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Let $V$ be an analytic subvariety of some open set of $\mathbb{C}^n$ (intersection of finitely many zero-locii of analytic functions). Hironaka's desingularization theorem provides a parameterization of $V$, that is: there exists a proper, onto map $M\rightarrow V$, where $M$ is a complex manifold, which is a biholomorphism outside the strict transform of the singular set of $V$. Therefore $V$ can be locally parameterized (in that sense), around anyone of its points, by an onto holomorphic mapping from an open set of some $\mathbb{C}^m$.

My question is the following: is it true that $V$ can be globally parameterized by an onto holomorphic mapping from an open set of $\mathbb{C}^m$?

The answer to this question boils down to knowing whether any complex manifold (obtained this way) can be the range of a holomorphic map from some open set of $\mathbb{C}^m$, which is a statement that I unfortunately don't know how to handle. Neither was I able to find in the litterature any specifics regarding the conformal nature of the manifold $M$ when $V$ lies in an affine space.

I thank the community in advance for any help/comment/rebuke !

Loïc Teyssier
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  • I don't understand what you mean by parametrization. I would guess you would require at least a $1-1$ map, which is not the case for a desingularization, even locally. – diverietti Jun 08 '12 at 12:26
  • Dear diverietti, In the context of this question, parametrization means something like surjective holomorphic map. Regards, – Emerton Jun 08 '12 at 13:08
  • Dear Emerton, thanks. It was not clear at all, was it? From the question you may understand also that proper is a condition. Is it? – diverietti Jun 08 '12 at 13:11
  • Emerton is right, by "parametrization" I mean the existence of a holomorphic and surjective (onto) map, which happens to be proper in Hironoka's theorem. It needs not indeed be one-to-one in general, but is except at the strict transfrom of the singular locus. In my question, though, I'm not interested in a proper map, just a holomoprhic surjective application (although properness should reasonably be expected if the answer is positive). – Loïc Teyssier Jun 08 '12 at 14:22
  • I edited the post correspondingly to remove any imprecision regarding this issue. – Loïc Teyssier Jun 08 '12 at 14:37
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    I don't understand why you insist with Hironaka's desingularization. I think the question is already non trivial in the smooth case, where Hironaka's is just the identity map. – diverietti Jun 08 '12 at 14:37
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    I agree. I insist on the singular case because it is were my first motivation lied, without having realized that it was already non trivial in the smooth setting. Yet the question remains. – Loïc Teyssier Jun 08 '12 at 19:30
  • A first perhaps trivial remark is that the answer is always positive if the dimension of $V$ is one, that is if $V$ is a Riemann surface. To see this, resolve the singularities, then take the universal cover. It cannot be $\mathbb P^1$, otherwise $V$ would be compact. But then it is either $\mathbb C$ or the unit complex disc. – diverietti Jun 11 '12 at 15:34
  • Of course in what I said in my previous cooment you have to suppose that your Riemann surface is irreducible and let's say connected. – diverietti Jun 11 '12 at 18:08

1 Answers1

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The answer to your question is yes, even in a much broader setting and moreover the parametrizing open set can be chosen always to be the unit polydisc!

Theorem. (Fornaess-Stout '77). If $X$ is an $n$-dimensional complex manifold, then there exists a locally biholomorphic map $\Phi$ from the open unit polydisc in $\mathbb C^n$ onto $X$ with the property that for each $x\in X$, the fiber $\Phi^{-1}(x)$ consists of no more than $(2n+1)4^n+2$ points.

Of course, if $X$ is just a complex space, you have to desingularize before, so that your map is no more finite, but still surjective of course.

Observe finally, that there is no assumption here on $X$, it works in the abstract case also, you don't need it to be an analytic subset of an open set of some complex vector space!

The paper where the theorem is has been indicated to me by T.-C. Dinh.

diverietti
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  • Thank you for the answer, it solves perfectly the question and, surprisingly, gives me an unhoped-for solution to the problem I was considering! Could you please possibly post the reference of said article? – Loïc Teyssier Jun 13 '12 at 09:31
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    Ok, never mind the reference, I got it from MathSciNet : \textbf{Fornaess, John Erik; Stout, Edgar Lee} : Spreading polydiscs on complex manifolds. Amer. J. Math. 99 (1977), no. 5, 933–960. – Loïc Teyssier Jun 13 '12 at 09:55
  • maybe you could consider to accept my answer if you think it is satisfactory for you. best, – diverietti Jun 13 '12 at 21:35
  • sorry, I didn't know you could do that. Of course I accept it and thank you again. Best regards. – Loïc Teyssier Jun 14 '12 at 14:56
  • I suspected you didn't know, that's why I allowed myself to suggest it to you, in order to "close" de question since you seemed satisfied! regards – diverietti Jun 14 '12 at 17:19