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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.
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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 $k[x_1, x_2, \ldots, x_d]$
.
1
vote
2
answers
118
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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 …
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 …
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.
0
votes
1
answer
166
views
Natural Poisson brackets on $S(V^*)$
Let $V$ be a finite dimensional vector space and $S(V)$ the corresponding symmetric algebra. Suppose that we have a Poisson bracket $\lambda = \{,\}: S(V) \otimes S(V) \to S(V)$. Let $V^*$ be the dual …
1
vote
1
answer
606
views
Questions about Sklyanin bracket
For every classical r-matrix $r$, there is a Poisson bracket called Sklyanin bracket associated to $r$. It is defined in (3.3) of page 5 in (http://arxiv.org/pdf/1101.0015v2.pdf) as follows.
\begin{al …
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} \ …
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, …
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
0
answers
73
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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 …
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, …
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
\ …
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 …
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 …