I’ve written about Möbius transformations many times because they’re simple functions that nevertheless have interesting properties.
A Möbius transformation is a function f : ℂ → ℂ of the form
f(z) = (az + b)/(cz + d)
where ad – bc ≠ 0. One of the basic properties of Möbius transformations is that they form a group. Except that’s not quite right if you want to be completely rigorous.
The problem is that a Möbius transformation isn’t a map from (all of) ℂ to ℂ unless c = 0 (which implies d cannot be 0). The usual way to fix this is to add a point at infinity, which makes things much simpler. Now we can say that the Möbius transformations form a group of automorphisms on the Riemann sphere S².
But if you insist on working in the finite complex plane, i.e. the complex plane ℂ with no point at infinity added, each Möbius transformations is actually a partial function on ℂ because a point may be missing from the domain. As detailed in [1], you technically do not have a group but rather an inverse monoid. (See the previous post on using inverse semigroups to think about floating point partial functions.)
You can make Möbius transformations into a group by defining the product of the Möbius transformation f above with
g(z) = (Az + B) / (Cz + D)
to be
(aAz + bCz + aB + bD) / (Acz + Cdz + Bc + dD),
which is what you’d get if you computed the composition f ∘ g as functions, ignoring any difficulties with domains.
The Möbius inverse monoid is surprisingly complex. Things are simpler if you compactify the complex plane by adding a point at infinity, or if you gloss over the fine points of function domains.
Related posts
- Transformations of Olympic rings
- Curiously simple approximations
- Solving for Möbius transformation coefficients
[1] Mark V. Lawson. The Möbius Inverse Monoid. Journal of Algebra. 200, 428–438 (1998).
It is also noteworthy that the operation you defined induce a morphism to the algebra of 2×2 complex matrices… which comes with a bunch of nice surprises too.