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For context, I am studying background material for as well as the basics of Teichmuller theory. I am currently struggling to understand Lehto's definition of quasiconformal in his text "Quasiconformal Mappings in the Plane". It is worth mentioning I am aware of the abundance of characterizations of QC mappings in terms of both geometry and analysis (e.g. stretching of modulus of rectangles, sending infinitesimal circles to infinitesimal ellipses, uniform bounds of $f_{\bar{z}} \leq k f_{z}$, etc.) The definition I am confused with goes back to Alfhors and Pfluger, as far as I know. So, for the present purposes Lehto wants to define a homeomorphism $f: \Omega \rightarrow \mathbb{C}$ as quasiconformal when it stretches/distorts rectangles only a finite amount. To mirror Lehto's terminology, I will call an image of a traditional euclidean rectangle under a conformal map a quadrilateral $Q$, where we keep track of the four distinguished vertices of $Q$. [I now only use the word rectangle when referring to a rectangle in the traditional sense - a polygon $R \subset \mathbb{C}$ with four sides, all of which are parallel to one of the coordinate axes, with interior angles of $\pi/2$]. To give a quantitative measure of the stretching of quadrilaterals under a mapping, we need to discuss the so-called modulus, or module of a rectangle. However, to define this, we must first recall both the Riemann Mapping Theorem and the Caratheodory Extension Theorem. The latter says that given a conformal bijection $f: \Omega \rightarrow \mathbb{D}$, we can extend $f$ to a homeomorphism $\bar{f}: \bar{\Omega} \rightarrow \bar{\mathbb{D}}$ when $\partial \Omega$ is a Jordan curve. From now on, I shall call a region $\Omega$, i.e. an open, connected subset of $\mathbb{C}$, a Jordan domain when it is a simply-connected, proper subset of $\mathbb{C}$, with boundary $\partial \Omega$ a Jordan curve. Lehto then more or less states the following fact as a consequence of Caratheodory's Theorem:

Lemma: Given two (proper) Jordan domains $\Omega, \Sigma$ in the complex plane and oriented triples $(u_1, u_2, u_3) \subset \partial \Omega$ and $(v_1, v_2, v_3) \subset \partial \Sigma$ respecting the orientations of $\partial \Omega, \partial \Sigma,$ there is a unique conformal bijection $f: \Omega \rightarrow \Sigma$ such that the homeomorphic Caratheodory extension $\bar{f} : \bar{\Omega} \rightarrow \bar{\Sigma}$ satisfies $\bar{f}(u_i) = v_i,$ for $i \in \{1,2,3\}$.

Now, let us step back for a moment. Given a rectangle $R$, it is easy to define the modulus. Translate the rectangle so that its vertices are $(0, a, a+bi, bi)$ and we set $Mod R:= a/b$. Then as a consequence of the aforementioned lemma -- two rectangles $R,S$, which are regarded as quadrilaterals by using their vertices listed counterclockwise starting from the bottom left, will be conformally equivalent iff they have the same modulus. This follows because we translate as necessary so that both rectangles have their first vertex sitting at the origin. From here, we see the only maps $g$ conformal on the interior of $R$ with extensions having $\bar{g}(R)$ also a rectangle will be an affine stretch $z \mapsto kz$, which of course preserves modulus. We learn there is a well-defined notion of modulus for conformal equivalence classes of rectangles. Ok, everything makes sense to me so far.

Where I get lost is in the claim that given an arbitrary quadrilateral $Q(v_1, v_2, v_3,v_4) \subset \mathbb{C} $, i.e. a Jordan curve with four ordered, distinguished points, can be mapped conformally to a rectangle. If this claim were to hold, then we would learn that defining $Mod Q:= Mod R$ where $R$ is any rectangle conformally equivalent to $Q$ is a well-defined notion of modulus for an arbitrary quadrilateral. I keep getting confused since we seem to have precisely 3 degrees of freedom according to the Lemma. If fixing the images of three boundary points under the extension determines the original conformal map, then how are we suddenly free to map all four vertices of our quadrilateral to vertices of a rectangle? To my mind, the best one can do is to use the RMT and Caratheodory to map the given quadrilateral $Q$ over to a quadrilateral $Q'$ consisting setwise of the closed unit disk $\bar{\mathbb{D}}$ with four distinguished points lying somewhere on $S^1 = \partial \bar{\mathbb{D}}$. Sure, we could then map the disk over to a rectangle conformally and extend the map homeomorphically to the boundary, but we only get to guarantee that 3 of our 4 distinguished vertices map to vertices of the rectangle in the image. I see no way to guarantee vertices mapping to vertices.

For concreteness, I can describe Lehto's approach more precisely. He wants to complete the conformal map of $Q$ over to a rectangle in two stages, the first of which I don't follow.

Step 1: Map $Q(v_1, v_2, v_3,v_4)$ over to a quadrilateral $Q'$ whose underlying set is the closure of the upper half plane $\overline{\mathbb{H}} \subset \hat{\mathbb{C}}$ and with distinguished vertices lying at $-1/k, -1,1,1/k$ for some $k \in (0,1)$, using the Riemann Mapping Theorem. My guess is here he wants to map over to the disk with RMT, use Blashke factors $ e^{i\theta} \frac{ z-\alpha}{1-\bar{\alpha}z} \in \text{Aut}(\mathbb{D})$ to do something nice on the boundary, then map back to the upper half plane via the standard map $g: \mathbb{D} \rightarrow \mathbb{H}$, $g(z) = i\frac{z-1}{z+1}$ and extend homeomorphically to the boundary.

If Step 1 is resolved, it is all clear to me. Step 2 is merely wield an appropriate Schwarz-Christoffel transformation to map this quadrilateral $Q'$ to a rectangle.

I feel as though I'm missing something simple, so help is much appreciated.

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  • $\begingroup$ note that the right map $g$ is $i\frac{1-z}{z+1}$ $\endgroup$
    – Conrad
    Commented Oct 15, 2019 at 0:21

1 Answer 1

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We want to get a conformal bijection $R$ from the quadrilateral $Q(v_1, v_2, v_3,v_4)$ (assumed counterclockwise as described) to the upper plane with $R(v_1)=-\frac{1}{k}, R(v_2)=-1, R(v_3)=1, R(v_4)=\frac{1}{k}$ for $0<k<1$ (since this is the counterclockwise orientaion here)

Let $F$ unique conformal bijection of $Q(v_1, v_2, v_3,v_4)$ onto the unit disc such that:

$F(v_2)=-i, F(v_3)=i, F(v_4)=-1$

Then $F(v_1)=w$ is on the arc from $-1$ to $-i$ by topology (by the orientation conventions above).

The automorphisms of the disc that fix $\pm i$ are precisely $\frac{z-ia}{1+iaz}$ for real $a$ in $(-1,1)$ so as we move $-1$ with such an $-1<a<0$ towards $i$ there is a unique $a$ for which the image of $-1$ and the image of $w$ above are conjugate and you compose $F$ with that getting a conformal map $G$ from the quadrilateral onto the unit disc s.t.:

$G(v_2)=-i, G(v_3)=i, G(v_4)=w_1, G(v_1)=\bar w_1$ where $\Re w_1 <0, \Im w_1 >0$

Apply now the standard Riemann map $i\frac{1-z}{z+1}$ from the unit disc to the upper plane and you are done since conjugate points on the circle go to $\pm$, while the right-hand side of the unit circle goes in between $-1,1$ so since we are on the left-hand side $v_1,v_4$ behave as they should (they are $\pm b, b>1$ with $v_1=-b$ so $k=\frac{1}{b}$

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