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Dressing actions

Take $\xi\in\gerg^*$ , and denote by $\xi^L$ the associated left invariant 1-form and by $\xi^R$ the associated right invariant 1-form, i.e.

$\begin{displaymath} \xi^L(g)=L_{g^{-1}}^*\xi\in T^*_gG;\qquad \xi^R(g)=R_{g^{-1}}^*\xi\in T^*_gG\, . \end{displaymath}$

$\begin{defn} Define $\la,\rho\:\gerg^* \to\gX(G)$ \begin{displaymath} \la(\xi):=... ...egin{displaymath} \rho(\xi):= -\char93 _{\Pi}(\xi^R) \end{displaymath}\end{defn}$

$\begin{lem} $\la$\ is a Lie algebra morphism, $\rho$\ is a Lie algebra antimorphism. \end{lem}$

$\begin{proof} \begin{align*} \la([\xi_1, \xi_2]) &= \char93 _{\Pi}([\xi_1, \xi_2... ... &= -[\char93 _{\Pi}(\xi_1^R), \char93 _{\Pi}(\xi_2^R)] \end{align*}\end{proof}$
Therefore $\la$ defines an infinitesimal left action of $\gerg^*$ on

and $\rho$ defines an infinitesimal right action of $\gerg^*$ on

. These are called infinitesimal dressing actions.
$\begin{exer} Prove that the inversion map $S\: g\mapsto g^{-1}$\ intertwines left and right infinitesimal dressing actions, i. e. $S_*\circ \la=\rho$. \end{exer}$

$\begin{defn} If the dressing action can be integrated to a global action of $G^*$ on $G$, the Poisson Lie group $G$\ is said to be \textbf{complete}. \end{defn}$
We recall that the notion of Poisson-Lie group is self dual, therefore the above defines also the left and right infinitesimal dressing actions of $\gerg$ on the dual Poisson-Lie group

.
$\begin{prop} Locally symplectic leaves of $G$\ coincide with the orbits of the l... ...oup is complete then the symplectic leaves coincide with such orbits. \end{prop}$

$\begin{proof} By definition left dressing vector fields are hamiltonian vector f... ... Thus $\sO$\ is a Poisson submanifold of $S$\ and therefore $\sO=S$. \end{proof}$
Dressing action is the most powerful tool for computing the symplectic foliation of Poisson Lie group.
$\begin{prop} Taking the derivative at $e$\ of left (resp. right) infinitesimal d... ...n you get (resp. minus) the coadjoint action of $\gerg^*$ on $\gerg$. \end{prop}$

$\begin{thm}[Semonov-Tian-Shansky,\cite{}] Left and right dressing actions are Poisson actions. \end{thm}$
How can one integrate the dressing action ? Recall the Drinfeld double $D\gerg = \gerg \oplus\gerg^*$ . Then locally (around $e\in DG$ )

$\begin{displaymath} DG\vert _U = GG^*\vert _U \end{displaymath}$

For any $d\in U$ denote with

its component in

, and with $d_{G^*}$ its component in

, such that $d=d_Gd_{G^*}$ .
$\begin{prop} The local action given by this splitting \begin{displaymath} \gerg^... ...f $G^*$\ on $G$\ integrating the infinitesimal dressing action $\la$. \end{prop}$
The proof relies on a characterisation of the dressing action we could not give.

Whenever holds globally you have the global dressing action.
$\begin{example} Standard Poisson Lie structure on $K$\ compact. $DK=G$\ complex ... ...refore symplectic leaves on $K$\ are orbits of an $AN_+$\ action. \end{example}$
Take $(G, \Pi=0)$ . Then $G^*\simeq \gerg^*$ abelian Lie group with Lie-Poisson bracket. The dressing action of on is given by

$\begin{displaymath} \la\:\gerg\to\gX(G^*) \end{displaymath}$

$\begin{displaymath} X\mapsto \char93 _{\mathrm{LP}}(X^L) \end{displaymath}$

where

is identified with an invariant 1-form on

(remark that $T_eG^*=\gerg^*$ , $T_e^* G^*=\gerg^{**}=\gerg$ ).

$\begin{displaymath} \scalar{\underbrace{\char93 _{\mathrm{LP}}(X^L)}_{\in\Om^1_{... ...iss{X^L}{Y^L}(\xi) =\scalar{\xi}{[X,Y]}=\scalar{\ad_X^*\xi}{Y} \end{displaymath}$

Therefore $\char93 _{\mathrm{LP}}(X^L)$ as vector field is the same as $-\ad^*_X$ . Thus locally it is given by coadjoint action of

on $\gerg^*$ . But this action is global. We recover the result that symplectic leaves for the Lie-Poisson structure are orbits of the coadjoint action.

How to integrate the dressing action ? Recall that the Drinfeld double is a Lie bialgebra on $D\gerg = \gerg \oplus\gerg^*$ which integrates to a Poisson Lie group . Then locally around $e\in DG$ we have

$\begin{displaymath} DG\vert _U = GG^*\vert _U \end{displaymath}$

Let $d\in U\subseteq DG$

$\begin{displaymath} d:=d_G\cdot d_{G^*} \end{displaymath}$

with obvious notation.
$\begin{prop} The local action given by the splitting \begin{displaymath} DG\vert... ...erg)_G \end{displaymath}integrates the infinitesimal dressing action. \end{prop}$

$\begin{remark} When you have a global splitting of the double, you have a global dressing action. \end{remark}$

$\begin{examples} \mbox{} \begin{enumerate} \item $K$\ compact with standard Pois... ... action of $G$\ on $G^*$\ is the coadjoint action. \end{enumerate}\end{examples}$

$\begin{thm} Let $g\in G$\ (around $e$). The leaf through $g$\ locally is the ima... ...d{displaymath}If the dressing action is global they are exactly those. \end{thm}$

Next: Quantization Up: Poisson actions Previous: Poisson homogeneous spaces Contents

Pawel Witkowski 2006-06-26