$\DeclareMathOperator\ad{ad}$Let $\mathfrak {g}$ be a Lie bialgebra. Then $\mathfrak {g}^{\ast}$ is also a Lie bialgebra which is dual to $\mathfrak {g}$. Let the brackets on $\mathfrak {g}$ and $\mathfrak {g}^{\ast}$ be denoted by $b$ and $b'$ respectively. Then how to define the coadjoint action of $\mathfrak {g}^{\ast}$ on $\mathfrak {g}$? I am familiar with the coadjoint action of $\mathfrak {g}$ on $\mathfrak {g}^{\ast}$ which is defined as follows:

Given $x \in \mathfrak {g}$ we define $\ad_{b}^{\ast} (x) = (\ad_{b} (-x))^{\ast}$. In terms of the pairing $(\cdot, \cdot)$ between $\mathfrak {g}$ and $\mathfrak {g}^{\ast}$ coadjoint action takes the following form:

$$\left (\ad_{b}^{\ast} (x) (\alpha), y \right ) = \left (\alpha, -\ad_{b} (x) (y) \right ) = - \alpha (b(x,y)).$$ Now how to define coadjoint action of $\mathfrak {g}^{\ast}$ on $\mathfrak {g}$? In one of the books on Poisson structures I have come across that it is being defined in terms of the pairing $(\cdot, \cdot)$ as follows: $$\left (b'(\xi, \eta), x \right ) = - \left (\eta, \ad_{b'}^{\ast} (\xi) (x) \right ).$$ Now $\ad_{b'} (\xi) \in \operatorname {End} (\mathfrak {g}^{\ast \ast})$. Then how can it act upon $x \in \mathfrak {g}$?

Any help in this regard would be warmly appreciated.

3more comments