Quasiconformal maps

Conformal mappings are often too rigid for us to use. We know that Riemann mappings, for example, are unique up to conformal automorphisms of the unit disk – the set of which forms a topological group homeomorphic to $S^1 \times \mathbb{D}$. In complex analysis, we often need maps which are more flexible and in many cases, what we are looking for are quasiconformal maps. While conformal maps preserve angle locally, quasiconformal maps distort angles locally up to some factor.

Throughout this post, we are only interested in orientation preserving homeomorphisms $f: U \to V$ between two non-empty open subsets $U$ and $V$ of $\mathbb{C}$.

$C^1$ $K$-quasiconformal maps

When $f=u+iv$ is $C^1$, its Jacobian $Jf$ must be positive. In terms of partial derivatives,

and so we must have the inequality $ \vert f_z \vert > \vert f_\bar{z} \vert $. We define the complex dilatation $\mu_f$ on $U$ to be $\mu_f(z) = \frac{f_\bar{z}(z)}{f_z (z)}$. Its values must lie in the unit disk $\mathbb{D}$.

Applying triangle inequality to the one-form $df = f_z dz + f_{\bar{z}} d\bar{z}$,

We then see that in general, the image under $f$ of the unit circle on the tangent space $T_z U$ is an ellipse of on the tangent space $T_{f(z)} V$. The eccentricity $D_f(z)$ of the ellipse, i.e. the ratio of the major axis to the minor axis, is given by:

We say that a $C^1$ map $f: U \to V$ is $K$-quasiconformal if $D_f$ is uniformly bounded above on $U$ by some constant $K\geq 1$. If so, the constant $K$ is called the dilatation of $f$. This definition is indeed a generalisation of conformal maps because when $K=1$, then $f_{\bar{z}} \equiv 0$.

One easy example is the following. For any real constant $a$, the shear transformation

is an $\mathbb{R}$-linear diffeomorphism from $\mathbb{C}$ onto itself. Its complex dilatation is the constant value $\frac{ai}{2-ai}$ and so $f$ is a $K$-quasiconformal map where $K = \frac{\sqrt{a^2 + 4} + a}{\sqrt{a^2+4}-a}$.

$K$-quasiconformal maps

Sometimes, $C^1$ is still too rigid. We can relax the regularity of $f$ and say that $f$ is $K$-quasiconformal if there is a constant $K\geq 1$ such that

• locally, $f$ is absolutely continuous along almost every horizontal and vertical line,
• $D_f(z) \leq K$ almost everywhere on $U$.

The second criterion is very much similar to the $C^1$ case. The first criterion is often called ACL (absolutely continuous on lines) and it simply states that $x \mapsto f(x+iy)$ (resp. $y \mapsto f(x+iy)$) is differentiable almost everywhere and satisfies the fundamental theorem of calculus for almost all $y$ (resp. $x$). With more careful analysis, we can show that $f$ must be (real) differentiable almost everywhere. (Refer to Ahlfors¹.) Since both conditions are local, we can generalise this definition to maps on Riemann surfaces as well.

There are of course plenty of examples of quasiconformal maps which are not $C^1$. One of them is the following. Pick any real number $a$ and let’s define $f : \mathbb{C} \to \mathbb{C}$ by

The function $f$ is a gluing of the identity map on the lower half plane and a shear map on the upper half plane. It is a homeomorphism which is differentiable everywhere except along the real axis $\mathbb{R}$, and its dilatation is again $K = \frac{\sqrt{a^2 + 4} + a}{\sqrt{a^2+4}-a}$.

A topological quadrilateral $Q = (Q,a_1,a_2,a_3,a_4)$ is a Jordan disk in $\mathbb{C}$ equipped with four vertices $a_i$ for $i=1,2,3,4$ lying on the boundary $\partial Q$ and labelled cyclically in a counterclockwise manner. By applying Riemann mapping theorem, Caratheodory’s theorem, and an appropriate Schwarz-Christoffel mapping, we can show that every quadrilateral $Q$ admits a unique positive number $m>0$ and a unique biholomorphism

such that $\phi$ maps the vertices $(a_1,a_2,a_3,a_4)$ to the tuple $(0,m,m+i,i)$. The (conformal) modulus $\text{mod}(Q)$ of $Q$ is the side length $m$.

I think it is clear from the definition that the modulus of a quadrilateral is invariant under conformal maps, i.e. $\text{mod} (f(Q)) = \text{mod} (Q)$ whenever $f$ is conformal on $Q$. Quasiconformal maps distort moduli up to some bounded constant; in fact, this can be taken as a geometric definition. (Refer to Ahlfors¹ Chapter 2. If you do so, notice that I am trying hard to avoid using the term ‘extremal length’.)

Theorem: An orientation preserving homeomorphism $f: U \to V$ between two open subsets $U$ and $V$ of $\mathbb{C}$ is $K$-quasiconformal if and only if for every quadrilateral $Q$ in $U$,

\[\frac{\text{mod}(Q)}{K} \leq \text{mod} (f(Q)) \leq K \text{mod}(Q).\]

An immediate consequence of the theorem is the following.

Corollary: The composition of a $C^1$ $K$-quasiconformal map and a $C^1$ $L$-quasiconformal map is $KL$-quasiconformal. The inverse of a $C^1$ $K$-quasiconformal map is again $K$-quasiconformal.

Note that in the $C^1$ case, the proof of the above can be done using complex chain rule. There are many other slick results on quasiconformal maps, but I will probably discuss this in another post.

References

1: L. Ahlfors. Lectures on Quasiconformal Mappings. Van Nostrand, 1966.