The Möbius Inverse Monoid

I've written about Mobius transformations many times because they're simple functions that nevertheless have interesting properties.

A Mobius transformation is a function f : of the form

f(z) = (az + b)/(cz + d)

where ad - bc 0. One of the basic properties of Mobius 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 Mobius transformation isn't a map from (all of) to 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 Mobius transformations form a group of automorphisms on the Riemann sphere S^2.

But if you insist on working in the finite complex plane, i.e. the complex plane with no point at infinity added, each Mobius transformations is actually a partial function on 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 Mobius transformations into a group by defining the product of the Mobius 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 Mobius 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 Mobius transformation coefficients

[1] Mark V. Lawson. The Mobius Inverse Monoid. Journal of Algebra. 200, 428-438 (1998).

The post The Mobius Inverse Monoid first appeared on John D. Cook.