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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.

4 votes
2 answers
486 views

Classifications of Lie bialgebras

What is the current status of the classifications of Lie bialgebras? In particular, has the following problem been solved? Let $gl_n$ be the general linear Lie algebra. Classify all Lie cobrackets $\d …
1 vote
1 answer
93 views

How to show that the structure constant on $\mathcal{G}^*$ is $C_{c}^{ab} = f_{cd}^b r^{ad} ...

Let $(\mathcal{G}, \mathcal{G}^*, \delta)$ be a Lie bialgebra. Suppose that the structure constant on $\mathcal{G}^*$ and $\mathcal{G}$ are \begin{align} & [t^a, t^b]_* = C_c^{ab} t_c, \\ & [t_a, t_b] …
1 vote
0 answers
73 views

Decompose elements in $SL_2$ as a pair of elements in $SL_2^*$.

I have a question about decomposing elements in $SL_2$ as a pair of elements in $SL_2^*$. Here $SL_2^*$ is the dual Poisson Lie group of $SL_2$ which is defined as follows. Let $G$ be a Poisson-Lie …
2 votes
1 answer
241 views

Trying to understand dressing actions

I am reading the lecture notes and trying to understand dressing actions. Let $G$ be a Poisson-Lie group and $G^*$ its dual Poisson-Lie group. In the lecture notes above, Proposition 5.22 on page 80 …
1 vote
1 answer
183 views

Is the action $T \times G \to G$ Poisson?

Let $G$ be a Poisson-Lie group. Let $M$ be a symplectic manifold. In the paper, the third paragraph of page 1238, it is said that an action $G \times M \to M$ is called Poisson if $G \times M \to M$ …
0 votes
Accepted

How to prove a bracket is super anti-commutative?

This question was solved by Vladimir Dotsenko in the comments of the question.
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0 votes
1 answer
185 views

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} \ …
1 vote
1 answer
189 views

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, …
1 vote
1 answer
395 views

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 \ …
1 vote

Is the antipode anti-bracketed?

This is also proved in Lemma 2.6 of the paper: CO-POISSON COALGEBRA AND CO-POISSON HOPF ALGEBRA STRUCTURES ON $k[x_1, x_2, \ldots, x_d]$ .
Jianrong Li's user avatar
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1 vote
2 answers
118 views

Is the antipode anti-bracketed?

In the book Algebras of Functions on Quantum Groups: Part I, Remark 3.1.4, we have the following result. Let $A$ be a Poisson Hopf algebra. That is, $A$ is both a Hopf algebra and a Poisson algebra …
2 votes
2 answers
1k views

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, …
1 vote
1 answer
190 views

Tensor product of two Poisson modules

Let $H$ be a Poisson algebra. A Poisson $H$-module is a vector space $V$ with two bilinear maps \begin{align} H \otimes V \to V \\ (h,v) \mapsto h.v, \end{align} \begin{align} H \otimes V \to V \\ (h, …
4 votes
1 answer
210 views

Cluster algebra structure compatible with Poisson brackets

Let $X$ be a Poisson variety. There is a concept "cluster algebra structure compatible with Poisson structure" introduced in the paper. Suppose that we construct a maximal independent set of functio …
3 votes
2 answers
284 views

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

By a theorem of Drinfeld, there is a one to one correspondence between Lie bialgebras and Poisson Lie groups. Therefore given a Lie cobracket $\delta: g \to \Lambda^2 g$, there is a Poisson bracket on …

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