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).
Let𝑈 ⊊ ℂbe a simply connected open subset which is notℂ.
Then there exists a biholomorphism𝑓 ∶ 𝑈 → 𝐷1(0)(i.e.𝑓 is holomorphic, bijective and𝑓−1is holomorphic).
We say that𝑈and𝐷1(0)areconformally equivalent.
Remark 2. Note that if𝑓 ∶ 𝑈 → 𝑉 is bijective and holomorphic then𝑓−1is holomorphic too.
Indeed, we proved that if𝑓 is injective and holomorphic then𝑓′never vanishes (Nov 30).
Then we can conclude using the inverse function theorem.
Note that this remark is false forℝ-differentiability: define 𝑓 ∶ ℝ → ℝby𝑓 (𝑥) = 𝑥3 then𝑓′(0) = 0and 𝑓−1(𝑥) =√𝑥3 is not differentiable at0.
Remark 3. The theorem is false if𝑈 = ℂ.
Indeed, by Liouville’s theorem, if𝑓 ∶ ℂ → 𝐷1(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 notℂ.
Otherwise stated, up to a biholomorphic transformation, there are only two open simply connected sets:
𝐷1(0)andℂ. Formally:
Corollary 4. Let𝑈 , 𝑉 ⊊ ℂbe two simply connected open subsets, none of which isℂ.
Then there exists a biholomorphism𝑓 ∶ 𝑈 → 𝑉 (i.e.𝑓 is holomorphic, bijective and𝑓−1is holomorphic).
Proof. By the Riemann mapping theorem, there exists biholomorphisms𝜑 ∶ 𝑈 → 𝐷1(0)and𝜓 ∶ 𝑉 → 𝐷1(0).
Then we can simply take𝑓 = 𝜓−1∘ 𝜑.
𝜑 𝜓
𝑓
𝑈 𝑉
𝐷1(0) Corollary 5. Let𝑈 ⊂ ℂbe an open subset. ■
Then𝑈is simply connected if and only if it is homeomorphic to𝐷1(0).
Proof.
⇒Assume that𝑈 ⊊ ℂis simply connected then there exists a biholomorphism𝑓 ∶ 𝑈 → 𝐷1(0). Particularly 𝑓 is a homeomorphism.
Note thatℂis also homeomorphic to𝐷1(0).
⇐Assume that there exists a homeomorphism𝑓 ∶ 𝑉 → 𝑈 where𝑉 = 𝐷1(0).
Since𝑉 is simply connected, we get that𝑈 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 byℍ = {𝑧 ∈ ℂ ∶ ℑ(𝑧) > 0}.
Then the mapping𝜑 ∶ ℍ → 𝐷1(0)defined by𝜑(𝑧) = 𝑧 − 𝑖
𝑧 + 𝑖is biholomorphic.
First check that𝜑is well-defined:∀𝑧 ∈ ℍ, 𝑧 ≠ −𝑖and𝜑(𝑧) ∈ 𝐷1(0).
Then note that𝜑is the restriction of a Möbius transformation𝜑 ∶ ̂̂ ℂ → ̂ℂ.
It is not too difficult to check that𝜑(ℝ ∪ {∞}) = 𝑆̂ 1(≔ {𝑧 ∈ ℂ ∶ |𝑧| = 1}).
The complement ofℝ ∪ {∞}inℂ̂has two connected components which areℍand−ℍ.
Andℂ ⧵ 𝑆̂ 1has two connected components: 𝐷1(0)and{𝑧 ∈ ℂ ∶ |𝑧| > 1} ∪ {∞}.
Since𝜑(𝑖) = 0 ∈ 𝐷1(0), we deduce that𝜑(ℍ) = 𝐷1(0).
ℜ ℑ
𝑖
Note that𝜑maps right angles to right angles!
𝜑
0
Example 8(A horizontal band).
We setℬ ≔ {𝑧 ∈ ℂ ∶ 0 < ℑ(𝑧) < 𝑎} , 𝑎 > 0.
We know that𝜓 ∶ ℬ → ℍdefined by𝜓(𝑧) = 𝑒𝜋𝑎𝑧is biholomorphic.
Hence𝑓 = 𝜑 ∘ 𝜓 ∶ ℬ → 𝐷1(0)is biholomorphic, where𝜑was defined in the previous example, i.e.
𝑓 (𝑧) = 𝑒𝜋𝑎𝑧− 𝑖 𝑒𝜋𝑎𝑧+ 𝑖
ℜ ℑ
ℑ(𝑧) = 𝑎 𝑓
0
1
Example 9. In practice the biholomorphism𝜑between𝑈and𝐷1(0)may be difficult to express explicitely.
For instance, the following set is simply connected 𝑈 = ((0, 1) × (0, 1)) ⧵
(⋃𝑛≥2{1
𝑛 }× (0,1 2 ))
but the behavior of𝜑around the boundary of𝐷1(0)is going to be quite complicated!
Or, even worse, take𝑈to be the interior of the Koch snowflake.
