December2 ,2020andDecember4 ,2020 THERIEMANNMAPPINGTHEOREM ComplexVariables

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MAT334H1-F – LEC0101

Complex Variables

THE RIEMANN MAPPING THEOREM

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The Riemann mapping 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 that𝑈 and𝐷₁(0)areconformally equivalent.

Remark

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 result is false forℝ-differentiability:

Define𝑓 ∶ ℝ → ℝby𝑓 (𝑥) = 𝑥³then𝑓^′(0) = 0and𝑓⁻¹(𝑥) =√𝑥³ is not differentiable at0.

Remark

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.

Jean-Baptiste Campesato MAT334H1-F – LEC0101 – Dec 2, 2020 and Dec 4, 2020 2 / 7

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The Riemann mapping theorem – 1

The Riemann mapping theorem

Remark

Note that if𝑓 ∶ 𝑈 → 𝑉 is bijective and holomorphic then𝑓⁻¹ is 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.

Remark

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The Riemann mapping theorem – 1

The Riemann mapping theorem

Remark

Note that if𝑓 ∶ 𝑈 → 𝑉 is bijective and holomorphic then𝑓⁻¹ is 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.

Remark

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The Riemann mapping theorem – 2

This 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

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)and 𝜓 ∶ 𝑉 → 𝐷₁(0). Then we can simply take𝑓 = 𝜓⁻¹∘ 𝜑.

𝜑 𝜓

𝑓

𝑈 𝑉

𝐷₁(0) ■

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The Riemann mapping theorem – 2

This 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

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)and 𝜓 ∶ 𝑉 → 𝐷₁(0). Then we can simply take𝑓 = 𝜓⁻¹∘ 𝜑.

𝜑 𝜓

𝑓

𝑈 𝑉

𝐷₁(0) ■

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The Riemann mapping theorem – 3

Corollary

Let𝑈 ⊂ ℂbe an open subset.

Then𝑈 is 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). Particularly𝑓 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

homeomorphisms. ■

Remark

Careful: the continuous image of a simply connected set may not be simply connected. For instanceexp(ℂ) = ℂ ⧵ {0}.

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The Riemann mapping theorem – 3

Corollary

Proof.

⇒Assume that𝑈 ⊊ ℂis simply connected then there exists a biholomorphism𝑓 ∶ 𝑈 → 𝐷₁(0).

Particularly𝑓 is a homeomorphism.

homeomorphisms. ■

Remark

Careful: the continuous image of a simply connected set may not be simply connected. For instanceexp(ℂ) = ℂ ⧵ {0}.

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The Riemann mapping theorem – 3

Corollary

Proof.

⇒Assume that𝑈 ⊊ ℂis simply connected then there exists a biholomorphism𝑓 ∶ 𝑈 → 𝐷₁(0).

Particularly𝑓 is a homeomorphism.

homeomorphisms. ■

Remark

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Example 1: the Poincaré half-plane

We define the Poincaré half-plane byℍ = {𝑧 ∈ ℂ ∶ ℑ(𝑧) > 0}.

The mapping𝜑 ∶ ℍ → 𝐷₁(0)defined by𝜑(𝑧) =𝑧 − 𝑖

𝑧 + 𝑖 is 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

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Example 2: a horizontal band

We set

ℬ ≔ {𝑧 ∈ ℂ ∶ 0 < ℑ(𝑧) < 𝑎} , 𝑎 > 0 We know that𝜓 ∶ ℬ → ℍdefined by𝜓(𝑧) = 𝑒^𝜋^𝑎^𝑧is biholomorphic.

Hence𝑓 = 𝜑 ∘ 𝜓 ∶ ℬ → 𝐷₁(0)is biholomorphic, where𝜑was defined in the previous slide, i.e.

𝑓 (𝑧) = 𝑒^𝜋^𝑎^𝑧− 𝑖 𝑒^𝜋^𝑎^𝑧+ 𝑖 ℑ

ℑ(𝑧) = 𝑎 𝑓

0 1

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Example 3

In practice the biholomorphism𝜑between𝑈 and𝐷₁(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!

Even worse, we can take𝑈to be the interior of the Koch snowflake.

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Example 3

In practice the biholomorphism𝜑between𝑈 and𝐷₁(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!