5

I'm trying to think of a quasi-projective variety that is not isomorphic to a quasi-affine one. I image that it must be $Y \subseteq \mathbb{P}^n$ of at least $n \geq 3$, and maybe $\operatorname{dim} Y \geq 2$ as well. I am also interested in finding a low (co)dimensional example of a quasi-projective that is not homeomorphic to a quasi-affine.

basket
  • 2,077
  • 2
    Clearly: you obviously also mean to rule out projective varieties from your consideration of quasi-projective varieties - but just to say it. – peter a g Oct 25 '16 at 00:18
  • Just to clarify: When you say homeomorphic, in what topology are you dealing? – Ted Shifrin Oct 25 '16 at 00:39
  • Zariski topology – basket Oct 25 '16 at 00:42
  • I just realized that my use of 'Hartog's Theorem' may cause people to believe this question refers to varieties over $\mathbb{C}$ in the Hausdorff topology. I would just like to clarify that by variety I mean in the sense of Hartshorne chapter 1, – basket Oct 25 '16 at 01:50
  • In regards to your answer to @TedShifrin's question: don't the arguments below apply then? Namely, if the variety $X={\mathbb P}^2 - pt$ were homeomorphic in the Zariski topology to a quasi-affine variety, it would have non-constant global functions. Am I missing something? – peter a g Oct 25 '16 at 11:55
  • @peterag That argument doesn't apply simply because a homeomorphism might not be an isomorphism of varieties. In other words, the rings of global functions might not be respected by the homeomorphism (so you can't say anything about global functions). In fact, $\mathbb P^2$ is homeomorphic to $\mathbb P^1 \times \mathbb P^1$ and $\mathbb P^1\times \mathbb P^1$ is homeomorphic to $\mathbb A^2$, because $\mathbb P^1$ is homeomorphic to ANY curve. So....homeomorphisms are vast in algebraic geometry, but they are not the right morphisms to consider! – Ariyan Javanpeykar Oct 25 '16 at 12:16
  • @AriyanJavanpeykar - 1) but $H^2({\mathbb P}^2 ({\mathbb C})) \not = 0$, whereas $H^2(\mathbb A^2({\mathbb C})) = 0$ - so the varieties are not homeomorphic, right? (singular co-homology, say). And differing $H^1$ for Riemann surfaces bar homeomorphisms. Agree? 2) your main point - my thought had been that a Zariski homeo would carry affine open sub. varieties to affine open sub. var's - but you're right - what does that say about the restriction maps? Not much I guess. So no guarantee that we can glue to create global functions. – peter a g Oct 25 '16 at 12:34
  • @AriyanJavanpeykar - thinking about it, I see that my alg. top. argument above might "not be convincing:" the Zariski top is coarser than the std. topology, so a Zariski homeo does not need to give rise to a map (let alone an iso) on the co-homology groups, as calculated in the standard topology. So! you are saying that any curve is homeomorphic, in the Zariski topology, to any other curve? because non-empty open = co-finite? Sorry for the stream of consciousness, everybody.... – peter a g Oct 25 '16 at 13:53
  • @peterag I disagree with 1), because you can't use an argument using the complex topology to say something about the Zariski topology. Anyway, your last comment is what I'm trying to say indeed. Let C and D be quasi-projective curves over an alg closed field. Let $C\to D$ be a bijection. Then it's clear that $C\to D$ is a homeomorphism because $C$ and $D$ have the co-finite topology. – Ariyan Javanpeykar Oct 25 '16 at 14:12

2 Answers2

6

I do not know of an easy way than appealing to Hartog's theorem. The example is take a high dimensional projective space and remove a point (or a line). It results in a quasi-projective variety as we are removing a Zariski closed subset. Now by Hartog's theorem (all varieties here are normal, in fact smooth), any global function on this subvariety will extend to the whole, hence a constant.

However a quasi-affine variety is rich with global regular functions. So this provides the example you are looking for.

P Vanchinathan
  • 19,504
  • 2
    I did not consider Hartog's theorem, but it provides a counter example for even $n = 2$ as David points out. I think I will refrain from accepting an answer for a while until someone addresses the homeomorphism problem though. – basket Oct 25 '16 at 00:09
  • Don't be in a hurry to accept an answer. Wait for a few hours. Also I do not understand much about topology of varieties; there are other experts who can answer with those aspects addressed. – P Vanchinathan Oct 25 '16 at 00:13
  • 1
    @PVanchinathan This provides an example of a quasi-projective non-projective variety which is not isomorphic to a quasi-affine variety in the category of varieties. But it doesn't necessarily give an example of a quasi-projective non-projective variety which is not homeomorphic to a quasi-affine variety. – Ariyan Javanpeykar Oct 25 '16 at 14:13
  • Though it is a rather old post, I would like to add a reference for the Hartog's theorem: see Proposition 9.1.4 of the wonderful lecture note http://www-personal.umich.edu/~mmustata/ag0523.pdf . – Hetong Xu Dec 26 '22 at 14:26
5

I think $\mathbb{P}^2 - pt$ is an example, but I would have to think longer about why this can't be quasi-affine.

(I think you can argue that if it were quasi-affine, it would have global functions, but if that were the case, you would be able to find global functions on $\mathbb{P}^2$, which can't be. Hartog's extension theorem works over $\mathbb{C}$.)

David Steinberg
  • 1,450