Möbius transformations
Recall that given
\[M = \begin{bmatrix}a&b\\c&d\end{bmatrix} \in \mat\]
such that $\det(M) = ad-bc \neq 0$, we define the corresponding Möbius transformation
\begin{align*}
f_M : \C_\infty &\longrightarrow \C_\infty \\
z &\longmapsto \frac{az+b}{cz+d}.
\end{align*}
Recall also that $\det(M)\neq 0$ if and only if $M$ is invertible.
Throughout this entire problem, all matrices are assumed to be invertible.
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Verify that for $A, B \in \mat$,
\[f_A\circ f_B = f_{AB}.\]
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Show that for any $M \in \mat$, $f_M$ is a bijection.
Since $f_M$ is holomorphic on its domain and its singularity is a pole, this tells us $f_M \in \aut(\C_\infty)$.
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Describe all $M$ for which $f_M(0) = 0$ and $f_M(\infty) = \infty$.
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Suppose that $\varphi \in \aut(\C_\infty)$ is such that $\varphi(0) = 0$ and $\varphi(\infty) = \infty$.
Show that for some $\lambda \in \C^\times$, $\varphi(z) = \lambda z$.
(Hint: show $\varphi(1/z)$ has a simple pole at zero; remove it.)
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Conclude that $\aut(\C_\infty) = \set{f_M \mid M \in \mat}$.
We can conclude that any automorphism of $\C_\infty$ is completely determined by its pole, its zero, and a scale.
Alternatively, given three input points and three output points, there is precisely one automorphism of $\C_\infty$ which takes each input point to its corresponding output.