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Grothendieck’s dessins d’enfants
Milan Wernke
July 13, 2012
Bachelor Thesis
Supervisor: Prof. dr. Sergey Shadrin
KdV Instituut voor wiskunde
Faculteit der Natuurwetenschappen, Wiskunde en Informatica
Universiteit van Amsterdam
AbstractA Riemann surface X together with a meromorphic functionf : X → C unramified outside {0, 1,∞} can be described us-ing a special type of embedded graphs, the dessins d’enfants.After a quick sidestep to Shabat polynomials, the above corre-spondence is made explicit. The usage of dessins d’enfants withinGalois theory is explained, and using the formalism of curves andvarieties, a proof of the Belyi theorem is given. This theoremtells us when a Riemann surface is defined over the field Q ofrational numbers. Lastly, a recursion formula for the number ofgenus g clean Belyi functions having n poles of fixed degree is de-rived. This formula is then shown to satisfy the Eynard-Orantintopological recursion.
Contact InformationTitle: Grothendieck’s dessins d’enfantsAuthor: Milan Wernke, [email protected], 6118704Supervisor: Prof. dr. Sergey ShadrinSecond assessor: Dr. Hessel PosthumaDate: July 13, 2012
Korteweg de Vries Instituut voor WiskundeUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math
Contents
Introduction 2
1 Preliminaries 41.1 Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Graphs and embeddings . . . . . . . . . . . . . . . . . . . . . 61.3 Riemann surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Dessins d’enfants and the Belyi Theorem 122.1 Shabat polynomials . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Examples of dessins d’enfants . . . . . . . . . . . . . . . . . . 212.3 Galois theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4 The Belyi theorem . . . . . . . . . . . . . . . . . . . . . . . . 28
3 Counting dessins d’enfants 363.1 The Eynard-Orantin recursion . . . . . . . . . . . . . . . . . . 363.2 A relation for the number of dessins . . . . . . . . . . . . . . . 393.3 The recursion formula for dessin d’enfants . . . . . . . . . . . 46
Summary 55
1
Introduction
Dessins d’enfants are embedded graphs that are used in modern day mathe-matics to study Riemann surfaces and to provide invariants for the action ofthe absolute Galois group. A dessin d’enfant is the French term for a ‘child’sdrawing’. These graphs have been given this name due to the fact that theyare rather simple of nature, at least on first sight.Proto-forms of dessins d’enfants were already used back in 1856 by WilliamHamilton in his work Icosian Calculus. In 1879 Felix Klein was the first touse the modern type of dessins, although he denoted them by Linienzuge,which means ‘line-track’ in German. After Klein, the dessins d’enfants wereforgotten for more than a century until Alexander Grothendieck rediscoveredthem in 1984, using them in his Esquisse d’un Programme. Grothendieck,who gave the dessins d’enfants their current name, was very intrigued by theusage within Galois theory of such an easy to understand object.Or, as he stated it himself:
“This discovery, which is technically so simple, made a very strongimpression on me, and it represents a decisive turning point in thecourse of my reflections, a shift in particular of my center of inter-est in mathematics, which suddenly found itself strongly focused.I do not believe that a mathematical fact has ever struck me quiteso strongly as this one, nor had a comparable psychological im-pact. This is surely because of the very familiar, non-technicalnature of the objects considered, of which any childs drawingscrawled on a bit of paper (at least if the drawing is made with-out lifting the pencil) gives a perfectly explicit example. To sucha dessin we find associated subtle arithmetic invariants, whichare completely turned topsy-turvy as soon as we add one morestroke.”
The main purpose of this thesis is to give the reader a better understandingof what is a dessin d’enfant, and show how a dessin d’enfant can be derivedfrom a Belyi pair and vice-versa. Once this is done, we will be discussing
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some important theorems and properties regarding dessins d’enfants. As wellas introducing the reader to some basic Galois theory.The plan of this thesis will be as follows: chapter one will contain the elemen-tary notions needed to define dessins d’enfants, as well as some propositionsthat will be used in proofs throughout the rest of the chapters. Most ofchapter one is based on the book of Lando and Zvonkin [1], which is in factthe primary source for this thesis. The proofs of the propositions (that areomitted in this chapter) all can be found in [1] as well. Chapter two alsorelies heavily on this source, in this second chapter the definition of a dessind’enfant will be given, together with examples intended to clarify the corre-spondence of Belyi pairs and dessins. Furthermore this chapter contains ashort elucidation on Galois theory and the chapter is concluded by the proofof the Belyi theorem. The second part of this proof is due to the paper ofBernhard Kock [5]. In chapter three we will find an effective recursion for-mula for the number of dessins d’enfants. This last chapter is based on thepaper of Dumitrescu and Mulase [8]. The formula for the number of dessinsis tested using the topological recursion of Eynard and Orantin [6].
Before we start with the first chapter, I would like to thank my supervisorSergey Shadrin for introducing me to this beautiful subject and express mygratitude for all the assistance he gave me writing this thesis. I would alsolike to thank Hessel Posthuma for his comments and advise in the final stagesof completion.
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Chapter 1
Preliminaries
The main goal of this chapter is to introduce the definitions, propositionsand theorems needed to properly discuss the theory of dessins d’enfants,which will be the subject of the second chapter. Note that the theorems andpropositions in this first chapter all are stated without proofs. (The proofsin question can be found in [1].)We start of this chapter by looking at a special type of continuous functionsbetween topological spaces.
1.1 Coverings
Definition 1.1. Let f : X → Y be a continuous map where X and Y arepath connected topological spaces. We call the triple (X, Y, f) an (unrami-fied) covering of Y by X if for every y ∈ Y there is a neighbourhood U 3 yand a discrete set S such that f−1(U) ⊂ X is homeomorphic to U × S.
The map f is called the covering map from X → Y .The connected components of f−1(U) are called the sheets of the coveringover U and f−1(y) is called the fiber over y.The degree of the covering deg(f) is equal to the cardinality of S. Whendeg(f) = n the covering map f is called n-sheeted. When n <∞ the cover-ing is finite-sheeted.
Definition 1.2. Let f1 : X1 → Y and f2 : X2 → Y be two unramifiedcoverings. They are said to be isomorphic if there exists a homeomorphismu : X1 → X2 such that the following diagram is commutative:
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X1
f1
u // X2
f2~~Y
Example 1.3. We start out with a simple example. Let both X and Y beequal to the unit circle S1. Describe a point in S1 by its angle ϕ mod 2π.The mapping
f : ϕ 7→ nϕ mod 2π
is then an example of an unramified covering of degree n. An arbitrary pointy ∈ Y then has a preimage f−1(y) consisting of n points in X. Below isshown an example for n = 8.
Figure 1.1: The unramified covering f : ϕ 7→ 8ϕ mod 2π.
Note that instead of describing points in S1 by means of their angular co-ordinate ϕ one can also describe them using the complex numbers z having|z| = 1. Using these coordinates we will now describe an unramified coveringof an annulus by another annulus.
Example 1.4. Let X = {(r, ϕ) | 0 ≤ r1 < r < r2} and Y = {(r, ϕ) | 0 ≤rn1 < r < rn2} be two annuli. The function
f : (r, ϕ) 7→ (rn, nϕ mod 2π)
is then an unramified covering of Y by X. In complex coordinates the cov-ering map can also be written in the form
f : z 7→ zn
In the figure below the preimage of a horizontal segment on Y is shown forn = 8.
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Figure 1.2: The unramified covering f : z 7→ z8.
Remark. For r1 = 0 the annuli are in fact punctured open disks. Adding thepoint with r = 0 to both X and Y one obtains a so called ramified coveringof an open disk by another open disk. The new mapping still is continuousand all points of Y have n preimages, except for the point in the center of Ywhich only has a single preimage, the point in the center of X. This preimagewill be called a critical point or ramification point of multiplicity n and thecenter of Y is called a critical value or branched point. So the set of criticalpoints of f are the points of X where f fails to be a covering of Y .If we denote the set of critical values of f with C we say that f is a coveringof Y by X unramified outside C.
From the discussion on coverings we turn to what seems a wholedifferent topic; namely that of graphs and embeddings. It will be shownin chapter two that the theory of dessins d’enfants will connect these twodistinct subjects of mathematics in a beautiful way.
1.2 Graphs and embeddings
Definition 1.5. A graph Γ is given by a triple (V,E, I) where V is a set ofvertices v, E is a set of edges e and I is an incidence relation between Vand E such that every edge e ∈ E is incident to either two different verticesv1, v2 ∈ V , or is incident twice to the same vertex v ∈ V in which case e iscalled a loop.
Definition 1.6. A path in a graph Γ is a sequence v0, e1, v1, e2, . . . , en, vnsuch that ei is incident to vi−1 and vi for i = 1, . . . , n. This path will be saidto connect v0 to vn.A graph Γ is connected if every two of its vertices can be connected by apath.
Definition 1.7. The number of edges incident to a vertex v will be calledthe degree, or valency, of the vertex and is denoted by deg(v). An important
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note is that every loop is counted twice.
Remark. From the above it immediately follows that∑
v∈V deg(V ) = 2|E|.Definition 1.8. Let Γ1 and Γ2 be graphs with corresponding vertex sets V1
and V2. A bijection f : V1 → V2 is called an isomorphism of graphs whenu, v ∈ V1 are adjacent in Γ1 ⇐⇒ f(u), f(v) ∈ V2 are adjacent in Γ2.Two vertices u, v are called adjacent in Γ when there is a single edge in Γconnecting them.
Now we have properly defined our graphs we want to perform acertain action on them, the action consists of placing a graph on a topologicalspace in a specific way. After performing the action on a graph, we say thatthe graph is embedded. We will now formalize the above notions.
Definition 1.9. A topological space X which is Hausdorff and has a count-able basis is called a manifold of dimension n when every point of X has aneighbourhood which is homeomorphic with an open set of Euclidian spaceof dimension n. A compact oriented two-dimensional manifold will be calleda surface.
Definition 1.10. An embedding M is a graph Γ embedded in a surface Xin such a way that:
• the vertices of Γ are represented as distinct points of the surface X
• the edges of Γ are represented as curves on X that intersect only at thevertices
• X \ Γ is a disjoint union of connected components called faces, whereeach face is homeomorphic to an open disk.
When speaking about the genus of the embedding M , the genus of the em-bedded surface X is meant. An embedding M is connected when the graphΓ is connected.
Example 1.11. We now give two clarifying examples:
Figure 1.3: Two embeddings of the same graph
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Figure 1.4: Not an embedding, the outer face is not homeomorphic to anopen disk
Definition 1.12. Two embeddings M1 ⊂ X1, M2 ⊂ X2 are isomorphic ifthere is an orientation preserving homeomorphism u : X1 → X2 such thatu|Γ1 is a graph isomorphism between Γ1 and Γ2.
Definition 1.13. The degree, or valency of a face f denoted deg(f) is thenumber of edges incident to this face. An edge is called incident to a facef when it belongs to the boundary of f . When both ‘sides’ of an edge areincident the edge is called an isthmus and the edge will be counted twice.
Proposition 1.14. Let F denote the set of faces of the embedding M withgraph Γ = (V,E, I). It holds that∑
f∈F
deg(f) = 2|E|.
Theorem 1.15. Given an embedding M as before with the additional prop-erty of being connected. We denote its Euler characteristic by χ(M) where
χ(M) = |V | − |E|+ |F |
It holds that χ(M) only depends on the genus g of the embedding M . Wehave the equality
χ(M) = 2− 2g
Example 1.16. We now give an example to illustrate the previous definitionsand theorem.Below is shown a connected embedding M with 4 faces corresponding to agraph (V,E, I) with: |V | = 5, |E| = 7.
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Figure 1.5: The numbers denote the degrees of the faces they are written in
The edge connecting the two most left vertices is an isthmus and the mostright edge is a loop. It holds that χ(M) = |V | − |E|+ |F | = 5− 7 + 4 = 2⇒g = 0. This is correct since the embedding is in the Riemann sphere, a spaceof genus zero that will be discussed soon.
Definition 1.17. An embedding of which the vertices are coloured in blackand white in such a way that each edge connects only vertices of differentcolours is called a hyperembedding.
Remark. Note that every hyperembedding is the preimage of the elementaryhyperembedding. This is the hyperembedding containing one face of degree 1and one edge connecting a single black vertex to a single white vertex.
Figure 1.6: The elementary hyperembedding
The graphs we study in this thesis will not be embedded into anyarbitrary surface. Only surfaces of a specific type, namely Riemann surfaces,will be considered. The definition and the properties of such surfaces will bethe subject of the next section.
1.3 Riemann surfaces
Definition 1.18. A complex analytical manifold of complex dimension oneis called a Riemann surface whenever it is a connected surface. A Riemannsurface is best visualized by a sphere with g handles on it, where g is thegenus of the Riemann surface.
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An example of an often used Riemann surface of genus zero is the Riemannsphere C which is obtained by adding the point ∞ to the complex plane.
Definition 1.19. We say that two Riemann surfaces are isomorphic whenthere exist a biholomorphic bijection between. The biholomorphic bijectionis then called a complex isomorphism.
Remark. Note that the isomorphism defined above is different from a home-omorphism. Two homeomorphic Riemann surfaces (meaning that they havethe same genus) may not be the same as complex manifolds as they canpossess different complex structures. This complex structure rises from thechoice of complex coordinates in a neighbourhood of every point.
Definition 1.20. Let X be a Riemann surface. A mapping f : X → C iscalled a meromorphic function on X if f is holomorphic on X \ P whereP = {x ∈ X : f(x) =∞} is a discrete set. P will be called the set of poles off . Another important notion is that of a zero of f , a point x ∈ X for whichf(x) = 0.
Before we continue, two important remarks about meromorphic functionsand Riemann surfaces must be made:
Remark. When X = C, then every meromorphic function is the quotient oftwo polynomial functions. In this case the meromorphic functions is called arational function. Meromorphic functions on C with only a single pole at ∞are polynomials themselves.
Remark. For each Riemann surface there are many non-constant meromor-phic functions defined on them.
Proposition 1.21. All Riemann surfaces can be realized as algebraic curvesin a complex projective plane. Therefore each Riemann surface can be givenby a system of polynomial equations with coefficients in C.
The above proposition is used in many occasions, in this thesis itwill for example be used to discuss the connection between Galois theoryand dessins d’enfants.
Definition 1.22. If a Riemann surface X can be realized by a system ofpolynomial equations with coefficients in a subfield K of C then we say thatX is defined over K.
Proposition 1.23. A nonconstant meromorphic function f : X → C is acovering of C.
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Definition 1.24. Let (X1, f1,C) and (X2, f2,C) be two complex coveringsof the Riemann sphere. They are said to be isomorphic if there exists abiholomorphic isomorphism u : X1 → X2 such that the following diagram iscommutative:
X1
f1 ��
u // X2
f2��C
Remark. Note that this is almost the same definition as given in 1.2, exceptfor the fact that u must be a complex isomorphism instead of a homeomor-phism.
Definition 1.25. Let Sn denote the symmetric group acting on n points. Ak-constellation is a sequence [g1, g2, . . . , gk] where gi ∈ Sn that satisfies thefollowing two properties:
• The group G = 〈g1, g2, . . . , gk〉 acts transitively on the set of n points.
• The product of the gi is the identity permutation: g1g2 . . . gk = id.
The cycle structure of a permutation g ∈ Sn is the partition λ ` n con-sisting of the lengths of the cycles of g. For example, let n = 12 andg = (1, 3, 5, 7)(2, 4, 6)(8, 9, 12). Then g contains one cycle of length 4, twocycles of length 3 and two cycles of lenght 1 (the fixed points 10 and 11 ).The partition λ of 12 = 4 + 3 + 3 + 1 + 1 will be denoted by (4,3,3,1,1) orjust simply by 43212.
Definition 1.26. Let C = [g1, g2, . . . , gk] be a constellation. The passport ofC is the sequence [λ1, . . . , λk] of partitions of n, where λi is the cycle structureof the permutation gi, i = 1, . . . , k.
Theorem 1.27 (Riemann’s existence theorem). Suppose a set R = {y1, . . . , yk}with yi ∈ C is fixed. Then for any constellation [g1, . . . , gk], gi ∈ Sn, thereexists a Riemann surface X and a meromorphic function f : X → C suchthat the critical values of f are y1, . . . , yk and the values g1, . . . , gk are the cor-responding monodromy permutations (which fix the structure). The ramifiedcovering f : X → C is unique up to isomorphism.
With this very important theorem we conclude the first chapter.
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Chapter 2
Dessins d’enfants and the BelyiTheorem
Definition 2.1. A Belyi function is a meromorphic function f : X → Cunramified outside {0, 1,∞}. A Belyi pair is a pair (X, f), where f is aBelyi function on the Riemann surface X.
Consider the segment [0, 1] ⊂ C, we denote 0 with • and 1 with ◦. Do-ing so the segment will look like the elementary hyperembedding as definedbefore (see the remark to definition 1.17). We then take the the preimageH = f−1([0, 1]) ⊂ X. H is then a hyperembedding on the Riemann surfaceX. This hyperembedding has the following properties: the white vertices ofH are the preimages of 1 and the black vertices are the preimages of 0. Also,the valencies of the vertices are equal to the multiplicities of the correspond-ing critical points. Note that vertices of valency one are not critical pointsof f but still may arise as the preimage of either 0 or 1. Each face of the hy-perembedding H contains exactly one pole. The poles in a hyperembeddingwill be denoted as the centers of faces. Since f is a Belyi function there willbe no other critical points of f on X besides the set of centers of faces and(black or white) vertices.A hyperembedding considered as a representation of a particular Belyi pairis called a dessin d’enfant. This representation is unique up to isomorphism.
Definition 2.2. A clean Belyi function is a special class of Belyi functionswhich have the property that all their critical points corresponding to thecritical value 1 have multiplicity 2. This implies that all clean Belyi functionare of even degree.
Remark. The dessin d’enfant corresponding to a clean Belyi function hassome very nice properties. The fact that every critical point corresponding
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to critical the value 1 has multiplicity 2 leads to a dessin which only haswhite vertices of degree 2. This means that every white vertex is the edgemidpoint of an edge connecting two black vertices (which may be the same).Therefore it is possible to draw the dessin d’enfant of a clean Belyi functionby only drawing the black vertices and the edges connecting them.
Before we give some examples of the correspondence of Belyi pairsand dessins d’enfants we first turn to a simpler case.
2.1 Shabat polynomials
Consider a polynomial P : C → C of degree n. The preimage P−1(y0) ={x ∈ C | P (x) = y0} of a fixed point y0 ∈ C usually consists of n points.However for some values of y0 this will not be the case since the equationP (x) = y0 may have multiple roots. For instance: let p(x) = x3 +2x2−7x+4and take a look at the set p−1(0) i.e. at the set of solutions of the equalityx3 +2x2−7x+4 = 0. Since p is a polynomial of degree 3 one expects this setto have three elements, however the only solutions to this problem are 1 and−4 so the set p−1(0) only has two elements. In this case 1 is a double root.This becomes clear by writing p(x) = x3 + 2x2 − 7x+ 4 = (x+ 4)(x− 1)2.It holds that a multiple root of the equation P (x) = y0 is also a root ofP ′(x) = 0. This leads us to the following definition:
Definition 2.3. Let P be a polynomial from C to C. A point x ∈ C for whichP ′(x) = 0 is called a critical point of P . The value y = P (x) ∈ C is calledthe critical value corresponding to the critical point x. The multiplicity ofthe critical point x is the value k ≥ 2 such that P (k)(x) 6= 0 while P ′(x) =P ′′(x) = . . . = P (k−1)(x) = 0.
In this section we will study a specific type of polynomials which will turnout to correspond to a simpler type of embeddings, namely trees. Here a treeis a connected embedding with just a single face.
Definition 2.4. A Shabat polynomial is a polynomial with at most twocritical values. Stated differently, for every Shabat polynomial P there existtwo distinct complex numbers y0, y1 such that
P ′(x) = 0⇒ P (x) ∈ {y0, y1}.
Remark. All Shabat polynomials P can be renormalized to have the criticalvalues 0 and 1 by the substitution P (x) −→ (P (x)− y0)/(y1 − y0).
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Suppose we are given a polynomial P of degree n and two values y0, y1 ∈ Cthat we connect with a straight line segment which we will denote by [y0, y1].To distinguish between the two points graphically we colour one in black andthe other in white. We will now look at the preimage P−1([y0, y1]). Whenthere are no critical values on the segment [y0, y1] the preimage of the segmentis the disjoint union of n “curvilinear segments”, as illustrated below.
Figure 2.1: Preimage of a segment with no critical values for n = 5
If we do allow the endpoints to become critical (so that there still are nocritical values inside the segment) the structure of the preimage will alter.Some of the curvilinear segments will be glued together at their endpoints.
Figure 2.2: Preimage of a segment with critical values on its endpoints, alsofor n = 5
The final question of course is, what happens when P is a Shabat polynomialand y0, y1 are its critical values?We will answer this question in the following proposition.
Proposition 2.5. Let P be a Shabat polynomial with corresponding criticalvalues y0 and y1. The preimage P−1([y0, y1]) is a bicoloured plane tree.
Proof. First note that since the preimage P−1([y0, y1]) is a (bicoloured) graphembedded in the complex plane C, a space of genus g = 0, it holds that (us-ing Euler’s characteristic) |V | − |E|+ |F | = 2.Now let P be of degree n, then the total number of solutions to the equationsP (x) = y0 and P (x) = y1 counted with multiplicities is equal to 2n. Thenumber of solutions to the equation P ′(x) = 0 (counted with multiplicities)is equal to the number of solutions to the previous equations with the multi-plicity of every solution diminished by one. (Since for example a triple root
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of P (x) = yi is a double root of P ′(x) = 0 .) Since P ′(x) is a polynomial ofdegree n − 1 the number of solutions to the equation P ′(x) = 0 is equal ton − 1. Then the total number of solutions to the two equations P (x) = y0
and P (x) = y1 counted without multiplicities is equal to 2n− (n−1) = n+1.This number is the amount of vertices the preimage P−1([y0, y1]) has. Sincethe degree of P is n we obtain an embedding with n+1 vertices and n edges.We then have two possibilities:
The embedding is connected, then by using the Euler characteristicabove we see that |F | = 2− |V | + |E| = 2− (n + 1) + n = 1. So we have aconnected (bicoloured) embedding with a single face, which is a tree.
If the embedding is not connected the embedding must possess acircuit as illustrated below.
Figure 2.3: A disconnected embedding with a circuit
We may assume both the points y0, y1 to be real (as we may apply an affinetransformation on C ). This will award us with a real-valued segment [y0, y1].Therefore the polynomial P only takes real values on the boundary of thedomain bounded by the circuit. Meaning that the imaginary part of ourpolynomial =P (x) is identically equal to zero on the boundary. =P (x) be-ing a harmonic function on R2 is therefore identically equal to zero insidethe boundary, which is impossible. The embedding therefore can not bedisconnected. This concludes the proposition.
Example 2.6. The polynomial P (x) = xn has a single critical point x = 0corresponding to the critical value y = 0, the order of this critical point is n.If we take the preimage under P of [0, 1] the corresponding bicoloured planetree has the shape of a “star-tree”.
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Figure 2.4: The tree corresponding to P (x) = xn for n = 8
Example 2.7. The function cosnϕ can be expressed as a polynomial ofdegree n in cosϕ. For example:
cos 2ϕ = 2 cos2 ϕ− 1
In general it holds thatcosnϕ = Tn(cosϕ),
where Tn(x) is the n-th Chebyshev polynomial. Tn(x) is defined by
Tn(x) = 2xTn−1(x)− Tn−2(x) with T0(x) = 1, T1(x) = x.
The graph of the polynomial Tn(x) on the segment [−1, 1] (that is Tn([−1, 1])and not T−1
n ([−1, 1]) ) resembles the graph of cosnϕ on [−π, 0]. This impliesthat all the maxima of Tn(x) are equal to 1 and all its minima to −1. SoTn(x) has n− 1 critical points of order 2, the points x = cos kπ
n, but only two
critical values (−1 and 1).The preimage under Tn(x) of the segment [y0, y1] = [−1, 1] therefore has theform of a “chain-tree”.
Figure 2.5: The tree corresponding to P (x) = Tn(x) for n = 8
Shabat polynomials are also called generalized Chebyshev polynomials.
We have already given the definition for equivalence of two mero-morphic functions, which become rational functions when we consider onlycoverings of C from C. However we are currently discussing polynomials,which means that we must preserve the point at ∞. That means that theonly allowed automorphisms must be affine. This leads us to the followingdefinition.
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Definition 2.8. Let P and Q be two Shabat polynomials, with critical valuesy0, y1 and z0, z1 respectively. We call the pairs (P, [y0, y1]) and (Q, [z0, z1])equivalent if there exist constants a, b, A,B ∈ C, a, A 6= 0 such that
Q(x) = AP (ax+ b) +B, and z0 = Ay0 +B, z1 = Ay1 +B.
When the pairs (P, [y0, y1]) and (Q, [z0, z1]) are equivalent we shall simplysay that the two polynomials P and Q themselves are equivalent.
Theorem 2.9. There is a bijection between the set of combinatorial bi-coloured plane trees and the set of equivalence classes of Shabat polynomials.
Proof. This theorem is a particular case of Riemann’s existence theorem (seetheorem 1.27) so there remains nothing to proof. We will follow here themain stages of the correspondence “tree 7→ polynomial”.Suppose we have a bicoloured plane tree, then we know that it is given by itspassport, which is a 3-constellation. Now this 3-constellation together withan arbitrary fixed pair of complex numbers y0 and y1 determines a ramifiedcovering. (Due to Riemann’s existence theorem.) Since the constellation isplanar this is a covering f : C→ C. Using the first remark to definition 1.20we see that f is a rational function. Since f only has a single pole at infinity,we see that f is a polynomial. And finally, Riemann’s existence theorem alsoassures the uniqueness of f .
Finding the Shabat polynomial corresponding to a given tree can be verycomplicated at times, but the above theorem assures us that a solution doesexist. This gives us the courage to carry on to some examples.
We will shortly give two examples on how to compute the Shabatpolynomial corresponding to a given tree. Before we do so however somegeneral notions need to be made.Given a tree where a1, a2, . . . , ap and b1, b2, . . . , bq are the coordinates of itsblack and white vertices respectively. We then denote this tree by its passport
[α, β, n] = [(α1, α2, . . . , αp); (β1, β2, . . . , βq), n].
Here it holds that αi is the valency of the vertex ai, βi the valency of vertexbi, and n is the number of edges of the tree. (Note that p+ q = n+ 1.) Letus take y0 = 0 and y1 = 1. This gives us the equalities
P (x) = C(x− a1)α1(x− a2)α2 . . . (x− ap)αp ,
P (x)− 1 = C(x− b1)β1(x− b2)β2 . . . (x− bq)βq .
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These equalities give rise to n algebraic equations in n+ 2 unknowns (C andthe n+ 1 a’s and b’s). This gives us two “degrees of freedom” which provideus the possibility of making an affine transformation of the complex plane.We will use this to either fix the position of two given vertices or make somedifferent choice which fixes the tree in an unambiguous position on the plane.
Example 2.10. Consider the tree with passport [(3, 1); (2, 1, 1), 4]. We willalso denote this passport more briefly as [31, 212, 4]. This passport describesa tree with one black vertex of degree 3, one black vertex of degree 1, onewhite vertex of degree 2, two white vertices of degree 1 and four edges. Thistree is shown below.
Figure 2.6: The tree corresponding to the passport [31, 212, 4]
We fix y0 = 0, doing so will make the coordinates of the black vertices theroots of the polynomial we are looking for, where the black vertex of degree3 is a triple root. The liberty of using affine transformations allows us toplace this vertex at the point x = 0, whilst placing the other black vertex atx = 1. This will give us the corresponding polynomial P (x) = x3(x− 1).To find the coordinates of the white vertices, we use the derivative P ′(x) =x2(4x − 3). This derivative has a double root at x = 0, which is to beexpected since we placed the vertex of degree 3 at x = 0, and has a singleroot at x = 3/4. This root then must be the position of the white vertex ofvalency 2. Having found this value we can compute the second critical valuey1, since
y1 = P (3
4) = −33
44= − 27
256.
To find the coordinates of the remaining two white vertices all we have to dois solve
x3(x− 1) = − 27
256.
This equation of course has a double root at x = 3/4, but it also has twosimple roots at (−1 ±
√−2)/4, which are the coordinates of the two white
vertices of valency 1.
18
From this result we can infer that it would have been more clever to posi-tion the black vertex of valency 1 at x = 4 and to change the sign of thepolynomial, changing it into P (x) = x3(4 − x). With this choice the whitevertex of degree 2 will be located at x = 3 and the second critical valuey1 = P (3) = 27. The coordinates of the other two white vertices would alsoalter to become −1±
√−2.
In the previous example we used an affine transformation to fix two of thevertices of the give tree. However, different approaches are also possible, asthe following example will illustrate.
Example 2.11. Consider the tree with passport [322, 2213, 7]. Again we willfix y0 = 0 and place the black vertex of degree 3 at x = 0. Instead of fixingthe location of yet another vertex, we place the two remaining black verticesin such a way that the (virtual) segment joining them has its midpoint atx = 1. This implies that the sum of the coordinates (as complex numbers)of these two black vertices equals 2, meaning that this assumption only usesone “degree of freedom”. This tree is shown below.
Figure 2.7: A tree corresponding to the passport [322, 2213, 7]
The black vertices of degree 2 then must be roots of a quadraticpolynomial x2 − 2x + a for some value of the unknown parameter a. Thisholds because the roots of x2 − 2x+ a are of the form (2±
√4− 4a)/2, and
the sum of these two roots equals 2 for all a. The Shabat polynomial maytherefore be chosen equal to
P (x) = x3(x2 − 2x+ a)2.
We would now like to find a, to do so however we need to find some moreconditions. Let us first take the derivative of our Shabat polynomial, we find
P ′(x) = x2(x2 − 2x+ a)(7x2 − 10x+ 3a).
19
Except for the expected double root at x = 0 and the two roots of x2−2x+athis polynomial has two other simple roots, those of the polynomial
Q(x) = 7x2 − 10x+ 3a.
These roots must be the coordinates of the two white vertices of degree 2.This gives us the following conditions:
• We need two vertices, so the roots of Q must be distinct, therefore
discriminant(Q) = 21a− 25 6= 0
• The two roots of Q must give the same value for P , this value is thesecond critical value y1.
Instead of just computing the roots and substituting them into P , we usehere a different, more interesting approach.We divide P by Q,
P = S ·Q+R,
where R is the remainder and therefore linear (as Q is quadratic). So we maywrite R(x) = Ax + B, A and B depending on a. When Q = 0, the value ofP equals the value of R, so the values of R at the roots of Q must be equalto each other. This implies that A = 0 and R is thus just a constant.If we do the computation we get the following results:
A = −16
76(21a− 25)(49a2 − 476a+ 400),
B = −196
76a(28a− 25)(7a− 10).
The equations A = 0 and 21a− 25 6= 0 now infer that
D(a) = 49a2 − 476a+ 400 = 0,
and therefore
a =1
7(34± 6
√21),
which gives the second critical value y1 = B = B(a).Only one of these values can be correct, and on first sight it is not clear whichvalue of a is the right one. If we do the computations for both values of ait turns out that for one value of a we obtain the tree shown in the figureabove and for the other value of a we obtain the tree shown below.
20
Figure 2.8: A tree corresponding to the second possible value of a
Note that this is a tree also corresponding to the passport [322, 2213, 7], thususing the method above we have covered multiple possibilities at once.
Before we continue, we first introduce some notions related to theabove example.
• Since the coefficients of the Shabat polynomial and its critical valuesbelong to the number field Q(
√21) we say that the corresponding tree
is defined over the field Q(√
21).
• The polynomial whose roots generate this field (in this case D(a) =49a2 − 476a+ 400 ), is called a defining polynomial.
• The two different trees are called conjugate since they arise from twoalgebraically conjugate values of the parameter a.
• The set consisting of the two above trees is a Galois orbit, which meansthat it is an orbit of the action of the universal Galois group Γ =Gal(Q | Q) on trees. These notions will be made more explicit later,for now it is enough to realize that when the Galois group Γ sends
√21
to −√
21, one of the Shabat polynomials is transformed into the otherand thus the corresponding tree is also transformed into the other tree.
It is now time to look at some general embeddings.
2.2 Examples of dessins d’enfants
Example 2.12. In this example we consider a simple embedding in theRiemann sphere, it is the preimage of the segment [0, 1] under an unknownBelyi function. Note that it is not a tree.
21
Figure 2.9: A simple embedding in C
This is an embedding corresponding to a clean Belyi function, recallthat this means that all the white vertices (that are not shown in the picture)are of degree 2 and are the edge midpoints of the edges between the twoblack vertices. (Remember that a midpoint does not have to lie exactly onthe middle of an edge.) Keeping this in mind we will now try to find thecorresponding Belyi function.Our embedding consists of a black vertex of degree 3 at x = 1, a face ofdegree 1 at x = 0, an outer face of degree 3 at x = ∞ and of course twowhite vertices with yet unknown coordinates. We also have a black vertex ofdegree 1 at x = a, with a a parameter that has to be determined.The information above gives us a function of the form
f(x) = K(x− 1)3(x− a)
x.
(Note that the numerator is of degree 4 and the denominator is of degree 1,so ∞ is indeed a pole of order 3.)We would now like to determine the parameters K and a. We do so bystudying the function f − 1, which should have two double roots as theembedding has two white vertices of degree 2. So f − 1 is of the form
f(x)− 1 = K(x2 + bx+ c)2
x.
This gives us the equality
K(x− 1)3(x− a)
x− 1 = K
(x− 1)3(x− a)− xx
= K(x2 + bx+ c)2
x
so we have four equations for the four unknown parameters K, a, b, c. Wecan however get some additional information by computing the derivative
f ′(x) = K(x− 1)2(3x2 − 2ax− a)
x2
and noting that the polynomial 3x2 − 2ax− a must be proportional to x2 +bx+ c . And thus we see that b = −2a/3 and c = −a/3. Substituting this in
22
the formulas above gives a = 9 and K = −1/64. The Belyi function is thusof the form
f(x) = −(x− 1)3(x− 9)
64x, f(x)− 1 = −(x2 − 6x− 3)2
64x.
The coordinates of the white vertices are the roots of x2−6x−3, so 3±2√
3.The embedding drawn with white vertices is shown below.
Figure 2.10: The complete embedding
So far all embeddings that we have discussed were of genus zero, inthe next example we will discuss an embedding in a surface of higher genus,genus one to be exact.
Example 2.13. Suppose we have an elliptic curve E defined by
y2 = x(x− 1)(x− 9),
and a meromorphic function on this curve which is the projection on the firstcoordinate:
p : E → C : (x, y) 7→ x.
For every finite x ∈ C, x 6= 0, 1, 9, there are two solutions of the equationdefining y and therefore there are two points (x, y) on the curve. This meansthat p is a covering of degree 2. For x = 0, 1, 9 there is only one solution fory and this solution will be of multiplicity 2. So the points 0, 1, 9 are criticalvalues of the projection p. It is not clear however, what happens at x =∞.
The equationy2 = x(x− 1)(x− 9) = x3 − 10x2 + 9x
is in fact just a simplified way of writing the equation
y2z = x3 − 10x2z + 9xz2
in homogeneous coordinates (x : y : z) of the complex projective plane CP 2.Together with a “projection” of the form
p : CP 2 → CP 1 : (x : y : z)→ (x : z)
23
which is defined everywhere except at (0 : 1 : 0). (As there is no point (0 : 0)in CP 1.) But this special point (0 : 1 : 0) does belong to E, which we cantell from the formula above.Now if a point (x : y : z) on E tends to (0 : 1 : 0), then (x : z) tends to(1 : 0). Because since y 6= 0 near (0 : 1 : 0) we may normalize using y = 1,then y2z = x3 − 10x2z + 9xz2 together with x = o(1) and z = o(1), showthat z ∼ x3 and therefore that z/x→ 0. It is thereby natural to extend theprojection p on E by continuity, setting
p : (0 : 1 : 0) 7→ (1 : 0),
where the point (1 : 0) represents ∞ ∈ C = CP 1. Furthermore if we fixx = 1 and consider small z, then y2 ≈ 1/z. So when a small value of z turnsaround 0 the value of y must change sign. This shows us finally that thepoint ∞ = (0 : 1) ∈ CP 1 is a critical value, and its preimage (0 : 1 : 0) ∈ Eis a critical point with multiplicity equal to 2.
We now know that p is not a Belyi function since it has the fourdistinct critical values 0, 1, 9 and ∞. But we do know a function which sendsall these values to 0, 1 and ∞, namely the function f of the previous example.We therefore create the composite function f ◦ p which takes (x, y) ∈ E tox ∈ C and then x to
z = −(x− 1)3(x− 9)
64x.
This composition gives us a Belyi function. Remember that the preimageunder f of 0 consists of a vertex of degree 3 at x = 1 and a vertex of degree 1at x = 9. Since p has a ramification of order 2 at these points, the preimageunder f ◦ p of 0 will consist of a black vertex of degree 6 at (1, 0) and a blackvertex of degree 2 at (9, 0). The same thing applies to the faces of degree 1and 3 of the embedding induced by f , since 0 and ∞ are also critical valuesof p. The coordinates of the white vertices corresponding to the function fare not critical values of p, this means that the embedding corresponding tof ◦ p will not have two white vertices of degree 4, but instead will containfour white vertices of degree 2. (As the preimage of 1 under f consisted ofthe points 3 ± 2
√3, both of degree 2.) So f ◦ p is a clean Belyi function
corresponding to the following embedding.
24
Figure 2.11: The embedding corresponding to f ◦ p on the elliptic curvey2 = x(x− 1)(x− 9)
We will now apply the same methodology to the curve E1 defined by
y2 = x(x− 1)(x− (3 + 2√
3)).
The projection p : (x, y) 7→ x now has the critical values 0, 1, 3+2√
3 and ∞.If we again take the composition with f , these values will all be send to0, 1 and ∞. In this case there is a curve E2 conjugate to the curve E1, E2 isdefined by
y2 = x(x− 1)(x− (3− 2√
3)).
This gives us the following two conjugate embeddings:
Figure 2.12: Two conjugate embeddings of genus 1
Note that we now show the white vertices explicitly as these images are notembeddings corresponding to clean Belyi functions. This is due to the factthat for both curves one of the white vertices became of degree 4, since thevalues 3+2
√3 and 3−2
√3 became critical values of p (where p is defined over
E1 and E2 respectively) . Although the embeddings look very different, beingconjugate embeddings does mean that they both have the same passport andare defined over the same number field. In this case they share the passport[612, 422, 8] and are both defined over Q(
√3).
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The following statement is the most fundamental thing to under-stand: not only a Belyi pair (a curve together with a Belyi function) pro-duces an embedding, but the correspondence also goes the other way around,meaning that an embedding fixes a curve. It is not possible to obtain thesame embedding on any other curve. To make this more clear we give thefollowing example: suppose an elliptic curve is fixed. Topologically speakingthis is just a torus and we can draw anything we like on its surface. Butif we want to obtain the embeddings shown above as the preimage of thesegment [0, 1] under a meromorphic function with critical values 0, 1 and ∞it will only work if the curve is E1 for the left embedding and is E2 for theright embedding.
Dessins d’enfants are often used in Galois theory. Although not much facetsof Galois theory are used in this thesis, we will now give, for the sake ofcompleteness, a brief introduction to the subject and explain the connectionwith dessins d’enfants.
2.3 Galois theory
Definition 2.14. An algebraic number is a number that is a root of a non-zero polynomial in one variable with rational coefficients. Let Q be thefield of all algebraic numbers. The universal Galois group is the group ofautomorphisms of this field Q that leave all rational elements (that is, allelements of Q) fixed. We have already seen before that we denote the absoluteGalois group by Γ = Gal(Q | Q). Two algebraic numbers a, b ∈ Q are calledconjugate if there exists an automorphism u : Q→ Q sending a to b.
Before showing the connection between dessins d’enfants and Galois theory,we will first present some facts which are important to keep in mind whendiscussing this theory.
Let k ⊂ Q be a number field, that is, a finite extension of Q.
Fact 2.15. Every automorphism of the number field k may be extended toan automorphism of the field Q.
For every finite extension k of Q there is a subgroup Γk < Γ offinite index that consists of all automorphisms of Q that leave the elementsof k fixed. The other way around, to any subgroup of finite index of Γ wemay associate a subfield of Q which contains all the algebraic numbers that
26
remain fixed under the action of the given subgroup. This leads us to thefollowing fact.
Fact 2.16. Subgroups of finite index of Γ are in one-to-one correspondencewith finite extensions of Q that are subfields of Q.
A subgroup N of Γ is called normal if it is invariant under conjugation, i.e.if αN = Nα, ∀α ∈ Γ. If N is a normal subgroup of Γ then the correspond-ing field K is called a Galois extension of Q. The following construction isoften used: let k be a number field and consider the corresponding subgroupΓk < Γ. Then take the maximal normal subgroup N /Γ contained in Γk. Thenumber field K that corresponds to this maximal normal subgroup N is thenthe minimal Galois extension of k. There are many possible extensions of anumber field k, but the minimal one is most often used. The Galois groupof the field k is the finite quotient group G = Γ/N . For example, let P bea polynomial that is irreducible over Q. When we add some of its roots toQ we obtain (after also applying all possible arithmetic operations) a finiteextension k of Q. If we add all roots of P to k we get a Galois extensionK. The field K will then be called the splitting field of P , and P will bedenoted as a defining polynomial of K. (This polynomial is not unique.) TheGalois group G of K will also be called the Galois group of the polynomialP . When P is of degree n then G will be a transitive permutation group ofdegree n. The following question now rises: Is every finite group the Galoisgroup of some finite extension of Q? This question is the famous “inverseGalois problem”. Up to this very day there is no conclusive answer to thisquestion yet. Dessins d’enfants may turn out to be an important tool in thesearch of an answer to this problem.
We will now explain how Gal(Q | Q) acts on dessins d’enfants.We first consider the planar case. Suppose we are given a dessin M with cor-responding Belyi function f , note that f is rational since it is a covering ofC by C. The Belyi theorem (which will be proved right after this subsection)tells us that f can be chosen in such a way that all its coefficients are alge-braic numbers. Finally, let K be the normal extension of Q that is generatedby these coefficients. We then can act on all these numbers simultaneouslyby an automorphism of K. In view of fact 2.15 this is the same as acting byan automorphism α of Q, i.e. an element α ∈ Γ. One fortunate result of thisaction is the fact that the newly created function fα is once again a Belyifunction. The dessin corresponding to this function is called Mα and is theresult of the action of α on M .For a dessin of higher genus (g ≥ 1) the procedure is similar, although this
27
time the action is on the Belyi pair as a whole. Let M be a dessin with cor-responding Belyi pair (X, f). According to proposition 1.21 the curve X canbe realized as an algebraic curve in CP k, meaning that it is a solution of asystem of algebraic equations in homogeneous coordinates (x0 : x1 : . . . : xk).According to Belyi’s theorem the coefficients of these equations can be cho-sen in such a way that they are algebraic numbers. Using Belyi’s theoremonce more, we see that the function f is a rational function in the variablesx0, x1, . . . , xk with coefficients that can be chosen to be algebraic numbers.Now the action of an automorphism α ∈ Γ is on all the above algebraic num-bers simultaneously, which results in a new Belyi pair (Xα, fα) correspondingto a new dessin Mα.A very important observation is that all orbits of the action of Γ on dessinsare finite. This holds, since the passport turns out to be an invariant underthe action, and the number of hyperembeddings with a given passport is fi-nite.
Let M be a dessin with corresponding stabilizer ΓM ≤ Γ. The orbitof M under the action of Γ is finite, so the subgroup ΓM is of finite indexin Γ. Now as before, let N ≤ ΓM be the maximal normal subgroup of Γcontained in ΓM and let K be its corresponding Galois extension of Q. Notethat N is the stabilizer of all elements of the orbit.
Definition 2.17. The field k corresponding to the group ΓM is the field ofmoduli of the dessin M . The field K is called the field of moduli of the orbitof M . The field of moduli is sometimes called the field of definition.
Remark. This definition in combination with fact 2.16 leads to the observa-tion that the field of moduli of an orbit can not be chosen arbitrarily, it is infact a characteristic of the orbit in question.
Example 2.18. A Galois orbit having n elements corresponds to a subgroupΓM of index n, and this implies that the field of moduli of the orbit is gen-erated by the roots of a polynomial of degree n. As a consequence of this, ifan orbit contains a single element, then its field of moduli is Q itself.
This concludes our exposition on Galois theory and dessins d’enfants.
2.4 The Belyi theorem
Theorem 2.19 (The Belyi Theorem). A Riemann surface X admits a modelover the field Q of algebraic numbers if and only if there exists a coveringf : X → C unramified outside {0, 1,∞}. In such case, the meromorphicfunction f can also be chosen in such a way that it will be defined over Q.
28
The Belyi theorem consists of two parts which are traditionally called the“obvious part” and the “difficult part”. It is the “if part” that is called the“obvious part”, but paradoxically enough it is not obvious at all. We will dothis part of the proof last. We first turn our attention to the “difficult part”,that is the “only if part”. We will give here the proof as formulated by Belyihimself, which is in fact a series of tricks that are quite easy to follow.
Theorem 2.20. If a Riemann surface X is defined over the field Q of alge-braic numbers, then there exists a meromorphic function f : X → C unram-ified outside {0, 1,∞}.
Proof. We will proof this theorem using three steps.
Step 1. Consider an arbitrary meromorphic function h : X → C definedover Q. For example, when the Riemann surface X is represented as analgebraic curve, then h may be chosen to be the projection to one of thecoordinates.Since h is arbitrary, some critical values of h will be rational and some willbe irrational but still algebraic. We turn our attention to just the irrationalcritical values as we temporarily forget about the rational critical values. LetS0 denote the set of all irrational critical values of h together with their al-gebraic conjugates. Let N = |S0|.
Step 2. The polynomial P0 that annihilates the set S0 (which means thatall elements of S0 are (the only) roots of P0 ) is defined over Q (as all elementsof S0 are algebraic) and is of degree N . The critical values of P0 are the valuesof this polynomial at the roots of its derivative. The derivative is of degreeN − 1 so the polynomial P0 has a maximum of N − 1 critical values. Wedenote the set consisting of all these values by S1. Since S1 already containsall the conjugates of its elements it holds that its annihilating polynomial P1
is defined over Q and is of degree at most N − 1. Continuing this process,we see that the set S2 of critical values of P1 has at most N − 2 values, andfrom this set we can obtain a new annihilating polynomial P2, etc.We finally obtain a composition of polynomials
PN−1 ◦ . . . ◦ P1 ◦ P0, degPm ≤ N −m,
which will send all the critical values of the initial function h to rationalnumbers. Note that all the rational critical values of h that we neglected atthe start are also send to rational numbers, because all the polynomials Pmhave rational coefficients.
29
Step 3. The final step is to send all the now rational critical values to0, 1 or ∞. Using an affine transformation we first place all of them (besidesinfinity) inside the segment [0, 1]. Then we apply the apply the (Shabat)polynomial
pm,n(x) =(m+ n)m+n
mmnnxm(1− x)n.
This mapping has the following properties: pm,n(0) = 0, pm,n(1) = 0, pm,n(∞) =∞ and pm,n([0, 1]) = [0, 1]. It also sends the rational number m/(m + n) to1, and all other rational values that remain to be changed to some otherrational numbers. Therefore, this mapping decreases the number of rationalnumbers under consideration. To be more specific, suppose that the criticalvalues are x1, . . . , xk with x1 = 0, x2 = 1, x3 = ∞ and that 0 < xi < 1 forall other i. Then writing xk = m/(m+ n) and applying the polynomial pm,ndiminishes k by one. We can keep applying these mappings as often as weneed until all remaining critical values have become 0, 1 or∞. The compositefunction created in steps one to three is therefore a Belyi function.This proofs the theorem.
Before starting with the “if” part of the proof we first have to intro-duce some new notions, since the approach for this part of the proof will bea more generalised one, based on the work of Bernhard Kock in [5].
Definition 2.21. Given a subfield D of a field C, we say that C is anextension of D and denote this by C/D. An element of C which is a root ofa nonzero polynomial with coefficients in D is called algebraic over D. Non-algebraic numbers are called transcendental. We say that D is algebraicallyclosed in C when C contains a root for every non-constant polynomial inD[x] (the ring of polynomials in the variable x with coefficients in D).
Remark. Given any field extension C/D, we can consider the automorphismgroup Aut(C/D), which consists of all field automorphisms α : C → C suchthat α(x) = x for all x ∈ D.
Lemma 2.22. Let D be a subfield of C. Any automorphism of D can beextended to an automorphism of C. Furthermore, we have:
CAut(C/D) = D
where CAut(C/D) is the set of elements in C fixed under all automorphisms inAut(C/D).
Proof. The first assertion is analogous to fact 2.15 and we will accept this asa fact as well.
30
For the second part of the lemma we can directly tell that D ⊆ CAut(C/D), asit is a tautology. The inclusion the other way is equivalent to the propertythat for any x ∈ C \ D, there is a α ∈ Aut(C/D) s.t. α(x) 6= x. When xis transcendent over D, then we could for example map x to −x which givesa D-automorphism of D(x) that does not fix x. This automorphism thencan be extended to the needed automorphism of C, by the first part of thelemma. When x is algebraic over D, we use a different approach. We choosean element y ∈ C \ {x} which is D-conjugate to x, meaning that there is apolynomial in D[x] that has both x and y as roots. We can then map x to yobtaining a D-embedding of D(x) into the normal closure L of D(x) over D.We can extend this embedding to a D-automorphism of L which in turn canbe extended to the desired D-automorphism of C, again by the first part ofthe lemma.
As usual, we denote the index of a subfield D in a field C by [C : D].
Lemma 2.23. For a field C, consider a subgroup U of Aut(C) and let V bea subgroup of U of finite index. Then the field extension CV /CU is finite. IfV is a normal subgroup of U then it holds that [CV : CU ] ≤ [U : V ].
Proof. It is a well known fact that there always exists a normal subgroupW of U of finite index that is contained in V . We then have a canonicalhomomorphism U/W → Aut(CW/CU). The field that is fixed under theimage of this homomorphism is CU . Thus CW/CU is a finite Galois extensionand hence we obtain:
[CW : CU ] = ord(Aut(CW/CU)) ≤ ord(U/W ) = [U : W ].
This proofs the lemma.
As said before, in the proof we will give momentarily, some moregeneral notions are used. We will therefore redefine some of our languagein terms of varieties and curves. Let C be a field, a curve over a field Cis a smooth projective geometrically connected variety of dimension 1 overC. Where a variety over C is an integral separated scheme X together witha morphism p : X → Spec(C) of finite type. For any α ∈ Aut(C) wedenote the variety that consists of the scheme X together with the structuremorphism Spec(α) ◦ p : X → Spec(C) with Xα/C. It is important to keepin mind that a curve X/C is nothing more than a finitely generated field Dof transcendence degree 1 over C, with the property that C is algebraicallyclosed in D. With these notions, the definition of a field of moduli will bethe following:
31
Definition 2.24. The field of moduli of a variety X/C is the field M(X) :=CU(X), where
U(X) := {α ∈ Aut(C) : Xα/C is isomorphic to X/C}.
Now let C be an algebraically closed field of characteristic 0 and let t : X → P1C
be a finite morphism from a curve X/C to the projective line P1C . The usual
notions of degree, critical points and critical values also apply for this generalmorphism t.
Definition 2.25. The moduli field of a morphism t is the field M(X, t) :=CU(X,t) fixed by the subgroup U(X, t) of U(X) that consists of all automor-phisms α ∈ Aut(C) such that there exists a isomorphism fα : Xα → X ofvarieties over C such that the following diagram is commutative:
Xα fα //
tα
��
X
t��
(P1C)α
Proj(α) // P1C
here Proj(α) is the automorphism of the scheme P1C = Proj(C[T0, T1]) in-
duced by the extension of the automorphism α to C[T0, T1]. (Which is de-noted by α again.)
Theorem 2.26. The curve X/C and the morphism t are both defined overa finite extension of M(X, t). If t is a Galois covering, that is, if the corre-sponding extension of function fields is Galois, then t and X/C are definedover M(X, t) itself.
Proof. First choose a rational point q of P1C which is not a critical value of
t, then choose a point r in t−1(q). Using the Riemann-Roch theorem (whichis closely related to Riemann’s existence theorem), we see that there is ameromorphic function z ∈ K(X)\C such that r is the only pole of z. Wherewe have that K(X) = C(t, z), since t is here considered as a meromorphicfunction on X and the field extension K(X)/C(t, z) is a subextension ofK(X)/C(z) and of K(X)/C(z), therefore the corresponding morphism ofcurves is both unramified and totally ramified at r. We may assume thatz is chosen such that the order m := −ordr(z) ∈ N, of its single pole r, isminimal in the set of meromorphic functions with a single pole in K(X) \C.Thus we see that
V := {x ∈ K(X) : ordr(x) ≥ −m} = C ⊕ Cz;
32
since, for any x1, x2 ∈ V with ordr(x1) = ordr(x2) = −m, there is a constantc ∈ C with −ordr(x1 − cx2) < m, and then x1 − cx2 must be a constantfunction, as m was minimal. Since q is not a critical value of t, the value ofthe meromorphic function t− q in r is a local parameter on X. If C = C thismeans that t − q gives a chart of X(C) in a neighbourhood of r that mapsr to zero. Clearly there is a unique function z′ ∈ V such that the leadingcoefficient in the Laurent expansion of z′ with respect to the local parametert − q (that is, the coefficient of (t − q)−m) is equal to 1 and the constantcoefficient (the coefficient of (t− q)0) is equal to 0. We may assume now thatz = z′. We now claim that the minimal polynomial of z over C(t) has itscoefficients in k(t), with k a finite extension of M(X, t). (And k = M(X, t),when t is a Galois covering.) This shows that the field extension K(X)/C(t)is defined over k. Then, by the correspondence between curves and functionfields, the theorem is proved.
Remark. We now give a proof to back up the claim above. We denote byU(X, t, r) the subgroup of automorphisms in U(X, t) with the additionalproperty that fα(rα) = r. (Here fα is the isomorphism as defined before,and rα denotes the point on Xα/c that corresponds to r.) Note that theisomorphism fα is unique since the action of Aut(t) on the fibre t−1(q) isfree. Therefore, mapping α to the automorphism of K(X) induced by fαgives an action of U(X, t, r) on K(X) by C-semilinear field automorphismsthat fix t ∈ K(X). Since U(X, t, r) is the stabilizer of [r] under the action(α, [r]) → [fα(rα)] of U(X, t) on t−1(q)/Aut(t), we see that U(X, t, r) hasfinite index in U(X, t). In fact, if t is a Galois covering, then it even holds thatU(X, t, r) = U(X, t) since t−1(q)/Aut(t) would consist of a single element.The function z ∈ K(X) is invariant under the action of U(X, t, r) as describedabove since the image α(z) for α ∈ U(X, t, r) has the same three definingproperties as z. And the invariance will thus also hold for the minimalpolynomial of z over C(t). Now lemma 2.23 proofs the claim.
Proposition 2.27. Let S be a discrete set of points of the Riemann sphereC = P1
C, and let d ≥ 1 be an integer. Then there are at most finitely manyisomorphism classes of pairs (X, t) where X/C is a curve and t : X → P1
Cis a finite morphism of varieties over C of degree d such that all its criticalvalues lie in S.
Remark. The definition for isomorphism of pairs is the natural one; twopairs (X1, t1), (X2, t2) are called isomorphic when there is an isomorphismf : X1 → X2 of varieties over C such that t2 ◦ f = t2.
Proof. If we pass from the finite morphism t : X → P1C to the continuous map
t(C) : X(C)→ P1(C) between the corresponding two Riemann surfaces and
33
restrict t(C) to the preimage of the punctured sphere P1(C) \ S, we create amap from the set of isomorphism classes of pairs to the set of homeomorphismclasses of unramified topological coverings of the sphere P1(C) \ S that areof degree d. We will denote this set by M. We will show that this map isinjective. Let (X1, t1) and (X2, t2) be two pairs as before, together with ahomeomorphism g : X1(C) \ t−1
1 (S)→ X2(C) \ t−12 (S), such that t2(C) ◦ g =
t1(C) on the domain of g. We then know that g is biholomorphic, since fori = 1, 2, we know that ti(C) |Xi(C)\t−1
i (S) is locally biholomorphic. Using anelementary fact in Complex Analysis, we can extend g to a biholomorphicmap h : X1(C) → X2(C) such that t2(C) ◦ h = t1(C). It is well known thatany biholomorphic map between complex curves is algebraic, so we obtainan isomorphism of varieties over C, which we denote by f : X1 → X2,such that t2 ◦ f = t1. This shows that (X1, t1) and (X2, t2) are isomorphicpairs. So all that remains to show is that M is a finite set. Any unramifiedcovering of P1(C) \ S is a quotient of the universal covering p by a subgroupof Aut(p) ∼= π1(P1(C)\S). So all that remains to show is that there are onlyfinitely many subgroups of index d of the fundamental group π1(P1(C) \ S).This follows directly from the fact that π1(P1(C) \ S) is finitely generated(as π1(P1(C) \ S) ∼= 〈γq, q ∈ S :
∏q∈S γq = 1〉 which is a free group of rank
|S| − 1) and the fact that a finitely generated group only has finitely manysubgroups of a given (finite) index. This concludes the proposition.
Theorem 2.28. Let X/C be a curve together with a finite morphism t :X → P1
C and let K be a subfield of C such that the critical values of t areK-rational. Then the field of moduli of t is contained in a finite extension ofK.
Proof. For any automorphism α ∈ Aut(C/K) it holds that the critical values
of t(α) : Xα tα→ (P1C)α
Proj(α)→ P1C all are in S too, and the degree of t(α) is the
same as the degree of t. So using proposition 2.27 we see that the orbit of theisomorphism class of the pair (X, t) under the action of Aut(C/K) is finite.Therefore the stabilizer is of finite index in Aut(C/K). Also, it is obviouslycontained in U(X, t). Now lemma 2.22 and lemma 2.23 together imply thatthe moduli field M(X, t) = CU(X,t) is contained in a finite extension of K =CAut(C/K).
We now have everything we need for proving the “obvious” part ofthe Belyi theorem, that is, the if part.
Theorem 2.29. A Riemann surface X admits a model over the field Q ofalgebraic numbers if there exists a covering f : X → C unramified outside{0, 1,∞}.
34
Proof. First assume we have a covering f with the above properties. Bytheorem 2.28 we see that M(X, f), the moduli field of f , is a number field.Therefore, by theorem 2.26, we know that X is also defined over a numberfield (which may be bigger). So indeed it holds that X admits a model overthe field Q. This proofs the Belyi theorem.
35
Chapter 3
Counting dessins d’enfants
In this chapter we will count the number of dessins d’enfants. To be moreprecise, we will be using the Eynard-Orantin topological recursion to testa generating function for the number of clean Belyi functions defined on aRiemann surface of genus g having n critical points corresponding to ∞ ofwhich the degrees are fixed.
3.1 The Eynard-Orantin recursion
Definition 3.1. A spectral curve E = (X, x, y), is the data of a Riemannsurface X together with two analytical functions x and y on some opendomain of X.
In some sense, we consider a parametric representation of a spectralcurve y(x), where the space of the parameter z is a Riemann surface X.
Definition 3.2. We say that a spectral curve is algebraic if X is a Riemannsurface of genus g, and if x and y are meromorphic functions on X. If inparticular, X is the Riemann sphere (so when g = 0 ), then we say that thespectral curve is rational.
The names rise from the fact that for an algebraic spectral curve onecan always find a polynomial relationship between x and y:
Pol(x, y) = 0.
If the curve is rational then this polynomial equation can be parametrizedwith two rational functions x(z), y(z) of some complex variable z.
Definition 3.3. A spectral curve (X, x, y) is called regular if:
36
• The differential form dx has a finite number of zeroes, and all zeroes ofdx are simple.
• The differential form dy does not vanish at the zeroes of dx.
We will now give the definition for the topological recursion of Eynard-Orantin for a spectral curve of genus 0 as stated in [8].
Definition 3.4. We start of with C where we have some preferred coordinatet. Now let S ⊂ C be a finite set of points and compact real curves such thatΣ = C\S is connected. The spectral curve of genus 0 consists of this Riemannsurface Σ together with a simply ramified holomorphic map
π : Σ→ C, t 7→ π(t) = x
where the differential dx only has simple zeroes. We denote byR = {p1, . . . , pr} ⊂Σ the set of ramification points, and by
U = trj=1Uj
the disjoint union of small neighbourhoods Uj around each pj such thatπ : Uj → π(Uj) ⊂ C is a double sheeted covering ramified only at pj. Lett denote the local Galois conjugate of t ∈ Uj. The canonical sheaf of Σ isdenoted by K. Since we have chosen a preferred coordinate t, we have apreferred basis dt for K and ∂/∂t for K−1.The meromorphic differential forms Wg,n(t1, . . . , tn), g = 0, 1, 2, . . ., n =1, 2, 3, . . ., are said to satisfy the Eynard-Orantin topological recursionif the following conditions are satisfied:
1. W0,1(t) ∈ H0(Σ,K).
2. W0,2(t1, t2) = dt1·dt2(t1−t2)2
− π∗ dx1·dx2(x1−x2)2
∈ H0(Σ × Σ,K⊗2(2∆)), where ∆ isthe diagonal of Σ× Σ.
3. The recursion kernel Kj(t, t1) ∈ H0(Uj × C, (K−1Uj⊗ K)(∆)) for t ∈ Uj
and t1 ∈ C is defined by
Kj(t, t1) =1
2
∫ ttW0,2(·, t1)
W0,1(t)−W0,1(t).
The kernel is an operator multiplying dt1 while contracting dt.
37
4. The general differential forms Wg,n(t1, . . . , tn) ∈ H0(Σn,K(∗R)⊗n) aremeromorphic symmetric differential forms with poles only at the rami-fication points in R for 2g−2 +n > 0. They are given by the recursionformula
Wg,n(t1, . . . , tn) =1
2πi
r∑j=1
∮Uj
Kj(t, t1)[Wg−1,n+1(t, t, t2, . . . , tn)
+
No (0,1) terms∑g1+g2=g
ItJ={2,...,n}
Wg1,|I|+1(t, tI)Wg2,|J |+1(t, tJ)].
The integral is taken with respect to t ∈ Uj along a positively ori-ented simple closed loop around pj, and tI = (ti)i∈I for a subsetI ⊂ {1, 2, . . . , n}.
5. The differential form W1,1(t1) requires a separate treatment becauseW0,2(t1, t2) is regular at the points of ramification but has poles else-where. This differential form is given by:
W1,1(t1) =1
2πi
r∑j=1
∮Uj
Kj(t, t1)
[W0,2(u, v) + π∗
dx(u) · dx(v)
(x(u)− x(v))2
]u=tv=t
.
Finally let y : Σ→ C be a holomorphic function defined by
W0,1(t) = y(t)dx(t).
Or equivalently, we could have defined the function by contraction, y =iXW0,1, where X is the vector field on Σ dual to dx(t) with respect to thecoordinate t. We then have an embedding
t : Σ→ C2, t 7→ (x(t), y(t)).
This concludes the Eynard-Orantin topological recursion.
Before we turn our attention to dessins d’enfants, one more notionhas to be introduced.
The Laplace transform is a special type of linear operator that transforms afunction f(t) with real argument t ≥ 0 into a new function F (s) with complexargument s. Note that there is essentially a bijective correspondence betweena function and its Laplace transform. The Laplace transform used in thissection will be defined shortly.
The authors of [8] proposed the following conjecture regarding theimportance of the Laplace transform within the formalism of the Eynard-Orantin recursion.
38
Conjecture 3.5 (The Laplace transform conjecture). If the unstable ge-ometries (g, n) = (0, 1) and (0, 2) make sense in a counting problem on theA-model side, then the Laplace transform of the solution to these cases de-termines the spectral curve and the recursion kernel of the Eynard-Orantinformalism, which is a B-model theory. The recursion then determines thesolution to the original counting problem for all (g, n).
It may now come as no surprise that the Laplace transform will beof high importance in the rest of this chapter.
3.2 A relation for the number of dessins
As usual, let f : X → C be a Belyi function on a Riemann surface X ofgenus g. Using the Belyi theorem, we see that X is defined over Q. Now letq1, . . . , qn be the set of poles of f and let their respective orders be given by(µ1, . . . , µn), which is a vector in Zn+. This vector containing the orders willbe called the profile of f (at infinity). When we start with our enumerationwe will label all the poles of f , thus an automorphism of a Belyi functionwill preserve the set of poles point-wise. Recall that a clean Belyi functiononly has critical points of order two corresponding to the critical value 1, aclean Belyi function will therefore have profile (2, 2, . . . , 2) at 1.We denote by Dg,n(µ1, . . . , µn) the number of clean Belyi functions of genusg with profile (µ1, . . . , µn) at∞. This number is the subject of our recursion.
The plan of this chapter is as follows. First we derive a recursionequation among the Dg,n(µ1, . . . , µn) for all (g, n). This equation will notbe an effective recursion formula yet, as Dg,n(µ1, . . . , µn) will appear in theequation in a complicated manner. We will therefore compute the Laplacetransform
FDg,n(w1, . . . , wn) =
∑µ1,...,µn>0
Dg,n(µ1, . . . , µn)e−(µ1w1+...+µnwn),
and then rewrite our recursion equation in terms of the newly found Laplacetransformed functions. After doing so we show that the symmetric differentialforms
WDg,n = d1 · · · dnFD
g,n
do satisfy the recursion formula of Eynard and Orantin.
In this section we will use a description of the clean Belyi pair (X, f)that is closely related to that of a dessin d’enfant. Instead of looking at the
39
preimage f−1 of the segment [0, 1] in C, which is an embedding Γ, we will lookat the preimage under f of the segment [1, i∞] = {1 + iy | 0 ≤ y ≤ ∞} ⊂ C,the vertical half-line with real part equal to one. This preimage f−1([1, i∞])will be denoted by Γ. Then the embedding Γ has n labeled vertices of de-grees (µ1, . . . , µn), which are the preimages under f of∞. Note that Γ is thedual embedding to the original embedding Γ; the black vertices of Γ (that is,the preimages of 0 ) are the centers of faces of Γ, while the centers of facesof Γ have become the (black) vertices of Γ. The preimages of 1 (which arethe white vertices) were the edge midpoints of the original embedding Γ andhave now become the edge midpoints of the dual embedding Γ. By abuse oflanguage, we will refer to the dual embedding Γ as a dessin.An important thing to keep in mind is that we now no longer count em-beddings of which the automorphism group preserves faces, but that we nowconsider embeddings of which the automorphism group preserves each vertexpoint-wise. Thus, the automorphism group can now permute faces.In this dual setting, we define the number of dessins (with the automorphismfactor) by
Dg,n(µ1, . . . , µn) =∑
Γ dessin of type (g,n)
1
|AutD(Γ)|,
where Γ is a dessin of genus g having n labeled vertices with degrees (µ1, . . . , µn),and AutD(Γ) is the automorphism of Γ that preserves every vertex point-wise.
Our goal is to find the spectral curve by looking at the problem forthe unstable curves (g, n) = (0, 1) and (g, n) = (0, 2). The dessins counted inD0,1(µ) for some integer µ ∈ Z+ are all spherical embeddings that have onlya single vertex of degree µ. Since any edge of such an embedding has to startand end in the sole vertex, it holds that every edge is a loop and thereforeµ must be even. We write µ = 2m. The contribution of an embedding Γto the number D0,1(µ) is counted with the weight 1/|AutD(Γ)|, which makesthe counting problem more difficult. Note however, that the automorphismgroup of a spherical dessin with a single vertex is a subgroup of Z/(2m)Zpreserving the graph. If we therefore place an outgoing arrow on one of the2m half-edges incident to the vertex (see the figure below), we can kill theautomorphism altogether.
40
Figure 3.1: An arrowed dessin of genus 0 containing a single vertex
There are 2m possibilities to place such an arrow, therefore the num-ber of arrowed graphs is equal to 2mD0,1(2m), which is now an integer. Usinga bijection argument with the number of possible arrangement of m pairs ofparentheses, we get the following result:
2mD0,1(2m) =1
m+ 1
(2m
m
)= Cm,
here Cm stands for the m-th Catalan number.
We define the Laplace transform of D0,1(µ) by
FD0,1 =
∞∑m=1
D0,1(2m)e−2mw.
Then the corresponding Eynard-Orantin differential
WD0,1 = dFD
0,1 = −∞∑m=1
2mD0,1(2m)e−2mwdw = −∞∑m=1
Cme−2mwdw
is a generating function of the Catalan numbers. A better choice would be
z(x) =∞∑m=0
Cm1
x2m+1=
1
x+
1
x3+
2
x5+
5
x7+
14
x9+
42
x11+ . . . .
The radius of convergence of this Laurent series is equal to 2, meaning thatfor |x| > 2 the series converges absolutely. Near (x, z) = (∞, 0) the inversefunction of z = z(x) is given by
x = z +1
z.
41
Which can be seen by solving the quadratic equation z2 − xz + 1 = 0 withrespect to z, as this is equivalent to the quadratic recursion
Cm+1 =∑i+j=m
Ci · Cj
of Catalan numbers. Now we define
x = ew
and allow the m = 0 term in the Eynard-Orantin diferential to obtain:
WD0,1 = −
∞∑m=0
Cmdx
x2m+1.
Then the Laplace transform of D0,1(2m) needs to be changed in:
FD0,1 =
∞∑m=1
D0,1(2m)e−2mw − w =∞∑m=1
D0,1(2m)1
x2m− log(x).
Although numerically it holds that D0,1(0) = 0, its behaviour around 0 isgiven by
limm→0
D0,1(2m)
x2m= − log(x),
which is in consistence with
limm→0
2mD0,1(2m) = C0 = 1.
From the above we see that
WD0,1 = −z(x)dx.
We have therefore identified the spectral curve for the counting problem ofthe dessins Dg,n(µ). The spectral curve is given by{
x = z + 1z
y = −z
To compute the recursion kernel we also need to compute D0,2(µ1, µ2) forthe other unstable curve (g, n) = (0, 2). In this dual setting, D0,2(µ1, µ2) isthe number of spherical dessins Γ having two vertices of degrees µ1 and µ2
counted with weight 1/|AutD(Γ)|. We will use here the result of Kodamaand Pierce (see [9], theorem 3.1).
42
Proposition 3.6. The number of spherical dessins Γ having two vertices ofdegrees µ1 and µ2, counted with weight 1/|AutD(Γ)|, is equal to
D0,2(µ1, µ2) =
12k
(2kk
)µ1 = 0, µ2 = 2k 6= 0
14
1j+k
(2jj
)(2kk
)µ1 = 2j 6= 0, µ2 = 2k 6= 0
1j+k+1
(2jj
)(2kk
)µ1 = 2j + 1, µ2 = 2k + 1
In all other cases D0,2(µ1, µ2) = 0.
Remark. The first case is not regular. When µ1 = 0, then the second vertexmust be of even degree, which gives us Ck/(2k) embeddings. Using theEuler characteristic we see that this embedding has k + 1 faces, since 2 =1− k+ (k+ 1). The vertex of degree 0 is located in one of these faces, so thetotal number of embeddings is equal to
k + 1
2kCk =
1
2k
(2k
k
).
But we are only counting connected embeddings, so the degree 0 vertices arenot allowed in our counting.
The number of dessins satisfies the following theorem:
Theorem 3.7. When g ≥ 0) and n ≥ 1 satisfy 2g− 2 +n ≥ 0, the followingrecursion equation holds for the number of dessins:
µ1Dg,n(µ1, . . . , µn) =n∑j=2
(µ1 + µj − 2)Dg,n−1(µ1 + µj − 2, µ[n]\{1,j})
+∑
α+β=µ1−2
αβ
Dg−1,n+1(α, β, µ[n]\{1}) +∑
g1+g2=gItJ={2,...,n}
Dg1,|I|+1(α, µI)Dg2,|J |+1(β, µJ)
,where µI = (µi)i∈I for a subset I ⊂ [n] = {1, 2, . . . , n}. The last sum onthe right-hand side is over all partitions of the genus g and the index set{2, 3, . . . , n} into two pieces.
Remark. Note that when g1 = 0 and I = ∅, then Dg,n(µ1, . . . , µn) appears onthe right hand-side of the equation as well. Thus this is merely an equationof the number of dessins, not an effective recursion formula.
Proof. We consider the set of genus g dessins with n vertices labeled by theindex set [n] = {1, 2, . . . , n} and of degrees (µ1, . . . , µn). The left hand side of
43
the above equation is the number of dessins with an outward arrow placed onone of the incident edges at the vertex 1. The above is based on the removalof this arrowed edge. There are two cases to be considered.
Case 1. The arrowed edge connects the vertex 1 and some other edge j > 1.We then remove this edge and put the vertices 1 and j together as we showin the below figure. We describe this process as shrinking the edge to a singlepoint. The resulting dessin will have one less vertex but the genus remainsthe same. The degree of the created vertex will be µ1 + µj − 2, and thedegrees of the other vertices remain unaffected.
Figure 3.2: The operation that shrinks the arrowed edge and joins the vertices1 and j
We want to make a bijection argument, so we need to be able toreconstruct the original dessin from the newly created one. Since both µ1
and µj were given as the input value, we have to specify which edges belongto 1 and which belong to j when separating the vertex of degree µ1 +µj − 2.To do so, we use a marker on one of the incident edges. We group this markededge together with µ1 − 2 edges following it according to cyclic order. Theother µj − 1 edges are also grouped. We then insert an edge and separatethe vertex into the two distinct vertices 1 and j, in such a way that the firstgroup of edges are incident to vertex 1 and the second group is incident toj, keeping in mind their cyclic orders. The contribution to the number ofdessins from this case is therefore
n∑j=2
(µ1 + µj − 2)Dg,n−1(µ1 + µj − 2, µ[n]\{1,j}).
Case 2. The arrowed edge is a loop of vertex 1. We then remove the loopfrom the dessin and separate the vertex into two vertices. The loop divides allthe remaining half-edges into two groups; one group consisting of all the half-edges that follow the arrowed half-edge in the cyclic order until the incoming
44
end of the marked loop, and the group consisting of all the other half-edges(see the figure below). Denote the number of half-edges in the first group byα, and let β be the number of half-edges in the other group. Then we havethe equality α + β = µ1 − 2, and we created two edges of degrees α and β.If we want to recover the original dessin from this new one, we need tomark one of the edges in each of the two groups (one half-edge from eachvertex) so that we can put the loop back in its original place. The numberof possibilities of these markings is αβ.
Figure 3.3: The operation removing one loop and separating one vertex intotwo vertices
The loop removal operation and the separation of the vertex in-creases the number of vertices from n to n + 1. The operation will alsochange the genus of the dessin. When the resulting embedding is connected,then the genus g changes into g − 1. If the resulting dessin is the disjointunion of two dessins of genera g1 and g2, then we know that g = g1 +g2. Thetotal contribution from this case is therefore
∑α+β=µ1−2
αβ
Dg−1,n+1(α, β, µ[n]\{1}) +∑
g1+g2=gItJ={2,...,n}
Dg1,|I|+1(α, µI)Dg2,|J |+1(β, µJ)
.Please note that the outward placed arrow determines the (two) groups ofhalf-edges uniquely, as one group is after the arrowed half-edge and the othergroup is before according to the cyclic order. Therefore there is no need tosymmetrize α and β. If we would have placed the arrow on the other end ofthe loop, then α and β would be interchanged.The sum of these two contributions gives the desired equation.
The relation given in theorem 3.7 becomes an effective recursionformula after taking the Laplace transform. We will do so in the next section,and afterwards we will verify that this generating function for the number ofdessins satisfies the Eynard-Orantin recursion formula.
45
3.3 The recursion formula for dessin d’enfants
The projection x = z+1/z of the spectral curve to the x-coordinate plane hastwo ramification points z = ±1, as they are both solutions to the equation1 − 1/z2 = 0. It is thereby natural to make the transition to a coordinatethat has these ramification points at 0 and ∞. We therefore define
z =t+ 1
t− 1.
Proposition 3.8. The Laplace transform of D0,2(µ1, µ2) is given by
FD0,2(t1, t2) :=
∑µ1,µ2>0
D0,2(µ1, µ2)e−(µ1w1+µ2w2)
= − log(1− z(x1)z(x2)) = log(t1 − 1) + log(t2 − 1)− log(−2(t1 + t2)),
here z(x) is the generating function of the Catalan numbers. We then have
WD0,2(t1, t2) = d1d2F
D0,2(t1, t2) =
dt1 · dt2(t1 − t2)2
− dx1 · dx2
(x1 − x2)2=
dt1 · dt2(t1 + t2)2
.
Proof. In terms of x = ew and using proposition 3.6 , the Laplace transformFD
0,2(t1, t2) is given by∑µ1,µ2>0
D0,2(µ1, µ2)e−(µ1w1+µ2w2)
=1
4
∞∑j,k=1
1
j + k
(2j
j
)(2k
k
)1
x2j1
1
x2k2
+∞∑
j,k=0
1
j + k + 1
(2j
j
)(2k
k
)1
x2j+11
1
x2k+12
.
Because
dx = (1− 1
z2)dz,
we see that
xd
dx=
z + 1z
1− 1z2
d
dz=z(z2 + 1)
z2 − 1
d
dz.
We now define
ξ0(x) =∞∑m=0
(2m
m
)1
x2m+1.
And we see that
1
2(1−x d
dx)∞∑m=0
1
m+ 1
(2m
m
)1
x2m+1=
∞∑m=0
1
2
1−−(2m+ 1)
m+ 1
(2m
m
)1
x2m+1= ξ0(x).
46
And in terms of z and t we thus write
ξ0(x) =1
2(1− z(z2 + 1)
z2 − 1
d
dz)z = − z
z2 − 1= −t
2 − 1
4t.
Furthermore note that
−(x1d
dx1
+ x2d
dx2
)(1
4
∞∑j,k=1
1
j + k
(2j
j
)(2k
k
)1
x2j1
1
x2k2
+∞∑
j,k=0
1
j + k + 1
(2j
j
)(2k
k
)1
x2j+11
1
x2k+12
)
=1
2(x1ξ0(x1)− 1)(x2ξ0(x2)− 1) + 2ξ0(x1)ξ0(x2) = 2z1z2
1 + z1z2
(z21 − 1)(z2
2 − 1)
= −(z1(z2
1 + 1)
z21 − 1
d
dz1
+z2(z2
2 + 1)
z22 − 1
d
dz2
)(− log(1− z1z2)).
Thus we have obtained the partial differential equation
(x1d
dx1
+ x2d
dx2
)(FD0,2(t1, t2) + log(1− z1z2)) = 0
for a holomorphic function in x1 and x2 defined for |x1| � 2 and |x2| � 2.Using the Laurent expansion of the Catalan numbers (as shown before), wesee that the first terms of the Laurent expansion of − log(1 − z(x1)z(x2))agree with the first few terms of the Laplace transform of D0,2(µ1, µ2) givenin the start of the proof. We therefore have the initial condition for theabove differential equation. The first part of the proposition now followsfrom the uniqueness of the solution to the Euler differential equation withinitial condition. The second part of the proposition follows directly bydifferentiation.
Speaking in terms of the t-coordinate as defined before, the localGalois conjugate of an element t ∈ Σ under the projection x : Σ → C isgiven by −t. This means that the recursion kernel for counting the numberof dessins is given by
KD(t, t1) =1
2
∫ −ttWD
0,2(·, t1)
WD0,1(−t)−WD
0,1(t)=
1
2(
1
t+ t1+
1
t− t1)
1t+1t−1− t−1
t+1
· 1
dx· dt1
= − 1
64(
1
t+ t1+
1
t− t1)(t2 − 1)3
t2· 1
dt· dt1
From (5) in the definition on the Eynard-Orantin recursion we obtain
WD1,1(t1) =
1
2πi
∫γ
KD(t, t1)[WD0,2(t,−t) +
dx · dx1
(x− x1)2]
= − 1
2πi
∫γ
KD(t, t1)dt · dt
4t2= − 1
128
(t21 − 1)3
t41dt1,
47
where the contour of integration γ consists of two concentric circles bothcentered around t = 0, where one has an infinitesimally small radius and theother has a large ( r > maxj∈N |tj| ) radius. The inner circle is positivelyoriented and the outer circle is negatively oriented. (See the below figure).
Figure 3.4: The integration contour γ
Using the general formula we can, for instance, compute the (g, n) = (0, 3)case.
WD0,3(t1, t2, t3) =
1
2πi
∫γ
KD(t, t1)[WD
0,2(t, t2)WD0,2(−t, t3) +WD
0,2(t, t3)WD0,2(−t, t2)
]=
1
64[
1
2πi
∫γ
(1
t+ t1+
1
t− t1
)(t2 − 1)3
t2
(1
(t+ t2)2(t− t3)2+
1
(t1 + t3)2(t1 − t2)2
)dt]dt1dt2dt3
= [− 1
32
(t21 − 1)3
t21
(1
(t+ t2)2(t− t3)2+
1
(t1 + t3)2(t1 − t2)2
)− 1
16
∂
∂t2
(t2
t22 − t21(t22 − 1)3
t22
1
(t2 + t3)2
)− 1
16
∂
∂t3
(t3
t23 − t21(t23 − 1)3
t23
1
(t2 + t3)2
)]dt1dt2dt3 = − 1
16
(1− 1
t21t22t
23
)dt1dt2dt3.
The next theorem concludes this chapter.
Theorem 3.9. We define the Laplace transform of the number of dessind’enfants by
FDg,n(t1, . . . , tn) =
∑µ∈Zn+
Dg,n(µ)e−(µ1w1+...+µnwn),
where the coordinate tj is related to the Laplace conjugate coordinate wj by
ewj =tj + 1
tj − 1+tj − 1
tj + 1.
48
The differential forms
WDg,n(t1, . . . , tn) = d1 · · · dnFD
g,n(t1, . . . , tn)
then satisfy the Eynard-Orantin topological recursion
WDg,n(t1, . . . , tn) =
1
64
1
2πi
∫γ
(1
t+ t1+
1
t− t1
)(t2 − 1)3
t2· 1
dt· dt1
× [n∑j=2
(WD
0,2(t, tj)Wg,n−1(−t, t2, . . . , tj, . . . , tn) +WD0,2(−t, tj)Wg,n−1(t, t2, . . . , tj, . . . , tn)
)+WD
g−1,n+1(t,−t, t2, . . . , tn) +stable∑
g1+g2=gItJ={2,...,n}
WDg1,|I|+1(t, tI)Wg2,|J |+1(−t, tJ)].
The last sum in the above recursion is restricted to the stable geometries.Differently stated, the partition should satisfy 2g1− 1 + |I| > 0 and 2g2− 1 +|J | > 0. The spectral curve of the Eynard-Orantin recursion is given by{
x = z + 1z
y = −z
with the preferred coordinate t such that
t =z + 1
z − 1.
Before starting with the proof of this theorem, we first proof thefollowing proposition.
Proposition 3.10. We use the variables xj defined by xj = ewj and write
WDg,n(t1, . . . , tn) = wg,n(x1, . . . , xn) dx1 · · · dxn.
The Laplace transform of the recursion equation 3.7 is the following differ-ential recursion:
− x1wg,n(x1, . . . , xn) =n∑j=2
∂
∂xj
(1
xj − x1
(wg,n−1(x2, . . . , xn)− wg,n−1(x1, x2, . . . , xj, . . . , xn))
)+ wg−1,n+1(x1, x1, x2, . . . , xn)
+∑
g1+g2=gItJ={2,...,n}
wg1,|I|+1(x1, xI)wg2,|J |+1(x1, xJ). (3.1)
49
Proof. We want to apply the operation
(−1)n∑
µ1,...,µn>0
µ2 · · ·µnn∏i=1
1
xµi+1i
to each side of 3.7. The left hand side then becomes wg,n(x1, . . . , xn). Thecomputation of the operation working on the second line of the equation 3.7is not too difficult, we consider here just the first term, as the second termis computed in the same way.
(−1)n∑
µ1,...,µn>0
µ2 · · ·µnn∏i=1
1
xµi+1i
∑α+β=µ1−2
αβDg−1,n+1(α, β, µ2, . . . , µn)
= − 1
x1
(−1)n+1∑
µ2,...,µn>0
∑α,β>0
αβµ2 · · ·µnDg−1,n+1(α, β, µ2, . . . , µn)1
xα+11
· 1
xβ+11
n∏i=2
1
xµi+1i
= − 1
x1
wg−1,n+1(x1, x1, x2, . . . , xn).
The second line of the equation thus gives
− 1
x1
wg−1,n+1(x1, x1, x2, . . . , xn) +∑
g1+g2=gItJ={2,...,n}
wg1,|I|+1(x1, xI)wg2,|J |+1(x1, xJ)
.
To compute the operation on the first line, we first fix j > 1 and set ν =µ1 + µj − 2 ≥ 0. Then
(−1)n∑
µ1,...,µn>0
µ2 · · ·µn(µ1 + µj − 2)×Dg,n−1(µ1 + µj − 2, µ2, . . . , µj, . . . , µn)n∏i=1
1
xµi+1i
= −∞∑ν=0
∑µ2,...,µj ,...,µn>0
(−1)n−1νµ2 · · · µj · · ·µn
×Dg,n−1(ν, µ2, . . . , µj, . . . , µn)1
xν+11
∏i 6=1,j
1
xµi+1i
ν+1∑µj=1
µjxµj−21
1
xµj+1j
. (3.2)
50
Now, while assuming |x1| < |xj|, we calculate
ν+1∑µj=1
µjxµj−21
1
xµj+1j
= − 1
x21
∂
∂xj
ν+1∑µj=0
(x1
xj
)µj
= − 1
x21
∂
∂xj
1
1− x1xj
−
(x1xj
)ν+2
1− x1xj
= − 1
x21
∂
∂xj
(1
1− x1xj
)+ xν1
∂
∂xj
(1
xj − x1
1
xν+1j
).(3.3)
Substituting 3.3 in 3.2 we obtain
(3.2) = wg,n−1(x1, x2, . . . , xj, . . . , xn)1
x21
∂
∂xj
(1
1− x1xj
)
− 1
x1
∂
∂xj
(1
xj − x1
wg,n−1(x2, . . . , xj, . . . , xn)
)= − 1
x1
∂
∂xj
(1
xj − x1
(wg,n−1(x2, . . . , xj, . . . , xn)− wg,n−1(x1, x2, . . . , xj, . . . , xn))
).
And this completes the proof of the proposition.
We will now proof the main theorem of this chapter.
Proof. When splitting the curve into two pieces, the second sum in proposi-tion 3.10 contains contributions from the unstable geometries (g, n) = (0, 1)and (0, 2). Our first step is to separate these contributions out. For g1 = 0and I = ∅, we have the contribution
2w0,1(x1)wg,n(x1, x2, . . . , xn).
For g1 = 0 and I = {j}, or g2 = 0 and J = {j}, we have the contribution
2n∑j=2
w0,2(x1, xj)wg,n−1(x1, . . . , xj, . . . , xn).
Both WD0,1 and WD
0,2 are defined on the spectral curve, so we may switch tothe preferred coordinate t. We then introduce
WDg,n(t1, . . . , tn) = wDg,n(t1, . . . , tn) dt1 · · · dtn = wg,n(x1, . . . , xn) dx1 · · · dxn.
51
Since w0,1(x) = −z(x) , we know that
w0,1(x) = − t+ 1
t− 1
w0,2(x1, x2) =1
(t1 + t2)2
(t21 − 1)2
8t1
(t22 − 1)2
8t2
wg,n(x1, . . . , xn) = (−1)nwDg,n(t1, . . . , tn)n∏i=1
(t2i − 1)2
8ti.
Proposition 3.10 is thus equivalent to
2
(t21 + 1
t21 − 1− t1 + 1
t1 − 1
)wDg,n(t1, . . . , tn) =
n∑j=2
[(t21 − 1)2(t2j − 1)2
16(t21 − t2j)2
8tj(t2j − 1)2
wDg,n−1(t1, . . . , tj, . . . , tn)
+∂
∂tj
((t21 − 1)(t2j − 1)
4(t21 − t2j)8t1
(t21 − 1)2
(t2j − 1)2
8tjwDg,n−1(t2, . . . , tn)
)]
+(t21 − 1)2
8t1
wDg−1,n+1(t1, t1, t2, . . . , tn) +stable∑
g1+g2=gItJ={2,...,n}
wDg1,|I|+1(t1, tI)wDg2,|J |+1(t1, tJ)
+2
n∑j=2
1
(t1 + tj)2
(t21 − 1)2
8t1wDg,n−1(t1, . . . , tj, . . . , tn)
=n∑j=2
[
(tj(t
21 − 1)2
2(t21 − t2j)2+
1
(t1 + tj)2
(t21 − 1)2
4t1
)wDg,n−1(t1, . . . , tj, . . . , tn)
+t1
t21 − 1
∂
∂tj
((t2j − 1)3
4tj(t21 − t2j)wDg,n−1(t2, . . . , tn)
)]
+(t21 − 1)2
8t1
wDg−1,n+1(t1, t1, t2, . . . , tn) +stable∑
g1+g2=gItJ={2,...,n}
wg1,|I|+1(t1, tI)wg2,|J |+1(t1, tJ)
.
Since
2
(t21 + 1
t21 − 1− t1 + 1
t1 − 1
)= − 4t1
t21 − 1,
we see that
wDg,n(t1, . . . , tn) =n∑j=2
[∂
∂tj
((t2j − 1)3
16tj(t21 − t2j)wDg,n−1(t2, . . . , tn)
)
+(t21 − 1)3
16t21
t21 + t2j(t21 − t2j)2
wDg,n−1(t1, . . . , tj, . . . , tn)]
52
− (t21 − 1)3
32t21
wDg−1,n+1(t1, t1, t2, . . . , tn) +stable∑
g1+g2=gItJ={2,...,n}
wg1,|I|+1(t1, tI)wg2,|J |+1(t1, tJ)
(3.4)
At this point we want to compute the integral in the main theorem
WDg,n(t1, . . . , tn) =
1
64
1
2πi
∫γ
(1
t+ t1+
1
t− t1
)(t2 − 1)3
t2· 1
dt· dt1
× [n∑j=2
(WD
0,2(t, tj)Wg,n−1(−t, t2, . . . , tj, . . . , tn) +WD0,2(−t, tj)Wg,n−1(t, t2, . . . , tj, . . . , tn)
)+WD
g−1,n+1(t,−t, t2, . . . , tn) +stable∑
g1+g2=gItJ={2,...,n}
WDg1,|I|+1(t, tI)Wg2,|J |+1(−t, tJ)].
Now recall that for 2g − 2 + n > 0, wDg,n(t1, . . . , tn) is a Laurent polynomialin t21, . . . , t
2n. So the third line of the above formula is immediately calculated
because the contour γ encloses both t1 and −t1 and contributes residues withthe negative sign. The result corresponds with the last line of 3.4. Likewise,since
WD0,2(t, tj)Wg,n−1(−t, t2, . . . , tj, . . . , tn) +WD
0,2(−t, tj)Wg,n−1(t, t2, . . . , tj, . . . , tn)
= −(
1
(t+ tj)2+
1
(t− tj)2
)wDg,n−1(t, t2, . . . , tj, . . . , tn) dt dt dt2 · · · dtj · · · dtn,
the residues at ±t1 contribute
−(t21 − 1)3(t21 + t2j)
16t21(t21 − t2j)2wDg,n−1(t1, . . . , tj, . . . , tn).
Which is the same value as the second line of 3.4.Within the contour there are second order poles at ±tj for each j ≥ 2, comingfrom WD
0,2(±t, tj). The calculations give us
1
64
1
2πi
∫γ
(1
t+ t1+
1
t− t1
)(t2 − 1)3
t2
n∑j=2
[WD0,2(t, tj)Wg,n−1(−t, t2, . . . , tj, . . . , tn)
+WD0,2(−t, tj)Wg,n−1(t, t2, . . . , tj, . . . , tn)]
= − 1
32
∂
∂tj
((1
tj + t1+
1
tj − t1
)(t2j − 1)3
t2jwDg,n−1(tj, t2, . . . , tj, . . . , tn)
)=
1
16
∂
∂tj
(1
t2j − t21(t2j − 1)3
tjwDg,n−1(tj, t2, . . . , tj, . . . , tn)
)53
And this gives the first line of 3.4, we have therefore completed the proof ofthe theorem.
With this theorem we have concluded our final chapter.
54
Summary
The complex plane, which we denote by C, plays a very important role inthis thesis. If we take the complex plane and deform it a bit, we obtain aRiemann surface. The number of ‘holes’ we get after deformation is calledthe genus of the surface.
The most left surface can be obtained by adding the point infinity (whichwe denote by ∞ ) to the complex plane, this space C is called the Riemannsphere. In this thesis we look at coverings of this Riemann sphere.A covering is a function f from a Riemann surface X to the Riemann sphereC, with a special property: there is an integer n such that for every value yin C there are n points in X that are send to this value y. Mathematicallywe say that the preimage f−1(y) consists of n points for each y in C.For some functions however, there are a few values in C that have a smallernumber of points in their preimage (smaller than the number n of points inthe preimage of all other values of C ). These values are called critical values.The points in the preimage of a critical value are called critical points. Themultiplicity of a critical point is ‘the number of solutions that are missing inits neighbourhood’. Mathematically, we denote the multiplicity of a criticalpoint x by k− 1, where k is the number of times we can differentiate f suchthat the value of this differentiated f in x still is equal to zero.For instance, look at a covering of the Riemann sphere by the Riemannsphere itself, that is defined by f : z → z2. Then the preimage of every valueconsist of two points, for example: f−1(1) = {1,−1}, f−1(−1) = {i,−i},f−1(2) = {
√2,−√
2}, f−1(i) = {(1 + i) · 1/√
2, (−1− i) · 1/√
2}.
55
To be exact, this holds for almost every value, you may have noticed that0 is the only value in C that has only one point in its preimage under f .(f−1(0) = {0} .) Thus 0 is a critical value of f . The corresponding criticalpoint is in this case also 0. The multiplicity of this point is 2.
In this thesis we investigate Belyi functions, coverings of which the onlycritical values are allowed to be 0 , 1 or ∞. Now let f be a Belyi functionfrom a Riemann surface X to the Riemann sphere. Consider the segment[0, 1] which is a subset of C. We denote 0 with • and 1 with ◦. Doing so thesegment will look like this:
We then take the the preimage f−1([0, 1]) ⊂ X. This gives us a graph drawnon the Riemann surface X. (A graph is a collection of dots, called vertices,and lines, which are called edges.) This graph has the following properties:the white vertices of the graph are the preimages of 1 and the black verticesare the preimages of 0. Also, the valencies (the number of edges going toa vertex) of the vertices are equal to the multiplicities of the correspondingcritical points. Note that vertices of valency one are not critical points of fbut still may arise as the preimage of either 0 or 1. The graph will separatethe surface in multiple areas, which are called faces. Each face of the graphH will contain exactly one pole (a point that is send to ∞ under f ), thesepoles are called centers of faces. Since f is a Belyi function there will be noother critical points of f on X besides the centers of faces and the (black orwhite) vertices.A graph that is drawn on a Riemann surface following the above recipe,resembles the given Belyi function and is called a dessin d’enfant.
The dessin d’enfant corresponding to f(x) = − (x−1)3(x−9)64x
, x in C.
56
Bibliography
[1] S. Lando en A. Zvonkin, Graphs on Surfaces and their Applications,Springer, 2004
[2] J. Betrema and A. Zvonkin, Plane trees and Shabat polynomials, DiscreteMathematics Volume 153, 1996
[3] A. Zvonkin, Belyi functions: Examples, Properties and Applications,http://.labri.fr/perso/zvonkin/Research/belyi.pdf, 2008
[4] L. Schneps and P. Lochak, Geometric Galois Action: Volume 1 and 2,Cambridge University Press, 1997
[5] B. Kock, Belyi’s theorem revisited, Contributions to Algebra and Geom-etry Volume 45, 2004
[6] B. Eynard and N. Orantin, Algebraic methods in random matrices andenumerative geometry, Springer, 2008
[7] B. Eynard and N. Orantin, Invariants of algebraic curves and topologicalexpansion, Communications in Number Theory and Physics Volume 1,2007
[8] O. Dumitrescu, M. Mulase, B. Safnuk and A. Sorkin, The spectralcurve of the Eynard-Orantin recursion via the Laplace transformation,http://arxiv.org/abs/1202.1159, 2012
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57
