Enumeration of plane hypermaps with a mixed boundary I

Jérémie Bouttier Thanks: Sorbonne Université and Université Paris Cité, CNRS, IMJ-PRG, F-75005 Paris, France Bertrand Eynard Thanks: Université Paris-Saclay, CNRS, CEA, Institut de physique théorique, 91191, Gif-sur-Yvette, France Thomas Lejeune¹¹footnotemark: 1

August 20, 2026

Abstract

Plane hypermaps are plane maps endowed with a proper coloration of their inner faces in black or white. We consider the problem of enumerating plane hypermaps with prescribed face degrees and a $k$ -alternating boundary condition: by this we mean the colors of inner faces incident to the outer face alternates at most $2k$ times when turning around the hypermap. The present paper deals with the cases $k=1,2$ , the general case being left to the forthcoming part II. Our approach relies on the so-called slice decomposition and uses crucially the notion of accessibility, which exploits the canonical orientation of hypermaps and the marking variable $t$ associated with vertices, to enumerate pointed hypermaps by decomposing them according to the set of all vertices that can access to the marked vertex. This process enables us to express the generating functions of hypermaps with mixed boundaries in terms of the generating functions of hypermap slices and to recover, in a purely combinatorial way, some formulas previously obtained through algebraic methods.

1 Introduction

1.1 Context and motivations

The study of maps, i.e. graphs drawn on surfaces, is a vibrant topic that connects combinatorics, probability theory, and theoretical physics, to name a few. See for instance [LZ04, Sch15, Eyn16, Cur23] and references therein. In this paper, we pursue the combinatorial study of hypermaps via the so-called slice decomposition, initiated in [AB26], with the goal of finding bijective proofs of the intriguing formulas given in [Eyn16, Chapter 8].

A hypermap can be seen as a properly face-bicolored map. Counting hypermaps with prescribed numbers of faces of each degree and color is an enumerative problem intimately connected with the so-called two-matrix model and the Ising model on random maps [Kaz86]. These are extremely interesting models from the point of view of theoretical physics. In particular, the Ising model on random maps is known to display a phase transition and a critical point [BK87, DHL25], at which the large-scale geometry of maps is expected to be described by the so-called Liouville quantum gravity metric at $\gamma=\sqrt{3}$ [DDG23], i.e. central charge $c=1/2$ . Confirming this expectation by rigorous mathematical results is a major open problem, and gaining a better combinatorial understanding of hypermaps will be helpful in such a task. Our focus is on hypermaps with a “mixed boundary”, as such objects arise when trying to apply the method of peeling [Cur23] to hypermaps.

Generating functions of these mixed boundary hypermaps were first studied in [Eyn03, Eyn03a] by turning Tutte’s edge removal recursions into algebraic equations, and solving these. All these generating functions are algebraic functions on a genus zero spectral curve $E(x,y)=0$ , whose rational parametrization $(x(z),y(z))$ is uniquely determined by equations (3) and (4) below. Further work was carried out in [EO05, Eyn16] in order to obtain explicit expressions using algebraic methods. It is natural to ask if these expressions can be obtained through a combinatorial approach, in order to obtain a better understanding of hypermaps. This is the question we solve in this paper and the next one [Lej26].

Acknowledgments.

We thank Marie Albenque and Emmanuel Guitter for useful discussions. This work is supported by the ERC-SyG project, Recursive and Exact New Quantum Theory (ReNewQuantum), which received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme under grant agreement No 810573, and by the Agence Nationale de la Recherche via the grant ANR-23-CE48-0018 “CartesEtPlus”.

1.2 Basic definitions and notations

We start by introducing some basic definitions, and refer to [Sch15] for more background. A plane map, hereafter called map for short, is a connected finite graph drawn in the plane¹¹ 1 The notion of plane map differs slightly from that of planar map, which is a graph drawn on the sphere. A plane map corresponds to a planar map with a distinguished face, chosen as the outer face when projecting the sphere on the plane. without edge crossings, and considered up to homeomorphism. Loops and multiple edges are allowed. A map consists of vertices and edges stemming from the graph structure, and of faces which are the connected components of the complement of the graph in the plane. Among these, the bounded ones are called inner faces, and the unbounded one is called the outer face. A corner is the angular sector between two edges appearing consecutively around a same vertex. It corresponds to an incidence between a vertex and a face. The degree of a vertex or face is its number of corners.

A bridge is an edge whose removal disconnects the map, or equivalently, since our map is plane, a bridge is an edge having the same face on its two sides. A path is a sequence of consecutive edges; it is said to be closed if it starts and ends at the same vertex. The contour of a face is the closed path formed by its incident edges. We consider below maps that are oriented, i.e. where each edge has an intrinsic orientation. A path is said to be directed if it respects the orientation of its edges.

A plane hypermap²² 2 What we call a plane hypermap here corresponds, in the terminology of [AB26], to a planar hypermap with one non-monochromatic boundary, which is selected as the outer face., hereafter simply called a hypermap, is a plane map whose edges are oriented in such a way that the contour of each inner face forms a closed directed path. We do not require the contour of the outer face to be directed. An inner face is called white if its contour is directed clockwise, and black if it is directed counter-clockwise. Note that adjacent inner faces always have opposite colours, and hence that an inner face cannot be incident to a bridge.

Given a hypermap and two corners $c,c^{\prime}$ incident to the outer face, the boundary interval $[c,c^{\prime}]$ is the portion of the contour of the outer face that lies between $c$ and $c^{\prime}$ , when turning counter-clockwise around the hypermap.

(a) A

5

-alternating hypermap contributing

\displaystyle\frac{t^{16}\,(t_{2}^{\circ})^{2}\,(t_{3}^{\circ})^{2}\,t_{2}^{\bullet}\,(t_{3}^{\bullet})^{3}\,t_{7}^{\bullet}}{x_{1}^{2}\,y_{1}^{2}\,x_{2}^{2}\,y_{2}^{2}\,x_{3}^{6}\,y_{3}^{3}\,x_{4}^{3}\,y_{4}^{2}\,x_{5}^{3}\,y_{5}^{3}}

H_{5}

(b) A

4

-alternating hypermap, accessibly pointed if

v_{1}

is marked and not if

v_{2}

is because of the presence of a bridge.

3

-alternating hypermap contributing

\displaystyle\frac{t^{10}\,t_{2}^{\circ}\,t_{3}^{\circ}\,t_{4}^{\circ}\,t_{7}^{\circ}\,t_{1}^{\bullet}\,t_{2}^{\bullet}\,(t_{3}^{\bullet})^{2}\,t_{5}^{\bullet}}{x_{1}^{2}\,y_{1}^{3}\,x_{2}^{2}\,y_{2}^{2}\,x_{3}^{1}\,y_{3}^{2}}

Figure 1: Examples of

k

-alternating hypermaps for different values of

k

. For readability, black faces are shown in blue and the boundary is shown in black. Also, marked corners with even (resp. odd) indices are shown in red (resp. blue).

For $k$ a positive integer, a hypermap $\mathfrak{m}$ is said to be $k$ -alternating if its outer face carries $2k$ marked incident corners $c_{0},c_{1},c_{2},\ldots,c_{2k-1},c_{2k}=c_{0}$ , appearing in counter-clockwise order and not necessarily distinct (as in Figure 1(c)), such that for all $i=1,\ldots,k$ , the boundary interval $[c_{2i-2},c_{2i-1}]$ is directed from $c_{2i-2}$ to $c_{2i-1}$ (i.e. counter-clockwise) and the boundary interval $[c_{2i-1},c_{2i}]$ is directed from $c_{2i}$ to $c_{2i-1}$ (i.e. clockwise). See Figure 1 for some examples. We denote the lengths (number of edges) of these boundary intervals by $\ell_{i}^{\bullet}(\mathfrak{m})$ and $\ell_{i}^{\circ}(\mathfrak{m})$ respectively.

Let $t,t_{1}^{\circ},t_{2}^{\circ},\ldots,t_{1}^{\bullet},t_{2}^{\bullet},\ldots$ be a collection of formal variables and $\mathcal{R}:=\mathbb{Q}[\![t,t_{1}^{\circ},t_{2}^{\circ},\ldots,t_{1}^{\bullet},t_{2}^{\bullet},\ldots]\!]$ the ring of formal power series in these variables. To a hypermap $\mathfrak{m}$ , we generally assign a weight

w(\mathfrak{m})=t^{\#\{\text{vertices of }\mathfrak{m}\}}\prod_{\begin{subarray}{c}f\text{ white}\\ \text{inner face}\end{subarray}}t_{\deg(f)}^{\circ}\prod_{\begin{subarray}{c}f\text{ black}\\ \text{inner face}\end{subarray}}t_{\deg(f)}^{\bullet}\in\mathcal{R},

(1)

where $\deg$ stands for the degree. Given extra formal variables $x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}$ , we then set

H_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k};t,(t_{i}^{\circ})_{i\geqslant 1},(t_{j}^{\bullet})_{j\geqslant 1}):=\delta_{k,1}+\sum_{\mathfrak{m}}\frac{w(\mathfrak{m})}{x_{1}^{\ell_{1}^{\bullet}(\mathfrak{m})+1}y_{1}^{\ell_{1}^{\circ}(\mathfrak{m})+1}\cdots x_{k}^{\ell_{k}^{\bullet}(\mathfrak{m})+1}y_{k}^{\ell_{k}^{\circ}(\mathfrak{m})+1}},

(2)

where the sum is over all $k$ -alternating hypermaps $\mathfrak{m}$ . This quantity is a formal power series in $t$ , the $t_{d}^{\circ}$ and $t_{d}^{\bullet}$ for $d\geq 1$ , and the $x_{i}^{-1}$ and $y_{i}^{-1}$ for $i\in[\![1,k]\!]$ , hence an element of $\big(\prod_{i=1}^{k}x_{i}^{-1}y_{i}^{-1}\big)\mathcal{R}[\![(x_{i}^{-1})_{1\leqslant i\leqslant k},(y_{j}^{-1})_{1\leqslant j\leqslant k}]\!]$ . Using inverse variables here, and adding the conventional term $\delta_{k,1}$ , are choices made to be consistent with [Eyn16].

In the present paper we will give formulas for $H_{1}$ and $H_{2}$ , leaving the case of general $k$ to a subsequent paper [Lej26]. However, we will obtain a general formula for a variant of $H_{k}$ involving so-called accessibly pointed hypermaps.

More precisely, we say that a vertex $v$ of a hypermap is accessible if, for every other vertex $u$ , there exists a directed path going from $u$ to $v$ . Note that we do not require the existence of a directed path from $v$ to $u$ . An accessibly pointed hypermap is a hypermap endowed with a distinguished accessible vertex. See Figure 1(b). We denote by $A_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})$ the generating function of accessibly pointed $k$ -alternating hypermaps, obtained by just changing the summation set in the right-hand side of (2) and removing the conventional term $\delta_{k,1}$ .

Finally, we define similarly the generating function $B_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})$ of strongly connected $k$ -alternating hypermaps. By strongly connected, we mean that for every pair $(u,v)$ of vertices, there exists a directed path from $u$ to $v$ ; equivalently, this means that every vertex is accessible, see Figure 1(c). As we will see in Section 2 below, a hypermap is strongly connected if and only if it contains no bridge.

We are now ready to state the main results of this paper.

1.3 Main results

Our first result is a general formula for $A_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})$ which, as in [Eyn03, AB26], is expressed in terms of some auxiliary quantities that we introduce first. Let us consider the formal Laurent power series

x(z):=\sum_{i\geq-1}a_{i}z^{-i}\quad\text{ and }\quad y(z):=\sum_{i\geq-1}b_{i}z^{i}

(3)

where the $a_{i}$ and $b_{i}$ are the elements of $\mathcal{R}$ determined recursively by the conditions

\begin{split}a_{i}=t\delta_{i,-1}+\sum_{d\geq 1}t^{\bullet}_{d}[z^{-i}]y(z)^{d-1}\qquad(i\geq-1)\\ b_{-1}=1,\qquad b_{i}=\sum_{d\geq 1}t^{\circ}_{d}[z^{i}]x(z)^{d-1}\qquad(i\geq 0).\end{split}

(4)

Here, the notation $[z^{i}]$ means that we extract the coefficient of $z^{i}$ in the Laurent power series on the right.

Theorem 1.

For any $k\geq 1$ , the generating function of accessibly pointed $k$ -alternating hypermaps reads

\begin{split}A_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})&=\sum_{\ell_{1}^{\bullet},\ell_{1}^{\circ},\ldots,\ell_{k}^{\bullet},\ell_{k}^{\circ}\geq 0}\frac{[z^{0}]x(z)^{\ell_{1}^{\bullet}+\cdots+\ell_{k}^{\bullet}}y(z)^{\ell_{1}^{\circ}+\cdots+\ell_{k}^{\circ}}}{x_{1}^{\ell_{1}^{\bullet}+1}y_{1}^{\ell_{1}^{\circ}+1}\cdots x_{k}^{\ell_{k}^{\bullet}+1}y_{k}^{\ell_{k}^{\circ}+1}}\\ &=[z^{0}]\frac{1}{\prod_{i=1}^{k}(x_{i}-x(z))(y_{i}-y(z))}.\end{split}

(5)

This theorem will be proved using slice decomposition in Section 3.2. The symmetry of $A_{k}$ under permutation of $x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}$ is remarkable and unexpected. Indeed, from the combinatorial definition of $H_{k}$ , $A_{k}$ and $B_{k}$ , we only expect symmetries such as invariance under rotation $(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})\mapsto(x_{2},\ldots,x_{k},x_{1};y_{2},\ldots,y_{k},y_{1})$ or invariance under reflection $(x_{1},\ldots,x_{k};\allowbreak y_{1},\ldots,y_{k})\mapsto(x_{k},x_{k-1},\ldots,x_{2},x_{1};y_{k-1},y_{k-2},\ldots,y_{1},\allowbreak y_{k})$ . This latter invariance corresponds to the weight-preserving involution on the set of plane hypermaps which consists in performing a reflection of the plane and reversing the orientation of the edges. Note that this involution does not preserve the orientation of edges, hence $A_{k}$ has a priori no reason to be invariant under such a transformation.

Now, we would like to deduce expressions for $H_{k}$ and $B_{k}$ from that of $A_{k}$ above. As said previously, we only consider the cases $k=1,2$ in the present paper. Let us first consider the case $k=1$ :

Proposition 2.

We have the relations

A_{1}(x;y)=\frac{\partial}{\partial t}\ln H_{1}(x;y)=\frac{\frac{\partial}{\partial t}B_{1}(x;y)}{1-B_{1}(x;y)}.

(6)

As a consequence, we have

H_{1}(x;y)=\frac{1}{1-B_{1}(x;y)}=\exp\left([z^{0}]z\ln\left(1-\frac{y(z)}{y}\right)\frac{\partial}{\partial z}\ln\left(1-\frac{x(z)}{x}\right)\right).

(7)

The expression (7) was previously given in [AB26]³³ 3 In this reference, $1$ -alternating hypermaps are called hypermaps with a Dobrushin boundary., where it is also explained how it matches with the results stated in [Eyn03, Eyn16]. Here, we give a new proof of this expression, which consists in integrating (6) and using Theorem 1 for $k=1$ .

Moreover, with the aim of giving various formulations for the central case $k=1$ , we will show the following identity thanks to Proposition 2,:

Proposition 3.

Assume white and black faces have bounded degrees, which means that we set $t_{i}^{\circ}$ and $t_{j}^{\bullet}$ to $0$ for $i$ and $j$ large enough so that $x(z)$ and $y(z)$ , defined in Equation (3), are now Laurent polynomials.
Let $E(x,y)$ be the resultant with respect to $z$ of $(x(z)-x,y(z)-y)$ . Also let $Y(x)$ , resp. $X(y)$ , be the unique Laurent power series in $x^{-1}$ such that $E(x,Y(x))=0$ (resp. the unique Laurent power series in $y^{-1}$ such that $E(X(y),y)=0$ ). Then we have the formula

A_{1}(x;y)=\frac{\partial}{\partial t}\ln\left(\frac{E(x,y)}{(x-X(y))(y-Y(x))}\right).

(8)

We will detail why $x(z)-x$ and $y(z)-y$ are Laurent polynomials, and no longer Laurent power series, in the case of bounded face degrees while proving this proposition in Section 4.2.4.

Finally, for the case $k=2$ , we will establish the following:

Theorem 4.

The generating function of $2$ -alternating hypermaps reads

H_{2}(x_{1},x_{2};y_{1},y_{2})=\frac{H_{1}(x_{1};y_{1})H_{1}(x_{2};y_{2})-H_{1}(x_{1};y_{2})H_{1}(x_{2};y_{1})}{(x_{1}-x_{2})(y_{1}-y_{2})}

(9)

and that of strongly connected $2$ -alternating hypermaps reads

B_{2}(x_{1},x_{2};y_{1},y_{2})=\frac{H_{1}(x_{1};y_{2})^{-1}H_{1}(x_{2};y_{1})^{-1}-H_{1}(x_{1};y_{1})^{-1}H_{1}(x_{2};y_{2})^{-1}}{(x_{1}-x_{2})(y_{1}-y_{2})}.

(10)

The first expression corresponds to [Eyn03, Eyn16, Corollary 8.4.1], and we here give a combinatorial proof, as asked in this reference.

1.4 Outline

In Section 3, we study the case of accessibly pointed hypermaps, where every vertex can be connected to a distinguished vertex, after redefining the essential tool for their enumeration: the hypermap slices. The realization of the link between accessible hypermaps and slice decomposition is expressed in Theorem 1.

We will then have the main tools required to prove relation (6) concerning $1$ -alternating hypermaps in Section 4.1. We will thus use these formulas to recover more classical results on this type of hypermaps, such as Equation (7), in Section 4.2, and Proposition 3 in Section 4.2.4. Section 4 as a whole will then establish Propositions 2 and 3.

Finally, in Section 5, we prove Theorem 4, which illustrates our general strategy: we will mark a vertex and then integrate in order to recover a formula in which $t$ does not appear explicitly.

The case of $k$ -alternating boundaries and the associated general formulas for $H_{k}$ and $B_{k}$ will appear in the second part of the article [Lej26].

2 Bridges and accessibility in plane hypermaps

Let us record here a few elementary observations related to the notion of accessibility and strong connectivity in plane hypermaps. We start with a simple but key lemma.

Lemma 5.

Given a plane hypermap, let $u$ and $v$ be two vertices such that there exists a non-directed path from $u$ to $v$ that does not contain a bridge. Then, there exists a directed path from $u$ to $v$ (and vice versa).

Proof.

By the assumption, there exists a not necessarily directed path $P$ from $u$ to $v$ which does not pass through a bridge. We may then construct a directed path $P^{\prime}$ from $u$ to $v$ as follows. Consider every edge $e$ that appears in the wrong direction along $P$ . Since $e$ is not a bridge, it is by planarity incident to at least one inner face $f$ . The contour of $f$ , being a directed cycle, provides a way to replace $e$ by a directed path (of possibly more than one edge) going in the right direction. ∎

As a consequence of our key lemma we get a full characterization of strong connectedness for plane hypermaps.

Lemma 6.

A plane hypermap is strongly connected if and only if it contains no bridge.

For instance, in Figure 1, only the third hypermap is strongly connected because of the existence of respectively three and one bridge in the first and second hypermaps.

Proof of Lemma 6.

Let $\mathfrak{m}$ be a plane hypermap. If $\mathfrak{m}$ has no bridge then it is strongly connected by Lemma 5. Conversely, if $\mathfrak{m}$ contains a bridge $b$ , then there exists no directed path going from the endpoint of $b$ to its origin. ∎

Remark 7.

Note that the proof of Lemma 5, and hence Lemma 6, rely crucially on planarity and on the fact that there exists only one face whose contour is not a directed cycle. Figure 2 displays two examples of oriented maps which are not strongly connected despite having no bridge.

Figure 2: Two maps with oriented edges which are not strongly connected despite having no bridge. The left one is planar and has two faces, none of which forms a directed cycle. The right one has genus one (we identify opposite sides of the dashed square) and one face.

Remark 8.

Another way to understand this lemma is to define an equivalence relation between the vertices of a hypermap. For $u,v$ vertices of a fixed hypermap, one sets $u\leftrightarrow v$ if, and only if, there exists a directed path from $u$ to $v$ and a directed path from $v$ to $u$ .

Then Lemma 5 and Lemma 6 imply that the equivalence class of a vertex with respect to this equivalence relation is a strongly connected hypermap. Hence any alternating hypermap can be divided into strongly connected maps. This will be useful in Sections 4.1.2 and 5.2.

Another consequence of our key lemma is the following:

Lemma 9.

For a vertex $v$ of a plane hypermap to be accessible, it is sufficient to ensure that it can be attained from any vertex incident to the outer face.

Proof.

This follows from the fact that, for any vertex $u$ of a plane hypermap, we may construct a directed path going from $u$ to a vertex incident to the outer face. To see this, recall that bridges may only be incident to the outer face, and apply Lemma 5. ∎

Remark 10.

(a) An inward corner

(b) An outward corner

Figure 3: Sketches of inward and outward corners. Inner faces are shown in light blue.

It is even sufficient to ensure accessibility from the inward corners, that is the outer corners having two incoming outer edges, see Figure 3. In an accessibly pointed $k$ -alternating hypermap, the inward corners are marked corners with odd indices, while those having two outgoing outer edges are called the outward corners and have even indices. Note that the even or odd marked corners are not all necessarily inward corners or outward corners: this situation occurs when some marked corners coincide, as illustrated on Figure 1(c) where $c_{4}=c_{5}$ is neither inward nor outward.

3 Enumeration of accessibly pointed $k$ -alternating hypermaps

The main goal of this section is to prove Theorem 1. To this end, we make use of the slice decomposition, introduced in the context of hypermaps in [AB26].

3.1 Slices

Reminders.

Let us start by recalling the definition of slices of types $\mathcal{A}$ and $\mathcal{B}$ given in [AB26]:

Definition 11.

A slice of type $\mathcal{A}$ or of type $\mathcal{B}$ is a plane hypermap with three distinguished corners, denoted $o$ , $l$ , and $r$ , appearing in counterclockwise order around the outer face, and satisfying the following conditions:

•

the boundary interval $[o,l]$ , called the left side, forms a directed path from $l$ to $o$ of minimal length, i.e. is a geodesic from $l$ to $o$ ,
•

the boundary interval $[r,o]$ , called the right side, forms the unique directed path from $r$ to $o$ of minimal length, i.e. is the unique geodesic from $r$ to $o$ ,
•

the left and right sides only meet at the vertex incident to $o$ , called the apex,
•

the boundary interval $[l,r]$ , called the base, forms a directed path from $l$ to $r$ for a slice of type $\mathcal{A}$ , and a directed path from $r$ to $l$ for a slice of type $\mathcal{B}$ .

When the base has length one, the slice is said to be elementary.

For any integer $i$ , let us denote by $a_{i}$ (resp. $b_{i}$ ) the generating function of elementary slices of type $\mathcal{A}$ (resp. $\mathcal{B}$ ), in which the difference between the length of the right side and the length of the left side is equal to $i$ (resp. $-i$ ). Here, the weight of a slice is defined by a slight variant of the formula (1), namely we do not attach a weight $t$ to the vertices belonging to the right side. It is straightforward to check that $a_{i}=b_{i}=0$ for $i<-1$ as there exists no slice contributing to these generating functions, and it was shown in [AB26] that the sequences $(a_{i})_{i\geq-1}$ and $(b_{i})_{i\geq-1}$ are determined recursively by (3) and (4).

Remark 12.

These recursive equations also appear in [Eyn16, Chapter 8]. Hypermaps are obtained by setting the variables denoted $a,b,c$ in this reference to $a=b=0$ , $c=-1$ . Our quantities $a_{k}$ and $b_{k}$ are related to $\gamma,\alpha_{k},\beta_{k}$ of this reference by

a_{-1}=\gamma^{2},\qquad a_{k}=\frac{\alpha_{k}}{\gamma^{k}},\qquad b_{k}=\beta_{k}\gamma^{k},

(11)

which essentially corresponds to a change of variable $z\mapsto\gamma z$ in $x(z)$ and $y(z)$ .

Generalized slices.

For our purposes it is convenient to consider a slightly generalized notion of slices:

Definition 13.

A slice is a plane hypermap with three distinguished corners, denoted $o$ , $l$ , and $r$ , appearing in counterclockwise order on its outer face, and satisfying the first three items in Definition 11, the fourth item being replaced by the mere requirement that the apex be accessible.

Remark 14.

Lemma 9 ensures that a slice of type $\mathcal{A}$ or $\mathcal{B}$ is still a slice according to the above generalized definition, since the contour of the outer face consists of two directed paths ending at the apex. In a generalized slice, the accessibility requirement forbids the base $[l,r]$ from containing a bridge that points away from the apex.

(a) A slice of type

\mathcal{A}

with
base word

a^{4}

(b) An elementary slice of type

\mathcal{B}

with base word

b

ab^{3}abab

Figure 4: Examples of slices of different types and their associated word in

\{a,b\}

. Here the left side is represented in green while the right side is red and the base is black, as in [AB26].

To each (generalized) slice, we associate its base word, which is a word in the alphabet $\{a,b\}$ , as follows. We consider the orientation of the arrows along the base, read from $l$ to $r$ : to each arrow oriented from $l$ to $r$ , we associate the letter $a$ , and to each arrow oriented from $r$ to $l$ , the letter $b$ . See Figure 4 for examples. Note that a slice has type $\mathcal{A}$ (resp. $\mathcal{B}$ ) if and only if its base word contains only $a$ ’s (resp. $b$ ’s). In an elementary slice the base word has a single letter.

Interestingly, the generating function of slices with a prescribed base word can be straightforwardly expressed in terms of the quantities $x(z)$ and $y(z)$ counting elementary slices.

Proposition 15.

Let $w$ be a word in the alphabet $\{a,b\}$ , and let $i$ be an integer. Then, the generating function of slices with base word $w$ , where the difference between the length of the left side and that of the right side is equal to $i$ , is given by

[z^{i}]x(z)^{n_{a}(w)}y(z)^{n_{b}(w)}

(12)

where $n_{a}(w)$ (resp. $n_{b}(w)$ ) denotes the number of occurrences of the letter $a$ (resp. $b$ ) in $w$ .

Proof.

The proof is a direct generalization of that of [AB26, Proposition 2.8 and Corollary 2.9]: a slice can be bijectively decomposed into a sequence of elementary slices by cutting along the leftmost geodesics going from each outer corner along the base to the apex. Note that such leftmost geodesics always exist by the requirement that the apex be accessible. The new feature here is that we may mix elementary slices of types $\mathcal{A}$ and $\mathcal{B}$ in the sequence. ∎

3.2 Bijection between slices and accessibly pointed hypermaps

Recall from Section 1.2 the definition of accessibly pointed $k$ -alternating hypermaps.

Proposition 16.

For any positive integer $k$ and any nonnegative integers $\ell_{1}^{\bullet},\ell_{1}^{\circ},\ldots,\ell_{k}^{\bullet},\ell_{k}^{\circ}$ , there is a weight-preserving bijection between the set of accessibly pointed $k$ -alternating hypermaps $\mathfrak{m}$ such that $\ell_{i}^{\bullet}(\mathfrak{m})=\ell_{i}^{\bullet}$ and $\ell_{i}^{\circ}(\mathfrak{m})=\ell_{i}^{\circ}$ for all $i=1,\ldots,k$ , and the set of slices with base word $a^{\ell_{1}^{\bullet}}b^{\ell_{1}^{\circ}}\cdots a^{\ell_{k}^{\bullet}}b^{\ell_{k}^{\circ}}$ where the left and right sides have the same length.

Proof.

Figure 5: Sketch of the proof of Proposition 16 in the case of a pointed

2

-alternating hypermap with

(\ell_{1}^{\circ},\ell_{1}^{\bullet},\ell_{2}^{\circ},\ell_{2}^{\bullet})=(1,1,2,2)

. The first step consists in taking the leftmost geodesic from the marked corner

c_{0}

(left figure) to the marked vertex

v

. Then one cuts along this geodesic,

c_{0}

is now decomposed into to vertices:

r

and

l

(second figure), where

r

and

l

is connected to

v

by resp. a unique geodesic and a generic geodesic. Then a slice of base word

aba^{2}b^{2}

rises by deforming continuously the second figure, which gives the third figure. Note that the marked vertex is now the apex.

The proof is illustrated on Figure 5 and is similar to that of [AB26, Proposition 3.4]: starting from an accessibly pointed $k$ -alternating hypermap with distinguished vertex $v$ , we cut along the leftmost geodesic going from the first marked corner $c_{0}$ to $v$ . Such a geodesic always exists since $v$ is assumed accessible. We may check that the resulting map is a slice, $v$ becoming the apex which remains accessible. Furthermore, the length constraints on the boundary intervals of the initial $k$ -alternating hypermap entail that the base word of the resulting slice is $a^{\ell_{1}^{\bullet}}b^{\ell_{1}^{\circ}}\cdots a^{\ell_{k}^{\bullet}}b^{\ell_{k}^{\circ}}$ . Finally, this construction is bijective as the inverse bijection consists in gluing the left and right sides of the slice. ∎

By combining Propositions 15 and 16 we obtain the following:

Proposition 17.

The generating function of accessibly pointed $k$ -alternating hypermaps $\mathfrak{m}$ such that $\ell_{i}^{\bullet}(\mathfrak{m})=\ell_{i}^{\bullet}$ and $\ell_{i}^{\circ}(\mathfrak{m})=\ell_{i}^{\circ}$ for all $i=1,\ldots,k$ is equal to

[z^{0}]x(z)^{\ell_{1}^{\bullet}+\cdots+\ell_{k}^{\bullet}}y(z)^{\ell_{1}^{\circ}+\cdots+\ell_{k}^{\circ}}.

(13)

Theorem 1 follows by summing over $\ell_{1}^{\bullet},\ell_{1}^{\circ},\ldots,\ell_{k}^{\bullet},\ell_{k}^{\circ}$ . Let us conclude this section by recording the following useful expression for $A_{k}$ in terms of $A_{1}$ :

Corollary 18.

For any $k\geq 1$ we have

A_{k}(x_{1},\ldots,x_{k};y_{1},\ldots,y_{k})=\sum_{1\leqslant i,j\leqslant k}\frac{A_{1}(x_{i};y_{j})}{\displaystyle\prod_{p\neq i}(x_{i}-x_{p})\prod_{q\neq j}(y_{j}-y_{q})}.

(14)

In particular for $k=2$ we have

A_{2}(x_{1},x_{2};y_{1},y_{2})=\frac{A_{1}(x_{1};y_{1})+A_{1}(x_{2};y_{2})-A_{1}(x_{1};y_{2})-A_{1}(x_{2};y_{1})}{(x_{1}-x_{2})(y_{1}-y_{2})}.

(15)

Proof.

Equation (14) follows directly from equation (5) by performing two partial fraction decompositions with respect to the variables $x(z)$ and $y(z)$ , and then applying the case $k=1$ to recover $A_{1}(x_{i};y_{j})$ . ∎

Note that the first formula of Corollary 18, made explicit in the case of accessibly pointed $2$ -alternating hypermaps, plays a key role when studying $k$ -alternating hypermaps, while Theorem 1 itself contains the full combinatorial interpretation of alternating hypermaps as slices. Both formulas are therefore essential, each in its own way.

4 Enumeration of $1$ -alternating hypermaps

In this section, we shall successively detail all the equalities appearing in Proposition 2. We begin with the combinatorial approach based on slices, in order to prove the first equation in (6). We then show how to derive an explicit expression for $H_{1}$ , as given in Equation (7), before returning to the case of bounded-face degrees and establishing the connection with the spectral curve.

4.1 How to use accessibility in pointed $1$ -alternating hypermaps

The purpose of this section is to establish Equation (6) in Proposition 2.

4.1.1 Generic case

Let us start by proving the first equality in (6) which amounts to

\frac{\partial H_{1}(x;y)}{\partial t}=A_{1}(x;y)H_{1}(x;y),

(16)

where we recall that $H_{1}$ is the generating function of $1$ -alternating hypermaps and $A_{1}$ that of accessibly pointed $1$ -alternating hypermaps.

We start by the observation that $\frac{\partial H_{1}}{\partial t}$ counts pointed, but not necessarily accessibly pointed, $1$ -alternating hypermaps. We shall therefore discuss what could prevent, in a pointed $1$ -alternating hypermap $\mathfrak{m}$ , the marked vertex $v$ from being accessible. By Remark 10, $v$ is accessible if and only if it is reachable from $c_{1}$ . Now, by Lemma 9, we see that the obstruction to accessibility is the existence of a bridge separating $v$ and the vertex incident to $c_{1}$ . See Figure 6 for an illustration of this situation and Figure 7 for a sketch of the proof.

Figure 6: Sketch of a

1

-alternating hypermap. One can observe that the marked vertex

v

is accessible from

c_{0}

but not from

c_{1}

because of the presence of bridges, dashed in green for readability, oriented toward

c_{1}

This suggests the following decomposition of $\mathfrak{m}$ when the obstruction occurs. Among all the bridges which separate $v$ from $c_{1}$ , let us denote by $b$ the one closest to $v$ . Removing $b$ splits $\mathfrak{m}$ into two components: let us denote by $\mathfrak{m}_{1}$ the component containing $v$ and $\mathfrak{m}_{2}$ the one containing $c_{1}$ . Note that $\mathfrak{m}_{1}$ and $\mathfrak{m}_{2}$ are both $1$ -alternating hypermaps. Furthermore, the inward corner $c_{1}^{\prime}$ of $\mathfrak{m}_{1}$ (at the position of the former bridge $b$ ) is by construction not separated from $v$ by a bridge. Hence, by Lemma 9 and Remark 10, $v$ is accessible in $\mathfrak{m}_{1}$ , which makes $\mathfrak{m}_{1}$ accessibly pointed.

We extend the decomposition of $\mathfrak{m}$ to the case where the obstruction does not occur by setting $\mathfrak{m}_{1}:=\mathfrak{m}$ and $\mathfrak{m}_{2}:=\emptyset$ . This construction defines a map $\mathfrak{m}\longmapsto(\mathfrak{m}_{1},\mathfrak{m}_{2})$ from the set of pointed $1$ -alternating hypermaps to the set of pairs consisting of an accessibly pointed $1$ -alternating hypermap and of a (possibly empty) unpointed $1$ -alternating hypermap. It is clear that the mapping is bijective: when $\mathfrak{m}_{2}$ is not empty, the inverse mapping consists in connecting the inward corner of $\mathfrak{m}_{1}$ to the outward corner of $\mathfrak{m}_{2}$ by a bridge directed from the former to the latter.
Translating this bijection into the language of generating functions, we get precisely Relation (16), upon noting that the case $\mathfrak{m}_{2}=\emptyset$ corresponds to the conventional term $\delta_{k,1}$ present in the definition (2) of $H_{k}$ (but absent in $A_{1}$ ), and that if $\mathfrak{m}_{2}\neq\emptyset$ then we have

\ell_{1}^{\bullet}(\mathfrak{m})=\ell_{1}^{\bullet}(\mathfrak{m}_{1})+\ell_{1}^{\bullet}(\mathfrak{m}_{2})+1,\qquad\ell_{1}^{\circ}(\mathfrak{m})=\ell_{1}^{\circ}(\mathfrak{m}_{1})+\ell_{1}^{\circ}(\mathfrak{m}_{2})+1

(17)

which ensures that the exponents of $x$ and $y$ match.

Figure 7: Sketch of the proof for the case

k=1

, inner faces are blurred in light blue here. On the left, a generic sketch of a pointed

1

-alternating hypermap. On the right, the last bridge

b

between

c_{1}

and

v

is dashed in green, and so

v

is accessible from the base vertex of

b

, which leads to the desired decomposition.

Remark 19.

Since $A_{1}$ has an explicit formula as a function of $x(z)$ and $y(z)$ derived from Theorem 1:

A_{1}(x;y)=[z^{0}]\frac{1}{(x-x(z))(y-y(z))},

we can finally use Equation (16) to write the following relation for $H_{1}$ :

H_{1}(x;y)=\exp\left([z^{0}]\int_{0}^{t}\frac{1}{x-x(z,\tau)}\frac{1}{y-y(z,\tau)}\,\mathrm{d}\tau\right).

(18)

The removal of the integral will be the main purpose of Section 4.2.

4.1.2 Strongly connected case

Let us now establish the second equality of Equation (6) in Proposition 2:

A_{1}(x;y)=\frac{\frac{\partial}{\partial t}B_{1}(x,y)}{1-B_{1}(x,y)},

(19)

where $B_{1}$ is the generating function of strongly connected $1$ -alternating hypermaps, which means maps with no bridge, in which every vertex is accessible by Lemma 6.

In fact, this relation is quite straightforward using Remark 8, and is illustrated in Figure 8. Indeed, consider an accessibly pointed $1$ -alternating hypermap $\mathfrak{m}$ , and denote by $v$ its marked vertex. Then the set of all vertices accessible from $v$ , i.e. the equivalence class of $v$ with respect to the relation $\leftrightarrow$ , is a strongly connected $1$ -alternating hypermap. We denote it by $\mathfrak{m}_{1}$ . Note that $c_{1}$ belongs to $\mathfrak{m}_{1}$ , otherwise there would exist a bridge in the wrong direction between $v$ and $c_{1}$ . Since $v$ is accessible, this bridge should be oriented toward $v$ , which leads to the existence of another inward corner that differs from $c_{1}$ , and this is not possible since the hypermap is $1$ -alternating.
Let us now consider $c_{0}$ . If $c_{0}$ is accessible from $v$ , then it belongs to $\mathfrak{m}_{1}$ , hence $v$ is accessible from any vertex and can access $c_{0}$ and $c_{1}$ , so it can be related to any other vertex of the hypermap by Lemma 9 and Remark 10, so $\mathfrak{m}=\mathfrak{m}_{1}$ . We then set $\mathfrak{m}_{2}:=\emptyset$ in this situation. However, if $c_{0}$ is not accessible from $v$ , then there should exist multiple bridges, oriented from $c_{0}$ toward $v$ , that obstruct $v$ from reaching $c_{0}$ . By Lemma 6, the component of $\mathfrak{m}$ that is between two consecutive bridges has to be a strongly connected $1$ -alternating hypermap. We denote by $\mathfrak{m}_{2}$ the whole component of $\mathfrak{m}$ which contains all the vertices that are not accessible from $v$ , which is a sequence of strongly connected hypermaps.
Then the map $\mathfrak{m}\longmapsto(\mathfrak{m}_{1},\mathfrak{m}_{2})$ from the set of accessibly pointed $1$ -alternating hypermaps to couples of pointed strongly connected $1$ -alternating hypermaps and a sequence of ordered unpointed strongly connected $1$ -alternating hypermaps is bijective. The inverse map is defined as in the last section, namely by putting a bridge relating the outward corner of one strongly connected hypermap to the inward corner of the following strongly connected hypermap (assuming $\mathfrak{m}_{1}$ is at the end of this sequence). Finally, using our convention on the exponents of $x$ and $y$ in alternating hypermaps, the language of generating functions translates this bijection into Relation (19).

Figure 8: Sketch of the proof of Equation (19)

Remark 20.

The present proof deals with the $1$ -alternating case, which only allows bridges oriented in a single direction: from the outward corner $c_{0}$ to the inward corner $c_{1}$ . We shall see in Section 5.2 that discussing more general cases in this regard will be more challenging.

Equations (16) and (19), when combined and integrated, yield the first equality of Equation (7):

H_{1}(x;y)=\frac{1}{1-B_{1}(x;y)}.

(20)

A direct proof of this formula can be found in [AB26], it is very similar to the above argument except that it does not involve a marked vertex.

4.2 Various way to integrate relations for $1$ -alternating hypermaps

In the previous section, we obtained various expressions for $H_{1}$ using the notion of accessibility. In particular, we made explicit Relation (18), which will now allow us, in this section, to derive new relations for $H_{1}$ with the aim of eliminating the integral with respect to the variable counting the number of vertices.

To this end, we shall introduce a simpler case of hypermaps than those previously considered, namely hypermaps with a monochromatic boundary. We start by studying this type of hypermaps, showing how they are strongly related to the slices defined earlier. This will enable us to prove a Poisson formula for $x(z,t)$ and $y(z,t)$ in Lemma 28, before providing proofs for Equations (7) and (8), which remain to be established.

4.2.1 Review of the monochromatic boundary case

In order to present a new relation between generating function of slices and of $H_{1}$ without any integral on $t$ , we need to introduce a few notations by recalling what has already been done for the monochromatic boundary case.

Definition 21.

Let $\mathfrak{M}^{\circ}$ (resp. $\mathfrak{M}^{\bullet}$ ) be the set of $1$ -alternating hypermaps $\mathfrak{m}$ such that $c_{0}$ and $c_{1}$ coincide and every edge on the boundary is oriented counterclockwise (resp. clockwise). This is equivalent to saying that $\ell^{\circ}(\mathfrak{m})=0$ (resp. $\ell^{\bullet}(\mathfrak{m})=0$ ). Each hypermap belonging to one of these two sets is said to have a monochromatic boundary, respectively white and black, see Figure 9 for an illustration.
One defines the corresponding generating functions as

W^{\circ}(x):=\sum_{\mathfrak{m}\in\mathfrak{M}^{\circ}}\frac{w(\mathfrak{m})}{x^{\ell^{\bullet}(\mathfrak{m})+1}}\in x^{-1}\mathcal{R}[\![x^{-1}]\!]\,\,\text{and}\,\,W^{\bullet}(y):=\sum_{\mathfrak{m}\in\mathfrak{M}^{\bullet}}\frac{w(\mathfrak{m})}{y^{\ell^{\circ}(\mathfrak{m})+1}}\in y^{-1}\mathcal{R}[\![y^{-1}]\!]

(21)

which are respectively exactly the coefficients $[y^{-1}]H_{1}(x;y)$ and $[x^{-1}]H_{1}(x;y)$ by definition. The weights $w$ are defined in Equation (1).

Remark 22.

This definition is equivalent to taking a planar hypermap with one white (resp. black) marked face as boundary in order to define a hypermap with a white (resp. black) boundary.

(a) a hypermap with white monochromatic boundary contributing to

W^{\circ}(x)

(b) a hypermap with black monochromatic boundary contributing to

W^{\bullet}(y)

Figure 9: Examples of hypermaps with monochromatic boundaries. Note that both marked corners are at the same position and that all faces incident to the boundary have the same color, hence the denomination "monochromatic".

Our first equation on these generating functions can shortly be found by a slice decomposition.

Proposition 23 ([AB26, Corollary 3.5]).

The generating functions of pointed hypermaps with monochromatic boundaries can be expressed as follows:

\frac{\partial W^{\circ}}{\partial t}(x,t)=[z^{0}]\frac{1}{x-x(z,t)}\,\,\,\text{and}\,\,\,\frac{\partial W^{\bullet}}{\partial t}(y,t)=[z^{0}]\frac{1}{y-y(z,t)}.

(22)

The proof of this proposition is based on slice decomposition and one can sketch it as follows: since we are in the plane case and the boundary is monochromatic, no bridge can appear, and so each pointed hypermap with a monochromatic boundary is accessibly pointed. Hence they can be decomposed as a sequence of slices of type $\mathcal{A}$ or $\mathcal{B}$ and the result follows from Proposition 17.
In order to integrate Equation (22) with respect to $t$ , one could think of performing a partial fraction decomposition of the denominator of the right-hand side. To do so, we surely need to find the zeros of the denominator, which motivates the following result.

Proposition 24 (see also [AB26, Section 5]).

The equations $x(z)=x$ (resp. $y(z)=y$ ) admits exactly one solution $z^{\circ}(x)$ (resp. $z^{\bullet}(y)$ ) such that $z^{\circ}(x)^{-1}$ belongs to $x^{-1}\mathcal{R}[\![x^{-1}]\!]$ (resp. $z^{\bullet}(y)$ belongs to $y^{-1}\mathcal{R}[\![y^{-1}]\!]$ ). In fact, $z^{\circ}(x)$ (resp. $z^{\bullet}(y)$ ) also verifies the equation $z^{\circ}(x(z))=z$ (resp. $z^{\bullet}(y(z))=z$ ).
Furthermore, in the case of bounded face degrees, the Laurent power series $Y(x)$ and $X(y)$ defined in Proposition 3 are given by

Y(x)=y(z^{\circ}(x))\,\,\,\text{and}\,\,\,X(y)=x(z^{\bullet}(y)).

(23)

Proof.

$\hookrightarrow$ We will only discuss the case of $y(z)=y$ ; the same discussion can be done for $x(z)=x$ by solving $x(z^{-1})=x$ . Recall that $y(z)\in z^{-1}\mathcal{R}[\![z]\!]$ so one can write the equation as $z=\frac{\phi(z)}{y}$ where $\phi(z):=zy(z)\in\mathcal{R}[\![z]\!]$ , such that $[z^{0}]\phi(z)=b_{-1}=1\neq 0$ . Hence the Lagrange’s inversion theorem gives the existence and uniqueness of a compositional inverse $z^{\bullet}(y)\in y^{-1}\mathcal{R}[\![y^{-1}]\!]$ of $y(z)$ . Let us emphasize that we do not need to extend $\mathcal{R}$ to a field in order to enable $z^{\bullet}(y)$ to have coefficients in $\mathcal{R}$ .
$\hookrightarrow$ The equations (23) come from the unicity of $Y(x)$ and $X(y)$ as Laurent power series in $x^{-1}$ and $y^{-1}$ respectively that satisfy $E(x,Y(x))=0$ and $E(X(y),y)=0$ . Using the characterization of $E$ as a resultant, meaning such that $E(x(z),y(z))=0$ , one only needs to take $z=z^{\circ}(x)$ (resp. $z=z^{\bullet}(y)$ ) and check that the corresponding functions $y(z^{\circ}(x))$ and $x(z^{\bullet}(y))$ are Laurent power series in $x^{-1}$ and $y^{-1}$ respectively to conclude. ∎

We are now ready to find an algebraic expression for $W^{\circ}(x)$ and $W^{\bullet}(y)$ .

Proposition 25 ([Eyn03, Eyn03a, AB26]).

The notations of the previous definitions lead to the following expressions for the generating functions of monochromatic boundary hypermaps:

W^{\circ}(x)=Y(x)-V_{\circ}^{\prime}(x)\,\,\,\text{and}\,\,\,W^{\bullet}(y)=X(y)-V_{\bullet}^{\prime}(y),

(24)

where $V_{\circ}$ and $V_{\bullet}$ are defined as

V_{\circ}(x)=\sum_{i\geqslant 1}t_{i}^{\circ}\frac{x^{i}}{i}\,\,\,\text{and}\,\,\,V_{\bullet}(y)=\sum_{i\geqslant 1}t_{i}^{\bullet}\frac{y^{i}}{i}.

(25)

We refer the reader to [Eyn16] for a computational proof of this formula and to [AB26] for a combinatorial proof. Here we will give a third proof based on the formula stated in Proposition 23 and the notations of Section 4.2.4 in Appendix A.

Remark 26.

Historically, the pair $(x(z),y(z))$ was introduced in [Eyn03] as a parametrization of the spectral curve $E(x,y)=0$ , where $E(x,y)\in\mathcal{R}[x,y]$ naturally appears when solving loop equations. In fact, one of the most surprising results of [AB26] was the combinatorial interpretation of this curve by proving that $x(z)$ and $y(z)$ are also the generating functions of elementary slices. So we already have a lot more information on slices through the algebraic geometry perspective, which is another argument to state slices as a good cornerstone of the hypermap theory.

In the spirit of the last proposition, another relation enables us to relate the generating functions of hypermaps with monochromatic boundaries and the specific roots $z^{\circ}$ and $z^{\bullet}$ .

Proposition 27 ([AB26]).

The following derivative relations hold:

\begin{split}\frac{\partial W^{\circ}}{\partial t}(x,t)&=\frac{\partial\ln(z^{\circ})}{\partial x}(x,t),\\ \frac{\partial W^{\bullet}}{\partial t}(y,t)&=-\frac{\partial\ln(z^{\bullet})}{\partial x}(y,t).\end{split}

(26)

These expressions can be deduced from Proposition 23 by a complex analytic argument: we extract the coefficient of $z^{0}$ by a contour integral and apply the residue theorem. Interestingly, the combinatorial approach taken in [AB26], and based on [BF02], is to interpret the generating function $\frac{\partial W^{\circ}(x,t)}{\partial t}$ (resp. $\frac{\partial W^{\bullet}(y,t)}{\partial t}$ ) as the generating function counting walks starting and ending at zero, such that each step of height $i$ is associated with a weight $a_{i}=[z^{-i}]x(z^{-1})$ (resp. $b_{i}=[z^{i}]y(z)$ ). In this frame, the quantity $z^{\circ}(x)$ (resp. $z^{\bullet}(y)$ ) is the generating functions of excursions, or positive walks starting and ending at $0$ .

4.2.2 Poisson formula for the parametrization $(x(z),y(z))$

Let us now examine an important formula, which first appeared in [Eyn03a, Eyn03] in the case of uncoloured planar maps, and which we shall here generalize.

Lemma 28 (Poisson formula for $x(z)$ and $y(z)$ ).

The following formula holds for the generating functions $x(z)$ and $y(z)$ of elementary slices:

\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)-\frac{\partial y}{\partial z}(z,t)\,\frac{\partial x}{\partial t}(z,t)=\frac{1}{z}.

(27)

Proof.

The first step of the proof comes from Propositions 24 and 25, which allows us to express $y(z,t)$ as a function of $x(z,t)$ . Indeed, one can write

y(z,t)=y(z^{\circ}(x(z,t),t),t)=Y(x(z,t),t).

By differentiating this relation using the chain rule, we obtain the two equations:

	$\displaystyle\frac{\partial y}{\partial z}(z,t)$	$\displaystyle=\frac{\partial x}{\partial z}(z,t)\cdot\frac{\partial Y}{\partial x}(x(z,t),t),$
	$\displaystyle\frac{\partial y}{\partial t}(z,t)$	$\displaystyle=\frac{\partial x}{\partial t}(z,t)\cdot\frac{\partial Y}{\partial x}(x(z,t),t)+\frac{\partial Y}{\partial t}(x(z,t),t).$

Substituting these two $y$ -derivatives into the Poisson bracket $\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)-\frac{\partial y}{\partial z}(z,t)\,\frac{\partial x}{\partial t}(z,t)$ , which we wish to simplify, gives:

\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)-\frac{\partial y}{\partial z}(z,t)\,\frac{\partial x}{\partial t}(z,t)=\frac{\partial Y}{\partial t}(x(z,t),t)\cdot\frac{\partial x}{\partial z}(z,t).

Recalling that $\frac{\partial Y}{\partial t}(x,t)=\frac{\partial W^{\circ}}{\partial t}(x,t)$ , we can simplify the right-hand side using Proposition 27, which provides a formula for this partial derivative. We thus find:

\frac{\partial Y}{\partial t}(x(z,t),t)=\frac{\partial W^{\circ}}{\partial t}(x(z,t),t)=\frac{\frac{\partial z^{\circ}}{\partial x}(x(z,t),t)}{z}.

Using the relation $z^{\circ}(x(z,t),t)=z$ from Proposition 24 and differentiating it with respect to $z$ in order to obtain $\frac{\partial z^{\circ}}{\partial x}(x(z,t),t)=\frac{1}{\frac{\partial x}{\partial z}(z,t)}$ , we finally get

\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)-\frac{\partial y}{\partial z}(z,t)\,\frac{\partial x}{\partial t}(z,t)=\frac{\frac{\partial z^{\circ}}{\partial x}(x(z,t),t)}{z^{\circ}(x(z,t),t)}\,\cdot\frac{\partial x}{\partial z}(z,t)=\frac{1}{z}.

This concludes the proof. ∎

It would be interesting to have a bijective proof of Lemma 28.

4.2.3 Application: an expression of $A_{1}$ as a $t$ -derivative

Now that we have all the necessary tools at our disposal, it remains only to combine them in such a way as to remove the integral with respect to $t$ in Equation (16). This is because it is difficult to interpret the algebraic nature of primitives, with respect to $t$ , of our generating functions. We shall therefore remedy this by showing that $A_{1}(x;y,t)$ can be expressed as a derivative with respect to $t$ .

Proposition 29.

The generating function of accessibly pointed $1$ -alternating hypermaps can be written as

A_{1}(x;y,t)=\frac{\partial}{\partial t}[z^{0}]\,z\ln\left(1-\frac{y(z,t)}{y}\right)\frac{\partial}{\partial z}\ln\left(1-\frac{x(z,t)}{x}\right).

(28)

This equation, completed with Equation (16), yields the last equality of Equation (6) in Proposition 2. Indeed, the remaining step is to check that both sides evaluate to $1$ at $t=0$ . In Appendix A, we prove that $[t^{0}]x(z,t)=V_{\bullet}^{\prime}(z^{-1})\in\mathcal{R}[z^{-1}]$ and $[t^{0}]y(z,t)=z^{-1}\in z^{-1}\mathcal{R}[z^{-1}]$ , which implies that the right-hand side indeed evaluate to $1$ as $t=0$ . Now let us prove Proposition 29.

Proof.

We start from the left-hand side by writing $A_{1}(x;y,t)$ in the form $[z^{0}]\frac{1}{(x-x(z,t))(y-y(z,t))}$ as given by Theorem 1 for $k=1$ . We then modify the numerator $1$ using Lemma 28 through the relation:

1=z\,\left(\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)-\frac{\partial y}{\partial z}(z,t)\,\frac{\partial x}{\partial t}(z,t)\right).

We can now proceed to the computation. Splitting the sum into two parts, we obtain:

[z^{0}]\frac{1}{(x-x(z,t))(y-y(z,t))}=[z^{0}]\,z\left(\frac{\frac{\partial x}{\partial z}(z,t)\,\frac{\partial y}{\partial t}(z,t)}{(x-x(z,t))(y-y(z,t))}-\frac{\frac{\partial x}{\partial t}(z,t)\,\frac{\partial y}{\partial z}(z,t)}{(x-x(z,t))(y-y(z,t))}\right).

Introducing logarithmic derivatives, we can rewrite the right-hand side as:

[z^{0}]\,z\left(\frac{\partial}{\partial z}\ln(x-x(z,t))\cdot\frac{\partial}{\partial t}\ln(y-y(z,t))-\frac{\partial}{\partial t}\ln(x-x(z,t))\cdot\frac{\partial}{\partial z}\ln(y-y(z,t))\right)=\\ \sum_{h\in\mathbb{Z}}h[z^{h}]\ln(x-x(z,t))\,[z^{-h}]\frac{\partial}{\partial t}\ln(y-y(z,t))-\sum_{h\in\mathbb{Z}}h[z^{-h}]\frac{\partial}{\partial t}\ln(x-x(z,t))\,[z^{h}]\ln(y-y(z,t)).

We then change $h$ to $-h$ in the second sum, which reveals the derivative of a product:

[z^{0}]\frac{1}{(x-x(z,t))(y-y(z,t))}=\frac{\partial}{\partial t}\sum_{h\in\mathbb{Z}}h[z^{h}]\ln(x-x(z,t))\,[z^{-h}]\ln(y-y(z,t)).

This relation can be rewritten as:

A_{1}(x;y)=\frac{\partial}{\partial t}[z^{0}]\,z\ln\left(1-\frac{y(z,t)}{y}\right)\frac{\partial}{\partial z}\ln\left(1-\frac{x(z,t)}{x}\right),

which is exactly what we wanted to show. ∎

4.2.4 A relation between $A_{1}$ and the spectral curve

As introduced in Section 4.2.1, hypermaps with a monochromatic boundary play a crucial role in the combinatorial interpretation of various algebraic quantities such as $X(y)$ and $Y(x)$ , hence in the understanding of the spectral curve $E(x,y)=0$ . The latter originally appeared in the context of solving Tutte’s equations, that is, by removing the marked edge of the outer face in order to derive a closed form for $H_{1}$ . This is how the following relation appeared in [Eyn03]:

H_{1}(x;y)=\frac{E(x;y)}{(x-X(y))(y-Y(x))}.

(29)

In fact, using Equation (6), one can rewrite this equation as an equation on the generating function of accessibly pointed $1$ -alternating hypermaps:

A_{1}(x;y)=\frac{\partial}{\partial t}\ln\left(\frac{E(x,y)}{(x-X(y))(y-Y(x))}\right).

The goal of this section is to prove this equation, which is the main result of Proposition 3, thereby giving another proof of Equation (29) without using Tutte’s method.

First, let us denote by $d^{\circ}$ (resp. $d^{\bullet}$ ) the maximal degree of white (resp. black) faces. In this case, the difference between the left and right sides of an elementary slice is bounded. We have already seen that it is bounded by $-1$ from below but, going around the face incident to the base edge of such a slice, we find an upper bound of $d^{\bullet}-1$ for slices of type $\mathcal{A}$ and of $d^{\circ}-1$ for slices of type $\mathcal{B}$ , since the left and right sides of the slice are geodesics. We therefore conclude that $z^{d^{\bullet}-1}(x(z)-x)$ is a polynomial in $z$ of degree at most $d^{\bullet}$ , while $z(y(z)-y)$ is a polynomial of degree at most $d^{\circ}$ . We may thus consider their resultant. This resultant is the minimal polynomial whose zeros are parametrized by $(x(z),y(z))$ , hence it is proportional to $E(x,y)$ . Note that the definition of $E$ assumes that the proportionality constant does not depend on $t$ , so it will not play any role, since its logarithmic derivative with respect to $t$ vanishes. From now on, we will say that $E$ is exactly the resultant of $(x(z)-x,y(z)-y)$ , for simplicity.
Since $E(x,y)$ is a resultant, we shall express it using all the roots of $x(z)-x$ and $y(z^{\prime})-y$ . We have already introduced one root $z^{\circ}$ and $z^{\bullet}$ for each of the two Laurent polynomials (which are in fact compositional inverses of each other); we now introduce the remaining ones. To this end, we apply the Newton–Puiseux theorem, which ensures that there exist $d^{\bullet}-1$ other roots of the equation $x(z)=x$ , denoted $z_{j}^{\circ}(x)$ for $1\leqslant j\leqslant d^{\bullet}-1$ , such that

z_{j}^{\circ}(x)\in e^{\frac{2ij\pi}{d^{\bullet}-1}}\left(\frac{x^{-1}}{a_{d^{\bullet}-1}}\right)^{\frac{1}{d^{\bullet}-1}}\,\left(1+x^{-\frac{1}{d^{\bullet}-1}}\mathbb{K}[\![x^{-\frac{1}{d^{\bullet}-1}}]\!]\right),

where $\mathbb{K}$ denotes the field of fractions of $\mathcal{R}$ . In particular, an interesting fact on the construction of these roots is that they are all distinct. Similarly, we introduce the other roots $z_{j}^{\bullet}(y)$ for $1\leqslant j\leqslant d^{\circ}-1$ of $y(z)=y$ , satisfying

z_{j}^{\bullet}(y)\in e^{\frac{2ij\pi}{d^{\circ}-1}}\left(\frac{y^{-1}}{b_{-1}}\right)^{-\frac{1}{d^{\circ}-1}}\,\left(1+y^{-\frac{1}{d^{\circ}-1}}\mathbb{K}[\![y^{-\frac{1}{d^{\circ}-1}}]\!]\right).

In what follows, we will call $z^{\bullet}(y)$ and all the $z_{j}^{\circ}(x)$ the small roots of the equations $y(z^{\prime})=y$ and $x(z)=x$ , as they tend to $0$ as $x,y$ goes to $\infty$ . Similarly, the other roots will be called the large roots since they tend to $\infty$ .

We now have all the ingredients needed to prove Equation (8). Let us begin by expanding the expression of the resultant $E(x,y)$ . Since the resultant of two Laurent polynomials $P$ and $Q$ is (proportional to) $\prod_{i}Q(\alpha_{i})$ , where the $\alpha_{i}$ are the roots of $P$ , we obtain the following explicit formula, with $Q(z):=x-x(z)$ and $P(z):=y-y(z)$ :

E(x,y)=(x-X(y))\prod_{i=1}^{d^{\circ}-1}(x-x(z_{i}^{\bullet}(y))),

where $X(y)=x(z^{\bullet}(y))$ . As explained earlier, the proportionality constant plays no role here, so we set it to $1$ . The latter relation can also be written as

\frac{E(x,y)}{(x-X(y))(y-Y(x))}=\frac{\prod_{j=1}^{d^{\circ}-1}(x-x(z_{j}^{\bullet}(y)))}{y-Y(x)}.

(30)

Taking the logarithm and differentiating then gives:

\frac{\partial}{\partial t}\ln\left(\frac{E(x,y)}{(x-X(y))(y-Y(x))}\right)=\frac{\frac{\mathrm{d}}{\mathrm{d}t}y(z^{\circ}(x,t),t)}{y-y(z^{\circ}(x))}-\sum_{j=1}^{d^{\bullet}-1}\frac{\frac{\mathrm{d}}{\mathrm{d}t}x(z_{j}^{\bullet}(y,t),t)}{x-x(z_{j}^{\bullet}(y))}.

(31)

Now that we have sufficiently developed the right-hand side of Equation (8), let us turn back to the left-hand side. Recall from Theorem 1 that we have an explicit form for the generating function of accessibly pointed $1$ -alternating maps. The next step consists in performing a partial fraction decomposition with respect to $z$ of this expression, before extracting the $z^{0}$ coefficient. We thus write:

	$\displaystyle A_{1}(x;y)$	$\displaystyle=[z^{0}]\frac{1}{(x-x(z))(y-y(z))}$
		$\displaystyle=[z^{0}]\frac{z^{d^{\bullet}}}{C(z-z^{\circ}(x))\prod_{j=1}^{d^{\circ}-1}(z-z_{j}^{\bullet}(y))\cdot(z-z^{\bullet}(y))\prod_{j=1}^{d^{\bullet}-1}(z-z_{j}^{\circ}(x))},$

where we place the large roots on the left and the small roots on the right of the product. The constant $C$ , independent of $z$ , will be irrelevant for our present discussion.
Performing the partial fraction decomposition leads to:

	$\displaystyle A_{1}(x;y)$	$\displaystyle=[z^{0}]\Bigg(\frac{1}{z-z^{\circ}(x)}\frac{(z^{\circ}(x))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z-z_{j}^{\bullet}(y)}\frac{(z_{j}^{\bullet}(y))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}$
		$\displaystyle+\frac{1}{z-z^{\bullet}(y)}\frac{(z^{\bullet}(y))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z^{\bullet}(y)}}+\sum_{j=1}^{d^{\bullet}-1}\frac{1}{z-z_{j}^{\circ}(x)}\frac{(z_{j}^{\circ}(x))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z_{j}^{\circ}(x)}}\Bigg).$

This expression simplifies slightly when differentiating the denominator. Indeed, let us simplify the first denominator:

	$\displaystyle\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z^{\circ}(x)}=$	$\displaystyle\underbrace{(x-x(z^{\circ}(x)))}_{=0}\frac{\partial z^{d^{\bullet}}(y-y(z))}{\partial z}_{\|z=z^{\circ}(x)}$
		$\displaystyle+(z^{\circ}(x))^{d^{\bullet}}(y-y(z^{\circ}(x)))\frac{\partial(x-x(z))}{\partial z}_{\|z=z^{\circ}(x)}$
	$\displaystyle=$	$\displaystyle-(z^{\circ}(x))^{d^{\bullet}}(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z^{\circ}(x)},$

and the three other $z$ -derivatives simplify the same way. This yields

A_{1}(x;y)=[z^{0}]\Bigg(\frac{1}{z-z^{\circ}(x)}\frac{-1}{(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{|z=z_{j}^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z-z_{j}^{\bullet}(y)}\frac{-1}{(x-x(z_{j}^{\bullet}(y)))\frac{\partial y}{\partial z}_{|z=z_{j}^{\bullet}(y)}}\\ +\frac{1}{z-z^{\bullet}(y)}\frac{-1}{(x-x(z^{\bullet}(y)))\frac{\partial y}{\partial z}_{|z=z^{\bullet}(y)}}+\sum_{j=1}^{d^{\bullet}-1}\frac{1}{z-z_{j}^{\circ}(x)}\frac{-1}{(y-y(z_{j}^{\circ}(x)))\frac{\partial x}{\partial z}_{|z=z_{j}^{\circ}(x)}}\Bigg).

We now extract the coefficient of $z^{0}$ from each term. Since the partial fraction decomposition was performed in a splitting field for $(x-x(z))(y-y(z))$ , recall that $A_{1}(x;y)$ belongs to $x^{-1}y^{-1}\mathcal{R}[\![x^{-1},y^{-1}]\!]$ . Thus, we are working in an algebraic extension of this ring, namely the field of Puiseux series in $x^{-1}$ and $y^{-1}$ with coefficients in $\mathbb{K}$ . This means that $\frac{1}{z-z^{\circ}(x)}$ should be expanded as a power series in $z$ , since its $z^{n}$ coefficient (for $n\geqslant 0$ ) is then $z^{\circ}(x)^{-n-1}$ , a power series in $x^{-1}$ . The same reasoning applies to the other small roots, so the fraction $\frac{1}{z-z_{j}^{\bullet}(y)}$ , $1\leqslant j\leqslant d^{\circ}-1$ , is expanded in the same way. In contrast, in the case of the large roots, $\frac{1}{z-z^{\bullet}(y)}$ and all the fractions $\frac{1}{z-z_{j}^{\circ}(x)}$ , $1\leqslant j\leqslant d^{\bullet}-1$ , need to be expanded as Laurent power series in $z^{-1}$ , so that their terms are series in $y^{-1}$ . Hence we need to write these fractions as $\frac{z^{-1}}{1-z^{-1}z^{\bullet}(y)}$ and expand for large $z$ , which finally have no $z^{0}$ -coefficient. So we write

	$\displaystyle A_{1}(x;y)$	$\displaystyle=[z^{0}]\Bigg(\frac{1}{z-z^{\circ}(x)}\frac{-1}{(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z-z_{j}^{\bullet}(y)}\frac{-1}{(x-x(z_{j}^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}$
		$\displaystyle+\frac{z^{-1}}{1-z^{-1}z^{\bullet}(y)}\frac{-1}{(x-x(z^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z^{\bullet}(y)}}+\sum_{j=1}^{d^{\bullet}-1}\frac{z^{-1}}{1-z^{-1}z_{j}^{\circ}(x)}\frac{-1}{(y-y(z_{j}^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}\Bigg)$
		$\displaystyle=\frac{1}{z^{\circ}(x)}\frac{1}{(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z_{j}^{\bullet}(y)}\frac{1}{(x-x(z_{j}^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}$
		$\displaystyle\overset{Y(x)=y(z^{\circ}(x))}{=}\frac{1}{z^{\circ}(x)\,\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x,t)}}\frac{1}{(y-Y(x))}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z_{j}^{\bullet}(y)\,\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}\frac{1}{(x-x(z_{j}^{\bullet}(y)))}.$

Finally, the following lemma allows us to relate this expression to Equation (31), which immediately yields the desired result.

Lemma 30.

We have the following formulas for derivatives of compositions:

\frac{\mathrm{d}}{{\mathrm{d}t}}y(z^{\circ}(x,t),t)=\frac{1}{z^{\circ}(x)\frac{\partial x}{\partial z}(z^{\circ}(x))}\,\,\,\text{and}\,\,\,\forall j\in[\![1,d^{\circ}-1]\!],\frac{\mathrm{d}}{{\mathrm{d}t}}x(z_{j}^{\bullet}(x,t),t)=-\frac{1}{z_{j}^{\bullet}(y)\frac{\partial y}{\partial z}(z_{j}^{\bullet}(y))}.

Proof.

Since the proof is the same for all formulas, we will only discuss the first one.

We start by applying the chain rule on the left-hand side:

\frac{\mathrm{d}}{{\mathrm{d}t}}y(z^{\circ}(x,t),t)=\frac{\partial y}{\partial t}(z^{\circ}(x,t),t)+\frac{\partial z^{\circ}}{\partial t}(y,t)\cdot\,\frac{\partial y}{\partial t}(z^{\circ}(x,t),t).

Hence we are led to make $\frac{\partial z^{\circ}}{\partial t}(x,t)$ explicit.

To this end, the idea is to differentiate with respect to $t$ the equation $x(z^{\circ}(x,t),t)=x$ . This gives

\frac{\partial z^{\circ}}{\partial t}(x,t)=-\frac{\frac{\partial x}{\partial t}(z^{\circ}(x,t),t)}{\frac{\partial x}{\partial z}(z^{\circ}(x,t),t)}.

We now only need to substitute this relation into the former one and conclude using Lemma 28:

	$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}y(z^{\circ}(x,t),t)$	$\displaystyle=\frac{\partial y}{\partial t}(z^{\circ}(x))+\left(-\frac{\frac{\partial x}{\partial t}(z^{\circ}(x))}{\frac{\partial x}{\partial z}(z^{\circ}(x))}\right)\cdot\,\frac{\partial y}{\partial t}(z^{\circ}(x))$
		$\displaystyle=\left(\frac{\partial x}{\partial z}\,\frac{\partial y}{\partial t}-\frac{\partial y}{\partial z}\,\frac{\partial x}{\partial t}\right)_{\|z=z^{\circ}(x)}\cdot\frac{1}{\frac{\partial x}{\partial z}(z^{\circ}(x))}$
		$\displaystyle=\frac{1}{z^{\circ}(x)\frac{\partial x}{\partial z}(z^{\circ}(x))}$

which concludes the proof of the lemma and hence the proof of Proposition 3.

∎

5 Enumeration of $2$ -alternating hypermaps

In this section, we establish Theorem 4. Our strategy relies on most of the concepts introduced earlier: we decompose a pointed $2$ -alternating hypermap according to the component containing the marked vertex and all the vertices that can reach it, called the accessible component, and the remaining part. We shall see that this decomposition follows naturally from a planarity argument, which allows us to separate the hypermap in a particularly straightforward way.

We then apply this result—namely Equation (9)—to derive the generating function of strongly connected $2$ -alternating hypermaps, which is Equation (10). This derivation does not explicitly rely on the notion of accessibility, although a similar argument could easily be adapted to the case of $B_{2}$ .

5.1 Expressing $2$ -alternating hypermaps with accessibly pointed hypermaps

Our approach follows the same direction as in the previous section during the proof of Equation (16), with one essential difference: we shall now investigate whether the outward corners are connected to the distinguished vertex of the hypermap or not. Note that, in the case of $1$ -alternating hypermaps, the marked vertex was always accessible from the outward corner, but here it will be a central element of our discussion.

The goal of this section is therefore to prove the following lemma:

Lemma 31.

A pointed $2$ -alternating hypermap can be decomposed as follows:

\begin{split}\frac{\partial H_{2}(x_{1},x_{2};y_{1},y_{2})}{\partial t}&=A_{2}(x_{1},x_{2};y_{1},y_{2})\,H_{1}(x_{1};y_{1})\,H_{1}(x_{2};y_{2})\\ &+\left(A_{1}(x_{1};y_{2})+A_{1}(x_{2};y_{1})\right)\,H_{2}(x_{1},x_{2};y_{1},y_{2}).\end{split}

(32)

Let us first see how this formula allows us to obtain Equation (9) of Theorem 4. To do so, we write, using Equation (16)

	$\displaystyle\frac{\partial}{\partial t}\left(\frac{H_{2}(x_{1},x_{2};y_{1},y_{2})}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}\right)=$	$\displaystyle\frac{1}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}\Bigg(\frac{\partial H_{2}(x_{1},x_{2};y_{1},y_{2})}{\partial t}$
		$\displaystyle\hskip 56.9055pt-(A_{1}(x_{1};y_{2})+A_{1}(x_{2};y_{1}))\,H_{2}(x_{1},x_{2};y_{1},y_{2})\Bigg).$

We then deduce, by substituting Equation (32):

\frac{\partial}{\partial t}\left(\frac{H_{2}(x_{1},x_{2};y_{1},y_{2})}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}\right)=A_{2}(x_{1},x_{2};y_{1},y_{2})\,\frac{H_{1}(x_{1};y_{1})\,H_{1}(x_{2};y_{2})}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}.

(33)

We can now use Equation (15) from Corollary 18. This allows us to recognize a total derivative on the right-hand side of Equation (33):

A_{2}(x_{1},x_{2};y_{1},y_{2})\,\frac{H_{1}(x_{1};y_{1})\,H_{1}(x_{2};y_{2})}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}=\frac{1}{(x_{1}-x_{2})(y_{1}-y_{2})}\,\frac{\partial}{\partial t}\left(\frac{H_{1}(x_{1};y_{1})\,H_{1}(x_{2};y_{2})}{H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})}\right),

by using the logarithmic derivative expression of $A_{1}$ in Equation (16). We then conclude by integrating and using the facts that $[t^{0}]\,H_{1}(t)=1$ and $[t^{0}]\,H_{2}(t)=0$ by definition of $H_{k}$ in Equation (2), which gives Equation (9).

We can now get back to the combinatorial proof of Equation (32).

Proof of Lemma 31.

Figure 10: Sketch of the proof of Lemma 31, from left to right the cases

1,2

and

3

according to whether

c_{0}

c_{2}

is related to

v

. We use here the shorthand notation

H_{i\mid j}:=H_{1}(x_{i};y_{j})

and

H_{12\mid 12}:=H_{2}(x_{1},x_{2};y_{1},y_{2})

for simplicity.

We begin the proof by considering a pointed $2$ -alternating hypermap, denoting by $v$ its marked vertex. We shall also use the corner notation $c_{0},c_{1},c_{2},c_{3}$ introduced in Section 1.2. We focus on the even-indexed corners $c_{0}$ and $c_{2}$ , since they are the ones most likely to access to $v$ .
Let us distinguish cases according to which of the corners $c_{0}$ and $c_{2}$ is connected to $v$ . Figure 10 gives a sketch of the proof.

$\hookrightarrow$ Case 1: $v$ is accessible from both $c_{0}$ and $c_{2}$ . In this case only $c_{1}$ and $c_{3}$ may fail to access to $v$ : there may exist a bridge on their way to $v$ . Taking the last such bridge for each of them isolates $c_{1}$ and $c_{3}$ into one $1$ -alternating hypermap each, of respective weight $H_{1}(x_{1};y_{1})$ and $H_{1}(x_{2};y_{2})$ . Note the presence of the additional $1$ in $H_{1}$ , which stands for the case where there is no bridge preventing $c_{1}$ or $c_{3}$ from being related to $v$ . What remains is an accessibly pointed $2$ -alternating hypermap component containing $v,c_{0}$ and $c_{2}$ , counted by $A_{2}(x_{1},x_{2};y_{1},y_{2})$ .
$\hookrightarrow$ Case 2: $v$ is accessible from $c_{2}$ but not from $c_{0}$ . Since $c_{0}$ is not related to $v$ , there must be a bridge oriented from $v$ to $c_{0}$ . Looking at the orientations of the boundary edges of the hypermap, this bridge cannot lie on $[c_{3},c_{1}]$ : otherwise, because of the orientation of the bridge, this boundary interval would have to contain an inward corner, hence a marked corner with odd index by Remark 10, which is not possible. So the bridge must belong to the boundary interval $[c_{1},c_{3}]$ . Taking the last of these bridges, we can split the pointed hypermap into two components: the first is not pointed and contains $c_{0},c_{1},c_{3}$ , while the second, containing $c_{2}$ and $v$ , is a pointed $1$ -alternating hypermap. The latter is also accessible since we took the last bridge, and $c_{2}$ must be an outward corner because the orientation of the bridge forces the existence of an outward corner belonging to $[c_{1},c_{3}]$ , which is then $c_{2}$ .
$\hookrightarrow$ Case 3: $v$ is accessible from $c_{0}$ but not from $c_{2}$ . By symmetry with Case 2, the corresponding term is $H_{2}(x_{1},x_{2};y_{1},y_{2})\,A_{1}(x_{1};y_{2})$ .
Note that the obstruction of $c_{0}$ from $v$ implies that $c_{2}$ can reach $v$ , and reciprocally. Hence the case where neither $c_{0}$ nor $c_{2}$ can reach $v$ does not occur. Finally, summing these three contributions gives exactly Equation (32). ∎

The idea of decomposing our proof according to which outward corners are related to the marked vertex stems from the fact that these are the boundary vertices most likely to be connected to any other vertex. Indeed, intuitively, inward corners have little chance of being connected to the marked vertex, since it suffices for a bridge formed by its two incident edges to obstruct such a connection—an event that is structurally inexpensive for the hypermap. It is also important to keep in mind that planarity and the uniqueness of the marked outer face are essential arguments here, since this enables us to decompose our $2$ -alternating hypermaps into independent components. These ideas form the foundation of the forthcoming generalization of Lemma 31 in the upcoming work [Lej26].

5.2 On strongly connected $2$ -alternating hypermaps

In order to study the generating series $B_{2}$ of strongly connected $2$ -alternating hypermaps, we start from a generic $2$ -alternating hypermap and ask what might prevent it from being strongly connected. As stated in Remark 10, it suffices to determine the conditions under which every vertex of the hypermap is accessible from the vertices incident to the corners $c_{1}$ and $c_{3}$ . We illustrate the following reasonning in Figure 11.
To that end, one can already see what could prevent corners $c_{1}$ and $c_{3}$ from being connected: the existence of a bridge belonging to $[c_{0},c_{2}]$ or a bridge belonging to $[c_{2},c_{0}]$ . By considering the last such bridge on the path connecting $c_{1}$ to $c_{3}$ , and then symmetrically on the path connecting $c_{3}$ to $c_{1}$ (which is possible since each bridge disconnects the hypermap), we observe the appearance of two components of weights $H_{1}(x_{1};y_{1})$ and $H_{1}(x_{2};y_{2})$ , to which our corners respectively belong. We may therefore remove the two bridges and examine what remains, renaming the corners incident to the endpoints of each bridge $c_{1}$ and $c_{3}$ as in the original definition, but now assuming the absence of bridges on the paths from $c_{1}$ to $c_{3}$ .
Let us now see how to connect $c_{1}$ and $c_{3}$ to $c_{0}$ . This is possible only if there is no bridge belonging to $[c_{3},c_{1}]$ . Taking the first such bridge produces a component containing $c_{0}$ of weight $H_{1}(x_{1};y_{2})$ . Symmetrically, for $c_{2}$ , we obtain a component of weight $H_{1}(x_{2};y_{1})$ .
By removing all these bridges, we can thus connect $c_{1}$ and $c_{3}$ to all boundary corners, and therefore every boundary vertex can be connected to any other boundary vertex. As a result, there cannot exist any remaining bridge in this component, by Lemma 9. The remaining component is therefore strongly connected, as stated in Lemma 6, and is thus counted by $B_{2}(x_{1},x_{2};y_{1},y_{2})$ once the bridge weights are added. Note that this decomposition is bijective, since one only needs to connect each $1$ -alternating hypermap to the corresponding marked corner of the strongly connected $2$ -alternating hypermap by a bridge.

Figure 11: Sketch of the decomposition of

H_{2}

with respect to

B_{2}

. The idea is simply to remove all bridges that prevent each inner vertex from being connected to the marked corners, since the accessibility criterion of Lemma 9 relies on accessibility from the boundary. The shorthand notations in the figure are the same as in Figure 10.

This leads to the formula

H_{2}(x_{1},x_{2};y_{1},y_{2})=H_{1}(x_{1};y_{1})\,H_{1}(x_{1};y_{2})\,H_{1}(x_{2};y_{1})\,H_{1}(x_{2};y_{2})\,B_{2}(x_{1},x_{2};y_{1},y_{2}).

(34)

This allows us to recover Equation (10) from Theorem 4, using the expression of $H_{2}(x_{1},x_{2};y_{1},y_{2})$ given in Equation (9). This finally completes the proof of Theorem 4.

6 Conclusion

The main objective of this work was to highlight a new way of enumerating plane hypermaps with a non-monochromatic boundary. To achieve this, we have shown that the variable $t$ , marking the vertices, plays a meaningful role. Indeed, marking a vertex gives rise to equations involving derivatives with respect to this variable. Moreover, studying the accessible component of the marked vertex—that is, the smallest hypermap in which all vertices can access the marked one—allows us to decompose pointed hypermaps. The rest of the work then consisted in understanding accessibly pointed hypermaps—made possible thanks to the notion of slices introduced in [AB26]—and in solving the resulting equations, either directly or by identifying a total derivative, in order to find algebraic expressions for alternating hypermaps. This approach enabled us to rederive several formulas appearing in [AB26] and [Eyn16].
The remaining open questions concern $k$ -alternating hypermaps for $k\geqslant 3$ —–a problem addressed in the forthcoming article [Lej26]—–as well as the extension to higher genus, in order to find new combinatorial interpretations of the algebraic results provided by topological recursion, as stated in [EO08]. Another direction for future work involves asymptotic considerations, which now appear more approachable thanks to a deeper understanding of the elementary building blocks—namely the generating functions of the elementary slices—bringing us closer to an understanding of distance distributions in hypermaps.

Appendix A Counting plane hypermaps with a monochromatic boundary via integration

In this appendix, we give a new proof of Proposition 25 using Proposition 23 which itself follows from slice decomposition. We shall first show that the derivatives with respect to $t$ of both sides of each equality in (24) are equal, before matching the integration constants.
We start from the case of pointed hypermaps with a white boundary, whose generating function is $\frac{\partial}{\partial t}W^{\circ}(x,t)$ . The first part of the following proof will be similar for pointed hypermaps with a black boundary.

Our idea is to use the expression $\frac{\partial}{\partial t}W^{\circ}(x,t)$ as $[z^{0}]\frac{1}{x-x(z,t)}$ and to perform a partial fraction decomposition. The issue is that this is only possible if we have finitely many roots for the equation $x=x(z,t)$ , which occurs only when the face degrees are bounded. Let us show that doing it for any bound for the face degrees is enough to conclude in the general case, thanks to the following lemma.

Lemma 32.

Let $f$ be an element of $\mathcal{R}$ , and let

f_{d}:=\Big[\prod_{i\geqslant d+1}(t_{i}^{\circ})^{0}\prod_{j\geqslant d+1}(t_{j}^{\bullet})^{0}\Big]f

be the restriction of $f$ to faces with degree bounded by $d\geq 1$ . Then the identity $f=0$ holds if, and only if, $f_{d}=0$ holds for every $d\geq 1$ .

Proof.

The direct implication is immediate by definition of $f_{d}$ .

For the converse, recall that a formal series in infinitely many variables (as $f$ which belongs to $\mathcal{R}$ ) is defined as a sum of monomials of finite degree-that is, involving finitely many variables. Hence one only need to check that each coefficient from $f$ in front of any of these finite-degree monomials vanish, which is easy by considering the identities $f_{d}=0$ because they are exactly restricted to the case of finite (face) degree. ∎

The usefulness of this formal lemma lies in the fact that Proposition 25 makes sense in $\mathcal{R}$ , while we wish to bound the degree of our faces (say, by $d^{\circ}$ for white faces and $d^{\bullet}$ for black faces) in order to speak of the roots of polynomials. However, this is merely a computational artifice, since the roots—besides the compositional inverses $z^{\circ}$ and $z^{\bullet}$ —no longer appear in the final result.
We then denote all our roots as in Section 4.2.4, and we can therefore write our partial fraction decomposition. We refer to the computation from Section 4.2.4, since the same ideas apply: performing the decomposition over the roots by turning the denominator into a polynomial, which leads to

\frac{1}{x-x(z)}=\frac{z^{d^{\bullet}}}{z^{d^{\bullet}}(x-x(z))}=\frac{1}{z-z^{\circ}(x)}\frac{(z^{\circ}(x))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))}{\partial z}_{|z=z^{\circ}(x)}}+\sum_{j=1}^{d^{\bullet}-1}\frac{1}{z-z_{j}^{\circ}(x)}\frac{(z_{j}^{\circ}(x))^{d^{\bullet}}}{\frac{\partial z^{d^{\bullet}}(x-x(z))}{\partial z}_{|z=z_{j}^{\circ}(x)}}.

This can be simplified by expanding the derivatives in the denominators as in Section 4.2.4:

\frac{1}{x-x(z)}=-\frac{1}{z-z^{\circ}(x)}\frac{1}{\frac{\partial x}{\partial z}_{|z=z^{\circ}(x)}}-\sum_{j=1}^{d^{\bullet}-1}\frac{1}{z-z_{j}^{\circ}(x)}\frac{1}{\frac{\partial x}{\partial z}_{|z=z_{j}^{\circ}(x)}}.

To extract the coefficient of $z^{0}$ as needed to compute $\frac{\partial}{\partial t}W^{\circ}(x,t)$ , only the first term contributes, since other terms are elements of $z^{-1}\mathcal{R}[\![z^{-1}]\!]$ because the partial fraction decomposition holds in $x^{-1}\mathcal{R}[\![x^{-1}]\!]$ , and thus vanish. We finally obtain:

[z^{0}]\frac{1}{x-x(z)}=\frac{1}{z^{\circ}(x)\,\frac{\partial x}{\partial z}_{|z=z^{\circ}(x)}}.

By Proposition 23 and by Lemma 30, we then have:

\frac{\partial}{\partial t}W^{\circ}(x,t)=\frac{\mathrm{d}}{{\mathrm{d}t}}y(z^{\circ}(x,t),t)=\frac{\partial}{\partial t}Y(x,t),

since $Y(x,t)=y(z^{\circ}(x,t),t)$ by Equation (23). By a dual argument for pointed hypermaps with a black boundary, we can show that

\frac{\partial}{\partial t}W^{\bullet}(y,t)=\frac{\partial}{\partial t}X(y,t).

Thanks to Lemma 32, the two last identities hold coefficientwise in $\mathcal{R}$ , and not just in the case of bounded face degrees.
It remains to perform the integration and so to determine the integration constants. We will then prove that

[t^{0}]W^{\circ}(x,t)-[t^{0}]Y(x,t)=-V_{\circ}^{\prime}(x)\quad\text{and}\quad[t^{0}]W^{\bullet}(y,t)-[t^{0}]X(y,t)=-V_{\bullet}^{\prime}(y).

First, note that there exists no hypermap with a monochromatic boundary and zero vertex. Hence, we directly have $[t^{0}]W^{\circ}(x,t)=0$ and $[t^{0}]W^{\bullet}(y,t)=0$ . So the problem reduces to showing that:

[t^{0}]Y(x,t)=V_{\circ}^{\prime}(x)\quad\text{and}\quad[t^{0}]X(y,t)=V_{\bullet}^{\prime}(y).

We will then use the expressions $X(y,t)=x(z^{\bullet}(y,t),t)$ and $Y(x,t)=y(z^{\circ}(x,t),t)$ from Proposition 24, from which we want to extract the $t^{0}$ coefficient. To do so, we must study elementary slices having no vertices, except those on the right boundary (since they carry no weight as defined in Definition 11).
$\hookrightarrow$ For slices of type $\mathcal{A}$ , the vertex incident to the corner $l$ is weightless only if it is the apex. Since there are no internal vertices, we are restricted to the case where the base is incident to a single black face, whose remaining edges form the right boundary of the slice. Computing the increment in each case gives $[t^{0}]x(z,t)=\sum_{k\geqslant 1}t_{k}^{\bullet}z^{1-k}=V_{\bullet}^{\prime}(z^{-1})$ . This is a formal series in $z^{-1}$ , which thus has no singular term in $z$ . Hence, $[t^{0}]x(z^{-1},t)$ is not invertible, implying $[t^{0}]z^{\circ}(x,t)^{-1}=0$ . See the proof of Proposition 24 for a clearer understanding of how $z^{\circ}(x,t)$ is defined.
$\hookrightarrow$ For slices of type $\mathcal{B}$ , identifying corners $l$ and $o$ restricts us to the trivial slice, consisting of one edge and two weightless vertices, with increment $-1$ . Thus, $[t^{0}]y(z,t)=z^{-1}$ . We then observe that $[t^{0}]z^{\bullet}(y,t)=y^{-1}$ , since it is the unique solution of $[t^{0}]y(z,t)=y$ .
These two observations allow us to conclude that $[t^{0}]z^{\bullet}(y,t)\neq 0$ and:

[t^{0}]X(y,t)=[t^{0}]x(z^{\bullet}(y,t),t)=\big([t^{0}]x(z,t)\big)_{|z=[t^{0}]z^{\bullet}(y,t)}=V_{\bullet}^{\prime}(z^{-1})_{|z=y^{-1}}=V_{\bullet}^{\prime}(y),

which is what we wanted.

On the other hand, things are more subtle in order to compute $[t^{0}]Y(x,t)=[t^{0}]y(z^{\circ}(x,t),t)$ due to the definition of $z^{\circ}(x)$ of which we only have information about its multiplicative inverse, which is ill-defined near $t=0$ as explained earlier. The idea is to make $x(z)$ appear and compose it with $z^{\circ}(x)$ on its right to make simplifications. We use then Equations (3) and (4):

y(z^{\circ}(x,t),t)=(z^{\circ}(x,t))^{-1}+\big([z^{\geqslant 0}]y(z,t)\big)_{|z=z^{\circ}(x,t)},

where $[z^{\geqslant}0]f(z)$ stands for considering only the nonnegative powers of $z$ in the Laurent polynomial $f(z)$ . The last term of the previous equation expands as

\big([z^{\geqslant 0}]y(z,t)\big)_{|z=z^{\circ}(x,t)}=\sum_{d\geqslant 1}t_{d}^{\circ}([z^{\geqslant 0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}.

Taking the $t^{0}$ coefficient and recalling that $[t^{0}]z^{\circ}(x,t)^{-1}=0$ , we simplify:

[t^{0}]y(z^{\circ}(x,t),t)=0+\sum_{d\geqslant 1}t_{d}^{\circ}[t^{0}]([z^{\geqslant 0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}.

Furthermore, since $([z^{<0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}$ is a polynomial in $z^{\circ}(x)$ , then one gets the equality $[t^{0}]([z^{<0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}=0$ since $[t^{0}]z^{\circ}(x,t)=0$ . We can therefore simplify the expression:

[t^{0}]([z^{\geqslant 0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}=[t^{0}](x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}=x^{d-1},

since $z^{\circ}(x)$ is the compositional inverse of $x(z)$ .
We finally conclude that:

[t^{0}]Y(x,t)=[t^{0}]y(z,t)_{|z=z^{\circ}(x,t)}=\sum_{d\geqslant 1}t_{d}^{\circ}([z^{\geqslant 0}]x(z,t)^{d-1})_{|z=z^{\circ}(x,t)}=\sum_{d\geqslant 1}t_{d}^{\circ}x^{d-1}=V_{\circ}^{\prime}(x),

which is precisely what we wanted to prove.

References

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[BF02] Cyril Banderier and Philippe Flajolet “Basic analytic combinatorics of directed lattice paths” Selected Papers in honour of Maurice Nivat In Theoretical Computer Science 281.1, 2002, pp. 37–80 DOI: https://doi.org/10.1016/S0304-3975(02)00007-5
[BK87] D.. Boulatov and V.. Kazakov “The Ising model on a random planar lattice: the structure of the phase transition and the exact critical exponents” In Phys. Lett. B 186.3-4, 1987, pp. 379–384 DOI: 10.1016/0370-2693(87)90312-1
[Cur23] Nicolas Curien “Peeling random planar maps. École d’Été de Probabilités de Saint-Flour XLIX – 2019” 2335, Lect. Notes Math. Cham: Springer, 2023 DOI: 10.1007/978-3-031-36854-7
[DDG23] Jian Ding, Julien Dubédat and Ewain Gwynne “Introduction to the Liouville quantum gravity metric” In International congress of mathematicians 2022, ICM 2022, Helsinki, Finland, virtual, July 6–14, 2022. Volume 6. Sections 12–14 Berlin: European Mathematical Society (EMS), 2023, pp. 4212–4244 DOI: 10.4171/ICM2022/40
[DHL25] Maurice Duits, Nathan Hayford and Seung-Yeop Lee “The Ising model coupled to 2D gravity: genus zero partition function” In SIGMA, Symmetry Integrability Geom. Methods Appl. 21, 2025, pp. paper 07990 DOI: 10.3842/SIGMA.2025.079
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[Lej26] Thomas Lejeune “Enumeration of plane hypermaps with a mixed boundary II” In preparation, 2026
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[Sch15] Gilles Schaeffer “Planar maps” http://www.lix.polytechnique.fr/˜schaeffe/Biblio/HB.pdf In Handbook of enumerative combinatorics, Discrete Math. Appl. (Boca Raton) CRC Press, Boca Raton, FL, 2015, pp. 335–395

	$\displaystyle\frac{\partial z^{d^{\bullet}}(x-x(z))(y-y(z))}{\partial z}_{\|z=z^{\circ}(x)}=$	$\displaystyle\underbrace{(x-x(z^{\circ}(x)))}_{=0}\frac{\partial z^{d^{\bullet}}(y-y(z))}{\partial z}_{\|z=z^{\circ}(x)}$
		$\displaystyle+(z^{\circ}(x))^{d^{\bullet}}(y-y(z^{\circ}(x)))\frac{\partial(x-x(z))}{\partial z}_{\|z=z^{\circ}(x)}$
	$\displaystyle=$	$\displaystyle-(z^{\circ}(x))^{d^{\bullet}}(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z^{\circ}(x)},$

	$\displaystyle A_{1}(x;y)$	$\displaystyle=[z^{0}]\Bigg(\frac{1}{z-z^{\circ}(x)}\frac{-1}{(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z-z_{j}^{\bullet}(y)}\frac{-1}{(x-x(z_{j}^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}$
		$\displaystyle+\frac{z^{-1}}{1-z^{-1}z^{\bullet}(y)}\frac{-1}{(x-x(z^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z^{\bullet}(y)}}+\sum_{j=1}^{d^{\bullet}-1}\frac{z^{-1}}{1-z^{-1}z_{j}^{\circ}(x)}\frac{-1}{(y-y(z_{j}^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}\Bigg)$
		$\displaystyle=\frac{1}{z^{\circ}(x)}\frac{1}{(y-y(z^{\circ}(x)))\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x)}}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z_{j}^{\bullet}(y)}\frac{1}{(x-x(z_{j}^{\bullet}(y)))\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}$
		$\displaystyle\overset{Y(x)=y(z^{\circ}(x))}{=}\frac{1}{z^{\circ}(x)\,\frac{\partial x}{\partial z}_{\|z=z_{j}^{\circ}(x,t)}}\frac{1}{(y-Y(x))}+\sum_{j=1}^{d^{\circ}-1}\frac{1}{z_{j}^{\bullet}(y)\,\frac{\partial y}{\partial z}_{\|z=z_{j}^{\bullet}(y)}}\frac{1}{(x-x(z_{j}^{\bullet}(y)))}.$

Enumeration of plane hypermaps with a mixed boundary I

Abstract

1 Introduction

1.1 Context and motivations

Acknowledgments.

1.2 Basic definitions and notations

1.3 Main results

Theorem 1.

Proposition 2.

Proposition 3.

Theorem 4.

1.4 Outline

2 Bridges and accessibility in plane hypermaps

Lemma 5.

Proof.

Lemma 6.

Proof of Lemma 6.

Remark 7.

Remark 8.

Lemma 9.

Proof.

Remark 10.

3 Enumeration of accessibly pointed kk-alternating hypermaps

3.1 Slices

Reminders.

Definition 11.

Remark 12.

Generalized slices.

Definition 13.

Remark 14.

Proposition 15.

Proof.

3.2 Bijection between slices and accessibly pointed hypermaps

Proposition 16.

Proof.

Proposition 17.

Corollary 18.

Proof.

4 Enumeration of 11-alternating hypermaps

4.1 How to use accessibility in pointed 11-alternating hypermaps

4.1.1 Generic case

Remark 19.

4.1.2 Strongly connected case

Remark 20.

4.2 Various way to integrate relations for 11-alternating hypermaps

4.2.1 Review of the monochromatic boundary case

Definition 21.

Remark 22.

Proposition 23 ([AB26, Corollary 3.5]).

Proposition 24 (see also [AB26, Section 5]).

Proof.

Proposition 25 ([Eyn03, Eyn03a, AB26]).

Remark 26.

Proposition 27 ([AB26]).

4.2.2 Poisson formula for the parametrization (x⁡(z),y⁡(z))(x(z),y(z))

Lemma 28 (Poisson formula for x⁡(z)x(z) and y⁡(z)y(z)).

Proof.

4.2.3 Application: an expression of A1A_{1} as a tt-derivative

Proposition 29.

Proof.

4.2.4 A relation between A1A_{1} and the spectral curve

Lemma 30.

Proof.

5 Enumeration of 22-alternating hypermaps

5.1 Expressing 22-alternating hypermaps with accessibly pointed hypermaps

Lemma 31.

Proof of Lemma 31.

5.2 On strongly connected 22-alternating hypermaps

6 Conclusion

Appendix A Counting plane hypermaps with a monochromatic boundary via integration

Lemma 32.

Proof.

References

3 Enumeration of accessibly pointed $k$ -alternating hypermaps

4 Enumeration of $1$ -alternating hypermaps

4.1 How to use accessibility in pointed $1$ -alternating hypermaps

4.2 Various way to integrate relations for $1$ -alternating hypermaps

4.2.2 Poisson formula for the parametrization $(x(z),y(z))$

Lemma 28 (Poisson formula for $x(z)$ and $y(z)$ ).

4.2.3 Application: an expression of $A_{1}$ as a $t$ -derivative

4.2.4 A relation between $A_{1}$ and the spectral curve

5 Enumeration of $2$ -alternating hypermaps

5.1 Expressing $2$ -alternating hypermaps with accessibly pointed hypermaps

5.2 On strongly connected $2$ -alternating hypermaps