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The following was already known to Riemann. Suppose that one is given a connected Riemann surface $X$, a finite set $\Delta \subset X$ and a homomorphism $\phi: \pi_1(X \backslash \Delta) \to S_d$ where $S_d$ is the permutation group on $d$ symbols. If this homomorphism is transitive, i.e. the image of $\phi$ acts transitively on $\{1, \ldots ,d\}$, then this data allows one to construct a unique connected Riemann surface $Y$ with a map $f: Y \to X$ which is a branched cover: when we restrict $f$ to $f^{-1}(X \backslash \Delta)$ it is a $d$-fold cover and around the branch points the monodromy is given by $\phi$. This is known as the Riemann existence theorem and it is proven by constructing the cover corresponding to kernel of $\phi$ and filling in the missing points with disks using $\phi$. If $X$ is compact, $Y$ will be as well.

My question is: given the Riemann sphere (or any other Riemann surface), which compact connected Riemann surfaces can one get if one is allowed to pick $\Delta$, $d$ and $\phi$ as above? This may seem like a trivial question: topologically any Riemann surface arises this way. But it is not clear to me what complex structures can arise. Alternatively, the question may be phrased as: is the map from such data to the disjoint union of moduli spaces of Riemann spaces of different genus surjective?

I would of course also be interested in literature discussing this or related questions.

Edit: the answers are correct that this question was easy. I was actually interested in the situation where we demand that $\phi$ assigns to a loop around a point in $\Delta$ permutation of consisting of disjoint cycles of length two, such that we get only branching of degree 2.

So I guess the updated question is: what restrictions can we place on $\phi$ such that we still get all compact connected Riemann surfaces this way?

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@skupers : The answer is ``all of them''. Indeed, any compact Riemann surface has a nonconstant meromorphic map (this follows eg from Riemann Roch, though often it is proven along the way to proving Riemann Roch), and thus is a branched cover of the sphere, and thus can be expressed via your construction. See any book on compact Riemann surfaces for the details (I recommend the books of Forester, Gunning, and Narasimhan). – Andy Putman Jan 25 '11 at 21:37
@Andy: your comment appeared exactly when I finished to type my answer... – Francesco Polizzi Jan 25 '11 at 21:48
up vote 7 down vote accepted

The answer is simple: any complex structure arises in this way.

Indeed, any compact Riemann surface $X$ admits a holomorphic cover $\phi \colon X \to \mathbb{P}^1$.

This is straightforward in genus $0$ and $1$. If the genus is at least $2$, then the linear system $|3K_X|$ is very ample, so it gives an embedding $\gamma \colon X \to \mathbb{P}^N$ and the map $\phi$ is obtained by composing $\gamma$ with a suitable projection.

We are using here the fact that every compact Riemann surface is a smooth complex projective curve; this follows from the existence of a meromorphic function on it, which is a non-trivial fact (see Andy Putman's comment).

EDIT. We can always chose $\phi$ so that the local monodromy is general, that is it is given by a single transposition around each branch point. In fact, projecting $X \subset \mathbb{P}^N$ from a general subspace of the right codimension, we obtain a plane curve $X' \subset \mathbb{P}^2$ whose singularities are at worst ordinary double points. Moreover, $X'$ has at most a finite number of bitangent lines; taking the projection $\pi_p \colon X' \to \mathbb{P}^1$ from a point $p$ not contained in any of the bitangent lines and composing with the map $X \to X'$ we obtain a finite cover $\phi \colon X \to \mathbb{P}^1$ whith the desired property.

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Minor nitpick: "recall that the linear system |3K_X| is always very ample" --- except for genus 0 or 1. – user5117 Jan 26 '11 at 9:12
@Artie: you are right. I was thinking in genus $\geq 2$, since $\mathbb{P}^1$ has no moduli and any genus $1$ curve is a double cover of $\mathbb{P}^1$ branched in four points. I fixed the answer, thank you. – Francesco Polizzi Jan 26 '11 at 9:23

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