# On the matrix representation of a composition of Möbius transforms

Let the Möbius transform associated to the matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ be defined as $\mu_A:\mathbb C\to\mathbb C:z\mapsto\frac{az+b}{cz+d}$ provided $\det A\neq 0$.
It is straightforward to verify that $\mu_A\circ\mu_B=\mu_{AB}$. I was wondering if there is a more intuitive (and preferably elementary) way to see why we have this identity without having to do the calculation.

I was thinking of viewing $\frac{az+b}{cz+d}$ as a 'formal' fraction; that is, just another notation for $\begin{pmatrix}az+b\\cz+d\end{pmatrix}$ and then trying to find out if $AB\begin{pmatrix}z\\1\end{pmatrix}$ corresponds to the usual composition $\mu_A\circ\mu_B$. I can't see it. There should be a deeper reason for this.

• What is the "algebra-precalculus" tag doing here? – Timbuc Jan 20 '15 at 21:16
• The description says "... and other symbolic-manipulation topics." That's why. I must say I hesitated about adding this tag, but it seemed more appropriate than matrices or abstract-algebra. Retag if you feel the need to. – punctured dusk Jan 20 '15 at 21:18
• It has to do with homogeneous coordinates in projective space. Section 3.VI in Needham's Visual Complex Analysis explains this. – Hans Lundmark Jan 20 '15 at 22:24

The group action of $\operatorname{GL}(2, \mathbb C)$ on $\mathbb C^2$ by left multiplication:
• stabilizes $(0,0)$, so it restricts to $\mathbb C^2\setminus 0$
• is linear, so descends to the projective space $\mathbb P^1(\mathbb C) = (\mathbb C^2\setminus 0)/\mathbb C^\times$
In $\mathbb P^1$, $[x, y] = [\frac xy,1]$ for $y\neq0$, so in the first coordinate map the group action reads $$\begin{pmatrix}a&b\\c&d\end{pmatrix}z = \frac{az+b}{cz+d}$$ for $z\neq\infty$ and $cz+d \neq 0$.