Let $C$ be a smooth projective connected curve of genus $g$ over $\bar{\mathbf{Q}}$. Fix a finite non-empty (Edit) set of closed points $S$ in $C$ and let $U$ be the complement of $S$ in $C$.

Q1. (Algebraic formulation) Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$ such that $\pi|_{U}$ is etale?

Equivalently, let $X$ be a compact connected Riemann surface of genus $g$ which can be defined over $\bar{\mathbf{Q}}$ and let $B$ be a finite set of of closed points in $X$ with complement $Y$.

Q1. (Analytic formulation ) Does there exist a finite topological cover $Y\longrightarrow \mathbf{P}^1(\mathbf{C})-\{0,1,\infty\}$ ?

The equivalence of these two questions follows from the proof of Belyi's theorem and Riemann's existence Theorem.

If the answer to Question 1 is positive, I would be very interested in knowing if the degree of $\pi$ can be bounded effectively.

Q2. Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1$ such that $\pi|_{U}$ is etale and $\deg \pi \leq c$, where $c$ is a constant depending only on $S$ and $g$?

Example. Suppose that $g=0$. Then, following Belyi's proof of his theorem, the answer to Question 1 is yes. The answer to Question 2 is also positive and an explicit upper bound for such a rational function is given by Khadjavi in An effective version of Belyi's Theorem.

I don't expect the answer to Question 1 to be easy. In fact, what I'm asking is to prove the existence of a Belyi morphism $\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$ with prescribed ramification. Now, that's probably very hard but definitely very interesting to find out.

Trivial Remark. Suppose that $g>1$. Then the automorphism group of $C$ is finite. Choose a Belyi morphism $\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$ and let $U_0\subset C$ be the complement of the ramification points of $\pi$. Then we see that Question 1 has a positive answer if we take $U$ to be $\sigma(U_0)$ with $\sigma$ an automorphism of $C$. But that's only finitely many examples.

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    $\begingroup$ This is clearly not possible if $S$ is empty (and $g>0$). $\endgroup$ Jun 21, 2011 at 10:39
  • $\begingroup$ Of course. I'll edit the question. $\endgroup$ Jun 21, 2011 at 11:12

1 Answer 1


No, it is easy to construct examples where this is not possible (aside from trivial ones with $|S| < 3$). For example, if $g(C)>0$ one can find $S$ arbitrarily large so that the points of $S$ give linearly independent elements in $Pic(C)$. For such an $S$ there can be no map of the kind you want since the elements of $S$ must be mapped to at least $2$ distinct points of $\mathbb{P}^1$ which would give a non-trivial relation on the classes of elements of $S$ in $Pic(C)$.

  • $\begingroup$ So if I understand correctly, for every integer n there exists a finite set of closed points S of cardinality n such that the points of S give linearly independent elements in Pic(C). For such an S, we can never find a morphism C-->P^1 of the kind as described in the question, right? $\endgroup$ Jun 21, 2011 at 11:15
  • $\begingroup$ So this raises another interesting question (in my opinion). Assume g(C) >0. Let n>2 be an integer. Does there exist a finite set of closed points S of cardinality n and a morphism pi:C--->P^1 satisfying the conditions of the question? Moreover, it would be nice to know the cardinality of the set I(C)={n: n>2 and there is a finite set of closed points S of cardinality n and pi:C--->P^1 as in question}. By Belyi's theorem, I is non-empty. Is it infinite? Finite? $\endgroup$ Jun 21, 2011 at 11:27
  • $\begingroup$ To answer your first comment, that is exactly what I claim. The inverse images of the points of $\mathbb{P}^1$, considered as divisors on $C$, are all linearly equivalent, so if $S$ is the union of the inverse images of more than one point then we get a non-trivial relation. $\endgroup$
    – naf
    Jun 21, 2011 at 12:09

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