Is the group-theoretic Grothendieck-Springer resolution Calabi-Yau? Any cotangent bundle is Calabi-Yau (by which I mean the canonical bundle is trivial), so the Springer resolution $T^*(G/B)$ is Calabi-Yau.  I think that the Grothendieck-Springer resolution $\tilde{\mathfrak{g}}$ is Calabi-Yau -- this is because a vector bundle over $X$ with sheaf of sections $\mathcal{E}$ has canonical bundle $\pi^* \omega_X \otimes \Lambda^{top} \pi^* \mathcal{E}^\vee$.  One can check that $\mathcal{E}$ is an extension of the cotangent bundle by trivial bundles.
Is the group-theoretic resolution
$$\tilde{G} = \{(x, g) \in G/B \times G \mid g \in xBx^{-1}\}$$
also Calabi-Yau, and how would one argue it?
 A: 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 $\mathcal{E}$.
