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Manin triples

$\begin{defn} Let $\gerg$\ be a Lie algebra with a non degenerate invariant symme... .... Then $(\gerg, \gerg_+, \gerg_-)$ is called a \textbf{Manin triple}. \end{defn}$
Using the form we can identify

$\begin{displaymath} \gerg_-\cong (\gerg_+)^*,\quad \gerg_+\cong (\gerg_-)^* \end{displaymath}$

In particular $\dim \gerg_+ = \dim\gerg_-$ .
$\begin{thm} \mbox{} \begin{enumerate} \item Suppose $(\gerg, \gerg_+, \gerg_-)$\... ...\gerg^*$\ with this form and bracket is a Manin triple. \end{enumerate}\end{thm}$

$\begin{proof} Let us rewrite the cocycle condition in a Lie bialgebra \begin{dis... ...a} \end{align*}and this is formula obtained before. This proves (1). \end{proof}$

$\begin{prop} Let $(\gerg, \dl)$\ be a Lie bialgebra, $(D\gerg, [-,-])$\ Lie alge... ...dl(X+\xi) = \dl(X)+ ^t[-,-](\xi) \end{displaymath}is a Lie cobracket. \end{prop}$

$\begin{example} $\gerg$\ complex simple Lie bialgebra, $\gerg\hookrightarrow \ge... ...isplaymath}Then $(\gerg\oplus\gerg, \gerg, S)$\ is a Manin triple. \end{example}$

$\begin{example} With the notation as before $(\gerg\oplus\gh, \mathfrak{b}_+, \mathfrak{b}_-)$ is a Manin triple. \end{example}$

Pawel Witkowski 2006-06-26