10 votes

Is there a classification of polynomial Poisson brackets?

I don't think we have anything like a classification. In the book Poisson Structures (2013) by Laurent-Gengoux, Pichereau, and Vanhaecke, it is written "For higher order Poisson structures, ...
9 votes
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Poisson cohomology

First of all, one can think of Poisson cohomology (or, rather, Poisson cochains) as functions on the $QP$-manifold $T^*[1]M$ with the differential as you describe. In particular, it has a natural odd ...
8 votes
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Two definitions of the super Jacobi identity

By super skew-symmetry $[x, y] = - (-1)^{|x||y|}[y, x]$. Also note that $\left|[x, y]\right| = |x| + |y| \bmod 2$, in particular, $(-1)^{|[x, y]|} = (-1)^{|x|+|y|}$. So we see that \begin{align*} [x, ...
8 votes
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Some elementary questions about deformation quantization

a lot of questions, let me try on some of them :) The bad news is that in most of the interesting situations the higher order terms of the star product, the $B_i$ will not vanish. Heuristically this ...
7 votes
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Finite-dimensional Poisson cohomologies

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 ...
7 votes
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How to solve this system of equations?

It seems this is a system of quadratic equations, so try to write it as an equation $z^T M z = 0$ with $z = (z_0,...,z_{\text{whatever}})$ and a symmetric matrix $M$ (it's always possible to choose $M$...
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7 votes

What is a Lagrangian submanifold intuitively?

Interestingly enough no one wrote about the first examples of Lagrangian manifolds in symplectic geometry. In the first papers on symplectic geometry (Theory of Systems of Rays, 1828), William R. ...
7 votes

Is every singular foliation induced by a Lie algebroid?

It is possible to define the Lie groupoid of a singular foliation and associates to it its Lie algebroid when it is smooth. This Lie algebroid satisfies the property 2. https://projecteuclid.org/...
6 votes
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Associativity of Kontsevich's star product up to second order

The answer is that the black underlined terms do not cancel. Instead, they contribute an extra term which gives precisely the Jacobi identity (times $2/3$). (The reason I missed this is that I ...
6 votes

Computations of certain Poisson cohomology groups

The Poisson cohomology associated to compact Lie algebras was computed by Ginzburg and Weinstein already in 1992: Lie Poisson structure on some Poisson Lie groups, Thm. 3.5 For the non-compact ...
5 votes

Is every singular foliation induced by a Lie algebroid?

For Stefan-Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample: http://users.uoa.gr/~iandroul/...
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5 votes
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Fixed point scheme definition

This might be an unhelpful answer, but have you already considered the paper John Fogarty - Fixed points schemes?
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4 votes
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Cluster algebra structure compatible with Poisson brackets

Yes. You might find the extended presentation of this work in the book by Gekhtman, Shapiro and Vainshtein more helpful. It is based on their papers in this area but with more examples. (I also ...
4 votes
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intersection of Voronoi cell and Circle

Condition upon the site we are interested in being the origin (that is a somewhat sloppy phrase, of course, but not more sloppy than "arbitrary cell region"; once you make sense of the latter, you can ...
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4 votes

Interpretation of the Schouten bracket as an integrability condition

One well-known example is the case of bivector fiels $\pi$. Then $[\pi, \pi] = 0$ is equivalent to say that $\{f, g\} = \pi(df, dg)$ is a Poisson bracket, i.e. satisfies the Jacobi identity. In this ...
3 votes

Why symplectic geometry gives Poisson geometry

Look up sub-Riemannian geometry. This is a quite developed theory now, but here the Riemannian metric is defined only along a subbundle of the tangent bundle which is assumed to be completely non-...
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3 votes

Two definitions of the super Jacobi identity

I believe these should be the same. The reason being, $J(x,y,z)$ reduces to the ordinary Jacobi identity in all cases except one: when any two of the elements in $\{x,y,z\}$ are Fermi and the third ...
3 votes
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Classifications of Lie bialgebras

Semisimple case: Belavin-Drinfeld result does not classify all lie bialgebras on a (finite-dim.) semisimple complex Lie algebra, but only so called quasi-triangular Lie bialgebras (those having a non ...
3 votes
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Kontsevich's flow on the space of Poisson structures

By looking at scaling, both $f$ and $g$ have to be cubic in $u$. Moreover, monomials in $f$ contain 3 derivatives in $y$ and 2 derivatives in $x$; similarly for $g$. There are 12 possible monomials ...
3 votes

What is a Lagrangian submanifold intuitively?

As mentioned in previous answers - Lagrangian submanifolds encode vast amount of information on the symplectic geometry of the ambient symplectic manifold $M$ they live in (like sheaves on an ...
3 votes

Integrating Poisson groups

I think this question was already answered in the comments, but I just wanted to point out Jiang Hua Lu's thesis (which can be found on her website http://hkumath.hku.hk/~jhlu/publications.html) talks ...
3 votes

Are symplectic realizations of a Poisson manifold unique?

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 ...
3 votes
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Questions about Sklyanin bracket

Perhaps you can use the method of section 4.3 (The second Russian formula (quadratic brackets)) in Kosmann-Schwarzbach, Lie Bialgebras, Poisson Lie Groups, and Dressing Transformations? They calculate ...
3 votes

Natural Poisson brackets on $S(V^*)$

Under the natural interpretation of the term "natural", the answer is negative. For example, consider a linear bracket corresponding to a Lie algebra that does not admit an invariant bilinear form (e....
3 votes
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Symplectic submanifolds of cotangent bundles of Lie groups

The preimage of a coadjoint orbit under a moment map is, under a mild transversality assumption, a coisotropic submanifold; so its null foliation $\smash{\ker(\omega_{|\Phi^{-1}(\mathcal O)})}$ is not ...
3 votes

Trying to understand dressing actions

I think that first you should understand that if your Poisson-Lie group has zero Poisson structure then the dressing action of $G$ on $G^*\simeq \mathfrak g^*$ is just the usual coadjoint action. Let ...
3 votes

If $A$ is a (shifted) Poisson algebra, what does $A[\varepsilon]$ represent?

I can sort of give an answer in the context of derived geometry. So let $A$,$B$ derived rings (i.e. dg-rings, simplicial rings, ${\mathbb{E}_\infty }$-rings). For a $B$-module $M$, we can define the ...
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3 votes
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Bracket systems (generalization of Poisson brackets)

I'm no expert on operads but it seems that a "bracket system" can be formalized as an operad; see e.g. the Poisson operad here, the "product-complete" condition could be related to having a Hopf ...
3 votes
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Lie bracket on the complex valued functions of the space of representations of a Riemann surface

In the complex case, Goldman's symplectic form is a $(2,0)$-form, and the Poisson bracket is in terms of that form (the bracket is determined by a bivector and the bivector corresponds to the form). ...
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2 votes

Classification of finite dimensional Lie subalgebras of $\mathbb R[q^1,\dots,q^n,p_1,\dots,p_n]$

Using the Leibniz rule and the induction on $deg(f) + deg (g)$, we can see that the braket on $A$ is exactely the Poisson bracket defined by him in 1809 :) That is: $$ \{f, \, g\} = \sum_{i=1}^r \, \...

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