4
$\begingroup$

By Riemann's existence theorem I mean this:

Let $X$ be some variety defined over $\mathbb{C}$, and let $Y$ be a topological covering space of $X$. Then $Y$ can be given the structure of a variety over $\mathbb{C}$, and furthermore this can be done so that the covering map will be algebraic.

It is frequently said that this theorem is not constructive. That is to say, that it is impossible to predict the polynomials that define $Y$ and the polynomial map that defines $Y\rightarrow X$. I want to read the proof, and understand for myself why this is true. Where can I find a good, preferably succinct, proof of this theorem in English?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ You need the covering to be finite, clearly. $\endgroup$
    – Ben McKay
    Commented Nov 12, 2011 at 20:08
  • 2
    $\begingroup$ See Expose XII, Theoreme 5.1 on page 332 of SGA1 available at arxiv.org/abs/math/0206203 $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 12, 2011 at 21:02
  • $\begingroup$ By the way, I'm not completely sure about what you're really looking for, but I think the proof of Theoreme 5.1 given in SGA1 of the essential surjectivity of $\psi$ is not constructive because the argument is partially a reduction via non-constructive reasoning to the case of an affine regular scheme. $\endgroup$
    – Ariyan Javanpeykar
    Commented Nov 12, 2011 at 21:06
  • $\begingroup$ @Ariyan: I doubt that it is constructive even for affine regular schemes. Otherwise one would have an algebraic proof that $\pi_1(\mathbb{P}^1\smallsetminus a_1,...,a_r)\cong$ the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1 ...\alpha_r =1\rangle$. $\endgroup$
    – James D. Taylor
    Commented Nov 13, 2011 at 0:32

1 Answer 1

Reset to default
2
$\begingroup$

See Finite covers of complex varieties (all but two questions answered!)

Improve this answer
$\endgroup$
1
  • $\begingroup$ Aha! I imagine you mean Brian's comment. Do you have a reference of where that is in SGA1? Even for a projective variety it is not perfectly obvious to me how GAGA implies it... $\endgroup$
    – Makhalan Duff
    Commented Nov 12, 2011 at 20:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .