# Maps of the Disc

What are the automorphisms of $${\mathbb{D}}, {\mathbf{C}}, {\mathbf{CP}}^1$$? For the disc, it’s known: \begin{align*} \mathop{\mathrm{Aut}}_{\mathbf{C}}({\mathbb{D}}) = \left\{{ z\mapsto e^{i\theta} \qty{\alpha - z \over 1 - \overline{\alpha} z} }\right\} .\end{align*}

Schwarz lemma.

\begin{align*} \mathop{\mathrm{Aut}}_{\mathbf{C}}({\mathbb{H}}) = \left\{{ z \mapsto {az+b \over cz+d} {~\mathrel{\Big\vert}~}a,b,c,d\in {\mathbf{C}}, ad-bc=1 }\right\} \cong{\operatorname{PSL}}_2({\mathbf{R}}) .\end{align*}

## Exercises

Show that $$\mathop{\mathrm{Aut}}({\mathbf{C}}) = \left\{{ z \mapsto az+b{~\mathrel{\Big\vert}~}a\in {\mathbf{C}}^{\times}, b\in {\mathbf{C}}}\right\}$$.

$$\supseteq$$: Clear, every affine function is bijective with inverse $$g(z) \coloneqq a^{-1}(z-b)$$, and holomorphic since \begin{align*} g(f(z)) = z \implies g'(f(z)) f'(z) = 1 \implies g'(w) = {1\over f'(f^{-1}(w))} .\end{align*}

$$\subseteq$$: Suppose $$f$$ is entire and bijective. If $$f$$ is bounded, then $$f(z) = b$$ is constant by Liouville and we’re done, so suppose not. Then $${\left\lvert {f(z)} \right\rvert}\to \infty$$ as $${\left\lvert {z} \right\rvert}\to \infty$$, making $$z=\infty$$ either an essential singularity or a pole, since it is isolated and not removable since $$f$$ is unbounded in every small enough neighborhood of $$\infty$$. If $$f(b) = 0$$ without loss of generality we can translate to assume $$f(0) = 0$$ by replacing $$f(z)$$ with $$f(z) - b$$. By bijectivity, $$z=0$$ must be a zero of order 1, so $${1\over f(z)}$$ has a pole of order 1 at $$z=0$$. So define $$G(z) \coloneqq{z\over f(z)}$$, which is now an entire function.

On a large enough disc, $${\left\lvert {f(z)} \right\rvert} > 1$$ for all $${\left\lvert {z} \right\rvert} > R$$, so $${\left\lvert {G(z)} \right\rvert} < 1R^{-1}$$ on $${\left\lvert {z} \right\rvert} > R$$. Since $$G$$ is holomorphic and thus continuous on $${\left\lvert {z} \right\rvert}\leq R$$, which is compact, $$G$$ is bounded here too. Thus $$G$$ is constant by Liouville, and $$G(z) = c \implies f(z) = c^{-1}z \coloneqq az$$. Unwinding the initial translation that ensured $$f(0)=0$$, we get $$f(z) = az + b$$.

Show that $$\mathop{\mathrm{Aut}}({\mathbf{CP}}^1) = \left\{{{az+b\over cz+d} {~\mathrel{\Big\vert}~}ad-bc\neq 0}\right\}$$.

Write $$I(z) \coloneqq{1\over z}$$ and note that \begin{align*} {az+b\over cz+d} = {a \over c} + {bc-ad\over c}{1\over cz+d} = (z\mapsto cz+d)\circ\qty{ z \mapsto {1\over z} }\circ\qty{{ a\over c} + {bc-ad\over c} z} ,\end{align*} which is $$I$$ composed with affine transformations. Note that the last map satisfies \begin{align*} {\frac{\partial }{\partial z}\,} {bc-ad \over c} z + {a\over c} = {bc-ad\over c} ,\end{align*} which is nonzero (making the map conformal) precisely when $$bd-ad\neq 0$$.

$$\supseteq$$: All LFTs are bijective, since if $$f$$ is an LFT then $$f = A_1 \circ I \circ A_2$$ with the $$A_i$$ affine and invertible, and $$I(z)$$ is invertible on $${\mathbf{CP}}^1$$, so $$f^{-1}= A_2^{-1}\circ I^{-1}\circ A_1^{-1}$$. Noting that the $$A_i$$ are conformal, if $$I$$ is conformal then by the chain rule $$f'\neq 0$$. That $$I$$ is conformal on $${\mathbf{CP}}^1$$: this clearly holds away from $$z=0$$ and $$z=\infty$$. Now use that $$I\qty{1\over w} = w$$ is conformal at zero, and $${1\over I(z)}$$ is conformal at $$\infty$$.

$$\subseteq$$: Suppose $$f\in \mathop{\mathrm{Aut}}({\mathbf{CP}}^1)$$. If $$f(\infty) = \infty$$, then the claim is that $$f$$ must be affine. Define $$F(z) \coloneqq{1\over f(1/z)}$$ satisfies $$F(0) = 0$$, and $$F$$ is conformal at zero since it is a composition of $$f$$ and $$I$$. So $$F(0) = 0$$ but $$F'(0) \neq 0$$, making zero a pole of $$F$$ of order one. So $$f$$ has a pole of order 1 at $$\infty$$, and by injectivity, $$f^{-1}(\infty) = \left\{{\infty}\right\}$$ has a single element. Now $$G(z) \coloneqq{f(z) - f(0) \over z}$$ is bounded on $${\mathbf{C}}$$ and thus constant, so $$f(z) - f(0) = cz \implies f(z) = cz + f(0)$$.

If $$f(\infty) = w < \infty$$, consider $$F(z) \coloneqq{1\over z-w}$$. By inspection, $$F$$ is an LFT, and $$G(z) \coloneqq{1\over f(z) - w} = (F\circ f)(z)$$ is an automorphisms of $${\mathbf{CP}}^1$$ satisfying $$G(\infty) = \infty$$. By the previous case, $$G$$ is affine, so \begin{align*} G(z) = az + b \implies {1\over f(z) - w} = az + b \implies f(z) = {1\over az + b } + w = {waz + wb\over az+b } ,\end{align*} which is visibly an LFT.

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