# On holomorphic branched coverings of a domain in the plane to the unit disk

This question is partly motivated by my answer to this question on math.stackexchange.

Let $\Omega$ be a bounded $n$-connected domain in the plane, bounded by $n$ pairwise disjoint Jordan curves.

It was proved by Ahlfors that by solving an extremal problem (related to so-called analytic capacity), one can obtain a function $f$ holomorphic in $\Omega$ with the following properties :

1. $f$ is an $n$-to-$1$ branched covering of $\Omega$ onto the unit disk $\mathbb{D}$,

2. $f$ extends continuously to the boundary of $\Omega$, and maps each boundary curve homeomorphically onto the unit circle.

See e.g. Krantz's Geometric function theory: explorations in complex analysis, theorem 4.5.9.

My question is :

Is there another (perhaps more intuitive) way to see that such a function necessarily exists?

• I don't know how to answer your question, but heuristically, this should hold true by consider branched covers over the disk with appropriate branching data. One may obtain an n-punctured sphere as a branched cover over a disk with n branched points. The dimension of Teichmuller space of an n-punctured sphere is 2n-3, which is the same as the dimension of the space of n points in a disk, modulo the mobius group. So this gives a map between Teichmuller spaces, which one would like to prove is onto. I'm not sure how to give an elementary proof of this though. Jun 18 '12 at 17:23
• I think you can do this by multiplying together functions coming from the Riemann mapping, one for each Jordan curve, then multiply by a term of the form $\exp(u(z)+iv(z))$ where $u$ is harmonic and chosen to make the Jordan curves map to the unit circle and $v$ is chosen to make $u+iv$ analytic. It could be that $v$ is multiple-valued (so, only defined on the covering surface) but, by combining the Riemann map terms with Mobius transformations on the unit disc, you can make sure that $v$ is singly valued. That's a rough sketch I have in my head anyway. Jun 18 '12 at 20:38
• Or, which amounts to the same thing, choose n points $P=\lbrace z_1,\ldots,z_n\rbrace$ in the domain and define a harmonic function on the domain minus P by $f(z)=\log\vert z-z_1\vert+\cdots+\log\vert z-z_n\vert+u(z)$ where $u$ is harmonic and chosen such that $f=0$ on the boundary. You can extend $f$ to a (multiple-valued) harmonic function $g(z)=f(z)+iv(z)$. Choosing $z_1,\ldots,z_n$ carefully, you can make this a singly valued function, and the mapping is given by $h(z)=\exp(g(z)$. Jun 18 '12 at 21:44
• @Geoge : I thought about doing this, but it's not clear to me how you can choose $z_1, \dots, z_n$ so that the resulting function will be single valued... Note that this requirement puts constraints on $z_1, \dots, z_n$, so there are choices of the $z_j$'s that won't work. Jun 19 '12 at 15:06
• @Agol: I certainly believe this would work, I think though that the Teichmuller spaces are not of the same dimension. The Teichmuller space of a sphere minus n disks has dimension 3n-6. An euler characteristic argument shows that the map to the disk is branched over 2n-2 points, so that Teichmuller space has dimension 4n-7. So the map will be surjective with a large dimensional fiber. Jun 20 '12 at 3:09

By applying the Riemann mapping theorem $n$ times, we can assume that $\Omega$ is bounded by analytic Jordan curves. In this case, we have the following theorem due to Bieberbach, which says that there are many proper holomorphic maps onto the unit disk.
Theorem Let $\Omega$ be a domain in $\mathbb{C}_{\infty}$ bounded by disjoint analytic Jordan curves $E_1, E_2, \dots, E_n$, and let $p \in X$. For each $j$, fix a point $\alpha_j \in E_j$. Then there is a unique proper holomorphic map $f:\Omega \to \mathbb{D}$ such that $f(\alpha_j)=1$ for all $j$ and $f(p)$=0.