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Puraṭci Vinnani
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Computing the Grothendieck-Springer resolution for $G = SL_2$

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.

Puraṭci Vinnani
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