## nowhereman 5 years ago Does the Kirillov-form depend on the point you choose from the coadjoint orbit?

1. Owlfred

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

2. nowhereman

On a Lie-group $$G$$ you have the conjugation action on itself: $g,h\in G, α_g(h) := ghg^{-1}$ and it's differential at the identity element $$e \in G$$ is an isomorphism of the Lie-algebra $$\mathrm{Ad(g)} := \mathrm d_e α_g \in \mathrm{GL}(\mathfrak g)$$ and the dual is called the coadjoint action: $\mathrm{Ad}^*(g) := \mathrm{Ad}(g^{-1})^* \in \mathrm{GL}(\mathfrak g^*)$ It admits an orbit at any $$α \in \mathfrak g^*$$:$\mathcal O_α := \{\mathrm{Ad}^*(g)(α)\mid g \in G\}\simeq G/G_α$ where $G_α := \{g \in G \mid \mathrm{Ad}^*(g)(α) = α\}$ So G acts on $$\mathcal O_α$$, creating fundamental vector fields $$\tilde X$$ for $$X \in \mathfrak g$$. Then you can define the sympletic form $ω(\tilde X, \tilde Y) = α([X,Y])$ So the question is: For $$β\in \mathcal O_α$$ do we have $ω(\tilde X, \tilde Y) = β([X,Y])\quad \text{?}$

3. amistre64

my guess is; it doesnt matter ....

4. nowhereman

I would like to see a proof. If you want to use the exponential map, that would be ok too ;-)