m

|»|=p+1

Assume ¬rst q > 0. If we choose 1 < i ¤ m, » = (1), then the variables ai do

»

not appear in the equations Es at all. Hence the choice i = m, » = (1) gives

(1)+1 (1)

(see 21.5.(ii)) 0 = ’2Cm m bm = ’2C1 bm ; and 1 < i < m, » = (1) now

(1)+1

yields ’Cm m bi = 0. Hence bi = 0 for all 1 < i ¤ m. Further, we take i = m,

» = (1) + 1v + 1m , v = m (note p ≥ q > 0), so that all the coe¬cients in F1 are

(1)

zero. In particular, v = 1 implies C1 a1 = 0 so that a1 = 0. Now, if 1 < v < m,

(1)+1v v (1)

a = C1 av = 0 and what remains are am and b1 , only. Taken

then Cv

(1) (1)+1 (1)

» = (1), i = 1, we see 0 = ’(q + 1)Cm am ’ qCm m b1 = C1 b1 and, ¬nally,

(1)+1

the choice i = m and » = (1) + 1m + 1m gives Cm m 2am = 0. This completes

the proof of the second lower bound in 21.6.

But if q = 0 and 21.5.(ii) holds, we can perform the above procedure after

forgetting all the equations Fk which are not de¬ned. We have only to notice

p + q = r ’ 1 > 0, so that |»| = p + 1 = r ≥ 2.

If n > 1, then the remaining three parts of the proof are complete recapitu-

lations of the above ones. This becomes clear if we notice, that we have used

± ±

neither any information on Cj , j > m, nor the fact that Cj = 0 if j ¤ m and

±i = 0 for some i > m. That is why we can go step by step through the above

proof on replacing 1 or m by m + 1 or m + n, respectively.

If n = 0, then neither 21.5.(iii) nor 21.5.(iv) can hold.

21.7. Proposition. Let h be a subalgebra of gr , m ≥ 1, n ≥ 0, r ≥ 2, which

m,n

does not contain br . Then

r’1

1

codim h ≥ (r ’ 1).

(1)

2

Moreover, if m > 1, n > 1, then

codim h ≥ min{r(m ’ 1), (r ’ 1)m, r(n ’ 1), (r ’ 1)n}

(2)

and if m > 1, n = 0, then

codim h ≥ min{r(m ’ 1), (r ’ 1)m}.

(3)

Proof. In 21.3 we deduced that we may suppose h is a subalgebra with grading

h = h0 • · · · • hr’1 , hi ‚ gi , hr’1 = gr’1 , and we proved the lower bound

(1). Let us assume m > 1, n = 0 and choose a non-zero form C on gr’1 with

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

21. Actions of jet groups 199

ker C ⊃ hr’1 . Then we know [hj , hr’j’1 ] ‚ hr’1 ‚ ker C and by lemma 21.5

there is a suitable coordinate expression of C satisfying one of the conditions

21.5.(i), 21.5.(ii). Therefore we can apply lemma 21.6.

±

Assume ¬rst Ci satis¬es 21.5.(i). Then for all j

codim hj + codim hr’j’1 ≥ 2m ’ 2

and consequently

r’1

codim hj ≥ r(m ’ 1).

codim h =

j=0

If 21.5.(ii) holds, then

codim h0 + codim hr’1 ≥ m

codim hj + codim hr’j’1 ≥ 2m

for 1 ¤ j ¤ r ’ 2, so that codim h ≥ m + (r ’ 2)m = (r ’ 1)m. This completes

the proof of (3) and analogous considerations lead to the estimate (2) if n > 1

and the coordinate expression of C satis¬es 21.5.(iii) or 21.5.(iv).

21.8. Examples.

1. Let h1 ‚ gr , m > 1, be de¬ned by

m

‚

; aj = 0 for j = 2, . . . , m, 1 ¤ |(1)| ¤ r}.

h1 = {ai x»

»

‚xi (1)

One sees immediately that the linear subspace h1 consists just of polynomial

vector ¬elds of degree r tangent to the line x2 = x3 = · · · = xm = 0, so that

h1 clearly is a Lie subalgebra in gr of codimension r(m ’ 1). Consider now the

m

subalgebra h ‚ gr consisting of projectable polynomial vector ¬elds of degree

m,n

r over polynomial vector ¬elds from h1 . This is a subalgebra of codimension

r(m ’ 1) in gr .

m,n

2. Consider the algebra h2 ‚ gr , n > 1, de¬ned analogously to the subal-

m+n

gebra h1

‚

; aj

h2 = {ai x» = 0 for 1 ¤ j ¤ m + n, j = m + 1, 1 ¤ |(m + 1)| ¤ r}

»

‚xi (m+1)

and de¬ne h = h2 © gr . Since every polynomial vector ¬eld in gr is tangent

m,n m,n

to the ¬ber over zero, this clearly is a Lie subalgebra with coordinate description

‚

; aj

h = {ai x» = 0 for m + 1 < j ¤ m + n, 1 ¤ |(m + 1)| ¤ r}

»

‚xi (m+1)

and codimension r(n ’ 1).

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

200 Chapter V. Finite order theorems

3. Let us recall that the divergence div X of a polynomial vector ¬eld X ∈ grm

r’1 ‚

can be viewed as the jet j0 (div X), see 13.6. So for an element a = ai x» ‚xi

»

we have

»i ai x»’1i .

div a = »

1¤i¤m

1¤|»|¤r

Let M be the line in Rm , m > 1, de¬ned by x2 = x3 = · · · = xm = 0 and

denote by h3 the linear subspace in g1 • · · · • gr’1 (note g0 is missing!)

‚ ‚

h3 = {ai x» ∈ g1 • · · · • gr’1 ; div(ai x» ) = 0}.

» » M

‚xi ‚xi

Of course, h3 is not a Lie subalgebra in gr . Let us further consider the Lie

m

subalgebra h4 ‚ gr’1 consisting of all polynomial vector ¬elds without absolute

m

’1

r

h4 ‚ gr and let us

terms and tangent to M , cf. example 1. Let h5 = πr’1 m

de¬ne a linear subspace

h6 = (h5 © g0 ) • (h3 © h5 ).

First we claim that h3 © h5 is a subalgebra. Indeed, if X, Y ∈ h3 © h5 , then

either the degree of both of them is less then r or their bracket is zero. But in

the ¬rst case, X and Y are tangent to M and their divergences are zero on M ,

so that 13.6.(1) implies div([X, Y ])|M = 0.

Now, consider a polynomial vector ¬eld X from the subalgebra h5 © g0 and a

¬eld Y ∈ h3 © h5 . Since every ¬eld from g0 has constant divergence everywhere

and X is tangent to M , 13.6.(1) implies div([X, Y ])|M = 0. So we have proved