Poisson algebra

Remarks

Clause 3 says that each map $b \mapsto \set{a, b}$ is a derivation of the associative product. Many authors, especially in geometry, also require the product to be commutative.

Example: $P = \R[x, y]$, or the smooth functions on $\R^2$, with the usual product and $\set{f, g} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x}$. The bracket is antisymmetric and satisfies the Jacobi identity, and the product rule for derivatives gives $\set{f, gh} = \set{f, g} h + g \set{f, h}$. Another example: every associative algebra with the commutator $\set{a, b} = ab - ba$ is a Poisson algebra, since $a(bc) - (bc)a = (ab - ba)c + b(ac - ca)$.

Non-example: $P = k[x]$ with the usual product and the bracket $\set{f, g} = f g' - g f'$, where $'$ is the derivative. This bracket is a Lie bracket, but clause 3 fails. With $b = c = 1$, clause 3 would give $\set{a, 1} = 2\set{a, 1}$, so $\set{a, 1} = 0$; yet $\set{x, 1} = x \cdot 0 - 1 \cdot 1 = -1$.