MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
This is clearly not possible if $S$ is empty (and $g>0$). – Torsten Ekedahl Jun 21 '11 at 10:39
Of course. I'll edit the question. – Ariyan Javanpeykar Jun 21 '11 at 11:12
up vote 3 down vote accepted

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 $Pic(C)$.

share|cite|improve this answer
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? – Ariyan Javanpeykar Jun 21 '11 at 11:15
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? – Ariyan Javanpeykar Jun 21 '11 at 11:27
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. – ulrich Jun 21 '11 at 12:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.