concerning coadjoint representation

Question

Let $\xi $ be the vector field on $\frak{g}^*$ (dual of Lie algebra) which correspond to element $X$ of the Lie algebra $\frak{g}$. Then why have we $\xi(F)=K_*(X)F$ where here $K=Ad^*(g)$ is coadjoint representation and $F\in \frak{g}^*$?.

Answer 1

Im not really sure to be able to answer exactly at your question, but i can try to help:

On $\mathfrak{g}^\ast$, you can define a Poisson bracket as follow: $$\forall \xi^\ast \in \mathfrak{g}^\ast, F,G \in \mathcal{C}^\infty(\mathfrak{g}^\ast), \lbrace F,G \rbrace(\xi^\ast):= <\xi^\ast,[dF(\xi^\ast),dG(\xi^\ast)>$$ Then for any fonction $H$n you define the Hamiltonian vector field $X_H$ by $$X_H.F(\xi^\ast):={F,H}(\xi^\ast)=<\xi^\ast,[dF(\xi^\ast),dH(\xi^\ast)]>=<ad^\ast_{dH(\xi^\ast)},dF(\xi^\ast)>$$ (Note that $dF(\xi^\ast) \in (\mathfrak{g}^\ast)^\ast=\mathfrak{g}$, with the canonical identification $\mathfrak{g}^\ast = T_{\xi^\ast}\mathfrak{g}^\ast$, then what i say is consistent...)

So we deduce that $$X_H(\xi^\ast)=ad^\ast_{dH(\xi^\ast)} \xi^\ast$$ with the previous identification.

In what you write, i don't understand who is $g$,but i think with what i say, and the fact that ad is the derivative of Ad, you can conclude, for a hamiltonian vector field...

I hope it helps...

Léo