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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 $C[G]$ corresponding to $\delta$, where $G$ is a Lie group and $g$ the Lie algebra of $G$. I would like to figure out how to compute the Poisson bracket on $C[G]$ using $\delta$ explicitly.

We can recover the cobracket on $g$ using the Poisson bracket on a Poisson Lie group $G$ easily. For example, consider $G = SL_2$. There is a Poisson bracket on $C[SL_2]$: \begin{align} & \{x_{11}, x_{12}\} = -1/2 x_{11} x_{12}, \\ & \{x_{11}, x_{21}\} = -1/2 x_{11} x_{21}, \\ & \{x_{11}, x_{22}\} = - x_{12} x_{21}, \\ & \{x_{12}, x_{21}\} = 0, \\ & \{x_{12}, x_{22}\} = -1/2 x_{12} x_{22}, \\ & \{x_{21}, x_{22}\} = -1/2 x_{21} x_{22}. \end{align} The corresponding cobracket $\delta$ on $sl_2 = \text{Span}(e,f,h)$ is \begin{align} & \delta(e) = -1/2 h \wedge e, \\ & \delta(f) = -1/2 h \wedge f, \\ & \delta(h) = 0. \end{align} The procedure is as follows (I learned this from Theo Johnson-Freyd). Choose local coordinates on $G$ centered at the identity. For example, choose: \begin{align} y_{11} & = x_{11} - 1, \\ y_{12} & = x_{12}, \\ y_{21} & = x_{21}, \\ y_{22} & = x_{22}-1. \end{align} Then \begin{align} & \{ y_{11}, y_{12} \} = \{ x_{11} - 1, x_{12} \} = \{ x_{11}, x_{12}\} = -1/2 x_{11} x_{12} = -1/2 (y_{11} + 1) y_{12} = -1/2 y_{12} - 1/2 y_{11}y_{12}, \\ & \{ y_{11}, y_{21} \} = -1/2 y_{21} + (\text{quadratic or higher}), \\ & \{ y_{12}, y_{21} \} = \text{quadratic or higher}. \end{align} We obtain the Lie bracket on $g^*$ if we take only linear terms. Dualizing the Lie bracket $\Lambda^2 g^* \to g^*$ gives the Lie cobracket on $g$.

Now I try to recover the Poisson bracket on $C[G]$ using a given cobracket $\delta: g \to \Lambda^2 g$. For example, let $G = SL_2$ and suppose that \delta(e) = -1/2 h \wedge e, \delta(f) = -1/2 h \wedge f, \delta(h) = 0. How to recover the corresponding Poisson bracket on C[SL_2]? I found that there is a formula in the book Introduction to quantum groups by Prof. Etingof and Prof. Schiffmann: $$p(e^a) = \sum_{n \geq 0} \frac{(- ad(a))^n}{(n+1)!} \delta(a), \ a \in g.$$ But it seems that it is not easy to use this formula to compute {x_{11}, x_{12}} = -1/2 x_{11} x_{12}?

Any help will be greatly appreciated!

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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 you need to decompose their differentials in terms of the basis of one forms that you wrote the bivector in. That way when you write $\pi ( df \wedge dg)$ you will just be reading off some coefficients. If you knew $f=a_{j_1}a_{j_2}\cdots$,$g=a_{i_1} \cdots$ were factorizations where the Poisson bivector was written with a sum of $\frac{\partial}{\partial a_k} \wedge \frac{\partial}{\partial a_l}$ kinds of terms, you would have written them in a dual pair of bases so the computation would be easier. See Gekhtman Shapiro Vainshtein.

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Not an answer, but too long for a comment.

The Poisson bracket on the group can be written in $r$-matrix notations:

$$ \{ M \frac{ \otimes}{,} M \} = [ r^{Group} , M \otimes M ] $$,

Taking $M = 1 + \epsilon L$, $r^{Group} = 1 + \epsilon r $ and looking on the terms in $\epsilon^2$ we get:

$$ \{ L \frac{ \otimes}{,} L \} = [r , L \otimes 1 + 1 \otimes L ] $$,

That is $r$-matrix bracket on Lie algebra. The point is that it is given by the same $r$-matrix. The only difference is that one bracket is quadratic (on the group), the other is linear (on Lie algebra). ($1+r$ or $r$ gives same bracket, because $1$ dissappears in RHS since it is given by commutator).

As far as I understand your question is how to go back from bracket on Lie algebra to bracket on Lie group. Well, as far as I understand that is somehow just saying that bracket on the group satisfies what we need: invariance and and limit to Lie algebra, the bracket satisfying such properties is unique and that is why it is what we need.

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  • $\begingroup$ thank you very much. If we have a Lie cobracket $\delta: g \to \Lambda^2 g$, how to compute $r^{Group}$? $\endgroup$ – Jianrong Li Feb 13 '16 at 2:32
  • $\begingroup$ @JianrongLi The point is that they are the same $r = r^{Group}$ , the transformations which I made $r = 1 + \epsilon r $ they do not change the bracket, because RHS of bracket is given by commutator $[r, M \otimes M]$ so it exactly equals to $ [ (1+ r) , M \otimes M]$ because $1$ commute with everything. At least that how I remember things - I have not thought on these for years, it might be I misremember something, but the idea is like that. $\endgroup$ – Alexander Chervov Feb 13 '16 at 9:53
  • $\begingroup$ thank you very much. I still have some questions. Using $\delta: L \to \Lambda^2 L$, we can obtain $[,]_{L^*}: \Lambda^2 L^* \to L^*$. But how to compute the r-matrix $r$ for the Lie algebra $L$? $\endgroup$ – Jianrong Li Feb 13 '16 at 14:25
  • $\begingroup$ @JianrongLi I guess you know that - if $\delta$ is coboundary it gives rise to $r$. The point is that in $r$-matrix notation corresponding bracket on $L^*$ will have the form $\{L \otimes L\} = [r , L\otimes 1 + 1 \otimes L ]$, $\endgroup$ – Alexander Chervov Feb 13 '16 at 15:52

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