Krzysztof Jarosz
Riemann Mapping Theorem in $\mathbb C^n$?
The classical Riemann Mapping Theorem states that a nontrivial simply connected domain $\Omega$ in $\mathbb C$ is holomorphically homeomorphic with the open unit disc $\mathbb D$. Furthermore, if the boundary of $\Omega$ is homeomorphic with the unit circle, then that homeomorphism from $\mathbb D$ onto $\Omega$ can be extended to the boundary. It is very well known that simply connected domains in $\mathbb C^n$, for $n>1$, are generally not holomorphically equivalent.
Are ''similar'' domains ''almost'' holomorphically equivalent? That may of course depend on the meaning of these two words. For Banach spaces $A,B$ the closeness is normally measured by the Banach-Mazur distance: $d_{B-M}\left(A,B\right)=\inf\left\{\|T\|\|T^{-1}\|:T:A\rightarrow B\right\}$. For domains $\Omega_1,\Omega_2$ in $\mathbb C$ the quasiconformal distance $d_q\left(\Omega_1,\Omega_2\right)$ is the most natural measure of closeness. For one dimensional surfaces two concepts coincide.
Theorem 1. Let $S_{i}$, $i=1,2$, be bordered one dimensional Riemann surfaces and $A\left(S_{i}\right)$ be the algebras of functions continuous on $S_{i}$ and analytic on $\text{int}S_{i}$. Then $d_{q}\left(S_{1},S_{2}\right)<1+\varepsilon$ iff $d_{B-M}\left(A\left(S_{1}\right),A\left(S_{2}\right)\right)<1+\varepsilon^{\prime}$, where $\varepsilon$ and $\varepsilon^{\prime}$ tend to zero simultaneously.
For example the domains $S_{\varepsilon}=\left\{z\in \mathbb{C}:1<\left\vert z\right\vert <2+\varepsilon \right\}$ are not holomorphically equivalent, but $d_{q}\left( S_{0},S_{\varepsilon}\right)\simeq d_{B-M}\left(A\left(S_{0}\right),A\left(S_{\varepsilon}\right)\right)$ as $\varepsilon\rightarrow0$.
We discuss the general setting of this problem for abstract uniform algebras and also what is known for the domains in $\mathbb{C}^{n}$.