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If $(M, P)$ is a (Hausdorff) Poisson manifold, then there exist a surjective Poisson submersion $\pi : (S, \omega) \to (M, P)$ with $(S, \omega)$ a symplectic manifold.

I am in a situation where I would like to study some objects on a Poisson manifold $M$ by pulling them back on some symplectic realization of $M$. The problem is that there might be many such symplectic realizations, and I do not know how any two of them are related to one another. I would like either to pick a "special", natural one among them, or to show that my construction doesn't depend on the choice of realization. From what I've read, I think that the first choice is out of the question.

From Weinstein's 1983 "The local structure of Poisson manifolds" I know only a local (and not really useful in my case) result (theorem 10.1, page 549). Can one hope for more? Are any two such symplectic realizations somehow isomorphic?

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    $\begingroup$ Well, there's the source simply connected symplectic groupoid if $S$ is integrable. But in general, if it's not integrable, then the symplectic groupoid structure is just local. $\endgroup$
    – user40276
    Commented Jan 22, 2016 at 19:41
  • $\begingroup$ You could look at Ivan Contreras' arxiv.org/pdf/1306.3943.pdf section 3.5. This is if you are okay with not doing the reduction. $\endgroup$
    – AHusain
    Commented Jan 23, 2016 at 0:42

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Any Poisson manifold $(M,P)$ admits a symplectic realisation $\pi : (S,\omega) \to (M,P)$ as defined above; this is Theorem 2 in this nice paper by Crainic and Marcut, generalising Weinstein's local existence argument. Their method does not construct all possible symplectic realisations, but just the one which comes from the local symplectic groupoid integrating the Lie algebroid associated to the Poisson structure (cf. @user40276's comment). In some sense, this is distinguished among all symplectic realisations because it comes from the (local!) symplectic Weinstein groupoid associated to the Poisson structure.

In general, a fixed Poisson manifold $(M,P)$ may admit several distinct symplectic realisations, e.g. the dimension of the symplectic manifold may vary. One useful way of looking at symplectic realisations comes from the work of Crainic and Fernandes. Any symplectic realisation $\pi : (S,\omega) \to (M,P)$ comes equipped with an action of the Lie algebroid $T^* M \to M$ associated to $(M,P)$ (cf. Step 1 in the proof of Theorem 8 of this great paper). In some sense, symplectic realisations of $(M,P)$ play the role of representations for this Lie algebroid (WARNING: this is not related to the notion of representation of Lie algebroids in the sense of, say, this MathOverflow question). For instance, a complete symplectic realisation $\pi : (S,\omega) \to (M,P)$, i.e. for any compactly supported $f \in C^{\infty}(P)$ the Hamiltonian vector field $X_{\pi^*f}$ is complete, plays the role of a faithful representation and allows to prove that $(M,P)$ is integrable, in analogy with the proof of Lie's third theorem using Ado's theorem (cf. Theorem 8 of the aforementioned paper by Crainic and Fernandes).

To conclude, I am not aware of a notion of equivalence of symplectic realisations for general Poisson manifolds. There is, however, a notable exception: if $(M,P)$ is integrable, then there is a notion of self Morita equivalence of $(M,P)$ which involves symplectic realisations with whose total spaces are the same. This gives rise to the Picard group of $(M,P)$ studied in detail in this recent paper. Not sure, however, if this goes in the direction the OP wanted.

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