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?

  • 2
    $\begingroup$ Your PS1 is only correct when you work sufficiently locally. Consider an irrational $\mathbb R^2$ wrapping around $T^n$ (like the irrational line on a torus). You should be able to arrange a Poisson structure with this plane (and its translates) as its symplectic leaves. Locally there are many Casimirs, but globally there are not. $\endgroup$ – Theo Johnson-Freyd Aug 12 '16 at 4:28
  • $\begingroup$ But PS2 can be generalized to the case when there is a point on a manifold in which $\Pi$ is of a maxical rang $\endgroup$ – anna abasheva Aug 12 '16 at 6:44

Consider on $\mathbb R^2$ the Poisson structure $\pi=(x^2+y^2)\partial_x\wedge\partial_y$. Then this Poisson manifold is not symplectic but with finite dimensional Poisson cohomology (It can be computed directly: the generator of nontrivial 1-dim Poisson cohomology is the modular vector field which is the rotation vector field $x\partial_y-y\partial_x$).

On the other hand if you consider $\pi_2=(x^2+y^2)^2\partial_x\wedge\partial_y$ which has the same symplectic foliation (a $0$-dim leaf and a symplectic complement), then Poisson cohomology is infinite-dimensional. This shows that finite-dimension of the Poisson complex cannot be read simply from the Poisson foliation, it rather tells you also how the Poisson structure degenrates at points in which the rank drops down.

More detailed results of this kind (on linear spaces) can be found in:

  1. Nakanishi, Poisson cohomology of plane quadratic structures Publ. RIMS Kyoto Univ. 33, 73-89 (1997).

  2. Roger-Vanhaecke, Poisson cohmology of the affine plane, Journ. Algebra 251, 448-480 (2002).

  3. Picherau, Poisson (co)homology and isolated singularities, Journ. Algebra 229, 747-777 (2005).

An interesting class of (non symplectic, not on affine space) Poisson manifolds with finite dimensional Poisson cohomology is described in the work of

  1. Radko: A classification of topologically stable Poisson structures on a compact oriented surface, Journ. Sympl. Geom. 1, 523-542 (2002).

As far as I know there is not known criterion to establish when a Poisson manifold has finite-dimensional Poisson cohomology. As clear from the above mentioned references it should contain informations on how the Poisson structures degenerates when its rank drops down.

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