Is any simply connected domain conformally equivalent to Cartesian product of unit disks?

Question

By Riemann mapping theorem, any simply connected domain is conformally equivalent to the unit disk. Is any simply connected domain in the complex plane conformally equivalent to the Cartesian product of an open unit disk and a closed unit disk?

Answer 1

No, this is only true for $\Bbb C^1 \leftrightarrow \Bbb R^2$. By (one of the many significant results called) Liouville's Theorem (in particular, not the well-known theorem from complex analysis that bears that name), for $n > 2$ any conformal isometry from one open subset of $\Bbb R^n$ to another is actually a restriction of a global isometry from the $1$-point compactification $\Bbb S^n$ of $\Bbb R^n$ to itself, and these in turn comprise an $\frac{1}{2}(n + 2)(n + 1)$-dimensional Lie group isomorphic to $SO(n + 1, 1)$. In particular there are "not enough" of these to map a given simply connected region in $\Bbb C^n$, $n > 1$ to another arbitrary simply connected region. (In fact, even the unit polydisk $\Bbb D \times \Bbb D \subset \Bbb C^2$ is not conformally equivalent to the unit ball in $\Bbb C^2 \leftrightarrow \Bbb R^4$.)

This contrasts dramatically with the R.M.T., which contrapositively implies that the family of conformal maps on (an open subset of) $\Bbb C \leftrightarrow \Bbb R^2$ is infinite-dimensional. This is largely responsible for the major qualitative difference between conformal geometry in $2$ dimensions and in higher dimensions.

Answer 2

As Travis pointed out, there are very few conformal maps in dimension higher than $2$. The natural generalization of Riemann's mapping theorem to higher dimension is to ask, not for a conformal bijection, but for a biholomorphic map, i.e. a bijective holomorphic map $\Omega \to \Omega'$ with a holomorphic inverse. Such maps do not preserve all angles, but are the natural isomorphisms from a complex-analytic point of view.

Even so, there is nothing like Riemann's mapping theorem for $n \ge 2$. For example, the ball and the polydisc are not biholomorphically equivalent (see for example this question).

Even more surprisingly, let $\mathbb{B}$ be the unit ball in $\mathbb{C^n}$. If you make a tiny perturbation of $\mathbb{B}$ you will almost certainly get a domain that is not biholomorphically equivalent to $\mathbb{B}$. This claim can be made precise in several way. For example, for any $\varepsilon > 0$ there are uncountably many pairwise non-biholomorphically equivalent domains with Hausdorff distance less than $\varepsilon$ from $\mathbb{B}$.