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If $X$ ad $Y$ are algebraic varieties over an algebraic closed field, what does exactly mean that $Y$ is a covering of $X$? Is $Y$ only a topological covering of $X$?

I will appreciate a reference.

Addenda in response to the above comments: Consider for example the following definition from the article "F.Catanese - Fibered Surfaces, Varieties Isogenous to a Product and Related Moduli Spaces":

$\quad$ Definition $3.1$. A surface $S$ is said to be isogenous to a product if $S$ admits a finite unramified covering which is isomorphic to a product of curves ($u:C_1\times C_2\to S$) of genera $g_i=\text{genus}(C_i)\ge1$. In the case where each $g_i\ge2$ we shall also say that $S$ is isogenous to a higher product.

Where by the word "surface" the author means a complex algebraic variety of dimension $2$ (that is also a complex manifold) . In this case what is a finite unramified covering of $S$? In the article there are no definitions for coverings, but it seems that the author refers to a well known concept. In literature I have found only two concepts of "covering":

  1. Topological coverings.
  2. Coverings for compact Riemann Surfaces, i.e. coverings for smooth projective algebraic curves.
Dubious
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  • I think that more context is needed. The common generalization of a covering map is that of an etale morphism. That said, if someone called a surjective finite map a "cover" I wouldn't be shocked. – Alex Youcis Dec 06 '13 at 09:07
  • The context are complex algebraic surfaces. An article that I'm reading cites a "cover" of a surface $X$. – Dubious Dec 06 '13 at 09:10
  • In the (connected) Riemann surface setting one usually thinks of a cover as being a proper holomorphic map. These are always a (topological) covering off of a discrete set. I'm not sure though which is more apropos for your paper. – Alex Youcis Dec 06 '13 at 09:20
  • So should I consider the complex manifold structure of my algebraic surface $X$? – Dubious Dec 06 '13 at 09:25
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    I mean, I have no idea. I'm not reading your paper, you are. It all depends on context. – Alex Youcis Dec 06 '13 at 09:25

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In this context (i.e., for finite degree maps of compact connected complex manifolds) a finite unramifield covering is a locally biholomorphic map $p: S' \to S$. Equivalently, $p$ is a holomorphic map of two compact complex manifolds which is also a topological covering. Equivalently, $p$ is a holomorphic fiber bundle with zero-dimensional (i.e., finite) fibers. The equivalences are not completely obvious. The two nontrivial ingredients you need are:

  1. Every bijective holomorphic map between two open subsets of ${\mathbb C}^n$
    is actually biholomorphic, see here. This implies that a holomorphic topological covering is a holomorphic fiber bundle with zero-dimensional fiber.

  2. Ehresmann's theorem which implies that a locally biholomorphic map between two connected compact complex manifolds is in fact a holomorphic fiber bundle with zero-dimensional fiber.

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Moishe Kohan
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