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.