Let $G$ be a finite subgroup of $U(m)$ such that $G$ acts freely on $\mathbb C^m \setminus \{0\}$.
If $\mathbb C^m/G$ has a crepant resolution, can we necessarily derive that $G \subset SU(m)$?
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$\begingroup$ You need to exclude the case that $m$ equals $1$. More precisely, the correct hypothesis is that the non-free locus has no irreducible component of codimension equal to $1$. $\endgroup$– Jason StarrCommented May 7, 2023 at 12:50
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