1 Introduction

# A combinatorial interpretation of the $κ^{\star}_{g}(n)$ coefficients

A combinatorial interpretation of the coefficients

Thomas J. X. Li, Christian M. Reidys

Institut for Matematik og Datalogi,

University of Southern Denmark,

Campusvej 55, DK-5230 Odense M, Denmark

Abstract

Studying the virtual Euler characteristic of the moduli space of curves, Harer and Zagier compute the generating function of unicellular maps of genus . They furthermore identify coefficients, , which fully determine the series . The main result of this paper is a combinatorial interpretation of . We show that these enumerate a class of unicellular maps, which correspond -to- to a specific type of trees, referred to as O-trees. O-trees are a variant of the C-decorated trees introduced by Chapuy, Féray and Fusy. We exhaustively enumerate the number of shapes of genus with edges, which is a specific class of unicellular maps with vertex degree at least three. Furthermore we give combinatorial proofs for expressing the generating functions and for unicellular maps and shapes in terms of , respectively. We then prove a two term recursion for and that for any fixed , the sequence is log-concave, where , for .

Keywords: unicellular map, fatgraph, O-tree, shape-polynomial, recursion

## 1. Introduction

A unicellular map is a connected graph embedded in a compact orientable surface, in such a way that its complement is homeomorphic to a polygon. Equivalently, a unicellular map of genus with edges can also be seen as gluing the edges of -gon into pairs to create an orientable surface of genus . It is related to the general theory of map enumeration, the study of moduli spaces of curves [11], the character theory of symmetric group [22, 13], the computation of matrix integrals [15], and also considered in a variety of application contexts [18, 2]. The most well-known example of unicellular maps is arguably the class of plane trees, enumerated by the Catalan numbers (see for example [21]).

In [11] Harer and Zagier study the virtual Euler characteristic of the moduli space of curves. The number counting the ways of gluing the edges of -gon in order to obtain an orientable surface of genus , i.e. , the number of unicellular maps of genus with edges turns out to play a crucial role in their computations and they discover the two term recursion

 (1.1) (n+1)ϵg(n)=2(2n−1)ϵg(n−1)+(n−1)(2n−1)(2n−3)ϵg−1(n−2).

Subsequently, they identify certain coefficients, , which they describe to “give the best coding of the information contained in the […] series”, [11]. The key link is the following functional relation between the generating function of and the generating function of :

 Cg(z)=1√1−4z K⋆g(z1−4z),

where .

The main result of this paper is a combinatorial interpretation of the coefficients discovered by Harer and Zagier. enumerates a class unicellular maps, which correspond -to- to certain O-trees. O-trees are a variant of the C-decorated trees introduced in [7], see Theorem 2. Using O-trees, we exhaustively enumerate a specific class of unicellular maps with vertex degree at least three, called shapes. We give combinatorial proofs for expressing the generating functions and for unicellular maps and shapes in terms of , respectively. In particular the are positive integers that satisfy an analogue of eq. (1.1)

 (1.2) (n+1)κ⋆g(n)=(n−1)(2n−1)(2n−3)κ⋆g−1(n−2)+2(2n−1)(2n−3)(2n−5)κ⋆g−1(n−3),

see Corollary 2 and Theorem 3. Eq. (1.2) has been independently discovered by Chekhov et al. [1] using the matrix model. We furthermore prove in Proposition 8 that for any fixed , the sequence is log-concave, where , for . We conjecture that the sequences and are infinitely log-concave.

## 2. Background

A map of genus is a connected graph embedded on a closed compact orientable surface of genus , such that is homeomorphic to a collection of polygons, which are called the faces of . Loops and multiple edges are allowed. The (multi)graph is called the underlying graph of and its underlying surface. Maps are considered up to homeomorphisms between the underlying surfaces. A sector of consists of two consecutive edges around a vertex. A rooted map is a map with a marked sector, called the root; the vertex incident to the root is called the root-vertex. In figures, we represent the root, by drawing a dashed edge attaching the root-vertex and a distinguished vertex, called the plant. By convention, the plant, plant-edge and its associated sector (around the plant) are not considered, when counting the number of vertices, edges or sectors. From now on, all maps are assumed to be rooted and accordingly the underlying graph of a rooted map is naturally vertex-rooted. A unicellular map is a map with a unique face. By Euler’s characteristic formula , a unicellular map of genus with edges has vertices. A plane tree is a unicellular map of genus .

We next introduce O-trees, which are directly implied by the concept of C-decorated trees by Chapuy et al. [7].

An O-permutation is a permutation where all cycles have odd length. For each O-permutation on elements, the genus of is defined as , where is the number of cycles of .

An O-tree with edges is a pair , where is a plane tree with edges and is an O-permutation on elements, see Figure 2(a). The genus of is defined to be the genus of . We canonically number the vertices of from to according to a left-to-right, depth-first traversal. Hence can be seen as a permutation of the vertices of , see Figure 2(b).

The underlying graph of is the (vertex-rooted) graph with edges, that is obtained from by merging the vertices in each cycle of (so that the vertices of correspond to the cycles of ) into a single vertex, see Figure 2(c).

###### Definition 1.

For nonnegative integers, let denote the set of unicellular maps of genus with edges, let be the set of O-permutations of genus on elements and let denote the set of O-trees of genus with edges, i.e., .

For two finite sets and , let denote their disjoint union and denote the set made of disjoint copies of . We write if there exists a bijection between and .

Let us first recall a combinatorial result of [6]:

###### Proposition 1 (Chapuy [6]).

For , let denote the set of maps from in which a set of vertices is distinguished. Then for and ,

 (2.1) 2g Eg(n)≃g⨄k=1E(2k+1)g−k(n).

In addition, if and are in correspondence, then the underlying graph of is obtained from the underlying graph of by merging the vertices in into a single vertex.

Remark: one key feature of this bijection is that it preserves the underlying graph of corresponding objects. By multiplying with the factor (which still preserves the underlying graph), we obtain

 (2.2) 2g 22gEg(n)≃g⨄k=122k⋅22(g−k)E(2k+1)g−k(n).

In analogy to the above decomposition of unicellular maps, there exists a recursive way to decompose O-permutations:

###### Proposition 2 (Chapuy et al.[7]).

For , let be the set of O-permutations from having labeled cycles. Then for and ,

 (2.3) 2g Og(n)≃g⨄k=122k⋅O(2k+1)g−k(n).

Furthermore, if and are in correspondence, then the cycles of are obtained from the cycles of by merging labeled cycles in into a single cycle.

Along these lines we furthermore observe:

###### Proposition 3 (Chapuy et al.[7]).

For , denote by the set of O-trees from in which a set of cycles is distinguished. Then for and ,

 2g Tg(n)≃g⨄k=122k⋅T(2k+1)g−k(n).

Furthermore, if and are in correspondence, then the underlying graph of is obtained from the underlying graph of by merging the vertices corresponding to cycles from into a single vertex.

The proofs of Proposition 2 and Proposition 3 are reformulations of those of [7] in the context of C-permutations and C-decorated trees. For completeness we give them in the Appendix.

Remark: the bijection for O-permutations preserves the cycles, which implies that the bijection for O-trees preserves the underlying graph of corresponding objects.

Combining Proposition 1 and Proposition 3, we inductively derive a bijection preserving the underlying graphs.

###### Theorem 1 (Chapuy et al.[7]).

For any non-negative integers and , there exists a bijection

 22gEg(n)≃Tg(n)=E0(n)×Og(n+1).

In addition, the cycles of an O-tree naturally correspond to the vertices of the associated unicellular map, such that the respective underlying graphs are the same.

Remark: in [7], Chapuy et al. prove the existence of a -to- correspondence between C-decorated trees and unicellular maps. The notion of C-permutation and C-decorated tree therein can be viewed as O-permutation and O-tree carrying a sign with each cycle, respectively. The reduction from C-decorated trees to O-trees allows us derive a -to- correspondence between O-trees and unicellular maps. Furthermore all the results in  [7] for C-decorated trees have an O-tree analogue.

The proof of Theorem 1 is a reformulation of that for C-decorated trees [7]. We give its proof in the Appendix.

## 3. Shapes

###### Definition 2.

A shape is a unicellular map having vertices of degree .

We adopt the convention that the plant-edge is taken into account when considering the degree of the root vertex.

###### Proposition 4.

[19] Given a shape of genus with edges, we have .

###### Proof.

By Euler’s characteristic formula, we have , where denotes the vertex set of a shape of genus with edges. On the one hand, any shape contains at least one vertex, which implies , i.e., . On the other hand, each vertex of a shape has . Then we derive , that is, . (Here we consider the plant and the plant-edge.) ∎

Let denote the set of -shapes, i.e. , shapes of genus with edges. Let denote the set of O-trees from such that each vertex in the underlying graph of the O-tree contains only vertices of degree , that is

 Rg(n)={(T,σ)∈E0(n)×Og(n+1)| each vertex of G(T,σ) has degree ≥3}.
###### Lemma 1.

For and , we have the bijection

 22gSg(n)≃Rg(n).

In addition, the cycles of an O-tree naturally correspond to the vertices of the associated unicellular map, in such a way that the respective underlying graphs are the same.

Note that a unicellular map is a shape if and only if each vertex in the underlying graph of the map has degree . Therefore, Lemma 1 follows directly from Theorem 1 by restricting the bijection to the set of shapes since the bijection therein preserves the underlying graph of corresponding objects.

Lemma 1 allows us to obtain deeper insight into shapes via O-permutations. To this end we consider the cycle-type of an O-permutation, i.e., a partition with parts of odd size. Given an O-permutation from , its cycle-type is a partition of with odd parts. We assume that , with . The partition naturally corresponds to the partition of . The fact that is a partition of follows from the identity . Here denotes the number of odd parts of , i.e., the number of parts of . Let denote number of parts of . It is clear that this a one-to-one correspondence. Therefore the cycle type of an O-permutation from can be indexed by an partition of .

The number of O-permutations of elements with cycle-type equal to is given by

 aγ(k)=(2g+t+k)!k!∏imi!(2i+1)mi,

where .

Let denote the set of O-permutations of genus with cycles of length and cycles of length . Note that the number of elements of an O-permutation from is . Then we have the following two cases:

1. For , the cardinality , denoted by , counts O-permutations of genus on elements without cycles of length (or cycle-type having the form ). Hence it is given by

 ag,t=∑γ⊢gℓ(γ)=taγ(0)=(2g+t)!∑γ⊢gℓ(γ)=t1∏imi!(2i+1)mi,

where runs over all partitions of with parts.

2. For arbitrary , each O-permutation in consists of an O-permutation from together with cycles of length . Then the set can be counted by first picking up elements from elements and then choosing an O-permutation from . Therefore

 |Og,t,k|=(2g+t+kk)ag,t=(2g+t+k)!k!∑γ⊢gℓ(γ)=t1∏imi!(2i+1)mi.

By definition,

 Og(n+1)=⨄t+k=n+1−2gOg,t,k.

Set . Let denote the set of O-trees from such that their associated O-permutation has cycles of length and cycles of length , i.e.,

 Rg,t,k={(T,σ)∈E0(n)×Og,t,k| each vertex of G(T,σ) has degree ≥3}.

Hence

 Rg(n)=⨄t+k=n+1−2gRg,t,k.
###### Lemma 2.

For , we have

 Rg,t,0=E0(2g+t−1)×Og,t,0.

Therefore

 |Rg,t,0|=Cat(2g+t−1)ag,t=(2(2g+t−1))!(2g+t−1)!∑γ⊢gℓ(γ)=t1∏imi!(2i+1)mi,

where is the -th Catalan number and runs over all partitions of with parts.

###### Proof.

By definition, . Given , each cycle of O-permutation has length . Then the underlying graph of , obtained from by merging into a single vertex the vertices in each cycle of , must have all vertices with degree . It implies that . Hence . ∎

To enumerate O-trees from for arbitrary , we observe that they can be reduced to O-trees from . The key idea is to eliminate the vertices corresponding to -cycles from an O-tree, thereby reducing to an O-tree without -cycles, i.e., O-tree from . This elimination on O-trees is reminiscent of Rémy’s bijection [20] on plane trees, which is briefly recalled below.

Rémy’s bijection reduces a plane tree with edges and a labeled vertex to a plane tree with edges and a sector labeled by or as follow

• if the labeled vertex is a leaf, is obtained from by contracting the edge connecting the labeled vertex and its father. Label by the sector associated with the labeled vertex,

• if the labeled vertex is a non-leaf, is obtained from by contracting the edge connecting the labeled vertex and its leftmost child. Label by the sector separating the leftmost subtree and the remaining subtree of the labeled vertex.

Therefore , see Figure 3. Rémy’s bijection has been applied in [7, 12, 16].

Given an O-tree , its traversal is defined as that of its underlying plane tree (traveling around the boundary of starting from the root-sector). A vertex of is called a -cycle if the corresponding element in is in a cycle of length . All sectors around the vertex are ordered according to the traversal of . A sector at in is called permissible if

• is not the last sector around according to the traversal of ,

• if vertex is -cycle, then is not the first two sectors around according to the traversal of .

###### Proposition 5.

Any O-tree has exactly permissible sectors.

###### Proof.

By definition of , each vertex in the underlying graph has degree . Then each -cycle vertex has degree in since it has the same degree as its corresponding vertex in the underlying graph.

Accordingly, any -cycle vertex has no less than sectors, whence its first two sectors never coincide with its last sector. Note that any has edges, sectors and -cycle vertices. Thus in , the set of permissible sectors is the set of all -sectors excluding all last sectors of vertices and all the first two sectors of -cycle vertices. Hence the number of permissible -sectors is given by . ∎

Let denote the set of O-trees with permissible, labeled sectors. By Proposition 5, and .

Let denote the set of O-trees with one labeled -cycle vertex. Since each O-tree has -cycle vertices, we have .

###### Lemma 3.

For , there exists a bijection between , the set of O-trees with one labeled -cycle vertex and , the set of O-trees with one permissible, labeled sector. Accordingly we have

 k|Rg,t,k|=(2g+t−k)|Rg,t,k−1|.
###### Proof.

Suppose we are given , and a -cycle vertex . has degree and is not a leaf.

We construct an O-tree from as follows: we apply Rémy’s bijection to the plane tree with respect to the non-leaf , i.e., contracting the edge connecting and its leftmost child. We obtain a plane tree together with a labeled sector . The correspondence between vertices in and those in gives us a canonical relabelling of elements of the permutation excluding the -cycle corresponding to . Let denote the permutation obtained from by deleting -cycle corresponding to and relabeling, see Figure 4.

We define the mapping

 Π:Rvg,t,k→R(1)g,t,k−1,(T,σ,v)↦(T′,σ′,s).

First we show that is well-defined. By construction, has edges and all cycles in have odd length, i.e., is an O-permutation. Further has elements, cycles of length and odd cycles of length . Hence . Let denote the vertex in , to which sector belongs to. is not the last sector around , since otherwise, by construction of Rémy’s bijection, the -cycle vertex in has degree at most two, a contradiction. If is a -cycle in , then the leftmost child of is a -cycle in and has degree at least in . By the way of contracting the edge connecting and and labeling the sector, is then not one of the first two sectors around . This shows that is permissible, whence and is well-defined.

To recover from , we apply the inverse of Rémy’s bijection to with respect to the sector , which is labeled and obtain a plane tree with non-leaf vertex . By construction of Rémy’s bijection and the definition of permissible sector, has degree . Set to be the permutation obtained from by adding the -cycle corresponding to and relabeling according to the correspondence between vertices in and those in , see Figure 4. It is clear that this is the inverse of , whence is bijective. ∎

By applying Lemma 3 successively, we derive

 |Rg,t,k| = 2g+t−kk|Rg,t,k−1| = 2g+t−kk⋅2g+t−k+1k−1⋅|Rg,t,k−2| = ⋯ = (2g+t−1k)|Rg,t,0|.

Accordingly, Lemma 3 induces a bijection from to :

###### Lemma 4.

For any , there exists a bijection from to , the set of O-trees with permissible, labeled sectors:

 |Rg,t,k|=(2g+t−1k)|Rg,t,0|.

Remark: given an O-tree, the number of -cycle vertices is bounded by .

In Figure 5, we show how to generate all and O-trees from O-trees.

Let

 κg,t=|Rg,t,0|22g=(2(2g+t−1))!22g(2g+t−1)!∑γ⊢gℓ(γ)=t1∏imi!(2i+1)mi,

where is a partition of with parts.

For , let be the number of shapes of genus with edges and denote the corresponding generating polynomial . Then

###### Lemma 5.

[12] For any , the generating polynomial of shapes is given by

 Sg(z)=g∑t=1κg,tz2g+t−1(1+z)2g+t−1.
###### Proof.

By Lemma 1, we have and furthermore

 22g6g−2⨄n=2gSg(n)≃6g−2⨄n=2gRg(n)=g⨄t=12g+t−1⨄k=0Rg,t,k.

Therefore, by Lemma 4,

 Sg(z) = 6g−2∑n=2g|Sg(n)|zn = 122gg∑t=12g+t−1∑k=0|Rg,t,k|z2g+t+k−1 = 122gg∑t=12g+t−1∑k=0(2g+t−1k)|Rg,t,0|z2g+t+k−1 = 122gg∑t=1|Rg,t,0|z2g+t−12g+t−1∑k=0(2g+t−1k)zk = 122gg∑t=1|Rg,t,0|z2g+t−1(1+z)2g+t−1 = g∑t=1κg,tz2g+t−1(1+z)2g+t−1.

###### Corollary 1.

We have

 (3.1) sg(n)=g∑t=1κg,t(2g+t−1n−(2g+t−1)),

where if or .

###### Proof.

By Lemma 5, we have

 6g−2∑n=2gsg(n)zn=g∑t=1κg,tz2g+t−1(1+z)2g+t−1=g∑t=12g+t−1∑i=0κg,t(2g+t−1i)z2g+t−1+i.

Set . By comparing both sides of the above identity, we obtain the corresponding formula for . ∎

###### Corollary 2.

The number is a positive integer.

###### Proof.

The positivity of is clear by definition. We proceed by induction on : assume that is an integer for and set . By eq. (3.1), we have

 sg(2g+t−1)=t∑j=1κg,j(2g+j−1t−j),

i.e.,

 κg,t=sg(2g+t−1)−t−1∑j=1κg,j(2g+j−1t−j).

Since and are integers for , is an integer. ∎

## 4. The coefficients κ⋆g(n)

Let denote the number of unicellular maps of genus with edges. In the following we derive an explicit formula for the generating function of unicellular maps of genus , which has the same coefficients as in the generating polynomial of shapes of genus in Lemma 5. This result has been observed in [12] by a different construction.

###### Lemma 6.

For any , the generating function of unicellular maps of genus is given by

 Cg(z)=g∑t=1κg,tz2g+t−1(1−4z)2g+t−12.
###### Proof.

Note that and , where . Thus . By Theorem 1, and we have

 ϵg(n) = 122gCat(n)g∑t=1(n+12g+t)ag,t = g∑t=1(2n)!22gn!(n+1−2g−t)!(2g+t)!ag,t.

Therefore using

 ∑n≥r−1(2n)!n!(n+1−r)!zn=(2(r−1))!(r−1)!zr−1(1−4z)r−12,

we compute

 Cg(z) = ∑n≥2gϵg(n)zn = ∑n≥2gg∑t=1(2n)!22gn!(n+1−2g−t)!(2g+t)!ag,tzn = g∑t=1ag,t22g(2g+t)!∑n≥2g(2n)!n!(n+1−2g−t)!zn = g∑t=1ag,t22g(2g+t)!⋅(2(2g+t−1))!(2g+t−1)!⋅z2g+t−1(1−4z)2g+t−12 = g∑t=1Cat(2g+t−1)ag,t22g⋅z2g+t−1(1−4z)2g+t−12 = g∑t=1κg,tz2g+t−1(1−4z)2g+t−12.

Let

 K⋆g(z)=3g−1∑n=2gκ⋆g(n)zn,

then [11] shows that

 Cg(z)=1√1−4z K⋆g(z1−4z).

In view of Lemma 1 and Lemma 6 this provides the following combinatorial interpretation of :

###### Theorem 2.

, where and counts the shapes of genus , which correspond to via the bijection in Lemma 1.

In [2], has been shown to have the form

 Cg(z)=Pg(z)(1−4z)3g−12,

where is a polynomial with integer coefficients.

Combining this with Lemma 6, we obtain an explicit formula for the polynomials in terms of :

###### Corollary 3.

For any , the polynomial is given by

 Pg(z)=g∑t=1κg,tz2g+t−1(1−4z)g−t.

Combining Lemma 5 and Lemma 6, we also derive the following functional relation between and

###### Corollary 4.

For , we have

 (4.1) Cg(z)=1+zC0(z)21−zC0(z)2Sg(zC0(z)21−zC0(z)2),

where the generating function of plane trees with edges is given by

This functional relation can also be derived via symbolic methods [9]. More precisely, we can construct a general unicellular map from a shape by first replacing each edge by a path and then attaching a plane tree to each sector.

We shall proceed by giving a bijective proof of a recurrence of .

###### Proposition 6.

For any , there exists a bijection

 Og,t,0≃(2g+t−1)(2g+t−2)(Og−1,t,0+