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Dec 16, 2016 at 19:37 vote accept agt
Jan 23, 2013 at 16:46 answer added Nicola Ciccoli timeline score: 2
Jan 23, 2013 at 12:14 comment added Joel Fine Thanks Robert, thats very clear. (And not so hard after all!)
Jan 22, 2013 at 22:38 comment added Robert Bryant If the leaves are $2$-dimensional, then the only obstruction is whether the leaves have a consistent orientation, i.e., whether there is a nonvanishing bi-vector $\pi$ on $M$ that is tangent to the leaves of $\mathcal{F}$. In this case, $[\pi,\pi]=0$ is an identity since $\pi$ is tangent to a foliation, so any such $\pi$ will be a Poisson structure.
Jan 22, 2013 at 21:53 comment added Joel Fine I agree with Tim that, as currently phrased, this question is an extremely hard open problem, way beyond current techniques. Perhaps though with different hypotheses, it might become a hard, open, but not completely impossible question. For example, what if the leaves are 2-dimensional? (Lefschetz fibrations play nicely with symplectic topology, after all.) What if we're allowed foliations and Poisson structures with singularities or degeneracies? (Perhaps along the lines of broken pencils on 4-manifolds?)
Jan 22, 2013 at 18:34 comment added Tim Perutz One of the main goals of symplectic geometry is to answer this question in the special case where $M$ is the unique leaf...
Jan 22, 2013 at 17:41 history asked agt CC BY-SA 3.0