Def: Let be regions. We say that and are biholomorphic or biholomorphically equivalent, if there's a bijective analytic function . We write it as . If we know the biholomorphism , then we can write it as
Obs: The is an equivalence relation.
, for the rest of the note.
We can see that since if there was a a function , analytic then would be constant.
We see that if is analytic and injective then , then is surjective. Meaning that if then .
The goal from Riemann was to characterize all the such that
Def: Let be a region. We say that satisfies the Riemann condition ( as a shorthand), if for every such that then , that
We see that , and we want to find all the biholomorphisms. We can use Schwarz Lemma to find them all.
Let , and we define
is a Möbius transformation. with , and . Since we can see that has a pole at , and , we see that is analytic on which contains . Thus is a biholomorphism.
Additionally,
We need to look at special properties of . $$\varphi_a'(z) = \frac{(1-\overline az) +(z-a)\overline a}{(1-\overline a z)}$$
and we see that$$\varphi_a'(a) = \frac{1}{1-|a|²}, \qquad\quad \varphi_a'(0) =1-|a|²$$ Lemma: Let be analytic, and , we call , then $$|f'(\alpha)| \le \frac{1-|\alpha|²}{1-|a|^2}$$Furthermore, if , then there's such that $$f(z) = \varphi_{-a}(c\varphi_a(z))$$in particular, is injective.
Th:, and , , then there's such that
Riemann Theorem
Let be a region. If satisfies the Riemann condition, for every , there exists a unique biholomorphism , such and .
The proof has 3 main parts:
Existence of biholomorphism(the hardest), requires the use of Normal families
Given that biholomorphism construct one that satisfies the additional constraints