Question concerning Schwarz-Christoffel Mappings and Conformal Modulus

Question

By the Riemann Mapping Theorem we know every region (open, connected subset of $\mathbb{C}$), that isn't the whole plane is conformally equivalent to the unit disk $\mathbb{D}$. By the Schwarz-Christoffel mapping we can conformally map the upper half plane $\mathbb{H} = \lbrace z : \Im(z) \geq 0 \rbrace$ to a polygon $P$, with interior mapping to interior and boundary mapping to boundary. Now the whole idea of Quasiconformal maps is that you can't conformally map a rectangle onto a square. Let $f_1 : \mathbb{H} \to P_1$ be a Schwarz-Christoffel mapping to a rectangle, $f_2: \mathbb{H} \to P_2$ to a square, then wouldn't $f_2^{-1} \circ f_1: P_2 \to P_1$ be a conformal map from square to rectangle? I realize that the boundary would be where we get problems, but is there some proof that this would always be non-conformal? Thank you.

Answer 1

First of all, if the sides of two rectangles $R$ and $\tilde{R}$ are proportional than in fact you can map one to the other conformally (just using a composition of a dilation, a translation and a rotation, in the most general case).

The problem that you have in mind when you are constructing your example was certainly the "Grötzsch problem": find a map $f:R\to \tilde{R}$ (of class $C^{1}$, for example) that is \emph{closest to a conformal map} (in a certain sense that is unnecessary for the discussion below) and map each vertice of $R$ onto a vertice of $\tilde{R}$ (respecting the order that they appear in each rectangle). We know by the Riemann mapping theorem that we could find maps $\psi:R\to \mathbb{D}$ and $\phi:\tilde{R}\to \mathbb{D}$ that are conformal and extends to a homeomorphism on the boundary (here in fact we are using Carathéodory theorem). This maps are unique under some choice of normalizations, that could be, for example, prescribing three points on the boundaries of each of this rectangles to be mapped in specific points on the boundary of the unit disk; after we do this, if there was a conformal solution to the Grötzsch problem, we just need to adjust the "extra point" that is missing. But we know from the theory of automorphisms of the unit disk that we cannot adjust four points on the boundary to be mapped on other four points, just because the automorphisms are Möbius transformations with a very specific shape: $$ T(z)=e^{i\theta}\frac{z-\alpha}{1-\overline{\alpha}z} $$
where $\theta\in\mathbb{R}$ and $\alpha\in\mathbb{D}$. Its quite easy to observe that we have just three degrees of freedom, so we cannot prescribe one more boundary correspondence (unless, of course, if your map is a rotation)