LLarge expanders in high genus unicellular maps
Baptiste LoufFebruary 26, 2021
Abstract
We study large uniform random maps with one face whose genus grows linearly withthe number of edges. They can be seen as a model of discrete hyperbolic geometry. In thepast, several of these hyperbolic geometric features have been discovered, such as their locallimit or their logarithmic diameter. In this work, we show that with high probability sucha map contains a very large induced subgraph that is an expander.
Combinatorial maps
Combinatorial maps are discrete geometric structures constructedby gluing polygons along their sides to form (compact, connected, oriented) surfaces. Theyappear in various contexts, from computer science to mathematical physics, and have beengiven a lot of attention in the past few decades. The first model that was extensively studiedis planar maps (or maps of the sphere), starting with their enumeration [26, 27] by generatingfunction methods. Later on, explicit constructions put planar maps in bijection with modelsof decorated trees [24, 25, 7, 5, 1], and geometric properties of large random uniform planarmaps have been studied [3, 13, 18, 22]. All these works were later extended to maps on fixedsurfaces of any genus (see for instance [28, 4] for enumeration, [12, 19] for bijections, and [6]for random maps).
High genus maps
Much more recently, another regime of maps has been studied: highgenus maps, that is (sequences of) maps whose genus grows linearly in the size of the map.The main goal is to study the geometric properties of a random uniform such map as the sizetends to infinity. By the Euler formula, the high genus implies that these maps have negativeaverage discrete curvature. They must therefore have hyperbolic features, some of whose havebeen identified in previous works [2, 23, 10, 9, 20].Mostly, two types of models of high genus maps have been dealt with. First, unicellularmaps , i.e. maps with one face, who are easier to tackle thanks to an explicit bijection [11], andthen more general models of maps like triangulations or quadrangulations. It is believed thatboth models have a similar behaviour. 1 a r X i v : . [ m a t h . C O ] F e b he local behaviour of high genus maps around their root is now well understood ([2] in theunicellular case, [10, 9] in the general case), and some global properties have been tackled: theplanarity radius [20] (see also [23] for unicellular maps) and the diameter ([23] for unicellularmaps, still open for other models). Large expanders: a result and a conjecture
In this paper we deal with yet anotherproperty: the presence of large expanders inside our map, in the case of unicellular maps.Contrary to the previous properties, this involves the whole geometric structure of the map.Expander graphs are very well connected graphs, in which every set of vertices has a largenumber of edges going out of it, which is a typical hyperbolic behaviour . Unfortunately, itis impossible that the whole map itself is an expander, since it can be shown that finite butvery large pending trees (which have very bad influence on the expansion of the graph) can befound somewhere in the map. However, it can be shown that most of the map is an expander,in the following sense.Let g n n → θ ∈ (0 , / U n , g n be a uniform unicellular map of size n and genus g n . Theorem 1.1.
For all ε > , there exists a κ > depending only of ε and θ such that thefollowing is true. With high probability , there exists an induced subgraph G n of U n , g n that hasat least (1 − ε ) n edges and is a κ -expander. It is natural to conjecture that a similar results holds for more general models of maps (i.e.,without a fixed number of faces). For instance, let T n , g n be a uniform triangulation of genus g n with 3 n edges. The following conjecture (and the present work) comes from a question ofItai Benjamini (private communication) about large expanders in high genus triangulations. Conjecture 1.
For all ε > , there exists a κ > depending only of ε and θ such that thefollowing is true. With high probability, there exists an induced subgraph G n of T n , g n that hasat least (1 − ε )3 n edges and is a κ -expander. This conjecture deals with the entire structure of the map, therefore we believe it is a veryambitious open problem about the geometry of high genus maps. Some other conjectures mightbe easier to tackle, see [20].
Structure of the paper
The proof of the main result involves the refinement/extension ofseveral known results. We chose to push all the technical proofs to the appendix, in order tomake it clear how we combine these results to obtain the proof of our main theorem.The main objects are defined in the next section, then we give an outline of the proof. Theproof consists roughly of two halves: showing that the core is an expander (Section 4) andshowing that a well defined “almost core” is still an expander while having a large proportionof the edges (Section 5). Finally, in the appendix, we give the proofs of the technical lemmas,as well as a causal graph of all the parameters involved (see Section D for more details). indeed, for each set, a large proportion of its mass is contained on its boundary throughout the paper, we will write with high probability or whp in lieu of with probability − o (1) as n → ∞ . Acknowledgements
The author is grateful to Thomas Budzinski, Guillaume Chapuy, SvanteJanson and Fiona Skerman for useful comments and discussions about this work, and to Guil-laume Conchon–Kerjan for pointing out an error in a previous version of this paper. Worksupported by the Knut and Alice Wallenberg foundation.
We begin with some definitions about graphs. Note that here we allow graphs to have loopsand multiple edges, such objects are also commonly called multigraphs. We will write e ( G )for the number of edges in a graph G . An induced subgraph of a graph G is a graph obtainedfrom G by deleting some of its vertices and all the edges incident to these vertices. Given asubset X of vertices of G , we write G [ X ] for the induced subgraph of G obtained by deletingall vertices that do not belong to X . A topological minor of a graph G is a graph obtained from G by deleting some of its vertices, some of its edges, and by “smoothing” some of its verticesof degree 2 as depicted in Figure 1.Given a graph G and a subset X of its vertices, we define vol( X ) = (cid:80) v ∈ X deg( v ) and ∂ G ( X ) as the number of edges of G with exactly one endpoint in X . Then we set h G ( X ) = ∂ G ( X )min(vol( X ) , vol( X )) , where X is the set of vertices of G that do not belong to X . A graph G is said to be a κ -expander if the following inequality holds for every subset X of vertices of G such that X (cid:54) = ∅ and X (cid:54) = ∅ : h G ( X ) (cid:62) κ. A map is the data of a collection of polygons whose sides were glued two by two to forma compact oriented surface. The interior of the polygons define the faces of the map. Afterthe gluing, the sides of the polygons become the edges of the map, and the vertices of thepolygons become the vertices of the map. Alternatively, a map is the data of a graph endowedwith a rotation system , i.e. a clockwise ordering of half-edges around each vertex. A unicellularmap of size n is the data of a 2 n -gon whose sides were glued two by two to form a compact,connected, orientable surface. The genus g of the surface is also called the genus of the map.We will consider rooted maps , i.e. maps with a distinguished oriented edge called the root . Let U n,g be the set of rooted unicellular maps of size n and genus g . A map of U n,g has exactly it is easily verified that one only needs to check this inequality for subsets X such that G [ X ] is connected. n + 1 − g vertices by Euler’s formula. We will denote by U n , g a random uniform element of U n,g .A tree is a unicellular map of genus 0. A doubly rooted tree is a tree with a ordered pairof distinct marked vertices (but without a distinguished oriented edge). The size of a doublyrooted tree is its number of edges. We set dt n to be the number of doubly rooted trees of size n . The core of a unicellular map m , noted core( m ), is the map obtained from m by iterativelydeleting all its leaves, then smoothing all its vertices of degree 2 (see Figure 2). We do notmake precise here how the root of core( m ) is obtained from the root of m , we will only explainit in Section C (the only place where the root matters, for enumeration purposes, everywhereelse in the paper we will only need the graph structure of the core).On the other hand, m can be obtained from core( m ) in a unique way by replacing eachedge of core( m ) by a doubly rooted tree. More precisely, given a doubly rooted tree t and itstwo distinguished vertices v and v , there is a unique simple path p going from v to v . Let e (resp. e ) be the edge of p that is incident to v (resp. v ), and let c (resp. c ) that comes rightbefore e (resp. e ) in the counterclockwise order around v (resp. v ). Now, we can removean edge e from core( m ) to obtain a map with a pair of marked corners c and c (cid:48) , and we canglue c on c and c on c (cid:48) (see Figure 3). The set of the doubly rooted trees used to construct m from core( m ) will be called the branches of m . We also define core Theorem 3.1 ([21], Theorem 1) . For all κ, α > , and for all < α (cid:48) < α , there exists a κ (cid:48) > such that the following holds for every (multi)graph G .If there exists a graph H satisfying the following conditions: • e ( H ) (cid:62) αe ( G ) , • H is a topological minor of G , • H is a κ -expander,then there exists a graph H ∗ satisfying the following conditions: • e ( H ∗ ) (cid:62) α (cid:48) e ( G ) , • H ∗ is an induced subgraph of G , • H ∗ is a κ (cid:48) -expander. Furthermore, we will prove the following: Proposition 3.2. For all ε > , there exist M and κ > that depend only on θ and ε suchthat the following is true whp: notice that if | d | is odd, or if | d | / k is even, then U ( d ) is empty. core There exists a universal δ > such that, whp, core ( U n , g n ) is a δ -expander. In the next section, we will prove Proposition 3.3, and Section 5 is devoted to the proof ofProposition 3.2. Here, we will prove Proposition 3.3 by comparing core( U n , g n ) to a well chosen map con-figuration model. We begin with two results about this model. We will consider sets d =( d , d , . . . , d k ) that do not contain any 1’s or 2’s, with | d | = 2 n , such that k + n is odd. Lemma 4.1. The map CM ( d ) is unicellular with probability greater than o (1)3 n as n → ∞ , where the o (1) is independent of d . The proof of this lemma is a little technical, it actually needs a refinement of an argumentof [8], it will be given in the appendix. This next proposition states that CM( d ) is an expanderwith very high probability. Proposition 4.2. The map CM ( d ) is a δ -expander (with the same δ as in Proposition 3.3)with probability − P ( CM ( d ) is disconnected ) − o (cid:18) n (cid:19) as n → ∞ , where the o (cid:0) n (cid:1) is independent of d . The proof is rather technical, but it is heavily inspired by [16, 17]. A careful analysis of thecases is needed, but there is no original idea involved, hence we delay it to the appendix.We are now ready to prove Proposition 3.3. Proof of Proposition 3.3. Let d = d (core( U n , g n )). The map core( U n , g n ) has genus g n → ∞ and only vertices of degree greater or equal to 3, hence | d | → ∞ . i.e. independent of θ . d , core( U n , g n ) is uniform in U ( d ). Also, CM( d ) conditioned onhaving one face is uniform in U ( d ). Hence, the probability of core( U n , g n ) not being a δ -expander is P (CM( d ) is not a δ -expander | CM( d ) is unicellular)which we can upper bound by P (CM( d ) is connected and not a δ -expander) P (CM( d ) is unicellular)(because all unicellular maps are connected). This is o (1) by Lemma 4.1 and Proposition 4.2.This o (1) does not depend on d , hence the proof is finished. In this section, we prove Proposition 3.2. Our strategy is the following: now that we know thatcore( U n , g n ) is an expander, we will add back to it the “small” branches of U n , g n to get veryclose to the size of U n , g n without penalizing the expansion too much. We start with technicallemmas. The first one states that replacing edges by small doubly rooted trees does not changethe Cheeger constant too much. Lemma 5.1. Let H be a graph and G be constructed by replacing each edge of H by a doublyrooted tree of size M or less. Then h G (cid:62) h H M + 1 . Proof. This proof is very similar to the proof of Lemma 5 of [21].In G , colour in red the vertices that come from H , and the rest in black. Let Y be a subsetof V ( G ) such that G [ Y ] is connected (recall that we only need to consider connected subsets).Let X be the set of red vertices in Y . We want to lower bound h G ( Y ) in terms of h H ( X ). SeeFigure 5 for an illustration.If X = ∅ , then G [ Y ] is a tree on at most M − G ( Y ) (cid:54) M − 1) and e G ( Y, Y ) = 2. Hence h G ( Y ) (cid:62) e G ( Y, Y )vol G ( Y ) (cid:62) M − . Similarly if X = ∅ then h G ( Y ) = h G ( Y ) (cid:62) / ( M − X (cid:54) = ∅ and X (cid:54) = ∅ . The number of edges of H which are incidentto a vertex of X is e H ( X ) + e H ( X, X ) (cid:54) vol H ( X ). Each edge of H is replaced by a tree withat most M edges, thus of volume at most 2( M − Y can be bounded above by 2( M − H ( X ). Hencevol G ( Y ) (cid:54) vol H ( X ) + 2( M − H ( X ) = (2 M − H ( X ) (5.1)and similarly vol G ( Y ) (cid:54) (2 M − H ( X ) . (5.2)7 X Figure 5: Comparing the edge expansions of Y in G and X in H . Here, M = 7.Now, each edge counted in e H ( X, X ) corresponds to a doubly rooted tree in G between Y and Y , therefore e G ( Y, Y ) (cid:62) e H ( X, X ) . (5.3)Hence, by (5.1), (5.2) and (5.3): h G ( Y ) (cid:62) M − h H ( X ) . This concludes the proof.The next lemma states that the big branches of U n , g n only make up for a very smallproportion of its size. Lemma 5.2. For all ε > , there exists a constant M such that, whp, the total size of thebranches of U n , g n that are bigger than M is less than εn . As for Proposition 4.2, we postpone the proof to the appendix. We need a precise estimationof the second moment of the size of large branches, and it makes it a bit technical.We are now ready to prove Proposition 3.2. Proof of Proposition 3.2. By Lemma 5.2, we know that there exists an M depending only on ε and θ such that core Recall that we have a set d = ( d , d , . . . , d k ) that does not contain any 1’s or 2’s, with | d | = 2 n ,such that k + n is odd. We want to show that CM( d ) has only one face with probability Θ (cid:0) n (cid:1) .8ur proof consists in estimating some quantities carefully in an argument of [8]. We howeverdo not know of a more direct proof.In [8], the authors consider a model that is dual to ours, i.e. they glue polygons together,with the condition that there are few one-gons and digons (our case fits into their assumptions,since we have none). More precisely, they have as a parameter a list P n of sizes of polygons thatsum to 2 n (this corresponds to our d ). The list P n contains P n elements (this correspondsto our k ). An important parameter in their proofs is a random number 0 (cid:54) τ n (cid:54) n .In section 4 of [8], they control the number of vertices of their map, which is the numberof faces in CM( d ). More precisely, equation 13 writes the number of vertices as X τ n + V n − τ n ) . (A.1)Let us define the notions used in (A.1). First, if we condition on τ n , then both terms in (A.1)are independent. From now on, we condition on τ n .It is shown in Section 4.3 that X τ n d TV = (1 + o (1))Poisson (cid:18) log (cid:18) nn − P n (cid:19)(cid:19) , (note that this o (1) can be made uniform in P n by a classical diagonal argument). In particular,since we have no one-gons or digons, we have P n (cid:54) n . This implies that P ( X τ n = 0) (cid:62) / o (1) . (A.2)Now let us turn to V n − τ n ) . For any p , V p is the number of vertices in a uniform unicellularmap on p edges. We can calculate P ( V p = 1) for even p : P ( V p = 1) = p edges with one vertex p edges . (A.3)The denominator in the formula above is easy to enumerate, it is (2 p − p -gon). To enumerate the numerator, we will use [15], more preciselyequation 14 for x = 1, and then Corollary 4.2 for g = p/ 2. It is equal to(2 p )!2 p p !( p + 1) . Therefore, we have exactly P ( V p = 1) = 1 p + 1if p is even. At the end of Section 4 in [8] (proof of Theorem 3), it is shown that X τ n + n − τ n + 1has the same parity as n + P n , which corresponds to n + k in our case (and we require it tobe odd), therefore n − τ n is even, if we condition on X τ n = 0. Hence P ( V n − τ n ) = 1 | X τ n = 0) (cid:62) n − τ n + 1 (cid:62) n + 1 . (A.4)9e are ready to conclude the proof of Lemma 4.1. Conditionally on τ n , by (A.2) and (A.4),the probability that CM( d ) has exactly one face is P ( X τ n = 0) P ( V n − τ n ) = 1 | X τ n = 0) (cid:62) o (1)3 n . This quantity is independent of τ n and uniform in d , hence it finishes the proof of Lemma 4.1. B Proof of Proposition 4.2 We recall that we work with a set of vertex degrees d such that | d | = 2 n (the dependence of d in n will be implicit). We want to prove that CM( d ) is a δ -expander with very high probability,for some universal δ > 0. For the sake of simplicity, we will not make δ explicit, but we willshow that it exists.In what follows, we will consider d as an ordered list ( d , d , . . . , d k ), and we will have alist of vertices ( v , v , . . . , v k ) equipped with distinguishable dangling half-edges, where v i hasdegree d i . For all I ⊂ [ k ], we set vol( I ) = (cid:80) i ∈ I d i , and N V ( d ) is the number of sets I ∈ [ k ]such that vol( I ) = V . We will also write v I = { v i | i ∈ I } .We begin by estimating N V ( d ). All fractions are to be understood as their floor values,which we do not write to make the notation less cumbersome. Lemma B.1. Let < V (cid:54) n , then N V ( d ) (cid:54) V / (cid:18) n/ V / (cid:19) . Proof. Since for all i we have d i (cid:62) 3, we have vol( I ) (cid:62) | I | , and thus if I is such that vol( I ) = V ,then | I | (cid:54) V / 3. Hence N V ( d ) (cid:54) V/ (cid:88) i =1 (cid:18) ki (cid:19) . But, since d i (cid:62) i , we have k (cid:54) n/ 3, thus N V ( d ) (cid:54) V/ (cid:88) i =1 (cid:18) n/ i (cid:19) . Finally, since V / (cid:54) (2 n/ (cid:0) n/ i (cid:1) is increasing in i in the range [1 , V / N V ( d ) (cid:54) V / (cid:18) n/ V / (cid:19) . Now, we will define a set of bad events. Let E V be the event that, in CM( d ), there existsa set I with vol( I ) = V and such that among all the dangling half-edges of v I , strictly less10han δV get paired with dangling half-edges of vertices outside v I . Notice that CM( d ) is nota δ -expander iff at least one of the E V happens. We will separate the analysis in two regimes,depending on the size of V . We introduce a universal, small enough η > 0. We do not make itexplicit, for the sake of simplicity, but we will show that it exists later on. Small subsets We will tackle the case V (cid:54) ηn . This proof follows the lines of [16][Theorem4.16] in the case of regular graphs.First we treat the case of V = 4 or V = 6. If we require that δ < , then if E or E happens, it implies that CM( d ) is disconnected. Hence P ( E ∪ E ) (cid:54) P (CM( d ) is disconnected) . (B.1)From now on, V (cid:62) 8. Given I ⊂ [ k ] of volume V and H a subset of the half-edges of v I wedefine the following event Y I,H := all half-edges of H are matched along themselves.If H has cardinality h , then P ( Y I,H ) = ( h − n − h − n − . By a union bound and Lemma B.1, we have P ( E V ) (cid:54) N V ( d ) (cid:88) (1 − δ ) V We will now care about bigger bad subsets, this time we need to control prob-abilities more carefully. The following proof is adapted from [17][Section 7].Given 0 (cid:54) y < u (cid:54) 1, let X u,y be the number of subsets I ⊂ [ k ] of volume un that haveexactly yn half-edges that are paired with half-edges outside I .We have (using Lemma B.1) E ( X u,y ) = N un ( d ) (cid:18) unyn (cid:19)(cid:18) n − unyn (cid:19) ( yn )!(2 n − un − yn − un − yn − n − (cid:54) un/ (cid:18) n/ un/ (cid:19)(cid:18) unyn (cid:19)(cid:18) n − unyn (cid:19) ( yn )!(2 n − un − yn − un − yn − n − . Therefore, by Stirling’s formula, we havelog E ( X u,y ) (cid:54) n ( f ( u, y ) + o (1)) , (B.3)where f ( u, y ) = log (cid:32)(cid:18) u u (2 − u ) − u (cid:19) / u u y y ( u − y ) u − y (2 − u ) − u y y (2 − u − y ) − u − y y y (cid:0) (2 − u − y ) − u − y ( u − y ) u − y (cid:1) / (cid:33) . Notice that f ( u, 0) = 16 log( u u (2 − u ) − u ) − 13 log 2 . It is easily verified that this function is decreasing and tends to zero as u → 0. Therefore,taking our η > − c = f ( η, / 2. We have, for all u (cid:62) η , f ( u, (cid:54) − c . Bycontinuity of f ( u, y ), there exists δ > u (cid:62) η and y < ηδf ( u, y ) < − c. Hence, by (B.3), the first moment method and a union bound, we have that (again, uni-formly in d ): (cid:88) V (cid:62) ηn P ( E V ) = o (cid:18) n (cid:19) . (B.4) Concluding the proof Combining (B.1), (B.2) and (B.4) yields the proof of Proposition 4.2.12 or Figure 6: Decomposing a doubly rooted tree along the path between its roots. The first (resp.second root) is a box (resp. square). C Proof of Lemma 5.2 This section is devoted to the proof of Lemma 5.2. The general idea of the proof is to approachthe sizes of the branches in U n , g n by random i.i.d. variables. This method is often called Poissonization by abuse of language, and it relies on the saddle point method. We will directlyapply the results of [14][Chapter VIII.8]. Counting doubly rooted trees We first need to compute the generating function of doublyrooted trees counted by edges. Let D ( z ) = (cid:88) s (cid:62) dt s z s , and let T ( z ) = 1 − √ − z z − D = T + T D by considering the path between the two roots of a doubly rooted tree (see Figure 6). Thisdirectly implies D ( z ) = − z + 1 − √ − z z − √ − z . (C.1)Also, C = z∂∂z D is the series of doubly rooted trees with a marked edge, and C ( z ) = z (2 − √ − z − z )(4 z − √ − z ) √ − z . (C.2) Studying the Poissonized law: defining the law For any 0 < β < / 4, we define thetwo random variables X β and Y β with laws P ( X β = k ) = ([ z k ] C ( z )) β k C ( β )13nd P ( Y β = k ) = ([ z k ] D ( z )) β k D ( β ) . Now, fix a constant θ (cid:54) c (cid:54) s n (cid:54) n − s n ∼ cn . We fix β such that E ( X β = k ) + s n E ( Y β = k ) = n. (C.3)Let us first show that this β exists. The equation above rewrites(1 + o (1)) c C ( β ) D ( β ) = 1Now, for 0 < c (cid:54) 1, the equation c C ( β ) D ( β ) = 1 has a root in [0 , / − c (cid:16) c + √ c +8 c (cid:17) − c . Hence, by continuity, for n large enough, (C.3) has a solution β ∈ [0 , / β isdecreasing in s n , hence there exists β ∗ < / s n (cid:62) θn , we have β (cid:54) β ∗ . Studying the Poissonized law: depoissonization probability Now, the probability gen-erating functions (in a variable u ) for X β and Y β are respectively C ( βu ) C ( β ) and D ( βu ) D ( β ) . Let Y , Y ,. . . , Y s n be i.i.d random variables distributed like Y β , and let S = X β + Y + Y + . . . + Y s n . We have E ( S ) = n , therefore, by [14][Corollary VIII.3], we have P ( S = n ) = Θ (cid:18) √ n (cid:19) (C.4)uniformly in s n ∈ [ θn, n − Studying the Poissonized law: large deviations Next we will estimate the large devia-tions of the Y i ’s. Fix an A > Aβ ∗ < / / D ). We have E ( A Y β ) = D ( Aβ ) D ( β ) (cid:54) D ( Aβ ) Aβ βD ( β ) . But D ( z ) /z is a series with positive coefficients, and hence increasing. When z → D ( z ) /z → E ( A Y β ) (cid:54) D ( Aβ ∗ ) Aβ ∗ =: W. Hence, by the Markov inequality, P ( Y β (cid:62) k ) (cid:54) WA k . (C.5)14igure 7: The core decomposition for the branch that contains the root. An edge is marked,and the root is obtained by orienting this marked edge towards the second root in the clockwiseexploration order around the branch.Now, let Y ( M ) i = Y i >M Y i and L ( M ) = X β >M X β + s n (cid:88) i =1 Y ( M ) i . Let us also fix 1 < B < A , and set r = B/A , we have E (cid:16) B Y ( M ) i (cid:17) = (cid:88) k (cid:62) M P ( Y β (cid:62) k ) B k (cid:54) W (cid:88) k (cid:62) M r k = W r M − r , where the inequality follows from (C.5). From now on, and until the end of the proof, let us fix M large enough such that W r M − r is small enough to guarantee E ( B Y ( M ) i ) c (cid:54) B ε / c ∈ [ θ, M is independent of s n as long as s n ∈ [ θn, n − E ( B Xβ>M X β ) = O (1), hence by the Markov inequality we obtain P ( L ( M ) > εn ) (cid:54) O (1) E (cid:16) B Y ( M ) i (cid:17) s n B εn (cid:54) − (1+ o (1)) n (C.6)uniformly in s n .Combining (C.6) with (C.4), we obtain the following for s n ∈ [ θn, n − 1] and n large enough P ( L ( M ) > εn | S = n ) (cid:54) √ − n . (C.7) Sizes of the branches To go from a map to its core, one removes each branch and replacesit by an edge. Since a rooted map has no automorphisms, these branches can be put in alist of doubly rooted tree whose total size is the number of edges in the map. One actuallyneeds to be a little more careful for the branch containing the root: mark the (unoriented)edge of the root in the doubly rooted tree, and replace it by a root edge with a coherentorientation (see Figure 7). This operation is bijective, therefore the list of sizes of the branchesof U n , g n , conditionally on core( U n , g n ) having s n + 1 edges is exactly the list of variables( X β , Y , Y , . . . , Y s n ), conditionally on S = n .We are ready to prove Lemma 5.2. 15 roof of Lemma 5.2. Since core( U n , g n ) is of genus g n and has one face, by the Euler formula,it has s n + 1 edges, with s n (cid:62) g n − > θn (for n large enough). The number of edges ofcore In this paper, we introduce several parameters, and it might not be clear that there is no“circularity” (especially since some of them are defined implicitly). 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