# A Generalization of the Ahlfors function to have varying degrees?

It's a classical result of Ahlfors that, for any sufficiently nice n-connected domain $\Omega \subset \mathbb C$ there is a holomorphic branched covering $f: \Omega \rightarrow \mathbb D$ to the disk $\mathbb D$, which extends continuously to the boundary and maps the boundary curves of $\Omega$ monotonically onto the boundary of the disk.

Now, suppose we have $n$ positive integers, $d_1,...,d_n$, and denote the boundary curves of $\Omega$ by $\gamma_1,...\gamma_n$. Can one find a holomorphic map $f: \Omega \rightarrow \mathbb D$ which extends continuously to the boundary of $\Omega$ and maps each $\gamma_i$ onto $S^1$ with degree $d_i$?

Even the case of $n=2$ seems difficult--one can readily achieve this with a real analytic function $f$ by a pretty basic interpolation.

The proof of the Ahlfors result also seems to be less than helpful--it involves finding an extremal for the analytic capacity functional and then analyzing the regularity of the optimal function. See, for example:

Krantz's Geometric function theory: explorations in complex analysis, Theorem 4.5.9.

• Note that one can obtain holomorphic branched covers of degree $n$ of $\Omega$ onto $\mathbb{D}$ without using Ahlfors functions. A theorem of Bieberbach states that for every $(b_1,\dots,b_n)$ where $b_j$ belongs to the curve $\gamma_j$, there is a degree $n$ branched cover $f:\Omega \to \mathbb{D}$ which maps each $b_j$ onto the point $1$. The proof involves taking linear combinations of Poisson kernels, and then showing that the coefficients can be chosen such that the resulting harmonic map has a well-defined conjugate. – Malik Younsi Aug 12 '15 at 18:04
• A key step is showing that a certain linear system has a unique solution, in order to "remove the periods" of the harmonic function. I don't know if this works with arbitrary degrees though... See The structure of the semi-group of proper holomorphic mappings of a planar domain to the unit disk by Bell and Kaleem. – Malik Younsi Aug 12 '15 at 18:05

The answer to your question is yes. Indeed, one can replace $\mathbb{D}$ with the right half plane, applying a Mobius transformation. Then the argument is quite simple if you are familiar with the following classical theorem of Bieberbach :

Theorem

Let $\Omega$ be a domain bounded by $n$ non-intersecting Jordan curves $\gamma_1,\dots,\gamma_n$. For $1\leq j \leq n$, fix one point $b_j$ in $\gamma_j$. Then there exists a degree $n$ holomorphic branched covering $F$ from $\Omega$ to the right half plane such that $F$ extends continuously to the boundary and $F(b_j)=\infty$ for each $j$. Moreover, $F$ is unique up to a positive multiplicative and an imaginary additive constant.

Using this, it is easy to prove the following, which implies the result that you want.

Theorem Let $\Omega$ be a domain bounded by $n$ non-intersecting Jordan curves $\gamma_1,\dots,\gamma_n$. For $1\leq j \leq n$, let $d_j$ be a positive integer. Then there exists a holomorphic branched covering $F$ from $\Omega$ to the right half plane which extends continuously to the boundary such that each restriction $F|_{\gamma_j}$ has degree $d_j$.

Proof Pick any points $a_1,\dots,a_{n-1}$ in $\gamma_1,\dots,\gamma_{n-1}$ respectively, and choose two distinct points $a_{n,1}$ and $a_{n,2}$ in $\gamma_n$. Let $F_1,F_2$ be branched coverings as in Bieberbach's theorem for the points $a_1,\dots,a_{n-1},a_{n,1}$ and $a_1,\dots,a_{n-1},a_{n,2}$ respectively. Then $F:=F_1+F_2$ is a holomorphic branched covering of $\Omega$ onto the right half-plane whose restriction to $\gamma_n$ has degree two and each other boundary restriction has degree one. We may continue this process and add other boundary points that map to infinity until we get the correct degrees. This completes the proof of the theorem.

All the details can be found in the paper The structure of the semi-group of proper holomorphic mappings of a planar domain to the unit disk by Bell and Kaleem.

See section 5, Proper holomorphic mappings of higher degree.

Let me explain a heuristic argument which is probably possible to convert into a proof. Let us deal with the ring. The covering will have degree $d=d_1+d_2$ and by Riemann-Hurwitz will have $r=d$ critical points. We can construct a Riemann surface spread over the disk with $r$ critical points which can be chosen arbitrary. This gives us $2r$ real parameters. Minus 3 parameters of the group of conformal automorphisms of the disc, $2r-3$ real parameters. But conformal classes of rings depend on 1 real parameter. Thus if $2r-3\geq 1$ we should be able to obtain every given ring in this way.

Similar parameter count works in general. This indicates that one can try to prove the result "by continuity".

Here are the details for the ring:

Imagine a two-sheeted Riemann surface over the unit disk, with branch points at $t,-t$, where $0<t<1$. This Riemann surface is conformally equivalent to a ring (by the Riemann Hurwitz) and the modulus of this ring is a function of $t$. When $t\to 0$, the modulus tends to $0$, when $t\to 1$, it tends to infinity. So for various $t$ we obtain all rings, and this construction gives the Ahlfors function of the ring.

Now modify our surface. On each of the two sheets make a vertical cut from $0$ to $i$. And glue a $d_1-1$ sheeted disk with the similar cut along the cut to the first sheet and $d_2-1$ sheeted disk to the second sheet. You obtain a Riemann surface whose boundary consists of a $d_1$ sheeted circle and a $d_2$-sheeted circle, both lying over the unit circle. It has $4$ branch points: two over $0$ connecting $d_1$ and $d_2$ sheets, respectively, and two over $t,-t$ connecting two sheets each.

It is clear that this new surface is a ring. (By Riemann-Hurwitz, or just by examining the gluing procedure).

Moving $t>0$ from $0$ to $1$, we obtain rings of all possible moduli, by an easy argument with extremal length.

This proves your conjecture for a ring. I believe that similar method will work in the general case.

• Even if all $d_k$'s are equal to one, the branched cover is not unique. Indeed, if $\Omega$ is $n$-connected and say, contains $\infty$, then the Ahlfors function is not unique as a degree $n$ branched cover of $\Omega$ onto $\mathbb{D}$ which vanishes at $\infty$. The family of such degree $n$ branched covers depends on $n$ parameters. – Malik Younsi Aug 13 '15 at 0:21
• @malik Younsi: thanks you are right, I corrected. – Alexandre Eremenko Aug 13 '15 at 2:50
• I think it would be fine (indeed preferable) to include your argument in the answer - it is not that long, after all! – Lasse Rempe-Gillen Aug 13 '15 at 11:17
• @LasseRempe-Gillen: what for? A complete solution is posted and accepted. – Alexandre Eremenko Aug 13 '15 at 15:53
• @AlexandreEremenko I happen to like your argument. :) And I think it is good that it is now on the page for the record, in case someone is interested in the future. – Lasse Rempe-Gillen Aug 14 '15 at 9:08