2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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