1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What is the "algebra-precalculus" tag doing here? $\endgroup$ – Timbuc Jan 20 '15 at 21:16
  • 1
    $\begingroup$ 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. $\endgroup$ – punctured dusk Jan 20 '15 at 21:18
  • $\begingroup$ It has to do with homogeneous coordinates in projective space. Section 3.VI in Needham's Visual Complex Analysis explains this. $\endgroup$ – Hans Lundmark Jan 20 '15 at 22:24
0
$\begingroup$

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 $cz+d \neq 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.