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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 !

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  • $\begingroup$ 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. $\endgroup$
    – diverietti
    Commented Jun 8, 2012 at 12:26
  • $\begingroup$ Dear diverietti, In the context of this question, parametrization means something like surjective holomorphic map. Regards, $\endgroup$
    – Emerton
    Commented Jun 8, 2012 at 13:08
  • $\begingroup$ 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? $\endgroup$
    – diverietti
    Commented Jun 8, 2012 at 13:11
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    $\begingroup$ 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. $\endgroup$
    – diverietti
    Commented Jun 8, 2012 at 14:37
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    $\begingroup$ 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. $\endgroup$
    – Loïc Teyssier
    Commented Jun 8, 2012 at 19:30

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Loïc Teyssier
    Commented Jun 13, 2012 at 9:31
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    $\begingroup$ 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. $\endgroup$
    – Loïc Teyssier
    Commented Jun 13, 2012 at 9:55
  • $\begingroup$ maybe you could consider to accept my answer if you think it is satisfactory for you. best, $\endgroup$
    – diverietti
    Commented Jun 13, 2012 at 21:35
  • $\begingroup$ sorry, I didn't know you could do that. Of course I accept it and thank you again. Best regards. $\endgroup$
    – Loïc Teyssier
    Commented Jun 14, 2012 at 14:56
  • $\begingroup$ 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 $\endgroup$
    – diverietti
    Commented Jun 14, 2012 at 17:19

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