 Solution

A Poisson manifold is a smooth manifold with an antisymmetric contravariant tensor field $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ such that the bracketsatisfies the Jacobi identity. Bilinearity, antisymmetry and the Leibniz rule follow directly from this formula. The Poisson bivector is allowed to be degenerate.

Apply the Jacobi identity to coordinate functions $x^i,x^j,x^k$. Necessity gives the coordinate Jacobi condition for a Poisson bivector:For sufficiency, expand $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. The terms differentiating $\pi$ are $J^{ijk}{f}_i{g}_j{h}_k$. All terms containing second derivatives cancel. For example, those containing $g_{j\ell }$ have coefficient $({\pi }^{i\ell }{\pi }^{jk}+{\pi }^{k\ell }{\pi }^{ij})f_i{h}_k$, which is antisymmetric under $j\leftrightarrow \ell$ after contraction with $f_i{h}_k$, whereas $g_{j\ell }$ is symmetric. The same argument handles the second derivatives of $f$ and $h$. Thus the displayed condition is both necessary and sufficient.

A symplectic manifold has a smooth nondegenerate closed differential form $\Omega$ of degree two. Write $\Omega =\frac{1}{2}{\Omega }_{ij}d{x}^i\wedge d{x}^j$ and choose the explicit inverse convention ${\pi }^{ik}{\Omega }_{kj}={\delta }_j^i$. This defines a smooth antisymmetric bivector. Differentiating the inverse matrix givesSubstitution into the Jacobi coefficient, followed by relabeling summed indices, givessince the parentheses are the components of $d\Omega$. Hence every symplectic manifold carries a Poisson structure. The opposite overall convention for the inverse bracket gives the same conclusion. Nondegeneracy and closure together are crucial; an arbitrary invertible antisymmetric tensor need not satisfy the Jacobi identity.

For a finite-dimensional Lie algebra $\mathfrak{g}$, its dual space ${\mathfrak{g}}^{\ast }$ carries the Lie-Poisson bracketwhere $d{f}_{\ell },d{g}_{\ell }$ are identified with elements of $\mathfrak{g}$. If $[e_i,e_j]=c_{ij}{}^k{e}_k$ and $x_i(\ell )=\ell (e_i)$, this reads $\{x_i,x_j\}=c_{ij}{}^k{x}_k$. The Jacobi coefficient iswhich vanishes by the Lie algebra Jacobi identity.

In particular, the dual of $\mathfrak{so}(3)$ is ${\mathbb{R}}^3$ with the rotational Lie-Poisson structure on R3,The Poisson bivector has rank two away from the origin and rank zero at the origin. The function $x^2+y^2+z^2$ is a Casimir function, since its bracket with every function vanishes. The centred spheres are its two-dimensional symplectic leaves, and the origin is a point leaf. This is a Poisson manifold but not a symplectic manifold: its bivector is not invertible, and an odd-dimensional manifold cannot have a nondegenerate alternating bilinear form.