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!