The Eynard-Orantin Recursion for the
Generalized Catalan Numbers
Patrick Buijsman
June 27, 2017
Bachelor Thesis
Supervisor: prof. dr. Sergey Shadrin
Abstract
We define the generalized Catalan numbers as an enumerative geometric problem (A-model), by counting graphs embedded on surfaces. We distinguish the generalized Cata-lan numbers by their type (g, n), where g is the genus of the surface and n is the number of vertices in the graph. Taking the Laplace transform of the generalized Catalan num-bers of type (g, n), we obtain a differential n-form. The differential n-forms satisfy a recursion equation known as the Eynard-Orantin recursion formula, which is a formula on the B-model side. We have thus found the mirror dual of the generalized Cata-lan numbers, and the Laplace transform takes the role of mirror symmetry. We prove that the Catalan partition function satisfies a Schr¨odinger-type equation, which is pre-dicted, since the B-model is mirror dual to an enumerative geometric problem. To define the Catalan partition function, we need primitive functions for each of the differential n-forms. Due to the A-model, we have natural zeroes for each of the primitive func-tions and therefore, the primitive funcfunc-tions are uniquely determined by the differential n-forms.
Title: The Eynard-Orantin Recursion for the Generalized Catalan Numbers Authors: Patrick Buijsman, patrick.buijsman@student.uva.nl, 10539417 Supervisor: prof. dr. Sergey Shadrin
Second grader: dr. Raf Bocklandt Date: June 27, 2017
Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math
Contents
1 Introduction 4
2 Preliminaries 5
2.1 Riemann surfaces 5
2.2 Functions on Riemann surfaces 6
3 The Catalan numbers and their generalization 8
4 The Eynard-Orantin recursion formula 12
5 The Catalan partition function 18
6 Consequences of the differential equation 22
7 Conclusion 26
Popular summary (in Dutch) 27
1 Introduction
Mirror symmetry describes an equivalence between mathematical objects. We call an object on one side of the equivalence an A-model, and an object on the other side a B-model. The symmetry allows us to calculate properties of an A-model by studying its dual B-model, and vice versa. Suppose we start with a counting problem on the A-model side, such that the unstable geometries (that is, (g, n) = (0, 1) and (0, 2)) make sense. Then, Dumitrescu, Mulase, Safnuk and Sorkin proposed [1] to take the Laplace transform of the solutions to these unstable cases, which (presumably) provides us with a spectral curve on the B-model side. Thus, the Laplace transform takes the role of mirror symmetry. The Eynard-Orantin recursion [2] then takes the spectral curve and defines, inductively on 2g − 2 + n, a meromorphic symmetric differential n-form Wg,n for each
(g, n) satisfying 2g − 2 + n > 0. This solves the counting problem on the A-model side. Moreover, since the spectral curve is mirror dual to an geometric enumerative problem, it is expected that the principal specialization of the partition function Z of the B-model satisfies a Schr¨odinger-type equation, P Z = 0, where P is a differential operator [3]. In recent publications, Dumitrescu, Mulase, Safnuk and Sorkin [1], and Mulase and Su lkowski [4] defined the Catalan numbers and their higher-genus generalization by counting graphs on surfaces, and could thus consider the counting problem as an A-model. They showed that its mirror dual exhibits the properties mentioned above. It is shown that the Laplace transform of the generalized Catalan numbers, defined as symmetric differential forms, satisfies the Eynard-Orantin recursion on the B-model side, and that the principal specialization of the partition function satisfies the following Schr¨odinger-type equation ~2 d 2 dx2 + ~x d dx+ 1 ZC(x, ~) = 0.
Here, we examine the mentioned properties for the Catalan numbers as follows. Chap-ter 2 introduces some basic preliminaries. The Catalan numbers and their higher-genus analogue are defined in Chapter 3, and the Eynard-Orantin theory for the generalized Catalan numbers is described in Chapter 4. In Chapter 5 we define the principal special-ization of the partition function, and prove an important theorem regarding the mirror dual of the generalized Catalan numbers. Chapter 6 then uses this theorem to proof several claims made in other chapters. The final chapter is to conclude.
2 Preliminaries
This chapter is based on definitions, described more thoroughly in [5, 6].
2.1 Riemann surfaces
We can think of a Riemann surface as a space that locally resembles an open set in C. To formalize this, let us first introduce complex charts. Recall that a topological space X is a Hausdorff space, if all distinct points in X are pairwise neighborhood-separable. Definition. Let U ⊂ X be an open set in a topological Hausdorff space X, and let V ⊂ C be an open set in the complex plane. A homeomorphism φ : U → V is called a chart on X with domain U . We say that two charts φ1: U1→ V1 and φ2: U2→ V2 are
compatible if either U1∩ U2= ∅, or the map
φ2◦ φ−11 : φ1(U1∩ U2) → φ2(U1∩ U2)
is holomorphic.
In the above definition, X locally resembles the complex plane for each point in U . Since we want X to locally resemble the complex plane everywhere, we need charts around every point in X, and these charts should be compatible. This brings us to the following definition
Definition. An atlas A on X is a collection A = {φα: Uα → Vα} of pairwise compatible
charts, such that X is the union of the domains, i.e., X =S
αUα. We say that two atlases
A and B are equivalent, if and only if their union A ∪ B is again an atlas. It follows that any atlas is contained in a (unique) maximal atlas. Such a maximal atlas is called a complex structure on X.
Recall that a topological space is second countable if there exists a countable basis for its topology. Let us now formally define the Riemann surface
Definition. A Riemann surface is a connected topological Hausdorff space that is second countable, together with a complex structure.
Examples of a Riemann surface include the complex numbers C, considered topologically as R2, the 2-sphere and the torus (see figure 2.1). Note that the torus has a ´hole´ in its surface, so we say that the torus has ´genus´ 1. The genus of a compact Riemann surface, denoted with integer g, is the amount of holes in the surface. We can think of a genus g surface as the surface that is obtained by attaching g ´handles´ to the 2-sphere.
Figure 2.1: Riemann surfaces of genus g = 0 (2-sphere, left), genus g = 1 (torus, middle) and genus g = 2 (right)
2.2 Functions on Riemann surfaces
Let p be a point on a Riemann surface X, and let f be a function on X defined near p. If we want to define or verify any particular property of f at p, we use the charts to transport the function to a neighborhood of p, and define or check the property there. Let us define when a function is meromorphic at a certain point or in a set of points. Definition. A function f on X is said to be holomorphic at p, if there exists a chart φ : U → V with p ∈ U , such that f ◦ φ−1 is holomorphic at φ(p). If f is holomorphic at every point in a set W , we say that f is holomorphic in W .
A function f is meromorphic at p, if it has either a removable singularity or a pole, or is holomorphic at p. We say that f is meromorphic in W if it is meromorphic at all points in W .
In order to define integration on the Riemann surface, we need objects to integrate. These objects are called forms.
Definition. A meromorphic differential form of order 1, also known as meromorphic 1-form on an open set V ⊂ C is an expression of the form w = f (z)dz, where f is a meromorphic function on V . We say that w is a meromorphic 1-form in the coordinate z.
For the extension of 1-forms onto Riemann surfaces, we refer to Miranda [5]. The idea is to define 1-forms for each of the charts in the atlas, in such a way that whenever two charts φ1, φ2 have overlapping domains, we can transform the associated 1-form wφ1 to
wφ2 under the change of coordinate mapping φ1◦ φ
−1 2 .
We now define differential forms of higher order, that is, k-forms.
Definition. Let X be a topological Hausdorff space of dimension n and let k be a positive integer. Let I = (i1, i2, . . . , ik) be an ordered k-tuple of integers such that 1 ≤ i1 < i2<
· · · < ik ≤ n. We call I an increasing k-index. Let us write dxI = dxi1∧ · · · ∧ dxik, where
∧ denotes the wedge product. These forms are called basic k-forms, and we can write any k-form as a linear combination of the dxI. Thus, w is a k-form on X, if it is of the
form w =P
IαI(~x)dxI, where we sum over all possible increasing k-indexes I, and ~x is
Note that, by definition of the Wedge product, dxi∧ dxi = 0 for any i. Therefore, a
k-form defined on a space of dimension n < k is always zero. We define the ´derivative´ of a k-form, which results in a (k + 1)-form.
Definition. Let w = P
JαJ(x1, . . . , xn)dxJ be a k-form. We define the operator di
by diw = PJ ∂x∂iαJ(x1, . . . , xn)dxJ ∧ dxi. In particular, if we start with a function
f (x1, . . . , xn) on X, and apply the operator d1· · · dn, we get
d1· · · dnf (x1, . . . , xn) =
∂n ∂x1· · · ∂xn
f (x1, . . . , xn)dx1∧ · · · ∧ dxn,
3 The Catalan numbers and their
generalization
Let us introduce cellular graphs [4], since we define the generalized Catalan numbers by counting cellular graphs on surfaces.
Let X be a connected, closed, oriented surface of genus g, and define the 0-skeleton of a cell-decomposition of X as the direct sum of the 0-cells of X. Attach copies of the 1-cells to the 0-skeleton, by a function that maps the boundary of each 1-cell into the 0-skeleton. The result is called the 1-skeleton of the cell-decomposition [7]. A cellular graph of type (g, n) is defined as a 1-skeleton consisting of n 0-cells, labeled by the index set [n] = {1, 2, . . . , n}. The 0-cells and 1-cells of a cellular graph are called the vertices and edges respectively, and the amount of half-edges joining in a vertex is called the degree of that vertex. We identify two cellular graphs of equal type, if there exists an orientation-preserving homeomorphism between the underlying surfaces, that maps one cellular graph to the other, and preserves the labeling of each vertex. The restriction of such a homeomorphism to the cellular graphs is called an isomorphism between cellular graphs [8]. An automorphism of a cellular graph is an isomorphism with itself.
Denote by Auth(Γ) the automorphism group of cellular graph Γ, and let Dg,n(µ1, . . . , µn)
be the number of connected cellular graphs Γ of type (g, n) with n labeled vertices of degree µ1, . . . , µn, counted with weight 1/|Auth(Γ)|. In general, Dg,n(µ1, . . . , µn) is a
rational number.
The weights 1/|Auth(Γ)| make counting more difficult. Let us therefore place an outgo-ing arrow to one of the µj half-edges incident to the j-th vertex for each j ∈ {1, . . . , n}
(see figure 3.1), which allows us to kill the automorphism altogether. Since there are µ1· · · µn choices to place the arrows, the number of arrowed graphs is given by
µ1· · · µnDg,n(µ1, . . . , µn). Let us denote this number by Cg,n(µ1, . . . , µn), i.e.,
Cg,n(µ1, . . . , µn) = µ1· · · µnDg,n(µ1, . . . , µn). (3.1)
Note that Auth(Γ) is a subgroup of Qn
i=1Z/µiZ that rotates each vertex, since an au-tomorphism fixes each vertex of Γ. Hence, the number Cg,n(µ1, . . . , µn) is always an
integer, which we call the generalized Catalan number of type (g, n). The naming of these integers comes from the following theorem.
Theorem 3.1 ([4]). For each g ≥ 0, n ≥ 1, the equation Cg,n(µ1, . . . , µn) = n X j=2 µjCg,n−1(µ1+ µj− 2, µ2, . . . ,cµj, . . . , µn) + X α+β=µ1−2 " Cg−1,n+1(α, β, µ2, . . . , µn) + X g1+g2=g ItJ ={2,...,n} Cg1,|I|+1(α, µI)Cg2,|J |+1(β, µJ) # (3.2) is satisfied. Here, cµj in the first line indicates that µj is omitted, and µI:= (µi)i∈I.
Note that this is not a recursion formula, since the term Cg,n(µ1, . . . , µn) appears
im-plicitly on the right-hand side as well. The proof is clearly stated by Mulase and Su lkowski [4]. The idea is to take an arrowed graph that is counted by the left-hand side, and to shrink the arrowed edge incident to p1, where p1 is one of the labeled vertices
{p1, . . . , pn}. First suppose the case where the arrowed edge connects p1 and pj for some
j > 1. Shrinking this edge, we merge p1 and pj and we thus obtain a graph with n − 1
vertices, where the merged vertex now has degree µ1 + µj − 2 (see figure 3.2). This
corresponds to the first line of equation (3.2). The second case is when the arrowed edge connects p1 with itself, and thus forms a loop. Then, the arrowed edge separates the
other half-edges into two groups containing α and β half-edges (see figure 3.2). When shrinking the arrowed edge, we split vertex p1 into two new vertices pα1 and p
β 1. The
vertex pα1 is connected to the first group of α half-edges, and pβ1 to the other group. This corresponds to the second line of equation (3.2). The shrinking may change the topology of the underlying surface. The first term counts the cases where the genus reduces to g − 1, and the second term counts the cases where it splits the surface into two pieces of genus g1 and g2 (where g1 or g2 is possibly zero).
Figure 3.2: The process of shrinking the arrowed edge incident to p1. The transformation
on the left illustrates the case where p1 connects to pj for some j > 1 and on
For (g, n) = (0, 1), there is only one vertex so the first line in equation (3.2) is zero. Moreover, the genus is already zero, so shrinking the arrowed edge can only result in two pieces of genus g1= 0 and g2 = 0. Therefore, equation (3.2) reduces to
C0,1(µ1) =
X
α+β=µ1−2
C0,1(α)C0,1(β). (3.3)
For a graph with only one vertex, we have the irregularity that a degree of 0 is allowed, because a single vertex is connected. Therefore, let us define C0,1(0) = 1. Since the only
vertex must have an even degree, we set µ = 2m. Then,
C0,1(2m) = X α+β=2(m−1) C0,1(α)C0,1(β) = 2(m−1) X α=0 C0,1(α)C0,1(2(m − 1) − α), (3.4)
which is the m-th Catalan number [9], and therefore explains the name of the generalized Catalan numbers.
We note that Zagier [8, Remark 3.1.7] states that there is no combinatorial interpretation of the recurrence relation
(m + 2)g(m + 1) = (4m + 2)g(m) + (4m3− m)g−1(m − 1), (3.5)
where g(m) denotes the number of ways to obtain a surface of genus g from a regular
2m-gon. Although the recurrence (3.5) is proven using closed formulas, a combinatorial proof might answer questions such as why the m-th Catalan number divides the integer 2gg(m) [8]. For g = 0, the number g(m) is equivalent to the m-th Catalan number Cm,
i.e., 0(m) = Cm= C0,1(2m). In this case, the recurrence (3.5) is given by (m+2)Cm+1 =
(4m + 2)Cm, which follows immediately from the closed formula Cm = m+11 2mm and
the identity 12 2(m+1)m+1
= 2m+1m+1 2mm
[10]. Here, we give a combinatorial proof of the recurrence for the g = 0 case, using the definition of the Catalan numbers in terms of cellular graphs given above.
Corollary 3.2. The Catalan numbers satisfy the recurrence equation
(m + 2)Cm+1 = (4m + 2)Cm. (3.6)
Proof. Take an arrowed graph with one vertex and examine the half-edges in counter-clockwise order, starting with the half-edge containing the arrow. If we encounter a pair of adjacent half-edges that belong to the same edge, we call this edge a hook (see figure 3.3). Note that, for m > 0, every graph contains at least one hook. Let us count the number of graphs with m + 1 edges, such that one of the hooks is marked.
Figure 3.3: An arrowed graph with one hook, colored in red.
Consider an arrowed graph with m edges. We obtain a marked arrowed graph with m + 1 edges, by placing a marked hook between any pair of adjacent half-edges. This gives us 2m choices for each graph, and we thus obtain 2m · Cm graphs with one marked
hook. However, this does not account for the possibility that the arrowed edge is marked. Therefore, consider now the graph with only 1 edge, which is arrowed and marked. There are Cm ways to complete the graph to a graph with m + 1 edges, such that the marked
edge remains a hook. Therefore, the total number of arrowed, marked graphs with m + 1 edges is (2m + 1)Cm.
On the other hand, we obtain a marked graph with m + 1 edges, by just choosing a hook in any graph with m + 1 edges. We claim that on average, a graph with m + 1 edges contains m+22 hooks. For this, let us first introduce a new notation.
Following Mulase [3], we denote each half-edge with a parenthesis, such that each edge corresponds to one opening ´(´ and one closing ´)´ parenthesis. Starting with the arrowed half-edge, we open the first bracket. Then we move in counterclockwise order through the graph. Each time a half-edge that starts a new loop is encountered, another bracket is opened. When a loop is closed, we close the bracket. For example, the graph in figure 3.3 corresponds to ´((()))´. Note that a hook in the graph corresponds to an adjacent pair of an opening and a closing parenthesis, i.e., the pair ´()´. Let us define the following involution. If γ1 and γ2 are (possibly empty) sets of brackets, we define
(γ1)γ2 = (γ2)γ1. For example, (())(()) = ((()))() = (()())() and (((()))) = ()((())) =
()()(()) = ()()()(). Note that, if k is the number of hooks in a set of e edges γ, then its involution γ contains e + 1 − k hooks. Thus, on average, a graph with m + 1 edges contains (k)+(m+2−k)2 = m+22 hooks. This proves the claim.
It therefore follows that the number of arrowed graphs with m+1 edges, with one marked hook, is given by m+22 Cm+1. Hence, we have the equality
m + 2 2
4 The Eynard-Orantin recursion formula
Let us introduce a generating function for the Catalan numbers, given by z = z(x) = ∞ X m=0 Cm x2m+1, (4.1)
where Cm= C0,1(2m) denotes the m-th Catalan number.
Lemma 4.1. The inverse function of z(x) that vanishes at x = ∞ is given by x = z + 1
z. (4.2)
Proof. Let us solve the equation x = z + 1z for z. Multiplying both sides with z yields z2− xz + 1 = 0, which has the solutions
z = 1 2 x ±px2− 4 = x 2 1 ± s 1 − 2 x 2 . (4.3)
Since z is should vanish at x = ∞, we need the solution with the minus sign. Using the binomial expansion, let us rewrite the square root appearing in the last line of the equation above. s 1 − 2 x 2 = 1 − 2 x 2! 1 2 = ∞ X n=0 1 /2 n − 2 x 2!n = ∞ X n=0 (−1)n+1 22n(2n − 1) 2n n (−1)n 2 x 2n = − ∞ X n=0 1 2n − 1 2n n x−2n = 1 − ∞ X n=1 1 2n − 1 2n n x−2n,
where we extracted the n = 0 term from the summation in the last equality. Substituting the above expression in the solution for z(x) given in equation (4.3) gives us
z = x 2 ∞ X n=1 1 2n − 1 2n n x−2n = ∞ X n=1 1 2 1 2n − 1 2n n x−2n+1.
Using the equalities 21 2nn = 2n−1n−1 and 2n−11 2n−1n−1 = 1n 2(n−1)n−1 [10], we obtain
z = ∞ X n=1 1 n 2(n − 1) n − 1 x−2n+1 = ∞ X m=0 1 m + 1 2m m x−(2m+1) = ∞ X m=0 Cm x2m+1.
In the second equality, the summation is re-indexed by m = n − 1, and the third equality uses that Cm= m+11 2mm [9].
The inverse of z(x) in equation (4.2) defines a Lagrangian immersion [4] Σ = C 3 z 7→ ((x(z), y(z)) ∈ T∗C,
(
x = z +1z
y = −z , (4.4)
which forms the spectral curve for the Eynard-Orantin recursion. The projection x = z + 1z has two ramification points z = ±1, so it is more convenient to use a coordinate that has these points at 0 and ∞. Therefore, introduce the coordinate t, related to z by
z = t + 1
t − 1, (4.5)
and define the mirror dual of the generalized Catalan numbers in terms of the preferred coordinate t, as the Laplace transform of the Cg,n, i.e.,
Wg,nC (t1, . . . , tn) = (−1)n
X
(µ1,...,µn)∈Zn+
Cg,n(µ1, . . . , µn)e−hw,µidw1. . . dwn. (4.6)
Here, hw, µi = Pn
i=1wiµi, and the Laplace transform coordinate w is related to the
coordinate t of the Lagrangian by ewi = x i = zi+ 1 zi = 2t 2 i + 1 t2 i − 1 , i ∈ {1, 2, . . . , n}. (4.7)
Note that Wg,nC is a symmetric differential form of degree n, defined as a formal power series in C[x−11 , . . . , x
−1
Theorem 4.2. The Wg,nC are meromorphic symmetric differential forms of degree n on Cn.
Note the irregularity for (g, n) = (0, 1), since a single vertex of degree 0 is connected. Therefore, we define W0,1C (t) = − ∞ X µ=0 C0,1(µ)e−wµdw,
which includes the case where µ = 0. If n = 1, the only vertex has an even degree, so we set µ = 2m. Then, in terms of the preferred coordinate t, we have
W0,1C (t) = − ∞ X m=0 C0,1(2m) x2m+1 dx = −zdx = 8t (t − 1)3(t + 1)dt. (4.8)
For (g, n) = (0, 2) we have [1, Proposition 4.1] W0,2C (t1, t2) =
dt1· dt2
(t1+ t2)2
. (4.9)
The recursion kernel for the generalized Catalan numbers is defined by [4]
KC(t, t1) = 1 2 Rs(t) t W C 0,2(·, t1) WC 0,1(s(t)) − W0,1C (t) ,
where the dot in W0,2C (·, t1) indicates that integration is done with respect to the first
argument of W0,2C (t, t1) and s(t) denotes the local Galois conjugate of t, that is, the
Galois conjugate of the projection x = 2tt22+1−1 of the immersion defined in equation (4.4).
Rewriting x = 2tt22+1−1 as a monic polynomial in t, we get
t2−x + 2
x − 2 = 0. (4.10)
For a given x, suppose that t is a solution to the above equation. Then the Galois conjugate s(t) of t is defined as the other solution to equation (4.10) for the same x [11]. Hence, it follows that s(t) = −t, and we calculate KC(t, t1) by
KC(t, t1) = 1 2 R−t t W0,2(·, t1) W0,1C(−t) − W0,1C (t) = − 1 64 1 t + t1 + 1 t − t1 (t2− 1)3 t2 · 1 dt · dt1 (4.11)
Theorem 4.3. The differential forms Wg,nC (t1, . . . , tn), subject to 2g − 2 + n > 0, satisfy
the Eynard-Orantin topological recursion formula. In this case, the recursion formula is given by Wg,n(t1, . . . , tn) = 1 2πi I γ KC(t, t1) n X j=2 W0,2C (t, tj)Wg,n−1C (−t, t2, . . . , ˆtj, . . . , tn) + W0,2C (−t, tj)Wg,n−1C (t, t2, . . . , ˆtj, . . . , tn) +Wg−1,n+1C (t, −t, t2, . . . , tn) + stable X g1+g2=g ItJ ={2,...,n} WgC 1,|I|+1(t, tI)W C g2,|J |+1(−t, tJ) . (4.12) Here, integration is taken with respect to t along a the closed loop γ (see figure 4.1) that consists of two circles, both centered at the origin, such that the inner circle has an infinitesimally radius and the outer circle has finite radius r > maxj∈{1,...,n}|tj|. The
inner circle is positively oriented, and the outer circle is negatively oriented. In the last summation, I t J is the disjoint union of I and J , |I| denotes the cardinality of I and tI := (ti)i∈I. By ¨stable” in the summation, we mean that the summation includes only
terms where 2g1− 1 + |I| > 0 or 2g2− 1 + |J | > 0.
Figure 4.1: The loop γ used for integration. The loop consists of two concentric cir-cles centered at the origin, where the inner circle has an infinitesimally radius and is positively oriented, while the outer circle has finite radius r > maxj∈{1,...,n}|tj| and is negatively oriented
Proof. With inspiration from [1], let us define the functions wCg,n(t1, . . . , tn) such that
Proposition 4.4. For each (g, n) subject to 2g − 2 + n > 0, the recursion equation wg,nC (t[n]) = − n X j=2 " ∂ ∂tj (t2j − 1)3 16tj(t21− t2j) wg,n−1C (t[ˆ1]) ! +(t 2 1− 1)3 16t21 t21+ t2j (t21− t2 j)2 wCg,n(t[ˆj]) # −(t 2 1− 1)3 32t21 wCg−1,n+1(t1, t1, t2, . . . , tn) + stable X g1+g2=g ItJ ={2,...,n} wCg 1,|I|+1(t1, tI)w C g2,|J |+1(t1, tJ) (4.14) is satisfied. Moreover, wg,nC (t1, . . . , tn) is a Laurent polynomial in t21, . . . , t2n.
Let us compute the contour integral of the Eynard-Orantin recursion in equation (4.12). Denote by h(t) the integrand of the contour integral, and by {p1, . . . , pr} the set of poles
of h(t) enclosed by γ. By the residue theorem (Cauchy) we have 1 2πi I γ h(t)dt = r X k=1 Rest=pk(h(t)).
Thus, the contour integral of the Eynard-Orantin recursion is found by summing the residues at the poles of the integrand. Substituting the recursion kernel for the expression given in (4.11), and rewriting the Wg,nC in terms of wg,nC , the second line of (4.12) becomes
−1 64 1 2πi I γ 1 t + t1 + 1 t − t1 (t2− 1)3 t2 · 1 dt · dt1· wCg−1,n+1(t, −t, t2, . . . , tn) + stable X g1+g2=g ItJ ={2,...,n} wCg1,|I|+1(t, tI)wCg2,|J |+1(−t, tJ) dt · dt · dt2· · · dtn
By Proposition 4.4, wg,nC (t1, . . . , tn) is a Laurent polynomial in t21, . . . , t2n, and therefore
has no poles within the contour γ. Hence, the only contributing poles are the simple poles in t = ±t1 that come from the terms t±t11. The sum of the residues at these poles
is equal to the second line of (4.14) times dt1· · · dtn.
For the first line of (4.12), we obtain 1 64 1 2πi I γ 1 t + t1 + 1 t − t1 (t2− 1)3 t2 · 1 dt · dt1· dt · Y i∈[n]\{1,j} dti · n X j=2 dt · dtj (t + tj)2 wg,n−1C (t, t2, . . . , ˆtj, . . . , tn) + dt · dtj (t − tj)2 wg,n−1C (−t, t2, . . . , ˆtj, . . . , tn) .
We have again two simple poles in t = ±t1. The contribution of the residues in these
In addition to the simple poles, we now have second order poles in t = ±tj for each
j ∈ {2, . . . , n}, due to the terms dt·dtj
(t±tj)2. The sum of the residues is given by the first
term of the first line in (4.14), again multiplied by dt1· · · dtn. Thus, by multiplying
both sides of equation (4.14) by dt1· · · dtn, the right-hand side becomes precisely the
right-hand side of equation (4.12), while the left hand side becomes the left-hand side of (4.12).
5 The Catalan partition function
The principal specialization of the partition function of the topological recursion in equation (4.12) is given by the following expression [4]
ZC(t, ~) = exp ∞ X g=0 ∞ X n=1 1 n!~ 2g−2+nFC g,n(t, t, . . . , t) , (5.1)
where each Fg,nC (t1, . . . , tn) is a symmetric meromorphic function on Cn, such that its
n-fold exterior derivative recovers Wg,nC (t1, . . . , tn). Thus,
Wg,nC (t1, . . . , tn) = d1· · · dnFg,nC (t1, . . . , tn). (5.2)
We call ZC the Catalan partition function, and Fg,nC a primitive function of the differ-ential form Wg,nC . Based on the conjecture of [1], let us define the primitive functions by taking the Laplace transform. Thus, for each g ≥ 0 and n ≥ 1 we define
Fg,nC (t1, . . . , tn) =
X
(µ1,...,µn)∈Zn+
Dg,n(µ1, . . . , µn)e−(w1µ1+···+wnµn), (5.3)
where the preferred coordinate t is related to the Laplace transform coordinate w by ewi = x i = zi+z1i = 2t 2 i+1 t2 i−1
, i ∈ {1, 2, . . . , n}. From equation (3.1) it follows that FC g,n
is indeed a primitive function for Wg,nC for each (g, n), since equation (5.2) is satisfied. In the next chapter, we prove that Fg,nC (t1, . . . , tn) is meromorphic on Cnfor each stable
(g, n).
Note that t = −1 =⇒ z = 0 =⇒ x = ∞. Therefore, we have the vanishing property Fg,nC (t1, . . . , tn)
tj=−1= 0 (5.4)
for any stable (g, n) and any j ∈ {1, . . . , n}. Thus, Fg,nC is uniquely determined by Wg,nC by the integration formula
Fg,nC (t1, . . . , tn) = Z t1 −1 · · · Z tn −1 Wg,nC (t1, . . . , tn). (5.5)
Since the unstable case (g, n) = (0, 1) has the irregularity that a degree µ = 0 is allowed, we modify the Laplace transform of D0,1(µ) accordingly. Thus,
F0,1C (t) = ∞ X µ=0 D0,1(µ)e−wµ− w = ∞ X µ=0 D0,1(µ) 1 xµ− log(x).
In terms of the t coordinate, we find F0,1C(t) = log t + 1 t − 1 −1 2 (t + 1)2 (t − 1)2. (5.6)
The other unstable case is given by [1] F0,2C (t1, t2) = − log − 2(t1+ t2) (t1− 1)(t2− 1) . (5.7)
The Catalan partition function ZC satisfies the following Schr¨odinger-type equation ~2 d 2 dx2 + ~x d dx + 1 ZC(t, ~) = 0, (5.8)
where, t is considered as a function of x by
t(x) = z(x) + 1 z(x) − 1,
and z(x) is given by the generating function for the Catalan numbers in equation (4.1). The operator ~dxd follows from y = ~dxd, where y is given by the Lagrangian immersion (4.4).
To prove this Schr¨odinger-type equation, the following theorem regarding the primitive functions is essential.
Theorem 5.1. For (g, n) subject to 2g − 2 + n > 0, the differential equation ∂ ∂t1 Fg,nC (t[n]) = − 1 16 n X j=2 " tj t21− t2 j (t21− 1)3 t21 ∂ ∂t1 Fg,n−1C (t[ˆj]) − (t2j− 1)3 t2j ∂ ∂tj Fg,n−1C (t[ˆ1]) !# −1 16 n X j=2 (t21− 1)2 t21 ∂ ∂t1 Fg,n−1C (t[ˆj]) −1 32 (t21− 1)3 t21 ∂2 ∂u1∂u2 Fg−1,n+1C (u1, u2, t2, t3, . . . , tn) u1=u2=t1 − 1 32 (t21− 1)3 t2 1 stable X g1+g2=g ItJ ={2,3,...,n} ∂ ∂t1 FgC1,|I|+1(t1, tI) ∂ ∂t1 FgC2,|J |+1(t1, tJ) (5.9)
holds. Herein, the index notations [n] = {1, 2, . . . , n} and [ˆj] = {1, 2, . . . , ˆj, . . . , n} are used.
Proof. In order to show that the above equation holds, multiply both sides of equation (3.2) with µ1Qni=1x
−µi i
µi and sum over all possible (µ1, . . . , µn) ∈ Z
n
+. On the left-hand
side of the equation, we get X (µ1,...,µn)∈Zn+ µ1 n Y i=1 x−µi i µi Cg,n(µ1, . . . , µn) = −x1 ∂ ∂x1 X (µ1,...,µn)∈Zn+ Cg,n(µ1, . . . , µn) µ1· · · µn n Y i=1 x−µi i = −x1 ∂ ∂x1 X (µ1,...,µn)∈Zn+ Dg,n(µ1, . . . , µn)e−hw,µi = −x1 ∂ ∂x1 Fg,nC (t1, . . . , tn) = t 4 1− 1 4t1 ∂ ∂t1 Fg,nC (t1, . . . , tn),
where the second and third equalities follow from equations (3.1) and (5.3) respectively. For the right-hand side, start with the first line of the right-hand side of equation (3.2). Setting ν = µ1+ µj− 2 we obtain X (µ1,...,µn)∈Zn+ µ1 n Y i=1 x−µi i µi n X j=2 µjCg,n−1(µ1+ µj− 2, µ2, . . . ,cµj, . . . , µn) = n X j=2 ∞ X ν=0 X (µ2,...,ˆµj,...,µn)∈Zn−2+ Cg,n−1(ν, µ2, . . . ,cµj, . . . , µn) µ2· · · µj−1µj+1· · · µn x−ν1 Y i6=1,j x−µi i ν+1 X µj=1 x1 xj µj = n X j=2 ∞ X ν=0 X (µ2,...,ˆµj,...,µn)∈Zn−2+ Cg,n−1(ν, µ2, . . . ,cµj, . . . , µn) µ2· · · µj−1µj+1· · · µn Y i6=1,j x−µi i ν+1 X µj=1 x−11 x−1j x1− xj h x−νj − x−ν1 i = n X j=2 1 x1− xj ∂ ∂x1 Fg,n−1C (t[ˆj]) − ∂ ∂xj Fg,n−1C (t[ˆ1]) = n X j=2 (t21− 1)(t2 j− 1) 4(t2j − t2 1) " −(t 2 1− 1)2 8t1 ∂ ∂t1 Fg,n−1C (t[ˆj]) + (t2j − 1)2 8tj ∂ ∂tj Fg,n−1C (t[ˆ1]) #
For the second line, let us compute only the second term and just state the first term, since the computation of the first term is similar. The first term is given by
X (µ1,...,µn)∈Zn+ µ1 n Y i=1 x−µi i µi X α+β=µ1−2 Cg−1,n+1(α, β, µ2, . . . , µn) = (t 2 1− 1)4 64t1 ∂2 ∂u1∂u2 Fg−1,n+1C (u1, u2, t2, t3, . . . , tn) u1=u2=t1
The computation of the second term is as follows X (µ1,...,µn)∈Zn+ µ1 n Y i=1 x−µi i µi X α+β=µ1−2 X g1+g2=g ItJ ={2,...,n} Cg1,|I|+1(α, µI)Cg2,|J |+1(β, µJ) = X g1+g2=g ItJ ={2,...,n} X (α,β,µ2...,µn)∈Zn+1+ x−(α+β−2)1 n Y i=2 x−µi i µi Cg1,|I|+1(α, µI)Cg2,|J |+1(β, µJ) = X g1+g2=g ItJ ={2,...,n} x21 X (α,µI)∈Z |I|+1 + x−α1 Y i∈I x−µi i µi Cg1,|I|+1(α, µI) X (α,µJ)∈Z |J |+1 + x−β1 Y j∈J x−µj j µj Cg2,|J |+1(β, µJ) = X g1+g2=g ItJ ={2,...,n} ∂ ∂x1 FgC 1,|I|+1(t1, tI) ∂ ∂x1 FgC 2,|J |+1(t1, tJ) = (t 2 1− 1)4 64t2 1 X g1+g2=g ItJ ={2,...,n} ∂ ∂t1 FgC1,|I|+1(t1, tI) ∂ ∂t1 FgC2,|J |+1(t1, tJ),
where the first equality uses the substitution α + β = µ1− 2. Note that the summation
also contains the unstable terms 2g1− 1 + |I| ≤ 0 and 2g2− 1 + |J | ≤ 0, so let’s calculate
them separately. For both combinations (g1 = 0, |I| = ∅) and (g2 = 0, |J | = ∅) the
summand is ∂ ∂t1 F0,1C (t1) ∂ ∂t1 Fg,nC (t1, . . . tn) = 8t1 (t21− 1)(t1− 1)2 ∂ ∂t1 Fg,nC (t1, . . . tn).
If g2 = 0 and J = {j} for some j ∈ {2, . . . , n}, the summand is given by
∂ ∂t1 F0,2C (t1, tj) ∂ ∂t1 Fg,n−1C (t[ˆj]) = tj+ 1 (t1− 1)(t1+ t2) ∂ ∂t1 Fg,n−1C (t[ˆj]).
The case where g1 = 0, I = {i} for some i ∈ {2, . . . , n} is identical. Separating the
unstable terms from the summation, the second term of the second line becomes (t21− 1)4 64t21 16t1 (t21− 1)(t1− 1)2 ∂ ∂t1 Fg,nC (t1, . . . tn) + n X j=2 2(tj+ 1) (t1− 1)(t1+ t2) ∂ ∂t1 Fg,n−1C (t[ˆj]) + stable X g1+g2=g ItJ ={2,...,n} ∂ ∂t1 FgC 1,|I|+1(t1, tI) ∂ ∂t1 FgC 2,|J |+1(t1, tJ) .
Combining the contributions of both lines and collecting similar terms, we obtain equa-tion (5.9).
6 Consequences of the differential
equation
The recursive differential equation in Theorem 5.1 allows us to prove Theorem 4.2 and Proposition 4.4. Let us start with the first part of Proposition 4.4.
Proof of Proposition 4.4, part 1. By definition of Fg,nC and wg,nC , we have that
wg,nC (t1, . . . , tn) = ∂n ∂t1· · · ∂tn Fg,nC (t1, . . . , tn) = ∂n−1 ∂t2· · · ∂tn ∂ ∂t1 Fg,nC (t1, . . . , tn)
Thus, if we apply the operator ∂t∂n−1
2···∂tn to both sides of (5.9), the left-hand side becomes
wCg,n(t1, . . . , tn). Collect on the right-hand side the terms with Fg,n−1C (t[ˆ1]). The result is precisely the right-hand side of equation (4.14). This proves the first part of Proposition 4.4.
In order to prove the second part of Proposition 4.4 and Theorem 4.2, we need some further preparation.
Theorem 6.1 ([12]). The function Fg,nC (t[n]) is uniquely determined by the differential
equation (5.9) of Theorem 5.1 and the vanishing property from equation (5.4). Proof. Suppose that FC
g,n(t[n]) is determined for all (g, n) subject to
0 < 2g − 2 + n < m − 1
for some m ≥ 2. Take any (g, n) such that 2g − 2 + n = m. Then ∂t∂
1F
C
g,n(t[n]) is given by
equation (5.9). Denote by r(t[n]) the right-hand side of (5.9), and integrate both sides
from −1 to t1 over dt1. We obtain the equality
Z t1 −1 ∂ ∂t1 Fg,nC (t[n])dt1= Z t1 −1 r(t[n])dt1.
By the vanishing property, the left-hand side is given by Z t1 −1 ∂ ∂t1 Fg,nC (t[n])dt1 = Fg,nC (t[n]) − Fg,nC (−1, t2, . . . , tn) = Fg,nC (t[n]), and thus, FC g,n(t[n]) is uniquely determined by Fg,nC (t[n]) = Z t1 −1 r(t[n])dt1. (6.1)
Let us use the above theorem to determine F1,1C (t1), one of the first stable cases. For
(g, n) = (1, 1), the summations appearing in r(t1) are empty, so r(t1) only consists of
the term with Fg−1,n+1C . We calculate
F1,1C (t1) = Z t1 −1 − 1 32 (t21− 1)3 t21 ∂2 ∂u1∂u2 F0,2C (u1, u2) u1=u2=t1 ! dt1.
From the definition of F0,2C given in equation (5.7) it follows that h ∂2 ∂u1∂u2F C 0,2(u1, u2) i u1=u2=t1 = 1 4t1, and therefore F1,1C (t1) = − 1 128 Z t1 −1 (t2 1− 1)3 t4 1 dt1 (6.2) = − 1 128 Z t1 −1 t21− 3 + 3t−21 − t−41 dt1 = − 1 384(t 3 1− 9t1− 16 − 9t−11 + t −3 1 ),
which corresponds to the value of [1], where they used the differential form W1,1C (t1) and
the Eynard-Orantin recursion formula (4.12). Since the calculation of F0,3C (t1, t2, t3) using
equation (6.1) is rather tedious, let us use the more compact formula [2, Theorem 4.1] W0,3C (t1, t2, t3) = 1 2πi Z γ W0,2C (t, t1)W0,2C (t, t2)W0,2C(t, t3) dx(t) · dy(t) , (6.3)
which follows from the Eynard-Orantin recursion formula (4.12). In terms of the t-coordinate, we have x(t) = 2t 2+ 1 t2− 1, y(t) = −t + 1 t − 1 by equations (4.4) and (4.5). Since W0,2C(t1, t2) = (tdt1·dt2
1+t2)2, we find W0,3C(t1, t2, t3) = − 1 16 1 2πi Z γ (t2− 1)2(t − 1)2 (t + t1)2(t + t2)2(t + t3)2 1 tdt dt1dt2dt3,
Summing the second order residues in t = −t1, −t2 and −t3, we find
W0,3C (t1, t2, t3) = − 1 16 1 − 1 t21t22t23 dt1dt2dt3. (6.4)
Thus, since F0,3C(t1, t2, t3) is uniquely determined by the integration formula (5.5), we
find F0,3C(t1, t2, t3) = − 1 (t1+ 1)(t2+ 1)(t3+ 1) 1 + 1 . (6.5)
Note that F1,1C (t1) and F0,3C (t1, t2, t3) are Laurent polynomials in t1, . . . , tn of degree 3,
and that every monomial appearing in F1,1C (t1) and F0,3C (t1, t2, t3) contains only an odd
power of each tj. By induction we show that a similar statement holds for all Fg,n in
the stable range.
Theorem 6.2 ([12]). For every (g, n) subject to 2g−2+n > 0, the function Fg,nC (t1, . . . , tn)
is a Laurent polynomial in t1, . . . , tn of degree 3(2g − 2 + n), and every monomial
ap-pearing in Fg,nC (t1, . . . , tn) contains only an odd power of each tj.
Proof. For (g, n) subject to 2g −2+n = 1, we have just shown that the statement is true. Note that the differential equation (5.9) allows us to determine Fg,nC for 2g −2+n > 1 one by one, only depending on F1,1C and F0,3C. Let us therefore suppose that the statement is true for all (g, n) subject to
0 < 2g − 2 + n < m − 1
for some m ≥ 2. Take any (g, n) such that 2g − 2 + n = m, and recall that r(t[n]) denotes
the right-hand side of (5.9). There are two issues we need to address. Firstly, r(t[n]) contains a factor t21
1−t2j
, which is not a Laurent polynomial. Secondly, the integration in (6.1) could produce logarithmic terms. Regarding both issues, let us use the following lemma.
Lemma. If f (x) is a Laurent polynomial in one variable of degree df that contains only
odd powers of x, then y x2− y2 (x2− 1)3 x2 ∂ ∂xf (x) − (y2− 1)3 y2 ∂ ∂yf (y) (6.6) is a Laurent polynomial in x and y of degree df+ 2, such that every monomial contains
only an even power of x and an odd power of y.
To prove this lemma, let h(x) be a Laurent polynomial in x2 of degree d
h. Then, each
term of h(x) − h(y) is of the form c(x2k− y2k) for some constant c and integer k ∈ Z.
Since x = y and x = −y are roots of h(x) − h(y), the polynomial h(x) − h(y) contains the factor (x2− y2), so h(x)−h(y)
x2−y2 is a Laurent polynomial in x2 and y2 of degree dh− 2.
Therefore, 1 x2− y2 (x2− 1)3 x2 ∂ ∂xf (x) − (y2− 1)3 y2 ∂ ∂yf (y)
is a Laurent polynomial in x2 and y2 of degree df + 1, which proves the lemma.
By the induction hypothesis and using the above lemma, we have that r(t[n]) is a Laurent
polynomial in t1, . . . , tn, such that each monomial contains only an even power of t1 and
an odd power of tj for each j ∈ {2, · · · , n}. Taking the integral with respect to t1, we
get that
Fg,nC (t[n]) = Z t1
−1
is a Laurent polynomial in t1, . . . , tn such that every monomial contains only an odd
power of tj for each j ∈ {1, . . . , n}. As for the degree of Fg,nC (t[n]), note that the term
of the highest order in r(t[n]) comes from the term Fg,n−1C . By the induction hypothesis,
Fg,n−1C has degree 3(2g − 1 + n), so using the lemma and noting the integration raises the degree by one, we have that Fg,nC (t[n]) has degree 3(2g − 2 + n). This completes the proof.
Thus, for 2g − 2 + n > 0, FC
g,n(t1, . . . , tn) is a Laurent polynomial in t1, . . . , tn, such
that each monomial contains only an odd power of each tj. Therefore, its n-fold
par-tial derivative wCg,n(t1, . . . , tn) = ∂
n
∂t1···∂tnF
C
g,n(t1, . . . , tn), and thus also Wg,nC (t1, . . . , tn) =
wCg,n(t1, . . . , tn)dt1· · · dtn is a Laurent polynomial in t21, . . . , t2n. This proves the second
part of Proposition 4.4. Moreover, both Fg,nC (t1, . . . , tn) and Wg,nC (t1, . . . , tn) are
mero-morphic on Cn, which establishes Theorem 4.2.
We remark that the generalized Catalan numbers are closely related to the counting of lattice points in the moduli space of curves. A precise definition of the lattice point counting function Ng,nis given in [13]. The definition is given in terms of ribbon graphs.
A ribbon graphs of type (g, n) has n labeled faces, rather than the n labeled vertices of a cellular graph. We define
Fg,nL (t1, . . . , tn) = X µ∈Zn + Ng,n(µ) n Y i=1 z−µi i
as the Laplace transform of the lattice point counting function, and the differential forms Wg,nL (t1, . . . , tn) = d1· · · dnFg,nL (t1, . . . , tn).
In [14, Theorem 7.2] it is shown that the Wg,nL satisfy exactly the same recursion formula (4.12) as the Wg,nC . Moreover, for the first stable cases, the differential forms agree, i.e. W1,1L(t1) = W1,1C (t1) and W0,3L (t1, t2, t3) = W0,3C (t1, t2, t3). Since the recursion formula
uniquely determines Wg,n from the initial stable cases W1,1 and W0,3, it follows that
Wg,nL (t1, . . . , tn) = Wg,nC (t1, . . . , tn) for all stable (g, n). An expression for the generalized
Catalan numbers in terms of the number of lattice points is given by [1]
Cg,n(µ1, . . . , µn) = X l1>µ12 · · · X ln>µn2 n Y i=1 (2li− µi) µi li Ng,n(2l1− µ1, . . . , 2ln− µn).
Recall that a ribbon graphs of type (g, n), has n labeled faces, rather than the n labeled vertices of a cellular graph. In [1], a cellular graph is described as the dual of a ribbon graph. We unsuccessfully attempted to provide a combinatorial proof for this duality.
7 Conclusion
We defined the generalized Catalan numbers as a geometric enumeration problem by counting cellular graphs, and therefore considered the problem as an A-model. The unstable cases (g, n) = (0, 1) and (0, 2) make sense in counting the generalized Catalan numbers. Therefore, as expected, the Laplace transform of the solutions to these cases provides a spectral curve for the Eynard-Orantin recursion on the B-model side. The Eynard-Orantin recursion thus defines meromorphic symmetric differential n-forms on Cn, for each (g, n). To show that the mirror dual of the generalized Catalan numbers satisfies the Eynard-Orantin recursion, we solved the counting problem by providing a non-recursive equation for the generalized Catalan numbers on the A-model side. Taking the Laplace transform of this equation, we obtained symmetric n-forms Wg,nC , and showed that these are meromorphic on Cnand satisfy the Eynard-Orantin recursion formula. Moreover, we defined the Catalan partition function ZC to show that it satisfies
a Schr¨odinger-type equation. For this, we needed a primitive function Fg,nC for the each n-form Wg,nC . In general, a primitive function is not unique due to the constants of integration. However, we showed that each Fg,n(t1, . . . , tn) has a natural zero in ti = −1
for each i ∈ {1, . . . , tn}, and therefore, Fg,nC is uniquely determined by Wg,nC . While the
Catalan numbers provide a relatively simple example, it has not yet been proven that the expected properties are satisfied in general. Moreover, taking the Laplace transform of the unstable cases provides the spectral curve if the unstable cases are defined on the A-model side. If the unstable cases do not make sense, then it remains a question if and how we can find the spectral curve on the B-model side.
Popular summary (in Dutch)
Spiegelsymmetrie is een equivalentie tussen wiskundige objecten. We gaan hier niet in op de precieze definitie van deze objecten, maar laten we een object aan de ene kant van de equivalentie een A-model noemen. Een object aan de andere kant van de equivalentie noemen we een B-model. Ieder A-model heeft een B-model als gespiegelde, en andersom. Spiegelsymmetrie stelt ons in staat om een type A-model probleem op te lossen door naar zijn gespiegelde B-model te kijken, en omgekeerd net zo. Als voorbeeld bekijken we de Catalan-getallen, die we kunnen zien als A-model.
De Catalan-getallen vormen een rij van positieve gehele getallen die begint met de getallen 1, 1, 2, 5, 14, 42, 132, 429, . . . . De Catalan-getallen komen voor bij verschillende
telproble-men. Hier bekijken we het aantal manieren waarop we
een ´graaf´ op het oppervlak van een bol kunnen teke-nen. Een graaf is een verzameling punten die met elkaar
verbonden zijn door lijnen. We kunnen ook een punt
met zichzelf verbinden, dan ontstaat er een lus. Zie
hiernaast de graaf die bestaat uit ´e´en punt met twee
lussen (de figuur 8), getekend op de bol. In het
ver-volg zullen we een graaf geven zonder de bol erbij te teke-nen.
Merk op dat een van de lussen in de figuur hierboven een pijl bevat. We gaan nu tellen op hoeveel ´verschillende´ manieren we een graaf met ´e´en punt op de bol kunnen tekenen, waarbij we steeds ´e´en lijn een pijl geven. Met verschillend of inequivalent bedoelen we dat de grafen niet op een bepaalde manier in elkaar getransformeerd kunnen worden. We mogen bijvoorbeeld de lijnen van een graaf tegen de klok in draaien. Hieronder staan links drie ´dezelfde´ grafen, en rechts een inequivalente graaf. We kunnen bij de tweede graaf de lijnen tegen de klok in draaien, dan krijgen we de eerste graaf. Bij de derde graaf kunnen we de grootste lus om de bol heen trekken, zodat we opnieuw de eerste graaf hebben. De vierde graaf is verschillend van de eerste drie.
De twee inequivalente grafen in de bovenstaande figuur zijn de enige grafen die we kun-nen tekekun-nen met ´e´en punt en twee lijnen. Een graaf met ´e´en punt en nul lijnen of met ´e´en lijn kunnen we allebei maar op ´e´en verschillende manier tekenen, en dus vinden we het rijtje getallen 1, 1, 2. Als we verder gaan met meer lijnen dan vinden we precies de Catalan-getallen.
Vervolgens kunnen we het telprobleem uitbreiden door het aantal punten in te graaf toe te laten nemen. We eisen echter wel dat de graaf samenhangend is. Dat wil zeggen dat er tussen elk tweetal punten in de graaf een wandeling mogelijk is. We kunnen dan in het ene punt beginnen, en via de lijnen van de graaf naar het andere punt lopen. Bovendien eisen we dat het oppervlak zonder de graaf een verzameling van schijven is. Intu¨ıtief kan dat worden voorgesteld door in het oppervlak te knippen. We knippen precies op de lijnen van de graaf, en wat overblijft moet een verzameling van schijven zijn. Naast het verhogen van het aantal punten in de graaf,
kunnen we ook de bol vervangen door een oppervlak met ´gaten´. Zie hiernaast een graaf getekend op de torus (ook wel ´donut´). De getallen die we verkri-jgen door de grafen met n punten op een oppervlak met g gaten te tellen, noemen we de gegeneraliseerde
Catalan-getallen van type (g, n). Het wordt echter
steeds moeilijker om de grafen te tellen, naarmate we n en g verhogen. Hiertoe kunnen we spiegelsymmetrie toepassen.
We kunnen namelijk de gespiegelde van de gegeneraliseerde Catalan getallen defini¨eren als een bepaald soort functies (meromorfe, symmetrische differentiaalvormen van orde n) die ´leven´ op n kopie¨en van het oppervlak van de bol. Deze functies (als B-model) bli-jken aan een ´recursievergelijking´ (de recursievergelijking van Eynard-Orantin) te vol-doen. Dat houdt in dat we alleen de gegeneraliseerde Catalan-getallen van type (0, 1) en (0, 2) hoeven bepalen, en vervolgens kunnen we voor ieder type (g, n) zodat 2g −2+n > 0 de gegeneraliseerde Catalan-getallen simpelweg berekenen, in plaats van grafen te tellen. Het vermoeden is dat we deze truc voor alle soortgelijke telproblemen kunnen toepassen, mits de eerste twee types (g, n) = (0, 1) en (0, 2) een betekenis hebben aan de kant van het A-model. Verder verwachten we dat het bijbehorende B-model voldoet aan een aantal eigenschappen, omdat dit afkomstig is van een A-model telprobleem. Het tellen van de (gegeneraliseerde) Catalan-getallen vormt een relatief eenvoudig voorbeeld dat blijkt te voldoen aan de verwachte eigenschappen.
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