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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.

azimut
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Bart Michels
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    What is the "algebra-precalculus" tag doing here? – Timbuc Jan 20 '15 at 21:16
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    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 [tag:matrices] or [tag:abstract-algebra]. Retag if you feel the need to. – Bart Michels 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

2 Answers2

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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 $cz+d \neq 0$.

Bart Michels
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Wikipedia has a quite nice section about the (projective) matrix representation of the Möbius transformation. The key insight is to realize that $z$ can be represented as a vector with homogeneous coordinates, $\vec{z}$, and that this basically turns the Möbius transformation into a matrix multiplication in linear algebra, and those multiplications are associative.

So, a chained Möbius transformation $\mu_A\circ(\mu_B\circ z)$ can be written as $A(B\,\vec{z})$ if we represent the Möbius transforations $\mu_A$ and $\mu_B$ as the two-by-two matrices $A$ and $B$, respectively, and $z$ as a vector with homogenuous coordinates, $\vec{z}$. Since matrix multiplications are associative, we can rewrite $A(B\,\vec{z})$ as $(AB)\vec{z}$, which is just another way to represent the Möbius transformation $\mu_{AB}\circ z$.

HelloGoodbye
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