Riemann Surfaces

Lectured by Prof Gabriel Paternain

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0 Brief Review of material from IB ComplexAnalysis

A smooth function f on an open subset U ∈ C is called a holomorphic functionor analytic if equivalently:

(i) f is complex differentiable. i.e. For each z0 ∈ U the derivative

f ′(z0) = lim f(z)−f(z0)z−z0 exists

(ii) f admits a power expansion, if we let D(a, r) = z : |z − a| < r thenf(z) =

∑∞n=0 cn(z − a)n valued for all z ∈ D(a, r)

(ii)⇒ We can write f(z) = g(z)(z − a)r locally with g(z) =∑∞r cn(z − a)n−r

analytic on D(a, r) and g(a) = cr 6= 0 This gives the ”principle of isolatedzeroes”. If f 6= 0 on a domain U ⊂ C the zeroes are isolated.Each point z has an open neighbourhood ∆ ⊂ U for which either f |∆ ≡ 0 orf is nowhere zero on ∆ \ z. This gives the identity pricniple. If two analyticfunctions f, g on a domain U ⊂ C agree on a non-discrete subset of U thenf ≡ g on U .

Definition. Domain=open and connected subset of C

Suppose f is analytic on a punctured disc D∗(a, r) = z ∈ C : 0 < |z−a| < rfor some r > 0. i.e. f has an isolated singularity at a.

There is a Laurent expansion of f at D ∗ (a, r)

f(z) =

∞∑−∞

cn(z − a)n

The singularity at a is:

• Removable if cn = 0 for all n < 0. i.e. f analytic at a.

• A Pole if ∃N > 0 : c−N 6= 0 but cn = 0 for all n < −N .N is the order of the pole.And we can write f(z) = (z−a)−Ng(z) with g analytic on D(a, r), g(a) =c−N 6= 0

• Essential if cn 6= 0 for infinitely many n < 0.

Removable Singularity theorem(RST): Singularity is removable⇔ f is boundedon some punctured neighbourhood of aCasaroti-Weierstrass Theorem(CWT): If f has an essential singularity at a, then∀w ∈ C∞ = C ∪ ∞ ∃ a sequence zn ∈ D∗(a, r) with zn → a and f(zn)→ wTogether these imply: f has a pole at a ⇔ f(z)→∞, z → a

Example. (i) 1ez−1 has a pole of order 1 at z = 0 while

zez−1 has a removable singularity.

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(ii) e1z has an essential singularity at z = 0.

Recall that a function f on an open set U ⊂ C is called meromorphic if it isanalytic apart from isolated singularities and they are poles.For example 1

e1z −1

is meromorphic in C∗ = C \ 0 but not meromorphic in C.

Poles are z = 12πin , n 6= 0, n ∈ Z

0.1 The Complex Logarithm

We will deal with ”multivalued functions” which are locally analytic but itcannot be defined globally.e.g. The complex logarithmFor a given z = reiθ 6= 0 possible values of log z are log r + i(2πn + θ) wheren ∈ Z (possible w ∈ C such that ew = zLet U1 = C \ R≥0. For each n ∈ Z, we can define log analytically on U bylog(reiθ) = log r + i(2πn+ θ) where 0 < θ < 2π. It is analytic because it is aninverse of the holomorphic map exp : x+iy : x ∈ R, 2πn < y < 2(n+1)π → Uand invoke the inverse function theorem.Alternatively, we can say h(z) =

∫ z−1

dww + (2n+ 1)πi and prove it is holomor-

phic with h′(z) = 1z

ln(z+τ)−ln(z)τ = 1

τ

∫ z+τz

dww =

∫ 1

0dtz+tτ = 1

z+t0τ→ 1

zClaim:h(z) is the required inverse for ez such that g(z) = ez

Proof. g(z) = eh(z)

z on U and g′(z) = 0 ⇒ g = constant g(−1) = 1 ⇒ eh(z) =z

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1 Analytic Continuation in the Plane

Definition. A function element is an ordered pair (f, U) where U is a domainand f is a function defined on it.

Definition. A function element in domain U is a pair (f,D) where D is asubdomain of U and f is holomorphic on D

Definition. Let (f,E), (g,E) be function elements in U .

(i) (g,E) is a direct analytic continuation of (f,D) written (f,D) ∼ (g,E) ifD ∩ E 6= φ and f = g on D ∩ E

(ii) (g,E) is an analytic continuation of (f,D) along a path γ : [0, 1] → U(Written (f,D) ≈γ (g,E)) if there are function elements(fi, Di)

ni=1 and o = t0 < t1 < ... < tn = 1 such that

γ([tj−1, tj ]) ∈ Dj j = 1, ..., n and(f,D) = (f0, D0) ∼ (f1, D1) ∼ ... ∼ (fn, Dn) = (g,E)(g,E) is an analytic continuation of (f,D) if ∃ such a path.Then (f,D) ≈ (g,E)

Remark. (i) If (f,D) ≈γ (g,E) and (f,D) ≈γ (h,E) for some γ ⇒ (g = h)We will prove this later when it is less messy and when we have moremachinery.

(ii) ∼ is not an equivalence relation. (Even if intersection is not empty.)However ≈ is

Definition. An equivalence class F under ≈ is called a complete analytic func-tion.

Example. We consider log from last time.U1 = C \ R≥0

U = CLet J = (α, β) = θ : R, α < θ < βDJ = z = reiθ 6= 0, θ ∈ JfJ(z) = log r + iθ

LJ = (fJ , DJ) is a function element(last time we saw DJ = U : J = (0, 2π)Consider the intervals A = (−pi2 , π2 )B = (π6 ,

7π6 )

C = ( 5π6 ,

11π6

and consider DA, DB , DC . We have a curve

γ : [0, 1]→ Ct 7→ e2πit

0 = t < 16 < 1

2 < 56 < 1 = t4 LA ∼ LB , LB ∼ LC But LA 6∼ LC Points in

DA ∩DC are of the form reiθ where θ ∈ (−π2 ,−π6 ) giving fA(z) = log z + iθ

And it is also of the form rei(θ+2π) with θ + 2π ∈ ( 3π2 ,

11π6 ) ⊂ C

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So fC(z) = log r + i(θ + 2π) = fA(z) + 2πiThis analytic continuation of LC on DA produces (DA, fA + 2πi) = LA+2π soLA+2πγγLA, so we can go around γ n times in both directions to produce allpossible values of logThis gives a F for example, the F determined by the equivalence class of(fA, DA)

1.1 The Riemann surface of log

Consider U = C \ R+fn(z) = log r + i(2πi+ θ) where n ∈ Z, θ ∈ (0, 2π)This glueing construction gives the ”Riemann Surface” R of log and is now aunivalued function on R.fn(reiθ) = log r + i(2πn+ θ) 0 < θ < 2π, n ∈ Z

More precisely let

R =∐k∈Z

(C∗ × k)

(the disjoint union)Topology: Open sets are unions of sets of the form

• D((η, k), r) where D((η, k), r) = (z, k) : |z − η| < rif η ∈ C× \ R≥0(r < dist(η,R≥0)

• A((η, b), r) := (z, b), |z − η| < r, Im(z) ≥ 0∐(z, b − 1), |z − η| <

r, Im(z) < 0if η ∈ R≥0 (r < |η|)

Check this makes R into a Hausdorff connected(also path-connected topologicalspace with π : R→ C∗, (z, b) 7→ z continuous.Obviously, π|D((η,b),r) : D((η, b), r)→ D(η, r)is a homeomorphism and the same holds forπ|A((η,b),r) : A((η, b), r)→ D(η, r)We capture this in the following definition.

Definition. A covering map of a topological space is a continuous map π :X → X where X,X Hausdorff path-connectedd topological spaces and π is alocal homeomorphism.i.e. For each x ∈ X then there is an open neighbourhood N of x such thatπ|N : N → N is a homeomorphism where N is an open neighbourhood ofx = π(x)X is called the base space.X is called the covering space.

Warning: Terminology, don’t confuse with the covering maps in AlgebraicTopology. This notion is weaker.The map π : R → C∗ from the example above has a property that for eachz ∈ C∗, there is a disc D ⊂ C∗ such that z ∈ D

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π−1(D) =∐i∈ZDi where π|Di : Di → D is a homeomorphism. These are called

regular and match the ones in Algebraic Topology.We will come back to this. Define f : R→ C f(eiθ, k) = log r+ i(2πk+θ) where0 ≤ θ < 2π

R C

C∗

f

πexp

To be upgraded to a holomorphic diagram Additional Example: We could dosomething quite similar with z

1n = exp( 1

n log z) on C∗, n ≥ 0(integers)

Possible branches are r1n e

iθn e

2iπkn where k = 0, 1, ..., n − 1 This time we only

need to glue n copies of C \ R≥0 and again we obtain a diagram

R C

C∗

f

πzn

1.2 Power Series

Recall that a power series about a ∈ C is uniformly and absolutely convergenton any closed disc inside its radius of convergence.If r <∞, what can we say about analytic continuations (f,D) where

f =

∞∑k=0

ak(z − a)k

where D = D(a, r)Without loss of generality, we will assume that a = 0, r = 1 so we work on∆ = D(0, 1) and Π := ∂D(0, 1) So we consider

f =

∞∑k=0

akzk

Definition. A point z ∈ Π is said to be regular if ∃g analytic on an openneighbourhood N of Z such that f = g on N ∩∆

Defining (f) =

f on D

g on N

we have a direct analytic continuation to ∆ ∪N If z is not regular, it is calledsingular.

Note: Regular points form open sets. Singular points form a closed set.

Example/Warning. 1. z regular 6⇒∑akz

k converges.e.g. f(z) =

∑∞0 zk = 1

1−z z ∈ ∆, Every point z ∈ Π \ 1 is regular but∑zk diverges for all z ∈ Π

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2.∑akz

k converges 6⇒ z regularf(z) =

∑∞2

zk(k−1)

Absolutely convergent if all z ∈ ∆.1 is a singular point.If f extends analytically to a neighbourhood of 1 then f ′′ =

∑∞0 zk = 1

1−zextends as well. Absurd

Theorem 1.1. Suppose f(z) =∑∞

0 akzk has radius of convergence 1 then

exists at least one singular point.

Proof. ∃ an open disc Dz such that f extends to a function fz on ∆ ∪Dz

If Dz1 ∩Dz2 6= φBut fz1 = f = fz2 on the open subset Dz1 ∩Dz2 ∩∆ 6= φ of Dz1 ∩Dz2 .By the identity principle fz1 = fz2 on Dz1 ∩ Dz2 . Since Π is compact we cancover it with finitely many discs Dz1 , ..., Dzn and f has an analytic extension to∆ ∩Dz1 ∩ ... ∩Dzn

However ∃δ > 0: D(0, 1 + δ) ⊂ ∆ ∪Dz1 ∪ ... ∪Dzn .And f has an analytic extension to D(0, 1 + δ). This is absurd since 1 is theradius of convergence.

Recall from last time that ∆ = D(0, r) and Π = ∂D(0, r)Π is called the natural boundary if every point is singular.

Example.

f(z) =

∞∑0

zk

has Π as a natural boundary.

Consider ω = e2πipq where p, q ∈ Z. We note that the set S = ω = e

2πipq : p, q ∈

Z is dense in Π. Since the singular points form a closed subset, it suffices toshow that every point in S is singular.Take r = 1.

f(r, ω) =

q−1∑0

rk!ωk! +

∞∑q

rk! →∞ as r → 1

⇒ Every point is singular.

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Variations. 1. f(z) =

∞∑1

zk!

k!also has Π as a natural boundary because f ′

does.

2. We can extend the notion of natural boundary to other curves(rather thanΠ). For example, the imaginary axis is a natural boundary for the analyticfunction. f(z) =

∑∞0 ek!z defined in Re(z) < 0

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2 Riemann Surfaces

Definition. A Riemann surfaceR is a connected Hausdorff topological space,togetherwith a collection of homeomorphisms φα : Uα → Dα ⊂ C with Uα open andφα(Uα) = Dα open in C such that

1. ⋃α

Uα = R

2. If Uα∩Uβ 6= φ then φβ φ−1α : φα(Uα∩Uβ)→ φβ(Uα∩Uβ) is holomorphic.

R

Uα Uβ

φα φβ

φβ φ−1α

Dα

φα(Uα ∩ Uβ)

Dβ

φβ(Uα ∩ Uβ)

The pair (Uα, φα) are called charts. The collection of all charts is called anatlas. The functions φβ φ−1

α are called the transition functions. (They areconformal equivalence.)

Remark. Each point z ∈ R has a neighbourhood holomorphic to a disc in C.Hence R assumed to be connected automatically gives path-connected. (R isconnected and locally path-connected as φ−1

α is continuous.)

Example. 1. R = C single chart φ(z) = z

2. R = C single chart φ(z) = z + 1

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Definition. Two atlases a1, a2 on a toplogical space R are equivalent if a1 ∩ a2

is also an atlas.i.e. For any charts (Uα, φα) ∈ a1 and (Vβ , ψβ) ∈ a2 we have ψβφ

−1α : φα(Uα ∩

Vβ)→ ψα(Uα ∩ Vβ) holomorphic.

Example. (See above) Atlases (1) and (2) are clearly equivalent but not equiv-alent to an atlas with the single chart φ(z) = z

Check: In definitions above equivalence of atlases, check that it is indeed anequivalence class.An equivalence class is called a conformal structure.

Remark. If R is a Riemann Surface with atlas (Uα, φα) and S is a connectedopen subset of R then S is a Riemann Surface with atlases (Uα ∩S, φα|Uα∩S)

Example. Every domain of C is a Riemann surface.

Note: Whenever we refer to C as a Riemann Surface, we mean C with thestandard conformal structure determined by (C, φ(z) = z)

Definition. Let R and S be Riemann surfaces with atlases Uα, φα and(Vβ , ψβ) respectively. A continuous map f : R → S is analytic/holomorphicif ∀α, β such that

• Uα ∩ f−1(Vβ) 6= φ

• ψβfφ−1α is analytic on φα(Uα ∩ f−1(Vβ))

f is called a conformal equivalence(analytic isomorphism/holomorphism) if ∃analytic inverse g : S → R

Example. (C, φ(z) = z) 7→ (C, ψ(z) = z)

Remark. f : R→ S holomorphic iff given z ∈ R, ∃(U, φ) with z ∈ U and (V, ψ)with f(z) ∈ V such that ψfφ−1 : φ(U ∩ f−1(V ))→ ψ(V ) is holomorphic. Thisis because if for example, (U ′, φ′) is another chart around z, then ψf(φ′)−1 =ψfφ−1φ(φ′)−1

Lemma 2.1. The composition of analytic Rf→ S

g→ G is analytic.

Proof. Write h = gf . Suppose charts are (Uα, φα), (Vβ , ψβ)(Wδ, θδ) Uα∩h−1(Wδ) is covered by Uα ∪ (f−1(Vβ) ∩ h−1(Wδ)).For β s.t. intersection is non-empty, we haveψβfφ

−1α : φα(Uα ∩ f−1(Vβ) ∩ h−1(Wδ))→ ψβ(Vβ ∩ g−1(Wδ))

θδgψ−1β : ψβ(Vβ ∩ g−1(Wδ))→ θδ(Wδ) both analytic.

f gUα Vβ Wδ

φα ψβ θδ

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So composition is analytic.⇒ θδgfφ

−1α : θδhφα is analytic on each φα(Uα ∩ f−1(Vβ) ∩ h−1(Wδ)) and these

cover φα(Uα ∩ h−1(Wδ))

Tautology: If a1, a2 are atlases for R and τ : (R, a1)→ (Ra2) holomoprhic⇔a1, a2 equivalent.If f : (R,α1)→ (S, α2) analytic then ai ∼ a′i then f ′ : (R, ai)→ (S, a′i) analytic.

f ′ : (R, a′1)τR→ (R, a1)

f→ (S, a2)τ−1S→ (S, a′2)

Can move between equivalent atlases, equivalence of atlas is an equivalencerelation.

Remark. Sufficient to check ∀z ∈ R, ∃ chart about z ∈ U , chart (U,ϕ) andchart (V, ψ) s.t. V 3 f(z), ψfϕ−1 holomorphic. ϕ(U ∩ f−1(N))→ ψ(V ). Thenif (U ′, ϕ′) another chart at z, (V ′, ψ′) at f(z) thenψ′f(φ′)−1 = (ψ′ψ−1)(ψfϕ−1)(ϕ(ϕ′)−1) is holomorphic.

Lemma 2.2. If R is a Riemann surface and π : R → R is a covering map oftopological spaces, then there exists a unique conformal structure on R suchthat π : R→ R is analytic.

Proof. Given a point z ∈ R, choose N s.t. z ∈ N , pi|N : N → π(N) is ahomeomorphism.Let (V, ψ) be any chart of R with π(z) ∈ V and take chart φ = ψπ : U →ψ(N ∪ V ).This chart (U, φ) form an atlas a of R transition functions are just restrictionof transitition fucntions of R.Let us check π analytic.

ψπφ−1 : ψπ(ψπ)−1 = ψππ−1ψ−1 = id

holomorphic on some appropriate domain.Uniqueness: Suppose a∗ = (Wα, θα is an atlas on R s.t.

pi : ((R), a∗)→ (R, a) is analytic.We wish to show that a∗ is equivalent to a.Given (U, φ) ∈ a as above. Assume thatπ|N : (N , a∗|N )→ (N, a|N ) is holomorphic.

Given (Wα, θα) ∈ a∗ s.t. Wα ∩ N ∩ π−1(V ) 6= φ have a diagram

Wα ∩ U N ∩ V

θα(Wα ∩ U) ψ(N ∩ V )

π

θαφ

ψ

By(*) ψπ︸︷︷︸φ

θ−1α is holomorphic.

This means φθ−1α is holomorphic. i.e. a∗ is equivalent to a.

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Example. Let us revisit log. Two lectures ago we constructed,

R C

C∗

f

πexp

R =∐k∈Z

C∗ × k By Lemma 2.2, ∃ unique conformal structure on R making π

holomorphic.(C∗, ψ(z) = z). Charts on R are making (U, φ) where φ = πNTo check that f is holomorphic we only need to check that f (π|N )−1 isholomorphic.But this is just a branch of log.In fact, f is a bijection and its inverse is also holomorphic.Since exp(f(z)) = π(z), locally (π|N )−1 exp and this is holomorphic.Similarly, with zn we get a n-folded cover.

R C

C∗

f

πzn

and using Lemma 2.2, we can upgrade the diagram to a diagram of holomorphicmaps between Riemann Surfaces.Important Examples: Riemann sphere, C∞ = C ∪ ∞Topology: Open subsets are either open subset of C or of the form∞ ∩ (C \ compact set )With this topology, C∞ is homeomorphic to S2 = x ∈ R3 : ||x|| = 1Homeomorphism is given by the stereographic projection π : S2 → C∞. Inparticular, C∞ is compact, connected and Hausdorff.Atlas: (C, φ1), φ1(z) = z, (C \ 0, φ2), φ2(z) = 1

zTransition Function:

C∗ → C∗

z 7→ 1

z

Alternatively, we can define charts in S2 directly.U1 = S2 \ (0, 0, 1), π1 projection from (0, 0, 1).U2 = S2 \ (0, 0,−1), π2 projection from (0, 0,−11).Recall π1(x)π2(x) = 1

π1 =x+ iy

1− z, π2 =

x− iy1 + z

Consider atlas (U1, π1), (U2, π2).The harmonic function is again.Complex Tori C Consider a lattice

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Λ = τ1Z + τ2Zτ1, τ2 6= 0, τ1/τ2 6∈ Rπ : C→ C/Λ We endow C/Λ with the quotient topology.This makes π continuous.C/Λ is a compact connected Hausdorff topological space.For any a ∈ T , choose w ∈ C s.t. π(U) 3 a and open neighbourhood N 3 Ws.t. π|N is a homeomorphism.Take as chart:φ = (π|N )−1 : U → NTwo such chartsφ1 : U1 → N1

φ2 : U2 → N2 with U1 ∩ U2 6= φgive rise to a transition function φ2φ

−1 : φ1(U1 ∩ U2) → φ(U1 ∩ U2) which islocally just a translation by some w ∈ Λ hence analytic. (Example sheet 1).

Theorem 2.3 (Inverse Function Theorem). Given any analytic function g onan open set V ⊂ C and a ∈ V such that g′(a) 6= 0, there exists an openneighbourhood N of a such that g|N : N → g(N) is a conformal equivlance ofN onto the open neighbourhood g(N) of g(a).

Proof. By considering g(z) − g(a), we may assume without loss of generalitythat g has a simple zero at a

Take a small disc D = D(a, ε) with D ⊂ V such that

• g is nowehre zero in D \ a

• g is nowhere zero in D

Principle of Argument ⇒ n(g γ, 0) = 1 = zeroes of g in Dwhere γ a closed loop around a.Note that g γ([0, 1]) compact hence closed.Therefore, C \ g(γ([0, 1])) is open and choose ∆ disc centred at 0 contained inthe connected component containing g(z).Hence n(g γ,w) = 1 for all w ∈ ∆By the principle of a argument again, N = (g|D)−1(∆) is an open neighbour-hood of a and g|N : N → ∆ is bijective.Then, the opening mapping theorem, (IB Complex Analysis)⇒ h = (g|N )−1) iscontinuous and hence g|N is a homeomorphism.Pick w0 ∈ ∆ and w 6= w0 and let z0 = h(w0)andz = h(w) and considerh(w0)−h(w)

w0−w = z0−zg(z0)−g(z) →

1g′(z)

Since w → w0 iff z → z0 as w → w0.(g|N is a homeomorphism.)

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2.1 Local Form of Holomorphic Function

Suppose we have a ∈ U ⊂ C and f 6∼= 0 is analytic in the domain U withf(a) = 0. Then we know we can write locally around a

f(z) = (z − a)rh(z)

where h is analytic, h′(a) 6= 0 and r > 0. We can find a disc D containing asuch that h(D) ⊂ C \ R≥0e

iα So on D we can choose an analytic function

l(z) = [h(z)]1r = exp( 1

r log(h(z)) Hence f(z) = [g(z)]r where g(z) = (z − a)l(z)By Theorem 2.3, g is locally a conformal equivalence which means if we siadw = g(z) then f(g−1(w)) = wr More generally, if f : R → C locally around apoint p0 ∈ R we can choose a chart φ : U → C and by replacing f by f(p)−f(p0)we may assume f(p0) = 0. If a = φ(p0) we can locally write the function fφ−1

as [g(z)]n as above.Locally around p0 we have a chart ψ = gφ and a diagram.

i.e. f(p) − f(p0) = ψr Finally, if f : R → S when R and S are both RiemannSurfaces take a chart (v, θ) around f(p0) with θ(f(p0)) = 0 and apply the aboveto θf so locally we have that f is of the form z 7→ zr

Theorem 2.4 (Open Mapping Theorem). Let f : R → S be a non-constantanalytic map of Riemann Surfaces. Then f is an open map.

Proof. Suppose W is open in R, z ∈ W . We are required to prove that f(W )contains an open neigbourhood of f(z). Choose charts(U, φ) in R with z ∈ U(V, ψ) in S with f(z) ∈ VLet D be an open disc with φ(D) s.t. φ−1(D) ⊂ U ∩W ∩ f−1(V )ψfφ−1 is non-constant.(If not by the identity principle of the Riemann Surfaces on Example Sheet 1Q 11, f is constant on R.)By the opne mapping theorem for plane domains(IB)⇒ ψfφ−1(D) is open.Have fφ−1(D) is open in S. Also, f(z) ∈ f(φ−1(D)) ⊂ f(W ) in S.Hence fφ−1(D) is the required open neighbourhood.

Corollary 2.5. Let f : R → S be analytic and non-constant. If R is compactthen f(R) = S and S is also compact.

Proof. By (2.4) and f(R) is open. Since R is compact, f(R) is compact andsince S is Hausdorff. f(R) is closed. Since S is connected and f(R) 6= φ wehave f(R) = S.

Corollary 2.6. Any analytic function f : R → S with R compact must beconstant.

Proof. Since C is not compact, thus it follows from (2.5)

Remark/Application: Suppose f : R \ a → C is analytic and a ∈ R.Taking local charts around a, we see f has a removable singularity ⇔ f is

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bounded in a neighbourhood of a.For instance, f : C → C holomorphic and bounded.Then f extends to a holo-morphic map f : C∞ → C and hence is constant.(i.e.Liouville’s theorem)

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2.2 Harmonic Functions

Definition. Let D be a plane domain and u : D → R is harmonic in u ∈ C2(D)∂2u∂x2 + ∂2u

∂y2 = 0 Laplace’s equation.

Remark. If f = u+ iv is analytic in D then u, v harmonic. (just use Cauchy-Riemann equations.)If D is a disc and u is harmonic then there is a holomorphic function, f : D → Cs.t. u = Re(f).(Example sheet 1, Q12)

Definition. A harmonic function h or a Riemann surface R is a continuousfunction f : R → R s.t. for any chart φ : U → φ(U) ⊂ C on R the functionhφ−1 is harmonic on φ(U).The remark above implies that this definition is independent for the choice ofcharts.Moreover, a continuous function h : R → R is harmonic iff each z ∈ R has aneigbourhood N s.t.h|N = Re(g) for some analytic function g on N .

Proposition 2.7. Non-constant harmonic functions h : R → R on a Riemannsurface are open maps. Hence if R is compact all harmonic functions are con-stant.

Proof. Suppose that U ⊂ R is open.For each z ∈ U , there exists open neighbourhood N ⊂ U s.t. z ∈ N andh = Re(g) where g : N → C holomorphic.g is non-constant.If g is constant ⇒ h is constant on open set.Example sheet 1 Q 13⇒ h globally constant.Open mapping theorem for g ⇒ ∃ neighbourhood g(z) = a+ ib of the form.(a− ε, a+ ε)× (b− ε, b+ ε) ⊂ g(N).Thus (a− ε, a+ ε) ⊂ h(N) ⊂ h(U)⇒ h(U) open.If R is compact and h is non-constant then h(R) is both open and closed.⇒ h(R) = R is compact.#

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2.3 Meromorphic functions

Definition. A meromorphic function f on a Riemann Surface R is an analyticmap f : R→ C∞

Lemma 2.8. If U ⊂ C is a plane domain, then this definition coincides withthe usual one of isolated poles.

Proof. Suppose f : U → C∞ analytic.If a ∈ U has f(a) ∈ C then on each neighbourhood N of a, f is an analyticfunction on N . If f(a) =∞ need to take a chart 1/w at ∞f analytic at a ⇒ g(z) = 1

f(z) is analytic in the neighbourhood of a with

g(a) = 0.Writing g(z) = (z − a)rh(z) h analytic around a and h(a) 6= 0 r > 0

f(z) = (z − a)−r1

h(z)with

1

h(z)holomorphic around a.

⇒ f has an isolated pole at a. Conversely if f s meromorphic in the usual sensewith a pole at a ∈ U we can writef(z) = (z − a)−rl(z) locally around a for some holomorphic function l withl(a) 6= 0.⇒ g(z) = 1

f(z) = (z − a)r 1l(z) is analytic around a.

f : U → C∞ holomorphic.

Example. (Related to Q 15 in Example Sheet 1)f(z) = z3 − zg(z) =

√z3 − z

(z, w)︸ ︷︷ ︸algebraic

curve

: w2 = f(z) ⊂ C︸ ︷︷ ︸cutting and

pasting

U = C \ [−1, 0] ∩ [1,∞) Locally around every point in C \ 0,±1, ∃ preciselytwo functions g such that [g(z)]2 = f(z)Claim: ∃g holomorphic on U s.t. g2(z) = f(z). We can do this by analyticcontinuation along paths.Principle of Argument: n(f γ, 0) = n(γ,−1) + n(γ, 0)︸ ︷︷ ︸

same

+n(γ, 1)︸ ︷︷ ︸→0

n(f γ, 0) = 2n(γ,−1) ∈ 2Zn(f γ, 0) even implies that as we analytically continue along a closed curve γwe return exactly where we started so such a g exists.R: Riemann surface obatined from glueing.

R is topologically a torus with four points removed.

17

3 The space of Germs and the Monodromy The-orem

First clear some topological preliminaries.

Definition. Suppose π : X → X is a covering map of topological spaces andγ : [0, 1]→ X is a path. A lift of γ is a path γ : [0, 1]→ X such that π γ = γ

Proposition 3.1. Uniqueness of Lifts Suppose γ1 and γ2 are lifts of γ withγ1(0) = γ2(0) then γ1 = γ2.

Proof. Consider I = t ∈ [0, 1] : γ1(t) = γ2(t)

• I is open If τ ∈ I choose open neighbourhood N of γ1(τ) = γ2(τ) suchthat π|N is homeorphic.By continuity there exists δ > 0: for t ∈ (τ − δ, τ + δ)⇒ γ1(t), γ2(t) ∈ NBut π(γ1(t) = π(γ2(t)) = γ(t)

⇒ ˜γ1(t) = ˜γ2(t) for all t ∈ (τ − δ, τ + δ)

• I is closed. If τ ∈ IC then γ1τ 6= γ2τX Hausdorff⇒ ∃ open sets U1, U2 ⊂ X such that γ1(τ) ∈ U1, γ2(τ) ∈U2, U1 ∩ U2 = φContinuity ⇒ ∃δ > 0 : |t− τ | < δ ⇒ γ1(t) ∈ U1, γ2(t) ∈ U2

Definition. A covering π : X → X is called regular if every x ∈ X has an openneigbourhood U such that

π−1(U) =∐ν

Uν

where Uν are disjoint open subsets of X such that π|Uν : Uν → U is a homeo-morphism for each ν.

Example. 1. Covering map

C→ C∗

z 7→ exp(z)

is regular but the restriction to −2π < =(z) < 2π is not.

2. π : C→ C/Λ, Λ lattice

Proposition 3.2. Let π : X → X be a regular covering and γ a path in X,z ∈ X such that π(z) = γ(0). Then ∃ a lift γ such that γ(0) = z.(Uniquely by (3.1))

Proof. Set I = t ∈ [0, 1] : ∃ a lift γ : [0, t]→ X with γ(0) = zClearly 0 ∈ I so I 6= φ.Set τ := sup I

18

Claim: ∃δ > 0 such that [0, τ + δ] ∩ [0, 1] ⊂ I. This implies τ = 1, 1 ∈ I andhence the result.Proof of claim: Choose an open neigbourhood of γ(τ)

with π−1(U) =∐ν

Uν as above

Continuity⇒ ∃δ > 0: t ∈ [τ − δ, τ + δ] ∩ [0, 1]⇒ γ(t) ∈ UBy definition of τ , ∃τ1 ∈ [τ − δ, τ ] : γ has a lift ˜gamma(0) = z

Now γ(τ1) ∈ π−1(U) =∐ν

Uν .

∃!ν s.t. γ(τ1) ⊂ UνDefining γ := (π|Uν )−1 γ on [τ, τ + δ] ∩ [0, 1] extends γ continuously to a lifton [0, τ + δ] ∩ [0, 1]

3.1 Homotopy of Lifts

Definition. Let X be a topological space and α, β : [0, 1] → X be paths withsame initial and final points.i.e. α(0) = β(0), α(1) = β(1)We say that α, β are homotpic in X if there exists a family (γs)s∈[0,1] of pathsin X s.t.

(i) γ0 = α, γ1 = β

(ii) γs(0) = α(0) = β(0), γs(1) = α(1) = β(1)

(iii)

(t, s) 7→ γs(t)

[0, 1]× [0, 1]→ X

is continuous

Definition. X is called simply connected if X is path-connected and everyclosed path γ : [0, 1]→ X is homotopic to the constant path γ(0).

Theorem. Monodrony TheoremLet π : X → X be a covering map and α, β paths in X. Assume

1. α, β homotopic in X

2. α, β have lifts α, β with α(0) = β(0)

3. Every path γ in X with γ(0) = α(0) = β(0) have a lift γ with γ(0) =α(0) = β(0).

Then α, β homotopic in X and in particular α(1) = β(1)

Proof. (Non-examinable) See Beardon p.106-107

19

Note: Regular⇒ (2) and (3) hold.Let G ⊂ C be a domain.

Definition. Let z ∈ G, (f,D), (g,E) be function elements in G. WriteWrite(f,D) ∼z (g,E) if z ∈ D ∩E,f = g on some open neighbourhood of z in D ∩EThis defines an equivalence relation and the equivalence classes containing (f,D)is called the germ of f at z denoted by [f ]z[f1]z1 = [f2]z2 ⇔ z1 = z2 = z, f = g on a neighbourhood of z.G = G (G) := all germs [f ]z over all z ∈ G Let (f,D) be a function element[f ]D := [f ]z : z ∈ D ⊂ G

1. Topology of GOpen sets are unions of set of the form [f ]DRequired to prove U1, U2 are open in G ⇒ U1 ∩ U2 is open

Proof. Suppose [f ]z ∈ U1∩U2 since U1, U2 are open, ∃(f1, D1) and (f2, D2)s.t.[f ]z ∈ [f1]D1

[f ]z ∈ [f2]D2

Choose representatives (f,D) of the germ [f ]z such that z ∈ D ⊂ D1∩D2

[f ]z = [f1]z = [f2]z ⇒ ∃ a neigbourhood of z: f = f1 = f2 on it. By theidentity principle f = f1 = f2 on DHence, [f ]z ∈ [f ]D ⊂ U1 ∩ U2

2. Topology is HausdorffSuppose [f ]z 6= [g]w.Choose domains D,E such that z ∈ D,w ∈ E f analytic on D, g analyticon E.If z 6= w choose D,E disjoint and [f ]D ∩ [g]E = φIf z = w choose D and E to coincide and claim [f ]D ∩ [g]E = φ.Otherwise [h]s ∈ [f ]D ∩ [g]E for s ∈ D.⇒ h = f = g is a neighbourhood of S.⇒ f = g on D by the identity principle.⇒ [f ]z = [g]z #Absurd

3. Projection map

π : G → G

[f ]z 7→ z

• π is continuous Given V ⊂ G open, note π−1(V ) =

• π|[f ]D : [f ]D → D is a homeomorphism.Certainly bijective and π|[f ]D is continuous.π|[f ]D is an open map.Take U open in [f ]D.

20

U open in [f ]D

⇔ U =⋃α

[f ]Dα where Dα are open in D

⇔ π(U) =⋃α

Dα

⇔ π(U) open in D.

4. Atlas on GOn each connected component π is a covering map so we use Lemma 2.2to give each connected component the structure of a Riemann surface forwhich π is analytic.

Charts are simply (U, φ), U = [f ]D

φ = π|[f ]D

5. Evaluation Map

E : G → C[f ]z 7→ f(z) This is analytic

This is analytic.Eφ−1(z) = E([f ]z) = f(z) holomorphic. (f,D), z ∈ D

Theorem 3.3. Let (f,D) and (g,E) be function elements in G and γ : [0, 1]→G be a path with γ(0) ∈ D, γ(1) ∈ E.Then (g,E) is an analytic continuation of (f,D) along γ ⇔ ∃ a path γ : [0, 1]→G such that πγ = γ and

γ(0) = [f ]γ(0)

γ(1) = [g]γ(1)

Proof. Assume (f,D) ∼γ (g,E)⇒ ∃(fj , Dj)j=1 and 0 = t0 < t1 < ... < tn = 1 with(f,D) = (f1, D1) ∼ (f2, D2) ∼ ... ∼ (fn, Dn) = (g,E)and γ([tj−1, tj ]) ⊂ Dj for j = 1, ..., n.Define γ(t) := [fj ]γ(t) for tj−1 ≤ t ≤ tjNote this is well-defined since fj = fj−1 on Dj−1 ∩ Dj (Thus take care ofendpoints.)Obviously πγ = γRequired to prove γ is continuous. γ : [0, 1]→ GConsider a basic open set [h]∆ and suppose γ(τ) ∈ [h]∆Without loss of generality, we may assume that τ ∈ [tj−1, tj ] soγ(τ) = [fj ]γ(τ) ∈ [h]∆i.e. γ(τ) ⊂ ∆ and fj = h is some open neighbourhood N of γ(τ) in ∆ ∩Dj

By continuity of γ, there is a δ > 0 s.t. if |t− τ |z < δ ⇒ γ(t) ∈ NThen γ(t) = [fj ]γ = [h]γ(t) ∈ [h]∆

21

⇒ γ is continuous.(⇐) Assume ∃γ → G is a path s.t. γ(0) = [f ]γ(0), γ(1) = [g]γ(0) and πγ = γClaim: We can find basic open sets [f ]D1

, ..., [f ]Dn in G for some n and 0 =t0 < t1 < ... < tn = 1 s.t. γ([tj−1, tj ]) ⊂ [fj ])Dj for each j.Proof of claim: Note that by continuity of γ for each t ∈ [0, 1] we can find abasic open set [ft]Dt 3 γ(t) in G and an open interval I(t) 3 t such thatγ(I(t) ∩ [0, 1]) ⊂ [ft]DThe intervals Ik cover [0, 1] hence by compactness we can find a finite subcover,which we can reorder to obtain intervals Ik = [ak, bk] with ak+1 < bk for allk = 1, ..., n.Choosing tk ∈ (ak+1, bk) we can arrange fk, γ([tj−1, tj ]) ⊂ [fj ]DjAdjustments: WLOG may assume that

1. Each Dj to be a disc.

2. Also, since γ(0) = [f ]γ(0) andγ(1) = [g]γ(1)

We may assume that D1 ⊂ D, Dn ⊂ E with f = f1 in D1 and g = fn inDn.For each 1 ≤ j ≤ n− 1, γ(tj) ∈ [fj ]Dj ∩ [fj+1]Dj+1 hence fj = fj + 1 on arange of γ(tj) and hence in all of Dj ∩Dj+1

(Identity Principle on intersection Dj ∩ Dj+1 is connected since Di aredisc(s).Thus, (f,D) ∼ (f,D1) = (f1, D1) ∼ ... ∼ (fn, Dn) = (g,Dn) ∼ (g,E)And we have γ([tj−1, tj ]) = πγ([tj−1, tj ]) ⊂ π([fj ]Dj ) = Dj

(We can always enlarge the dissection0 < t1

2 < t1 < ... < tn−1 <tn−1+1

2 < 1to account for his first and last continuation.)

Corollary 3.4. Let F be a complete analytic function in G and put GF =⋃(f,D)∈F [f ]D. Then GF is a connected component of G .

Proof. (3.3)⇒ GF is path-connected, hence connected.GF is open by definition.Also GF = G

⋂F ′6=F GF ′.

Hence GF is a connected component.

Summary: G ⊂ C, F complete analytic function then we have constructed aRiemann Surface GF and a diagram.π : GF → G projection map which is analytic covering.GF is the Riemann Surface associated with F .

22

GF

G

D

[f ]D

C

ε

f

Theorem 3.5 (Uniqueness of Continuation along paths). If (g,E) and (h,E)are analytic continuations of (f,D) along some path γ in G then g = h in E.

Proof. (3.3)⇒ ∃γ, γ∗ s.t. γ0 = γ∗(0) = [f ]γ(0) inGF andγ(1) = [g]γ(1)

γ∗(1) = [h]γ(1)

Uniquness of lifts (3.1)⇒ γ(1) = γ∗(1)⇒ g = h in a neighbourhood of γ(1)⇒ g = h on E by the identity principle.

Theorem 3.6 (Classical Monodromy Theorem). Suppose (f,D) can be contin-ued along all paths in G which start in D. If (g,E) and (h,E) are continuationsof (f,D) along paths α, β which are homotopic in G, then g = h on E.

Proof. (3.3)⇒ ∃ lifts α and β s.t.α(0) = [f ]α(0) , β(0) = [f ]β(0)

α(1) = [f ]α(1) , β(1) = [f ]β(1)

Since α(0) = β(0) we have α(0) = β(0).So by Monodromy Theorem(Topological) now lifts α, β are homotopic and inparticular, α(1) = β(1) Hence as before g = h on E.

Corollary 3.7. Suppose G ⊂ C is simply-connected and (f,D) is a functionelement in G which can be analytically continued along all paths in G whichstart in D. Then, ∃ a single-valued analytic continuation f of (f,D) to all G.

Proof. If (g,E) and (h,E) are two analytic continuations of (f,D) then the twoanalytic continuations of each other along some closed path γ in G.Since G is simply-connected, γ is homotopic to a point, hence (3.6)⇒ g = h onE.Deforming f = g on E(for all (g,E) ∈ F) gives an analytic extension of (f,D)to all of G.

23

4 Multiplicities and Degrees

f : R→ S with R,S compact Rimeann surfaces, f non-constant.Given z ∈ U ⊂ C(domain) and a non-constant holomorphic function f , we canwrite locallyf(w) = f(z) + (w − z)mf (z)f1(w)where f1(z) 6= 0We call mf the multiplicity or valency of f at z.

Lemma 4.1. Suppose g and h are analytic and non-constant on domains in Cwith image(h) ⊂ image(g) then,

mgh(z) = mg(h(z))mh(z)

Proof. 1. h(w) = h(z) + (w − z)mh(z)h1(w) locally with h1(z) 6= 0

2. g(η) = g(η0) + (η − η0)mg(η)g1(η) locally with g1(η) 6= 0 and η0 = h(z).Substitute (1) into (2).g(h(w)) = g(h(z)) + (w − z)mh(z)mg(η0)(h1(w))mg(η0)g1(h(w))locally with g1(h(z)) = g(η0) 6= 0 and h1(z) 6= 0Consider f : R→ S analytic and non-constant and z ∈ R.Take charts (U, φ) around z and (V, ψ) around f(z) and definemf (z) = mψfφ−1(φ(z))This does not depend on the choice of charts.If we choose (U , φ), (V , ψ) instead then,

ψf ˜φ−1 = ψψ−1︸ ︷︷ ︸ψfφ−1φ︸ ︷︷ ︸local conformal equivalence

φφ−1︸ ︷︷ ︸Since mψψ−1(ψ(f(z))) = 1 and mφφ−1(φ(f(z))) = 1

(4.1)⇒ mψfφ−1(φ(z)) = mψfφ−1(φ(z))

Given any chart θ : V → C in S, where f(z) ∈ V ⊂ S with θ(f(z)) = 0, wesaw in §2 that there is a chart ψ : N → C s.t. ψ(z) = 0, z ∈ N ⊂ f−1(V ) andθfψ−1 : ψ(N)→ C is just z 7→ zr where r = mf (z)

N ⊂ R S

C C

f

ψ θ

zr

Remark. 1. mf (z) > 1 only at isolated points of R. (f non-constant.)Clear from the plane case given by vanishing of f ′.

2. Points z ∈ R s.t. mf (z) > 1 are called ”Ramification Points” and mf (z)is called the ramification index. The image of a ramification is called abranch point.

24

Example. p(z) =

d∑0

akzk polynomial of degree d, ad 6= 0.

Define p(∞) =∞. Then p becomes a holomorphic map C∞ → C∞.Take charts φ(w) = ψ(w) = 1

w , 1w on C∞ \ 0.

ψpφ−1(z) =1

p(1/z)=

zd∑d0 akz

d−k= zdg(z) with g(0) 6= 0

⇒ mp(∞) = d

Theorem 4.2 (Valency Theorem). Suppose f : R→ S is a non-constant holo-morphic map with R compact(⇒ f(R) = S and S compact.)Then ∃ an integer n ≥ 1 s.t. f is an n to 1 map of R onto S counting multiplic-ities.

n =∑

z∈f−1(w)

mf (z)

∀w ∈ S

Definition. n is called the degree of f , also denoted n(f), deg(f).

Proof. f−1(w) is finite. Note f−1(w) is compact and discrete.(Identity Princi-ple)Define n(w) :=

∑z∈f−1(w)mf (z) for w ∈ S.

Our aim is to prove that w 7→ n(w) is constant.Take w0 ∈ S and write f−1(w0) = z1, ..., zqFor any point z0 ∈ R, we know that ∃ charts

ψ : N 7→ Cθ : N 7→ C

with f(n) ⊂ V and θfψ−1 is just z 7→ zmp(z0)

Hence ∃ disjoint neighbourhoods N1, ..., Nq of z1, ..., zq s.t. f is an mf (zj) to 1map. (Counting multiplicities) of Nj onto f(Nj) for each j − 1, ...q.Include imageClaim.:There exists neighbourhood M of w0 s.t. f(R \ ∪q1Nj) ∩M = φ

Proof of claim.: R \q⋃1

Nj closed in R.

Hence compact ⇒ f(R \q⋃1

Nj) is compact hence closed.

Then M = S \ f(R \q⋃1

Nj) is open containing w0 and M ∩ f(R \j⋃1

Nj) = φ

Corollary 4.3 (Fundamental Theorem of Algebra, one more time!). Let P bea polynomial of degree d, then P has d zeroes counted with multiplicities.

25

Proof 1. We have seen that P gives rise to P : C∞ → C∞ holomorphic with

P (∞) =∞ and mP (∞) = d ⇒(4.2)

# of zeroes of P =∑

z∈P−1(0)

mP (z) = d

Proof 2. By (2.5) p is onto. Hence there exists a zero of p. Use remaindertheorem to get d zeroes.

4.1 Rational Maps

Proposition 4.4. f : C∞ → C∞ holomorphic and non-constant

⇔ f(z) =c(z − α1)...(z − αn)

(z − β1)...(z − β2)a non-constant rational function in C.

Proof. (⇐)f(z) and f( 1z ) are meromorphic and so define a holomorphic map of

C∞.(⇒) W.l.o.g assume f(∞) ∈ C. Otherwise consider 1

f . Then f−1(∞) is a finite

subset of C say z1, ..., zq and f is meromorphic in C as each zj is a pole.Near zj , f has the form,

f(z) =

∞∑−kj

ai(z − zj)i

and set Qj =

−1∑−kj

ai(z − zj)i︸ ︷︷ ︸rational function

Consider f = f − (Q1 + ... + Qq). f has now only removable singularity at

z1, ..., zq. Hence f gives rise to a holomorphic map f : C∞ → C∞. By (2.6) fis constant and thus f = Q1 + ...+Qq+constant.

Remark. f(∞) ∈ C ⇔ n ≤ m and in this case deg(f) = m = # of polescounting multiplicities.(Example Sheet 2) deg(f) = maxm,nIf f is an analytic isomorphism, deg(f) = 1⇒ m = n = 1.

Hence f(z) =az + b

cz + dwith ad− bc 6= 0, a Mobius map.

So Aut(C∞) =Mobvious group.

4.2 Quick review of triangulations and Euler characteris-tic(from IB Geometry)

S compact surface.Topological triangle on S=homeomorphic image of a closed triangle T ⊂ R2.

Definition. A topological triangulation consists of a finite collection of topo-logical triangles whose union is all of S s.t. (i) Two triangles are either disjointor their intersection is a common vertex or common edge.

26

F = # of trianglesV = # of verticesE = # of edges

e := F − E + VEuler-number of triangulation

Fact. Fact 1 One can always find a triangulation.

Fact. Fact 2 e does not depend on the chosen triangulation; it is a topologicalinvariant χ(S).

Example. 1. S2

F = 8E = 12V = 6χ(S2) = 2

2. Torus T 2 F = 18, E = 27, V = 9χ(T 2) = 0

Fact. A compact Riemann Surface R is topologically a sphere with g handles.e.g.g=0.χ(R) = 2− 2g g genus of R.f : R → S be a non-constant holomorphic map between compact RiemannSurfaces. Given p ∈ R, let ep be the ramification index mf (p). Recall ep >1 ⇔ p is a ramification point and the set of these points i actually finite. Inparticular, the following sum∑

p∈Rep − 1 is finite.

Theorem 4.5 (Riemann Hurwitz). If f has degree n, then

χ(R) = nχ(S)−∑p∈R

(ep − 1)

Proof. Let Q1, ..., Qr denote the branch points of f . Suppose we have a trian-gulation on S by subdividing it we can assume that the Qi are vertices.

The proof of the valency theorem theorem (4.2) gives us an open coverU1, ..., Ur, Ur+1, ..., Um of S such that

(i) If j > r, f−1(Uj) has a disjoint union of open subset of V for whichf |V : V → U is a conformal equivalence so U contains no branch points.

(ii) If j ≤ r Qj ∈ Uj and for each point p ∈ f−1(Qj) there exists a uniquecomponent V of f−1(Uj) so that f |V : V → Uj is an eP to 1 map overUj \ Qj given in terms of suitable charts by z 7→ zep

Now we ???? our trinagulation by means of Euclidean triangles as follows.

27

...to ensure that each topological triangle is centred in some Uj .T then lifts to a triangle on R if no vertex of T is a branch point, these trianglesare disjoint; either T ⊂ Uj if j > r(when it is obvious) or T ⊂ Uj if j < r andstill okay. If T has a vertex at a branch point Qi,it still lifts to n triangles of Rbut there will be ep vertices in common at each ramification point p ∈ f−1(Qj).We therefore get at triangulation on R corresponding to the one in S s.t.F, F

′, V, V

′, E,E

′denoted the # of faces, vertices and edges respectively on

S and R and we see thatF′

= uF, E′

= uE, V′

= nV −∑p∈R

ep − 1.

Hence χ(R) = nχ(S)−∑p∈R

ep − 1.

Trivial Remark.

2− 2g(R) = n(2− 2g(S))−∑p∈R

ep − 1

⇒ g(R) ≥ g(S)

Example. R ⊂ C× C R = (z, w)|w2 = f(z) where f(z) = z3 − z

π : R→ C (0, 0) = π−1(0)

(z, w) 7→ z (1, 0) = π−1(1)

(−1, 0) = π−1(−1)

g : R→ C(z, w) 7→ w

g around these 3 points is an inverse. g−1(w) = (f−1(w2), w)Look at π around (0, 0) and (±1, 0)

πg−1(w) = f−1(w2)

dπg−1

dw= (f−1)

′(w2) · 2w

Assume R can be compactified. We can add a point c s.t. R = R ∪ c is acompact Riemann Surface and π(c) = ∞ so that π : R → C∞ is holomorphic.deg(π) = 2, branch points 0,±1 ramification points (0, 0), (±1, 0) with ep = 2.Since π−1(∞) = c then ∞ is also a branch point and ec = 2.

Riemann Hurwitz χ(R) = 2χ(C∞)− 4 = 2× 2− 4 = 0⇒ g(R) = 1Compactificaiton R is a torus.

One could push this further and consider f a polynomial with 2g+ 1 simpleroots to obtain a Riemann Surface R with genus g.

28

5 Meromorphic Periodic Functions

f : C→ C∞ analytic and non-constant. We say w is a period for f if f(z+w) =f(z) for all z ∈ C.The set of period Ω satisfies

1. Ω is an additive subgroup of C.

2. Ω contains only isolated points.

Example sheet #3 Q1 ⇒ Only 3 possibilities for Ω.

1. Ω = 0 non-periodic

2. Ω = Zw1 w1 6= 0 simply periodic

3. Ω = Zw1 + Zw2 w1, w2 6= 0, w1

w26∈ R Doubly periodic or Elliptic.

5.1 Simply Periodic Functions

Without loss of generality, assume Ω = Z. i.e. f(z + 1) = f(z) for all z ∈ C

Proposition 5.1. Define p : C → C∗ by p(z) = e2πiz. If f is meromorphic inC with periods Z then there exists a unique f meromorphic on C∗ s.t. f = f p

C C∞

C∗

f

Pf

Proof. Define f(e2πiz) = f(z).Well defined: If e2πiz1 = e2πiz2 ⇔ z1−z2 ∈ Z⇔ f(z1) = f(z2) Clearly f = f pand f is unique.If V is open in C∞ then p−1 f−1V︸ ︷︷ ︸

open

= f−1V is open in C

⇒ f−1V open

f meromorphic locally with f(w) = f(

log(w)2πi

)Corollary 5.2. Suppose f has no poles on the strip S = z : α < Im(z) < βThen f(z) =

∑∞−∞ ake

2πikz is locally uniformly convergent on S.

Proof. f is analytic on the annulus, p(S) = w : e−2πβ < |w| < e−2πα Lau-rent’s theorem

⇒ f(w) =

∞∑−∞

akwk locally uniformly convergent on p(S). ⇒ f(z) =

∑∞−∞ ake

2πzk

(Fourier expansion of simply periodic functions)

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5.2 Doubly Periodic Functions

Let Λ = Zw1 + Zw2 be a lattice in C. A meromorphic function f on C withperiod Λ is called elliptic.These functions form a field.An elliptic function is determined by its value on any parallelogramP = z + t1w1 + t2w2 : t1, t2 ∈ [0, 1)If it has no poles ⇒ bounded in C. ⇒ constant by Liouville’s theorem.Now we look at the complex torus π : C→ C/Λ.

Proposition 5.3. If f is an elliptic function with period Ω then there exists aunique meromorphic map f on C/Λ s.t. f = f π

C C∞

C/Λ

f

πf

Proof. Same as Proposition 5.1(Exercise).

Corollary 5.4. The degree of f is n ≥ 2.

Proof 1. If deg(f) = 1 then f : C/Λ→ C∞ is a conformal equivalence. But thisis absurd since they have different genus. (C/Λ and C∞)

Proof 2. Consider a period parallelogram P such that ∂P contains no poles off .∑z∈P

Reszf =1

2πi

∫∂P

f(z)dz = 0

⇒ f cannot just have a simple pole.

5.3 Weierstrass ℘ function

Definition. For a given lattice Λ define ℘(z) = ℘λ(z) = 1z2 +

∑z∈Λ\0

1(z−w)2 −

1w2

Convergence?

Lemma.∑

w∈Λ\0

1

|w|t<∞⇔ t > 2

Proof. The function (t1, t2)∈R2

7→ t1w1 + t2w2 is continuous and non-zero on the

compact set (t1, t2) : |t1|+ |t2| = 1And hence its modulus is bound above and below by constants C, c > 0.Set t1 = k

|k|+|l| and t2 = l|k|+|l|

(k, l) ∈ Z2 \ 0, 0⇒ c(|k|+ |l|) ≤ |kw1 + lw2| ≤ C(|k|+ |l|)

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Hence,∑

(k,l)∈Z2\0,0 1|kw1+lw2|

t

<∞

⇔∞∑q=1

∑|k|+|l|=1

1

qt<∞

⇔∞∑q=1

4

qt−1<∞⇔ t > 2

Theorem 5.5. ℘ is elliptic with periods Λ Moreover, ℘ is an even function. ℘is 2 to 1 on a fundamental/period parallelogram.

Proof. Let |z| < r and |w| > 2r.∣∣∣∣ 1

(z − w)2− 1

w2

∣∣∣∣ =

∣∣∣∣ 2wz − z2

(w − z)2w2

∣∣∣∣≤ 2r|w|+ r2

|w|4/4︸ ︷︷ ︸|z|<r, and |w−z|> |w|2

So then,

℘(z) =1

z2+

∑w∈Λ,0<|w|<2r

(1

(z − w)2− 1

w2

)+∑|w|≥2r

(1

(z − w)2− 1

w2

)︸ ︷︷ ︸converges uniformly for |z| < r

So indeed ℘ is meromorphic.Tirvially ℘ is even.By uniform convergence, we can differentiate termwise to see,

℘′(z) =∑w∈Λ

(−2)

(z − w)2

so clearly we have,℘′(z + w) = ℘′(z)

for all w ∈ ΛSo for fixed w ∈ Λ, ℘(z+w)−℘(z) = c is a constant (independent of z). Replacez by −z − w in the last line.

℘(z + w)− ℘(z) = c = ℘(−z) = −℘(−z − w) = −c

⇒ c = 0 So Λ is contained in the periods of ℘ and since ℘ only has roots on Λ,these are all the periods.In a fundamental parallelogram, ℘ has a double pole.So by the valency theorem, this implies that ℘ defines a 2 to 1 map C/Λ→ C∞

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Remark. ℘ is characterised by having periods Λ, poles only on Λ and ℘(z) −1

z2→ 0 as z → 0

Why? Let f be another meromorphic function. Then f − ℘ is periodic with atworst simple poles...It’s constant(deg ≥ 2)Now considering

f(z)− ℘(z) =

(f(z)− 1

z2

)−(℘(z)− 1

z2

)We see f − ℘ ≡ 0.The function ℘′ has triple poles on Λ and valency 3. It is clearly odd and

℘′(w1

z

)=↑

periodicity

℘′(−w1

z

)= −℘′

(w1

z

)

⇒ w1

z is a zero of ℘′/Similarly, w2

2 and w1+w2

2 are zeroes of ℘′(z).Therefore the analytic map ℘ : C/Λ→ C∞ is ramified at the points 0, w1

2 ,w2

2 ,w1+w2

2and it has branch points

∞, ℘(w1

2

)=e1

℘(w2

2

)=e3

℘

(w1 + w2

2

)=e2

Check: 4 branch points is compatible with degree-genusRiemann-Hurwitz

formula

Note: The points e1, e2 and e3 are all distinct because otherwise ℘ would havevalency at least 4.

Proposition 5.6. There is an algebraic relation

(℘′(z))2

= 4 (℘(z))3 − g2℘(z)− g3

for constant g2, g3 ∈ C depending only on Λ.

Proof. Using

℘(z) =1

z2+

∑w∈Λ\0

(1

(z − w)2− 1

w2

), the Laurent expansion of ℘(z) over 0 has

℘(z) =1

z2+ az2 +O(z4)

So

℘′(z) =−2

z3+ 2az +O(z3)

(℘′(z))2 − 4(℘(z))3 =−g2

z2+ (locally holomorphic function)

⇒ ℘′(z)− 4℘(z) + g2℘(z) is constant by periodicity. (Liouville’s theorem)

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Remark. Since ℘′(z) = 0 if z =w1

2,w2

2,w1 + w2

2,

we find that the cubic equation,

4x3 − g2x− g3 = 4(x− e1)(x− e2)(x− e3)

(℘′(z))2 = 4(℘(z)− e1)(℘(z)− e2)(℘(z)− e3)

In particular, note that e1 + e2 + e3 = 0

Remark. Another interpretation of Proposition 5.6.Consider the map (℘(z), ℘′(z)) as defining an analytic map.

F : (C/Λ) \ 0 → C2

The image of F lies on the cubic curve, y2 = 4x3 − g2x− g3.This can be compactified (cf. Example Sheet 1 Q15) to a curve in P3(C) definedby the homogeneous equation

y2z = 4x3 − g2xz2 − g3z

3

[x : y : z] coordinate on P3(z)

Last Task: We want to understand all elliptic functions on C/Λ

Theorem 5.7. Let f be an elliptic function with period Λ. Then there arerational functions on Q1, Q2.

f = Q1(℘) + ℘′Q2(℘)

In particular, if f is even then you can take Q2 ≡ 0

Proof. We first treat the case when f is assumed to be even.

Let E =

t ∈ C|z ∈ 1

2Λ or f ′(z) = 0

Then f(E) is finite, so we can choose distinct , d ∈ C \ f(E).

Let g(z) =f(z)− cf(z)− d

so at poles of f , g(z)→ 1

Then g is elliptic and its zeroes,(where f(z) = c) and poles (f(z) = d) are allsimple as f ′(z) = 0 at each point and g is even since f is.Let us list the zeroes and poles of g as a1, ..., am,−a1, ...,−am has zeroes(since f(aj) = c ⇒ aj 6∈ 1

2Λ,aj = −aj)Similarly list the poles of g as b1, ..., bn,−b1, ...,−bn.Let

h(z) =(℘(z)− ℘(a1))(℘(z)− ℘(a2))...(℘(z)− ℘(am))

(℘(z)− ℘(b1))...(℘(z)− ℘(bn))

Then h is elliptic with the same zeroes and poles as g, all of which are simple.

Sog

his an elliptic function without poles so is constant A.

g

h= A⇒ Ah =

f(z)− cf(z)− d

33

⇒Algebraic

Rearrangement

f = Q1(℘)

If f is odd then f/℘′(z) is an elliptic function s.t. f/℘′(z) = Q2(z) and hencef = ℘′ ·Q2 In general we can write

f(z) =f(z) + f(−z)

2+f(z)− f(−z)

2

and apply the previous cases.

Remark. The upshot of the previous results is that the field of meromorphicfunctions on C/Λ is just the field of fractions

C[x, y]/(y2 = 4x3 − g2x− g3)

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6 The Uniformization Theorem

Definition. A group of homeomorphism of a topological group space X is saidto act properly discontinuously if every x ∈ X has an open neighbourhood Us.t. all g(U) for g ∈ G are pairwise disjoint.

Consider the quotient map π : X → X/G. Continuous wrt the quotienttopology.

Given x ∈ X and a neighbourhood U as above π−1(πU) =∐g∈G gU ⇒ πU

is a bijection and hence a homeomorphism. Thus if X is path-connected, the πis a regular covering.

Remark. If X is a Riemann Surface and G ⊂ Aut(X) is a subgroup of ana-lytic automorphisms automorphisms acting discontinuously, then X/G has thestructure of a Riemann Surface; we may assume that (U, φ) is a chart with Uas above then φ (π|U )−1 is the region contained on X/G

This is a conformal strucutre: Transition functions are analytic since G ⊂Aut(X).

Example. Λ ⊂ C is a lattice by translations ⇒ C/Λ is a Riemann Surface.

Note: If X → X is a regular covering, the group G of ”deck transformations”acts properly and discontinuously and X = X/G

6.1 Uniformization Theorem

1. (Hard) Any simply connected Riemann Surface is conformally equivalentto one of C∞,C and ∆.

Note that C homeomorphic to ∆ but not conformally equivalent by Liou-ville.

2. Easier Any Riemann Surface R is conformally equivalent to R/G for R toone of the Riemann Surface in (1) and G ⊂ Aut(R), a group of analyticautomorphism acting properly and discontinuously. We say that R isuniformized by R

Example. S2 has only one confromal structure; if R is homeomorphic to S2

it is simply connected and hence by (1) is conformally equivalent to C∞ Nowconsider the 3 possibilities for R in turn

(i) R = C∞ Aut(C∞) consists of Mobius transformations z 7→ az+bcz+d ad−bc 6=

0 and they have 1 or 2 fixed points ⇒ if G acts properly discontinuously⇒ G = id ⇒ R = C∞

(ii) R = C , Aut(C) consists of maps of the form az + b, a ∈ C∗. If a 6= 1,∃ a fixed point so G consists of translations and so G → C g 7→ g(0).Moreover G consists only of isolated points(Check!). Example Sheet 3Q1⇒ G = id. Zw1 on λ lattice. No R is conformally equivalent toC,C∗ on C/Λ(complex torus)

35

(iii) R = ∆, R is not conformally equivalent to R,C,C∗. Indeed suppose itis. Get f and using the Monodromy theorem and Liouville, f is constant.Hence any Riemann Surface which is not C∞,C,C∗ or C/Λ is uniformizedby ∆

Recall: Aut(∆) consists of Mobius transformations z 7→ eiθz − a1− az

where a ∈∆ Large literature concerning G ⊂ Aut(∆) acting properly discontinuouslyFuschian GroupsConsequences:

(a) Compact Riemann Surfaces of genus 1 are uniformized by ∆

(b) Riemann Mapping Theorem: If U 6⊂ C domain and simply connected,then it is conformally equivalent to ∆. Just need to show 6 ∃ conformalequivalence f : C→ U . This is the standard arugement saying that f can’thave an essential singularity. If it did the Casorati-Weierstrass contradictsthe injectivity of f .

Hence f extends to an analytic map f : C∞ → f(C∞) ⊂ C∞. This ishowever not surjective.

(c) If a Riemann Surface R is uniformized by ∆ ,∃ a hyperbolic metric on Rsince Aut(∆) consists of isometries of the hyperbolic metric on ∆.Hence if R is compact, Guass-Bonnet⇒ χ(R) < 0 so a compact surface ofgenus 1 must be uniformized by C.⇒ It is a complex torus C/Λ!

(d) Picard Theorem If f : C→ C \ 0, 1 analytic, f is constant.

Proof. Example Sheet 3 Q9⇒ C\0, 1 is uniformized by ∆. MonodromyArgument f lifts up to f (cf. Example Sheet 2 Q3)Hence by Liouville, f is constant.⇒ f is constant.

Any abstract Reimann Surface can be realised as an algebraic object somewhere.

36