Hamiltonian vector fields on Weil bundles

Let M be a paracompact smooth manifold, A a Weil algebra and M^{A} the associated Weil bundle. In this paper, we give a characterization of hamiltonian field on M^{A} in the case of Poisson manifold and of Symplectic manifold.


Introduction
In what follows, we denote M, a paracompact differentiable manifold of dimension n, C ∞ (M) the algebra of smooth functions on M and A a local algebra in the sense of André Weil i.e a real commutative algebra of finite dimension, with unit, and with an unique maximal ideal m of codimension 1 over R [14].. In this case, there exists an integer h such that m h+1 = (0) and m h (0). The integer h is the height of A. Also we have A = R ⊕ m.
We recall that a near point of x ∈ M of kind A is a morphism of algebras such that for all g in C ∞ (N), is differentiable, [7].. Thus, for f ∈ C ∞ (M), the map is a commutative algebra over A with unit and the map is an injective morphism of algebras. Then, we have [2].: In [2]. and [9]., we showed that the following assertions are equivalent: In all what follows, we denote X(M A ) the set of vector fields on M A and such that X(ϕ · ψ) = X(ϕ) · ψ + ϕ · X(ψ), for any ϕ, ψ ∈ C ∞ (M A , A).
Then [9]., The map is skew-symmetric A-bilinear and defines a structure of an A-Lie algebra over X(M A ). If is a vector field on M, then there exists one and only one A-linear derivation, called prolongation of the vector field θ, such that The map is an injective morphism of R-Lie algebras.

Structure of A-Poisson manifold on M A when M is a Poisson manifold
We recall that a Poisson structure on a smooth manifold M is due to the existence of a bracket {, } on C ∞ (M) such that the pair (C ∞ (M), {, }) is a real Lie algebra such that, for any f ∈ C ∞ (M) the map In this case we say that M is a Poisson manifold and C ∞ (M) is a Poisson algebra [12], [13]..

We denote
the adjoint representation and d ad the operator of cohomology associated to this representation. For any p ∈ N, When M is a smooth manifold, A a weil algebra and M A the associated Weil bundle, the A-algebra be the prolongation of the vector field ad( f ) and let the unique A-linear derivation (vector field) such that and for ϕ, ψ ∈ C ∞ (M A , A) and for a ∈ A, In [1]., we have show that this bracket defines a structure of A-Poisson algebra on C ∞ (M A , A). A). We denote d A the cohomology operator associated to this adjoint representation [9].. For any p ∈ N, Λ p For where ϕ i means that the term , ϕ i is omitted.

Hamiltonian vector fields on weil bundles
When M is a Poisson manifold with bracket {, }, a vector field 1. is locally hamiltonian if θ is closed for the cohomology associated with the adjoint representation i.e. d ad θ = 0.
2. is globally hamiltonian if θ is exact for the cohomology associated with the adjoint representation Thus, a vector field 1. is locally hamiltonian if X is closed for the cohomology associated with the adjoint representation i.e d A X = 0.
2. is globally hamiltonian if X is exact for the cohomology associated with the adjoint representation i.e. there exists ϕ ∈ C ∞ (M A , A) such that X = d A (ϕ).

is locally hamiltonian if and only if the vector field
is locally hamiltonian.
Proof. Indeed, for any η ∈ Λ p Pois (M), we have In particular, for p = 1, we have Thus, d ad θ = 0 if and only if d A (θ A ) = 0.

Proposition 3. When M A is a A-Poisson manifold with bracket {, } A , then, a vector field
locally hamiltonian is a derivation of the Poisson A-algebra C ∞ (M A , A).
and if d A X = 0, then for any ϕ, ψ ∈ C ∞ (M A , A), That ends the proof.

Proposition 4. Let M be a Poisson manifold with bracket {, }. If a vector field
is globally hamiltonian then the vector field is globally hamiltonian.
Proof. Based on the assumptions, there exists f ∈ C ∞ (M) such that θ = d ad ( f ). Thus, is globally hamiltonian.

Proposition 5. When M A is a A-Poisson manifold with bracket {, } A , then a vector field
globally hamiltonian is the derivation interior of the Poisson A-algebra C ∞ (M A , A).
Proof. If the vector field X : Thus, X is globally hamiltonian if there exists ϕ ∈ C ∞ (M A , A) such that X = τ ϕ = ad(ϕ) i.e. X is the interior derivation of the Poisson A-algebra C ∞ (M A , A).

Hamiltonian vector fields on M A when M is a symplectic manifold
When (M, Ω) is a symplectic manifold, then (M A , Ω A ) is a symplectic A-manifold [1].
For any f ∈ C ∞ (M), we denote X f the unique vector field on M such that is the operator of de Rham cohomology. We denote the operator of cohomology associated with the representation For ϕ ∈ C ∞ (M A , A), we denote X ϕ the unique vector field on M A , considered as a derivation of The bracket defines a structure of A-Poisson manifold on M A and for any f ∈ C ∞ (M), We deduce that [1].: Therefore, for any ϕ, ψ ∈ C ∞ (M A , A), we have

Proposition 7.
If ω is a differential form on M and if θ is a vector field on M, then Proof. If the degree of ω is p, then (i θ ω) A is the unique differential A-form of degree p − 1 such that A for any θ 1 , θ 2 , ..., θ p−1 ∈ X(M). As i θ A (ω A ) is of degree p − 1 and is such that = ω(θ, θ 1 , ..., θ p−1 ) A for any θ 1 , θ 2 , ..., θ p−1 ∈ X(M), we conclude that (i θ ω) When (M, Ω) is a symplectic manifold, 1. a vector field θ on M is locally hamiltonian if the form i θ Ω is closed for the de Rham cohomology and θ is globally 2. a vector field X on M A is locally hamiltonian if the form i X Ω A is d A -closed and X is globally hamil- Proof. For any θ ∈ X(M), we have Thus, θ is locally hamiltonian, i.e d(i θ Ω) = 0 if and only if, is a locally hamiltonian vector field. Proof. Let (M A , Ω A ) be a symplectic manifold. For any ϕ, ψ ∈ C ∞ (M A , A), If X is locally hamiltonian vector field, we have d A (i X Ω A ) = 0 i.e. for any X and Y ∈ X(M A ), In particular, for any X ϕ and X ψ , we have Hence X({ϕ, ψ} Ω A ) = {X(ϕ), ψ} Ω A + {ϕ, X(ψ)} Ω A .
That ends the proof. is globally hamiltonian then the vector field is globally hamiltonian.