ar X iv : m at h / 06 02 24 6 v 1 [ m at h . R A ] 1 1 Fe b 20 06 Poisson algebras

Poisson algebras are usually defined as structures with two operations, a commutative associative one and an anticommutative one satisfying the Jacobi identity. These operations are tied up by a distributive law, the Leibniz law. We present Poisson algebras as algebras with one operation which enables one to study them as part of nonassociative algebras. We study the algebraic and cohomological properties of Poisson algebras, their deformations as nonassociative algebras and we give the classifications in low dimensions. 1 Poisson algebras presented as nonassociative algebras 1.1 Definition A Poisson algebra over a field K is a Lie algebra g whose bracket is denoted [ , ], with in addition an associative commutative multiplication, denoted ◦, satisfying the Leibniz condition : [X ◦ Y, Z] = X ◦ [Y, Z] + [X,Z] ◦ Y, for all X,Y, Z ∈ g. Such a definition is equivalent [11] to the following : Definition 1 A Poisson algebra P over the field K of characteristic different of 2 or 3 is a nonassociative algebra with multiplication · satisfying 3A(X,Y, Z) = (X · Z) · Y + (Y · Z) ·X − (Y ·X) · Z − (Z ·X) · Y (1) where A(X,Y, Z) is the associator of the product · that is A(X,Y, Z) = (X ·Y ) · Z −X · (Y · Z). Such a nonassociative multiplication is Lie-admissible, that is [X,Y ] = 1 2 (X · Y − Y ·X) is a Lie bracket. We denote by gP this Lie algebra associated to P . Moreover X ◦ Y = 1 2 (X · Y + Y ·X) 1 is a commutative associative multiplication. We denote AP this associative algebra. The Leibniz condition between the Lie bracket and the associative multiplication ◦ is equivalent to (1). Notations. In the following, we will denote sometimes (when no confusion is possible) the Poisson product by by XY instead of X · · ·Y . The corresponding associative product will be always denoted by X ◦ Y . Proposition 2 A Poisson algebra P is a power associative algebra. Proof. Recall that a nonassociative algebra is called power associative if every element generates an associative subalgebra. For any element X of the Poisson algebra P with product ·, we have A(X,X,X) = 0 and the algebra is thirdpower associative. We define the power of X by X = X, X i+1 = X ·X i and we have to prove that X X = X X = X i+j for all i, j ≥ 1 and 0 ≤ n < j, 0 ≤ p < i. First we prove that X i ·X = X ·X . We have for j > 1 3A(X,X, X) = 3(XX −XX) = 0 so for any j > 1, XX = XX. Let i ≥ 1 and suppose that XX = XX for p ≤ i. Then [X , X] = X ◦ [X , X] +X i ◦ [X,X] But by definition [X , X] = 1 2 (X X −XX ) = 0 and we have [X , X] = 0 andX X = XX . Then for any j ≥ 1, i ≥ 1X X = XX . Now we have to prove that X X = X i+j . From X X = XX i = X i+1 and X X = XX i we obtain : 3A(X , X,X) = 3(X i+j −X X) = X X −X i+j that is X X = X i+j . Let us suppose that X X = X i+j . Then 3A(X , X,X) = 3(X i+j −X X) = X X −X i+j this gives X X = X i+j and the algebra P is power associative. Proposition 3 A Poisson algebra is flexible. Proof. The algebra is flexible if A(X,Y,X) = 0. Here we have 3A(X,Y,X) = XY + (Y X)X − (Y X)X −XY = 0. We deduce easily from the definition of Poisson product the following identities related to the associator: A(X,Y, Z) +A(Z, Y,X) = 0 (flexibility) (2) A(X,Y, Z) +A(Y, Z,X)−A(Y,X,Z) = 0. (3) 2 1.2 Poisson algebras as K [Σ3]-associative algebras In [8],we have studied a large class of nonassociative algebras. In this section we show that Poisson algebras belongs to this category of algebras. Let Σ3 be the 3-order symmetric group and K [Σ3] its K-group algebra. A (nonassociative) K-algebra is called a K [Σ3]-associative algebra if there exists v ∈ K [Σ3], v 6= 0, such that Aμ ◦ Φv = 0 where Aμ is the associator of the considered algebra A and φv : A ⊗3 → A defined by Φσ(v1, v2, v3) = (vσ−1(1), vσ−1(2), vσ−1(3)) for all σ ∈ Σ3. Now suppose that A is a Poisson algebra. Thus, from (3) we see that the associator of the multiplication satisfies Aμ ◦Φv1 = 0 for v1 = Id− τ12 + c1 where τ12(1, 2, 3) = (2, 1, 3) (more generaly τij exchanges the elements i and j) and c(1, 2, 3) = (2, 3, 1). The flexibility identity (2) can be written Aμ ◦ Φv2 = 0 for v2 = Id + τ13. Comparing to the classification of [8] we deduce that any Poisson algebra is an algebra of type (IV ) 1.3 Pierce decomposition As P is a power associative algebra, we can consider a notion of nilalgebra. We will say that P is a nilalgebra if any element X is nilpotent, i.e. ∀X ∈ P , ∃r ∈ N such that X = 0 Proposition 4 Any finite dimensional Poisson algebra which is not a nilalgebra contains an idempotent. This is a consequence of the power associativity of a Poisson algebra. Let e be an idempotent. Then e = e and e ◦ e = e. Thus e is an idempotent of the associative algebra AP . The Leibniz identity implies [e, x] = [e ◦ e, x] = 2e ◦ [e, x] then [e, x] is zero or eigenvector of the operator L◦e : x → e ◦ x in AP associated to the eigenvalue 1 2 . As e is an idempotent, the eigenvalues associated to Le are 1 or 0. Then [e, x] = 0. Proposition 5 Let P be a Poisson algebra such that the center of the associated Lie algebra gP is zero. Then P has no idempotent different from zero. If P is of finite dimension, it is a nilalgebra. 3 Suppose that there exists an idempotent e 6= 0. As P is flexible, the operators Le and R ◦ e defined by L ◦ e(x) = e ◦ x and R ◦ e(x) = x ◦ e are commuting. Then P has the decomposition P = P0,0 ⊕ P0,1 ⊕ P1,0 ⊕ P1,1 with Pi,j = {xi,j ∈ P such that exi,j = ixi,j , xi,je = jxi,j} . From the previous proposition, e ∈ Z(gP). Then [e, x] = 0 for any x, that is ex = xe and P0,1 = P1,0 = {0} . Proposition 6 If the Poisson algebra P has a non zero idempotent, it admits the following Pierce decomposition P = P0,0 ⊕ P1,1 and P0,0 and P1,1 are also Poisson algebras. Proof. We have to show that P0,0 and P1,1 are two Poisson sub-algebras. Let x, y ∈ P0,0. Then ex = ey = xe = ye = 0. From (1), we have