Problem in understanding the coadjoint action of $\mathfrak {g}^{\ast}$ on $\mathfrak {g}$

Question

$\DeclareMathOperator\ad{ad}$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.

Answer 1

As @Michael Barz pointed out in his comment above it turns out that $$\text {ad}_{b'}^{\ast} (\xi) = \varphi^{-1} \circ \left (\text {ad}_{b'} (-\xi) \right )^{\ast} \circ \varphi$$ where $\varphi : \mathfrak {g} \longrightarrow \mathfrak {g}^{\ast \ast}$ is the canonical isomorphism given by $\varphi (x) (f) = f(x)$ for all $x \in \mathfrak {g}$ and $f \in \mathfrak {g}^{\ast}.$ Now let us compute $\left (\eta, \text {ad}_{b'}^{\ast} (\xi) (x) \right ).$ We have $$\begin{align*} \left (\eta, \text {ad}_{b'}^{\ast} (\xi) (x) \right ) & = \eta \left (\text {ad}_{b'}^{\ast} (\xi) (x) \right ) \\ & = \eta \left (\varphi^{-1} \left (\varphi (x) \circ \text {ad}_{b'} (-\xi) \right ) \right ) \end{align*}$$

Now $\varphi (x) \circ \text {ad}_{b'} (-\xi) \in \mathfrak {g}^{\ast \ast}.$ Since $\varphi$ is an isomorphism (in particular, onto) there exists $y \in \mathfrak {g}$ such that $\varphi (y) = \varphi (x) \circ \text {ad}_{b'} (-\xi).$ So we have $$\begin{align*} \left (\eta, \text {ad}_{b'}^{\ast} (\xi) (x) \right ) & = \eta (y) \\ & = \varphi (y) (\eta) \\ & = \left (\varphi (x) \circ \text {ad}_{b'} (-\xi) \right ) (\eta) \\ & = \varphi (x) (b'(-\xi, \eta)) \\ & = - b'(\xi, \eta) (x) \\ & = - (b'(\xi, \eta), x) \end{align*}$$ as required.