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.