Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathfrak{g}$ be a simple Lie algebra; let $R = \mathfrak{g}^{*, reg}$ denote the regular locus in the dual Lie algebra. Consider the vector bundle $\mathfrak{z}$ over $R$, whose fiber over a point $\xi \in R$ is the stabilizer of $\xi$ in the dual to $\mathfrak{g}$. Using the isomorphism $R/G = \mathfrak{t}/W$, we have a map $p: R \rightarrow \mathfrak{t}/W$; let $\mathcal{T}'$ denote the co-tangent bundle to $R$.

Question: Why is $p^* \mathcal{T}' = \mathfrak{z}$?

It should suffices to show that the fibers of these two vector bundles over a point in the base, $\xi$, are the same, but I was having trouble computing the fibers of $p^* \mathcal{T}'$.

(This fact was mentioned in the second paragraph of Section $2.6$, pg 6 of this paper.) Sorry about my poor choice of notation, I was having trouble using Latex on MO.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

I think this can be understood in terms of Hamiltonian reduction.

The group $G$ acts on $\mathfrak g^{\ast,reg}$ inducing an action on $T^\ast \mathfrak g^{\ast,reg}$. The associated moment map $\mu: T^\ast \mathfrak g^{\ast,reg} \to \mathfrak g^\ast$ is:

$\mu(\xi,x) = coad(x)(\xi) - \xi$.

A basic result of Hamiltonian reduction is that a covector at $[\xi]$ on the quotient $\mathfrak g^{\ast,reg}/G = \mathfrak t^\ast /W$ is given by an element $x \in T^\ast _\xi \mathfrak g^{\ast,reg}$ such that $\mu(x,\xi)=0$. Noting that $\mu^{-1}(0) \to \mathfrak g^{\ast,reg}$ is equal to the bundle $\mathfrak z$ of centralizers implies your result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.