A non-Poisson bivector cannot be a first-order star commutator

On ${\mathbb{R}}^3$ with coordinates $(x,y,z)$, take the smooth bivector $\pi =x{\mathrm{\partial }}_y\wedge {\mathrm{\partial }}_z+y{\mathrm{\partial }}_z\wedge {\mathrm{\partial }}_x+xy{\mathrm{\partial }}_x\wedge {\mathrm{\partial }}_y\mathrm{.}$ By [F1], its coordinate brackets are $\{y,z{\}}_{\pi }=x$, $\{z,x{\}}_{\pi }=y$, and $\{x,y{\}}_{\pi }=xy$.

Using the biderivation rule, its Jacobiator on these coordinate functions is $\{\{x,y{\}}_{\pi },z{\}}_{\pi }+\{\{y,z{\}}_{\pi },x{\}}_{\pi }+\{\{z,x{\}}_{\pi },y{\}}_{\pi }=\{xy,z{\}}_{\pi }+\{x,x{\}}_{\pi }+\{y,y{\}}_{\pi }=x^2-y^2\mathrm{.}$ At $(x,y,z)=(1,0,0)$ this is nonzero; by [F2], $\pi$ is not Poisson.

If a unital associative formal deformation had this first-order skew bracket, [F3] would force that bracket to satisfy Jacobi, contradicting step 2.1. Therefore this smooth bivector is not the first-order antisymmetric commutator of any such deformation, refuting the universal claim.