Let $X_t$ be a family of algebraic varieties (my interest is CalabiYau varieties, but I don't think that's important) over $\mathbb{C}$, smooth for $t \neq 0$, on which a group $G$ acts fibrewise. Suppose further that $X_0$ admits at least one crepant resolution. Does there always exist an equivariant crepant resolution? If not, are there conditions under which such exists?
1 Answer
Consider $\mathbb Z/2$ acting on $\{xyzw=t\}$ by $x\leftrightarrow y$. This swaps the two small resolutions of the central fibre (the 3fold ordinary double point $xy=zw$ in $\mathbb C^4$). So there can't be an equivariant small resolution.
(A formal proof might go along these lines: $\mathbb Z/2$ does act on the blow up of the ODP, swapping the two rulings of the $\mathbb P^1\times\mathbb P^1$ exceptional divisor. The small resolutions are contractions of this blow up. If $\mathbb Z/2$ acted on one of them, it would act on its $H^2$. Pulling back, its action on $H^2(\mathbb P^1\times\mathbb P^1)\cong\mathbb Z\oplus\mathbb Z$ would be the identity on the contracted $\mathbb Z$ summand, contradicting the fact that it swaps the summands.)
However if your (finite?) group action permutes the singular loci with no stabilisers then there would surely be a crepant resolution.

$\begingroup$ By the way, this example doesn't preserve the holomorphic 3form, so might not interest you. Can anyone think of an example where the group preserves the 3form ? $\endgroup$ Commented Apr 17, 2012 at 21:39

$\begingroup$ Thanks Richard, that's a very clear counterexample. It's not a problem that the holomorphic 3form is noninvariant; in fact, for the work which prompted this question, I'm interested in exactly such examples. $\endgroup$ Commented Apr 18, 2012 at 10:52