# Factorization of conformal maps between annuli

Consider two doubly-connected open subsets $A$ and $A'$ of the Riemann sphere. We assume these two domains to be of same modulus (the moduli space being one real parameter), i.e. we assume that there exists a holomorphic bijection $\phi:A\rightarrow A'$. Note that the map $\phi$ is then unique up to precomposition by the automorphisms of the annulus $A$, loosely speaking the choice of whether we switch the two boundary components, plus a rotation.

Question
Is it possible to factor $\phi$ as a composition of holomorphic maps $\phi=f_1\circ g_1 \circ \cdots \circ f_n \circ g_n$, where the maps $f_i$ and $g_i$ are defined on simply-connected domains?
If yes, can we find such a factorization with $n=1$?

Motivation: case $\phi=f_1$.
To fix ideas, suppose the annuli $A$ and $A'$ separate $0$ and $\infty$, and that the map $\phi$ sends outer boundary to outer boundary. Assume that the map $\phi$ can be extended to the inside of $A$, i.e. $\phi$ is the restriction to $A$ of a holomorphic function defined on a simply-connected domain containing $0$ and bounded by the outer boundary of $A$. I claim that this gives no information on the outer boundary of the domain $A'$. However, once we fix the outer boundary of $A'$, its inner boundary has to live in a (small) family of three real parameters. Inded, fix the outer boundary of $A'$ to what you wish, and see by Riemann uniformization that the possible maps $\phi$ form a three real parameters family.

Thinking in terms of degrees of freedom, it seems it could be possible, for any couple of annular domains $A$ and $A'$, to write $\phi$ as a composition $f\circ g$, where $f$ extends to the inside, and $g$ extends to the outside: either one of these maps $f$ and $g$ allows complete freedom on either the exterior or the interior boundary of its image.

Note: uniqueness if $n=1$.
In the $n=1$ case, the factorization, if it exists is unique up to Moebius transformation.

Indeed, suppose $\phi = f\circ g = \tilde f \circ \tilde g$. Let us consider the holomorphic function $\tilde f^{-1} \circ f = \tilde g \circ g^{-1}$, a priori a holomorphic bijection $g(A)\rightarrow \tilde g(A)$. However, $\tilde f^{-1} \circ f$ extends inside $g(A)$ i.e. it is actually defined on a simply-connected domain bounded by the outer boundary of $g(A)$. Whereas $\tilde g \circ g^{-1}$ extends outside of $g(A)$. The function $\tilde f^{-1} \circ f$ can thus be extended to the whole sphere, and hence is a Moebius transformation.

There is such a factorization, with $n=1$.
Let us suppose first that the annuli are non-degenerate, and the boundaries are sufficiently smooth, and $\phi$ sends the outer boundary of $A$ to the outer boundary of $A'$. Then $\phi$ has a quasiconformal extenson, say $\Phi$ which is a homeomorphism between the spheres. Let $\mu=\Phi_{\overline{z}}/\Phi_z$ be the Beltrami coefficient of $\Phi$. It naturally splits into two parts $\mu^+$ and $\mu^-$ supported on the outside and on the inside. Now consider a homeomorphic solution $\Phi^+$ of the Beltrami equation $\Phi^+_{\overline{z}}=\mu^+\Phi_z$. It is a homeomorphism of the sphere, conformal inside the outer component of $A$. It maps $A$ onto some $A_1$ conformally. Using this map, we push forward $\mu^-$ to the inside component of $A_1$ and obtain a new Beltrami coefficient $\nu$ on the inside component of $A_1$. Then solve the Beltrami equation ${\Phi^-}_{\overline{z}}=\nu{\Phi^-}_z$. The composition $\psi=\Phi^-\circ\Phi^+$ is a quasiconformal map with the same Beltrami coefficient as $\phi$. Therefore $\psi=L\circ\phi$ where $L$ is fractional-linear, and we obtain a required factorization: $\Phi^+$ is conformal on the complement of the outside component of $A$ and $\Phi^-$ is conformal on the complement of the inside component of $A_1$.
To do the general case, consider an exhaustion of $A$ by annuli with anaytic boundaries, say $A=\lim A_n$.Then $A_n'=\phi(A)_n$ also have analytic boundaries, and the previous argument applies. Now, it is not difficult to prove, using a normal family argument that the limit of these factorizations will give the required factorization of $\phi$.