Analogue of the Harer-Zagier formula

# An analogue of the Harer-Zagier formula for unicellular maps on general surfaces

## Abstract.

A unicellular map is the embedding of a connected graph in a surface in such a way that the complement of the graph is simply connected. In a famous article, Harer and Zagier established a formula for the generating function of unicellular maps counted according to the number of vertices and edges. The keystone of their approach is a counting formula for unicellular maps on orientable surfaces with  edges, and with vertices colored using every color in  (adjacent vertices are authorized to have the same color). We give an analogue of this formula for general (locally orientable) surfaces.

Our approach is bijective and is inspired by Lass’s proof of the Harer-Zagier formula. We first revisit Lass’s proof and twist it into a bijection between unicellular maps on orientable surfaces with vertices colored using every color in , and maps with vertex set  on orientable surfaces with a marked spanning tree. The bijection immediately implies Harer-Zagier’s formula and a formula by Jackson concerning bipartite unicellular maps. It also shed a new light on constructions by Goulden and Nica, Schaeffer and Vassilieva, and Morales and Vassilieva. We then extend the bijection to general surfaces and obtain a correspondence between unicellular maps on general surfaces with vertices colored using every color in , and maps on orientable surfaces with vertex set  with a marked planar submap. This correspondence gives an analogue of the Harer-Zagier formula for general surfaces. We also show that this formula implies a recursion formula due to Ledoux for the numbers of unicellular maps with given numbers of vertices and edges.

Supported by the ANR project A3 and the European project ExploreMaps - ERC StG 208471.

## 1. Introduction

A map is a cellular embedding of a connected graph on a surface considered up to homeomorphism (see Section 3 for definitions). A planar map is a map on the sphere. A map is said unicellular if it has a single face. For instance, the planar unicellular maps are the plane trees. A map is said orientable if the underlying surface is orientable.

In [7] Harer and Zagier considered the problem of enumerating orientable unicellular maps according to the number of edges and vertices (or, equivalently by Euler formula, according to the number of edges and the genus). The keystone of their approach is an enumerative formula for unicellular maps with colored vertices (these colorings are not necessarily proper, that is, two adjacent vertices can have the same color): they proved that the number of rooted unicellular maps on orientable surfaces having edges and vertices colored using every color in is

 (1) Tn(q)=2q−1(nq−1)(2n−1)!!,

where .

From (1), it follows that the numbers of unicellular maps on orientable surfaces with edges and vertices satisfy

 (2) n+1∑v=1ϵv(n)Nv=n+1∑q=1(Nq)2q−1(nq−1)(2n−1)!!.

since both sides of the equation represent the number of unicellular maps with vertices colored using some of the colors in (since maps with edges have at most vertices, and the index on the right-hand-side corresponds to the number of colors really used). From this equation, Harer and Zagier obtained a recurrence relation for the numbers of orientable unicellular maps with edges and vertices:

 (n+1)ϵv(n)=(4n−2)ϵv−1(n−1)+(n−1)(2n−1)(2n−3)ϵv(n−2).

The proof of (1) in [7] used a matrix integral argument (see [9] for an accessible presentation). A combinatorial proof was later given by Lass [10], and subsequently a bijective proof was given by Goulden and Nica [6].

The goal of this paper is to give an analogue of Equation (1) for unicellular maps on general (i.e. locally orientable) surfaces, and to describe the bijections hiding behind it (see Equation (4) below). In order to do so, we first simplify slightly the proof of Lass for the orientable case, and then show how to extend it in order to deal with the general case.

In Section 3, we revisit Lass’s proof of (1) in order to obtain a bijection between orientable unicellular maps colored using the every colors in and orientable tree-rooted maps (maps with a marked spanning tree) with vertex set . This bijective twist given to Lass’s proof turns out to simplify certain calculations because tree-rooted maps are easily seen to be counted by the right-hand-side of (1). As mentioned above, a bijection was already exhibited by Goulden and Nica for unicellular maps on orientable surfaces in [6]. However, this bijection is presented in terms of the permutations (in particular the image of the bijection is not expressed in terms of maps) which makes its definition and analysis more delicate. By contrast, it is immediate to see that the bijection satisfies the following property:

1. for any unicellular map colored using every colors in and for any colors , the number of edges of with endpoints of color is equal to the number of edges between the vertices in the tree-rooted map .

This property leads in particular to a bijective proof of the following formula by Jackson [8] (closely related to another formula found independently by Adrianov [1]) for the number of orientable bipartite unicellular maps with the sets of vertices colored with colors and colors respectively forming the bipartition (the root-vertex being colored in ):

 (3) Tn(p,q)=n!(n−1p−1,q−1,n−p−q+1).

A bijective proof of (3) was already given by Schaeffer and Vassilieva [15]. Again, the presentation there is in terms of permutations. In [13], Morales and Vassilieva analyzed the bijection in [15] in order to enumerate bipartite unicellular maps colored in according to the sum of degrees of the vertices of each color. These refined enumerative results can be obtained using the property (). Thus, in the orientable case, the bijection gives a unified ways of obtaining the bijective results in [6, 15, 13].

In Section 4, we turn to general surfaces. We extend the bijection into a correspondence between unicellular maps colored using every color in and orientable planar-rooted maps (maps with a marked spanning connected planar submap1) with vertex set . One of the new ingredients used to prove the correspondence is a bijection between planar maps and certain “decorated” plane trees (which is closely related to a bijection by Schaeffer [14] for Eulerian planar maps). Proofs concerning this bijection are given in Section 5. The correspondence between colored unicellular maps and planar-rooted maps is not one-to-one, but rather -to-, where is the number of faces of the planar submap. This yields the following formula for the number of rooted unicellular maps colored using every color in :

 (4) Un(q)=n−q+2∑r=1q!r!2r−1Pq,r(2n2q+2r−4)(2n−2q−2r+1)!!,

where is the number of (unlabeled) rooted planar maps with vertices and faces. More refined enumerative results (in the spirit of [13]) can be obtained from the correspondence because this bijection satisfies:

1. if a unicellular map and a planar-rooted map are in correspondence by , then for any color the sum of the degrees of the vertices colored in is equal to the degree of the vertex in .

Let us situate our work among some other enumerative results for unicellular maps on general surfaces. In [5] Goulden and Jackson (using representation theory techniques) gave a way of expressing the generating functions of maps on general surfaces in terms of matrix integrals over the Gaussian Orthogonal Ensemble. This expression allowed them to prove a formula for the number of unicellular maps on general surfaces with vertices colored using some of the colors in :

 (5) ^Un(N)=n!n∑k=022n−kn∑r=0(n−12n−r)(k+r−1k)(N−12r)+(2n−1)!!N−1∑q=12q−1(N−1q)(nq−1),

where for any real number and any integer the notation stands for . Given that , it is natural to try to prove (4) starting from (5) or the converse, but we have failed to do so. In [11], Ledoux showed that the matrix integral considered by Goulden and Jackson satisfies a differential equation, implying a recursion formula for the numbers of unicellular maps with edges and vertices:

 (6) (n+1)ηv(n)=(4n−1)(2ηv−1(n−1)−ηv(n−1))+(2n−3)((10n2−9n)ηv(n−2)+8ηv−1(n−2)−8ηv−2(n−2))+5(2n−3)(2n−4)(2n−5)(ηv(n−3)−2ηv−1(n−3))−2(2n−3)(2n−4)(2n−5)(2n−6)(2n−7)ηv(n−4),

valid for all . We explain in Section 6 how to prove this relation starting from (4).

We mention lastly that a different bijective approach for orientable unicellular maps was introduced by Chapuy [4] (giving (2) and some new results). In [2], this approach was partially adapted to general surfaces, giving an exact formula for the number of unicellular precubic maps (maps with vertices of degree either 1 or 3). However, no exact formula for non-precubic unicellular maps on general surfaces has been derived so far by this method.

## 2. Unicellular maps and Eulerian tours

In this section we set our definitions and extend the correspondence used in [10] between colored unicellular maps and certain Eulerian tours on graphs.

Our graphs can have loops and multiple edges. A graph with edges is said edge-labelled if its edges are labelled with distinct numbers in . A map is a cellular embedding of a connected graph in a (2-dimensional, smooth, compact, boundaryless) surface considered up to homeomorphism. By cellular we mean that the faces (connected components of the complement of the graph) are simply connected. Thus, maps are the gluings of polygons obtained by identifying the edges of the polygons two by two, considered up to homeomorphism. In particular, the unicellular maps (maps with a single face) with edges are the gluings obtained from a -gon (polygon with edges); see Figure 1. A polygon is rooted by orienting one of its edges and calling it the root-edge. A rooted-map is obtained by gluing polygons one of which is rooted; this distinguish one of the edges as the root-edge and specifies a direction and a side for the root. For instance, the root-edge, its side and direction are indicated by an arrow labelled 1 in Figure 1. The root-vertex is the origin of the root-edge.

Given two edges of a polygon, there are two ways of gluing them together: we call orientable gluing the identification of the edges giving a topological cylinder, and non-orientable gluing the identification giving a Möbius band. It is easy to see that a unicellular map obtained by gluing the edges of a polygon in pairs is orientable if and only if each of the gluing is orientable. Hence there are rooted unicellular maps with edges and orientable ones (since is the number of ways of partitioning a set of elements in non-ordered pairs). Observe that if the edges of the rooted -gon are oriented into a directed cycle, then the orientable gluings are the one for which any pair of edges glued together have opposite orientations along the gluing.

We call -colored a map in which the vertices are colored using every color in (adjacent vertices can have the same color). A key observation made in [10] is that making the tour of the face of a -colored unicellular map starting from the root induces a tour on a graph having vertex set . We now make this statement precise. Les be an undirected edge-labelled graph. We call bi-Eulerian tour of a directed path starting and ending at the same vertex called the origin and using every edge exactly twice. If has loops, the convention is that tours can either use the loop twice in the same direction, or in the two opposite directions (so that there are 2 different bi-Eulerian tours on the graph consisting of a single loop, and more generally, there are bi-Eulerian tours on an edge-labelled graph consisting of loops).

###### Lemma 1.

There is a bijection between rooted edge-labelled -colored unicellular maps with edges and the set of pairs , where is an edge-labelled connected graph with edges and vertex set , and is a bi-Eulerian tour of . Moreover, if then

• the unicellular map is orientable if and only if the tour never uses an edge twice in the same direction,

• for any colors in the number of edges in having endpoints of colors is equal to the number of edges in having endpoints .

Lemma 1 is illustrated by Figure 2.

###### Proof.

A -colored unicellular map is obtained by coloring the vertices of a rooted -gon with every color in and then choosing a gluing of the edges respecting the colors. We consider the orientation of the -gon into a directed cycle respecting the orientation of the root-edge. Choosing a coloring of the vertices of the -gon using every colors in is the same as choosing a connected directed graph with vertex set together with an Eulerian tour of (a directed path starting and ending at the same vertex and using every edge twice). Indeed, the th edge of the Eulerian tour identifies with the th edge around the -gon (starting from the root-edge).

Now, given a colored -gon and the associated pair , the gluings of the edges of the polygon respecting the colors are clearly in bijection with the partitions of the arcs of into pairs of arcs having the same endpoints, with the convention that two loops incident to the same vertex can be paired in two ways (either paired “in opposite direction” or “in the same direction”). Also, labelling the edges of the unicellular map obtained is the same as labelling the pairs in . Lastly, the triples clearly identify with the pairs , where is an edge-labelled connected graph with edges and vertex set , and is a bi-Eulerian tour of . The bijection established has all the claimed properties. ∎

## 3. Orientable unicellular maps

In this section, we twist Lass’s proof [10] of (1) into a bijection between colored unicellular maps and tree-rooted maps. The bijection obtained can be seen as the (unified) “map version” of the bijections presented in terms of permutations in [6] and [15] to prove respectively (1) and (3).

We first recall the so-called BEST Theorem. An Eulerian tour of a directed graph is a directed path starting and ending at the same vertex called the origin and using each arc exactly once. A spanning tree of is directed toward a vertex if the path between any vertex vertex and in is directed toward .

###### Lemma 2 (BEST Theorem).

Let be an edge-labelled directed graph such that any vertex is incident to as many ingoing and outgoing arcs, and let be a vertex. Then, the set of Eulerian tours of with origin is in bijection with the set of pairs , where is a spanning tree directed toward , and is a choice for each vertex of a total order of the outgoing arcs not in .

The proof of the BEST Theorem can be found in [16]2. Following the method of Lass we will now use the BEST Theorem to count bi-Eulerian tours. But we will shortcut this counting by considering rotation systems of graphs. Let be a graph. A half-edge is an incidence between an edge and a vertex (one can think of half-edges as obtained from cutting an edge at a midpoint); by convention a loop corresponds to two half-edges both incident to the same vertex. A rotation system of is a choice for each vertex of a cyclic ordering of the half-edges incident to . A rooted rotation system with root-vertex is a choice for each vertex of a cyclic order of the half-edges incident to and the choice of a total order of the half-edges incident to . We now recall how to associate a rooted rotation system to a rooted orientable map . Given a rooted map of underlying graph on an orientable surface , one can define the “positive orientation” of the surface by requiring that the marked side of the root-edge precedes (the outgoing half-edge of) the root-edge in counterclockwise order around the root-vertex. Then one defines the rotation system of as follows: around any non-root vertex the rotation system is the cyclic order of the half-edges in counterclockwise direction around , and around the root-vertex the rotation system is the total order obtained by breaking the cyclic counterclockwise order in order to make the outgoing half-edge of the root-edge be the smallest element. The proof of the following classical result can be found in [12].

###### Lemma 3 (Embedding Theorem).

For any connected graph , the mapping is a bijection between the rooted orientable maps of underlying graph and the rooted rotation systems of .

We are now ready to state the relation between bi-Eulerian tours and tree-rooted maps (rooted maps with a marked spanning tree).

###### Lemma 4.

Let be a connected edge-labelled undirected graph. The set of bi-Eulerian tours of that never take an edge twice in the same direction are in bijection with pairs made of a spanning tree and a rooted rotation system . Consequently, this set of bi-Eulerian tours is in bijection with the set of tree-rooted maps on orientable surfaces having underlying graph .

###### Proof.

Let be a vertex of . We will establish a bijection between the set of bi-Eulerian tours of of origin that never take an edge twice in the same direction and the set of pairs made of a spanning tree and a rooted rotation system with root-vertex .

Let be the directed graph obtained from by replacing each edge by two arcs in opposite directions taking the same label as . Observe that the arcs of are all distinguishable except for pairs of arcs corresponding to a loop of . The set of bi-Eulerian tours of is clearly in bijection with the set of Eulerian tours of having origin . Hence, by the BEST Theorem, the set is in bijection with the set of pairs , where is a spanning tree of directed toward and is a choice of a total ordering of the outgoing arcs of not in at each vertex. Again here if two arcs comes from a loop of , the total order does not distinguish between these two arcs.

It only remains to show that the set is in bijection with the set . First of all, the spanning trees of directed toward are in natural bijection with the spanning trees of . Moreover, the arcs of and the half-edges of are in natural correspondence (with arcs of indistinguishable if and only if they come from the same loop, and half-edges of indistinguishable if and only if they come from the same loop). Therefore, for the vertex , the total orderings of the outgoing arcs of incident to correspond bijectively to the total orderings of the half-edges of incident to . Similarly, for any vertex and any spanning tree directed toward , the total orderings of the outgoing arcs of not in (that is, the total orderings of all the outgoing arcs except the one leading to its parent in ) corresponds bijectively to the cyclic orderings of the half-edges of incident to . This gives a bijection between and . ∎

We now summarize the consequences of Lemmas 1 and 4.

###### Theorem 5.

There is a bijection between -colored rooted unicellular maps with edges on orientable surfaces and tree-rooted maps with edges and vertex set on orientable surfaces. Moreover for all the number of edges with endpoints of colors in a -colored unicellular map is equal to the number of edges between vertices and in the tree-rooted map .

###### Proof.

Lemmas 1 and 4 imply a bijection having the claimed properties between edge-labelled maps. However, the edges of a non-labelled rooted maps are all distinguishable. Hence each non-labelled rooted maps with edges correspond to different edge-labelled rooted maps. ∎

Theorem 5 is illustrated in Figure 3. Before closing this section we explain how to obtain (1) and (3) from Theorem 5. We start with the proof of (1). By Theorem 5, is the number of tree-rooted maps with edges and vertex set on orientable surfaces. One can think of a tree-rooted maps in terms of its rotation system: one can first choose the tree together with its rooted rotation system (that is, a rooted plane tree) and then add the other edges. It is well known that the rooted plane trees with unlabeled vertices are counted by the Catalan number , hence the number of trees with vertex set is . In order to add the edges not in the tree, we need first to choose where to insert the half-edges around the tree (equivalently, how to complete the rooted rotation system of the tree with new half-edges). A rooted plane tree with vertices has positions where a half-edge can be inserted, hence there are ways to add the additional half-edges around the tree. Lastly there are ways of pairing these half edges. This gives (1):

 Tn(q)=(2q−2)!(q−1)!(2n2q−2)(2n−2q+1)!!=2q−1(nq−1)(2n−1)!!.

We prove (3) in a similar fashion. By Theorem 5, is the number of (bipartite) tree-rooted maps on orientable surfaces with edges and vertex set such that any edge joins a vertex in to a vertex in , and the root-vertex is in . It is well known that the bicolored trees with black unlabeled vertices, and white unlabeled vertices (with a black root-vertex and every edge joining a black vertex to a white vertex) are counted by the Narayama number . Hence, there are ways to choose a vertex-labelled tree such that any edge joins a vertex in to a vertex in , and the root-vertex is in . The other edges have one half-edge incident to a vertex in and one half-edge incident to a vertex in . There are positions to insert a half-edge around vertices in , and positions to insert a half-edge around vertices in . Hence, there are ways to add the half-edges around the tree. Lastly, there are ways of pairing these half-edges. This gives (3).

## 4. General unicellular maps

In this section we extend the bijection to unicellular maps on general surfaces. We start with some definitions. A balanced orientation of an undirected graph is an orientation of a (possibly empty) subset of edges of such that for any vertex the numbers of ingoing and outgoing half-edges incident to are equal. If has a vertex distinguished as the root-vertex, then we say that a balanced orientation and a spanning tree are compatible if any oriented edge in has an orientation that coincides with the orientation of toward . The external weight of the pair is the number of oriented edges not in . We call compatibly-oriented tree-rooted map, a rooted map with a marked spanning tree and a compatible balanced orientation. For an edge-labelled graph , we denote by the set of edge-labelled compatibly-oriented tree-rooted maps with underlying graph , and we denote by the set of bi-Eulerian tours of .

###### Lemma 6.

Let be a connected edge-labelled undirected graph. There exists a surjective mapping between and satisfying: for any bi-Eulerian tour in there is an integer such that the preimage consists of compatibly-oriented tree-rooted maps of external weight . Moreover if and only if the bi-Eulerian tour never takes an edge twice in the same direction

###### Proof.

Observe that to any bi-Eulerian tour of , one can associate a balanced orientation of by orienting the edges of which are taken twice in the same direction during the tour according to this direction. We now fix a vertex and a balanced orientation and show that there is a mapping with the desired properties between set of compatibly-oriented tree-rooted maps with root-vertex and balanced orientation , and the set of bi-Eulerian tours with origin such that .

Let be the directed graph obtained by replacing each edge of by two arcs with the same label as : the non-oriented edges of are replaced by two arcs in opposite directions, while the oriented edges of are replaced by two arcs in the direction of . In the two arcs that come from a loop of or from an oriented edge of are indistinguishable. Clearly the set of bi-Eulerian tours of is in bijection with the set of Eulerian tours of origin on . Hence, by the BEST Theorem, the set is in bijection with the set of pairs made of a spanning tree of directed toward and a choice for each vertex of a total order on the outgoing arcs not in incident to . Again here if two arcs comes from a loop of or from a directed edge of , the spanning tree and total orders do not distinguish between these two arcs.

It remains to establish the relation between and . Observe first that the spanning trees of directed toward are in bijection with the spanning trees of that are compatible with . Let be a spanning tree of directed toward and let be the external weight of the associated spanning tree of . We only need to show that the number of rooted rotation systems of with root-vertex is times the number of choices for each vertex of of a total order on the outgoing arcs not in . We first consider a vertex . The half-edges of incident to are all distinguishable, except for pairs of half-edges belonging to a non-oriented loop. Hence, there are different cyclic orderings of the half-edges incident to , where is the degree of and is the number of incident non-oriented loops. On the other hand, there are outgoing arcs incident to not in , and these arcs are all distinguishable except for pairs of half-edges belonging to a loop or to an oriented edge of not in . Hence, there are different total orderings of the outgoing arcs not in incident to , where is the number of edges of not in oriented away from . A similar reasoning applies for the root-vertex , and shows that there are times more rooted rotation systems of with root-vertex than of choices for each vertex of a total order on the outgoing arcs not in . Moreover, the sum is equal to the external weight . Lastly, the external weight is 0 if and only if there is no oriented edge in (equivalently the tours in never take an edge twice in the same direction). ∎

Putting Lemmas 1 and 6 together (and getting rid of the edge-labellings as in the previous section) one gets:

###### Theorem 7.

Let be a positive integer. There is a surjective mapping from the set of compatibly-oriented tree-rooted maps with vertex set on orientable surfaces to the set of -colored unicellular maps on general surfaces satisfying: for all -colored unicellular map ,

• there is an integer such that the preimage consists of compatibly-oriented tree-rooted maps of external weight ; moreover if and only if is orientable,

• for all the number of edges with endpoints of colors in is equal to the number of edges with endpoints in any compatibly-oriented tree-rooted map in .

We now proceed to count compatibly-oriented tree-rooted maps according to their external weight. We first focus on the simpler structure of near-Eulerian trees. A near-Eulerian tree is a rooted plane tree with a (possibly empty) subset of edges oriented toward the root-vertex, together with some dangling half-edges, called buds, which are either ingoing or outgoing (at each vertex the rotation system includes both the half-edges in the tree and the buds) such that that at any vertex the numbers of ingoing and outgoing half-edges (including the buds) are equal. An example of near-Eulerian tree is given in Figure 4(left). It is easily seen that near-Eulerian trees have the same numbers of ingoing and outgoing buds; we call external weight this number.

Observe that near-Eulerian trees of external weight are the maps obtained from compatibly-oriented tree-rooted maps by deleting all the non-oriented edges not in the spanning tree, and cutting in two halves the oriented edges not in the spanning tree. We will now exhibit a bijection between near-Eulerian trees and rooted plane maps, that is, rooted planar maps (in the sphere) with a marked face3. It will be convenient to identify rooted plane maps with rooted maps embedded in the plane (the marked face become the infinite face of the plane). If a near-Eulerian tree is embedded in the plane, an edge drawn from an outgoing bud to an ingoing bud of is said counterclockwise around if it has the infinite face on its right. There is a unique way to draw a set of non-crossing counterclockwise edges around joining each outgoing bud to an ingoing bud; see Figure 4. We call this process the closure of , and denote by the rooted plane map obtained.

###### Theorem 8.

The closure is a bijection between near-Eulerian trees and rooted plane maps.

The proof of Theorem 8 is delayed to the next section. We now state its consequences for unicellular maps. A spanning submap of a map is a subset of edges giving a connected subgraph containing every vertex. A spanning submap of an orientable map is said planar if the rotation system of inherited from makes a planar map. A planar-rooted map is a rooted map on an orientable surface with a marked spanning planar submap. A -externally-labelled planar-rooted map is a pair where is a planar-rooted map whose marked submap has faces, and is a permutation of .

###### Theorem 9.

Let be a positive integer. There exists a surjective mapping from the set of externally-labelled planar-rooted maps with vertex set on orientable surfaces, to the set of -colored rooted unicellular maps on general surfaces satisfying: for all -colored unicellular map ,

• there is an integer such that the preimage consists of -externally-labelled planar-rooted maps; moreover if and only if is orientable,

• for all the sum of degrees of the vertices of colored is equal to the degree of the vertex in any planar-rooted map in .

Theorem 9 is illustrated in Figure 5. Observe that tree-rooted maps can be seen as 1-externally-labelled planar-rooted maps, and that the mapping given in Theorem 9 can be seen as the extension of the bijection given in Theorem 5 for orientable unicellular maps.

###### Proof.

Let us call near-Eulerian tree-rooted maps of external weight a map obtained by adding some non-oriented edges to a near-Eulerian tree of external weight . Near-Eulerian tree-rooted maps are obtained from compatibly-oriented tree-rooted maps by cutting in two halves the oriented edges not in the spanning tree. This cutting process is -to-1 between compatibly-oriented tree-rooted of external weight and near-Eulerian tree-rooted maps of external weight (since there is of pairing ingoing buds with outgoing buds). Moreover, the cutting process is degree preserving (the degree of any vertex is preserved).

The bijection between near-Eulerian trees and plane maps can be seen as a -to-1 correspondence between near-Eulerian trees with external weight and rooted planar maps with faces (since plane maps are planar maps with a marked face). Moreover, the correspondence can easily be extended to a degree-preserving -to-1 correspondence between near-Eulerian tree-rooted maps of external weight and planar-rooted maps whose planar submap has faces. Indeed, one can just apply the closure to the near-Eulerian tree-rooted maps by ignoring the non-oriented edges not in the spanning tree (see Figure 5). Hence combining the cutting process and the closure gives a degree-preserving -to-1 correspondence between compatibly-oriented tree-rooted maps of external weight and planar-rooted maps whose planar submap has faces. Combining this correspondence with Theorem 7 completes the proof. ∎

Theorem 9 gives the following enumerative results.

###### Corollary 10.

The number of -colored rooted unicellular maps with edges on general surfaces is

 (7) Un(q)=n−q+2∑r=1q!r!Pq,r2r−1(2n2q+2r−4)(2n−2q−2r+3)!!,

where is the number of unlabeled rooted planar maps with vertices and faces. Consequently, the series and are related by:

 (8) U(t,w)=12exp(t2)Q(t2,2w).
###### Proof.

By Theorem 9, the proof of (7) amounts to showing that there are planar-rooted maps with edges, vertex set and such that the submap has faces. One can first choose the planar submap, and then insert the additional edges. There are choice for the planar submap with vertex set and faces. The planar submap has edges by Euler relation, hence there are position to insert a half-edge around it. Thus there are ways of inserting the additional half-edges around the submap and then ways of pairing them together. This proves (7). From this equation, one gets

 U(t,w) = ∑n≥0,q>0tnwqn+q−2∑r=1q!r!Pq,r2n−q+1(2q+2r−4)!(n−q−r+2)! = 12∑q>0wq∑r>0q!r!2qPq,r(2q+2r−4)!∑n≥q+r−2tn2n(n−q−r+2)!.

Using gives (8). ∎

We do not know any closed formula for the numbers . However, standard techniques dating back to Tutte [17] (recursive decompositions of maps and the quadratic method; see e.g. [3]) give the following algebraic equation:

 (9) 27P4−(36x+36y−1)P3+(24x2y+24xy2−16x3−16y3+8x2+8y2+46xy−x−y)P2+xy(16x2+16y2−64xy−8x−8y+1)P−x2y2(16x2+16y2−32xy−8x−8y+1)=0

characterizing the series . As explained in Section 6, the algebraic equation (9) yields differential equations characterizing the series and .

## 5. Proof of the bijectivity of Γ.

This section is devoted to the proof of Theorem 8. Recall that a rooted plane map is a rooted map drawn in the plane (this is equivalent to a planar map on the sphere with a marked face). The infinite face of a plane map is called outer face and the others are called inner faces. A dual-path for a face of a plane map is a path in the plane starting inside and ending in the outer face which avoids the vertices of ; its length is the number of edges of crossed by the path. The dual-distance of a face is the minimal length of a dual-path for . The dual distance of the inner faces are represented in Figure 6. The dual-distance orientation of is defined as the orientation of the subset of edges of which are incident to two faces of different dual-distances in such a way that the face with largest dual-distance lies on the left of the oriented edge . The dual-distance orientation is shown in Figure 6.

###### Lemma 11.

The dual-distance orientation of a plane map is such that, at any vertex the numbers of incident ingoing and outgoing half-edges are equal.

###### Proof.

Turning around in counterclockwise direction the dual-distance increases by 1 each time an outgoing half-edge is crossed and decreases by 1 each time an ingoing half-edge is crossed. Hence the number of ingoing and outgoing half-edges crossed must be equal for a complete tour around . ∎

Let be a rooted plane map. For a face of dual-distance 1, we consider the edges incident to both and the outer face. By deleting these edges one gets connected components, one of which contains the root-vertex. We call breakable the edge incident to both and the outer face that follows when turning around in counterclockwise direction. The breakable edges of a map are represented in Figure 6. We now consider the dual-distance orientation of . Observe that all the breakable edges of are oriented, hence cutting the breakable edges in two halves produces an ingoing and an outgoing bud. Moreover, cutting all the breakable edges of does not disconnects . Indeed, no cocycle of contains only breakable edges since a cocycle containing a breakable edge of a face also contains another edge of which is either not incident to the outer face (hence not breakable) or incident to the outer face but not breakable (since it is not the breakable edge of ). Thus cutting all the breakable edges of produces a connected map with some ingoing and outgoing buds and decreases by 1 the dual-distance of every inner face. If has some inner faces, then we can define the breakable edges of as before and consider the map with buds resulting from cutting these edges, etc. We denote by the rooted plane tree with partial orientation and buds resulting from this “opening” process.

###### Lemma 12.

For any rooted plane map , the opened map is a near-Eulerian tree and .

###### Proof.

Let . By Lemma 11, each vertex of is incident to as many ingoing and outgoing half-edges. We now consider an edge of , and want to prove that it is not oriented in the wrong direction in . By definition, if is oriented, then is is incident to two faces of different dual distances in . We consider the step of the opening process at which both sides of became equal to the outer face. Just before this step, the edge was incident to the outer face and to an inner face of dual-distance 1 (and had on its left by definition of the dual-distance orientation). Since is in , it means that is not the breakable edge of . Therefore, the definition of the breakable edge of implies that cutting this breakable edge makes a bridge oriented toward the component containing the root-vertex. Hence the orientation of coincides with the orientation toward the root-vertex of . Thus, is a near-Eulerian tree.

We now need to show that . Clearly the pairs of outgoing and ingoing buds of which came from cutting breakable edges of can all be joined in a non-crossing way (since is planar). Moreover, it is clear from the definition of the dual-distance orientation that any breakable edge is counterclockwise around . Thus by definition of the closure . ∎

To complete the proof of Theorem 8 it only remains to prove that the closure  is injective. Let be a plane map and let be a near-Eulerian tree such that . The tree induces an orientation of a subset of edges of : the subset of oriented edges of (oriented toward the root-vertex of ), together with the set of edges formed during the closure (oriented from the outgoing bud to the ingoing bud). We call this partial orientation the orientation of induced by . We now show that any induced orientation of is equal to the dual-distance orientation.

###### Lemma 13.

Let be a near-Eulerian tree and let be its closure. The orientation induced by coincide with the dual-distance orientation of . Moreover, for any inner face of , there is a dual-path for of minimal length (equal to the dual-distance of ) which does not cross any edge of .

###### Proof.

We call -distance of a face of the minimal length of a dual-path for which does not cross any edge of . We also define the -distance orientation of as the orientation of the subset of edges of which are incident to two faces of different -distances in such a way that the face with largest -distance lies on the left of the oriented edge .

We first show that the -distance orientation coincides with the orientation induced by . Let be an edge not in . Clearly, the edge separates two faces whose -distances differ by 1. Moreover, the -distance orientation of coincides with the orientation induced by : in both of these orientations the face with largest -distance is on the left of (since is counterclockwise around ). We now consider an edge in the tree . Deleting gives two subtrees with buds, say an with containing the root-vertex. Let be the number of outgoing buds in paired with ingoing buds in during the closure of . The number of outgoing buds in paired with ingoing buds in is equal to if is non-oriented and is equal to if is oriented in (since is oriented toward in this case). Since all the pairing are counterclockwise around , the -distance of the two faces of incident to are respectively and . Hence the -distance orientation of coincides with the orientation induced by : the edge is oriented if and only if , and in this case the face of largest -distance is on the left of .

We have shown that the orientation induced by coincides with the -distance orientation. Thus it only remain to show that -distance is equal to the dual-distance. Let be a face of and let be respectively its dual-distance and its -distance. Clearly, . We now consider a dual-path of minimal length for . The dual-path starts inside the face which has -distance and ends in the outer face which has -distance 0. Moreover, we have shown above that for any edge of the difference of -distance between the two faces separated by is at most 1. Hence, the -distance decreases by at most 1 each time the path crosses an edge. This shows that crosses at least edges, so that . ∎

We now complete the proof of the injectivity of . Let be a plane rooted map, and let be a near-Eulerian tree such that . We want to prove that . Given Lemma 13, it suffices to prove that and have the same edges (since the orientation of the edges and buds of are uniquely determined by the dual-orientation of ). We denote , and for we denote by the map with buds obtained after steps of the opening process. We will now show by induction on that the edges of are in for all  (hence that has the same edges as ). The case is obvious, and to prove the induction step we can suppose that the edges of are in and show that none of breakable edges of is in . Let be a face of dual-distance 1 in . Let be the set of edges incident to both and the outer face of . The second statement in Lemma 13 shows that at least one edge in is not in (because is the set of edges incident to and faces of smaller dual-distance, so that any dual-path for of minimal length crosses one of these edges). We call the edge which is not in and observe that the edges in are all in (because cutting two edges in would disconnect hence also ). Moreover, the first statement in Lemma 13 shows that the dual-distance orientation of the edges in coincide with the orientation of toward its root-vertex. Therefore, the only possibility is that is the breakable edge of . This shows that the breakable edge of is not in . This proves the induction step, thus . This completes the proof of the injectivity, hence bijectivity of .

## 6. Recursion formula for unicellular maps

In this section we sketch a proof of Ledoux’s recursion formula (6) starting from (8). Some of the calculations require a computer algebra system (or a lot of patience).

The number of unicellular maps with edges and vertices colored using some colors in can be seen as a polynomial whose coefficients are the numbers : . Hence, proving (6) amounts to proving for all .

 (n+1)^Un(x)=(4n−1)(2x−1)^Un−1(x)+(2n−3)(10n2−9n+8x−8x2)^Un−2(x)+5(2n−3)(2n−4)(2n−5)(1−2x)^Un−3(x)−2(2n−3)(2n−4)(2n−5)(2n−6)(2n−7)^Un−4(x).

Equivalently, one has to prove for all ,

 (10) −(2n)(2n−1)(2n−2)(n+1)^Vn(x)+(4n−1)(2n−2)(2x−1)^Vn−1(x)+(10n2−9n+8x−8x2)^Vn−2(x)+5(1−2x)^Vn−3(x)−2^Vn−4(x)=0,

where for all . We now translate (10) in terms of the numbers . Recall that (because for any positive integer , counts unicellular maps with edges and vertices colored using some colors in , the index representing the number of colors really used).

###### Lemma 14.

Let be polynomials in . The polynomial decomposes on the basis and we denote by the coefficients: . Let be bivariate polynomials. Then, the polynomials satisfy a recursion formula

 (11) for all n≥0,    k∑i=0ci(n,x)Rn−i(x)=0,

(with the convention for ) if and only if the formal power series satisfies

 (12) k∑i=0ci(Δt,Δw)(tiR(t,w))=0,

where and are the operators on formal power series defined by

 Δt:F(t,w)↦t∂∂tF(t,w),     Δw:F(t,w)↦w∂∂w((1+w)F(t,w)),

and is the operator obtained by composing these two commuting operators according to .

###### Proof.

Pascal’s rule for binomial coefficients implies that for any polynomial , one has . This relation gives the equivalence between (11) and a (linear) recursion formula for the numbers with coefficients in . Writing this relation in terms of operators acting on gives (12). ∎

By Lemma 14, Formula (10) is equivalent to a linear differential equation on the power series . This series is equal to by (8), hence by factorizing out in the differential equation one obtains a (linear) differential equation for . The differential equation obtained for is:

 (6tw+4w2−36t−7w−6)Q(t,w)−(12t2+8tw+7w−25)∂∂tQ(t,w)+w(w+2)(8w+6t−7)∂∂wQ(t,w)+(2t2−4tw+37t+9)∂2∂t2Q(t,w)−w(w+2)(8t+7)∂2∂w∂tQ(t,w)+2w2(w+2)2∂2∂w2Q(t,w)+t(8t+11)∂3∂t3Q(t,w)−4wt(w+2)∂3∂w∂t2Q(t,w)+2t2∂4∂t4Q(t,w)=0.

In terms of coefficients, this equation reads: for all integers

 (13) 2(q+1)(q+2)^Pq,r+3+6(q+2)^Pq+1,r+1−(q+2)(7+8r)^Pq+1,r+2−(4q+4r+11)(q+r+2)(q+2)^Pq+1,r+3−12(r+1)^Pq+2,r−2(3q2+6qr+18q+15r+25−r2)^Pq+2,r+1+(q+r+2)(8qr+8r2+7q+29r+18)^Pq+2,r+2+(q+r+4)(q+r+3)(q+r+2)(2q+2r+5)^Pq+2,r+3=0

where if and otherwise.

So far we have proved, using (8), that Ledoux’s formula (6) is equivalent to Equation (13) about the numbers of planar maps. We will now prove (13) using the algebraic equation (9) satisfied by the generating function . First of all, the cases or of (13) are trivial. The case is easily checked using the fact that for all , (because