Questions tagged [poisson-geometry]

Poisson geometry is the study of varieties endowed with a Poisson structure, which is a certain kind of 2-tensor. This is closely related to symplectic geometry.

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What is a Lagrangian submanifold intuitively?

What are good ways to think about Lagrangian submanifolds? Why should one care about them? More generally: same questions about (co)isotropic ones. Answers from a classical mechanics point of view ...
Jan Weidner's user avatar
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21 votes
1 answer
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Why symplectic geometry gives Poisson geometry

One way I've learned to understand Poisson geometry is to consider it as symplectic geometry with no open conditions - i.e. no condition of nondegeneracy. This idea can be applied to many other ...
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Construct super Poisson brackets on the coordinate rings of Lie super groups

On line 7 of page 61 of the book a guide to quantum groups, a Poisson bracket is defined on $\mathbb{C}[GL_n]$ for every classical $r$-matrix as follows. Let $V$ be a vector space with a basis $v_1, \...
Jianrong Li's user avatar
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Absent 2nd order terms in deformation quantization of Poisson manifolds

I am reading Kontsevich' famous paper on deformation quantization of Poisson manifolds. In section 1.4.2 on page 4 he gives the general formula for the star product associated to a Poisson structure ...
miramo's user avatar
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2 answers
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Does the 4-sphere have a nonzero Poisson structure as a Poisson homogeneous space?

It is known that the 4-sphere does not have a symplectic structure. However, it does admit Poisson structures, for example the zero Poisson structure, which is quite boring. Does it have other, more ...
Manuel Bärenz's user avatar
2 votes
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Two definitions of the super Jacobi identity

In this paper, page 149, the super Jacobi identity is given by \begin{align} J(x, y,z) := (-1)^{|x||z|}[[x, y],z] +(-1)^{|z||y|}[[z,x], y]+(-1)^{|y||x|}[[y,z],x] = 0. \end{align} But in this article, ...
Jianrong Li's user avatar
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1 vote
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Super version of Poisson brackets of tensor products

Let $A$ be a Poisson super algebra ($A$ is a super algebra and $A$ satisfies super Jacobi identity, super commutativity, super Leibniz rule). Super version of the product of two tensor products is \...
Jianrong Li's user avatar
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Finite-dimensional Poisson cohomologies

Suppose we have a poisson manifold $M$ whose Poisson cohomology is finite-dimensional in each degree? Does it mean that our manifold is symplectic? Are there other cohomological criteria of ...
anna abasheva's user avatar
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How to prove a bracket is super anti-commutative?

On page 12 of the paper, there is a formula about super Poisson bracket on a Lie super group $G$: \begin{align} \{\phi, \psi\} = \sum_{\mu, \nu} (-1)^{|\phi||\nu|} r^{\mu \nu} ( R_{\mu} \phi R_{\nu} \...
Jianrong Li's user avatar
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