Poisson pencil (${\Pi }_{\lambda }={\Pi }_1+\lambda {\Pi }_2$)

Two Poisson structures are compatible when every linear combination ${\Pi }_1+\lambda {\Pi }_2$ is a Poisson structure. This family is a Poisson pencil. In terms of the Schouten-Nijenhuis bracket, compatibility means $[{\Pi }_1,{\Pi }_2]=0$, in addition to the two individual Jacobi identities. Constant Poisson bivectors are compatible. Over a real manifold, complex parameter values refer to the complexification of its function algebra.

Start from a Casimir function $H_0$ of the first Poisson structure. Whenever possible, choose $H_n$ whose first Hamiltonian vector field is minus the second Hamiltonian vector field of $H_{n-1}$. A sequence extending indefinitely is the coefficient sequence of a formal Casimir function for the Poisson pencil. Its functions pairwise commute for both Poisson brackets. Existence at each stage is an additional solvability requirement; compatibility alone does not guarantee it.

The next step of a Lenard-Magri recursion requires $-{\Pi }_{2}d{H}_{n-1}$ to be a Hamiltonian vector field for ${\Pi }_1$. It must be tangent to the symplectic leaves, and its contraction with the leaf symplectic form must have a global potential, with a compatible smooth extension between leaves. Even the first step may fail: on ${\mathbb{R}}^2$, take ${\Pi }_1=0$, ${\Pi }_2={\partial }_x\wedge {\partial }_y$ and $H_0=x$. Both Poisson structures are compatible, but ${\Pi }_{1}d{H}_1=-{\Pi }_{2}dx$ cannot hold, because its left side is zero and its right side is nonzero.

For a Lenard-Magri recursion, antisymmetry shifts an index: $\{H_i,H_j{\}}_1=\{H_{i-1},H_{j+1}{\}}_1$ when $i\ge 1$. Repeating reaches $\{H_0,H_{i+j}{\}}_1=0$, because $H_0$ is a Casimir function. The recursion then gives $\{H_i,H_j{\}}_2=-\{H_{i+1},H_j{\}}_1=0$. This proof applies to every finite initial portion for the indices that can be shifted inside it, and to all pairs in an infinite sequence.