möbius transformation | Samin Riasat's Blog

A Möbius transformation is a map $f:\mathbb C_\infty\to\mathbb C_\infty$ of the form

$\displaystyle f(z)=\frac{az+b}{cz+d},\quad a,b,c,d\in\mathbb C,\quad ad-bc\neq 0$

where $\mathbb C_\infty:=\mathbb C\cup\{\infty\}$ is the extended complex plane and the ‘point at infinity’ $\infty$ is defined so that

if $c\neq 0$ then $f(\infty)=a/c$ and $f(-d/c)=\infty$ ;
if $c=0$ then $f(\infty)=\infty$ .

The following video gives a very illuminating illustration. The sphere in the video is called the Riemann sphere, which in a sense ‘wraps up’ the extended complex plane into a sphere. Each point on the sphere corresponds to a unique point on the plane (i.e. there is a bijection between points on the extended plane and points on the sphere), with the ‘light source’ being the point at infinity. This bijective correspondence is the main reason for including the point at infinity.

According to the video any Möbius transformation can be generated by the four basic ones: translations, dilations, rotations and inversions:

Translation: $f(z)=z+b$ , $b\in\mathbb C$
Dilation: $f(z)=az$ , $a\in\mathbb R$
Rotation: $f(z)=e^{i\theta}z$ , $\theta\in[0,2\pi]$
Inversion: $f(z)=1/z$

Exercise. Show that any Möbius transformation is a composition of these four operations.

The Möbius transformations in fact form a group $\mathcal M$ under composition which acts on $\mathbb C_\infty$ . Moreover, we have a surjective homomorphism

$\displaystyle\begin{matrix}\hbox{GL}_2(\mathbb C)\to\mathcal M,\quad\left(\begin{matrix} a & b\\ c & d\end{matrix}\right)\mapsto\displaystyle\frac{az+b}{cz+d}\end{matrix}$ .

Möbius transformations exhibit very interesting properties, some of which are:

Proposition 1. Given distinct $z_1,z_2,z_3\in\mathbb C_\infty$ , there is a unique Möbius map $f\in\mathcal M$ such that

$\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$

Proof. It is not difficult to work out that the unique $f$ is given by

$\displaystyle f(z)=\frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}$ . $\square$

Proposition 2. The action of $\mathcal M$ on $\mathbb C_\infty$ is sharply triply transitive: if $z_1,z_2,z_3\in\mathbb C_\infty$ are distinct and $w_1,w_2,w_3\in\mathbb C_\infty$ are distinct, then there eixsts a unique $f\in\mathcal M$ such that $f(z_i)=w_i$ for $i=1,2,3$ .

Proof. By proposition 1, there is a unique $g\in\mathcal M$ such that

$\displaystyle g:\left(\begin{matrix} z_1\\ z_2\\ z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$

and a unique $h\in\mathcal M$ such that

$\displaystyle h:\left(\begin{matrix} w_1\\ w_2\\ w_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$ .

Then $f=h^{-1}\circ g$ is the unique map satisfying the required property. $\square$

The cross-ratio $[z_1,z_2,z_3,z_4]$ of four distinct points $z_1,z_2,z_3,z_4\in\mathbb C_\infty$ is defined to be the unique $\lambda\in\mathbb C_\infty$ such that if $f\in\mathcal M$ is the unique map satisfying

$\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$

then $f(z_4)=\lambda$ , i.e.

$\displaystyle [z_1,z_2,z_3,z_4]=\frac{z_1-z_4}{z_3-z_4}\frac{z_2-z_3}{z_2-z_1}$ . $(*)$

One nice thing about cross-ratios is that they are preserved by Möbius transformations.

Proposition 3. If $f\in\mathcal M$ , then $[z_1,z_2,z_3,z_4]=[f(z_1),f(z_2),f(z_3),f(z_4)]$ .

Proof. Let $g\in\mathcal M$ such that

$\displaystyle g:\left(\begin{matrix} z_1\\ z_2\\ z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$ .

Then $[z_1,z_2,z_3,z_4]=g(z_4)$ . Likewise, if $h\in\mathcal M$ satisfies

$\displaystyle h:\left(\begin{matrix} f(z_1)\\ f(z_2)\\ f(z_3)\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$

then $h\circ f=g$ by proposition 1, so $[f(z_1),f(z_2),f(z_3),f(z_4)]=h(f(z_4))$ $=g(z_4)=[z_1,z_2,z_3,z_4]$ . $\square$

From $(*)$ we observe that some permutations of $1,2,3,4$ leave the value of the cross-ratio $[z_1,z_2,z_3,z_4]$ unaltered, e.g. $(1\ 3)(2\ 4)\in S_4$ is one such. What about the others?

Let $S_4$ act on the indices of the cross-ratio $[z_1,z_2,z_3,z_4]$ . The permutations $\sigma\in S_4$ that fix $[z_1,z_2,z_3,z_4]$ form the stabiliser subgroup of $S_4$ of this action. Using transitivity and invariance (propositions 2 and 3), the orbit of $[z_1,z_2,z_3,z_4]$ is just the different assignments of the values $0,1,\infty$ to $z_1,z_2,z_3$ ; i.e. the distinct cross-ratios that we get by permuting the indices are just

$[z_1,z_2,z_3,z_4]$ , $[z_1,z_3,z_2,z_4]$ , $[z_2,z_1,z_3,z_4]$ ,

$[z_2,z_3,z_1,z_4]$ , $[z_3,z_1,z_2,z_4]$ , $[z_3,z_2,z_1,z_4]$ .

Let $f\in\mathcal M$ such that

$\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)$ .

Then writing $\lambda=f(z_4)=[z_1,z_2,z_3,z_4]$ shows that the values of the above cross-ratios are (not necessarily in this order—too lazy to work out the precise order)

$\displaystyle \lambda,\frac 1\lambda, 1-\lambda,\frac{1}{1-\lambda},\frac{\lambda}{\lambda-1},\frac{\lambda-1}{\lambda}$

and they (as functions of $\lambda$ ) form the subgroup of $\mathcal M$ that fixes the set $\{0,1,\infty\}$ . This group is isomorphic to $S_3$ .

So there are in fact four permutations $\sigma\in S_4$ such that

$[z_1,z_2,z_3,z_4]=[z_{\sigma(1)},z_{\sigma(2)},z_{\sigma(3)},z_{\sigma(4)}]$

and they form a subgroup of $S_4$ that is isomorphic to the Klein four-group

$V_4=\{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}$ .

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