Through the Looking-Glass, and What Ahlfors Found There

There are two ways to extend a function f\colon [0,+\infty) to the negative semi-axis:

  • Even reflection f(-x)=f(x)
  • Odd reflection f(-x)=-f(x)
Reflection in 1D

As the sketch shows, the even reflection is more likely to yield a continuous function. The odd reflection has no chance of being continuous unless f(0)=0. On the other hand, it has the same slope to the left and to the right of 0, so if f happens to be differentiable with f(0)=0, then the odd reflection is differentiable too.

Most importantly for this post, the odd extension has a chance of being a bijection. For example, f(x)=\sqrt{x} extends to f(x)=x/\sqrt{|x|}, which is a homeomorphism of \mathbb R onto itself.

Let’s move one dimension up. It’s more convenient to place the looking-glass onto the horizontal axis. We have a map that f=(u,v) is defined in the upper halfplane y\ge 0, and would like to extend its definition to y<0. If f sends the line y=0 into v=0, we can extend it using the mixed even-u/odd-v method: u(x,-y)=u(x,y), v(x,-y)=-v(x,y). This is easier to write down in complex notation: f(\bar z)=\overline{f(z)}. If f was a homeomorphism of the upper halfplane onto itself, the extension will be a homeomorphism of the entire plane. The process looks like this:

But in general, the image of the horizontal axis is not a straight line. How can we find a reflection of f in such a strange mirror?

Curved looking-glass

Going back to the origins of geometry, we can use similar triangles to locate the image of a point:


Denoting the purple point by z and the blue one by z+h, we see that the (complex) scaling factor of the triangle is \displaystyle \frac{f(z+h)-f(z)}{h}, and therefore the image of the red point is

\displaystyle f(\bar z)=f(z)+\frac{f(z+h)-f(z)}{h}(\bar z-z)

This looks promising, but there is no obvious reason why f(\bar z) should fall into the region behind the looking-glass, or why this extension should be a homeomorphism. It is also unclear how to choose h: should it be real, as on the picture above? should its size be fixed or depend on z? Or maybe… get rid of h be letting h\to 0? Let’s do that. The extension becomes

\displaystyle f(\bar z)=f(z)+f'(z)(\bar z-z)

Can’t get much simpler than that. The identity f(z)=z extends to f(\bar z)=z+1(\bar z-z)=\bar z, which is encouraging but not very interesting. Let’s try other powers f(z)=z^p. The extension \displaystyle f(\bar z)=z^p+pz^{p-1}(\bar z-z) is homogeneous of degree p, so concentric circles around 0 are mapped onto similar curves. The plots below show the images of concentric circles for several value of p: the image of upper semicircle is green, the image of the bottom (where the extension formula is used) is in red.

Let’s begin with p>1:

p=5/4, extension is a homeomorphism
p=3/2, extension forms cusps
p=5/3, injectivity is lost -- the extended map overlaps itself

And it gets worse as p increases. In the opposite direction, consider powers less than 1:

p=2/3: looking good
p=1/2: no longer injective
p=1/3: horrible, the extension overlaps the green part

It’s surprisingly easy to get to the root of the problem. Take the complex derivatives of the extended map: f(\bar z)_{z}=f''(z)(\bar z-z) and f(\bar z)_{\bar z}=f'(z). To preserve orientation, the extension must satisfy |f(\bar z)_{z}|<|f(\bar z)_{\bar z}|, which requires the original map to satisfy 2|f''(z)|\mathrm{Im}\,z<|f'(z)|. With f(z)=z^p this boils down to |p-1|<1/2.

This extension (in a more general form) was given by Lars Ahlfors in "Sufficient conditions for quasiconformal extension" (1973). I learned about it in when working on quasisymmetric graphs. The plots were created with Maple:

with(plots): p:=3/2:
curves:={seq(complexplot((1+k/5)^p*exp(I*p*t),t=0..Pi,color=green,thickness=2), k=-4..4)} union {seq(complexplot((1+k/5)^p*((1-p)*exp(I*p*t)+p*exp((p-2)*I*t)),t=0..Pi,thickness=2) ,k=-4..4)}:

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