Today I showed, using some ad hoc algebraic topology, that if $\Sigma$ is a Riemann surface and $\mathfrak{p} \subset \Sigma$ is a finite set of points, then there is another Riemann surface $S$ and a branched covering $\phi : S \to \Sigma$ which has non-trivial branch points EXACTLY on the inverse image of $\mathfrak{p}$.
I'm suspicious that this was known or used elsewhere, and that it possibly follows trivially from some more sophisticated algebraic geometric mechanism. Does this ring a bell to anybody?
Best Answer
Yes. The fundamental group of this Riemann surface minus those branch points is $< a_1, b_1, ..., a_g, b_g, c_1, ..., c_r| [a_1,b_1]...[a_g,b_g]c_1...c_r=1>$ (where $g$ is the genus, and $r$ is the number of the soon-to-be branch points). We have to guarantee that those will be branch points. Take any group generated by $r-1$ non trivial elements (such that their product isn't $1$). Map $\pi_1$ to that group such that each $a_i$ and $b_i$ go to 1, and each $c_i$ ($i$ going from $1$ to $r-1$) will go to said generators of the chosen group. Let $c_r$ go the the inverse of what $c_1...c_{r-1}$ goes to. Then this will correspond to some topological cover of $\Sigma$. Riemann's existence theorem says that we can make any (finite) topological cover into an algebraic cover. So this topological cover corresponds to a Riemann surface dominating it. It's an easy exercise to show that the ramification index at each of the preimages of your branch points (let's say branch point number $i$) is the order of the image of $c_i$ in the chosen group.