MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X_t$ be a family of algebraic varieties (my interest is Calabi-Yau varieties, but I don't think that's important) over $\mathbb{C}$, smooth for $t \neq 0$, on which a group $G$ acts fibre-wise. 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?

share|cite|improve this question
up vote 5 down vote accepted

Consider $\mathbb Z/2$ acting on $\{xy-zw=t\}$ by $x\leftrightarrow y$. This swaps the two small resolutions of the central fibre (the 3-fold 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.

share|cite|improve this answer
By the way, this example doesn't preserve the holomorphic 3-form, so might not interest you. Can anyone think of an example where the group preserves the 3-form ? – Richard Thomas Apr 17 '12 at 21:39
Thanks Richard, that's a very clear counter-example. It's not a problem that the holomorphic 3-form is non-invariant; in fact, for the work which prompted this question, I'm interested in exactly such examples. – Rhys Davies Apr 18 '12 at 10:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.