Let $M$ be a symplectic manifold with symplectic form $\Omega$ The Poisson bracket is a bilinear operation on the set of differentiable functions on $M$ In terms of local Darboux coordinates $p_1, \ldots, p_n, q_1, \ldots, q_n$ the Poisson bracket of two functions is defined as follows: $$[f,g] = \sum_{i=1}^n {\partial f \over \partial q_i} {\partial g \over \partial p_i} - {\partial f \over \partial p_i} {\partial g \over \partial q_i}$$ It can be shown that the value of $[f,g]$ does not depend on the choice of Darboux coordinates. Therefore, the Poisson bracket is a well-defined operation on the symplectic manifold. Also, some authors use a different sign convention -- what they call $[f,g]$ is what would be referred to as $-[f,g]$ here.

The Poisson bracket can be defined without reference to a special coordinate system as follows: $$[f,g] = \Omega^{-1} (df, dg) = \sum_{i=1}^{2n} \Omega^{ij} {\partial f \over \partial x_i} {\partial g \over \partial x_j}$$ Here $\Omega^{-1}$ is the inverse of the symplectic form, and its components in an arbitrary coordinate system are denoted $\Omega^{ij}$

The Poisson bracket sastisfies several important algebraic identities. It is antisymmetric: $$[f,g] = -[g,f]$$ It is a derivation: $$[fg,h] = f[g,h] + g[f,h]$$ It satisfies Jacobi's identitity: $$[f,[g,h]] + [g,[h,f]] + [h,[f,g]] = 0$$

The Hamilton equations can be expressed elegantly in terms of the Poisson bracket. If $X$ is a smooth function on $M$ we can describe the time-evolution of $X$ by the equation $${dX \over dt} = [X,H]$$ If $X$ is a smooth function on $\mathbb{R} \times M$ we can describe the time-evolution of $X$ by the more general equation $${dX \over dt} = {\partial X \over \partial t} - [X,H]$$