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Consider a finite set (of cardinality $\ge 2$) $S \subseteq \mathbb{C}$ and the holomorphic universal covering map $\pi: \ \mathbb{H} \rightarrow \mathbb{C} \setminus S$, where $\mathbb{H}$ denotes the upper half-plane. Take a half-line $l$ starting from an element of $S$ which goes to infinity without intersecting $S$. The question is: what is $\pi^{-1}(l)$?

Thanks in advance.

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    $\begingroup$ $l$ is simply-connected, so its inverse image is a countable union of disjoint copies of $\Bbb{R}$. What else do you want to know? $\endgroup$
    – abx
    Commented Nov 29, 2017 at 14:03
  • $\begingroup$ I would like to draw them explicitly on the half-plane. Thanks for the answer. $\endgroup$
    – gm01
    Commented Nov 29, 2017 at 15:00
  • $\begingroup$ @gm01: With a computer? You need to solve a differential equation for this. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 29, 2017 at 15:01
  • $\begingroup$ Even in the simpler case $\#S=1$, say $S=\{0\}$ and $\pi\colon\mathbb{C}\to\mathbb{C}\setminus\{0\}=\mathbb{C}^\times$ is given the complex exponential map, I don't think there's a nice friendly description to be found. (To put it differently, this is asking what is the Mercator projection of a small circle through the north pole, and I don't think it's a particularly remarkable curve.) $\endgroup$
    – Gro-Tsen
    Commented Nov 29, 2017 at 15:03

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For the case where $|S|=2$ you can assume that $S=\{0,1\}$. It is then well-known that you can take $\pi$ to be the elliptic modular function $\lambda$. It follows easily that it works equally well to take $\pi(z)=1/\lambda(z)$. This induces a conformal isomorphism $\mathbb{H}/G\to\mathbb{C}\setminus S$, where $G$ is the group of matrices $g=\left[\begin{array}{cc}a&b \\ c &d\end{array}\right]\in SL_2(\mathbb{Z})$ with $g=1\pmod{2}$. The action on $\mathbb{H}$ is by $z\mapsto (az+b)/(cz+d)$. If we let $l$ denote the ray $(1,\infty)$ then $\pi^{-1}(l)=\lambda^{-1}((0,1))$, and one can check that this is the union of the $G$-translates of the positive imaginary axis.

If you have Maple you can enter

pi  := (z) -> EllipticModulus(exp(Pi*I*z))^(-2);
phi := (w) -> log(EllipticNome(1/sqrt(w)))/(Pi*I);

Then $\pi$ is as discussed above, and $\phi$ is a local inverse for $\pi$. The group $G$ is generated by the transformations $z\mapsto z+2$ and $z\mapsto z/(2z+1)$. Using these you can plot lots of other things if you want.

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