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

I'm aware of the existence of complete (abstract) algebraic varieties that are not projective but, probably due to my ignorance, I have the impression that they arise only as very particular examples constructed just with the purpose of finding such an example. My question (perhaps a bit vague) is:

Are there exemples in the literature in which complete non-projective varieties appear without "being expected" from the beginning or without just being the goal of the construction or proof?

share|cite|improve this question
There is lots of non-projective complete toric varieties... – Piotr Achinger Nov 5 '12 at 2:05
I think there is an example of a projective variety $X$ with an action of a finite group $G$ such that the quotient $X/G$ is not projective. Would this fit your question? – J.C. Ottem Nov 5 '12 at 2:59
@J.C. Ottem: There is certainly no such example; for a finite group acting on a quasi-projective variety the quotient is also quasi-projective. What you are perhaps thinking of is that there is an action of a finite group on a non quasi-projective variety such that the quotient (which always exists as an algebraic space) is not a variety. – ulrich Nov 5 '12 at 4:23
@Dima Pasechnik: If a quotient of a projective variety exists as a quasi-projective variety then the quotient is obviously projective (being the image of a projective variety). – ulrich Nov 5 '12 at 6:13
@JC Ottern: ... or you might be thinking of a non-liftable (from char. $p>0$ to $0$) variety. There is an example of Serre involving a quotient construction. – Damian Rössler Nov 5 '12 at 7:35

Assume $X$ is a projective threefold with $n$ ordinary double points. Then $X$ has $2^n$ small resolutions of singularities. Usually, almost of all of them are non-projective.

share|cite|improve this answer
How is that possible? I thought that a birational map to something normal is a blow-up and a blow-up is a projective morphism, so a resolution of something projective should be projective as well. Why am I wrong? – Piotr Achinger Nov 5 '12 at 6:08
@Piotr: I don't see why the target is normal (but I am inexperienced with singular 3-folds). – S. Carnahan Nov 5 '12 at 6:30
@Piotr Aichinger: a birational morphism between two regular surfaces is a sequence of blow-ups but this is not true in higher dimensions. In general, there is the "strong factorisation conjecture", to the effect that there are morphisms $X'\to X$, $X'\to Y$, which are sequences of blow-ups along regular centers, and such that as a birational map (not morphism) $X\to Y$ is equal to the inverse of $X'\to X$ composed with $X'\to Y$. See the papers by Wlodarczyk, Abramovich et al. for this and the "weak factorisation conjecture". – Damian Rössler Nov 5 '12 at 7:47
Piotr, the "standard example" of a proper, non-projective toric variety is also one of Sasha's examples. I'm thinking of the fan on p. 71 of Fulton, where, there are 3 diagonal lines arranged in a spiral which give the non-projectivity. If you remove these lines, you get a toric variety with 3 ordinary double point singularities. A small toric resolution of this singularity can be gotten by blowing up a T-invariant divisor. However, any such divisor passes through 2 of the singular points. Thus, you can get only 6 of the 8 possible combinations by blow-ups. The other 2 are not projective. – Dustin Cartwright Nov 5 '12 at 15:44
Since producing an example ordinarily proceeds by consciously causing some necessary criterion for projectivity to fail, you seem to be asking whether some natural construction has led to examples which were later found to be non projective. The toric examples here seem to qualify. To me the most natural case is complete moduli spaces. Indeed the D-M-M compact moduli space of curves was unknown to be projective when constructed. Kolla'r has shown that all smooth toric varieties are moduli spaces, hence some of the examples given here occur this way. – roy smith Nov 9 '12 at 19:19

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.