In Ahlfors' "Lectures on Quasiconformal Mappings," he shows that, using the geometric definition of K-quasiconformal, if a map between regions is locally K-quasiconformal then it is globally K-quasiconformal. At a critical step in the proof he assumes that one can find a suitably fine subdivision to make the proof work. However, as pointed out in the Editor's notes, page 83, this itself requires proof, and this is what I am having a hard time understanding.

Everything in the Editor's notes makes sense until it says "so one can show by using the Teichmüller extremal problem in Chapter IIIA that it contains a horizontal line segment". I just don't see how this follows. Can someone please help me? I've been beating my head against this.


1 Answer 1


Let's review from the beginning. Shishikura's suggestion is to begin by dividing the rectangle $Q$ into small subrectangles by vertical and horizontal lines (with respect to the structure of $Q$). When these are mapped to $Q'$ we get something like this, where the curves of subdivision need not be horizontal or vertical with respect to $Q'$. I'll call them pseudo-horizontal and pseudo-vertical.


We focus on each pseudo-vertical strip $Q_i'$ and try to change the subdivision within it so that it's made by horizontal curves with respect to $Q_i'$. Let's conformally map $Q_i'$ onto a rectangle:


We'd like to draw a horizontal line within each small quadrilateral here, and then erase the original pseudo-horizontal curves. This will be the new subdivision. Potential obstacle: maybe there is no suitable horizontal segment, like in the quadrilateral with sides highlighted in blue.

This is where a modulus estimate is needed. I will argue that a quadrilateral of the kind drawn in blue cannot have modulus less than $1$.

Suppose it does. Map the infinite vertical strip obtained by vertically extending the above rectangle onto upper half-plane using the exponential map. The horizontal segments are transformed into half-circles centered at $0$. Adding to this quadrilateral its reflection in the real axis, we obtain an annulus $\Omega$ that

  1. Does not contain any circle centered at the origin
  2. Has conformal modulus greater than $1/2$ (because the modulus is the reciprocal of extremal length of separating curves, which is twice the extremal length of the curves joining the blue sides of the quadrilateral).

The combination of 1 and 2 contradicts the solution of the aforementioned Teichmüller extremal problem. Indeed, let $C_1$ and $C_2$ be the bounded and unbounded components of $\Omega^c$. Note that $0\in C_1$. Let $z_1$ be a point of $C_1$ with maximal absolute value, and let $z_2$ be a point of $C_2$ with minimal absolute value. By assumption 1 above, $|z_2|\le |z_1|$.

By scaling and rotating $\Omega$, we may assume that $z_1=-1$ and consequently, $|z_2|\le 1$. Hence, the modulus of $\Omega$ does not exceed the modulus of the Teichmüller ring $T = \mathbb{C}\setminus ([-1,0]\cup [1,\infty))$.

But $\operatorname{Mod} T=1/2$. To prove this, cut $T$ in half and note that the extremal length of the curves connecting $(-\infty,-1]$ to $[0,1]$ in the upper halfplane is $1$ is $1$, as this quadrilateral is isometric to its conjugate. This contradiction completes the proof.

(The passage from quadrilaterals to annuli could be avoided if one formulated and solved a version of the Teichmüller extremal problem for quadrilaterals contained in the upper half plane, with two sides on the real axis.)

  • 1
    $\begingroup$ First of all, thank you for your response. I think I get what you are saying, but I don't think that just because the blue segments themselves can be joined by horizontal lines, the curvy pseudo-horizontal curves joining the blue segments at the ends won't be warped enough to prevent a horizontal line within the quadrilateral. $\endgroup$ Feb 25, 2015 at 2:32
  • $\begingroup$ @AaronCohen I corrected the argument. $\endgroup$
    – user147263
    Apr 12, 2015 at 3:28

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