Automorphisms on upper half plane

Question

I am supposed to prove that: \begin{align} Aut(\mathbb{H})= \left\{ T:\mathbb{H}\to \mathbb{H}, T(z)=\frac{az+b}{cz+d} \vert a,b,c,d\in\mathbb{R}; ad-bc>0 \right\} \end{align}. where $\mathbb{H}$ denotes the upper half plane of the complex plane. I was able to show "$\supseteq$", but a have a hard time for the other direction. My idea was to look at a function $g:= C\circ f\circ C^{-1}$ where $C$ is the Cayley function and $f \in Aut(\mathbb{D})$ and $\mathbb{D}$ as the unit disc and then use maybe schwarzes Lemma. the problem is, I didn't even get $g(0)=0$, so I couldn't apply schwarz Lemma and even then I'm not sure if this would even work. Any suggestions?

Answer 1

Ok, here is my try; let me preface this by saying there there are probably many standard proofs of this fact in textbooks, and I am not sure that what I write below is 100% correct, so if anyone wants to improve/correct something, feel free to do so!

Anyway, let $f:\mathbb{H}\to\mathbb{H}$ be a biholomorphism, and denote by $\zeta$ the complex number $f(i)$. For the moment, assume there is a matrix $\begin{pmatrix}a & b\\ c &d\end{pmatrix}\in\mathrm{GL}_2(\mathbb{R})$ with positive determinant, such that the induced Möbius transformation $$\varphi:z\mapsto\frac{az+b}{cz+d}$$ sends $\zeta$ to $i$, so that $\varphi\circ f$ is a biholomorphism of $\mathbb{H}$ such that $\varphi\circ f(i)=i$. With this in place, we can use your idea: consider the Cayley transform $C:\mathbb{CP}^1\to\mathbb{CP}^1$, defined on $\mathbb{C}=\mathbb{CP}^1\setminus\{\infty\}$ by $$C(z)=\frac{z-i}{z+i},$$ and consider the composition $\psi=C\circ\varphi\circ f\circ C^{-1}$. This is a biholomorphism of the disk, since $C$ maps the upper-half plane to the disk, and as $\varphi\circ f(i)=i$, we deduce that $\psi(0)=0$. Also, notice that $\lvert\psi'(0)\rvert\not=0$, as the maps $C$, $\varphi$, and $f$ are all holomorphic and injective. By the Schwarz Lemma (or the Riemann mapping Theorem) then we deduce that there is some $\alpha\in\mathbb{C}$, $\alpha\not=0$, such that $\psi(z)=\alpha z$. Notice also that $\lvert\alpha\rvert=1$, as $\psi$ preserves $\partial\mathbb{D}$.

Then, for the original biholomorphism $f$ we get $$f(z)=\varphi^{-1}\left(C^{-1}(\alpha\,C(z))\right).$$ Now, we just need to show that $C^{-1}(\alpha\,C(z))$ is a Möbius transform, as $\varphi$ is a Möbius transformation and the Möbius transformations form a group (exercise!)

So, let's compute: \begin{equation} C^{-1}(\alpha\,C(z))=\frac{\alpha\,C(z)+1}{i \alpha\,C(z)-i}=\frac{1}{i}\frac{\alpha\frac{z-i}{z+i}+1}{\alpha\frac{z-i}{z+i}-1}=\frac{1}{i}\frac{\alpha(z-i)+z+i}{\alpha(z-i)-(z+i)}=\frac{z(\alpha+1)+i(1-\alpha)}{z\,i(\alpha-1)+(1+\alpha)}. \end{equation} This is not quite a Möbius transform, as the coefficients are not real. However, we can rewrite it (if $\alpha\not=-1$, but in that case we already have a Möbius transform) as $$\frac{z(\alpha+1)+i(1-\alpha)}{z\,i(\alpha-1)+(1+\alpha)}=\frac{z\lvert 1+\alpha\rvert^2+i(1-\alpha)(\bar{\alpha}+1)}{z\,i(\alpha-1)(\bar{\alpha}+1)+\lvert 1+\alpha\rvert^2},$$ that is, since $\alpha$ has norm $1$, \begin{equation} C^{-1}(\alpha\,C(z))=\frac{z\lvert 1+\alpha\rvert^2+2\,\Im(\alpha)}{-2\,\Im(\alpha)\,z+\lvert 1+\alpha\rvert^2} \end{equation} is the Möbius transformation associated to the matrix $$\begin{pmatrix}\lvert 1+\alpha\rvert^2 & 2\Im(\alpha) \\ -2\Im(\alpha) & \lvert 1+\alpha\rvert^2\end{pmatrix}.$$

So, it remains to prove the claim laid out at the beginning, that for each $\zeta$ there is a Möbius transformation sending $\zeta$ to $i$. This can be obtained as the composition of two different transformations: first find a matrix whose associated map $\varphi_0$ translates $\zeta$ to the imaginary axis $i\mathbb{R}_{>0}$, then compose it with a second Möbius transformation $\varphi_1$ that rescales the imaginary axis. I think it would be a good exercise for you to try to find these two transformations, let me know if you need help for that!