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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?

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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 $\mathcal{E}$.

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