Let $f:\mathbb{H}\rightarrow\mathbb{H}$ be any analytic automorphism of the upper half plane $\mathbb{H}=\{z\in \mathbb{C}:\text{Im}z>0\}$. Then can it be written by the form
$$f(z)=\frac{az+b}{cz+d}$$
for which $a, b, c, d\in \mathbb{R}$ with $ad-bc=1$?
Motivation of this problem. I am reading Seere's A Course in Arithmetric. In the page 77, I think that what he is saying "…it is the group of all analytic automorphism of H.", to summarize, is
$$\text{Aut}(\mathbb{H})=\text{PSL}_2(\mathbb{R}).$$
It is clear that $\text{Aut}(\mathbb{H})\supset \text{PSL}_2(\mathbb{R}),$ but I can not understand the reason why the converse inclusion is also true.
My Attempt. I have already known the following statement:
Theorem. The only analytic automorphisms $f:\mathbb{C}\rightarrow \mathbb{C}$ are the functions of the form
$$f(z)=az+b,$$
where $a,b\in \mathbb{R}$ with $a\neq 0$.
I expect any analytic automorphism of the upper half plane to be a meromorphic function of some analytic automorphisms of the complex plane. Is it correct? And how do I show the relation $ad-bc=1$?
Best Answer
Compose $f\in Aut(\Bbb{H})$ with some Möbius transformations to get $g = \phi \circ \rho \circ f \circ \psi \in Aut(\Bbb{D})$ such that $g(0)=0$.
Both $g(z)/z,z/g(z)$ are analytic on $\Bbb{D}$. By the maximum modulus principle $|g(z)/z|\le 1$ and $|z/g(z)|\le 1+\epsilon$.
It implies that $|g(z)/z|$ is constant $= 1$, so $g(z)/z$ is constant and $g(z)=e^{i\phi} z$. From this you got that $f$ was itself a Möbius transformation, in $PSL_2(\Bbb{R})$.