Poisson cohomology Let $\Pi$ be a Poisson structure on a manifold $M$. Then we can define a differential $d$ on the complex $\Lambda^{\bullet}M$
$$
C^{\infty}M \to TM \to...\Lambda^kTM \to...
$$
in the following way:
$$
d = [\Pi,\cdot],
$$
where $[\cdot,\cdot]$ is the Schouten-Nijenhuis bracket. The cohomology of this complex is called Poisson cohomology.
I have several question about this construction:


*

*Suppose $M=\mathbb R^{2d+s}$ with a Poisson structure $\sum\limits_{i=1}^{d}\frac{\partial}{\partial q_i}\wedge \frac{\partial}{\partial p_i}$. Is it true that Poisson cohomology of $M$ are zero?

*Is this construction functorial? I mean, let $f\colon M\to N$ be a poisson map, then do we have a map of complexes $f_*\colon \Lambda^{\bullet}M \to \Lambda^{\bullet}N$?

*Let $F\colon M\times I \to N$ be a homotopy such that for each $t$ $F_t$ is a Poisson map. Is it true that $F_0$ and $F_1$ induce the same map from Poisson cohomology of $M$ to Poisson cohomology of $N$?
 A: First of all, one can think of Poisson cohomology (or, rather, Poisson cochains) as functions on the $QP$-manifold $T^*[1]M$ with the differential as you describe. In particular, it has a natural odd Poisson ($\mathbb{P}_2$) structure given by the Schouten bracket.


*

*You can split $T^*[1](\mathbb{R}^{2d+s}) = T^*[1](\mathbb{R}^{2d})\times T^*[1](\mathbb{R}^s)$ and so the same decomposition applies to Poisson cohomology. Since $\mathbb{R}^{2d}$ is symplectic, its Poisson cohomology coincides with the de Rham cohomology which is quasi-isomorphic to $\mathbb{R}$ in degree 0. Therefore, Poisson cohomology of $\mathbb{R}^{2d+s}$ coincides with the Poisson cohomology of $\mathbb{R}^s$ with the zero Poisson structure, i.e. functions on $T^*[1](\mathbb{R}^s)$.

*Given a map of manifolds $M\rightarrow N$, there is no natural map on their cotangent bundles. Instead, there is a Lagrangian correspondence
$$T^* M \leftarrow T^*N\times_N M \rightarrow T^* N.$$
The same construction works with a shift and so instead of a map one has a dg bimodule between Poisson cohomologies of $M$ and $N$. The underlying graded bimodule is $C^\infty(T^*[1] N\times_N M)$. The differential is constructed as follows. Recall that for any coisotropic submanifold $C\hookrightarrow X$ the graded commutative algebra $\bigwedge^\bullet N_{C/X}$ carries a natural differential (a version of the differential you wrote). The map $M\rightarrow N$ is Poisson iff its graph $M\rightarrow M \times N$ is coisotropic and taking the exterior algebra on the normal bundle of this coisotropic you recover $C^\infty(T^*[1] N\times_N M)$ and that's the required differential.

*Given such a homotopy, you get a homotopy of (dg) bimodules from point 2, which seems to be the best you can hope for.
