# Uniformizing variable for branched covering of the Riemann sphere

Suppose I have a function $Q(z)$ of a complex variable $z\in\mathbb P^1$, possessing square root type branch points at the positions $\left\{z_i\right\}_{i=1}^{2M}$. I know that the Riemann surface $\mathcal M$ on which $Q$ lives has genus $0$.

I have the following questions:

• Aside from the knowledge of the poisitions $\left\{z_i\right\}_{i=1}^{2M}$, which additional informations on $Q$ do I need in order to uniquely define the uniformizing variable $\omega(z)$?
• Does an explicit general formula for $\omega(z)$ exists?
• If so, can the limit $M\rightarrow\infty$ be obtained?

I remember seeing an expression of the type $\omega(z)=\frac{1}{z}+\sum_{k=1} a_k z^k$, but I am not sure the context corresponded to mine. Moreover there was no mention of how to fix the coefficients $a_k$ in terms of the positions of the branch points or informations on $Q$.

Thanks a lot!

Stefano

1. Besides the position of ramification points $z_j$ you need monodromy of $Q$ to determine $M$. Once $M$ is defined, you need a normalization of your uniformizing function: it is defined up to a conformal automorphism of the sphere $L^1$.
2. No general formula exists. To find the uniformization in the case of genus $0$, you have to find a rational function with simple critical points and prescribed critical values. The degree of your function is $M+1$ by Riemann-Hurwitz formula. You write it with undetermined coefficients, add normalization, for example $f(0)=a,f(1)=b,f(\infty)=c$, where $a,b,c$ are some points distinct from your $z_j$, and write the condition that critical values are $z_j$. You obtain $2M+3$ algebraic equations with $2M+3$ unknowns, this system has finitely many solutions. Different solutions correspond to different possible monodromies. The system is difficult to solve when the degree is high.
• Thanks a lot! This is already quite useful. So the ansatz I should make is a function of the type $\omega(z)=\frac{a_0 + a_1 z + \cdots + a_{M+1} z^{M+1}}{1 + b_1 z + \cdots + b_{M+1} z^{M+1}}$, right? – Stefano Mar 5 '17 at 12:26
• Yes, but you can normalize it by setting $a_0=b_{M+1}=0$ and $f(1)=1)$, for example. This reduces the number of unknowns and equations. – Alexandre Eremenko Mar 5 '17 at 13:43