2
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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

3
$\begingroup$

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 $Pic(C)$.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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