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Apr 9, 2018 at 13:09 comment added Alex M. @NicolaCiccoli: Yes, this explicit notation makes things clear.
Apr 9, 2018 at 6:23 comment added Nicola Ciccoli @Alex M. Yes they are the coordinate functions and no I do not see any problem with antisymmetry. It is the Poisson bracket given by the (evidently) antisymmetric bivector $\pi=xy\partial_x\wedge\partial_y$. Is this clearer?
Apr 8, 2018 at 15:10 comment added Alex M. @NicolaCiccoli: I apologize for the intrusion, but do $x$ and $y$ represent the coordinate functions on $\mathbb R^2$? If so, your Poisson bracket is not antisymmetric, so I'm obviously misinterpreting your notations.
Jul 30, 2014 at 8:25 comment added Nicola Ciccoli You also asked for references, I guess that here mathnet.or.kr/mathnet/paper_file/glasgow/Gordon/poisord.pdf you may find some pieces fo the general theory...
Jul 30, 2014 at 6:37 comment added Nicola Ciccoli Maybe more than an analogue; what is around here is the infinitesimal action of all the Hamiltonian vector fields...
Jul 29, 2014 at 18:51 comment added Allen Knutson Okay, so the group-action analogue is that a $G$-invariant subvariety may fail to be the intersection of a number of $G$-invariant hypersurfaces.
Jul 29, 2014 at 15:57 vote accept Allen Knutson
Jul 29, 2014 at 10:38 comment added Nicola Ciccoli To put it another way: being a symplectic leaf is still quite far from being a level set of Casimir functions. There are many examples of algebraic Poisson manifolds with dense leaves, or leaves which are dense inside closed submanifolds.
Jul 29, 2014 at 8:27 comment added Stefan Waldmann Being a Poisson ideal, $I$ is also a associative ideal. So to make things interesting, it should not contain the constants. I think your example completely settles the question (in the negative).
Jul 29, 2014 at 8:10 history edited Nicola Ciccoli CC BY-SA 3.0
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Jul 29, 2014 at 7:55 history answered Nicola Ciccoli CC BY-SA 3.0