Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 $X$ be a compact connected Riemann surface, and let $S\subset X$ be a finite subset.

Does there exist a morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ which is unramified at the points of $S$?

I'm interested in the case where $X$ is of genus at least $2$. (The genus zero case is trivial: take $f$ to be the identity.)

The answer is trivial when $S$ is empty. (Any morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ will do.)

Let $h:X\to \mathbf{P}^1(\mathbf{C})$ be a morphism with ramification locus $R(h)$. Then, if $S\subset X\backslash R(h)$, the answer is yes.

How effective can our answer be? That is, suppose that there exists such an $f$. Then, can we bound its degree?

The title is a special case of the above question: take $S=\{\textrm{pt}\}$.

share|cite|improve this question
The answer to your first question is yes by a (a weak version of) Riemann-Roch. Pick a point $P \notin S$ and write $\ell(nP)$ for the dimension of the space of meromorphic functions unramified outside of $P$ and with a pole of order at most $n$ at $P$. Then $\ell(nP) \underset{n\to +\infty}{\longrightarrow} +\infty$. Regarding your second question, the Riemann-Hurwitz formula relates the degree to the local ramification indexes. – AFK Dec 25 '11 at 0:45
@YBL: the condition "unramified outside of $P$" is not linear - how does Riemann-Roch say anything about such a space of functions? – user8268 Dec 25 '11 at 21:20

If $S=\lbrace P\rbrace $, put an algebraic structure $X^{alg}$ on $X$ and take a uniformizing parameter $ t\in \mathcal O_{X^{alg},P}$ at $P$.
Since $\mathcal O_{X^{alg},P}\subset Rat(X^{alg})=\mathcal Mer(X)$, the meromorphic function $t$ seen as a map $X\to \mathbb P^1(\mathbb C)$ solves your problem.
You can generalize that to the case where $S$ is an arbitrary finite set, again by putting an algebraic structure on the Riemann surface $X$: look up Corollary 1.16 in Chapter VI of Miranda's book, which solves your problem.

Edit: Let me however give a self-contained proof using Riemann-Roch.

Fix an arbitrary point $x_0\in X\setminus S$ outside of $S=\lbrace x_1,...,x_n \rbrace$.
Consider the divisors (where $N$ will be determined later)
$$D_1=(-2)\cdot x_1+...+(-2)\cdot x_n+N\cdot x_0\quad \text {and} \quad D_2=(-1)\cdot x_1+...+(-1)\cdot x_n+N\cdot x_0$$ and their associated sheaves (=line bundles) $\mathcal O(D_1), \mathcal O(D_2)$.
They give rise to a short exact sequence of sheaves $$ 0\to \mathcal O(D_1)\to \mathcal O(D_2)\to \mathcal Q\to 0 $$ where the quotient sheaf $\mathcal Q$ is a finite sum of sky-scraper sheaves.
Taking cohomology we get $$ ...\to H^0(X, D_2)\to H^0(X,\mathcal Q) \to H^1(X,D_1)\to ...$$
Now $H^1(X,D_1)$ is dual to $H^0(X,K_X(2\cdot x_1+...+2\cdot x_n-N\cdot x_0))$ (by Serre duality) and is thus zero for $N\gt 2n+2g-2 $.
This choice of $N$ implies that the morphism $\gamma : H^0(X, D_2)\to H^0(X,\mathcal Q)$ is surjective.

And what has this got to do with your problem? It solves it!
Indeed, if you choose a coordinate $z_i$ near $x_i$, the stalk $\mathcal Q_{x_i}$ is identified with the complex line $\mathbb C\cdot z_i$ and by choosing a section $s\in H^0(X, \mathcal O(D_2))$ mapping to a $\gamma(s)\in H^0(X,\mathcal Q)$ non-zero at every $x_i$ you obtain the required meromorphic function $s$: its only pole is at $x_0$ and it has zeros of order exactly $1$ at the $x_i$'s.

share|cite|improve this answer
Dear Georges, I'm trying to see the geometry in your answer. Correct me if I'm wrong, but are we saying something along the lines of: We embed $X$ in $\mathbb P^3$ (i.e. put an algebraic structure on $X$). Given a line $L \simeq \mathbb P^1$ in $\mathbb P^3$ we consider a linear projection $f : \mathbb P^3 \to L$. The claim is now that given a point $p$ on $X$, we can choose $L$ in such a way that $f|_X$ is non-ramified at $p$? – Gunnar Þór Magnússon Feb 7 '12 at 13:34
Dear @Gunnar, I might be misunderstanding you but in principle you must project from a point of projective space onto a plane. A variant of what you suggest would be to choose a plane $F$ with linear eqution $f$ not going through $p$ and a plane $G$ with linear eqution $g$ going through $p$ but transversal to $X$ at $p$ (that is not containing the tangent projective line $\mathbb T_p(X)$ to $X$ at $p$) . The required rational function would then be the restriction $(g/f)|X$. – Georges Elencwajg Feb 7 '12 at 14:22

You can always get an $f$ of degree $\max(g+1,n)$, where $g$ is the genus and $n$ the number of points in $S$. I don't think it is a good estimate (the problem is that I construct $f$ so that every point of $S$ is a (simple) pole; there should be $f$'s with lower degrees if the values at points of $S$ are different)

Here is how to see it. Let $P_1,\dots,P_n$ be the points in $S$ and let $z_i$ be a local coordinate around $P_i$ (s.t. $z_i(P_i)=0$) on some disc $D_i\subset \Sigma$. Suppose the discs don't overlap. The function $1/z_i$ on $D_i\cap (\Sigma -P_i)$ gives a class $\alpha_i\in H^1(\Sigma,\mathcal{O})\cong\Omega^1(\Sigma)^*$ (use $D_i$ and $\Sigma-P_i$ as an open cover of $\Sigma$; $1/z_i$ is a function on $D_i\cap(\Sigma-P-i)$). We know that $\alpha_i\neq0$ (if $\alpha_i$ were a coboundary then $1/z_i=h-k$, where $h$ is holomorphic on $\Sigma-P_i$ and $k$ on $D_i$, but that means that $h$ is meromorphic on $\Sigma$ with a unique pole at $P_i$, which is impossible if $g>0$).

We can suppose $n>g$ (if not then add some more points to $S$). As $\dim H^1(\Sigma,\mathcal{O})=g$ and all $\alpha_i$'s are non-$0$, there are $c_1,\dots,c_n\in\mathbb{C}$, all non-$0$, such that $\sum c_i\alpha_i= 0\in H^1(\Sigma,\mathcal{O})$. If you write $\sum c_i\alpha_i$ as a coboundary for the open cover $\Sigma-S,D_1,D_2,\dots,D_n$ of $\Sigma$, the holomorphic function on $\Sigma-S$ is a meromorphic function on $\Sigma$ with a simple pole at every point of $S$, hence non-ramified at $S$, and its degree is the number of poles (i.e. $n$).

edit I changed my answer completely as it contained a mortal gap

share|cite|improve this answer

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.