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Recently, when writing a review for MathSciNet, the following question arose:

Is it true that two smooth complex varieties that are biholomorphic are algebraically isomorphic? The converse is true just because polynomials are holomorphic.

I was mainly interested in the affine case since that was the context of the review I was writing.

I knew about exotic affine spaces, so two smooth complex affine varieties that are real analytically isomorphic do not have to be algebraically isomorphic. But real analytic maps between complex manifolds need not be holomorphic in general (just think of complex conjugation on $\mathbb{C}$). So that is not enough.

I posted the question on my Facebook page, and it got answered:

  1. I was reminded that Serre's GAGA Theorem implies that it is true for projective varieties. But there are quasiprojective counterexamples provided on MO. See the answer of Georges Elencwajg given here.

  2. Then it was pointed out that the answer in the link above is a manifold which is both affine and non-affine. So what about two affine varieties?

  3. I then found a reference for such an example. Two exotic affine spaces that are biholomorphic yet not algebraically isomorphic. See here. This is a very nice result. Since it is in dimension 3, it pretty much shows that at the very first point where something can go wrong, it does. A similar result for exotic affine spheres (apparently due to appear in J of Alg. Geo.) is here.

So I finally come to my questions.

Question 1: Is the example in the reference I found the first example of two affine varieties that are biholomorphic but not algebraically isomorphic?

Question 2: Is there some general reason to believe that all exotic affine spaces should be biholomorphic?

Comment: It seems to me within context to ask similar questions about objects, not only maps between objects. Here is some of what I have learned in that regard: all smooth complex varieties are complex manifolds since they are covered by smooth affine open sets with polynomial (hence holomorphic) transition charts. Conversely, a closed analytic subspace of projective space is algebraic by Chow's Theorem. But there are compact complex manifolds that are not algebraic (see here), which also shows that a closed analytic subspace of affine space need not be algebraic.

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1 Answer 1

up vote 4 down vote accepted

Question 1: The first example published seems to be the following (Corollary 4 in "Embeddings of Danielewski surfaces", G. Freudenburg and L. Moser-Jauslin, Math.Z. 2003 ) For any $a\in \mathbb{C}^*$, the surfaces in $\mathbb{C}^3$ given by $x^2z-y^2-a$ and $x^2z-(1+x)y^2-a$ are algebraically not isomorphic, but holomorphically isomorphic.

Question 2: The answer should be no because of the strong analytic cancellation theorem of Zaidenberg (see http://arxiv.org/pdf/alg-geom/9506005v1.pdf page 5). It says that if $X,X'$ are contractible non-biholomorphic surfaces of general-type, then $X\times \mathbb{A}^1$ and $X'\times\mathbb{A}^1$ are not biholomorphic but are both exotic affine spaces.

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First, thank you for the answer. I am tempted to accept it, but have a couple questions first. With respect to (1), I agree that it is an earlier example. However, how do you know it is the first published answer? I could not find in that article a claim by the authors to that effect. If you did a MathSciNet and arXiv search, that is fine, and I would accept that. But I just would like to know to what extent this is an "earlier find" versus a "scholarly fact". –  Sean Lawton May 30 '13 at 18:12
    
With respect to (2), I just looked at the paper. Nice! Prop. 2.4 directly answers my question it appears. It also shows that there exists examples of exotic affine spaces that are not biholomophic to affine space (analytically exotic); which is interesting too. I have probably a dumb question with respect to your example: how do you know that there exists appropriate X and X' that are not biholomophic to begin with? The example you give is two different exotic products of affine 3-space. The example in Prop. 2.4 seems to compare an exotic product with an exotic affine arising differently. –  Sean Lawton May 30 '13 at 19:20
    
Ah, I see, my second question is irrelevant. There certainly are appropriate affine surfaces that are not isomorphic, and that suffices by the Strong Analytic Cancellation Theorem. Great! I will accept your answer, even without my queries resolved. Thank you again! –  Sean Lawton May 30 '13 at 19:26
    
Here is the link for the first article you cited: link.springer.com/content/pdf/10.1007%2Fs00209-003-0572-5.pdf –  Sean Lawton May 30 '13 at 20:23
1  
Dear Sean, yes for the first question I am not sure that there is no previous example, but at least I talked with some specialists and did not get any other one. This is of course not a proof, that's why I said "seems to be". For (2), it is true that to complete the argument you need to find at least two surfaces which are not biholomorphic. You probably could find these in the work of Ramanujan for example. –  Jérémy Blanc May 31 '13 at 10:19

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