For a smooth manifold $M$, there is a natural $1$-form $\theta$ on $T^*M$ such that $\Bbb d \theta$ is a symplectic form. Somewhat symetrically, on $TM$ there is a natural tangent field $V$. Is it possible, then, to act symetrically and construct some "exterior derivative" $\Bbb d$ such that $\Bbb d V$ should be a Poisson bivector?
I know that, if $P$ is a Poisson bivector, then such an exterior derivative exists, using the Schouten-Nijenhuis braket, as $[P, \cdot]$ - but it is exactly the Poisson bivector that I am missing!