aa r X i v : . [ m a t h . C O ] F e b GENERALIZED CATALAN NUMBERS FROM HYPERGRAPHS
PAUL E. GUNNELLS
Abstract.
The Catalan numbers ( C n ) n ≥ = 1 , , , , , , . . . form one of themost venerable sequences in combinatorics. They have many combinatorial in-terpretations, from counting bracketings of products in non-associative algebra tocounting plane trees and noncrossing set partitions. They also arise in the GUEmatrix model as the leading coefficient of certain polynomials, a connection closelyrelated to the plane trees and noncrossing set partitions interpretations. In thispaper we define a generalization of the Catalan numbers. In fact we define aninfinite collection of generalizations C ( m ) n , m ≥
1, with m = 1 giving the usualCatalans. The sequence C ( m ) n comes from studying certain matrix models attachedto hypergraphs. We also give some combinatorial interpretations of these numbers,and conjecture some asymptotics. Introduction
The
Catalan numbers ( C n ) n ≥ , , , , , , , , , . . . , form one of the most venerable sequences in combinatorics. They have many combi-natorial interpretations, far more than can be reproduced here. We only mention afew: (i) A plane tree is a rooted tree with an ordering specified for the children ofeach vertex. Then C n counts the number of plane trees with n + 1 vertices.(ii) A Dyck path of length n is a directed path from (0 ,
0) to ( n,
0) in R thatonly uses steps of type (1 ,
1) and (1 , −
1) and never crosses below the x -axis.Then C n counts the number of Dyck paths of length 2 n .(iii) A ballot sequence of length 2 n is a sequence ( a , . . . , a n ) with a i ∈ {± } with total sum 0 and with all partial sums nonnegative. Then C n counts thenumber of ballot sequences of length 2 n .(iv) A binary plane tree is the empty graph or a plane tree in which every node N has at most two children, which are called the left and right children of Date : 2 April 2019.2010
Mathematics Subject Classification.
Primary 05A10, 11B65.
Key words and phrases.
Matrix models, hypergraphs, Catalan numbers.The author was partially supported by NSF grant DMS 1501832. We thank Dan Yasaki and tworeferees for many helpful comments. N . Furthermore, if a node has only one child, then it must be either a left orright child. Then C n counts the number of binary plane trees with n vertices.(v) A regular ( n + 2)-gon can be subdivided into triangles without adding newvertices by drawing n − C n counts these subdivisions.(vi) Let Π be a polygon with 2 n sides. A pairing of Π is a partition of the edgesof Π into blocks of size 2 and a choice of relative orientations for the edgesin each block. Any pairing π of Π determines a compact topological surfaceΣ π of some genus: one identifies the edges together according to the pairing.One can show that the surface Σ π is orientable if and only if the edges ineach pair are have opposite orientations as one walks around the boundaryof Π (cf. [11, Ch. 1]). Then C n counts the number of pairings of the sidesof Π such that Σ π is orientable and has genus 0, i.e. is homeomorphic to the2-sphere.The definitive reference for combinatorial interpretations of Catalan numbers isRichard Stanley’s recent monograph [17]. It contains no fewer than 214 differentinterpretations of the C n .( ) Indeed, the first five interpretations given above appearin [17, Ch. 2] as items (6), (25), (77), (4), and (1) respectively. The last interpretation(counting genus 0 polygon gluings) is unfortunately not in [17]. However, it is easilyseen to be equivalent to [17, Ch. 2, (59)], which counts the number of ways to draw n nonintersecting chords joining 2 n points on the circumference of a circle. Anotherresource is OEIS [14], where the Catalans are sequence A000108 . The goal of this paper is to give a family of generalizations of the C n . Foreach integer m ≥
1, we define a sequence of integers ( C ( m ) n ) n ≥ ; for m = 1 we have C (1) n = C n . Here are some further examples: m = 2 : 1 , , , , , , , , , , , . . .m = 3 : 1 , , , , , , , , , . . .m = 4 : 1 , , , , , , , , . . .m = 5 : 1 , , , , , , , . . .m = 6 : 1 , , , , , , . . .m = 7 : 1 , , , , , , , . . . The C ( m ) n are defined in terms of counting walks on trees, weighted by the orders oftheir automorphism groups. For m = 1 the resulting expression is not usually given An earlier version of this list is contained in [16], with additions available on Stanley’s website[15].
YPERGRAPH CATALAN NUMBERS 3 as a standard combinatorial interpretation of the Catalan numbers, but it is knownto compute them; we will prove it in the course of proving Theorem 4.9. In fact,from our definition it is not clear that the C ( m ) n are actually integers , even for m = 1,although we will see this by giving several combinatorial interpretations of them.Here is the plan of the paper. In § C ( m ) n , and in § n and any m . In § C ( m ) n based on six standardinterpretations of the Catalan numbers. In § C ( m ) n , and conjecture some asymptotics of C ( m ) n for fixed m as n → ∞ . Finally, in § Hypergraph Catalan Numbers
We begin by giving the description of the Catalan numbers that we wish togeneralize. Let T n be the set of unlabeled trees on n vertices. The sequence | T n | appears on OEIS as sequence A000055 , and begins1 , , , , , , , , , , , . . . , where | T | := 1 by convention.Let T ∈ T n , and for each vertex v ∈ T , let a T ( v ) be the number of walks thatbegin and end at v and traverse each edge of T exactly twice. Note that any suchwalk visits each other vertex at least once, and may do so multiple times. Let Γ( T )be the automorphism group of T , and let | Γ( T ) | be its order.Figure 1 shows an example of the numbers a T ( v ) for the 3 trees in T . The outernumbers on the leaves of the upper left tree are 1 because the only possible walk isto go from one end of the tree to the other, then back to the beginning. The innernumbers on the same tree are 2, because one must first choose a direction in whichto head, then must go all the way to that end, then back through the starting pointto the other end, then back to the initial vertex. The Catalan number C n is given by (1) C n = X T ∈ T n +1 X v ∈ T a T ( v ) | Γ( T ) | . For example, using Figure 1 we have C = 82 + 4824 + 162 = 4 + 2 + 8 = 14 . We defer the verification of Proposition 2.2 until §
4, when we discuss combinatorialinterpretations. The proposition will be proved in the course of Theorem 4.9.
PAUL E. GUNNELLS
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11 222 222 4 6 6 66 6 24
Figure 1.
The set T with vertices labeled by a T ( v ). Going clockwisefrom the upper left, the orders of the automorphism groups are 2 , , Now let m ≥ C ( m ) n is defined essentially as in(1), but we modify the definition of the numbers a T ( v ): Let m ≥
1, let T ∈ T n +1 , and let v ∈ T . Then an a ( m ) T -tour beginning at v is a walk that begins and ends at v and traverses each edge of T exactly 2 m times. We denote by a ( m ) T ( v ) the number of a ( m ) T -tours beginning at v .We note that a (1) T ( v ) = a T ( v ). As before, in an a ( m ) T -tour each vertex of T will bevisited at least once. Furthermore, since T is a tree, each edge is visited m timeswhile going away from v and m times while coming back to v . The hypergraph Catalan numbers C ( m ) n are defined by(2) C ( m ) n = X T ∈ T n +1 X v ∈ T a ( m ) T ( v ) | Γ( T ) | . For example the numbers a (2) T ( v ) are shown in Figure 2 for the three trees in T .The numbers are larger now, since walks have many more options. Using the numbersin Figure 2 we find C (2)4 = 2162 + 504024 + 7202 = 108 + 210 + 360 = 678 . Computing the C ( m ) n In this section we show that once one has sufficient knowledge of the trees in T n +1 to compute C n , one can easily compute C ( m ) n for any m . In other words, if onewants to extend the data in § n and let m grow. Let T ∈ T n +1 , and let v ∈ T have degree d ( v ) . Then we have (3) X v ∈ T a ( m ) T ( v ) = 2 nm n +1 ( m !) n Y v ∈ T ( md ( v ) − . YPERGRAPH CATALAN NUMBERS 5
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Figure 2.
The set T with vertices labeled by a (2) T ( v ).For example, for the tree on the right of Figure 2, we find2 · · (2!) (2 · − (2 · − , which agrees with the data in the figure. Proof.
We use the results in [18, Ch. 10], which treat Eulerian tours in balanceddigraphs, and so we begin by recalling some notation. Let G be a connected digraphand let ˜ G be the associated undirected graph. Suppose G is balanced ; this meansthat the outdegree o ( v ) of each vertex v is equal to its indegree i ( v ). Given a vertex v of G , an oriented spanning tree with root at v is a subgraph T ⊂ G such that ˜ T isa spanning tree of ˜ G in the usual sense, and such that all the edges of T are orientedtowards v . Then since G is connected and balanced it is Eulerian [18, Theorem 10.1],and according to [18, Theorem 10.2] given an edge e the number ε ( G, e ) of Euleriantours of G beginning with the (directed) edge e is(4) ε ( G, e ) = τ ( G, e ) Y v ∈ G ( o ( v ) − , where τ ( G, e ) is the number of oriented spanning trees of G with root at the initialvertex of e . Furthermore, it is known that τ ( G, e ) is independent of e [18, Corollary10.3].Now let T be a tree with n +1 vertices and fix m . The left of (3) is the total numberof a ( m ) T -tours on T (Definition 2.4). We will count a ( m ) T -tours by first counting Euleriantours on the canonical balanced digraph T m built from T by replacing each edge with2 m edges, m oriented in one direction and m oriented in the other. Let v ∈ T .Then it is clear that an a ( m ) T -tour contributing to a ( m ) T ( v ) determines (non-uniquely)an Eulerian tour on T m starting and ending at v . Hence we can use (4) to compute a ( m ) T ( v ). In particular for an edge e ∈ T m we have(5) ε ( T m , e ) = τ ( T m , e ) Y v ∈ T ( md ( v ) − , PAUL E. GUNNELLS where d ( v ) is the degree of v in T .Now we go from (5) to (3). First, the number τ ( T m , e ) of oriented spanning treesin T m is m n (after fixing a root we pick one of m possible properly oriented edges foreach edge in T ). Next, to get the total number of Eulerian tours of T m , we multiply ε ( T m , e ) by the number of edges of T m , which is 2 mn . The result is(6) (cid:12)(cid:12)(cid:12)n Eulerian tours of T m o(cid:12)(cid:12)(cid:12) = 2 mn · m n · Y v ∈ T ( md ( v ) − . Now let π be the map π : n Eulerian tours of T m o −→ n a ( m ) T -tours of T o that replaces each edge of an Eulerian tour of T m with the corresponding edge in T . It is clear that π is surjective. Furthermore, each a ( m ) T -tour w of T has precisely( m !) n preimages under this map, since w traverses each edge of T in each directionprecisely m times, and we get to choose in which order these m traversals correspondto the m corresponding edges of T m . Thus(7) (cid:12)(cid:12)(cid:12)n Eulerian tours of T m o(cid:12)(cid:12)(cid:12) = ( m !) n · (cid:12)(cid:12)(cid:12)n a ( m ) T -tours of T o(cid:12)(cid:12)(cid:12) . Comparing (6) and (7) yields (3), and completes the proof. (cid:3)
Thus to compute C ( m ) n for any m one only needs the trees in T n +1 together withtheir vertex degrees and orders of their automorphism groups. This can be done, atleast for reasonable values of n , using the software nauty [12]. For example there are751065460 ≈ trees in T ; nauty is able to compute them on a laptop in less than14 seconds. On the other hand, computing the orders of all the automorphism groupstakes longer. For instance there are only 823065 ≈ trees in T , and computingall their automorphism groups takes just over 4 hours. Combinatorial interpretations
As it turns out, the numbers C ( m ) n have a variety of combinatorial interpreta-tions, in fact in terms of objects used to count the usual Catalan number C nm . As weshall see, only certain of these will contribute to C ( m ) n , and in general a given objectmay contribute in several different ways. It will also be evident that any standardCatalan interpretation can be turned into one for the C ( m ) n . We begin by introducingsome notation. Let X be an combinatorial object; we do not give a precise definition of X ,but the reader should imagine that X is something used in a standard Catalaninterpretation, such as those in § X will be an aggregate of smaller elements, and we say that a level structure for X isa surjective map ℓ from these elements to a finite set [[ N ]] := { , , . . . , N } for some YPERGRAPH CATALAN NUMBERS 7 N ∈ Z > . We will say x ∈ X is on a higher level than x ′ ∈ X if ℓ ( x ) > ℓ ( x ′ ), withsimilar conventions for same and lower level. The i th level X i ⊂ X with respect to ℓ will be the preimage ℓ − ( i ) ⊂ X . In some interpretations, we will also have a zerothlevel X ; these will usually be combinatorial objects that are naturally rooted. If X has a zeroth level, we will require | X | = 1. Note that in the pictures that follow,the function ℓ will not typically correspond to the height of elements of X in theirpositions in the figures. The object X will be a poset. If x, x ′ ∈ X and x covers x ′ , we will saythat x is a parent of the child x ′ . The level structures we consider will always becompatible with the poset structure, in that any child x ′ of a given parent x willsatisfy ℓ ( x ′ ) = ℓ ( x ) + 1. In other words, children will always lie one level above theirparents. We remark the poset structure determines the level of each element as longas a level-0 element is present. Finally, let m be a positive integer. We will consider m -labeling the positivelevels of X , which means the following. First fix an infinite set L of labels. Let X = X ⊔ G i ≥ X i be the disjoint decomposition of X into levels, where X may be empty. For each positive level X i , i >
0, we choose a disjoint decomposition of X i into subsets of order m ; in particular, this implies | X i | ≡ m for i > m -labeling is a map assigning an element of L to each of these subsets. We willsay that an m -labeling is admissible if the following are true: • Distinct subsets receive distinct labels. • The labeling is compatible with the poset structure, in the following sense: iftwo elements x, x ′ share a given label, then the labels of their parents agree.In other words, an admissible m -labeling is a partition of the set X r X of all non-level-0 elements of X into disjoint size m blocks such that if two elements u, v ∈ X r X have children in the same block, then u must belong to the same block as v .We also consider two labelings to be equivalent if one is obtained from the other bypermuting labels. Note that all m -labelings are admissible if m = 1. We give an example to clarify this terminology. Let X be a plane tree. Theelements of X are its vertices. Let v be the root. We can define a level structure ℓ : X → Z ≥ by setting ℓ ( x ) to be the distance in X to v . Note that X = { v } has size 1, but of course the positive levels X i can be bigger. Given two vertices x, x ′ , one is a parent or a child of the other if it is in the usual sense of trees: x is aparent (respectively, child) of x ′ if x and x ′ are joined by an edge and x lies closer to(respectively, further from) the root than x ′ . Figure 3 shows two plane trees X, Y . PAUL E. GUNNELLS
Each tree has four levels. Parents appear above their children, and levels increase aswe move down the figure.Next we consider labelings. Figure 4 shows the two rooted trees
X, Y equippedwith 2-labelings, with labeling set L = { a, b, c, d, e } . We have arbitrarily orderedthe labels as indicated; one would obtain an equivalent labeling after permuting thelabels. The left tree X is admissibly 2-labeled: if two vertices have the same label,so do their parents. The right tree Y , however, is not admissibly 2-labeled: the twovertices at the bottom have the same label e , but their parents have two differentlabels c, d . Now let X nm be a set of objects constituting a combinatorial interpretationof C nm (we will say which we consider in a moment). Then we will define a posetstructure and levels on each X ∈ X nm , and show(8) C ( m ) n = X X ∈ X nm N m ( X ) , where N m ( X ) is the number of admissible m -labelings of X . As mentioned before,an object X ∈ X nm cannot have N m ( X ) = 0 unless its level structure satisfies anobvious congruence condition: the size of a nonzero level must be divisible by m .This condition is not sufficient, however, as one can see in Figure 4. Both trees havepositive levels of even size, but there is no way to give the right tree an admissible2-labeling: the parents of the vertices labeled e are forced to have different labels.On the other hand, if m = 1 then any m -labeling is admissible. Hence when m = 1each X ∈ X nm contributes to C nm with N m ( X ) = 1, and one recovers a usualcombinatorial interpretation of the Catalan numbers.PSfrag replacements X X X X Y Y Y Y Figure 3.
Two rooted trees
X, Y with their levels.
We are now ready to give our combinatorial interpretations. For each weexplain their level and hierarchical structures. Examples of all the objects are shownin Figures 6–7.
YPERGRAPH CATALAN NUMBERS 9
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Figure 4. X and Y . The labeling of Y is not admissible. (i) Plane trees. The set X nm is the set of plane trees on nm + 1 vertices. Thelevels are the distance to the root, and the vertices are parents/children of each otherif they are in the usual sense. (ii) Dyck paths. The set X nm is the set of Dyck paths from (0 ,
0) to (2 mn, π are its slabs , defined as follows. Let R π be the interior of theregion bounded by π and the x -axis. Then a slab is a connected component of theintersection of R π with an open strip Y k = { ( x, y ) ∈ R | k − < y < k } , where k ≥ k if it lies in Y k ; the zeroth level is empty. A slab S is a parent of a slab S ′ if S sits in a lower level and sits under S ′ . (iii) Ballot sequences. The set X nm consists of the ballot sequences B = ( a , . . . , a nm ).Let s k = P ki =1 a i be the k th partial sum. Let i < j . We say i, j are a pair if (i) a i = 1, a j = −
1; (ii) s i = s j + 1; and (iii) j is the minimal index greater than i forwhich these conditions are true. Then the elements of a ballot sequence are its pairs.The level of a pair ( i, j ) is the value of s i ; the zeroth level is empty. A pair p = ( i, j )is the parent of q = ( k, l ) if the level of p is one less than that of q and i < k and l > j . (iv) Binary plane trees. The set X nm consists of all binary plane trees with nm vertices. The zeroth level is empty. The i th level for i ≥ i − v ∈ X i is a parent of v ′ ∈ X i +1 if there is apath from v to v ′ with exactly one left step. (v) Triangulations. Let Π = Π nm +2 be a regular polygon with nm + 2 sides. Then X mn consists of triangulations ∆ of Π that do not have new vertices. The elements of∆ are its triangles, and the levels of ∆ are given by the sets of left-turning triangles ,which means the following. Fix once and for all an edge e of Π and let T be thetriangle of ∆ meeting e . As one enters T across e there is a unique exiting edge e R to the right and one e L to the left. We say the triangle T R meeting T at e R , if thereis one, is obtained by turning right, and the triangle T L across e L , if there is one,is obtained by turning left. Then the first level ∆ of ∆ consists of T and all the triangles that can be reached from T by turning left. The second level ∆ consists ofthe triangles that can be reached by turning right once from a triangle on the firstlevel, and then turning left an arbitrary number of times. Continuing this process,each triangle in ∆ gets placed into a unique positive level. The zeroth level is empty.A triangle T ∈ ∆ i is the parent of T ′ ∈ ∆ i +1 if T ′ can be reached from T by a singleright turn followed by any number of left turns.PSfrag replacements eeT L T R Figure 5.
The left and right triangles determined by an edge e , anda triangulation of a polygon with its level structure. The light greytriangles, which are obtained by turning left after entering across theedge e , are on level 1. The white triangles are on level 2, and the darkgrey triangles are on level 3. (vi) Polygon gluings. Let Π = Π mn be a regular polygon with 2 mn sides with adistinguished vertex. An oriented gluing of Π is a partition of its sides into n blocksof size 2 m , with the sides oriented such that as one moves clockwise around Π, theorientations in a given block alternate (cf. the right column of Figure 7). We alsoassume that the first edge in any of these blocks is oriented such that clockwise ispositive. The labeling of the edges induces an equivalence relation on the vertices ofΠ, after one performs the identifications. Then X nm is the set of such gluings withthe number of equivalence classes of vertices maximal . A vertex b is a child of a ifthe edge joining them is positive from a to b . The distinguished vertex is at level 0,and the levels of the others are determined by requiring that passing from parent tochild increases the level. Figures 6–7 illustrate the combinatorial interpretations used in the computa-tion of C (2)2 = 6. For this number each object that affords an admissible 2-labelinghas at most two levels; we show the first level using light grey and the second level YPERGRAPH CATALAN NUMBERS 11 using white. Labelings are indicated by the letters a, b . Note that only one objecthas more than one 2-labeling, namely the one appearing in the first three lines ofeach figure.PSfrag replacements aa aa aa aa aa aa aa aa a aa aaa aaa aaaa aaa aa aa bb bbbb bb bb bb bb bbb bbb b bbbbb bbbb bb bb bbcdef
11 111111 1 111 1 1 11 1 111 1 11 1 11 − − − − − − − − − − − − − − − − − − − − − − − − − −
111 1111 − − − − − − Figure 6.
Plane trees, Dyck paths, ballot sequences.
The interpretations (i)–(vi) given in § C ( m ) n ,i.e. (8) holds.Proof. First we claim that all the interpretations (i)–(vi) give the same counts. To seethis, note that if X nm is any of the above sets (with no labelings), then | X nm | = C nm ,and there are known bijections between the different objects (cf. [17, Theorem 1.5.1]).For the convenience of the reader, we recall these bijections. To simplify notation weset m = 1. Plane trees and Dyck paths.
Let T be a plane tree with n + 1 vertices with rootvertex v . Then T has n edges, and one can build a length 2 n Dyck path π ( T ) asfollows. Recall that a plane tree either consists of the single vertex v , or else v has asequence of subtrees T , . . . , T k , each of which is a plane tree. To construct the Dyckpath attached to T , one performs the preorder tree traversal : this is the traversalthat begins by visiting the root vertex, then recursively visits the subtrees of the root PSfrag replacements aaaaaa aaaa aa aa a aaa aa aa aa b bbb bbb b bb bb bb bb bbb b bb bbcd eeeeeef ∗∗∗∗∗∗
Figure 7.
Binary trees, triangulations, polygon gluings. T , . . . , T k , in order. The traversal returns to the root along the root-incident edgeevery time it is done traversing a subtree.During the traversal one crosses each edge of T exactly twice, once going down,and once going up. For every move down one appends the step (1 ,
1) to π ( T ), andfor every move up one appends (1 , − n steps, never goes below the x -axis, and ends at (2 n, π , one builds a plane tree T ( π ) whose preordertraversal is encoded by π . YPERGRAPH CATALAN NUMBERS 13
Dyck paths and ballot sequences.
Let π be a Dyck path of length 2 n . Wecreate a sequence b ( π ) = ( a , . . . , a n ) with a i ∈ {± } by projecting onto the secondcoordinate: (1 , , −
7→ −
1. Thus b ( π ) records the change in heightabove the x -axis as one moves along π . Then b ( π ) is clearly a ballot sequence:the condition that π never goes below the x -axis ensures that the partial sums arenonnegative. It is also clear that this process can be reversed to give a map fromballot sequences to Dyck paths. Plane trees and polygon gluings.
Let T be a plane tree. We can regard T asembedded in the sphere S , for example by applying stereographic projection to theplane. If one cuts the sphere open along the edges of T , one obtains a polygon Π n with 2 n sides together with data of an oriented gluing: the edges come naturallyin pairs, and the root becomes the distinguished vertex of Π n . To go backwards,if one starts with a polygon Π n and identifies the edges in pairs, one obtains atopological surface S together with an embedded connected graph G ( G may haveloops or multiple edges). The surface S is orientable if and only if the gluing data isoriented (as one goes around the boundary of the polygon, every pair of edges to beglued must appear with opposite orientations). If S is orientable, then the graph G is a tree if and only if S is a sphere; this happens if and only if the number of verticesof G is maximal after gluing; this follows from considering the Euler characteristic of S , computed as 1 − | edges | + | vertices | . Triangulations and binary plane trees.
Let ∆ be a triangulation of the polygonΠ n +2 with a distinguished edge e . One can make a binary plane tree B (∆) by takingthe dual of ∆ as follows (cf. Figure 8). The vertices of B (∆) are the triangles of∆. Two vertices are joined by an edge if and only if they correspond to adjacenttriangles in ∆. The distinguished edge e sits in the boundary of one triangle, whichdetermines the root of B (∆) (in Figure 8 the root is indicated by a doubled circle).The resulting tree is embedded in the plane and is a binary tree exactly because eachtriangle has three sides. It is also easy to see that this construction is reversible. Plane trees and binary plane trees.
Finally we come to this bijection, due tode Bruijn–Morselt [4], which is the most interesting of all. We follow the presentationin the proof of [17, Theorem 1.5.1] closely. Let T be a plane tree on n + 1 vertices. We will construct a binary tree B ( T )on n vertices; in fact the vertices of B ( T ) are the non-root vertices of T . First we As mentioned before ( § C n is the number of ways to draw n nonintersectingchords joining 2 n points on the circumference of a circle. If one starts with a connected orientedpolygon gluing, and draws chords between the centers of the edge pairs, one obtains n chords as in(59). The condition that these chords do not intersect is exactly equivalent to the resulting surfacebeing a sphere. In particular we follow the paragraph labeled (iii) → (ii) at the bottom of p. 8. delete the root from T and all edges incident to it. Then we remove all edges that arenot the leftmost edge from any vertex. In other words, if the children of v in T are v , . . . , v k , then we remove the edges { v, v } ), . . . , { v, v k } and leave the edge { v, v } untouched. After this the remaining edges become the left edges in the binary tree B ( T ).To construct the right edges in B ( T ), we create edges horizontally across thevertices of T . In particular, given a vertex v in T , we draw edges from each child w of v to the child of v immediately to the right of w , if this child exists. Finally theroot of B ( T ) is the leftmost child of the root of T . (See Figure 9 for an example.)This process is easily seen to be reversible.PSfrag replacements e Figure 8.
Making a binary tree B (∆) from a triangulation ∆. Thevertices of B (∆) are white and its edges are dashed. The root vertexis circled.We now return to the original discussion. We claim that for m >
1, the samebijections prove that all interpretations give the same counts. One needs only tocheck that the definitions of admissible labelings are compatible. For instance, topass from a plane tree T to a Dyck path π , one starts at the root of the T andtraverses it in preorder (process the node, traverse the left subtree, then traverse theright subtree). As one descends T along an edge, one steps up in π by (1 , , − T coincide with the slabs of π . Indeed, the slabsof π are constructed so that the corresponding vertices of T are at the same distancefrom the root, and the poset structure in the slabs mirrors that of the vertices of T . It is also easy to see that the different notions of admissible labelings agree. Theverifications for the other interpretations are similar. YPERGRAPH CATALAN NUMBERS 15
PSfrag replacements −→−→
Figure 9.
Making a binary tree B ( T ) from a plane tree T . We firstdelete the root of T and all edges incident to it. Then we delete theedges in T that are not furthest to the left. The deleted edges areshown in gray, and the remaining edges are drawn with a wider penwidth. We then create horizontal edges, shown as dashed lines, bydrawing horizontally from the leftmost child of a vertex v through theremaining children of v , in order. The original (respectively, horizontal)edges become the left (resp., right) edges of B ( T ).To complete the proof, we must check that any one of them actually computes C ( m ) n . We will use admissibly labeled plane trees (i).Let C = C n +1 be the set of pairs n ( T, w ) (cid:12)(cid:12)(cid:12) T ∈ T n +1 , w an a ( m ) T -tour o modulo the equivalence relation ( T, w ) ∼ ( T ′ , w ′ ) if T = T ′ and there is an automor-phism of T taking w to w ′ . Then | C | = C ( m ) n . Let P = P n be the set of plane treeson n vertices, and let A = A nm +1 be the set of pairs(9) n ( T, l ) (cid:12)(cid:12)(cid:12) T ∈ P nm +1 and l is an admissible m -labeling of T o . Then | A | is the right hand side of (8). We will prove that (8) holds by constructinga bijection between C and A .We begin by defining two maps α : C −→ A , β : A −→ C . First we define α . Given ( T, w ) ∈ C , let v be the vertex where w begins and ends.Then α ( T, w ) is the plane tree ˜ T whose Dyck word π ( ˜ T ) is determined by the followingrule: the i th step of π ( ˜ T ) is an up-step if the i th step of w moves towards v , and isa down-step otherwise. We equip ˜ T with a labeling l as follows: the i th vertex inpreorder traversal of ˜ T is given the label of the starting point of the i th edge of w . This labeling l is an admissible m -labeling: if x , x are two vertices of ˜ T with thesame labels, then the parents y , y of x , x receive the label of the unique vertex y ∈ T that is one step closer to v . Thus α ( T, w ) ∈ A . We also claim α is well-defined.Suppose α ( T, w ) = ( ˜
T , l ) and (
T, w ) ∼ ( T ′ , w ′ ). Then α ( T ′ , w ′ ) is the pair ( ˜ T , l ′ ),where the labels of l ′ are a permutation of those of l . Figure 10 shows an example ofthe computation of α for m = 2.Now we define β . Let ( S, l ) ∈ A . Let ¯ S be the graph obtained from S by firstidentifying any two vertices of S that share the same label, and then by replacingparallel edges in the resulting multigraph by single edges. We note that ¯ S is actuallya tree. Indeed, the admissibility of the labeling implies that the map E ( S ) → E ( ¯ S )on edges is m : 1, and the map on vertices is m : 1 away from the root and 1 : 1 onthe root. Thus ¯ S has n + 1 vertices and n edges, and since it is clearly connected ¯ S is a tree. Finally, we define a walk w = w ( ¯ S ) on ¯ S by writing the sequence of labelsencountered during the preorder traversal of S . The m -admissibility of the labelingimplies that w is an a ( m ) S -tour, and thus we have defined an element β ( S, l ) = ( ¯
S, w ).Figure 11 shows an example of this construction.To complete the proof, we must show that α , β are bijections. First we show β ◦ α = 1 C . We claim that if ( ¯ S, w ( ¯ S )) = ( β ◦ α )( T, w ), then ¯ S = T and w ( ¯ S ) = w (in other words, the representative of an equivalence class in C is taken to itself).This is clearly true if T has one vertex. Suppose the claim is true for all pairs ( T, w )with ≤ n vertices. Let T be a tree with n + 1 vertices, let w be an a ( m ) T -tour of T beginning at v , let x be a leaf of T , and let y be the neighbor of x in T . Deleting x and modifying w appropriately we obtain a pair ( T ′ , w ′ ). In ( S ′ , l ′ ) = α ( T ′ , w ′ ) wehave m vertices y , . . . , y m corresponding to y . Then ( S, l ) = α ( T, w ) is obtained from( S ′ , l ′ ) by placing m new vertices x , . . . , x m under the y i according to where theyappear in the walk w . By induction β ( S ′ , l ′ ) = ( T ′ , w ′ ). When we construct β ( S, l )the only differences are that now we collapse the m edges of the form ( x i , y j ) to asingle edge in ¯ S joining x to y , and that we build w ( ¯ S ) from w ( ¯ S ′ ) by incorporatingthe vertex x . Clearly ¯ S = T . Moreover w ( ¯ S ) = w , since the edges in S were built tomake preorder traversal in S match the original walk w . This shows β ◦ α = 1 C .Now we show α ◦ β = 1 A . The proof is similar. Clearly this is true for all( S, l ) ∈ A with one vertex. Suppose that for all ( S, l ) with < nm + 1 vertices wehave ( α ◦ β )( S, l ) = (
S, l ). Let (
S, l ) have nm + 1 vertices and choose a leaf x of S of highest level. The admissibility of the labeling l implies that there are m − x , . . . , x m with the same label as x , and since x lies in thehighest level, the vertices x , . . . , x m must also be leaves of S . Let y , . . . , y m be thevertices of S with label equal to any parent of x i . In particular, the parents of the x i are a subset of the y j , but not every y j is necessarily a parent of an x i . Deleting the x i we obtain a labeled plane tree ( S ′ , l ′ ). If ( ¯ S ′ , w ( ¯ S ′ )) = α ( S ′ , l ′ ), then by induction wehave β ( ¯ S ′ , w ( ¯ S ′ )) = ( S ′ , l ′ ), and moreover the tree ¯ S ′ is obtained from ¯ S by deleting YPERGRAPH CATALAN NUMBERS 17 x (the image of the x i ). The walk w ( ¯ S ) is obtained from w ( ¯ S ′ ) by inserting stepsalong the edge ( x, y ) exactly according to the up- and down-steps in S along theedges of the form ( x i , y j ), where y is the image of the y j . After applying α we exactlyrecover the edges ( x i , y j ) in S . This shows α ◦ β = 1 A , and completes the proof ofthe theorem. (cid:3) The interpretations in Theorem 4.9 make it possible to define varioushigher analogues of other standard numbers, such as Narayana numbers, and higher q -analogues. We have not pursued these definitions.PSfrag replacements a aa bbb ccc ddd vv T ˜ T −→ Figure 10.
Applying the map α to a pair ( T, w ) gives an admissibly m -labeled plane tree ˜ T . The label set is { a, b, c, d } , and the walk is w = vadavbvadavcvbvcv . If we apply the automorphism of T thatswaps vertices b and c , we obtain an equivalent admissibly m -labeledplane tree.PSfrag replacements a aa a b bbb c ccc d ddd v vv S ¯ S −→−→ Figure 11.
The map β is the composition of these two arrows; appliedto a pair ( S, l ) gives a tree ¯ S and a walk w . The resulting walk is thesame as that in Figure 10. Generating functions and asymptotics
Let(10) F m ( x ) = X n ≥ C ( m ) n x n be the ordinary generating function of the C ( m ) n . In this section we explain how touse the combinatorial interpretations in § x . Then we will give a conjecture of the asymptotic behavior of C ( m ) n that generalizes the famous formula C n ∼ n n / √ π , ( n → ∞ ) . We will compute F m ( x ) by showing that counting admissibly labeled planetrees on nm + 1 vertices is equivalent to counting colored plane trees on n + 1 vertices.To count the latter, we use standard generating function techniques as described inFlajolet–Sedgewick [8]. For the benefit of the reader and to keep our presentation asself-contained as possible, we recall the results we need here. (i) Let A ( x ) = P k ≥ a k x k ∈ Z [[ x ]] be an integral formal power series with a k ≥ .Let P A be the set of all colored plane trees such that any vertex with k childrencan be painted one of a k colors. Let P A ∈ Z [[ x ]] be the ordinary generatingfunction of P A , so that the coefficient [ x n ] P A ( x ) of x n counts the trees in P A with n vertices. Then P A satisfies the functional relation (11) P A = xA ( P A ) . (ii) Let B ( x ) = P k ≥ b k x k ∈ Z [[ x ]] be another integral formal power series with b k ≥ . Let P A,B be the set of all colored plane trees such that (a) anynon-root vertex with k children can be painted one of a k colors, and (b) ifthe root vertex has degree k then it can be painted any one of b k colors. Let P A,B ∈ Z [[ x ]] be the ordinary generating function of P A,B . Then P A,B satisfiesthe functional relation (12) P A,B = xB ( P A ) , where P A satisfies (11) .Proof. These results are proved in [8]. Specifically (i) is [8, Proposition I.5] and (ii)follows from [8, Example III.8]. We sketch the proof here.For (11), recall that a plane tree is a rooted tree with an ordering specified for thechildren of each vertex. This is equivalent to the recursive specification that a planetree is a root vertex v with a (possibly empty) sequence of plane trees T , . . . , T k attached in order to v , with the root of T i becoming the i th child of v . If P is the YPERGRAPH CATALAN NUMBERS 19 ordinary generating function of plane trees, this description leads to the well-knownfunctional relation P = x (1 + P + P + P + · · · ) , where the right hand side encodes the process of attaching an arbitrary sequence ofplane trees to an initial root vertex.Now consider the collection P A of plane trees colored according to A . In termsof the recursive specification above, one can think of building a tree in P A by (i)picking a root vertex v and deciding on the number k of its children, (ii) coloring v one of a k different colors, and (iii) continuing the process recursively to each of the k children of v . This leads to the functional relation P A = x ( a + a P A + a P A + a P A + · · · ) , which proves (11).To see (12), we note that a tree in P A,B is built similarly. We (i) choose a rootvertex v and decide on the number k of its children, (ii) color v one of b k differentcolors, and (iii) place trees from P A under v with their roots as v ’s children. Thisleads to P A,B = x ( b + b P A + b P A + b P A + · · · ) , which proves (12). (cid:3) If S is a plane tree appearing in P A,B , then we say S has beenequipped with an ( A, B ) -coloring . Now we define the generating functions for the colorings we will need. For anypair g ≥ , r ≥
1, let λ ( r, g ) be the dimension of the space of degree g homogeneouspolynomials in r variables. We have λ ( r, g ) = (cid:18) r − gr − (cid:19) . For any k ≥
0, let W m ( k ) be the number of partitions of a set of size k into blocks ofsize m . Then W m ( k ) = 0 unless m divides k , and in this case we have W m ( k ) = k !( m !) k/m ( k/m )! . Put ℓ m ( x ) = X d ≥ W m ( dm ) λ ( m, dm ) x d and h m ( x ) = X d ≥ W m ( dm ) x d . Let f m ( x ) ∈ Z [[ x ]] satisfy the functional equation (13) f m ( x ) = xℓ m ( f m ( x )) . Then the generating function F m ( x ) = P n ≥ C ( m ) n x n +1 satisfies (14) F m ( x ) = xh m ( f m ( x )) . Proof.
Let us fix m ≥ F , f , ℓ , h instead of F m , f m , ℓ m , h m . Then by Proposition 5.3 F is the ordinary generating function of the( ℓ, h )-colored plane trees P ℓ,h . Recall (9) that A is the set of pairs ( T, l ) where T ∈ P nm +1 and l is an admissible m -labeling of T . We will construct a surjectivemap(15) ρ : A −→ P show that it induces a bijection between trees ( ℓ, h )-colored plane trees with n + 1vertices and admissibly m -labeled plane trees A nm +1 on nm + 1 vertices. First weassume that the labeling set L is the positive integers Z ≥ , with its standard totalorder. Then we assume that ( S, l ) ∈ A is labeled so that the following propertieshold:(i) If S has nm + 1 vertices, then l is surjective onto [[ n ]] ⊂ L .(ii) If x, y ∈ S are two vertices and y is in a higher level than x (i.e. y is furtherfrom the root than x ), then l ( x ) < l ( y ).(iii) If x, x ′ ∈ S are on the same level with l ( x ) < l ( x ′ ), and y (respectively, y ′ ) isa child of x (resp., x ′ ), then l ( y ) < l ( y ′ ).(iv) If L ′ ⊂ L is the subset of labels appearing in a fixed level of S , then theelements of L ′ are ordered compatibly with when they first appear whenread from left to right: the leftmost vertex receives the smallest label from L ′ , the first new label seen when reading to the right is next largest availablelabel from L ′ , and so on until the final label in L ′ is seen.It is clear that, given any admissibly m -labeled tree, one can permute the labels tosatisfy these properties, and that the resulting labeling is unique.Recall that in the proof of Theorem 4.9, we built a topological tree ¯ S on n + 1vertices from S by identifying vertices with the same labels and replacing paralleledges with single edges. After fixing a total order on L and requiring that the labelsof S satisfy the above, we obtain a canonical plane tree structure on ¯ S . The root of¯ S is the image of the root of S , and each non root vertex of ¯ S receives a unique labelfrom 1 to n . The children of any vertex are ordered using the order in [[ n ]] ⊂ Z > . Inthe resulting plane tree, which we also denote by ¯ S , the labels in a given level increasewhen read from left to right, and labels in higher levels are larger than those in lowerlevels. In particular, the labeling of ¯ S is uniquely determined; for this discussion,we will say that the plane tree ¯ S has been canonically labeled. Figure 12 shows anexample of an admissibly 2-labeled plane tree S satisfying the conditions (i)–(iv),and the resulting canonically labeled plane tree ¯ S .We have thus constructed the map ρ in (15). We claim ρ is surjective. We let¯ S be a canonically labeled plane tree, and we build an admissibly m -labeled tree YPERGRAPH CATALAN NUMBERS 21 ( S, l ) ∈ ρ − ( ¯ S ) as follows. Suppose ¯ v ∈ ¯ S is the root with children x , . . . , x d , readfrom left to right. Under the root v ∈ S we place dm vertices(16) x (1)1 , . . . , x ( m )1 , . . . , x (1) d , . . . , x ( m ) d , from left to right with l ( x ( j ) i ) = l ( x i ), and put ρ ( x ( j ) i ) = ρ ( x i ). Now suppose ¯ v ∈ ¯ S is anon root vertex with d children. By induction on distance to the root, we may assumethat ¯ v has already been lifted to m vertices v (1) , . . . , v ( m ) with l ( v ( i ) ) = l (¯ v ), andthese vertices have been placed in their level in S in some order. We lift the children x , . . . , x d of ¯ v to dm vertices as before, again with l ( x ( j ) i ) = l ( x i ) and ρ ( x ( j ) i ) = ρ ( x i ),and we make them all children of v (1)1 in the order (16). Continuing in this way weobtain an admissibly m -labeled ( S, l ) ∈ ρ − ( ¯ S ) with l satisfying (i)–(iv), which shows ρ is surjective.To complete the proof, we claim that the inverse image ρ − ( ¯ S ) is in bijection withthe ( ℓ, h )-colorings of ¯ S . To see this we revisit the proof of surjectivity and see whatadditional choices one could make along the way to build ( S, l ). First consider theroot vertex ¯ v ∈ ¯ S and suppose it has d children x , . . . , x d . As before we lift thesechildren to dm vertices x (1)1 , . . . , x ( m ) d under the root v ∈ S , but we are not obligatedto order them as in (16). We can use any order compatible with the induced orderingof the labels of the x , . . . , x d and with the requirements (i)–(iv). In particular thisimplies that these orders are in bijection with set partitions of [[ dm ]] into d blocks B , . . . , B d of size m : if the vertices x ( j ) i are ordered under v as y , . . . , y dm , then l ( y i ) = k if and only if i ∈ B k . Since there are W m ( dm ) such set partitions, we see that the root must be coloredaccording to h .Now consider a non root vertex ¯ v with d children x , . . . , x d . Again by inductionwe know that the vertices v (1) , . . . , v ( m ) mapping to ¯ v have been placed in order intheir level in S . To place the dm vertices x (1)1 , . . . , x ( m ) d in their level we must do twothings:(A) We must choose an order of the x ( j ) i in their level, and this order must becompatible with l ( x ( j ) i ) = l ( x i ) and the requirements (i)–(iv).(B) After fixing the order of the x ( j ) i , we must decide how to place them undertheir (potential) parents v (1) , . . . , v ( m ) .As before (A) corresponds to a set partition of [[ dm ]] into d blocks of size m . Thedata in (B) corresponds to an order preserving map from [[ dm ]] to [[ m ]]: if x ∈ { x ( j ) i } is a child of v ∈ { v ( j ) } , then any x ′ > x (according to the order chosen in (A)) cannotbe a child of any v ′ < v . Such maps are counted by λ ( m, dm ). Indeed, a degree dm monomial z e · · · z e m m encodes that the first e in [[ dm ]] map to 1 ∈ [[ m ]], the next e map to 2, and so on. This means the total number of choices is W m ( dm ) · λ ( m, dm ), which implies that any non root vertex with d children must be colored according to ℓ . We have thus shown that any ( S, l ) in the inverse image ρ − ( ¯ S ) correspondsuniquely to an ( ℓ, h )-coloring of ¯ S . By Proposition 5.3, this completes the proof. (cid:3) PSfrag replacements
11 1 22 2 333 444 555 666 777 888 −→ S ¯ S Figure 12.
The map ρ takes the plane tree S on nm + 1 vertices toa canonically labeled plane tree ¯ S on n + 1 vertices. At this point the reader may be nostalgic for the recurrence relationsatisfied by the classical Catalan numbers, which corresponds to the equation(17) xF − F + 1 = 0 . For m >
1, the power series F m ( x ) does not appear to be algebraic, so unfortunatelyone does not have such a simple recurrence relation on its coefficients. However,experimentally one finds that there is an algebraic relation satisfied by f m and F m :(18) f m − xF m + x = 0 . Relation (18) is easily checked when m = 1. We have f = xF , so (18) is reallythe same as (17). For m > ℓ m and h m are related by xℓ m ( x ) = h m ( x ) − , from which (18) easily follows.The relation (18) gives a connection between pairs of objects computing hypergraphCatalan numbers in the spirit of that encoded by (17). Consider m = 2. We have f ( x ) = x + 3 x + 24 x + 267 x + · · · , and f ( x ) = x + 6 x + 57 x + 678 x + · · · = x ( F ( x ) − . YPERGRAPH CATALAN NUMBERS 23
We see C (2)3 = 57 as the coefficient of x in f ( x ) , which comes from the coefficientsof x, x , x in f ( x ) via(19) 57 = 24 · · · . This is certainly reminiscent of the classical Catalan relation, although there is animportant difference. The numbers involved on the right of (19) are connected with C (2)2 , C (2)3 , C (4)2 , but they enumerate proper subsets of the associated objects, not thefull sets. Indeed, this is obviously true, since both sides of (19) involve the samenumber C (2)3 ! We have not attempted to explore this connection further. We conclude this section by discussing asymptotics for the C ( m ) n . The resultshere are purely experimental; none have been proved, although based on our numeri-cal experiments we are confident in them. We learned this technique from Don Zagier,who calls it multiplying by n ; an excellent lecture by him at ICTP demonstratingthe method can be found online [20].Suppose one has a sequence a = { a n } n ≥ that one believes satisfies an asymptoticof the form(20) a n ∼ C + C /n + C /n + · · · ( n → ∞ ) . It may be difficult to extract C to high precision even when n is large, since C /n might still be non-negligible. However it is possible to wash away the contributionsof the nonconstant terms C k /n k , k ≥
1. One multiplies both sides of (20) by n (or any other reasonable even power of n ), and then applies the difference operator(8!) − ∆ to both sides, where ∆ a is the sequence (∆ a ) n := a n +1 − a n . The operator∆ k annihilates any polynomial p ( n ) of degree < k , takes n k to k !, and takes n − l to arational function in n of degree − l − k . Let b = { b n } be the sequence resulting frommultiplying a n by n /
8! and applying the difference operator ∆ eight times. Thenwe have(21) b n ∼ C + C p − ( n ) + C p − ( n ) + · · · , where p − k ( n ) denotes a rational function in n of degree − k . If one then evaluatesthe left of (21) at a large value of n , the effects of C k , k ≥ C . One can then repeat the process with the sequence n ( a n − C ) to find C , and so on. We illustrate with the series F ( x ) = 1 + x + 6 x + 57 x + 678 x + · · · . Playingwith the data one makes the ansatz(22) C (2) n ∼ KA n n ! n ρ Here the degree of a rational function
P/Q in one variable means deg P − deg Q . for some constants K, A, ρ . Indeed, apart from the n !, the right of (22) is typical forthese kinds of problems, and was our initial guess; it quickly became evident that n !needed to be included. The series a n := C (2) n /n !should then appear to grow exponentially, and the sequence of ratios { a n /a n − } should satisfy (20) with C = A . Indeed, using 100 terms of F and ∆ we get A ≈ . · · · . Next we consider the sequence a n := C (2) n / (2 n n !) , which we expect to be asymptotic to Kn ρ . We can detect ρ using the sequence b n = (∆ log a ) n / (log( n + 1) − log n ), which satisfies (20) with C = − ρ . Again with100 terms and ∆ we get − ρ ≈ . · · · . Hence we have C (2) n ∼ K · n n ! √ n and we must find the constant K . We consider b n = C (2) n √ n/ n n ! and look for C . This time finding K is more difficult. Taking 200 terms and applying ∆ ,this number appears to be 5 . · · · . We use the Inverse SymbolCalculator [2], which attempts to symbolically reconstruct a given real number usingvarious techniques, and find K ≈ e / / √ π. (The e / is surprising, but is apparently correct. Using 600 terms of the sequence,we find that K √ π/ e / with relative error < − .) The conclusion is C (2) n ? ∼ e / · n +1 n ! √ πn , ( n → ∞ ) . We present asymptotics for the C ( m ) n as a conjecture: Let m > . Then as n → ∞ , we have C ( m ) n ∼ K m · (cid:0) m m − ( m − (cid:1) n +1 ( n !) m − ( πn ) ( m − / , where the constant K m is defined by K m = e / if m = 2 , (cid:0) (cid:1)(cid:0) (cid:1) · · · (cid:0) m − (cid:1) /m (2 m − / if m ≥ and is odd, and √ (cid:0) (cid:1)(cid:0) (cid:1) · · · (cid:0) m − (cid:1) /m (2 m − / if m ≥ and is even. YPERGRAPH CATALAN NUMBERS 25
We have tested Conjecture 5.10 numerically using 100 terms of F m ( x ) for all m ≤
30. We have not systematically tried to find higher terms in theasymptotic expansion of C ( m ) n , as in (20). Connection with matrix models
We finish by explaining how the numbers C ( m ) n are related to hypergraphs andmatrix models [5]. We first explain the connection between graphs, matrix models,and the usual Catalan numbers. For more information, we refer to Harer–Zagier [9],Etingof [6, § Let dµ ( x ) be the measure on polynomial functions on R with moments h x r i := Z R x r dµ ( x ) = W ( r ) , where W ( r ) is the number of pairings on a set of size r . It is well-known that dµ ( x )is essentially the Gaussian measure, up to normalization: we have h x r i = (2 π ) − / Z R x r e − x / dx, where dx is the usual Lebesgue measure on R .Let g , g , . . . be a family of indeterminates, and let S ( x ) be the formal power series P r ≥ g r x r /r !. We can compute the expectation h exp( S ( tx )) i as a formal power seriesin t with coefficients in the polynomial ring Q [ g , g , . . . ]. We have(23) h exp S ( tx ) i = 1 + A t / A t / A t /
48 + · · · where A = g + g , A = g + 6 g g + 4 g g + 3 g + g ,A = g + 15 g g + 20 g g + 45 g g + 15 g g + 60 g g g + 6 g g + 15 g + 15 g g + 10 g + g . The series (23) can be interpreted as a generating function for graphs weighted bythe inverse of the orders of their automorphism groups (cf. [6, Theorem 3.3]). Let n = ( n , n , . . . ) be a vector of nonnegative integers, with n i nonzero only for finitelymany i . Let | n | = P n i . We say a graph γ has profile n if it has n i vertices ofdegree i . Let G ( n ) be the set of all graphs of profile n , up to isomorphism (we allowloops and multiple edges). By an automorphism of a graph, we mean a self-mapthat permutes edges and vertices. In particular, automorphisms include permutingmultiedges between two vertices, and flipping loops at a vertex (exchanging the two half-edges emanating from the vertex that form the loop). For any γ ∈ G ( n ), letΓ( γ ) be its automorphism group. Then we have h exp S ( tx ) i = X n t | n | X γ ∈ G ( n ) Q i g n i i | Γ( γ ) | . For example, consider the term 5 g /
24 from A in (23). There are two graphs withthis profile, shown in Figure 13. The left has 2 · · · / /
12 = 5 / Figure 13.
The two graphs with profile g . Now we want to replace the Gaussian measure, which is connected to countingpairings of a set, with something that is connected to the numbers W m ( r ), whichcount set partitions of [[ r ]] with blocks of size 2 m . Let dµ m ( x ) be the “measure”on polynomial functions on R that gives the monomial x r the expectation W m ( r ).More precisely, we consider the function taking x r to W m ( r ) and extend linearlyto polynomials. This is not a measure in the usual sense, although formally we canregard it as such. The “expectation” h exp S ( tx ) i m is then a well-defined power seriesin t , and has a combinatorial interpretation via hyperbaggraphs.Recall that a hypergraph on a vertex set V — a notion due to Berge [1] — is acollection of subsets of V , called the hyperedges . The degree of a vertex is the numberof hyperedges it belongs to, and a hypergraph is regular if these numbers are the samefor all vertices. The order of a hyperedge is its number of vertices. If all hyperedgeshave the same order, we say that the hypergraph is uniform .Now suppose we allow V to be a multiset, in other words a set with a multiplicitymap V → Z ≥ . Then these constructions lead to hyperbaggraphs , due to Ouvrard–LeGoff–Marchand-Maillet [13].( ) We extend the notions of regularity and uniformityabove by incorporating the multiplicity in an obvious way (the order of a subset of amultiset is sum of the multiplicities of its elements).With these definitions, the expectation h exp S ( tx ) i m now enumerates uniformhyperbaggraphs of all profiles weighted by the inverses of their automorphism groups,where each hyperedge has 2 m elements. For example,(24) h exp S ( tx ) i = 1 + B t /
24 + B t / · · · In the CS literature, multisets are sometimes called bags . YPERGRAPH CATALAN NUMBERS 27 where B = g + 6 g g + 4 g g + 3 g + g , B = g + 28 g g + 56 g g + · · · + 35 g + g . The computation of the contribution 35 g / B in (24) is as follows. Thereare three hyperbaggraphs of this profile, each with two hyperedges. The underlyingset of vertices has 2 elements a , b , and we represent a hyperedge by a monomial inthese variables. The profile g means that each vertex has degree 4, and since 2 m = 4we must have uniformity 4. Thus we want pairs of monomials in a, b of total degree4. This gives(25) { a , b } , { a b, ab } , { a b , a b } . The orders of the automorphism groups are(26) 2 · (4!) , · (3!) , · · (2!) (2!) . For example, the automorphisms of the last hyperbaggraph come from interchangingthe vertices, interchanging the two hyperedges, and the internal flips within thehyperedges; the last type of automorphism cannot occur for graphs. Adding theinverses of these orders, one finds 1 / /
72 + 1 /
64 = 35 / B above.As a final remark, we note that A = B . This is a general phenomenon: onecan show that the coefficient of t n in h exp S ( tx ) i m is the complete exponential Bellpolynomial Y n ( g , . . . , g n ), divided by (2 m )! d d !, where d = n/ m . We refer to [3,p. 134, eqn. 3b] for the definition of these; the coefficients of the Y n can be found onOEIS as sequence A178867 . Now we pass to matrix models. Let V = V N be the real vector space of N × N complex Hermitian matrices. The space V has real dimension N . For anypolynomial function f : V → R , define(27) h f i = C − Z V f ( X ) exp( − Tr X / dX, where Tr( X ) = P i X ii is the sum of diagonal entries and the constant C is determinedby the normalization h i = 1. The measure exp( − Tr X / dX is essentially theproduct of the Gaussian measures dµ ( x ) from § V . The only difference is that for any off-diagonal entry Z ij = X ij + √− Y ij , wehave rescaled the measure so that for even r we have (cid:10) X rij (cid:11) = (cid:10) Y rij (cid:11) = W ( r ) / r/ .Now consider (27) evaluated on the polynomial given by taking the trace of the r th power:(28) P ( N, r ) = h Tr X r i . For r odd (28) vanishes for all N . On the other hand, for r even and N fixed, itturns out that P ( N, r ) is an integer, and as a function of N is a polynomial of degree r/ Furthermore, the number P ( N, r ) has the following remarkable combinatorial in-terpretation. Let Π r be a polygon with r sides. Any pairing π of the sides of Π r determines a topological surface Σ( π ) endowed with an embedded graph (the imagesof the edges and vertices of Π r ). Let v ( π ) be the number of vertices in this embeddedgraph. Then we have(29) P ( N, r ) = X π N v ( π ) , where the sum is taken over all oriented pairings of the edges of Π r such that theresulting topological surface Σ π is orientable. For example, we have(30) P ( N,
0) =
N, P ( N,
2) = N , P ( N,
4) = 2 N + N,P ( N,
6) = 5 N + 10 N , P ( N,
8) = 14 N + 70 N + 21 N. The pairings yielding P ( N,
4) are shown in Figure 14.PSfrag replacements N N N Figure 14.
Computing P ( N,
4) = 2 N + N . One can see from (30) that the leading coefficient of P ( N, r ) is none other thanthe Catalan number C r/ . This follows easily from the interpretation of the Catalannumbers in terms of polygon gluings ((vi) from § r such that the number of vertices v ( π )in the orientable surface Σ r is maximal ; this is exactly the interpretation above. Now we modify the matrix model. We replace the Gaussian measureexp( − Tr X / dX by the product of the formal measures dµ m ( x ) taken over the real coordinates;again we rescale on the off-diagonal coordinates so that for r ≡ m we have (cid:10) X rij (cid:11) m = (cid:10) Y rij (cid:11) m = W m ( r ) / r/ (2 m ) . We write the corresponding formal measure by dµ m ( X ). Then we obtain a new matrix model where polygons are glued by groupingtheir edges into subsets of size 2 m instead of pairs. As above one can see that for r ≡ m , the integrals(31) Z V Tr X r dµ m ( X ) YPERGRAPH CATALAN NUMBERS 29 are polynomials P m ( N, r ) in the dimension N . For example, when 2 m = 4 we have P ( N,
4) = N , P ( N,
8) = 6 N + 21 N + 8 N,P ( N,
12) = 57 N + 715 N + 2991 N + 2012 N. The hypergraph Catalan numbers C ( m ) r are the leading coefficients of these polyno-mials: we have P m ( N, mr ) = C ( m ) r N r +1 + · · · . A direct computation with the definition (31) shows that these numbers are computedvia Definition 2.5. For more details about these matrix models and the geometry ofthe polynomials P m ( N, mr ), see [5]. References [1] C. Berge,
Graphs and hypergraphs , North-Holland Publishing Co., Amsterdam-London; Amer-ican Elsevier Publishing Co., Inc., New York, 1973, Translated from the French by EdwardMinieka, North-Holland Mathematical Library, Vol. 6.[2] J. M. Borwein, P. B. Borwein, and S. Plouffe,
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Nauty , available at http://pallini.di.uniroma1.it .[13] X. Ouvrard, J. L. Goff, and S. Marchand-Maillet,
Adjacency and tensor representation ingeneral hypergraphs. part 2: Multisets, hb-graphs and related e-adjacency tensors , CoRR abs/1805.11952 (2018), arXiv:1805.11952.[14] N. J. A. Sloane,
The On-Line Encyclopedia of Integer Sequences , oeis.org .[15] R. P. Stanley, .[16] R. P. Stanley, Enumerative combinatorics. Vol. 2 , Cambridge Studies in Advanced Mathemat-ics, vol. 62, Cambridge University Press, Cambridge, 1999, With a foreword by Gian-CarloRota and appendix 1 by Sergey Fomin.[17] R. P. Stanley,
Catalan Numbers , Cambridge University Press, New York, 2015. [18] R. P. Stanley,
Algebraic combinatorics , Undergraduate Texts in Mathematics, Springer, Cham,2018, Walks, trees, tableaux, and more.[19] M. Wilson, 2019, personal communication.[20] D. Zagier,
ICTP Basic Notions Seminar Series: “Asymptotics” , 2014, . Department of Mathematics and Statistics, University of Massachusetts, Amherst,MA 01003-9305
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