Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 63 / 5

A symplectic manifold is a smooth manifold $P$ with a closed, nondegenerate two-form $\omega$. Nondegeneracy forces $\dim P=2n$. Examples are ${\mathbb{R}}^{2n}$ with $\omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, an oriented surface with a nowhere-zero area form, and every cotangent bundle with its canonical form.

A Poisson manifold has a bivector $\Pi$ such that $\{f,g\}=\Pi (df,dg)$ is an antisymmetric bilinear Poisson bracket, is a derivation in each argument, and satisfies the Jacobi identity. Its rank need not be full or constant. The zero bivector gives a Poisson manifold on any smooth manifold, including an odd-dimensional or nonorientable one. A nonzero example on ${\mathbb{R}}^3$ is the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$: it has rank two off the origin and rank zero at the origin, so it does not arise from a symplectic form on the whole three-dimensional space. Even on ${\mathbb{R}}^2$, $\Pi =x{\partial }_x\wedge {\partial }_y$ is Poisson, since its Jacobi condition is a three-vector and hence vanishes, but its degeneracy on $x=0$ prevents it from being the inverse of a global symplectic form. A manifold carrying a degenerate Poisson structure might separately admit some other symplectic structure; these are distinct claims.

At each point of a symplectic manifold, linear nondegeneracy gives a basis with $\omega ={\sum }_i{e}^i\wedge f^i$. Its top wedge product of differential forms isThus ${\omega }^n/n!$ is a global nowhere-zero volume form and selects a consistent symplectic orientation. This proves every symplectic manifold is orientable without choosing a global frame.

Let $\pi :T^{\ast }M\to M$ and let a point of the cotangent bundle be $(x,p)$ with $p\in T_x^{\ast }M$. Define the canonical Liouville one-form intrinsically by ${\theta }_{(x,p)}(V)=p(d\pi (V))$. In local coordinates it is $\theta =p_{i}d{q}^i$, soThis definition is global, $d\omega =0$ by $d^2=0$, and its coordinate matrix has invertible blocks. Hence $T^{\ast }M$ is symplectic for every smooth $M$, even if $M$ itself is not orientable.

The coordinate Jacobi condition for a Poisson bivector isNecessity follows by applying the Jacobi identity to three coordinate functions. For arbitrary functions, the second-derivative terms in the Jacobiator cancel by antisymmetry of $\Pi$; the terms left are exactly this coefficient multiplying $f_i{g}_j{h}_k$, proving sufficiency.

For a symplectic form, take ${\iota }_{X_f}\omega =df$ and $\{f,g\}=\omega (X_f,X_g)$. With these signs the matrix of $\Pi$ is $-{\omega }^{-1}$. Differentiation gives ${\partial }_{\ell }\Pi =\Pi ({\partial }_{\ell }\omega )\Pi$. The cyclic expression above consequently equalswhich vanishes because $d\omega =0$. This directly proves that a symplectic manifold defines a Poisson manifold.