Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 74 / 5

A Poisson manifold has a skew bilinear bracket on smooth functions which is a derivation in each argument and satisfies the Jacobi identity. Locally a Poisson bivector $\Pi$ represents it byThe derivation property is automatic for this expression. On applying the Jacobi identity to coordinate functions, and then using the derivation property for arbitrary functions, the necessary and sufficient condition isEquivalently its Schouten-Nijenhuis bracket $[\Pi ,\Pi ]$ vanishes.

A symplectic manifold has a closed nondegenerate two-form $\omega$. Write its coefficient matrix as ${\Omega }_{ij}$ and set ${\Pi }^{ij}=({\Omega }^{-1}{)}^{ij}$. It is skew and defines the bracket above. Differentiating $\Pi \Omega =I$ gives ${\partial }_{\ell }\Pi =-\Pi ({\partial }_{\ell }\Omega )\Pi$. Substitution into the Jacobi condition giveswhere the last equality is $d\omega =0$. Thus every symplectic manifold is Poisson. In this convention the Hamiltonian vector field satisfies ${\iota }_{X_f}\omega =-df$, and $\{f,g\}=X_{g}f$; this will also be the convention below.

The converse fails because a Poisson bivector need not be invertible. On ${\mathbb{R}}^2$, take $\Pi =0$. Its bracket is identically zero and satisfies every Poisson axiom, but it cannot be the inverse of any nondegenerate two-form. This is an even-dimensional counterexample, so odd dimension is not the only obstruction.