2 Quasi-iso is an iso

It seems to me that there is a quasi-isomorphism (quasi-)isomorphism between the de Rham algebra and the dg algebra of polyvector fields equipped with the differential $[\pi,-]$ (where $\pi$ is the Poisson structure corresponding to the symplectic form).

Through this isomorphism the equation $d\omega=0$ is sent to $[\pi,\pi]=0$, and the equation $\omega=d\lambda$ is sent to $\pi=[\pi,V]$, where $V$ is a vector field.

On the level of the Poisson algebra of functions it tells you that for any two functions $f,g$, we have (up to a sign) $$\{f,g\}=V(\{f,g\})-\{V(f),g\}+\{f,V(g)\}$$

Algebraically you can say that there is a derivation $V$ for the product such that the Poisson bracket is its own derived bracket w.r.t. $V$.

1

It seems to me that there is a quasi-isomorphism between the de Rham algebra and the dg algebra of polyvector fields equipped with the differential $[\pi,-]$ (where $\pi$ is the Poisson structure corresponding to the symplectic form).

Through this isomorphism the equation $d\omega=0$ is sent to $[\pi,\pi]=0$, and the equation $\omega=d\lambda$ is sent to $\pi=[\pi,V]$, where $V$ is a vector field.

On the level of the Poisson algebra of functions it tells you that for any two functions $f,g$, we have (up to a sign) $$\{f,g\}=V(\{f,g\})-\{V(f),g\}+\{f,V(g)\}$$

Algebraically you can say that there is a derivation $V$ for the product such that the Poisson bracket is its own derived bracket w.r.t. $V$.