4 added 573 characters in body; edited title; edited title

# Finite covers of complex varieties (allbuttwoquestionsanswered!)

If

EDIT: Thanks to several people I almost have a finite group complete answer. BCnrd pointed out that the generalized Riemann Existence Theorem shows that $G$ acts on \tilde{Y}$can be uniquely given the structure of a complex variety$X$which is projective/quasiprojective\tilde{Y}'$. Also, then it is elementary to show Georges pointed out that the Kodaira embedding theorem implies that $G/X$ \tilde{Y}'$is a complex variety which projective if$Y'$is projective/quasiprojectiveprojective. This leaves two parts open to my original question. If$Y'$is quasiprojective, then must$\tilde{Y}'$be quasiprojective? Also, if$Y'$is affine, then must$\tilde{Y}'$be affine? The impression I was wondering whether you could go in get from reading the other direction. Here comments is that the answers are "yes" and that everyone but me is able to easily prove them given what has already been said. Can anyone give me a precise questionhint or a reference as to how to proceed? Thanks to everyone for all your help so far. ORIGINAL QUESTION: Let$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'$is projective/quasiprojective/affine? This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties. Since such things are not available in higher dimensions, I'm stuck. I should maybe remark that I don't even know how to do the above for affine varieties. Thanks! 3 added 14 characters in body If a finite group$G$acts on a complex variety$X$which is projective/quasiprojective, then it is elementary to show that$G/X$is a complex variety which is projective/quasiprojective. I was wondering whether you could go in the other direction. Here is a precise question. Let$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}'$to be projective/quasiprojective projective/quasiprojective/affine if$Y'$is projective/quasiprojectiveprojective/quasiprojective/affine? This kind of thing is true for Riemann surfaces, but even there I don't know how to prove it except by going through the whole machinery showing that all compact Riemann surfaces are projective varieties. Since such things are not available in higher dimensions, I'm stuck. I should maybe remark that I don't even know how to do the above for affine varieties. Thanks! 2 deleted 24 characters in body; added 90 characters in body If a finite group$G$acts on a complex variety$X$, X$ which is projective/quasiprojective, then it is elementary to show that $G/X$ is a complex variety . Taking quotients like this preserves nice properties like being projective or quasiprojectivewhich is projective/quasiprojective. I was wondering whether you could go in the other direction.

Here is a precise 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 if $Y'$ is projective/quasiprojective?

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!

1