10 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, ...
Michael Albanese's user avatar
10 votes
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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, ...
John Machacek's user avatar
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 ...
Pavel Safronov's user avatar
8 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. Debord - Holonomy Groupoids of ...
Tsemo Aristide's user avatar
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 ...
Stefan Waldmann's user avatar
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 ...
Nicola Ciccoli's user avatar
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. ...
alvarezpaiva's user avatar
  • 13.2k
6 votes

Is every singular foliation induced by a Lie algebroid?

$\newcommand\cF{\mathcal F}$For Stefan–Sussmann singular foliations, the answer is negative: See Prop. 1.3 in the following paper, for the construction of an explicit counterexample: Androulidakis and ...
Iakovos's user avatar
  • 61
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 ...
Florian Zeiser's user avatar
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?
Aurelio's user avatar
  • 271
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 ...
Jan Grabowski's user avatar
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 ...
fedja's user avatar
  • 59.3k
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 ...
Stefan Waldmann's user avatar
4 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 ...
Ricardo Buring's user avatar
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 ...
T. Amdeberhan's user avatar
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-...
Peter Michor's user avatar
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3 votes

Classifications of Lie bialgebras

Sorry for the self publicity, but for the especific example of $gl_n(k)$, you can view it as $gl_n(k)\cong sl_n(k)\times k$ and we did some work for trivial central extensions that allow you to ...
Marco Farinati's user avatar
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 ...
Nicola Ciccoli's user avatar
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 ...
Daniele Sepe's user avatar
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 ...
Didier Collard's user avatar
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 ...
Yochay Jerby's user avatar
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 ...
Francois Ziegler's user avatar
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 ...
Nicola Ciccoli's user avatar
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 ...
JJJ's user avatar
  • 167
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). ...
Sean Lawton's user avatar
  • 8,384
3 votes
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Nonstandard Podles spheres as $U_c(\frak{h})$ invariants

Yes, there is a one parameter family of coideal subalgebra of $U_q(\mathfrak{sl}_2)$ that give the PodleĹ› sphere algebras as their coinvariants. The generators of those coideals are given in [1]. More ...
Makoto Yamashita's user avatar
2 votes

Recover Poisson bracket on $C[G]$ using the Lie cobracket $\delta: g \to \Lambda^2 g$

You have the Poisson bivector explicitly at every point $exp(a)$ and you know that it is multiplicative. When $\delta=dr$ was exact you wrote it like $exp(a).r-r.exp(a)$. You have some functions and ...
AHusain's user avatar
  • 973
2 votes
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Poisson structure on the dual Lie algebroid

Depending on you sign convention, this goes as follows. First you denote the bundle projection by $pr\colon E^* \longrightarrow X$. For a section $s \in \Gamma^\infty(E)$ you have a linear function $J(...
Stefan Waldmann's user avatar
2 votes
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Is the antipode anti-bracketed?

Yes. Suppose $A$ is a bracked Hopf algebra. Equip $A [\hbar]/\hbar^2$ with the multiplication $a \cdot_\hbar b = ab + \hbar \{a,b\}$. That this is associative follows from the Leibniz rule --- you ...
Theo Johnson-Freyd's user avatar
2 votes
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coisotropic submanifolds on poisson manifolds

The answer is 'Yes', at least with some additional transversality conditions. Corollary (1.2.6) in Weinstein's Coisotropic calculus and Poisson groupoids states that Let $M$ be a submanifold of ...
José Figueroa-O'Farrill's user avatar

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