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Hi,

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 :

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

Note :

The case $n=1$ is of the above result is essentially the Riemann mapping theorem (for Jordan domains) + Carathéodory's theorem on the boundary behaviour of conformal mappings.

edited Jun 18 at 16:40

asked Jun 13 at 14:39

Malik Younsi
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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. – Agol Jun 18 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. – George Lowther Jun 18 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 $h(z)=\exp(g(z)$. – George Lowther Jun 18 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. – Malik Younsi Jun 19 at 15:06

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@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. – unknown (google) Jun 20 at 3:09

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

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