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Lie bialgebras

$\begin{defn} A \textbf{Lie bialgebra} is a pair $(\gerg, \dl)$\ where $\gerg$\ i... ...(X))=0$) \item $\dl([X, Y])=\ad_X\dl(Y) - \ad_Y\dl(X)$ \end{enumerate}\end{defn}$
We have just proven that the tangent space of a Poisson Lie group has a Lie bialgebra structure. To what extend is the converse true ?
$\begin{example} $\gerg$\ abelian Lie algebra. Thus any $\dl\: \gerg\to \La^2\ger... ...algebra which does not integrate to a Poisson Lie group structure. \end{example}$
The point here is
$\begin{lem} Given a 1-cocycle $\gerg\to \gerg\wedge\gerg$\ there is a unique 1-c... ...imply connected Lie group integrating $\gerg$\ (i.e. $\Lie(G)=\gerg$). \end{lem}$
Basically this is all you need to prove
$\begin{thm}[Drinfel'd] The correspondence $G\mapsto \gerg$\ gives you a 1:1 correspondence between Lie bialgebras and Poisson Lie groups. \end{thm}$
Given any Poisson Lie group $(G, \Pi)$ consider its Lie bialgebra $(\gerg, \dl)$ . Then $(\gerg^*, ^t[-,-])$ is a Lie bialgebra. Therefore it integrates to a unique connected, simply connected Poisson Lie group

called the dual Poisson Lie group of

Lie bialgebras form a category. Morphisms are those homomorphisms which respect $\dl$

$\begin{displaymath} \xymatrix{ \gerg\wedge\gerg \ar[r]^{\chi\tensor\chi} & \gerg... ... \\ \gerg \ar[u]^{\dl} \ar[r]_{\chi} & \gerg' \ar[u]_{\dl'} } \end{displaymath}$

$\begin{prop} Given a Lie bialgebra $(\gerg, \dl)$, the vector space $\gerg^*$ ha... ...$\gerg$, and the bracket $[-,-]'$\ in $\gerg^*$\ being dual to $\dl$. \end{prop}$

$\begin{defn} $\gerg^*$\ is called dual bialgebra of $\gerg$. \end{defn}$

$\begin{examples} \mbox{} \begin{enumerate} \item Any Lie algebra with $\dl=0$. \... ...ndard structures are equivalent up to conjugation. \end{enumerate}\end{examples}$

Pawel Witkowski 2006-06-26