We know the automorphism group of the upper-half plane $\mathbb{H} \subset \mathbb{C}$ is given by $\Big\{ \frac{az+b}{cz+d} \quad \big|\quad a,b,c,d \in \mathbb{R} \quad \text{and } ad-bc>0 \Big\}$. This is usually proved in textbooks by mapping this set to $\text{Aut}(\mathbb{D})$ and then using our explicit knowledge of $\text{Aut}(\mathbb{D})$.
I was trying to prove this result using the fact that $\text{Aut}(\mathbb{C}_{\infty})=\Big\{ \frac{az+b}{cz+d} \quad \big| \quad a,b,c,d \in \mathbb{C} \quad \text{and } ad-bc \neq 0 \Big \}$, where $\mathbb{C}_{\infty}$ is the Riemann sphere. The idea is:
It is clear that $\Big\{ \frac{az+b}{cz+d} \quad \big| \quad a,b,c,d \in \mathbb{R} \quad \text{and } ad-bc>0 \Big\}$ are in $\text{Aut}(\mathbb{H})$.
For the other way, let $\varphi \in \text{Aut}({\mathbb{H}})$. Suppose we can extend $\varphi$ to an automorphism of $\mathbb{C}_{\infty}$ and call this extension $\Phi$. Now,
\begin{equation}
\Phi(z)=\frac{az+b}{cz+d} \quad a,b,c,d \in \mathbb{C} \quad \text{and } ad-bc \neq 0
\end{equation}
As $\Phi \big| _{\mathbb{H}}=\varphi \in \text{Aut}(\mathbb{H})$, we must have $\Phi(\mathbb{R}) \subseteq \mathbb{R}$. So $a,b,c \text{ and }d$ are in fact real. Now
\begin{equation}
\text{Im}(\Phi(z))= \text{Im}\left(\frac{az+b}{cz+d}\right)=\text{Im}\left(\frac{(az+b)(c\bar{z}+d)}{|cz+d|^2}\right)=\text{Im}\left(\frac{adz+bc\bar{z}}{|cz+d|^2}\right)=\frac{(ad-bc)y}{|cz+d|^2}
\end{equation}
where $y=\text{Im}(z)$. We need $\text{Im}(\Phi(z))>0$ whenever $\text{Im}(z)>0$. Hence, $ad-bc>0$.
But it is not at all clear to me how to extend $\varphi$ to an automorphism of $\mathbb{C}_{\infty}$. I know it can be continuously extended to $\mathbb{R}$ and then to the whole of $\mathbb{C}$ by a reflection but this is not enough. Any help would be appreciated.
Best Answer
Knowing that there exist functions analytic on the upper half plane that cannot be made continuous at any real value (j-invariants), leads me to agree the most important part of your proof is
I agree that is an absolutely non-obvious result. It would probably require demonstrating that analytic functions have particular restrictions and requirements, like the maximum modulus principle. From there we would probably need to demonstrate that under certain restrictions, like one-to-one and onto, that analytic functions must either be a constant/linear/polynomial/rational, etc. Theorems like Liouville's, Schwarz inequality, uniqueness of Blaschke products, etc. Eventually we arrive at the far reaching result that any analytic function that maps $\mathbb{D}$ to itself that is one-to-one and onto, must be special class of Mobius transformations.
To somehow bypass "using the $Aut(\mathbb{D})$" result would probably require re-proving all the lemmas that came before it for the upper half plane instead. For example, the "Schwarz inequality for half planes".
What I am trying to summarize is that because j-invariants exist, your proof essentially demonstrates that NO j-invariant can be one-to-one and onto $\mathbb{H}$; which is absolutely true, but I don't see where that is carefully demonstrated in your statements.
Now if you assume, a priori, that any function in $Aut(\mathbb{H})$ is a mobius transformation, then your arguments follow just fine. But if by $Aut(\mathbb{H})$, you simply mean the collection of analytic functions that map $\mathbb{H}$ to itself, then I don't think the composition with $Aut(\mathbb{D})$ can be discarded so easily.