University of Toronto – MAT334H1-F – LEC0101 Complex Variables 18 – The Riemann mapping theorem Jean-Baptiste Campesato December 2nd, 2020 and December 4th, 2020 Theorem 1 (The Riemann mapping theorem). Let 푈 ⊊ ℂ be a simply connected open subset which is not ℂ. −1 Then there exists a biholomorphism 푓 ∶ 푈 → 퐷1(0) (i.e. 푓 is holomorphic, bijective and 푓 is holomorphic). We say that 푈 and 퐷1(0) are conformally equivalent. Remark 2. Note that if 푓 ∶ 푈 → 푉 is bijective and holomorphic then 푓 −1 is holomorphic too. Indeed, we proved that if 푓 is injective and holomorphic then 푓 ′ never vanishes (Nov 30). Then we can conclude using the inverse function theorem. Note that this remark is false for ℝ-differentiability: define 푓 ∶ ℝ → ℝ by 푓(푥) = 푥3 then 푓 ′(0) = 0 and 푓 −1(푥) = √3 푥 is not differentiable at 0. Remark 3. The theorem is false if 푈 = ℂ. Indeed, by Liouville’s theorem, if 푓 ∶ ℂ → 퐷1(0) is holomorphic then it is constant (as a bounded entire function), so it can’t be bijective. The Riemann mapping theorem states that up to biholomorphic transformations, the unit disk is a model for open simply connected sets which are not ℂ. Otherwise stated, up to a biholomorphic transformation, there are only two open simply connected sets: 퐷1(0) and ℂ. Formally: Corollary 4. Let 푈, 푉 ⊊ ℂ be two simply connected open subsets, none of which is ℂ. Then there exists a biholomorphism 푓 ∶ 푈 → 푉 (i.e. 푓 is holomorphic, bijective and 푓 −1 is holomorphic). Proof. By the Riemann mapping theorem, there exists biholomorphisms 휑 ∶ 푈 → 퐷1(0) and 휓 ∶ 푉 → 퐷1(0). Then we can simply take 푓 = 휓−1 ∘ 휑. 푓 푈 휑 휓 푉 퐷1(0) ■ Corollary 5. Let 푈 ⊂ ℂ be an open subset. Then 푈 is simply connected if and only if it is homeomorphic to 퐷1(0). Proof. ⇒ Assume that 푈 ⊊ ℂ is simply connected then there exists a biholomorphism 푓 ∶ 푈 → 퐷1(0). Particularly 푓 is a homeomorphism. Note that ℂ is also homeomorphic to 퐷1(0). ⇐ Assume that there exists a homeomorphism 푓 ∶ 푉 → 푈 where 푉 = 퐷1(0). Since 푉 is simply connected, we get that 푈 is too since simple connectedness is preserved by homeomor- phisms. ■ Remark 6. Careful: the continuous image of a simply connected set may not be simply connected. For instance exp(ℂ) = ℂ ⧵ {0}. 2 The Riemann mapping theorem Example 7 (The Poincaré half-plane). We define the Poincaré half-plane by ℍ = {푧 ∈ ℂ ∶ ℑ(푧) > 0}. 푧 − 푖 Then the mapping 휑 ∶ ℍ → 퐷 (0) defined by 휑(푧) = is biholomorphic. 1 푧 + 푖 First check that 휑 is well-defined: ∀푧 ∈ ℍ, 푧 ≠ −푖 and 휑(푧) ∈ 퐷1(0). Then note that 휑 is the restriction of a Möbius transformation ̂휑∶ ℂ̂ → ℂ̂. It is not too difficult to check that ̂휑(ℝ∪ {∞}) = 푆1(≔ {푧 ∈ ℂ ∶ |푧| = 1}). The complement of ℝ ∪ {∞} in ℂ̂ has two connected components which are ℍ and −ℍ. ̂ 1 And ℂ ⧵ 푆 has two connected components: 퐷1(0) and {푧 ∈ ℂ ∶ |푧| > 1} ∪ {∞}. Since 휑(푖) = 0 ∈ 퐷1(0), we deduce that 휑(ℍ) = 퐷1(0). ℑ 휑 푖 0 ℜ Note that 휑 maps right angles to right angles! Example 8 (A horizontal band). We set ℬ ≔ {푧 ∈ ℂ ∶ 0 < ℑ(푧) < 푎} , 푎 > 0. 휋 푧 We know that 휓 ∶ ℬ → ℍ defined by 휓(푧) = 푒 푎 is biholomorphic. Hence 푓 = 휑 ∘ 휓 ∶ ℬ → 퐷1(0) is biholomorphic, where 휑 was defined in the previous example, i.e. 휋 푧 푒 푎 − 푖 푓(푧) = 휋 푧 푒 푎 + 푖 ℑ ℑ(푧) = 푎 푓 1 0 ℜ Example 9. In practice the biholomorphism 휑 between 푈 and 퐷1(0) may be difficult to express explicitely. For instance, the following set is simply connected 1 1 푈 = (0, 1) × (0, 1) ⧵ × 0, ( ) ⋃ { } ( ) (푛≥2 푛 2 ) but the behavior of 휑 around the boundary of 퐷1(0) is going to be quite complicated! Or, even worse, take 푈 to be the interior of the Koch snowflake..