Let $G = SL\_2, \mathfrak{g} = \mathfrak{sl}\_2$, $B$ the Borel subgroup, and $\mathfrak{u}$ the unipotent radical; so that $G/B = \mathbb{P}^1$; how does $\widetilde{\mathfrak{g}}$ decompose as a vector bundle over $\mathbb{P}^1$? Recall the definition:


$\widetilde{\mathfrak{g}} = $ {$(X, gB) \in \mathfrak{g}^\* \times \mathbb{P}^1 | X|\_{g \mathfrak{u}} = 0$}


This should be simple, but I'm having trouble. 



Computing the Grothendieck-Springer resolution for $G = SL_2$
Let $\frak{b}\subseteq\frak{g}$ denote the Lie algebra of your Borel $B$. There is a natural $G$-equivariant isomorphism $\tilde{\frak{g}}\cong G\times\_B\frak{b}$ of vector bundles over $G/B$, where $G\times\_B\frak{b}$ is the associated bundle arising from the adjoint representation of $B$ on $\frak{b}$. Let us take $B$ to be the standard Borel of upper-triangular matrices of unit determinant, and consider the maximal torus $T\subseteq B$ of diagonal matrices. Under the adjoint representation restricted to $T$, $\frak{b}$ decomposes into irreducible subrepresentations of weights $0$ and $2$. Conjecturally, anyway, I would not be surprised if you could use this to show that $\tilde{\frak{g}}\cong\mathcal{O}(0)\oplus\mathcal{O}(2)$. Still, I am not sure about this, and there are a few details to be resolved. I hope this is useful.


