# Biholomophic non-Algebraically Isomorphic Varieties

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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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.