$\DeclareMathOperator\ad{ad}$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.

  • 3
    $\begingroup$ Are your algebras finite-dimensional? Then $\mathfrak g \to \mathfrak g^{**}$ is an isomorphism. $\endgroup$
    – LSpice
    Jan 30, 2023 at 20:17
  • $\begingroup$ @LSpice$:$ I think it is implicitly assumed in the context. $\endgroup$ Jan 30, 2023 at 20:41
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    $\begingroup$ @AnilBagchi. There is a canonical isomorphism between a finite-dimensional vector space and its double dual; yes, there are many isomorphisms between a space and its double dual (vector spaces have heaps of automorphisms), but there is one especially natural one, and it is that one you use to identify the two spaces in this context. $\endgroup$ Jan 30, 2023 at 21:00
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    $\begingroup$ This is not completely explicit but $\xi\cdot g\in\mathfrak g$, for $\xi\in\mathfrak g^*$ and $g\in\mathfrak g$, is uniquely determined by the condition $\forall\eta\in\mathfrak g^*\ \eta(\xi\cdot g)=b'(\eta,\xi)(g)$. $\endgroup$ Jan 31, 2023 at 7:23
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    $\begingroup$ Given that you understand the action of $\mathfrak{g}$ on $\mathfrak{g}^*$, this means that you understand the action of $\mathfrak{g}^*$ on $\mathfrak{g}^{**}$. This then suggests the following question. Suppose that $\mathfrak{g}$ isn't finite dimensional. Then the natural map $\mathfrak{g}\to\mathfrak{g}^{**}$ is injective, but not surjective. Is $\mathfrak{g}$ then an invariant subspace under the action of $\mathfrak{g}^*$? Or do you know of an example where it is not invariant? Or is the problem that if $\mathfrak{g}$ is not finite dimensional then $\mathfrak{g}^*$ isn't a Lie bialgebra? $\endgroup$ Oct 29, 2023 at 21:07

1 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.

  • $\begingroup$ @Michael Barz $:$ I think this is what you meant to say. Am I right? $\endgroup$ Feb 1, 2023 at 6:10

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