By a Calabi-Yau variety I mean a smooth projective variety over the complex numbers with numerically trivial canonical divisor.

For each postive integer $n$ does there exist a finite group $G$ (possibly depending on $n$) which is not the fundamental group of a Calabi-Yau variety of dimension $n$?

For $n=1,2$ this follows easily from the classification of Calabi-Yau varieties of these dimensions, the only non-trivial finite fundamental group being $\mathbb{Z}/2\mathbb{Z}$ (for Enriques surfaces). For $n=3$ some finite non-abelian groups are known to occur as fundamental groups but I do not know of any non-existence results.

If there are only finitely many families of Calabi-Yau varieties of a given dimension then the question would clearly have a positive answer. However, this is far from being known so I am interested in other possible approaches.


2 Answers 2


If n is even, then I believe you can use the Atiyah-Bott fixed point formula to rule out many cases. For instance, let G be a simple, non-cyclic group. Consider the action of G on the Hodge group $H^{0,n}(X)$. Since G has only the trivial character, this action must be trivial. Then for every element g in G, the holomorphic Lefschetz number is 2 (if n is odd, the number is 0, which doesn't help). Therefore g has a fixed point.

  • $\begingroup$ Of course I should hasten to add, I (as most complex geometers) define "Calabi-Yau" to mean that $h^{0,p}(X)$ equals $0$ except for $p=0$ and $p=n$ (use Berger's theorem to break up into Calabi-Yau, hyper-Kaehler and Abelian variety factors). $\endgroup$ Commented May 17, 2011 at 19:03

This is definitely not an easy question. For $n=3$ checkout the paper Calabi-Yau Threefolds of Quotient Type by Oguiso and Sakurai. They are ("only") concerned with whether the fundamental group is finite. However, that should perhaps be the first step towards answering your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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