Is the group-theoretic Grothendieck-Springer resolution Calabi-Yau?

Solution 1:

I think I have an answer. Let $p: E \rightarrow X$ be a vector bundle over $X$. Since this is a smooth map, we have a short exact sequence $$0 \rightarrow p^* \Omega^1_X \rightarrow \Omega^1_E \rightarrow \Omega^1_{E/X} \rightarrow 0$$ and taking top exterior powers, $$\omega_E \simeq p^* \omega_X \otimes \Lambda^n \Omega^1_{E/X}$$ Checking the definitions gives us $\Omega^1_{E/X} \simeq p^* \mathcal{E}^\vee$.

For the Lie algebra Grothendieck-Springer resolution, we take $\mathcal{E}$ to be a trivial extension of the cotangent sheaf. Fix this notation. For the group-theoretic version $p^*: \tilde{G} \rightarrow G/B$ is a $B$-bundle over $G/B$, but it is still smooth, and the short exact sequence still splits. The result we want follows from the claim that $\Omega^1_{\tilde{G}/(G/B)} \simeq p^* \mathcal{E}^\vee$. As a warmup, we can compute $\Omega^1_B$; as an equivariant sheaf, this is $\mathcal{O}_B \otimes \mathfrak{b}^*$. By $G$-equivariance, $\Omega^1_{\tilde{G}/X} = p^*\mathcal{F}^\vee$ where the total bundle of $\mathcal{F}$ is $G \times_B \mathfrak{b}$, but this is just $\mathcal{E}$.