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EDIT: Thanks to several people I almost have a complete answer. BCnrd pointed out that the generalized Riemann Existence Theorem shows that $\tilde{Y}$ can be uniquely given the structure of a complex variety $\tilde{Y}'$. Also, Georges pointed out that the Kodaira embedding theorem implies that $\tilde{Y}'$ is projective if $Y'$ is projective.

This leaves two parts open to my original question. If $Y'$ is quasiprojective, then must $\tilde{Y}'$ be quasiprojective? Also, if $Y'$ is affine, then must $\tilde{Y}'$ be affine?

The impression I get from reading the comments is that the answers are "yes" and that everyone but me is able to easily prove them given what has already been said. Can anyone give me a hint or a reference as to how to proceed?

Thanks to everyone for all your help so far.

ORIGINAL QUESTION:

Let $Y$ be a complex manifold that can be given the structure of a complex variety $Y'$. Let $\pi:\tilde{Y} \rightarrow Y$ be a finite, unramified cover of $Y$. Can $\tilde{Y}$ be given the structure of a complex variety $\tilde{Y}'$ such that there is a finite map $\pi' : \tilde{Y}' \rightarrow Y'$ making the obvious diagram commute? If the answer is yes, then can we take $\tilde{Y}'$ to be projective/quasiprojective/affine if $Y'$ is projective/quasiprojective/affine?

This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties. Since such things are not available in higher dimensions, I'm stuck.

I should maybe remark that I don't even know how to do the above for affine varieties.

Thanks!

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    $\begingroup$ This is an excellent question, and I am looking forward to hearing the answers. However, I want to warn you that the statement in the first sentence is false. If $X$ is the variety I discussed at mathoverflow.net/questions/36189/… , then the quotient of $X$ by $G$ is an algebraic space, not a variety. $\endgroup$
    – David E Speyer
    Commented Oct 1, 2010 at 21:19
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    $\begingroup$ Yes to all. For the projectivity, you need to know about ample line bundles and the fact that ampleness is preserved under pullbacks along finite maps. $\endgroup$
    – Donu Arapura
    Commented Oct 1, 2010 at 21:19
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    $\begingroup$ Nikita, the phrase to enter into Google is "Riemann Existence Theorem". Without any smoothness hypotheses or anything else, if $X$ is an arbitrary $\mathbf{C}$-scheme locally of finite type and $X^{\rm{an}}$ denotes the associated complex-analytic space, then the functor $E \rightsquigarrow E^{\rm{an}}$ from finite etale covers of $X$ to finite covering spaces of $X^{\rm{an}}$ is an equivalence of categories. The proof of Grauert-Remmert was very difficult. Grothendieck's proof in SGA1 is a marvelous application of Hironaka to reduce to the projective case, where GAGA applies. $\endgroup$
    – BCnrd
    Commented Oct 2, 2010 at 2:29
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    $\begingroup$ If $\tilde{Y}' \to Y'$ is a finite unramified (i.e. etale) cover, then this map is finite in the sense of algebraic geometry (i.e. preimage of any Spec $A$ is Spec $B$ where $B$ is finite over $A$) and hence if $Y'$ is affine so is $\tilde{Y}'$. Also, since a finite morphism is in particular projective, if $Y'$ is (quasi-)projective, so will $\tilde{Y}'$ be. (The projective case is also answered by Georges's answer, but once one has the Riemann existence theorem, the more general quasi-projective case follows by the argument indicated, namely that the composite of (quasi-)projective ... $\endgroup$
    – Emerton
    Commented Oct 3, 2010 at 3:44
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    $\begingroup$ Just to add a bit to Emerton's comment, for many properties P of morphisms that make sense algebraically and analytically (smooth [= "submersion"], etale, finite fibers, separated, flat, open immersion, closed immersion, proper,...), if $f:X \rightarrow Y$ is map between finite type $\mathbf{C}$-schemes then $f$ satisfies P if and only if $f^{\rm{an}}$ does; see SGA1, Exp. XII. But both analytically and algebraically, "finite" (= classified by coherent sheaf of algebras) is same as "proper with finite fibers" (deeper fact on algebraic side), so $f$ finite iff $f^{\rm{an}}$ is. $\endgroup$
    – BCnrd
    Commented Oct 3, 2010 at 11:08

3 Answers 3

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Dear Nikita, Kodaira has proved the following fantastic embedding theorem for complex manifolds (conjectured by Hodge in 1950):

A compact complex manifold X is projective algebraic if and only if it has a closed positive (1,1) form whose cohomology class is rational.

Succinct explanation : The positivity is a complex differential geometric notion: locally a (1,1) form can be written $ i \Sigma h_{jk} dz_j d\bar {z_k}$ and the matrix $(h_{jk})$ is required to be positive definite.This closed form represents a class in singular cohomology by De Rham's theorem and this class should actually be in $H^2(X,\mathbb Q)$

It follows easily from this that

Given a holomorphic finite unramified covering $\tilde X \to X$ of compact holomorphic manifolds, the manifold $X$ is projective algebraic if and only if $\tilde X$ is

You can read the proofs in Griffiths-Harris's Principles of Algebraic Geometry.

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  • $\begingroup$ Donu, I hope you won't mind my answering before you give your complete answer: I had the feeling that it would be of a more algebraic nature, given your reference to SGA (to which I don't have access right now). $\endgroup$
    – Georges Elencwajg
    Commented Oct 1, 2010 at 22:53
  • $\begingroup$ Georges, of course I don't mind. It would have been a while before I had a chance to say anything substantial. Nice answer by the way. $\endgroup$
    – Donu Arapura
    Commented Oct 2, 2010 at 11:47
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Georges' answer is very nice so I will just add some comments.

In the curve case the assumption "unramified" is not necessary; in fact, every finite cover of a Riemann surface is still a Riemann surface (this is essentially Riemann Existence Theorem).

In higher dimension the situation is more complicate and "unramified" is definitely necessary. In fact, Donaldson and Auroux proved that every real, symplectic 4-manifold $(X, \omega)$ can be realized as a finite branched cover of $\mathbb{CP}^2$, and moreover such a cover $f \colon X \to \mathbb{CP}^2$ is completely determined, up to symplectomorphisms, by

  • the branch curve $D \subset \mathbb{CP}^2$ and
  • the monodromy representation $\theta \colon \pi_1(\mathbb{CP}^2 -D) \to S_{N}$, where $N:= \deg f$.

Finally, $X$ is complex - projective if and only if $D$ is isotopic to a complex curve.

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A finite unramified map mapping to an algebraic variety is finite in the algebraic sense and hence affine, so the preimage of an affine subset is affine. (I suppose this assumes that by affine you mean affine algebraic. I am not entirely sure what happens if the target is an affine analytic subset of $\mathbb C^n$. I assume there may be an analytic equivalent of a finite map being affine, but I don't know).

A finite map is also projective, hence if the target is quasi-projective, then the source is projective over something quasi-projective so itself is quasi-projective.

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