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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.

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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.

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