Graphs on Surfaces and Arithmetical Applications

Iulia - Catalina Plesca

Chapter 0

Introduction

0.1 Summary

In the present report, I try to present some topics suggested by Belyi’s Theorem: Riemann surfaces,algebraic curves, covering spaces, dessins d’enfants.I start the first chapter, Riemann Surfaces, by presenting some elements of complex analysis inseveral variables: holomorphy, analyticity, Cauchy - Riemann equations, the inverse function theo-rem, and the implicit function theorem. The results in this Section are needed in order to introducethe concepts in the following Section, Complex Manifolds. Here, I give some examples of manifolds:the Riemann sphere, the real sphere and the Poincare half-plane. In order to introduce a particularlyimportant example, the complex projective space, I show the necessary results related to quotientmanifolds. After this discussion in general context, I define Riemann surfaces as unidimensionalcomplex manifolds. I present concepts related to morphisms between Riemann surfaces. Inthe second chapter, Covering Spaces, I introduce unramified coverings in the general case oftopological spaces and gave some examples of coverings of the circle. I include results linked to pathsand coverings in order to introduce universal coverings that allow to characterize for coverings.I continue then with presenting ramified coverings.In the third chapter, Belyi’s Theorem, I presented the statement of the theorem for Riemannsurfaces. I presented the proof of the direct implication, starting with an example.The Appendix on graph theory introduces some concepts needed in the following chapter. Thechapter Dessins d’enfants is the one that gives the name of the paper. I begin with presentinggraphs on surfaces and give some examples, introducing dessins d’enfants. Next, I presented thepermutation representation of a dessin which gives a correspondence between dessins and tran-sitive permutation groups with two generators. Then, I continued with another correspondence, theone between dessins and Belyi function (meromorphic functions with at most three critical values).Using the triangle decomposition, I presented the construction of the Belyi function associated ofa dessin. Also, using the proprieties of ramified coverings, I showed the construction of the dessinassociated to a Belyi function.

0.2 Short History of the Topic

Gennadii Vladimirovich Belyi proved in 1979 the direct implication of what is known today as Belyi’sTheorem. His two proofs are relatively elementary and and had a profound impact on AlexanderGrothendieck (1928 - 2014).Starting from this result, Grothendieck established in Esquisee d’un Programme in 1984 a series ofideas for mathematical research. Among others, he suggested to study the Galois group Gal(Q/Q).On this note, he introduced dessins d’enfants and the correspondence between these and transitivegroups of permutation with three generators. He also presented the correspondence between dessinsand what is known today as Bely functions [9].The first examples of dessins had appeared in William Rowan Hamilton’s work (1805 - 1865) regard-ing Hamiltonian circuits in icosahedral graphs. He had studied the permutation representation of

1

this graph and in general of regular polyhedra. The general algorithm for this correspondence wasdeveloped simultaneously by Jean Malgoire and Claire Voisin on one side and by Gareth Jones andDavid Singerman on the other [13]. Other early examples belong to Felix Klein (1849 - 1925).Belyi’s Theorem was stated in the context of algebraic curves, but it can also be stated for Riemannsurfaces. Thus, Belyi’s Theorem unifies many topics. Recently, a version for complex surfaces, i.e.2-dimensional complex manifolds has been proven [6]. Furthermore, there are some versions of Belyi’sTheorem in characteristic p [5]. A current problem is to compute the Belyi functions for particulardessins. At this moment, there is no complete and coherent algorithm for these computations [24].In this thesis, I tried to give the necessary theoretical background in order to understand the currentresearch.

0.3 Notations

We shall use the following notations throughout the text:

• If V is a topological space, by U ⊂d V we mean that U is an open subset of V .

• If n ∈ N and a ∈ Cn (or Rn), we denote by aj ∈ C (or R), j = 1, n, the complex (real) coordinatesof a, i.e. a = (a1, a2, . . . , an).

• If m,n ∈ N and f : Ω ⊂ Cm → Cn, we denote by fj : Ω→ C (j = 1, n) the coordinates of f , i.e.f = (f1, f2, . . . , fn).

• If n ∈ N, we denote by Idn the n× n identity matrix: Idn ∈Mn×n(C).

• If A is a set, we denote by |A| its cardinal.

• If a ∈ C, r ∈ R∗+, we denote by C(a, r) = z ∈ C||z − a| = r.

• If r1 < r2 ∈ R+, we denote the annulus of radii r1 and r2 by D(r1; r2) = (r, ϕ)|r1 < r < r2, ϕ ∈[0, 2π) (we have taken polar coordinates) which can be seen as a subset of C.

• If n ∈ N∗, we denote by Sn the permutation group of order n.

2

Contents

0 Introduction 10.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Short History of the Topic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Riemann Surfaces 51.1 Complex Analysis in Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Complex Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Topological Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 The Definition of a Complex Manifold . . . . . . . . . . . . . . . . . . . . . . . 101.2.3 Quotient Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Compact Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.1 Morphisms between Riemann Surfaces . . . . . . . . . . . . . . . . . . . . . . . 14

2 Covering Spaces 182.1 Unramified Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.1.1 Examples and Properties of Covering Spaces . . . . . . . . . . . . . . . . . . . 182.1.2 Paths and covering spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.3 Universal Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2 Ramified Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Belyi’s Theorem 273.1 Statement of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 The Direct Implication of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.2.1 An Example of Belyi Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2.2 The Actual Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 Dessins d’enfants 314.1 Graphs on Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 The Permutation Representation of a Dessin . . . . . . . . . . . . . . . . . . . . . . . 324.3 The Correspondence between Dessins and Belyi Function . . . . . . . . . . . . . . . . 34

4.3.1 The Belyi Function Associated to a Dessin . . . . . . . . . . . . . . . . . . . . 344.3.2 The Dessin Associated to a Belyi Function . . . . . . . . . . . . . . . . . . . . 36

A Graph Theory Concepts 38

3

List of Figures

1.1 The torus, surface of genus 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2 The change of coordinates between two local charts . . . . . . . . . . . . . . . . . . . . 111.3 The real sphere as complex manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1 The covering of S1 with R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.2 The Covering of S1 with S1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 The unramified covering of the annulus D(rn1 ; rn2 ) by the annulus D(r1; r2) . . . . . . . 24

4.1 Example of map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.2 Counterexample of map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 K4 embedded in the torus and in the sphere . . . . . . . . . . . . . . . . . . . . . . . . 324.4 Tree embedded in the sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.5 Example of a Dessin from a Tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . 334.6 The Permutation Representation of a Dessin . . . . . . . . . . . . . . . . . . . . . . . 334.7 The Correspondence between the Product σbσa and the Faces of the Dessin . . . . . . 344.8 The Triangle Decomposition of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4

Chapter 1

Riemann Surfaces

1.1 Complex Analysis in Several Variables

In order to introduce complex manifolds, we need some results concerning functions of several complexvariables.

Definition 1.1. 1. [11] Let n ∈ N∗ and Ω ⊂d Cn. Let f : Ω ⊂d Cn → C. The function f is said tobe holomorphic on Ω if

• f is continuous on Ω and

• ∀j = 1, n and ∀z1, z2, . . . , zj−1, zj+1, . . . , zn ∈ C, the function φj : Ωj → C, given by:φj(ζ) = f(z1, z2, . . . , zj−1, ζ, zj+1, . . . , zn, is holomorphic on Ωj , where

Ωj = ζ ∈ C|(z1, z2, . . . , zj−1, ζ, zj+1, . . . , zn) ∈ Ω.

2. Let f : Ω ⊂d Cn → Cm. The function f said to be holomorphic on Ω if ∀k ∈ ←−−1,m the k-thcomponent of f , fk, is holomorphic on Ω.

Remark 1.1. • The continuity condition in Definition 1.1 is required in order to simplify someproofs, but Hartogs’ Theorem shows it can be left out [8].

• The second condition in the Definition of holomorphy requires each f to be holomorphic in eachvariable.

Example 1.1. Let f : C2 → C2, f(z1, z2) = (f1, f2), f1 = z1z2 and f2 = z21 + z2

2 .f1 and f2 are polynomials, therefore holomorphic in each variable ⇒ f holomorphic.

Counterexample 1.1. Let f : C2 → C2, f(z1, z2) = (f1, f2), f1 = z1z2, f2 = z1.f2 seen as function in z2 is not holomorphic, because it does not check the Cauchy - Riemann conditionon one variable, therefore f is not holomorphic.

Definition 1.2. [8] Let n ∈ N∗ and Ω ⊂d Cn and f : Ω→ C. The function f is said to be analyticalon Ω if ∀a ∈ Ω,∃U ⊂d Ω, a ∈ U such that f has a power series expansion:

f(z) =∞∑

v1,v2,...,vn=0

av1,v2,...,vn(z − a1)v1(z − a2)v2 . . . (z − an)vn

which converges ∀z ∈ U .

Proposition 1.1. Let n ∈ N∗ and Ω ⊂d Cn. If f : Ω → C is analytical on Ω, then f is holomorphicon Ω.

Proof.

S =∞∑

v1,v2,...,vn=0

av1,v2,...,vn(z − a1)v1(z − a2)v2 . . . (z − an)vn .

5

From the definition of analytical functions, a series of the form S is absolutely convergent on opensets U , neighbourhoods for each z ∈ Ω. It follows that f is continuous on Ω.Furthermore, anyway we rearrange the terms of S, it stays absolutely and uniformly convergent.Therefore, if we fix all coordinates of z without zj : x1, x2, . . . , xj−1, xj+1, . . . , xn, the terms of theseries in the expression of S can be rearranged such that S is an absolutely uniformly convergentseries on a neighbourhood V ∈ Ωj in zj . From the case of a single variable, it follows that φj : Ωj → Cis holomorphic on Ωj , ∀j = 1, n thus f is holomorphic.

Definition 1.3. Let n ∈ N∗, r ∈ Rn+ and z0 ∈ Cn. The open polydisc of centre z0 and radius r isdefined as:

D(z0, r) := z ∈ Cn||zj − z0j | < rj , j = 1, n

and the closed polydisc of centre z0 and radius r as:

B(z0, r) := z ∈ Cn||zj − z0j | ≤ rj , j = 1, n

Proposition 1.2. (Osgood’s Lemma)[8] Let n ∈ N∗ and Ω ⊂d Cn. If f : Ω → C is holomorphic,then it is analytical on Ω.

Proof. Let a ∈ Ω and B(a, r) ⊂ Ω be a closed polydisc. Because f is holomorphic in each variablein an open neighbourhood of B(a, r), we can apply Cauchy’s Integral Formula for one variable. Letz ∈ B(a, r). Let us consider the one variable function: g1(w) = f(w, z2, . . . , zn) and apply Cauchy’sIntegral Formula:

f(z1, z2, . . . , zn) = g1(w) =1

2πi

∫C(a1,r1)

g1(w)

w − z1dw =

1

2πi

∫C(a1,r1)

f(ζ1, z2)

ζ1 − z1dζ1.

For ζ1 ∈ C(a1, r1), function g2(w) = f(ζ1, w) has a complex derivative in a neighbourhood of B(a2, r2)in C. Applying, we get:

f(ζ1, z2, . . . , zn) = g2(w) =1

2πi

∫C(a2,r2)

g2(w)

w − z2dw =

1

2πi

∫C(a2,r2)

f(ζ1, ζ2)

ζ2 − z2dζ2.

After applying it repeatedly:

f(z) =

(1

2πi

)n ∫C(a1,r1)

dζ1

ζ1 − z1. . .

∫C(an,rn)

f(ζ)dζnζn − zn

f continuous, we apply Fubini’s formula==========================

(1

2πi

)n ∫∂B(a,r)

f(ζ)dζ1 . . . dζn(ζ1 − z1) . . . (ζn − zn)

, ∀z ∈ B(a, r).

Let us try to obtain a series for the term under the integral, for each j = 1, n:

1

ζj − zj=

1

ζj − aj − (zj − aj)=

1

ζj − aj1

1− zj−ajζj−aj

=∞∑vj=0

(zj − aj)vj(ζj − aj)vj+1 .

Multiplying these, we get:

1

(ζ1 − z1) . . . (ζn − zn)=

∞∑v1,...,vn=0

(z1 − a1)v1 . . . (zn − an)vn

(ζ1 − a1)v1+1 . . . (ζn − an)vn+1.

The series in the right-hand side converges uniformly and absolutely in ζ and z. Substituting in theprevious formula, we get:

f(z) =

(1

2πi

)n ∫|w−ζ|=r

∞∑v1,...,vn=0

(z1 − a1)v1 . . . (zn − an)vn

(ζ1 − a1)v1+1 . . . (ζn − an)vn+1f(ζ)dζ1 . . . dζn

the integral commutes===============

with the series

∞∑v1,...,vn=0

((1

2πi

)n ∫|w−ζ|=r

f(ζ)dζ1 . . . dζn(ζ1 − a1)v1+1 . . . (ζn − an)vn+1

)(z1 − a1)v1 . . . (zn − an)vn .

We get a series for the function f , and it follows that f analytical, which concludes the proof.

6

Remark 1.2. Osgood’s lemma gives a generalization of the Cauchy - Riemann conditions. Let n ∈ N∗,Ω ⊂d C and f : Ω→ C. For z ∈ Ω, we denote with xj , respectively yj the real part and the imaginarypart of zj : zj = xj + iyj . Introduce the following differential operators: ∂

∂zjand ∂

∂zj. We require

the following conditions: ∂∂zjzj = 1, ∂

∂zjzj = 1, ∂

∂zjzj = 0 and ∂

∂zjzj = 0 and we get the following

expressions:∂

∂zj=

1

2

(∂

∂xj− i ∂

∂yj

)and

∂

∂zj=

1

2

(∂

∂xj+ i

∂

∂yj

).

Theorem 1.1. [8](Cauchy - Riemann Conditions) Let f : Ω ⊂d Cn → C such that it is contin-uously differentiable in the real coordinates of Cn. The function f is analytical on Ω if and only if itverifies the system:

∂

∂zjf(z) = 0, j = 1, n. (1.1)

Proof. Consider the following decomposition for the function f : f = u + iv, where u is the real partand v is the imaginary part. On the other hand, apply the differential operators:

∂f(z)

∂zj=

(∂u

∂xj− ∂v

∂yj

)− i(∂u

∂yj+

∂v

∂xj

).

Equation 1.1 becomes equivalent with Cauchy - Riemann equations with holomorphy in each variable.From Proposition 1.1 and 1.2’s Theorem, it follows that the concepts of holomorphic and analyticalfunction are equivalent; in this way, we get that the equations 1.1 are equivalent with the analyticitywith f .

The Cauchy - Riemann Conditions allow us to prove that the composition of two holomorphicfunctions is holomorphic which is a very important result when studying complex manifolds.

Proposition 1.3. [8] Let m,n ∈ N∗, Ω1 ⊂d Cn and Ω2 ⊂d Cm. Let f : Ω1 → Ω2 and g : Ω2 → C. Iff and g are holomorphic on Ω1 respectively Ω2, then the composition g f : Ω1 → C is holomorphicΩ1.

Proof. Consider the real and imaginary part of the coordinate functions of f : fj = uj + ivj , ∀j = 1, n.We denote by z = (z1, z2, . . . , zn) the coordinates in D and with w = (w1, . . . , wm) the coordinates inD′. Since the function are differentiable, we can derive the composition of functions:

∂g(f(z))

∂zj=

m∑k=1

(∂g

∂uk

∂uk∂zj

+∂g

∂vk

∂vk∂zj

)=

m∑k=1

(∂g

∂uk

∂(fk + fk)

2∂zj− i ∂g

∂vk

∂(fk − fk)2∂zj

)=

m∑k=1

(1

2

(∂g

∂uk− i ∂g

∂vk

)∂fk∂zj

+1

2

(∂g

∂uk+ i

∂g

∂vk

)∂fk∂zj

)=

m∑k=1

(∂g

∂wk

∂fk∂zj

+∂g

∂wk

∂fk∂zj

)f holomorphic⇒ ∂fk

∂zj=0; g holomorphic⇒ ∂gk

∂wk=0

================================ 0,

which by the Cauchy - Riemann conditions gives the holomorphy of the composition.

Remark 1.3. In the proof above, we obtained the expression for the derivative for g f :

∂g(f(z))

∂zj=

m∑k=1

(∂g

∂wk

∂fk∂zj

+∂g

∂wk

∂fk∂zj

)Theorem 1.2. [3][7](Inverse function theorem)

Let U, V ⊂d Cn such that 0 ∈ U . Let f : U → V be a holomorphic function. If J (f) = (∂fj∂zk

)1≤j,k≤nis non-singular at 0, then ∃U0 neighbourhood a lui 0 such that f |U0 : U0 → f(U0) bijection and f−1

holomorphic in f(0).

7

Proof. Let fj = uj + ivj , 1 ≤ j ≤ n, uj , vj : Cn → R and zk = xk + iyk, 1 ≤ k ≤ n. Because f isholomorphic, fj are holomorphic and it follows that uj , vj are smooth real functions.

Let us make the following notation for the real Jacobian f : det |JR(f)| =∣∣∣∣A C

B D

∣∣∣∣, where A =∂uj∂xk

,

B =∂vj∂xk

, C =∂uj∂yk

and D =∂vj∂yk

.

From the Cauchy - Riemann Conditions applied to the functions fj , it follows that∂uj∂xk

=∂vj∂yk

and

∂uj∂yk

= − ∂vj∂xk

, therefore det |JR(f)| =∣∣∣∣A −BB A

∣∣∣∣ = A2 +B2.

∂fj∂zk

= 12

(∂∂xk− i ∂

∂yk

)(uj + ivj) = 1

2

((∂uj∂xk

+∂vj∂yk

)+ i(∂vj∂xk− ∂uj

∂yk

))Conditiile Cauchy - Riemann

===================∂uj∂xk

+

i∂vj∂xk

= Ajk + iBjk ⇒ J (f) = A+ iB.

Since J (f) is non-singular in 0⇒ det(J (f))|0 6= 0⇒ det(JR(f))|0 > 0Teorema de inversare locala ın cazul real========================⇒

f has a real differential inverse around 0, therefore:

f−1(f(z)) = z ⇒ ∂f−1(f(z))

∂z= 0

Observatia de mai sus=============⇒

n∑k=1

(∂f−1

l

∂zk

∂fk∂zj

+∂f−1

l

∂zk

∂fk∂zj

)= 0

f holomorphic========⇒

n∑k=1

(∂f−1

l

∂zk

(∂fk∂zj

))= 0, ∀j, l.

Since (∂fk∂zj) 6= 0⇒ ∂f−1

l∂zk

= 0, ∀j, k, therefore f−1 is holomorphic in a neighbourhood of f(0).

Theorem 1.3. (The Implicit Function Theorem)[7] Let n, l ∈ N∗ and f : Cn → Cl holomorphic.If

det

(∂fj∂zk

(0)

)1≤j,k≤l

6= 0,

then ∃w : Cn−l → Ck such that on a neighbourhood of 0 ∈ Cn:

f1(z) = · · · = fl(z) = 0⇔ zr = wr(zl+1, . . . , zn), ∀r ∈ 1, l. (1.2)

Proof. From the Implicit Function Theorem in the real case, we get the functions w1, . . . , wl that check1.2. Only their holomorphy needs to be proven. We want to check the Cauchy - Riemann Conditions.Applying the differential operators ∂

∂zαwhere α ∈ l + 1, n, we get that:

0 =∂

∂zα(fj(w1(zl+1, . . . , zn), . . . , wl(zl+1, . . . , zn), zl+1, . . . , zn))⇒

0 =∂fj(w1(zl+1, . . . , zn), . . . , wl(zl+1, . . . , zn), zl+1, . . . , zn)

∂zα+

l∑k=1

∂fj(w1(zl+1, . . . , zn), . . . , wl(zl+1, . . . , zn), zl+1, . . . , zn)

∂wk(zl+1, . . . , zn)· ∂wk(zl+1, . . . , zn)

∂zα+

l∑k=1

∂fj(w1(zl+1, . . . , zn), . . . , wl(zl+1, . . . , zn), zl+1, . . . , zn)

∂wk(zl+1, . . . , zn)· ∂wk(zl+1, . . . , zn)

∂zα

fholomorphic========⇒

0 =

l∑k=1

∂fj(w1(zl+1, . . . , zn), . . . , wl(zl+1, . . . , zn), zl+1, . . . , zn)

∂wk(zl+1, . . . , zn)· ∂wk(zl+1, . . . , zn)

∂zα.

Since det(∂fj∂zk

(0))

1≤j,k≤l6= 0, it follows that

∂wk(zl+1,...,zn)∂zα

, therefore the functions wr are holomorphic,

∀r ∈ 1, l.

Proposition 1.4. [7] Let n ∈ N∗ and U, V ⊂d Cn. If f : U → V is a holomorphic, bijective function,then f−1 is holomorphic.

8

Proof. We want to check the conditions in the hypothesis of Theorem 1.2, that is to show thatJ (f) 6= 0.We shall prove this fact by induction on n.For n = 1, from the holomorphy of f it follows that ∂f

∂z 6= 0⇒ J (f) 6= 0.The inductive step: assume that ∀m ∈ N∗, ∀U, V ⊂d Cm, if f : U → V is a holomorphic, bijectivefunction, then f−1 is holomorphic.Let Z,W ⊂d Cn and F : Z →W be a holomorphic function. Denote with z the coordinates on Z andwith w the coordinates on W . Assume, without loss of generality, that 0 ∈ Z. Let rk(J (f))|z=0 =

k ∈ N. Assume, also without loss of generality, that(∂fj∂zl

)1≤j,l≤k

|z=0 6= 0.

By Theorem 1.3, we can find the following system of coordinates:z′j = fj(z), j ∈ 1, k

z′α = zα, α ∈ k + 1, n.

Seeing f as a function in the variable z′, f sends z′1 = · · · = z′k = 0 to w1 = · · · = wk. Furthermore,(∂fα∂z′β

)|(z′1=···=z′k=0) is singular. Applying the inductive hypothesis, it follows that k = n (which means

that det(J (f)) 6= 0 and concludes the proof) or k = 0. If k = 0⇒ ∂fj∂zl

= 0,∀j, l ∈ 1, n⇒ f is constanton each connected component of U . Since U is open it follows that each connected component of U isopen, therefore it has a positive Lebesgue measure. The fact that f is bijective gives a contradiction.Therefore, indeed |J (f)| 6= 0.Now, using Theorem 1.2, it follows that f admits locally a holomorphic inverse and since the inverseof a function is unique, it follows that f−1 is holomorphic.

1.2 Complex Manifolds

1.2.1 Topological Manifolds

Definition 1.4. [20] Let M be a topological space. M is said to be an m-dimensional localeuclidean space (m ∈ N∗) if ∀P ∈ M , ∃U ⊂d M , P ∈ M and ϕ : U → Rm such that ϕ(U) ⊂ Rmand ϕ : U → ϕ(U) homeomorphism.

Definition 1.5. [20] Let M be a topological space. M is an m-dimensional topological manifoldif

• M is an m-dimensional local euclidean space,

• M is Hausdorff

• M has a countable base of the topology.

Definition 1.6. A topological surface is a 2-dimensional topological manifold.A differentiable surface is a 2-dimensional differentiable manifold.

Definition 1.7. A differentiable manifold is said to be orientable if ∃A = (Uα, ϕα)|α ∈ A an atlassuch that for any two maps (Uα, ϕα), (Uβ, ϕβ) ∈ A with Uα ∩ Uβ 6= ∅ the Jacobian of the change ofcoordinates is strictly positive: detJ (ϕβ ϕ−1

α ) > 0.

Definition 1.8. [10] Let X be a 2-dimensional differentiable orientable, connected manifold. Thegenus of X is the natural number representing the maximal number of cuts along simple, closedcurves such that the manifold stays connected.

Example 1.2. The real sphere

S2 = (x, y, t) ∈ R3|x2 + y2 + t2 = 1

has genus 0.

9

Figure 1.1: The torus, surface of genus 1

Example 1.3. The torus is a rotation surface generated by the rotation of a circle in space on an axisthat is coplanar with the circle. Therefore, the torus T can be seen as the product of two circles:T ' S1 × S1.We can cut along a ”small” circle, without loosing connectedness. The torus has genus 1.

Remark 1.4. [10] The concept of genus can be visualised intuitively as the number of ”holes” of asurface. This vision can be formalised as follows. Each connected closed surface, can be obtained asthe connected sum S2 and g tori T = S1 × S1 where g is the genus of the surface.

Definition 1.9. [4] Let X be a differentiable oriented surface (i.e. X an orientable surface on whichwe chose an orientation). Let γ be a Jordan curve, differentiable around a point P of X, that isγ = ∂Ω, where Ω is diffeomorphic with a disc. Then the orientation of X induces an orientation on γ.Define an orientation on γ as follows: let Q ∈ ∂Ω and (U,ϕ) be a local chart with ϕ(U) = D(0, 1) ⊂ R2

and let ϕ(Q) = 0 such that ϕ(U ∩ Ω) = H ∩ D(0, 1). Choose the positive orientation as the onethat ”travels” on U ∩ ∂Ω from ϕ−1(−ε) to ϕ−1(ε).

1.2.2 The Definition of a Complex Manifold

Definition 1.10. [3] Let m ∈ N∗. Let M be a 2m dimensional topological manifold. We call localcomplex chart on M (chart on M) a pair (U,ϕ) where U ⊂d M and ϕ : U → Cm such thatϕ : U → ϕ(U) is a homeomorphism.Let P ∈M . We say that a complex local chart on M , (U,ϕ), is centred in P if P ∈ U and ϕ(U) = 0.We say that two complex local charts on M , (Uα, ϕα) and (Uβ, ϕβ), are compatible if either Uα∩Uβ =∅, either Uα ∩ Uβ 6= ∅ and fβα := ϕβ ϕ−1

α : ϕα(Uα ∩ Uβ)→ ϕβ(Uα ∩ Uβ) is holomorphic. We call thefunction fβα change of coordinates.

Remark 1.5. If two charts are compatible, it follows from Proposition 1.4, that the change of coordi-nates, if it exists, is actually biholomorphic.

Definition 1.11. [3] A holomorphic atlas on M is a collection A = (Uα, ϕα)α∈A of charts on Msuch that

⋃α∈A Uα = M and ∀α, β ∈ A, the charts (Uα, ϕα) and (Uβ, ϕβ) are compatible.

A holomorphic atlas is maximal if ∀(U,ϕ) local chart on M such that (U,ϕ) compatible with(Uα, ϕα), ∀α ∈ A, then (U,ϕ) ∈ A.

Proposition 1.5. A holomorphic atlas determines a unique maximal holomorphic atlas.

Proof. Let A = (Uα, ϕα)α∈A be an atlas and define Ac = (Vβ, φβ)β∈B|(Vβ, φβ) compatibila with(Uα, ϕα), ∀α ∈ A. Let us observe that A ⊂ Ac ⇒

⋃β∈B Vβ = M .

Let (Vβ1 , φβ1), (Vβ2 , φβ2) ∈ Ac with Vβ1 ∩ Vβ2 6= ∅. We want to show that they are are compatible.The change of coordinates is given by fβ2β1 := ϕβ2 ϕ

−1β1

: ϕβ1(Vβ1 ∩ Vβ2) → ϕβ2(Vβ1 ∩ Vβ2). Let

P ∈ (Vβ1 ∩ Vβ2)

⋃α∈A Uα=M

========⇒ ∃α0 ∈ A : P ∈ Uα0 . Let Y = Vβ1 ∩ Vβ2 ∩ Uα0 . It follows thatP ∈ Y . On Y , we can write the change of coordinates as a composition of holomorphic functions:

10

Figure 1.2: The change of coordinates between two local charts

fβ2β1 := (ϕβ2 φ−1α ) (φα ϕ−1

β1) ⇒ fβ2β1 holomorphic on W ⇒ fβ2β1 holomorphic around each

P ∈ Vβ1 ∩ Vβ2 ⇒ fβ2β1 holomorphic in Vβ1 ∩ Vβ2 .

Let (X, τ) local chart on M compatible with all charts in Ac A⊂Ac

===⇒ (X, τ) local chart on M compatiblewith all the charts in A ⇒ (X,ψ) ∈ Ac. We get that A is a maximal atlas.Consider a maximal holomorphic atlas A′ = (Wγ , ψγ)γ∈C such that A ⊂ A′. Let (W,ψ) ∈ A′, thuscompatible with (Uα, zα), ∀α ∈ A ⇒ (W,ψ) ∈ Ac ⇒ A′ ⊂ Ac. On the other hand, A′ is maximal⇒ Ac = A′ which proves the uniqueness.

Definition 1.12. An m-dimensional complex manifold is an m-dimensional topological manifoldM endowed with a maximal atlas.

Example 1.4. The simplest example of a complex manifold is the complex space: Cn. An atlas canbe given with a single local chart: (Cn, Id).We can consider as manifolds examples connected subsets U of Cn. Also, an atlas can be given witha single local chart, given by: (U, Id). Particularly important examples are:

the unit poydisc: D(0, 1) = z ∈ Cn||zi| < 1.

the Poincare half-plane: H = z ∈ C|Im(z) > 0.

Example 1.5. Another example of complex manifold is the 3-dimensional real sphere:

S2 = (x, y, t) ∈ R3|x2 + y2 + t2 = 1

As a difference from the previous cases, a single local chart is not enough because the sphere is nothomeomorphic with the real plane. See the plane t = 0 as C and consider the stereographic projections.Consider the atlas consisting of two charts: A = (UN , zN ), (US , zS).UN = S2 \ (0, 0, 1), zN : UN → C given by: zN (P (xP , yP , tP )) = NP ∩ t = 0.NP : x

xP= y

yP= t−1

tP−1 . Intersecting with t = 0, we obtain: x = xP1−tP and y = yP

1−tP . Simplifying the

notation and going to complex coordinates, we get that: zN (x, y, t) = x1−t + i y

1−t .

Analogously, let US = S2 \ (0, 0,−1), and zS : US → C given by: zS(P (xS , yS , tS)) = SP ∩ t = 0.SP : x

xP= y

yP= t+1

tP+1 . Intersecting with t = 0, we get that: x = xP1+tP

and y = yP1+tP

. Simplifying

the notation and going to complex coordinates, we get: zS(x, y, t) = x1+t − i

y1+t .

UN ∪ US = S2. By the geometrical construction, it follows that: zN : UN → zN (UN ) = C \ 0 and

11

Figure 1.3: The real sphere as complex manifold

zS : US → zS(US) = C \ 0 are homeomorphisms.

Let (x, y, t) ∈ S2. Let us observe that zN (x, y, t)zS(x, y, t) = (x+iy)(x−iy)(1+t)(1−t) = x2+y2

1−t2x2+y2+t2=1

========== 1.

The change of coordinates is given by: zS z−1N : zN (UN ∩ US) = C \ 0 → zS(UN ∩ US) =

C \ 0, zS z−1N (z) = 1

z which is a holomorphic function.We can conclude that S2 is a 1-dimensional complex variety.

Remark 1.6. 1. Identify C with R × R, we can see an m-dimensional complex manifold withthe atlas A = (U,ϕ) as a 2m-dimensional differentiable manifold 2m with the atlas B =(U,Re(ϕ))|(U,ϕ) ∈ A ∪ U, Im(ϕ))|(U,ϕ) ∈ A.

2. Riemann surfaces are orientable.Consider the Jacobian of the change of coordinates fβα = uβα+ ivβα, it follows from the Cauchy- Riemann relations:

∂fβα∂(x, y)

=

∣∣∣∣∣∂uβα∂x

∂vβα∂x

∂uβα∂y

∂vβα∂y

∣∣∣∣∣ conditiile Cauchy - Riemann===================

∣∣∣∣∣ ∂uβα∂x∂vβα∂x

−∂vβα∂x

∂uβα∂x

∣∣∣∣∣ =

(∂uβα∂x

)2

+

(∂vβα∂x

)2

> 0

1.2.3 Quotient Manifolds

These manifolds can be introduced through an equivalence relation. That is why we need sometheoretical results to see in what conditions a factor space can be organized as a variety.

Remark 1.7. Let X be a non-empty topological space and ∼ an equivalence relation on X. Denote by[x] the equivalence class of element x ∈ X. Consider the canonical projection π : X → X/ ∼, givenby π(x) = [x]. Give X/ ∼ the topology induced by projection: D ⊂ X/ ∼ is open in X/ ∼ if π−1(D)is open in X.

Definition 1.13. [20] Let A ⊂ X. The saturated of the set A is the set

[A] := y ∈ X|∃x ∈ A : y ∼ x.

Definition 1.14. [20] An equivalence relation ∼ is open if the following implication holds:

∀D ⊂d X → [D] ⊂d X.

12

Definition 1.15. [20] If ∼ is an equivalence relation, its graph is the set:

G∼ = (x, y)| ∈ X ×X|x ∼ y.

The following results hold:

Proposition 1.6. The projection application π : X → X/ ∼ is an open application if and only if therelation ∼ is open.

Proof. ”⇒” Let π be an open application. Let D ⊂d Xπ open application===========⇒ π(D) ⊂d X/ ∼

definition of the==========⇒quotient topology

π−1(π(D)) ⊂d Xπ−1(π(D))=[D]=========⇒ [D] ⊂d X ⇒∼ open.

”⇐” Let∼ be an open relation. LetD ⊂d X ⇒ [D] ⊂d Xπ−1(π(D))=[D]=========⇒ π−1(π(D)) ⊂d X

definition of the==========⇒quotient topology

π(D) ⊂d X/ ∼ ⇒ π open.

Proposition 1.7. Let ∼ be an open equivalence relation on X. If X admits a countable base of thetopology, then X/ ∼ admits a countable base of the topology.

Proof. Let B = Di|i ∈ N ⊂ N be a base of the topology on X. Let us show that B = π(Di)|Di ∈ Bis a base of the topology on X/ ∼.Let D ⊂d X/ ∼⇒ π−1(D) ⊂d X ⇒ ∃M ⊂ N such that π−1(D) =

⋃j∈N Bj ⇒ π(π−1(D)) =

π(⋃j∈N Bj) ⇒ D =

⋃j∈N Bj . On the other hand |B| ≤ |B|, therefore B countable, which concludes

the proof.

Proposition 1.8. Let X be a Hausdorff topological space and ∼ open equivalence relation on X.X/ ∼ is Hausdorff if G∼ is a closed set.

Proof. Let [x], [y] ∈ X/ ∼ such that [x] 6= [y]⇒ x y ⇒ (x, y) /∈ G∼(X×X)\G∼ open==========⇒ ∃U ⊂d (X ×X)

such that (x, y) ∈ U ⊂ (X × X) \ G∼. Since on X × X we have the product topology, it followsthat ∃Ux ⊂d X, x ∈ Ux, ∃Uy ⊂d X, y ∈ Uy with Ux × Uy ⊂ U , that is (Ux × Uy) ∩ G∼ = ∅ ⇒∀x ∈ Ux,∀y ∈ Uy : (x, y) /∈ G∼ ⇒ ∀x ∈ Ux, ∀y ∈ Uy : [x] 6= [y] ⇒ π(Ux) ∩ π(Uy) = ∅. Butπ open relation ⇒ π(Ux), π(Uy) ⊂d X/ ∼. Since [x] ∈ π(Ux) and [y] ∈ π(Uy), it follows that the twoelements [x], [y] are Hausdorff separated, which concludes the proof.

Example 1.6. The projective complex space: Pn(C).Let n ∈ N∗. Consider on Cn+1 \0 the following relation: Let u, v ∈ Cn+1 \0, u ∼ v if ∃λ ∈ C\0such that u = λv.

• It is to prove that ∼ is an equivalence relation:

Reflexivity: Let u ∈ Cn+1 \ 0. Take λ = 1 ∈ C∗, u = 1 · u, which proves that ∼ is transitive.

Symmetry: Let u, v ∈ Cn+1 \ 0 such that u ∼ v ⇒ ∃λ ∈ C∗ such that u = λv. Sinceλ 6= 0⇒ 1

λ ∈ C∗. v = 1

λu⇒ v ∼ u⇒ ∼ is symmetrical.

Transitivity: Let u, v, w ∈ Cn+1 \ 0 such that u ∼ v and v ∼ w ⇒ ∃λ, δ ∈ C∗ such thatu = λv and v = δw ⇒ u = λδ ⇒ u ∼ w ⇒ ∼ is transitive.

• Denote by (u0 : u1 : · · · : un) the equivalence class of the element (u0, u1, . . . , un). Define Pn(C) =(Cn+1 \ 0)/ ∼ and the projection application π : (Cn+1 \ 0) → Pn(C): π(u0, u1, . . . , un) =(u0 : u1 : · · · : un).Pn(C) has the quotient topology: U ⊂d Pn(C) if π−1(U) ⊂d Cn+1.

• Let us prove that ∼ is an open relation.Let λ ∈ C∗. Define Fλ : Cn+1 → Cn+1: F (u) = λu. Let us observe that ∀λ ∈ C∗, Fλ smoothbijection (it is a polynomial function on the coordinates). Furthermore, F−1

λ = F 1λ

which is a

smooth bijection ⇒ Fλ homeomorphism ⇒ Fλ open application.Let D ⊂d Cn+1 ⇒ Fλ(D) ⊂d X, ∀λ ∈ C∗. On the other hand, [D] = v ∈ Cn+1|∃v ∈ D : v ∼u = λu|λ ∈ C∗, u ∈ D =

⋃λ∈C∗ λD =

⋃λ∈C∗ Fλ(D) ⊂d Cn+1. According to Proposition 1.6,

it follows that the relation ∼ is open.

13

• Cn+1\0 has a countable base for the topology, thus, from Proposition 1.7 it follows that Pn(C)has a countable base of the topology.

• ”∼” is closed in (Cn+1 \ 0) × (Cn+1 \ 0). Define f : (Cn+1 \ 0) × (Cn+1 \ 0) → C,

f(u, v) =∑

j,k

∣∣∣∣∣∣∣∣uj vjuk vk

∣∣∣∣∣∣∣∣2. Let us notice that: f(u, v) = 0⇔ u ∼ v.

”⇐” u ∼ v ⇒ ∃λ ∈ C∗ : uj = λvj , ∀j = 0, n⇒∣∣∣∣uj vjuk vk

∣∣∣∣ = 0⇒ f(u, v) = 0.

”⇒” f(u, v) = 0f(u,v) suma de numere reale pozitive======================⇒ . Each determinant in the sum is 0, i.e.

∣∣∣∣uj vjuk vk

∣∣∣∣ =

0, ∀j, k = 0, n∃uj0 6=0====⇒ vk =

ukvj0uj0

, ∀k = 0, n ⇒ v = λu where λ =vj0uj0

. If vj0 = 0 ⇒ v = 0

contradiction ⇒ vj0 6= 0⇒ λ ∈ C∗ ⇒ u ∼ v.

f is continuous ⇒ f−1(0) = G∼ closedProposition 1.8=========⇒ Pn(C) Hausdorff.

• Consider the following sets: Uj := (u0 : u1 : · · · : un)|uj 6= 0, j = 1, n. Let us notice thatπ−1(Uj) = (u0, u1, . . . , un)|uj 6= 0 is an open set in Cn+1 \ 0 ⇒ Uj is an open set in Pn(C).

Consider the functions zj : Uj → Cn, j = 1, n, given by zj(u0 : u1 : · · · : un) = (u0uj , . . . ,ujuj, . . . , unuj ).

It is easy to prove that zj are well defined. Indeed, let (u0 : u1 : · · · : un) = (λu0 : λu1 : · · · :

λun) ∈ Pn(C), λ ∈ C∗. zj(λu0 : λu1 : · · · : λun) = (λu0λuj, . . . ,

λujλuj

, . . . , λunλuj) = (u0uj , . . . ,

ujuj, . . . , unuj ) =

zj(u0 : u1 : · · · : un).The inverses of the functions zj are: z−1

j : Cn → Uj , z−1j (t1, . . . , tn) = (t1 : · · · : tj : 1 : tj+1 : · · · :

tn). Let us notice that zj are homeomorphisms.∪nj=0Uj = Pn(C). It still needs to be proven that the changes of coordinates are holomorphic.

Let j, k ∈ 0, n and t ∈ Cn.zk(z

−1j (t)) = zk(t1 : · · · : tj : 1 : tj+1 : . . . tn)

=

(

t1tk+1

, . . . , tktk+1

, 1,tk+2

tk+1, . . . ,

tjtk+1

, 1tk+1

,tj+1

tk+1, . . . , tn

tk+1

), if k + 1 ≤ j(

t1tk+1

, . . . ,tjtk+1

, 1tk+1

,tj+1

tk+1, . . . , tk

tk+1, 1,

tk+2

tk+1, . . . , tn

tk+1

), if k > j

which are holomorphic functions. This ends the description of the projective space as a complexmanifold.

1.3 Compact Riemann Surfaces

1.3.1 Morphisms between Riemann Surfaces

Definition 1.16. We call Riemann surface a 1-dimensional complex variety.

Example 1.7. We can consider as Riemann surfaces all the connected subsets of C.

Example 1.8. The real sphere S2 is a Riemann surface.

Example 1.9. P1(C) is a Riemann surface.

Example 1.10. [4] We call the Riemann sphere or the extended complex plane the followingRiemann surface. Add an infinity point to the set of complex numbers: C = C ∪ ∞. Define afundamental family of neighbourhoods: V (∞, R) = z ∈ C, |z| > R ∪ ∞, where R ∈ R+. TheRiemann surface structure is given by the atlas: A = (U1, z1), (U2, z2), where U1 = C, ϕ1 : U1 → C

given by ϕ1(z) = z and U2 = C \ 0, ϕ2 : U2 → C given by ϕ2(z) =

1z , If z 6=∞0, If z =∞

.

The union of the chart domains covers the whole C: U1 ∪U2 = C. The change of coordinates is givenby ϕ2 ϕ−1

1 : ϕ1(U1 ∩ U2) = C∗ → ϕ(U1 ∩ U2) = C∗, ϕ2 ϕ−1(z) = 1z , holomorphic function.

Example 1.11. (Hyperelliptic curves)[4] Let g ∈ N and A = ak ∈ C|k ∈ 1, 2g + 1. We want toshow that S = (x, y) ∈ C2|y2 =

∏2g+1k=1 (x − ak) is a Riemann surface. The topology is that of a

subspace of C2.Let us define the charts. For each point P = (x0, y0) ∈ S we identify two cases:

14

1. x0 /∈ A (therefore y0 6= 0). Define ϕP (x, y) = x − x0 on an pathwise-connected neighbourhoodof P , UP ⊂d S, such that x ∈ C|∃y ∈ C such that (x, y) ∈ UP ∩A = ∅.

2. ∃j ∈ 1,2g + 1 such that x0 = aj and y0 = 0. For the function x − x0, exists a holomorphicradical function. Therefore, we can define ψP (x, y) =

√x− x0 on an pathwise-connected neigh-

bourhood of P , VP ⊂d S, such that x ∈ C|∃y ∈ C such that (x, y) ∈ VP ∩ (A \ aj) = ∅.

Since we defined maps around each S, it follows that the union of the chart domains is the whole S.Let us check that the change of coordinates are compatible. Let Q1(x1, y1), Q2(x2, y2) ∈ S.

1. If Q1, Q2 have charts of the first type, then the change of coordinates is ϕQ1 ϕ−1Q2

: ϕQ2(UQ1 ∩UQ2)→ ϕQ1(UQ1 ∩ UQ2) given by ϕQ1 ϕ−1

Q2(z) = z + x2 − x1, obviously holomorphic.

2. If Q1 has chart of the first type and Q2 of the second, then the change of coordinates is ϕQ1ψ−1Q2

:

ψQ2(UQ1 ∩ VR)→ ϕQ1(UQ1 ∩ VR) given by ϕQ1 ψ−1Q2

(z) = z2 + x2 − x1, holomorphic.

3. If Q1 has chart of the second type and Q2 of the first, then the change of coordinates is ψQ1ϕ−1Q2

:

ϕR(VQ1 ∩UQ2)→ ψQ1(VQ1 ∩UQ2) given by ψQ1 ϕ−1Q2

(z) =√z + x2 − x1, which is holomorphic.

4. We chose the chart domains such if Q1 and Q2 have charts of the second type, their chartdomains do not intersect.

Furthermore, we can prove that S is pathwise-connected. Let R1(xR, yR) and R2(aj , 0) ∈ S whereaj ∈ A. If the continuous application t : [0, 1] → C is a path with t(0) = xR, t(1) = aj , we build

the continuous application T : [0, 1] → S given by T (α) =

(t(α),

√∏2g+1k=1 (t(α)− ak)

)is a path

in S from R1 to R2 (T [0, 1] ⊂ S because C is an algebraically closed field, therefore the equationy2 −

∏2g+1k=1 (x− ak) = 0 has always at least one solution).

We can compactify S, by adding an infinity point: ∞: Sc = S ∪ ∞, the same way as in Example1.10. We can define a fundamental system of neighbourhoods for ∞: WR = (x, y) ∈ C||x| > R, y2 =∏2g+1k=1 (x− ak) ∪ ∞.

A chart domain that includes ∞ is given by WR0 , pathwise-connected neighbourhood of ∞ such thatx ∈ C|∃y ∈ C such that (x, y) ∈ WR0 ∩ A = ∅. We define φ : WR0 → C such that: φ(∞) = 0 andφ(x, y) = 1√

x. We have coordinate changes just for maps of the first type. Let P (x0, y0). The change

of coordinates is given by ϕP φ−1(z) = 1z2− x0 which is holomorphic because z = 0 does not belong

to the domain.Sc is a compact Riemann surface. This results from the fact that Sc can be written as the union oftwo compact sets: Sc = (x, y) ∈ S||x| ≤ R0 + 1 ∪ φ−1(D(0, R0 + 1)).

Definition 1.17. Let S o surface Riemann and f : S → C. The function f is said to be holomorphicif ∀P ∈ S, ∃(U,ϕ) chart on S, P ∈ U , such that the function f ϕ−1 : ϕ(U) ⊂ C→ C is holomorphicon a neighbourhood around ϕ(P ).

Proposition 1.9. Let S be a Riemann surface and f : S → C. The function f is holomorphic if andonly if for any chart on S, (V, ψ), the composition f ψ−1 is holomorphic on ψ(V ).

Proof. ”⇐” Let (V, ψ) be a chart on S. Let P ∈ V ⇒ ∃(U,ϕ) chart on S, P ∈ U , such that thefunction f ϕ−1 is holomorphic on a neighbourhood around ϕ(P ). The change of coordinates ϕ ψ−1

is holomorphic on U ∩V ⇒ f ψ−1 = (f ϕ−1) (f ϕ−1) is holomorphic on a neighbourhood aroundψ(P ). Since the result checks for any P , the conclusion follows.

”⇒” Consider the atlas A = (Uα, ϕα)α∈A on S. Let P ∈ S A atlas====⇒ ∃α0 ∈ A such that P ∈ Uα0 . The

composition f ϕ−1α0

is holomorphic on ϕα0(Uα0), therefore f is holomorphic.

Definition 1.18. Let S be a surface Riemann and f : S → C. The function f is said to be mero-morphic if for an chart on S, (V, ψ), the composition f ψ−1 : ψ(V ) ⊂ C is meromorphic.

15

Definition 1.19. Let S and S′ be two Riemann surfaces. A morphism between S and S′ is acontinuous function f : S → S′ such that ∀P ∈ S, ∃(U,ϕ) chart on S, P ∈ U and ∃(V, ψ), F (U) ⊂ Vsuch that ψ f ϕ−1 : ϕ(U)→ ψ(V ) is a holomorphic function on a neighbourhood around ϕ(P ).If the morphism f : S → S′ is bijective, f is said to be an isomorphism.An isomorphism that has the same domain and codomain f : S → S is called an automorphism.

Example 1.12. Let S, S′ be two connected compact Riemann surfaces and f : S → S′ be a non-constant morphism. From the Open Application Principle, it follows that f(S) is simultaneouslyclosed and open, therefore f(S) = S′.

From Riemann’s Theorem of Conformal Representation, any connected sets U, V (d C are isomorphic.If we choose: U = H and V = D(0, 1), f : U → V , f(z) = z−i

z+i is an isomorphism.

Indeed, for z ∈ H Im(z)>0=====⇒ z + i 6= 0⇒ f holomorphic.

Let z1, z2 ∈ H, such that f(z1) = f(z2)⇒ z1−1z1+1 = z2−1

z2+1 ⇒ (z1− 1)(z2 + i) = (z1 + 1)(z2− i)⇒ z1 = z2,thus f injective.Let w ∈ D(0, 1). We search for z ∈ H such that f(z) = w ⇒ z−i

z+i = w ⇒ z = iw+11−w . We want to show

that z obtain this way is in H, i.e. Im(z) > 0⇔ 1i (z − z) > 0⇔ w+1

1−w + w+11−w > 0⇔ w+1

1−w + w+11−w > 0⇒

2(1−ww)1−w−w+ww > 0

w∈D(0,1)⇒ww<1===========⇒ 1− w − w − ww > 0⇒ (Re(w)− 1)2 + Im(w)2 > 0, thus f is indeed

surjective, which concludes the proof.

Proposition 1.10. The set of automorphisms on a Riemann surface S, Aut(S), with the usualcomposition of functions is a group.

Definition 1.20. Let S be a Riemann surface and f : S → C be a meromorphic function. Let P ∈ Sand (U,ϕ) chart on S centred at P ∈ U . Consider the Laurent series for f ϕ−1 around 0:

f ϕ−1(z) =∞∑−∞

anzn.

Let n0 ∈ N such that: an0 6= 0 and an0−k = 0, ∀k ∈ N∗. The number n0 is called the order of f in Pand is denoted by ordf (P ).

Remark 1.8. [4] Definition 1.20 is correct, more precisely, it does not depend on the choice of the local.Indeed, let (V, ψ) be another local chart on S with P ∈ V , ψ(P ) = 0. There are the two Laurentseries:

f ϕ−1(z) =

∞∑n≥n0

anzn

f ψ−1(z) =

∞∑m≥m0

bmzm.

The change of coordinates ϕ ψ−1 is biholomorphic. Furthermore, ψ−1(0) = P ⇒ ϕ(ψ−1(0)) = 0 ⇒ϕ ψ−1 has a Taylor series around 0 of the form ϕ ψ−1(z) =

∑∞k=1 ckz

k with c1 6= 0.There is the identity f ψ−1 = (f ϕ−1) (ϕ ψ−1) ⇒

∑∞n≥n0

anzn =

∑∞m≥m0

bm(∑∞

k=1 ckzk)m ⇒

m0 = n0, therefore the order of f in P does not depend on the choice of chart.

Definition 1.21. Let S be a Riemann surface, f : S → C be a meromorphic function and P ∈ S.

• If ordf (P ) > 0, P is said to be a zero for f .

• If ordf (P ) > 0, P is said to be a pole for f .

Definition 1.22. Let f : S → S′ be a morphism of Riemann surfaces, P ∈ S and (V, ψ) be a localchart on S centred at P . The number:

mP (f) := ordψf P

16

is called the multiplicity order of f in P .If mP (f) ≥ 2, P is said to be a branching point with multiplicity order mP (f). The value of f inthe branching point, f(P ) ∈ S′, is called a critical value. The set of critical values of the morphismf will be denoted by Bf .If the morphism f : S → S′ has branch points, the morphism is said to be ramified.

Remark 1.9. The multiplicity of a morphism does not depend on the choice of maps, the proof beingsimilar with the one in the case of a meromorphic function. Furthermore, let us observe that if

mP (f) ≥ 2⇒ (ψ ϕ−1)′(ϕ(P )) = 0holomorphic functios have isolated zeroes=========================⇒ the sets of branching points and

critical values are finite for compact Riemann surfaces.

Example 1.13. [4] Consider Sc the Riemann surface in Example 1.11 and the Riemann sphere. Considerthe application πx : Sc → C given by πx(x, y) = x and πx(∞) =∞. We keep the notations in Examples1.10 and 1.11.The following cases occur:

• Let P (x0, y0) ∈ S where x0 /∈ A. P has a chart of the first type. The composition is given byϕ1 πx ϕ−1

P (z) = z + x0, which is holomorphic.

• Let P (aj , 0) ∈ S. P has a chart of the second type. The composition is given by ϕ1πxψ−1P (z) =

z2 + aj , which is holomorphic.

• Let ∞ ∈ S. In this case, ϕ2 πx φ−1(z) = z2, holomorphic.

The computations above show that πx is a morphism of Riemann surface. The branching points(of order 2) are in the set (ak, 0)|k ∈ 1, 2g + 1 ∪ ∞ and the corresponding critical values area1, . . . , ak,∞.

17

Chapter 2

Covering Spaces

2.1 Unramified Coverings

2.1.1 Examples and Properties of Covering Spaces

Definition 2.1. [17] Let X, Y be topological spaces and f : Y → X be a continuous function. Thefunction f is said to be a local homeomorphism if ∀y ∈ Y , ∃V open neighbourhood of y such thatf |V : V → f(V ) is a homeomorphism.

Definition 2.2. [16] Let X be a topological space. A covering of X is a pair (Y, p) where

• Y is a topological space

• p : Y → X a continuous function

such that ∀x ∈ X, ∃U ⊂d X a pathwise-connected neighbourhood of x such that p|V : V → Uhomeomorphism, ∀V pathwise-connected component of p−1(U).The space Y is called a covering space of X, and the function p is called a projection from Y toX. The set U is called an elementary neighbourhood. The connected components of p−1(U) arecalled sheets of the covering over U . For y ∈ Y , p−1(y) is called the fibre over y.

Example 2.1. [12] Let X = S1 and its parametrisation X = (cos t, sin t)|t ∈ [0, 2π). Actually, we

Figure 2.1: The covering of S1 with R

can consider the parameter t in any interval of the type [a, b) where b− a = 2π.

18

With this idea in mind, consider: Y = R, p : Y → X, p(t) = (cos t, sin t). p is obviously continuous.Let x ∈ X,x = (cos t0, sin t0), t0 ∈ [0, 2π). Consider the open interval I = (t0 − π, t0 + π), which givesa pathwise-connected open neighbourhood of x: U = (cos t, sin t)|t ∈ I. p−1(U) = ∪k∈Z(t0 + 2kπ −π, t0 + 2kπ + π). A pathwise-connected component of p−1(U) has the form Vk = (t0 + 2kπ − π, t0 +2kπ + π), and p|Vk : Vk → U is a homeomorphism.Actually, instead of I we can consider any open interval J such that t0 ∈ J .

Example 2.2. [12] Consider another covering of X = S1, this time with Y = S1. Consider theparametrization S1 = (cos t, sin t)|t ∈ [0, 2π). Let the projection be p : Y → X, p(t) = nt (mod 2π)where n ∈ N∗ fixed. p is obviously continuous.Let x be given by t0 ∈ [0, 2π). We can consider as elementary neighbourhood any open interval that

Figure 2.2: The Covering of S1 with S1

contains x. p−1(x) has n points.If we see the sphere as S1 = z ∈ C||z| = 1, the same projection can be written as p(z) = zn.

Remark 2.1. If (Y, p) covering of X and U ⊂d X elementary neighbourhood of x ∈ X, it follows thatU is an elementary neighbourhood ∀z ∈ U .

Proposition 2.1. Let X topological space and (Y, p) a covering of X. p is a local homeomorphism.

Proof. Let y ∈ Y . From the Definition of the covering, ∃U pathwise-connected open neighbourhoodof f−1(x) such that p|V : V → U is a homeomorphism, where V is the pathwise-connected componentof p−1(U) that contains y. It follows that V is connected, thus open, which concludes the proof.

Counterexample 2.1. [17] Keeping in mind Example 2.1, let us consider again the circle, X = S1,and its parametrisation, X = (cos t, sin t)|t ∈ [0, 2π). Pick Y = (0, 3π), p : Y → X, p(t) =(cos t, sin t). Let x = (1, 0) ∈ X. An open neighbourhood U of x contains a set of the form J =(cos t, sin t)|t ∈ (−r, r) ⇒ p−1(U) contains the connected component V = (0, r) and p|V : V → U isnot a homeomorphism.

Proposition 2.2. Let X be a topological space and (Y, p) be a covering of X. If X is pathwise-connected, it follows that |p−1(x)| is constant, ∀x ∈ X.

Proof. ∀z ∈ X we denote by Uz its elementary neighbourhood from the Definition of covering. For afixed z ∈ X, |p−1(z)| is constant in Uz.Let x, x′ ∈ X. Since X is pathwise-connected, it follows that ∃f : [0, 1] → X continuous such that

19

f(0) = x and f(1) = x′. ∀z ∈ Im f consider the set Mz = t ∈ [0, 1]|f(t) ∈ Uz obviously non-empty.Furthermore, let us observe that if z 6= x′, supMz > max f−1(z).”Following” Gf , we start from x, we choose y = f−1(supMx) and observe that: |p−1(x)| = |p−1(z)|.Send y to x and get at each step a new y, until reaching x′. Since f continuous and [0, 1] compact,it follows that f([0, 1]) is compact, therefore the algorithm ends in a finite number of steps. We getthat |p−1(x)| = |p−1(z)|.

Definition 2.3. Let X be a topological space, x ∈ X and (Y, p) a covering of X. If X is pathwise-connected, then |p−1(x)| is called the degree of the covering. If it is finite, then the covering iscalled finite.

Definition 2.4. [17] Let X be a topological space. Two coverings of X, (Y1, p1), (Y2, p2), are said tobe homomorphic if ∃u : X1 → X2 continuous such that: p2u = p1. If p is a homeomorphism, thecoverings (Y1, p1) and (Y2, p2) are said to be isomorphic.

Definition 2.5. Let X be a topological space and (Y, p) a covering of X. If f : Y → Y homeomor-phism, f is called automorphism of the covering. The set of automorphisms of the covering (Y, p)is denoted by Aut(Y, p).

Proposition 2.3. Let X be a topological space and (Y, p) a covering of X. Aut(Y, p) together withthe usual composition of functions is a group.

Proof. The set of automorphisms of the topological space Y , Aut(Y ), is a group and Aut(Y, p) ⊂Aut(Y ). What is left to be shown is that Aut(Y, p) ≤ Aut(Y ). Let f ∈ Aut(Y, p) ⇒ pf = p ⇒ p =pf−1 ⇒ f−1 ∈ Aut(Y, p).Let g, h ∈ Aut(Y, p) ⇒ pg = p and pf = p ⇒ pgf = pf = p ⇒ gf ∈ Aut(Y, p) which concludes theproof.

Proposition 2.4. Let X be a topological space and (Y, p) a covering of X. If X is Hausdorff, thenY is Hausdorff.

Proof. Let y1, y2 ∈ Y , y1 6= y2. We distinguish two cases:

I. p(y1) = p(y2) = x ∈ X.Consider Ux the elementary neighbourhood of x. Let p−1(Ux) =

⊔j∈J Vj , Vj ⊂d Y . Since

f |Vj : Vj ⇒ Ux homeomorphism for each j, it follows that ∃k, l ∈ J , k 6= l such that y1 ∈ Vk andy2 ∈ Vl. Vk ∩ Vl = ∅, therefore we got two disjoint neighbourhoods for the two points.

II. x1 := p(y1) 6= x2 := p(y2).Since X Hausdorff, it follows that ∃W1,W2 ⊂d X, x1 ∈ W1, x2 ∈ W2 such that W1 ∩W2 = ∅.Consider U1 and U2 the elementary neighbourhoods of x1 respectively x2. Let R1 = U1∩W1 and

R2 = U2 ∩W2W1∩W2=∅======⇒ R1 ∩R2 = ∅. Consider p−1(R1) =

⊔a∈AOa and p−1(R2) =

⊔b∈B Pb and

pick a0 ∈ A and b0 ∈ B such that y1 ∈ Oa0 and y2 ∈ Pb0 . The neighbourhoods Oa0 and Pb0 givethe separation of the points y1 and y2.

2.1.2 Paths and covering spaces

Definition 2.6. [17] Let X be a metric space and Ujj∈J an open cover of X. A Lebesgue numberof the cover Ujj∈J is a real positive number ε such that ∀V ⊂ X with diam(V ) < ε ⇒ ∃j ∈ Jsuch that V ⊂ Uj .

Proposition 2.5. [19] Any covering of a metric compact space admits a Lebesgue number.

Proof. Let X be a metric compact space and Ujj∈J an open cover of X. Since X is compact,∃K ⊂ J , K finite, such that Ujj∈K cover of X. To simplify the writing, we can denote the sets inK with U1, U2, . . . , Un, where n ∈ N∗.

20

Define f : X → R, f(x) := 1n

∑nj=1 d(x,X \ Uj). Since f is a continuous function on a compact set,

∃ε := min(f). We want to show that ε > 0. Let x ∈ X. Since Ujj∈1,n is a cover of X, it follows

that ∃j0 ∈ 1, n such that x ∈ Uj0 ⇒ d(x,X \ Uj0) > 0⇒ ε > 0.If V ⊂ X with diam(V ) < ε ⇒ ∃x0 ∈ X such that V ⊂ D(x0, ε). Since f(x0) ≥ ε, it follows that∃j ∈ 1, n such that d(x0, X \ Uj) ≥ ε⇒ D(x0, ε) ⊂ Uj ⇒ V ⊂ Uj .

Proposition 2.6. [17] Let X be a topological space, (Y, p) a covering of X, y ∈ Y and x := p(y).Then, ∀t : I → X path in X with t(0) = x, ∃T : [0, 1]→ Y path in Y with T (0) = y such that pT = t.

Y

p

y

p

[0, 1]

T

==

t

!!

X x

Proof. Let U = Ujj∈J be a cover with elementary neighbourhoods of X. We get t−1(Uj)j∈J coverof [0, 1]. Since [0, 1] is compact, we can extract a finite subcover of [0, 1]: t−1(Uk)k∈1,m where m ∈ Nand Uk ∈ U . Since [0, 1] compact metric space, it follows from Proposition 2.5 that there exists εLebesgue number of the cover. Let n ∈ N such that 1

n < ε. Consider the subintervals [ ln ,l+1n ], where

l ∈ 0, n− 1. It follows that ∀l ∈ 0, n− 1,∃kl ∈ 1,m such that [ ln ,l+1n ] ⊂ t−1(Ukl). Let V0 ⊂ Y the

connected component of p−1(Uj0) such that y ∈ V0. Since p|V0 : V0 → Uj0 homeomorphism, it followsthat we can define uniquely T (z) = p|−1

V0(t(z)) where z ∈ [0, 1

n ]. Let V1 ⊂ Y be the pathwise-connected

component of p−1(Uj1) such that T ( 1n) ∈ V1. Since p|V1 : V1 → Uj1 homeomorphism, it follows that

we can define uniquely T (z) = p|−1V1

(t(z)) where z ∈ [ 1n ,

2n ]. T is well defined and continuous. Repeat

the procedure for all l ∈ 0, n− 1.

Definition 2.7. Let X be a topological space, (Y, p) a covering of X and t : I → X path in X. Apath S in Y with pS = t is called a lift of the path t.

Proposition 2.7. Let X be a pathwise-connected topological space, (X1, p1), (X1, p1) two coverings ofX and f : X1 → X2 a morphism of coverings. It follows that (X1, f) is a covering of X2.

Proof. • We will show that for any point x ∈ X we can choose an elementary neighbourhood Uthat works for both coverings. Indeed, let U1 respectively U2 be the elementary neighbourhoodsfor x in the coverings (X1, p1) and (X2, p2). Let U be the connected component of U1 ∩U2 thatcontains x. U is an elementary neighbourhood for x for both coverings.

• We shall show that f is surjective. Let y ∈ X2. Let x1 ∈ X such that x2 := f(x1) ∈ X2

is in the same pathwise - connected component with y. Denote by x0 := p1(x1)f morphism

=========p2(f(x1)) = p2(x2) ∈ X. Let t be a path in X2 with t(0) = x2 and t(1) = y and let r = p2 t.From Proposition 2.6, it follows that ∃!s path in X1 starting in x1 such that sp1 = r. Letx be the final point of s. Let us observe that the paths fs and t have the same startingpoint and both check p2fs = p2t = r. The uniqueness of the lift in Proposition 2.6 gives thatfs = t⇒ f(s(1)) = t(1)⇒ f(x) = y, therefore, indeed, f is surjective.

x // y 1t

oo

X1 3 x1f

//

p1

&&

x2 ∈ X2

p2

xx

0too

rtt

svv

x0 ∈ X

21

• Let z ∈ X2. We want to show that we can construct an elementary neighbourhood for z. Fromthe first part, it follows that we can choose U , an elementary neighbourhood for p2(z) for bothcoverings. We choose W the pathwise connected component of p−1

2 (U) that contains z. LetV be a pathwise connected component of f−1(W ) (which is nonempty). f |V : V → W is ahomeomorphism being a composition p−1

2 |U p1|V .

X1 ⊃ Vf

//

p1

''

X2 ⊃W 3 z

p2vv

X ⊃ U 3 p2(z)

2.1.3 Universal Coverings

Definition 2.8. The topological space X is called simply connected if it is pathwise-connected andthe fundamental group π1(X) is trivial.

Proposition 2.8. [4] Let X be a topological space and (Y, p) a covering of X. Let X0 be a simplyconnected topological space. If f : X0 → X continuous, then exists f : X0 → Y continuous such thatp f = f .

Proof. Let x0 ∈ X0. Fix y ∈ p−1(f(x0)). Since X0 is simply connected, thus pathwise-connected, itfollows that ∀x ∈ X0, there exists a path t : [0, 1] → X0 such that t(0) = x0 and t(1) = x. We lift

the path f t in Y with starting point y and we get (f t) path in Y with p (f t) = f t. Define

f(x) = (f t)(1).

[0, 1](ft)

t

Y

p

X0

f||

foo

X

Let us prove now that f is well - defined. Let s : [0, 1] → X0 be another path such that such thats(0) = x0 and s(1) = x. Since X0 is simply connected, it follows that there is ∃H : [0, 1]× [0, 1]→ X0

homotopy such that H(0)(α) = t(α) and H(1)(β) = s(β) where α, β ∈ [0, 1]. We want to show that

f t(1) = f s(1). The function α→ ˜f H(α)(1) is continuous and has values in the fibre p−1(f(x)),

which is a discrete set, making the function continuous. In particular, f t(1) = ˜f H(0)(1) =˜f H(1)(1) = f s(1).

Definition 2.9. [4] In the context of Proposition 2.8, the function f is called a lift of the function f .

Remark 2.2. From the Proof of Proposition 2.8, it follows that fixing a point of a lift determines ituniquely.

Definition 2.10. [2] Let X be a topological space. A covering (Y, p) of X is said to be universal ifY is simply connected.

Example 2.3. The universal covering of the pointed disc D(0; 1) seen as a subset of C is given by(H, p), p : H→ D(0; 1) where p(z) = e2πiz.

Proposition 2.9. [4] Let X be a pathwise-connected topological space, x0 ∈ X and (X, p) a universalcovering of X. Then the following hold

1. X is unique up to isomorphism,

22

2. the group of automorphisms of the universal covering Aut(X, p) can be identified with the fun-damental group π1(X,x0)

3. there exists a homeomorphismX/Aut(X, p)→ X

[x]→ p(x)

4. each covering of X is isomorphic with a covering of the form X/G→ X/Aut(X, p) ' X whereG ≤ Aut(X, p).

Proof. 1. Let (X1, p1) and (X2, p2) be two universal coverings of X. Define p1 as a lift of p1 suchthat: p2p1 = p1. Let x ∈ X1. We can construct p2 as the unique lift of p2 in X1 such thatp2 p1(x) = x.

X1

p1

p1//X2

p2~~

p2oo

X

2. Let x0 := p−1(x0) ∈ X.We want to construct θ : Aut(X, p)→ π1(X,x0). Let f ∈ Aut(X, p) and tf : [0, 1]→ X a pathwith tf (0) = x0 and tf (1) = f(x0). σf := p tf is a path in X. σf (0) = p(tf (0)) = p(x0) = x0

and σf (0) = p(tf (1)) = p(f(x0))f∈Aut(X,p)

========= p(x0) = x0. Thus σf is a loop in X with origin inx0. Define θ(f) = [σf ] ∈ π1(X,x0).

[0, 1]tf//

σf((

Xp

f// X

p

X

We want to define ψ : π1(X,x0) → Aut(X, p). Let [t] ∈ π1(X,x0). Consider t the unique liftof t in X with t(0) = x0. Define ft ∈ Aut(X, p) as the unique lift of p such that ft(x0) = t(1).Take ψ([t]) = ft. Repeating the same reasoning in the Proof of Proposition 2.8, it follows thatψ is well-defined.

[0, 1]

t!!

t // X

p

X

3. The elements of Aut(X, p) are lifts of the function p : X → X and are in correspondence withpoints din fibre. The application X/Aut(X, p)→ X is a homeomorphism.

4. If (Y, q) is a covering of X consider p the lift of the continuous function p. From Proposition 2.7,(X, p) is a universal covering of Y . Using the previous point , it follows that Y ' X/Aut(X, p).To conclude the proof, let us notice that: Aut(X, p) ≤ Aut(X, p). Indeed, let f ∈ Aut(X, p)⇒p f = p⇒ q p f = q p qp=p

===⇒ p f = p⇒ f ∈ Aut(X, p).

X

p

p// Y

q

X

Example 2.4. Starting from Example 2.3 and using Proposition 2.9, we obtain that the coverings ofthe pointed disc D(0; 1) with D(0; 1) are isomorphic with coverings of the type D(0; 1) → D(0; 1),z → zn where n ∈ N∗.

23

2.2 Ramified Coverings

We shall begin with an example that is most eloquent.

Example 2.5. [12] Consider the annulus: X = D(rn1 ; rn2 ) (where r1 < r2 ∈ R+, n ∈ N∗) and thecovering given by Y = D(r1; r2), p : Y → X, p(r, ϕ) = (rn, nϕ), where nϕ is taken modulo 2π. Seeing

in polar coordinates, the projection is given by: zf−→ zn.

Taking r1 = 0, the annuli become pointed discs. Adding the centre to X and to Y , (Y, p) becomes a

Figure 2.3: The unramified covering of the annulus D(rn1 ; rn2 ) by the annulus D(r1; r2)

ramified covering of a pointed disc by a pointed disc: p is still continuous and all the points of Yexcept one have n preimages, while the centre has a single preimage (the centre of X). The centre ofY is a ramification point of order n and the centre of X is a critical value.

Definition 2.11. Let X be a topological space. A ramified covering of X is a pair (Y, p) where Yis a topological space and p : Y → X a continuous function such that ∃B ⊂ X nowhere dense suchthat (Y \ p−1(B), p|Y \p−1(B) : p−1(B) → X \ B) is a covering. If B is minimal with this propriety, itis called the set of critical values and is denoted by Bf . Its counterimage, p−1(Bf ), is called theramification set.

In this paper, we are interested in Riemann surfaces, therefore we shall study ramified coveringsfor Riemann surfaces.

Theorem 2.1. [4] Let f : Y → X be a nonconstant morphism of compact connected Riemann surfaces.

1. Let Σf ⊂ X be the set of critical values of f , as defined in Definition 1.22. Let Y ∗ := Y \f−1(Σ)and X∗ := S \ Σf . Then the function

f∗ : Y ∗ → X∗

given by f∗ = f |Y ∗ is a projection as in Definition 2.2, thus (Y, f) is a ramified covering of X.

2. The sum∑

y∈f−1(x)my(f) does not depend on x ∈ X.

Proof. 1. First of all, let us notice that f∗ is a local homeomorphism.Indeed, let y ∈ Y ∗. Since f is a morphism between Riemann surfaces, ∃(U,ψ) local chart on Y ,y ∈ U , ∃(V, ϕ) local chart on X with f(U) ⊂ V such that ϕf ψ−1 : ψ(U)→ ϕ(V ) holomorphic.Choose W ⊂ U open neighbourhood of y such that W ∩ f−1(Σ) = ∅ and Z = f(W )⇒ W ⊂ Vand Z∩Σ = ∅. Furthermore, since (U,ψ), (V, ϕ) are local charts, it follows that ψ|W : W → ψ(W )and ϕ|Z : Z → ϕ(Z) homeomorphisms. It follows that ϕ|f ψ|−1 : ψ(W )→ ϕ(Z) holomorphic,therefore f |W : W → f(W ) homeomorphism, making f∗ homeomorphism local.Now let us show that f∗ is a covering. Let x ∈ X∗. It follows from Remark 1.9, that f−1(x)is a finite set, f−1(x) = y1, y2, . . . , yd, where d ∈ N∗. Let Q be a neighbourhood of x andV1, V2, . . . , Vd neighbourhoods of y1, y2, . . . , yd such that f |Vj : Vj → Q are homeomorphisms,

j ∈ 1, d.

24

We want to prove that Q can be chosen such that f−1(Q) is a disjoint union of Vj , f−1(Q) =⊔d

j=1 Vj . If this were not true, then ∃(xn) ⊂ X,xn → x such that each fibre f−1(xn) has at

least one point yn /∈⋃dj=1 Vj . Let y′ be a limit of this sequence. From the continuity of f it

follows that f(x) = y′, therefore y′ ∈ y1, y2, . . . , yd. It follows that ∃j = 1, d,∃n ∈ N such thatyn ∈ Vj , which is absurd.

2. Let x ∈ X. From Remark 1.9, it follows that f−1(x) is a finite set. Let this be denotedf−1(x) = yjj∈1,d, where d ∈ N∗. Let Ux be a neighbourhood of x, isomorphic with a disc,

such that Ux ∩ f−1(Σ) ⊂ x. Then f−1(Ux) =⊔dj=1 Vj , where each Vj is a neighbourhood of

yj homeomorphic with a disc. Let V ∗x := Vx \ x and U∗j := Uj \ yj. Then U∗j∼= D(0; 1) and

f−1(V ∗x ) =⊔j∈J U

∗j .

Each restriction f∗|Vj : Vj → Ux is a covering. We have the following diagram:

V ∗j

ψ

f∗// U∗x

ϕ

D(0; 1) D(0; 1)

We get a covering of D(0; 1) with D(0; 1). These are of the form z → zmj , where mj ∈ N∗. Weget that f is of order mj in yj and that Vj contains mj of the d preimages of each value in Vx.In particular,

∑j=1mj = d.

Definition 2.12. Let X,Y be two compact pathwise - connected Riemann surfaces. Let f : Y → Xbe a nonconstant morphism and let x ∈ X.

deg(f) =∑

y|f(y)=x

my(f)

is called the degree of f .

Theorem 2.2. [4] Let X be a compact Riemann surface, Σ ⊂ X finite. Let X∗ = X \ Σ and(Y ∗, f∗ : Y ∗ → X∗) a finite covering of X∗. Then ∃!Y compact Riemann surface such that Y ∗ ⊂ Yand ∃!f∗ : Y → X holomorphic such that f∗|Y ∗ = f .

Proof. Let x ∈ Σ and (Ux, ϕx) be a local chart on Y centred at x. Let U∗x = Ux \ x and f−1(U∗x) =⊔rj=1 V

∗x,j its decomposition into pathwise-connected components. Each of them is a covering space of

U∗x , therefore from Proposition 2.9, there exists a commutative diagram:

V ∗x,jf

//

ϕ∗j

U∗x

ϕx

D(0; 1)z→zmj// D(0; 1)

where ϕj and ϕx are holomorphic bijections.

We create an additional point P jx which will become the centre of V ∗x,j . Let Vx,j = V ∗x,j ∪ Pjx and

extend ϕ∗j : V ∗x,j → D(0; 1) to a bijection ϕj : Vx,j → D(0, 1). Set ϕj |V ∗x,j = ϕ∗j and ϕj(Pjx) = 0 and

give a topology on Y := Y ∗ ∪ P jxx∈Σ,j∈1,r establishing ϕj homeomorphisms. We want to set on Ya Riemann surface structure. Let (Ua, ψa)a∈A be an atlas on Y . If Ua ∩ Σ = ∅, consider the maps:(V ba , ϕ

ba) where V b

a pathwise - connected component of f−1(U ba) and ϕba = ψa f : U ba → ψ(V ba ). The

changes of coordinates are holomorphic: ϕba (ϕdc)−1 = ψa (ψc)

−1.Consider A = (V b

a , ϕba) ∪ Vx,j . The change of coordinates are also holomorphic. Extend f∗ to

f : Y → X by setting f |Y ∗ = f∗, f(P jx) = x.

25

Proposition 2.10. [4] Let K ⊂ C be a field and F ∈ K[X,Y ] an irreducible polynomial. Considerthe writings for F : F (X,Y ) =

∑0j=n pj(X)Xj =

∑0k=m qk(Y )Xk, where m,n ∈ N and pj ∈ K[X],

qk ∈ K[Y ], j ∈ 0, n, k ∈ 0,m.If n ≥ 1, define:

SXF = (x, y) ∈ C2|F (x, y) = 0, FX(x, y) 6= 0, pn(x) 6= 0

SYF = (x, y) ∈ C2|F (x, y) = 0, FY (x, y) 6= 0, qm(y) 6= 0.

Then:

1. SXF and SYF are connected Riemann surfaces on which the coordinate functions πx and πy areholomorphic,

2. ∃!SF compact connected Riemann surface such that SXF , SYF ⊂ SF ,

3. the coordinate functions πx and πy can be extended to meromorphic functions SF ,

4. the branching points of πx are in the set SF \ SXF , and the ones of πy are in the set SF \ SYF .

Proof. (Idea) The structure of Riemann surface of SxF comes from solving y in terms of x, which

is possible by The Implicit Function Theorem. We get that πx : SxF → C is holomorphic. Since

F and FY have just a finite number of common zeroes, it follows that C \ πx(SXF ) is finite. Let

C \ πx(SXF ) = a1, . . . , ar,∞. Let W be a connected component of SXF . πX |W : W → C \ πx(SXF ) isa covering and thus, it can be extended to a morphism of Riemann surfaces.

26

Chapter 3

Belyi’s Theorem

3.1 Statement of the Theorem

Definition 3.1. [4] Let S be a Riemann surface and K ⊂ C be an infinite field. We say that S isdefined over K if it ∃F ∈ K[X,Y ] irreducible polynomial such that S ' SF as Riemann surfaces,where SF is given by Proposition 2.10.

Example 3.1. Consider the polynomial F ∈ Q(e)[X,Y ], given by F (X,Y ) = Y 2 − X3 + e3. SF isobviously defined over Q(e).On the other hand, consider G ∈ Q(e)[X,Y ], given by G(X,Y ) = Y 2−X3 +1. SG is defined obviouslyover Q. Let us notice the following isomorphism: f : SF → SG, f(x, y) = (xe ,

y√e3

), therefore SF is

defined also over Q.

Remark 3.1. We will be interested if a Riemann surface is defined over Q. In order to do this, westudy the main theorem of this thesis: Belyi’s Theorem. Belyi’s Theorem was stated initially in thecontext of the projective curves.

Theorem 3.1. [15] Let X be an algebraic projective curve defined over C. The following are equiva-lent:

1. X is defined over Q

2. There is a finite morphism π : X → P1(C) with three critical values.

Remark 3.2. Reformulating for Riemann surfaces, we get Theorem 3.2.

Theorem 3.2. (Belyi’s Theorem)[4] Let S be a compact Riemann surface. The following areequivalent:

1. S is defined over Q

2. There is a meromorphic function f : S → C such that Bf ⊂ 0, 1,∞.

Definition 3.2. Let S be a compact Riemann surface. A Belyi function is a function f : S → Csuch that |Bf | ≤ 3.

Remark 3.3. Composing maybe with Mobius transformations, we can consider that Bf ⊆ 0, 1,∞.Remark 3.4. The proof of the inverse implication is relatively complicated and is based on a moregeneral criterion that gives an equivalence criterion for a Riemann surface to be defined over Q.

3.2 The Direct Implication of the Theorem

3.2.1 An Example of Belyi Function

[4] Let us begin with an example in the case derived from Example 1.11: Let F ∈ C[X,Y ] be apolynomial, given by

F (X,Y ) = Y 2 −X(X − 1)(X − λ)

27

where λ = mm+n ,m, n ∈ N

∗. The compactified Riemann surface given by F (X,Y ) = 0, Sc, is obviously

defined clearly over Q. According to Belyi’s Theorem, we should find f : Sc → C with only criticalvalues. The morphism in Example 1.13 πx : Sc → C given by πx(x, y) = x, π(∞) =∞ has four criticalvalues 0, 1, λ,∞.

Definition 3.3. [4] The Belyi polynomial Pm,n : C→ C is given by:

Pm,n(x) =(m+ n)m+n

mmnnxm(1− x)n

Proposition 3.1. [4] The Belyi polynomial Pm,n : C→ C satisfies the following proprieties:

1. Pm,n(0) = 0, Pm,n(1) = 0, Pm,n(∞) =∞, Pm,n(λ) = 1.

2. BPm,n = 0, 1,∞.

Proof. 1. • It is obvious that 0 and 1 are roots of Pm,n.

• lim|x|→∞ Pm,n(x) = lim|x|→∞(m+n)m+n

mmnn |x|m|1− x|n =∞.

• P (λ) = (m+n)m+n

mmnn

(m

m+n

)m (1− m

m+n

)n= (m+n)m+n

mmnnmm

(m+n)mnn

(m+n)n = 1.

2. We obtain the set of branching points by making the derivative of the function Pm,n equal to 0:

P ′(x) = 0⇔ (m+n)m+n

mmnn [mxm−1(1−x)n−nxm(1−x)n−1] = 0⇔ (m+n)m+n

mmnn xm−1(1−x)n−1[m(1−x)− nx] = 0⇔ (m+n)m+n

mmnn xm−1(1− x)n−1[m− (m+ n)x] = 0.We can choose m,n ≥ 1⇒ the set of branching points is given by: 0, 1, λ,∞.

The composition Pm,n πx : Sc → P1(C) has branching points (0, 0), (1, 0), (λ, 0),∞ ∈ Sc withcritical values 0, 1,∞, therefore it is a Belyi function.

3.2.2 The Actual Proof

Proposition 3.2. Let X,Y, Z be Riemann surfaces and f : X → Y , g : Y → Z morphisms ofRiemann surfaces. The following relation takes place:

Bgf = Bg ∪ g(Bf )

Proof. There is the composition:

Xf−→ Y

g−→ Z.

Let Q ∈ Im(g f)⇒ ∃P ∈ X such that g(f(P )) = Q⇒ ∃(V, ψ) local chart centred at Q,∃(U,ϕ) localchart centred at P such that ψ (g f) ϕ−1 has a Laurent series around z = 0: ψ (g f) ϕ−1 =amz

m + am+1zm+1 + . . . where m = ord(ψf) P .

Let R = f(P ) and (W, τ) local chart centred at R. Consider the Laurent series around 0:

ψ g τ−1 = bnzn + an+1z

n+1 + . . .

andτ f ϕ−1 = cpz

p + cp+1zp+1 + . . .

where n = ordψgτ−1 R and p = ordτfϕ−1 P .The equality ψ (g f) ϕ−1 = (ψ g τ−1) (τ f ϕ−1) gives

amzm + am+1z

m+1 + · · · = bn(cpzp + cp+1z

p+1 + . . . )n + bn+1(cpzp + cp+1z

p+1 + . . . )n+1 + . . . .

It follows that m = pn. The following equivalences take place: Q ∈ Bgf ⇔ ordψ(gf)ϕ−1 P ⇔ m ≥2 ⇔ pn ≥ 2 ⇔ p ≥ 2 or n ≥ 2 ⇔ ordτfϕ−1 P ≥ 2 or ordψgτ−1 R ≥ 2 ⇔ f(P ) = R ∈ Bf org(R) = Q ∈ Bg ⇔ Q ∈ Bg ∪ g(Bf ), which concludes the proof.

28

Proposition 3.3. [4] Let f : S → P1(C) with the set of critical values 0, 1, λ1, . . . , λn,∞ ⊂ Q∪∞.Then there it exists F : S → P1(C) Belyi function.

Proof. We want to obtain only three critical values of f , reducing the number of critical values at eachstep.If λ1 < 0, we compose with the Mobius M1(x) = 1 − x and obtain the new set of critical values0, 1, 1 − λ1, 1 − λ2, . . . , 1 − λn,∞. Now 1 − λ1 > 0. Redenote 1 − λj with λj , j ∈ 1, n where nowλ1 > 0. If λ1 > 1, we compose with the Mobius transformation M2(x) = 1

x and we obtain a new setof critical values 0, 1, 1

λ1, 1λ2, . . . , 1

λn,∞. Now 1

λ1∈ (0, 1). Redenote 1

λjwith λj , j ∈ 1, n where now

λ1 ∈ (0, 1)∩Q⇒ ∃m,n ∈ N \ 0, 1 such that λ1 = mm+n . The composition with the Belyi polynomial

Pm,n has strictly less critical values: 0, 1, Pλ1(λ2), . . . , Pλ1(λn),∞ ⊂ Q ∪ ∞.Since the set of critical values is finite, we can reduce it to three values applying inductively the stepsabove.

Proposition 3.4. Let S be a compact connected Riemann surface. If S is defined over Q, then thereexists a morphism f : S → C such that Bf ⊂ Q ∪ ∞, where n ∈ N.

Proof. S is defined over Q, thus it exista F ∈ Q[X,Y ] such that S ' SF . We can write F as:F (X,Y ) =

∑nj=0 pj(X)Y j , where pj ∈ Q[X]. Consider the morphism πx : SF → C and the set of

critical values Bπx = µ1, . . . , µS. From Theorem 2.10, ∀k ∈ 1, s it follows that one of the followingcases takes place:

• pn(µk) = 0pn∈Q[X]=====⇒ µk ∈ Q.

• µk =∞.

• ∃α such that F (α, µk) = FY (α, µk) = 0. From Bezout’s Theorem, it follows that, in particular,µk ∈ Q.

From the above, it follows that Bπx ⊂ Q∪ ∞. If Bπx ⊂ Q∪ ∞, we conclude the proof. If not, wedo the following inductive reasoning. Let m1 ∈ Q[T ] be the minimal polynomial of the set Bπx \ ∞.Let B1 = β1, . . . , βd be the set of roots of m′1 and p the minimal polynomial of B1. By the propertiesof the minimal polynomial, deg(p) ≤ deg(m′1).We have the composition:

SFπx−→ C m1−−→ C

(x, y)→ x→ m1(x).

By Proposition 3.2, it follows that Bm1πx = m1(B1) ∪ 0,∞.If Bm1πx ⊂ Q ∪ 0,∞, which concludes the proof. If not, let m2 ∈ Q[T ] be the minimal polynomialof Bm1 = m1(β1), . . . ,m1(βd). The following inequality takes place [Q(m1(β1)) : Q] ≤ [Q(β1) : Q].We have:

deg(m2) ≤ deg(p) ≤ deg(m′1) < deg(m1). (3.1)

We get a composition of morphisms:

SFπx−→ C m1−−→ C m2−−→ C

(x, y)→ x→ m1(x)→ m2(m1(x)).

Denote by B2 the set of roots of m′2. The set of critical values Bm2m1πx is Bm2m1πx = m2(B2) ∪m2(Bm1πx). By construction, m2(Bm1πx) ⊂ Q∪∞. If Bm2m1πx ∈ Q∪∞ the proof is concluded.If not, we continue the inductive process and construct m3 the minimal polynomial of the set B2. Theprocess continuous until Bk ⊂ Q ∪ ∞ which will happen after a finite number of steps, because, ateach step: deg(mk+1) ≤ deg(mk),∀k.

29

Example 3.2. [4] Consider the Example in Subsection 3.2.1 for λ =√

2 and construct the associatedBelyi function following the algorithm in the proof of the Theorem.

y2 = x(x− 1)(x−√

2) ∪ ∞

Let F = Y 2−X(X−1)(X−√

2) ∈ Q[X,Y ]. Bπx = 0, 1,√

2,∞,√

2 /∈ Q. We get f(x, y) = −4(x2−1)(x2−2)2

.

Sc

πx(x,y)=x

f

C

m1(t)=t2−2(polinomul minimal al lui√

2)

C

u→−1/u

C

P1,1(z)=4z(1−z)

C

30

Chapter 4

Dessins d’enfants

4.1 Graphs on Surfaces

Definition 4.1. A topological disc is a topological surface homeomorphic to a disc.

Example 4.1. Example 1.12 shows that H is a topological disc.

Definition 4.2. [12] A map is a pair (X,D) of a finite graphD = (V,E, I) embedded in the topologicalsurface X such that:

• X \ Γ has a finite number of connected components, called faces of D,

• each face of D is a topological disc.

Example 4.2. (T,D) (where T is the torus) is a map.

Figure 4.1: Example of map

Counterexample 4.1. (T,D) where T is the torus is not a map because face 1 is not homeomorphic toa disc.

Figure 4.2: Counterexample of map

Definition 4.3. [4] The genus of the map (X,D) is the genus of the topological manifold X.

Definition 4.4. [4] Two maps (X1,D1), (X2,D2) are equivalent if ∃f : X1 → X2 homeomorphism

keeping the orientation such that the restriction of f to D1 induces an isomorphism: D1

f∼= D2.

31

Counterexample 4.2. [12] Consider K4 embedded in the sphere S2 and in the torus T. The two mapsare not equivalent because the sphere and the torus are not homeomorphic (it follows from the factthat: H1(S2) = 0, and H1(T) = Z).

Figure 4.3: K4 embedded in the torus and in the sphere

Definition 4.5. [4] A dessin d’enfant (dessin) is a map (X,D), such that X is a topologicalorientated compact manifold and D is a graph such that:

• D is simple

• D is bicoloured

Remark 4.1. The requirement for the graph to be bicoloured implies that each face has an even numberof edges. We shall consider the two colours to be black and white.

Example 4.3. From Proposition A.5, it follows that any tree is bicoloured. Each tree embedded inthe sphere is a dessin with a single face. Moreover, the only manifold X in which a tree T can beembedded is the sphere. T is homeomorphic to a point, therefore since X \ T is homeomorphic to acircle, it follows that X is homeomorphic to S2.

Figure 4.4: Tree embedded in the sphere

Example 4.4. [4] The tetrahedron T can be seen naturally as a graph embedded in the sphere. Fur-thermore, S2 \ T is the union of a finite number of topological discs. On the other hand, being acomplete graph with 4 vertices, it is not bicoloured. To get a bicoloured graph, let all the vertices ofT be black and ”add” vertices in the middle of each edge, getting the dessin below.

4.2 The Permutation Representation of a Dessin

Algorithm 4.1. [4] Let (X,D) be a dessin with N edges, denoted from 1 to N . Consider two permu-tations σa, σb ∈ SN defined as follows. For every i ∈ 1, N , draw a topological circle around the whitevertex of edge i and set σw(i) = j, if j is the edge that follows i after a positive rotation. Analogously,σb is obtained applying the same construction to the black vertices. The pair (σw, σb) is called thepermutation representation of the dessin D.

32

Figure 4.5: Example of a Dessin from a Tetrahedron

Example 4.5. In Example 4.4, the permutation representation of the dessin is given by:

σw = (1, 2)(3, 4)(5, 6)(7, 8)(9, 10)(11, 12) and σb = (1, 11, 10)(2, 3, 8)(4, 9, 5)(6, 12, 7).

Figure 4.6: The Permutation Representation of a Dessin

Remark 4.2. [4] Consider a dessin (X,D) with the permutation representation (σw, σb). The cyclesof σw are in bijection with the white vertices of D, the length of each cycle being the degree of thecorresponding white vertex. The cycles of σb are in bijection with black vertices of D, the length ofeach cycle being the degree of the corresponding black vertex.The cycles of σbσw (or the cycles of σwσb) are in bijection with the faces of D. More precisely, a cycleof length m:

i, (σbσa)(i), (σbσa)2(i), . . . , (σbσa)

m−1(i)

enumerates half of the 2m edges of one of the faces that contains edge i. Indeed, let ABCD . . . bethe polygonal border of a face (which has at least four edges, since the graph is bicoloured) with Aa black vertex. Starting from edge i = AB, σa(i) represents the edge BC (since it is the edge thatfollows AB after a positive rotation around the white vertex B), (σbσa)(i) represents the edge CD(since it is the edge that follows BC after a positive rotation around the black vertex C) and so on.In Example 4.4, σbσw = (1, 3, 9)(2, 11, 7)(4, 8, 6)(5, 12, 10). On the face ABCDEF , the cycle (1, 3, 9)generates edges: AB,CD,EF .

Proposition 4.1. [4] Let (X,D = (V,E)) be a dessin and (σa, σb) its permutation representation.The subgroup generated by the two permutations 〈σa, σb〉 ⊂ S|E| is transitive.

Proof. Denote by N = |E| and choose i 6= j ∈ 1, N . We want to show that ∃σ ∈ 〈σa, σb〉 such thatσ(i) = j.Consider the edges i and j in graph D. Since D is connected, it follows that there exists a fi-nite path connecting them: µ = (v1, i, v2, k1, v3, . . . , vt−2, kt−3, vt−1, j, vt) where v1, . . . , vt ∈ V and

33

Figure 4.7: The Correspondence between the Product σbσa and the Faces of the Dessin

k1, . . . , kt−3, i, j ∈ E.Assume, without loss of generality, that v1 is white, therefore v2 is black. We do the following reasoning(1). Let dD(v2) = m0 ∈ N \ 0, 1. It follows that there is p0 ∈ N∗, p0 ≤ m0 − 1 such that σp0b (i) = k1.If j = k1, σ = σp0b which concludes the proof. Otherwise, v3 is white. Let dD(v2) = m1 ∈ N \ 0, 1.It follows that there exists p1 ∈ N∗, p0 ≤ m1 − 1 such that σp1a (i) = k2. If j = k2, σ = σp1a σp0bwhich concludes the proof. Otherwise, we repeat reasoning (1) until we get in j and obtain σ as thecomposition of permutations at each step.

Remark 4.3. [4] Let (X,D) be a dessin with a given numbering of the edges and consider the per-mutation representation of the dessin: (σa, σb). If we renumber the edges in another way, we obtaina new permutation representation of the dessin: (σ′a, σ

′n)). There is a permutation σ ∈ Sn such that

σa = σσ′a and σb = σσ′n.

Proposition 4.2. [4] Let σa, σb ∈ Sn be two permutation such that 〈σa, σb〉 is a transitive group. Thenthere exists a dessin (X,D) such that its permutation representation is (σa, σb).

Proof. Consider the permutation representation σaσb: σaσb = σ1 . . . σk where ord(σj) = nj , j ∈ 1, k.

Obviously,∑k

j=1 nj = n. Create k disjoint polygons with 2nj edges, j ∈ 1, k, which will representthe faces of the dessin. Colour the vertices of each polygon in white and black and number the edgeswith numbers from 1 to n the edges using the information given by σwσb (which helps us label halfthe edges) and by σbσw (which helps us label the other half edges). Obviously, each number appearstwice. Now we glue along the edges labelled with the same number and we get a connected compactsurface because it is formed by a sequence of edges pairwise identified. The transitivity of the group〈σa, σb〉 insures that all the faces are connected. The orientability of the surface comes from the factthat the graph is bicoloured which insures the fact that the resulting surface does not contain Mobiusbands.

Remark 4.4. Proposition 4.1 and 4.2 give a bijective correspondence between dessins with n verticesand the transitive subgroups with two generators of Sn.

4.3 The Correspondence between Dessins and Belyi Function

4.3.1 The Belyi Function Associated to a Dessin

Definition 4.6. [4] Let (X,D) be a dessin. The decomposition in triangles of the surface X isa collection of triangles that covers all X such that the intersection of two triangles is either empty,either consisting of vertices or vertices and edges.

We present an algorithm for constructing a triangle decomposition of the surface X.

Algorithm 4.2. [4] Let (X,D) = (V (D), E(D)) be a dessin.

1. We choose a centre for each face of X \ D, and we mark it with ×.

34

2. Let v ∈ V (D) be a white vertex. Let d = dD(v).

3. The d edges of D incident to v divide a disc around v in d open disjoint sectors, i.e. D \ D =∪dj=1A

vj .

4. Draw a topological segment γvj that starts in v, crosses through Avj and ends in the centre of theface of D that contains Avj without intersecting any edge except at the vertex v.

5. We repeat the same procedure for every white vertex and then for every black vertex.

6. This construction associates to each edge j = [v, w] ∈ E(D) of D two triangles: T 1j and T 2

j . thetwo triangles have a common edge j, and the other two edges are in the sets γvk respectively

γwk . In this way, we described a triangle decomposition of T = T δj where j ∈ 1, |E(D)| andδ ∈ 1, 2. Let us observe that if dΓ(v) = 1, then the two triangles can have more than one edgein common.

7. The orientation of X allows us to classify the triangles T δj into two types which will be denoted

by: Twj (white) if the circuit → • → × → follows the positive orientation of the curve ∂T δjand T bj (black) otherwise. By construction, the adjacent triangles are of different colours; inparticular each face of D has the same number of black and white triangles.

Example 4.6. We will follow the Algorithm 4.2 on one face of the map in Example 4.4.

Figure 4.8: The Triangle Decomposition of Surfaces

Remark 4.5. [4] Let (X,D) be a dessin. Applying Algorithm 4.2, we get a decomposition of the surfaceX verifying the following properties:

1. Each triangle has vertices of the tree types: •, and ×.

2. Each triangle is completely determined by the following two pieces of information: its colourand the edge that is common with D.

3. Each edge j of D belongs to two triangles: T aj and Tnj .

4. Two adjacent triangles (with at least one common edge) are of different colours.

5. Each face f of D is decomposed into an union of 2mf triangles (2mf ∈ N∗), mf white and mf

black.

The purpose of this section is to describe an algorithm that associates a Belyi function to a dessin.

Algorithm 4.3. [4] Let (X,D = (V (D, E(D)) be a dessin with triangle decomposition: T (D) =T δj |δ ∈ a, n, j ∈ 1, |E|.We make the following notations: H+ = z ∈ C|Im(z) > 0 and H− = z ∈ C|Im(z) < 0.

35

1. We start from a white triangles: T aj . They represent the image of an injective continuousfunction: F aj : taj → X where taj are triangles in the plane that sends vertices to vertices and

edges to edges. Consider its inverse: (F aj )−1 : T aj → taj . From the Schwarz - Cristoffel Theorem

it follows that there exists a homeomorphism φaj : taj → H+that checks:

φaj (∂taj ) = R

φaj () = 0

φaj (•) = 1

φaj (×) =∞

. (4.1)

Composing the two functions, we get a homeomorphism: faj : T aj → H+, faj := φaj (F aj )−1 which

checks: faj (∂taj ) = Rfaj () = 0

faj (•) = 1

faj (×) =∞

. (4.2)

Analogously, let fnj : Tnj → H− which checks:fnj (∂Tnj ) = Rfnj () = 0

fnj (•) = 1

fnj (×) =∞

. (4.3)

These morphisms can be glued together to construct a continuous function: f : X → C with itsrestriction fT (D) : X∗ = X \ f−1

T (D) → C0, 1,∞ is a covering, which gives X∗ the Riemann surface

structure making fT (D) holomorphic (this follows from Theorem 2.2). The function fT (D) becomes a

Belyi function from X to C.

4.3.2 The Dessin Associated to a Belyi Function

Proposition 4.3. [4] Let f : X → C be a Belyi morphism and Df := f−1([0, 1]). If we see Df asa bicoloured graph embedded in X its white vertices being given by f−1(0), and the black by f−1(1),then:

1. (X,Df ) is a dessin,

2. the sets f−1([∞, 0]), f−1([0, 1]) and f−1([1,∞]) are union of topological segments, the unionbecoming the edges of a triangle decomposition T (Df ).

Proof. 1. Applying Theorem 2.1 for the function f : S → C, it follows that it is a ramified covering.The edges of the graph are the counterimage of the interval [0, 1], thus the graph is bicoloured.Furthermore, we get that f−1(C \ 0, 1,∞ is a disjoint union of topological discs, each of themcontaining one point of the counterimage f−1(∞). These will be the centres of the faces.

2. Consider the diagram

X \ f−1(0, 1,∞)

f |X\f−1(0,1,∞)

H+ \ 0, 1,∞

ik

55

i // C \ 0, 1,∞

where i is the inclusion and ik is the lifting of i. This lift is possible because H+ \ 0, 1,∞

is simply connected and the number of possible lifts is equal with the degree of f , each being

36

determined by the choice for a fixed point ε ∈ (0, 1). Furthermore, ik is injective and can be

continuously extended to H+. Indeed, by Theorem 2.1, ∀P ∈ f−1(0, 1) admits a neighbour-

hood DP homeomorphic with a disc on which f is of the form z → zmP where mP ∈ N∗.Thus, f−1([0, ε]) ∩ DP is a union of mP segments that intersect only in in P . Define ik(0) as

the point P0 ∈ f−1(0) determined by ik(ε) ∈ DP0 . The extension ik : H+ → X stays con-

tinuous and bijective. Furthermore, it is actually a homeomorphism, because H+is compact.

Therefore, T+k := (H+

) can be seen as a triangle ik([−∞, 0]), ik([0, 1]) and ik([1,∞]), where

k ∈ 1, deg(f) and −∞ = ∞. Analogously, we build T−k = ik(H−

). All these triangles give atriangle decomposition T (Df ) of X associated to D.

Remark 4.6. We showed a correspondence between Belyi functions and dessins. Summing up thisinformation with the permutation representation of a dessin, we get the following connections:

The group 〈σa, σb〉 Dessin Belyi function withBf = 0,1,∞

Cycles of σw White vertices of the graph f−1(0)Cycles of σb Black vertices of the graph f−1(1)The length of the corre-sponding cycle of σw

The degree of the white vertex vw ∈ V The branching order of the branch-ing point Pva ∈ f−1(0)

The length of the corre-sponding cycle of σb

The degree of the black vertex vb ∈ V The branching order of the branch-ing point Pvb ∈ f−1(1)

The order of the permu-tation group

The number of edges of the graph The degree of f

The cycles of σb σw The faces of the graph f−1(∞)The permutations in thesubgroup

The edges of the graph f−1([0, 1])

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Appendix A

Graph Theory Concepts

This chapter introduces several concepts of graph theory, in order to fix the notations and the terms.

Definition A.1. Let V 6= ∅. The set of pairs of V is the set of multisets of V with two elements:P2(V ) = [u, v]|u, v ∈ V (the same element can appear more than once).

Definition A.2. A undirected graph is an ordered triplet Γ = (V,E, I), where V and E are setsand I : E → P2(V ). The elements of V are called vertices of Γ, the elements of E are called edgesof Γ and the function I is called function of incidence. An edge e such that I(e) = [u, u] is called aloop. A graph Γ = (V,E, I) is finite if |V |, |E| ∈ N.

Remark A.1. Because we shall not study any other type of graphs, we shall call the objects in DefinitionA.2 graphs.

Definition A.3. Let Γ = (V,E, I) be a graph.

1. The vertices u, v ∈ V are said to be adjacent if ∃e ∈ E such that I(e) = [u, v]. A vertex isadjacent to itself if it exists e ∈ E such that I(e) = [u, u].

2. The edges e, f ∈ E are said to be adjacent if |I(e) ∩ I(f)| = 1.

3. The distinct edges e, f ∈ E are said to be parallel if |I(e) ∩ I(f)| = 2.

4. Let e ∈ E and u ∈ I(E). The vertex u is said to be incident to the edge e and the other wayround.

Definition A.4. [12] Let u ∈ V . The neighbourhood of u is the set of all vertices adjacent to it.The degree (the valency) is the number of edges adjacent to it (any loop is counted twice).

Proposition A.1. [25] Let Γ = (V,E, I) be a graph. The following result holds:∑v∈V

dΓ(v) = 2|E|.

Proof. Let M = (v, e)|v ∈ V, e ∈ E, v ∈ e.∑v∈V dΓ(v) =

∑v∈V |e ∈ E|v ∈ e| = |M | =

∑e∈E |v ∈ V |v ∈ e| =

∑e∈E 2 = 2|E|.

We shall distinguish some particular graphs.

Definition A.5. Let Γ = (V,E, I) be a graph.

1. If E = ∅, Γ is said to be the empty graph.

2. If Γ has no loops and no parallel edges, Γ is said to be a simple graph.

Remark A.2. If Γ = (V,E, I) is a simple graph, the edges of the graph can be identified with theirimages which are pairs of distinct elements of V . Consider P2(V ) = W ⊂ V ||W | = 2. If the graphis simple, then I(E) ⊂ P2(V ).

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Definition A.6. A simple graph Γ = (V,E) is a complete graph (all vertices are adjacent) ifI(E) = P2(V ). The complete graph with n vertices is usually denoted by Kn.

Definition A.7. [25] Let Γ = (V,E, I) be a graph.

1. The graph Γ′ = (V ′, E′, I ′) is a subgraph of Γ if V ′ ⊆ V , E′ ⊆ E ∩ P2(V ′) and I ′ = I|E′ .

2. Let W ⊂ V . The subgraph of Γ induced by W is the subgraph ΓW (W,E ∩ P2(W ), IW ).

3. Let V ′ ⊂ V . Γ \V′ is the subgraph induced by V \ V ′.

Definition A.8. Let Γ = (V,E, I) be a graph. A walk in Γ is a sequence of vertices and edges:µ = (v1, e1, v2, . . . , vk, ek, vk+1), where vj ∈ V, ej ∈ E, I(ej) = vj , vj+1.

Definition A.9. Let Γ = (V,E, I) be a graph and µ = (v1, e1, v2, . . . , vk, ek, vk+1) a walk in Γ.

1. The length of the walk is the number of edges in the walk: l(µ) = k.

2. The first vertex, v1, and the last vertex, vk+1, are called ends of the walk and µ connects v1

and vk+1.

3. If the vertices are all distinct: vj 6= vl,∀j, l ∈ 1, n+ 1, j 6= l, µ is a path.

4. If vj 6= vl, ∀j, l ∈ 1, n, j 6= l and v1 = vk+1, µ is a trail.

5. If µ is a trail, the vertices v1, v2, . . . , vk+1 form a cycle.

Definition A.10. Let Γ = (V,E, I) be a graph.

1. Γ is connected if ∀x, y ∈ E, if x 6= y then exists µ walk connecting x and y.

2. A connected component of Γ is a maximal subgraph of Γ′ such that it is connected as agraph.

Definition A.11. [25] Let k ∈ N∗ and Γ = (V,E) be a simple graph. Γ is said to be k-partite ifV =

⊔kj=1 Vk such that ∀e ∈ E, ∃j, l ∈ 1, n, j 6= l such that |I(e) ∩ Vj | = |I(e) ∩ Vl| = 1.

Remark A.3. If k = 2, we usually say that the graph is bipartite.

Definition A.12. [22] Let k ∈ N∗ and Γ = (V,E, I) be a simple graph. A k-proper colouring ofthe vertices Γ is a function, f : V → 1, 2, . . . , k, such that ∀u, v ∈ I(E), f(u) 6= f(v).

Remark A.4. [22] The set 1, 2, . . . , k can be seen as a set of colour, the k-colouring give each vertexthe value f(v), the colour of the vertex v.

Example A.1. Let Γ = (V,E, I) be a finite simple graph and k = |E|. Any injective function f : V →1, 2, . . . , k is a k-proper colouring.

Definition A.13. Let k ∈ N∗, Γ = (V,E, I) be a simple graph and f : V → 1, 2, . . . , k a k-colouring.If f is surjective, the graph Γ is said to be k-coloured.

Remark A.5. 1. If the graph is 2-coloured, the graph is usually said to be bicoloured.

2. A k-coloured graph can be seen as a k-partite graph and viceversa. Indeed, if Γ = (V,E, I) isk-coloured, with the colouring function f : V → 1, 2, . . . , k, then V =

⊔kj=1 f

−1(k). The other

way round, if Γ is k−partite: V =⊔kj=1 Vk, we build f : V → 1, 2, . . . , k where we define f(v)

as the index of the subset in which v is.

Definition A.14. Let Γ be a simple graph. The chromatic number of Γ is the smallest naturalnumber k ∈ N∗ for which Γ admits a k-proper colouring.

Definition A.15. [25] A tree is a connected simple graph without cycles.

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Proposition A.2. [25] Let Γ = (V,E, I) be a graph. If |E| ≥ |V |, then G has cycles.

Proposition A.3. [25] Let Γ = (V,E, I) be a simple graph with |V | = n, |E| = m. Let p be thenumber of connected components of Γ. Γ does not have cycles if and only if m− n+ p = 0.

Proposition A.4. A tree has at least one vertex of degree 1.

Proof. Let Γ = (V,E, I) be a tree with |V | = n, |E| = m. Since Γ is connected ⇒ ∀v ∈ V , dΓ(v) ≥ 1.

Assume that ∀v ∈ V , dΓ(v) ≥ 2. Γ is a treeProposition A.3=========⇒ m = n− 1. From Proposition A.1 it follows

that,∑

v∈V dΓ(v) = 2m = 2(n − 1). On the other hand, ∀v ∈ V , dΓ(v) ≥ 2 ⇒∑

v∈V dΓ(v) ≥ 2n,which is absurd ⇒ ∃v0 ∈ V such that dΓ(v0) = 1.

Proposition A.5. [22] A tree with at least two vertices is bicoloured.

Proof. We shall prove this result by induction on n the number of vertices.For n = 2, let V = v1, v2 be the set of vertices. Define the colouring function f : V → 1, 2 givenby: f(v1) = 1 and f(v2) = 2.Inductive step: P (n) → P (n + 1). Assume that each tree with n vertices is bicoloured. Let Γ bea graph with n + 1 vertices. From Proposition A.4 it follows that ∃v0 ∈ V such that dΓ(v0) = 1.Γ \ v0 is a tree with n vertices. By the inductive hypothesis, it is bicoloured. Colour v0 with theopposite colour of the only adjacent vertex and we get a 2-surjective colouring for Γ, which concludesthe proof.

Definition A.16. Let Γ = (V,E, I) and Γ′ = (V ′, E′, I ′) be two graphs. We call a morphism ofgraphs from Γ to Γ′ a function f : V → V ′ such that ∀u, v ∈ V if [u, v] ∈ I(E), then [f(u), f(v)] ∈I(E′).

Definition A.17. Let Γ = (V,E, I) and Γ′ = (V ′, E′, I ′) be two graphs. An isomorphism ofgraphs between Γ and Γ′ is a function f : V → V ′ such that ∀u, v ∈ V , [u, v] ∈ I(E) if and only if[f(u), f(v)] ∈ I(E′).

Definition A.18. [18] A graph Γ is embedded in a topological space X if the vertices of Γ are distinctpoints in X and every edge of Γ is a simple arc connecting in X the two vertices which it joins in Gsuch that its interior is disjoint from other edges and vertices.

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