Biholomorphism

Subjects: Complex Analysis

Def: Let U,VC be regions. We say that U and V are biholomorphic or biholomorphically equivalent, if there's a bijective analytic function f:UV. We write it as UV. If we know the biholomorphism f, then we can write it as UfV

Obs: The is an equivalence relation.

D:=B1(0), for the rest of the note.

We can see that CD since if there was a a function f:CD, analytic then f would be constant.

We see that if f:CC is analytic and injective then f(z)=az+b, then f is surjective. Meaning that if UC then U=C.

The goal from Riemann was to characterize all the UC such that UD

Def: Let UC be a region. We say that U satisfies the Riemann condition (URC as a shorthand), if for every fH(U) such that Z(f)= then gH(U), that g²=f

We see that DD, and we want to find all the biholomorphisms. We can use Schwarz Lemma to find them all.

Let aD, and we define

φa(z)=za1az

is a Möbius transformation. with φa(a)=0, and φa[D]=D. Since we can see that φa has a pole at 1/a, and |1/a|>1, we see that φa is analytic on B|1/a|(0) which contains D. Thus φa:DD is a biholomorphism.

Additionally, φa1=φa

We need to look at special properties of φa. $$\varphi_a'(z) = \frac{(1-\overline az) +(z-a)\overline a}{(1-\overline a z)}$$
and we see that$$\varphi_a'(a) = \frac{1}{1-|a|²}, \qquad\quad \varphi_a'(0) =1-|a|²$$
Lemma: Let f:DD be analytic, and aD, we call α=f(a), then $$|f'(\alpha)| \le \frac{1-|\alpha|²}{1-|a|^2}$$Furthermore, if |f(α)|=1|α|²1|a|2, then there's cD such that $$f(z) = \varphi_{-a}(c\varphi_a(z))$$in particular, f is injective.

Th: DfD, and aD, f(a)=0, then there's cD such that f=cφa

Riemann Theorem

Let UC be a region. If U satisfies the Riemann condition, for every aU, there exists a unique biholomorphism f:UD, such f(a)=0 and f(a)>0.

The proof has 3 main parts:

  1. Existence of biholomorphism(the hardest), requires the use of Normal families
  2. Given that biholomorphism construct one that satisfies the additional constraints
  3. Uniqueness of the biholomorphism