Suppose we have a poisson manifold $M$ whose Poisson cohomology is finite-dimensional in each degree? Does it mean that our manifold is symplectic? Are there other cohomological criteria of symplecticity? For example, suppose that all Casimir functions on a manifold are constant, does it imply that this manifold is symplectic?

PS1. For example, if the maximal rang of the poisson bracket is not $\dim M/2$ the Casimir functions are not just constants. So, near each point in an open set of a Poisson manifold the Poisson form is written in some coordinates $p_1,...p_m, q_1,...q_m,y_1,...y_l$ like this

$$ \Pi =\sum\limits_{i=1..m} \frac{\partial}{\partial q_i}\wedge\frac{\partial}{\partial p_i} $$.

Now we can take any of the function $f(y_i)$ . It will be Casimir but not constant '

PPS1 It's only the receipt for generating local Casimir function. So, does anyone know an example of a Poisson manifold not of the maximum rang, whose Casimir functions are only constant?

PS2. I've made up an example of non-symplectic manifold with only constant Cazimir functions. It is $\mathbb R^2$, $\{x,y\}=x$.

So now I have another questions 1. What are the obstacles in a Poissonian manifolds to have only constant Cazimir functions? 2. What are the obstactles in Poissonian manifolds to have finite-dimensional Poisson cohomology in each degree?