In the case of $sl_2$, there are three Lie bialgebra structures. We have three cobrackets $\delta: sl_2 \to \Lambda^2 sl_2$. Each $\delta$ can be written as $\delta=d r$ for some matrix $r$. Therefore all the Lie bialgebra structure on $sl_2$ are coboundary. Are all the Lie bialgebra structure on $sl_n$ coboundary for each $n$?
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asked Feb 16, 2016 at 2:41
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1$\begingroup$ As Makoto Yamashita writes it is because H^2(semisimpleLieAlg) = 0 - "Whitehead's lemma" en.wikipedia.org/wiki/Whitehead%27s_lemma_(Lie_algebras) $\endgroup$– Alexander ChervovCommented Feb 16, 2016 at 7:26
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The compatibility between bracket and cobracket $\delta$ can be interpreted as $\delta$ being a $1$-cocycle. Then Whitehead's lemma implies the existence of such $r$.
