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A Poisson manifold is a smooth manifold with a smooth antisymmetric bivector $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ whose function bracket $\{f,g\}={\pi }^{ij}({\partial }_{i}f)({\partial }_{j}g)$ satisfies the Jacobi identity. The bracket is automatically a derivation in each argument. A symplectic manifold has a nondegenerate differential two-form $\omega$ satisfying $d\omega =0$, that is, a closed differential form; nondegeneracy forces its dimension to be even. It determines a nondegenerate Poisson structure. We use matrix convention $\pi =-{\omega }^{-1}$, so that ${\iota }_{X_H}\omega =dH$ corresponds to $X_H(f)=\{f,H\}$.

Examples of genuinely nonsymplectic Poisson geometries include the zero bracket on ${\mathbb{R}}^3$ and the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$ on ${\mathbb{R}}^3$. Their three-dimensional underlying manifold cannot carry a nondegenerate two-form. In the latter case the Poisson rank is two away from the origin and zero at the origin, with spheres and the origin as symplectic leaves. Even in even dimension a given Poisson structure need not come from a symplectic form: $\pi =x{\partial }_x\wedge {\partial }_y$ on ${\mathbb{R}}^2$ satisfies Jacobi but vanishes along $x=0$. The statement concerns this degenerate structure, not a claim that ${\mathbb{R}}^2$ admits no other symplectic structure.

For any candidate bivector, applying the Jacobi identity to coordinate functions gives the necessary conditionIt is sufficient as well: expanding $\{f,\{g,h\}\}$ and its cyclic companions cancels all terms with second derivatives by antisymmetry of $\pi$. What remains is $J^{ijk}({\partial }_{i}f)({\partial }_{j}g)({\partial }_{k}h)$. This is the coordinate Jacobi condition for a Poisson bivector.

For a symplectic form, differentiate $\omega \pi =-I$ to obtain ${\partial }_{\ell }\pi =\pi ({\partial }_{\ell }\omega )\pi$. Substituting this into the cyclic expression givesbecause $d\omega =0$. Thus the inverse symplectic structure always satisfies the Jacobi identity. Conversely, for a nondegenerate Poisson structure, multiplying by three inverse matrices proves closure of its associated two-form.

In momentum space, choose Hamiltonian $H(\mathbf{p})=E(\mathbf{p})$ and the constant Poisson bivectorIts Poisson bracket is $\{f,g\}=e\mathbf{B}\cdot (\nabla f\times \nabla g)$. The Hamiltonian vector field therefore givesexactly the prescribed motion. The sign here uses the signed charge $e$ as printed; an electron's physical charge may be negative. Since the magnetic field is uniform, every derivative of ${\pi }^{ij}$ vanishes, so every term in $J^{ijk}$ is zero. Jacobi holds independently of the dispersion function $E$. This is the momentum-space Poisson structure for a uniform magnetic field.

For $e\mathbf{B}\ne 0$ this bivector has rank two and $C(\mathbf{p})=\mathbf{B}\cdot \mathbf{p}$ is a Casimir function of a Poisson manifold: $\{C,f\}=0$ for every $f$. Its symplectic leaves are the planes $C=c$. For axes with $\mathbf{B}=B_0{\mathbf{e}}_3$, the leaf form is $\omega =(e{B}_0{)}^{-1}d{p}_1\wedge d{p}_2$ in our inverse-sign convention. Also $\dot{E}=\{E,E\}=0$, so trajectories on a Fermi surface lie on its intersections with these planes, typically curves. The energy surface is invariant for this Hamiltonian, but need not be a Poisson submanifold; the bracket has been defined on the whole momentum space. If $e\mathbf{B}=0$, the Poisson structure and motion vanish.