The Möbius Inverse Monoid

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 adbc ≠ 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 fg 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.

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

One thought on “The Möbius Inverse Monoid

  1. 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.

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