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Poisson actions

Recall some notations. Let $G$ be a Lie group, $\gerg=\Lie(G)$ its Lie algebra, $L_g, R_g\: G\to G$ left and right translations with derivatives $L_{g,*}\: T_hG\to T_{gh}G$, $R_{g,*}\: T_hG\to T_{hg}G$.

Let $(G, \Pi)$ be a Poisson Lie group, i.e.

\begin{displaymath}
\Pi(g_1\cdot g_2)=L_{g_1, *}\Pi(g_2) + R_{g_2, *}\Pi(g_1)
\end{displaymath}

and let $\eta\: G\to \La^2 T_e G = \La^2\gerg$ be

\begin{displaymath}
\eta(g)=R_{g^{-1}, *} \Pi(g).
\end{displaymath}

Then $\eta$ is a 1-cocycle of $G$ with respect to adjoint action on $\La^2\gerg$, i.e.

\begin{displaymath}
\eta(g_1g_2)=\eta(g_1)+\Ad_{g_1}\eta(g_2)
\end{displaymath}

Let $\dl\: \gerg\to\La^2\gerg$,

\begin{displaymath}
\dl(X)=\ddt{t}\eta(e^{tX})\big\vert _{t=0}
\end{displaymath}

Then $(\gerg, \dl)$ is a Lie bialgebra
\begin{align*}
(\gerg, [-,-]) &\quad \text{is Lie} \\ [1ex]
(\gerg^*, ^t \dl) &\quad \text{is Lie} \\
\end{align*}
satisfying compatibility

\begin{displaymath}
\dl([X, Y])=\ad_X\dl(Y) - \ad_Y \dl(X).
\end{displaymath}

The Lie algebra $\gerg^*$ integrates to a (unique) connected (simply connected) Poisson Lie group $G^*$. Furthermore on $\gerg\oplus\gerg^*$ we have the following Lie bracket

\begin{displaymath}[X+\xi, Y+\eta]= ([X, Y] + \ad_X^*\eta - \ad_Y^*\xi, [\xi, \eta] + \ad_{\xi}^* Y
-\ad_{\eta}^* X)
\end{displaymath}

and Lie cobracket

\begin{displaymath}
\dl_D(X+\xi) = \dl(X) + \dl^*(\xi)
\end{displaymath}

This makes $\gerg\oplus\gerg^*$ a Lie bialgebra, which is called Drinfeld double of a Lie algebra $\gerg$. It integrates to (a unique sonnected, simply connected) Poisson Lie group $DG$ called Drinfeld double of a Lie group $G$.



Subsections
next up previous contents
Next: Poisson actions Up: From Poisson to Quantum Previous: Manin triples   Contents
Pawel Witkowski 2006-06-26