There are two ways to extend a function to the negative semi-axis:

- Even reflection
- Odd reflection

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 . On the other hand, it has the same slope to the left and to the right of , so if happens to be differentiable with , then the odd reflection is differentiable too.

Most importantly for this post, the odd extension has a chance of being a bijection. For example, extends to , which is a homeomorphism of 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 is defined in the upper halfplane , and would like to extend its definition to . If sends the line into , we can extend it using the mixed even-u/odd-v method: , . This is easier to write down in complex notation: . If 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 in such a strange mirror?

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

Denoting the purple point by and the blue one by , we see that the (complex) scaling factor of the triangle is , and therefore the image of the red point is

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

Can’t get much simpler than that. The identity extends to , which is encouraging but not very interesting. Let’s try other powers . The extension is homogeneous of degree , so concentric circles around 0 are mapped onto similar curves. The plots below show the images of concentric circles for several value of : 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 :

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

It’s surprisingly easy to get to the root of the problem. Take the complex derivatives of the extended map: and . To preserve orientation, the extension must satisfy , which requires the original map to satisfy . With this boils down to .

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)}:

display(curves);