Journal of Lie Theory Volume 11 (2001) 129–134 ◯^{C}2001 Heldermann Verlag

Ideals of finite codimension in contact Lie algebra

Mohammed Benalili and Azzedine Lansari

Communicated by J. Faraut

Abstract. Ideals of germs of vector fields leaving 0 fixed in R^{2n+1} , of finite codimension in contact Lie algebra contain the ideal of germs infinitely flat at 0. We give an application.

1991 Mathematics Subject Classification: 17B66 Keywords and phrases: Contact Lie algebra, Jacobi bracket, pseudogroup action.

1. Introduction

In this paper, we characterize ideals of germs of vector fields X with X(0) = 0 in R^{2n+1}, of finite codimension in a contact Lie algebra. Let α = dx_{2n+1} + \frac{1}{2} ∑_{i=1}^{n}(x_{i}dx_{n+i} − x_{n+i}dx_{i}) denote the canonical contact form on R^{2n+1}, and let χ_{α} denote the contact Lie algebra of germs of vector fields leaving 0 fixed. Also, denote by χ_{α}^{∞} the ideal of χ_{α}, of infinitely flat germs at 0. Finally, let F denote the space of germs of smooth functions. The main theorem is:

Theorem 1.1. Let I be a finite codimension ideal of χ_{α}. Then I contains χ_{α}^{∞} .

As a consequence of this result we obtain a reduction of the action of the group of germs of origin preserving contact diffeomorphisms to the action of the group of infinite jets ( at the origin ) on a Lie group; we pass to the linear version. We encounter such an action when we deal with the theory of natural fiber bundles ( see [1] , [2] [3]). In [2] and [3], the authors used Borel lemma and Whitney extension theorem to prove the reduction of the above action. In the category of manifolds endowed with geometric structures, Borel lemma and Whitney extension theorem fail. The interest of finding appropriate methods for this cases was raised in [2] . Our result applies in the category of contact manifolds.

2. Fundamental lemma

The proof of Theorem1 is based on the following lemma.

ISSN 0949–5932 / $2.50 ◯^{C} Heldermann Verlag Lemma 2.1. Let V be a space of finite codimension in a real linear space E and ψ an endomorphism of E such that:

1) ψ(V) ⊂ V

2) for any b ∈ R, ψ + bI is onto in E.

3) for any b, c ∈ R with b^{2} − 4c < 0, ψ^{2} + bψ + cI is onto in E. Then V = E.

Proof. Since the subspace V of E is invariant by the endomorphism ψ: E → E, there exists a unique endomorphism ψ: E/V → E/V such that πoψ =ψ oπ, where π: E → E/V denotes the canonical projection. If V is of non zero finite codimension in E, the quotient space E/V is of non zero finite dimension and ψ admits an eigenvalue. If this eigenvalue is real, there exists b ∈ R such that ψ +bI_{E/V} is not onto in E/V. So ψ + bI_{E} cannot be onto in E, since if it is this contradicts the property πoψ =ψ oπ. And if this eigenvalue is complex, there are b and c ∈ R such that b^{2} − 4c < 0 and ψ^{2} +b ψ +cI_{E/V} is not onto in E/V . The same argument as above shows that ψ^{2} + bψ + cI_{E} cannot be onto in E. This proves Lemma 2.

3. Preparatory lemmas

We know that φ: χ_{α} → F defined by φ (X) = i_{X}α ( where i_{X}α denotes the interior product by X) is an isomorphism. φ induces a Lie algebra structure on F, of which the bracket operation is called a Jacobi bracket and will be denoted by { }, it has the following representation

{f, g} = ∑_{i=1}^{n}(δf/δx_{i}.δg/δx_{n+i} − δf/δx_{n+i}.δg/δx_{i}) − g∂f/δx_{2n+1} + f∂g/∂x_{2n+1}

where δ/δx_{i} = ∂/∂x_{i} + \frac{1}{2}x_{n+i}∂/∂x_{2n+1}, δ/δx_{n+i} = ∂/∂x_{n+i} − \frac{1}{2}x_{i}∂/∂x_{2n+1}.

Let k(t) be a bounded continuous function for all 0 ≤ t ≤ 1 and x = (x_{1}, …, x_{2n+1}).

Lemma 3.1. For any b ∈ R and h ∈ F with J_{o}^{∞}h = 0, the germ of the function g defined by the integral

g(x) = ∫_{0}^{1} t^{b}k(t)h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt  (1)

is smooth and infinitely flat at 0.

Proof. Since h is infinitely flat at 0, for any positive integer m, there exist constants δ > 0 and M > 0 such that for all x with |x| ≤ δ, we have

|h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1}))| ≤ Mδ^{m}t^{\frac{1}{2} m}.

By hypothesis there exists a constant c such |k(t)| ≤ c for any 0 ≤ t ≤ 1; then

\begin{matrix}|k(t)t^{b}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})| ≤ \\ cMt^{b+ \frac{m}{2}} = f_{m}(t). \end{matrix}

Since m is arbitrary, we choose m_{o} such that b + \frac{m_{o}}{2} − 1 > 0; consequently the integral defining the function g converges uniformly in a neighborhood of 0. Now, we shall show that the function g is C^{∞} and infinitely flat at 0. For any multiindice α ∈ N^{2n+1} we have

D_{x}^{α}(h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})) = t^{|α|/2}(D_{x}^{α}h)(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})

where |α| means the length of α. Since D_{x}^{α}h is infinitely flat at 0, there exist constants δ > 0 and M > 0 such that:

|(D_{x}^{α}h)(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})| ≤ Mt^{1/2(m+|α|)}δ^{m}

provided that |x| ≤ δ. Since m is arbitrary, we choose m_{o} such that |α| + m_{o} + b − 1 > 0. Consequently, for any multi-indice α ∈ N^{2n+1},

∫_{0}^{1} t^{b}k(t)D_{x}^{α}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt

converges uniformly in a neighborhood of 0, so the function g defined by (1) is smooth. Now since D_{x}^{α}h(0) = 0 and D_{x}^{α}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1}) converges uniformly to D_{x}^{α}h(0) with respect to x, we pass to the limit and obtain D_{x}^{α}g(0) = 0 . This proves Lemma 3.

Let z be the (2n + 1)^{th} coordinate x_{2n+1}.

Lemma 3.2. For any b ∈ R and h ∈ F with J_{0}^{∞}h = 0, there exists g ∈ F with J_{0}^{∞}g = 0 such that

{z, g} + bg = h  (2)

Proof. Equation (2) is equivalent to

\frac{1}{2} ∑_{i=1}^{n}(x_{i}∂g/∂x_{i} + x_{n+i}∂g/∂x_{n+i}) + x_{2n+1}∂g/∂x_{2n+1} + (b − 1)g = h

Consider the function g defined by

g(x) = ∫_{0}^{1} t^{b−2}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt.

By Lemma 3, g(x) is smooth and infinitely flat at 0. Since

t\frac{dh}{dt}(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1}) = \frac{1}{2} ∑_{i=1}^{n} t^{1/2}(x_{i}∂h/∂x_{i} + x_{n+i}∂h/∂x_{n+i}) + tx_{2n+1}∂h/∂x_{2n+1}

it follows that

\begin{matrix}\frac{1}{2} ∑_{i=1}^{n}(x_{i}∂g/∂x_{i} + x_{n+i}∂g/∂x_{n+i}) + x_{2n+1}∂g/∂x_{2n+1} = \\ ∫_{0}^{1} t^{b−1} \frac{dh}{dt}(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt \end{matrix}  (3)

= [t^{b−1}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})]_{0}^{1} − (b − 1) ∫_{0}^{1} t^{b−2}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt.

The infinitely flatness of h at 0 leads us to write for any b ∈ R

lim_{t→0} t^{b−1}h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1}) = 0.

Hence

\frac{1}{2} ∑_{i=1}^{n}(x_{i}∂g/∂x_{i} + x_{n+i}∂g/∂x_{n+i}) + x_{2n+1}∂g/∂x_{2n+1} = h − (b − 1)g(x).

This checks Lemma 4.

Lemma 3.3. For any b, c ∈ R with b^{2} − 4c < 0 and any h ∈ F with J_{0}^{∞}h = 0, there exists g ∈ F with J_{0}^{∞}g = 0 such that

{z, {z, g}} + b {z, g} + cg = h.  (4)

Proof. Similarly as in Lemma 4, we must find g ∈ F with J_{0}^{∞}g = 0 satisfying equation(4). Let

k(t) = \frac{−2t^{−b/2}}{\sqrt{4c − b^{2}}} sin(\sqrt{4c − b^{2}}/2) log |t| .

Note that k(t) is the solution of the Cauchy problem

\{  \begin{matrix}t^{2}k′′(t) + (1 + b)tk′(t) + ck(t) = 0 \\ k′(1) = −1 k(1) = 0. \end{matrix}  (5)

Consider the function g defined by the integral

g(x) = ∫_{o}^{1} t^{b−2}k(t)h(t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1})dt.  (6)

For simplicity we write (.) for (t^{1/2}x_{1}, ..., t^{1/2}x_{2n}, tx_{2n+1}). By Lemma 3 g(x) given by (6) is smooth and infinitely flat at 0. On the other hand from formula (3) we obtain

\begin{matrix}\{z, g\} + g = \frac{1}{2} ∑_{i=1}^{n}(x_{i}∂g/∂x_{i} + x_{n+i}∂g/∂x_{n+i}) + x_{2n+1}∂g/∂x_{2n+1} \\ = ∫_{o}^{1} t^{b−1}k(t)\frac{dh}{dt}(.)dt, \end{matrix}

and

\{z, \{z, g\}\} = g + ∫_{o}^{1} t^{b−1}k(t)\frac{d}{dt}(t\frac{dh}{dt} − 2h) (.) dt.

An integration by parts gives us

\{z, \{z, g\}\} = [t^{b−1}k(t)(t\frac{dh}{dt} − 2h) (.)]_{0}^{1} +
∫_{0}^{1} t^{b−2} ((2b − 1)k(t) + 2tk′(t)) h(.)dt−
∫_{0}^{1} t^{b−2} ((b − 1)k(t) + tk′(t)) t\frac{dh}{dt}(.)dt,

and

\begin{matrix}∫_{0}^{1} t^{b−1}((b − 1)k(t) + tk′(t))\frac{dh}{dt}(.)dt = \\ [t^{b−1}((b − 1)k(t) + tk′(t))h(.)]_{0}^{1} − ∫_{0}^{1} t^{b−2} (t^{2}k′′(t) + (2b − 1)tk′(t) + (b − 1)^{2}k(t)) h(.)dt. \end{matrix}

Then

\begin{matrix}\{z, \{z, g\}\} = h(x) + ∫_{0}^{1} t^{b−2} (t^{2}k′′(t) + (2b + 1)tk′(t) + b^{2}k(t)) h(.)dt \\ = h(x) − b \{z, g\} − cg + ∫_{0}^{1} t^{b−2} (t^{2}k′′(t) + (b + 1)tk′(t) + ck(t)) h(.)dt. \end{matrix}

Since k(t) is solution of (5), Lemma 5 is proved.

4. Proof of the main result

We apply Lemma 2 to E = χ_{α}^{∞} and I^{∞} = χ_{α}^{∞} ∩ I. From well known facts in linear algebra, we obtain

dim(χ_{α}^{∞}/χ_{α}^{∞} ∩ I) = dim((χ_{α}^{∞} + I)/I) ≤ dim(χ_{α}/I) < +∞.

So I^{∞} is of finite codimension in χ_{α}^{∞} . The endomorphism ψ = [X_{o}, .] , where X_{o}(x_{1}, ..., x_{2n+1}) = ∑_{i=1}^{2n+1} x_{i} \frac{∂}{∂x_{i}} (X_{o} = φ^{−1}(z)), satisfies obviously the first assumption of Lemma 2, and by Lemma 4 and 5, the conditions of Lemma 2 are fulfilled. Consequently χ_{α}^{∞} = I^{∞} i.e. χ_{α}^{∞} ⊂ I.

5. Application

Let Γ be a pseudogroup of local diffeomorphisms of R^{n}.

Definition 5.1. A left action of a pseudogroup Γ on a Lie group G, is a functorial assignment to each f ∈ Γ with domain U a smooth map f: U ×G → G such that the following axioms are satisfied

1) For any ξ ∈ U and any a, b ∈ G,

f (ξ, ab) =f (ξ, a)b.

2) For any open set V of U

\overline{f |_{V}} =f |_{V ×G} .

3) For any f, g ∈ Γ,

\overline{g ∘ f}(ξ, y) =g (f(ξ), f (ξ, y)).

4) Let I be an open interval of the real line R. If f: I × U → R^{n} is a smooth map such that ∀t ∈ I, f_{t} ∈ Γ, where f_{t}(x) = f(t, x), the map

\begin{matrix}I × U × G → G \\ (t, (ξ, y)) → f_{t} (ξ, y) \end{matrix}

is smooth.

Remark. It is obvious from axiom 2) that the action is local, i.e. depends only on germs.

Let P be the pseudogroup of contact diffeomorphisms and P_{o} the group of those fixing 0 and L(G) be the Lie algebra of a Lie group G. Suppose that P acts on G; let X ∈ χ_{α} and ϕ_{t} = exp(tX) be the flow generated by X. So ϕ_{t} ∈ P_{o} for any t ∈ I ( since X leaves 0 invariant ). Let (ϕ_{t})_{t} be the flow of diffeomorphisms ( translations ) induced by the action of the pseudogroup P on G. Consider the vector field defined by:

X= \frac{d}{dt} |_{t=0} (ϕ_{t})_{o}

where (ϕ_{t})_{o}(x) =ϕ_{t} (0, x), then we obtain a Lie algebra homomorphism:

H : χ_{α} → L(G), H(X) =X

Proposition 5.2. The homomorphism H depends only on the infinite jet J_{o}^{∞}X of X at 0.

Proof. The kernel I of H is a finite codimension ideal of χ_{α}, so by Theorem1 its contains χ_{α}^{∞} and then H depends only on J_{o}^{∞}X.

Acknowledgments. The authors wish to thank the referee for his careful reading and helpful suggestions on the improvement of the manuscript.

References

[1] Bénalili, M., Fibrés naturels définis sur la catégorie des Γ−variétés, Circolo Mat. di Palermo 13 (1994), 309–328.

[2] Epstein, D. B. A., and W. P. Thurston, Transformation groups and natural bundles, Proc. London Math. Soc. 38 (1979), 219–237.

[3] Palais, R. S., and C. L. Terng, Natural bundles have finite order, Topology 16 (1977), 271–277.

Mohammed Benalili and Azzedine Lansari Département de Mathématiques Faculté des Sciences BP 119 Université de Tlemcen Algérie m-benalili@mail.univ-tlemcen.dz

Received October 22, 1999 and in final form May 24, 2000