15 Geometry of the Unit Disk Importance of ∆(1) It is known that any one complex dimensional simply connected (i.e., there is no “holes”) domains (or complex manifold) are, up to a one-to-one-on-to analytic map: the complex plane C, the Riemann Sphere Cˆ and the unit disk ∆(1). It is important to understand geometry on these three spaces. There is a geometric fact: for any topological space X, there is a unique covering map π : Y → X 50 where Y is simply connected (i.e., there is no “holes”) topological space such that the quotient map Y/G → X is homeomorphic, where G is a certain subgroup of Aut(Y ). Similar statements are true in differential geometry and complex geometry. In complex geometry, for any domain X in C (or any 1 dimensional manifold), there is a unique covering map π : Y → X where Y is simply connected complex manifolds (i.e., either Y = C, or Y = Cˆ, or Y = ∆(1)) such that the quotient map Y/G → X is 1-to-1 and onto analytic function where G is a certain subgroup of Aut(Y ). We has showed that any analytic map f : Cˆ → C must be constant. Also by Liouville theorem, 51 any analytic map f : C → X, where X is a bounded domain in C, must be constant. Therefore, for majority of X, the covering space Y is the unit disk ∆(1). In order to study domains, we need to study Aut(∆(1)). Automorphism Group Aut(∆(1)) Let us claim a result without proof here: any 1-to-1 and onto analytic map f : ∆(1) → ∆(1) must be a linear transformation which sends the boundary ∂∆(1) onto it self. By assuming this result, we have the following. Theorem 15.1 (i) A linear transformation f ∈ Aut(∆(1)) is of the form z − a f(z)= eiθ , ∀z ∈ ∆(1) 1 − az for some a ∈ ∆(1) and θ ∈ R. (ii) ∀a, b ∈ ∆(1), there is an f ∈ Aut(∆(1)) such that f(a)= b. 50 By “covering map”, it means that for any point x ∈ X, there is neighborhood Ux of x in X such that −1 π (Ux) is the union of disjoint open sets ∪j Vx,j and that the restriction π|Vx,j is homeomorphic. 51If f : C → C is an analytic function and is bounded, then f is constant. 98 Proof: (i) Step 1. For any given point a ∈ ∆(1), we claim: ∃ φa ∈ Aut(∆(1)) such that φa(a)=0. (60) 1 Notice that a and a are mutual reflection with respect to the unit circle ∂∆(1), and that 0 and ∞ are also mutual reflection with respect to the unit circle ∂∆(1). If such Φa exists, by Theorem 14.1, we must have 1 φ (a)=0, φ = ∞. (61) a a a Then it is easy to find a linear transformation satisfying (61): z − a φ (z)= . a 1 − az Check: The map φa sends the unit circle to itself, i.e., to show: |φa(z)| =1, ∀|z| = 1. z − a i.e., to show : =1, ∀|z| =1, 1 − az 1 Since z = z , the above is equal to z − a =1, ∀|z| =1. 1 − a 1 z i.e., to show: z − a =1, ∀|z| =1 a − z which is trivially true. Claim (60) is proved. Step 2. Next we claim: If g ∈ Aug(∆(1)) such that g(0) = 0, then g(z)= eiθz forsomeθ ∈ R. (62) In fact, since g(0) = 0, by Theorem 14.1, =⇒ g(∞)= ∞ because 0 and ∞ are mutual reflection with respect to ∂∆(1). az+b a a =⇒ Write g(z)= cz+d to imply g(z)= d z = αz where α = d . =⇒ By g(∂∆(1)) ⊂ ∂∆(1), it implies |αz| =1, ∀|z| = 1. 99 i.e., |α| = 1, so that α = eiθ =⇒ Claim (62) is proved. Step 3. Now we consider general case: for any f ∈ Aut(∆(1)). Let a := f −1(0). =⇒ By Step 1, ∃φa ∈ Aut(∆(1)) such that φa(a) = 0. −1 =⇒ φa (0) = a. −1 −1 =⇒ f ◦ φa ∈ Aut(∆(1)) such that f ◦ φa (0) = 0. −1 iθ R =⇒ By Step 2, f ◦ φa (w)= e w for some θ ∈ , ∀w. iθ iθ z−a =⇒ f(z)= e φa(z)= e 1−az . (ii) ∀a, b ∈ ∆(1), by (i), ∃φa,φb ∈ Aut(∆(1)) such that φa(a) = 0 and φb(b) = 0. −1 =⇒ f = φb ◦ φa is the desired one. Relationship between the Upper Half Plane H and the unit Disk ∆(1) H := {z ∈ C | Im(z) > 0} is the upper half plane. Its boundary ∂H is the real line {z ∈ C | Im(z)=0}. Since a line or a circle in C corresponds a circle in Cˆ, the line line ∂H is a circle in Cˆ so that H is a disk in the Riemann Sphere. Now both the unit disk ∆(1) and H are disks in the Riemann Sphere, we expect to identify H and ∆(1) by a linear transformation. Claim: There is a linear transformation Ψ such that Ψ(H) = ∆(1). (63) To do that, we need to find a linear transformation Ψ sending the boundary onto the boundary: Ψ(∂H)= ∂∆(1). (64) Taking two mutual reflection points i and −i with respect to the line ∂H = R and 0, ∞ with respect to the circle ∂∆(1), by Theorem 12.2, there is a unique linear transformation Ψ such that Ψ(i)=0, Ψ(−i)= ∞. (65) Take z − i Ψ(z)= z + i which clearly satisfies (65). To verify (64), we need to show: |Ψ(z)| =1, ∀z ∈ R, i.e., to show: |Ψ(z)|2 =1, ∀z ∈ R, 100 i.e., to show: z − i z + i · =1, ∀z ∈ R, z + i z − i i.e., to show: |z|2 − iz + iz +1|z|2 + iz − iz +1, ∀z ∈ R, i.e., iz = iz, ∀z ∈ R, i.e., z = z, ∀z ∈ R, R z−z which is true because a complex number z ∈ if and only if Im(z)= 2i = 0. Finally, since Ψ(i) = 0, an interior point i of H has been mapped to an interior point 0 of ∆(1), Claim (63) is proved. Let Aut(H) := {all linear transformations f such that f(H)= H}. Proposition 15.2 Aut(H) ∼= Aut(∆(1)). Proof: Let Ψ : H → ∆(1) as constructed above. Consider the diagram f H −→ H Ψ ⇓ ⇓ Ψ g ∆(1) −→ ∆(1) We define a map Aut(H) → Aut(∆(1)), f 7→ g := Ψ ◦ f ◦ Ψ−1, which is one-to-one and onto, identifying these two groups. H az+b R Proposition 15.3 f ∈ Aut( ) ⇐⇒ f(z)= cz+d with ad − bc > 0 and a,b,c,d ∈ . Proof: (⇐=) Such f satisfies f(R) ⊂ R, f maps the boundary of H onto itself. To prove such f maps the interior points of H into interior points, we notice 1 az + b az + b (ad − bc)(z − z) Im(z) Im(f(z)) = − = =(ad − bc) , 2icz + d cz + d 2i|cz + d|2 |cz + d|2 where (ad − bc) > 0 and a, b, c and d are real, so that Im(f(z)) > 0 whenever Im(z) > 0. This proves f(H)= H. az+b (=⇒) Let f(z)= cz+d with ad − bc =6 0 be a linear transformation. 101 Suppose f(∞)= ∞ and f(0) = 0. Then it implies f(z)= az. (66) Since f(1) ∈ R, a ∈ R. We are done. For a general f, since f(0), f(1), f(∞) ∈ R ∪ {∞}, we find a linear treansformation αz+β R Ψ(z)= γz+κ where α,β,γ,κ ∈ such that Ψ(f(0)) = 0, Ψ(f(∞)) = ∞. =⇒ f ◦ Ψ satisfies f ◦ Ψ(0) = 0 and f ◦ Ψ(∞)= ∞. =⇒ By (66), f ◦ Ψ(z)= az where a ∈ R. −1 az+b =⇒ f = aΨ so that f(z)= cz+d where a,b,c,d are all real. =⇒ Since Im(f) > 0 whenever Im(z) > 0, together with the identity below: 1 az + b az + b (ad − bc)Im(z) Im(f(z)) = − = , 2icz + d cz + d |cz + b|2 we conclude ad − bx > 0. By above, we can identify with the disk ∆(1) with the upper half plane H. 102.