Möbius transformations

Teaching note: Möbius Transformations

This short note is written for those taking my second-year analysis course.

Möbius transformations are brilliant. They are conformal (=angle preserving) maps from the extended complex plane to itself.

Definition 1 (The extended complex plane). The extended complex plane is {\overline{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}}, i.e. {\mathbb{C}} together with an entirely new element, denoted {\infty}.

For now, we won’t extend the domains of {+} and {\cdot} to include {\infty}; so the extended complex plane has no more algebraic structure than the field of complex numbers.

Definition 2 (Möbius transformations). A Möbius transformation is a function {f:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}} such that

\displaystyle  f(z)=\left\{\begin{array}{ll}\frac{az+b}{cz+d}&z\neq-\frac{d}{c},\infty\\\infty&z=-\frac{d}{c}\\\frac{a}{c}&z=\infty,\end{array}\right.

for some {a,b,c,d\in\mathbb{C}} such that {ad-bc\neq0}.

The second and third cases in the definition above are important, but we specify a Möbius transformation simply by writing e.g.

\displaystyle f(z):=\frac{az+b}{cz+d},

and leave the other two cases implicit.

Nevertheless, such a Möbius transformation does not determine the quadruple {(a,b,c,d)}, but does determine the tuple {[a;b;c;d]} in projective space. For a trivial example: the identity map {z} is a Möbius transformation, equal to {\frac{2z}{2}}.

Example 1. Consider the Möbius transformation {g:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}} defined by {g(z):=\frac{1}{z-1}.} Note that {g} really is a Möbius transformation since {-1\neq0}. A few example evaluations: {g(0)=-1}, {g(1)=\infty}, and {g(\infty)=0}.

Proposition 3. The composition of two Möbius transformations is a Möbius transformation.

Proof: For {i=1,2}, consider a Möbius transformation {f_{i}(z):=\frac{a_{i}z+b_{i}}{c_{i}z+d_{i}}}, where {a_{i}d_{i}-b_{i}c_{i}\neq0}. The composition is \displaystyle \begin{array}{rcl} f_{2}\circ f_{1}(z)&=&\frac{(a_{1}a_{2}+c_{1}b_{2})z+b_{1}a_{2}+d_{1}b_{2}}{(a_{1}c_{2}+c_{1}d_{2})z+b_{1}c_{2}+d_{1}d_{2}},\end{array} which is a Möbius transformation since \displaystyle (a_{1}a_{2}+c_{1}b_{2})(b_{1}c_{2}+d_{1}d_{2}) - (a_{1}c_{2}+c_{1}d_{2})(b_{1}a_{2}+d_{1}b_{2}) = (a_{1}d_{1}-b_{1}c_{1})(a_{2}d_{2}-b_{2}c_{2}) is not zero. \Box

Proposition 4. Möbius transformations are bijections.

Proof: Let {f(z)=\frac{az+b}{cz+d}}, with {ad-bc\neq0}. We find the candidate for its inverse, and check it really is. Define {g:\overline{\mathbb{C}}\longrightarrow\overline{\mathbb{C}}}
by

\displaystyle g(w)=\frac{dw-b}{-cz+a}.

We compose:

\displaystyle \begin{array}{lll} 	g\circ f(z)	&=&	\frac{d\frac{az+b}{cz+d}-b}{-c\frac{az+b}{cz+d}+a}\\ 	&=&	\frac{(ad-bc)z}{ad-bc}\\ 	&=&	z, \end{array}

the identity! The composition {f\circ g} is the identity by the same calculation. \Box

Definition 5 (Möbius group). The set of Möbius transformations, equipped with composition, forms the Möbius group, denoted {\mathbf{M}}.

We are able to represent Möbius transformations as matrices, in the following way.

Definition 6. Define the map

\displaystyle \begin{array}{rcl} 	\Phi:\mathbf{M} &\longrightarrow& \mathrm{PGL}_{2}(\mathbb{C})\\ 	\frac{az+b}{cz+d} &\longmapsto& \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{array}

Naturally we wonder whether, under this representation of Möbius transformations as matrices, the composition of functions corresponds to the multiplication of matrices.

Proposition 7. {\Phi} is an isomorphism.

Proof: It is clear that {\Phi} is a bijection. For {i=1,2}, consider {f_{i}\in\mathbf{M}} defined by {f_{i}(z):=\frac{a_{i}z+b_{i}}{c_{i}z+d_{i}}}, where {a_{i}d_{i}-b_{i}c_{i}\neq0}.
The composition is

\displaystyle \begin{array}{rcl} f_{2}\circ f_{1}(z)&=&\frac{(a_{1}a_{2}+c_{1}b_{2})z+b_{1}a_{2}+d_{1}b_{2}}{(a_{1}c_{2}+c_{1}d_{2})z+b_{1}c_{2}+d_{1}d_{2}},\end{array}

and we have

\displaystyle \begin{array}{rcl} \Phi(f_{2})\Phi(f_{1})	&=&	\begin{pmatrix}a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}\begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}\\&=& \begin{pmatrix} a_{1}a_{2}+c_{1}b_{2} & b_{1}a_{2}+d_{1}b_{2} \\ a_{1}c_{2}+c_{1}d_{2} & b_{1}c_{2}+d_{1}d_{2} \end{pmatrix}\\&=& \Phi(f_{2}\circ f_{1}),\end{array}

which shows that {\Phi} is a homomorphism. \Box

Proposition 8. Möbius transformations act {3}-transitively on {\overline{\mathbb{C}}}.

Proof sketch: To see this, let {w_{0},w_{1},w_{\infty}} be three distinct elements of the extended complex plane. We wish to choose {[a;b;c;d]} such that

\displaystyle f(z)=\frac{az+b}{cz+d}

satisfies

\displaystyle f(0)=w_{0},\,f(1)=w_{1},\,f(\infty)=w_{\infty}.

Equivalently, such that the three equations

  • {\frac{b}{d}=w_{0}}
  • {\frac{a+b}{c+d}=w_{1}}
  • {\frac{a}{c}=w_{\infty}}

hold. It suffices to satisfy \displaystyle w_{\infty}c+w_{0}d=w_{1}c+w_{1}d, which can indeed be accomplished. \Box

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