Poisson bracket

Remarks

Equivalently $\{f, g\} = df(X_g) = X_g(f) = -X_f(g)$. The bracket is bilinear and antisymmetric, satisfies the Leibniz rule $\{f, gh\} = \{f, g\} h + g \{f, h\}$, and satisfies the Jacobi identity because $d\omega = 0$. With these conventions $X_{\{f, g\}} = -[X_f, X_g]$.

Example: on $\R^{2n}$ with $\omega = \sum_i dq_i \wedge dp_i$, $$\{f, g\} = \sum_{i=1}^n \left( \frac{\partial f}{\partial q_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q_i} \right).$$ For instance $X_{p_j} = \partial / \partial q_j$, so $\{q_i, p_j\} = X_{p_j}(q_i) = \delta_{ij}$, while $\{q_i, q_j\} = \{p_i, p_j\} = 0$.

Non-example: on $\R^4$ the closed $2$-form $dq_1 \wedge dp_1$ is degenerate, so it does not define a bracket in this way: the function $q_2$ has no Hamiltonian vector field for it.

Along an integral curve $\gamma$ of $X_H$, a function $g$ satisfies $\frac{d}{dt} g(\gamma(t)) = \{g, H\}(\gamma(t))$. Some authors use the opposite sign convention for $\{f, g\}$.