SStokes posets and serpent nests
F. Chapoton ∗ May 15, 2018
Abstract
We study two different objects attached to an arbitrary quadrangulation of a regular polygon. Thefirst one is a poset, closely related to the Stokes polytopes introduced by Baryshnikov. The second oneis a set of some paths configurations inside the quadrangulation, satisfying some specific constraints.These objects provide a generalisation of the existing combinatorics of cluster algebras and nonnestingpartitions of type A . keywords: poset, quadrangulation, Tamari lattice MSC:
Introduction
The research leading to this article started with the desire to understand the apparition of associahedra inan article of Baryshnikov [4] and in an article of Kapranov and Saito [20]. Associahedra, also known asStasheff polytopes, are classical in algebraic topology, where they are used to define and study associativityup to homotopy [29]. More recently, they have made their way into the theory of cluster algebras, in whichthey control the combinatorial aspects of the type A cluster algebras [14]. It was not clearly obvious how torelate the contexts used by Baryshnikov and Kapranov-Saito to either associativity or cluster algebra theory.Moreover both articles were seeing associahedra as members of a larger family of polytopes, all closely tiedtogether, whereas associahedra usually stand alone.Another strand of research then came into play, centred on the Tamari lattices [24]. The Tamari latticesare closely related to associahedra and associativity, and can be defined by orienting the associativity relationfrom left-bracketing to right-bracketing. They are also well-understood in the context of cluster algebras,where they are just one among several Cambrian lattices of type A , depending on the choice of an orientationof the Dynkin diagram [25, 26]. It is remarkable that all the Cambrian lattices attached to type A n sharethe same underlying graph, which is nothing else that the graph of vertices and edges of the associahedra.This is also the flip graph of triangulations and the mutation graph of cluster algebras of type A .The Tamari lattices have yet another, maybe less well-known, interpretation. They can be seen as aspecial case of the partial orders defined by Happel-Unger [15] and Riedtmann-Schofield [28] on the sets oftilting modules over quivers.One can consider the notion of module over the Tamari lattices, defined for example using their incidencealgebras. It turns out that there is a very interesting subset of modules, indexed by the set of quadrangu-lations of a regular polygon. It would be too long to tell here how precisely these modules were stumbledupon, but it involved an anticyclic operad made with all Tamari lattices, and a mysterious map from the K of Tamari lattices to rational functions [9]. All this seems in fact to take place naturally in the Tamarilattices seen as partial orders on tilting modules.Anyway, these modules indexed by quadrangulations can be seen as living in the derived categories ofmodules over the Tamari lattices. A natural question to ask was then: what are the morphisms betweenthese modules ? It appeared that they may be described using flips of quadrangulations, under the conditionthat a flip should not be repeated twice. ∗ L’auteur a b´en´efici´e d’une aide de l’Agence Nationale de la Recherche (projet Carma, r´ef´erence ANR-12-BS01-0017). Ilremercie aussi l’Institut Mittag-Leffler pour son accueil chaleureux et ses excellentes conditions de travail. a r X i v : . [ m a t h . R T ] M a y t this point, the connection was made with the work of Baryshnikov, which also involved quadrangu-lations and their flips. It turned out that the match was perfect. This is essentially the story behind thecontent of the article, which concentrates on the combinatorial side and leaves most of the representationtheory aspects aside.Let us now describe what is done, and then comment on the results.In section 1, a poset is associated with every quadrangulation in a regular polygon. This is done usingthe compatibility relation between quadrangulations introduced by Baryshnikov. First, one defines directedgraphs (digraphs) by orienting the allowed flips in a specific way. Then it takes some work to show thatthese digraphs are Hasse diagrams of posets, that we choose to call the Stokes posets. The main tool isan inductive description of the digraphs, relating a quadrangulation and a smaller one with a leaf squareremoved. One also shows that the associated undirected graphs are connected and regular.In the articles of Baryshnikov and his collaborators on the subject [4, 5, 16], no digraphs were considered,but instead polytopes were associated with quadrangulations. Their graphs of vertices and edges coincidesby definition with the associated undirected graphs of our digraphs.In the rest of the first section, one proposes a conjectural manner to define the same digraphs withouthaving to use the compatibility condition. The proof that this works is sadly missing.In section 2, one considers examples and properties of the Stokes posets. In particular, one proves thatthe Tamari lattice is recovered as a special case, for a very regular quadrangulation.It is then shown that the posets attached to some quadrangulations can be written as Cartesian productsof smaller posets of the same kind. This happens under the condition that there is a bridge in the quadran-gulation. One then says a few words about the simplicial complexes attached to the quadrangulation, whichcan be thought of as the dual of Baryshnikov’s Stokes polytopes. In particular, enumerative aspects of thissimplicial complex can be encoded into an F -triangle.Next, a conjecture is proposed about what happens when a quadrangulation is twisted along an edge,namely that the undirected graph, simplicial complex and F -triangle should not change.Then section 3 is a rather sketchy proposal for a vast generalisation of the theory of Stokes posets to otherroot systems and finite Coxeter groups. This is stated first in the setting of exceptional sequences, and thenusing factorisation of Coxeter elements into reflections, or equivalently maximal chains in the noncrossingpartition lattices.In the next section 4, one introduces combinatorial configurations fitting inside quadrangulations. Theyare made of elementary pieces called serpents , and therefore named serpent nests . These combinatorialobjects should be closely related to the posets introduced in the first section. For every quadrangulation Q ,there should be as many elements in the poset attached to Q as there are serpent nests in Q . The serpentnests should be thought of as playing the same role as nonnesting partitions play for classical root systems.After their definition, the section contains various results about serpent nests, that sometimes are similarto some statements or conjectures about the Stokes posets made in the previous sections.First, it is shown that serpent nests do factorise in presence of a bridge, just as Stokes posets do. Thenan enumerative study is made, by introducing h -vectors and H -triangles that counts serpent nests accordingto their ranks. A duality is obtained, that implies a symmetry of the h -vector. The H -triangle is conjecturedto be related to the F -triangle, by the same formula that work between clusters F -triangles and nonnestingpartitions H -triangles for all root systems (see for example [31] and the references therein).It is then shown that, for the same regular quadrangulation already considered in the first section, theserpent nests are in a simple bijection with nonnesting partitions of type A . Another interesting family ofexamples is considered, for which the number of serpent nests is given by a Lucas sequence. This givesessentially the (bridge-less) quadrangulations with the smallest possible numbers of serpent nests.A paragraph is devoted to an interesting strategy to count serpent nests, namely cut quadrangulations intwo parts along one edge, compute something related to open serpent nests for every half, and reconstitutethe result from that. This seems also to be related to the Dynkin type B and the usual folding procedure ofsimply-laced Dynkin diagrams.Then section 5 deals with an algebraic structure that gathers all the quadrangulations together, namelya commutative algebra endowed with a derivation. Product is disjoint union or union along a bridge andthe derivative is given by cutting at inner edges. This is analogue to the parabolic structure on quivers andDynkin diagrams.The last section 6 contains elementary enumerative results about various sorts of quadrangulations. The2quivalence classes under twisting are also counted, as they should correspond to the distinct undirectedgraphs of Stokes posets and do describe quadrangulations with the same serpent nests. For some reasonrelated to fans and not studied in this article, quadrangulations that do not contain a cross are also considered.Let us now summarise the article and give further comments. Taking as input a quadrangulation, onedefines a poset and a finite set. For some special quadrangulations, one recovers the Tamari lattices and thenonnesting partitions associated to the Dynkin type A by the theory of cluster algebras and root systems. Itis in fact expected that all the posets will be lattices, and also that other Cambrian lattices can be obtainedfor other ribbon quadrangulations.It is therefore very tempting to make an analogy where quadrangulations play the role of quivers, andquadrangulations up to twist that of Dynkin diagrams. Then one gets analogues of clusters, positive clusters,exchange graphs, but only at a combinatorial level. It seems that most of the properties known in thecombinatorial side of cluster theory do extend. It would be interesting to see if analogues of cluster variablescould make sense.As explained before, the article [20] of Kapranov and Saito was one source of motivation. The relationshipwith our results is not as direct as in the case of Baryshnikov, but should exist. It would probably be bestseen using the noncrossing trees and exceptional sequences point of view explained in section 3.Flips in quadrangulations have been considered also in the context of m -analogues of cluster theory for m = 2. The 2-Cambrian lattices introduced recently in [30] are defined by allowing two consecutive flipsinside an hexagon, but not three. The precise relationships with this article remain to be explored.As a side remark, one can note that very similar combinatorial objects as the ones involved here (quad-rangulations and noncrossing trees) have appeared in the study of multiple zeta values [13, 12]. Whetherthere is a connection or not remains to be settled.The name of Stokes polytopes was used by Baryshnikov because of the relation of his work with theStokes phenomenon. In the article [19], the Stokes phenomenon is also connected to cluster algebras. It isnot clear if these connections are the same or at least could fit in a common framework.Let us finish by saying that there are several aspects missing in the present article, the most importantones being the in-depth study of related fans and polytopes, only sketched in [4], and also the closely relatedquestion to check that the posets are indeed lattices. These questions are left for future work. Informally, a quadrangulation is a partition of the interior of a regular polygon into quadrilaterals. This ispossible if and only if the regular polygon has an even number of vertices.More formally, one considers a regular polygon with vertices numbered from 0 to 2 n + 1 in the negative(clockwise) direction. A quadrangulation is defined as a collection of pairs of (not consecutive) vertices ofthis polygon, such that the associated edges do not cross and define only quadrilateral regions.Figure 1: Two views of the same quadrangulation ( n = 6)It will sometimes be convenient to depict quadrangulations by deforming the quadrilaterals so that theybecome more regular and of similar area. The boundary is then no longer a regular polygon, but a simplepolygonal closed curve, as in the right of Figure 1. 3he number of distinct quadrangulations inside the polygon with 2 n + 2 vertices is also the number ofternary trees with 2 n + 1 leaves and is classically given by12 n + 1 (cid:18) nn (cid:19) . (1)The bijection is the obvious planar duality, after a boundary edge has been chosen as root.Let us fix some terminology. A quadrangulation in the regular polygon with 2 n + 2 vertices is made of2 n + 2 boundary edges and n − inner edges . The n quadrilateral regions will be called squares . The word edge will generally be used to designate both inner and boundary edges, unless the context make it clearthat it must be of one kind only. Y. Baryshnikov has introduced a compatibility condition between quadrangulations. Let us describe thisprecisely.It will be convenient for us to consider a fixed quadrangulation Q drawn inside a regular polygon with2 n + 2 vertices, and to define only compatibility with Q . Let us choose an alternating black and whitecolouring of the vertices of the regular polygon. This choice is arbitrary, and the compatibility conditionwill not depend on it.The quadrangulation Q is seen as a (blue and dashed) theatre backdrop, on which one superimposesanother (red) regular polygon, slightly rotated in the negative (clockwise) direction by an angle of π n +4 , sothat the vertices interlace. The black-and-white colouring of the vertices of the rotated polygon is inheritedfrom the initial colouring.Figure 2: Backdrop Q (blue and dashed) and two Q -compatible Q (cid:48) and Q (cid:48)(cid:48) (red)By convention, all the edges (red or blue) are oriented from white vertices ◦ to black vertices • .An edge i in the red polygon is said to be compatible with an edge j of Q if the pair (red oriented edge i , blue oriented edge j ) defines the positive orientation of the plane.An edge i in the red polygon is said to be Q - compatible if it is compatible with all edges of Q . This isequivalent to say that i is compatible with all inner edges of Q , because the condition is always satisfiedwith respect to boundary edges of Q .Moreover, compatibility with Q is also automatic for the boundary edges of the red polygon.A quadrangulation Q (cid:48) is Q - compatible if and only if all its edges are Q -compatible. This is equivalent torequire that every inner edge of Q (cid:48) is compatible with all inner edges of Q .Note that Q -compatibility is unchanged under switching the choice of the black and white colouring ofthe vertices, as this changes the orientation of all edges.It is clear that there is a finite number of Q -compatible quadrangulations. This number depends on Q .There are always two simple ones. First, Q (seen in the red polygon after a negative rotation by π n +4 )4s clearly compatible with itself. The other one is the quadrangulation obtained from the backdrop Q bypositive rotation by π n +4 (or from the red Q by positive rotation by π n +2 ). Let us call it τ ( Q ) . Proposition 1.1
Let Q be a fixed quadrangulation. Let Q (cid:48) be a Q -compatible quadrangulation. For everyinner edge i of Q (cid:48) , there exists exactly one other inner edge j such that Q (cid:48) \ { i } ∪ { j } is a Q -compatiblequadrangulation. Proof.
This follows from Lemma 1.2 below.
Lemma 1.2
Let Q be a fixed quadrangulation. Consider an hexagon H made of six Q -compatible edges.Then there are exactly two Q -compatible inner edges in the interior of H . Proof.
Let a , b , a , b , a , b be the vertices of the hexagon H in clockwise order, with indices in Z / Z .One can assume without restriction that the colours are white for a vertices and black for b vertices. Itfollows from the Q -compatibility of H that inner edges of Q can only enter H by crossing a inner edge a i +1 → b of H and can only exit by crossing a inner edge a i → b i of H . If there is no inner edge in Q entering H through a i +1 → b i and leaving H through a i +2 → b i +2 for some i (call it a long inner edge ),then Q contains an empty hexagon, which is absurd. Therefore there is at least a long inner edge in Q , sayfrom a → b to a → b . All long inner edges must then enter and exit H by crossing the same edges of H ,for otherwise they would cross. Then among the three possible inner edges inside the hexagon H , one cancheck that a → b and a → b are Q -compatible, but that a → b is not. Proposition 1.1 implies that one can move between Q -compatible quadrangulations by removing any inneredge and inserting another one. These moves between quadrangulations will be called flips .This defines a graph with vertices the Q -compatible quadrangulations and edges the flips. This is clearlya regular graph where every vertex has arity n −
1. Let us denote it by γ Q and call it the flip graph of Q -compatible quadrangulations. Later on, one will show that this graph is connected.Let us now describe how one can orient the edges of the flip graph γ Q .Figure 3: Simple example of flip, oriented from left (id) to right (op) Consider a flip relating two quadrangulations Q (cid:48) and Q (cid:48)(cid:48) . Only the inner edge inside the modified hexagon H is changed. The two possible inner edges i (cid:48) and i (cid:48)(cid:48) can be distinguished as follows. Recall from the proofof Lemma 1.2 that there must be at least one long inner edge j in Q entering the hexagon and leaving itthrough the opposite side. The starting vertex of the inner edge of H is adjacent This notation is chosen to remind of the Auslander-Reiten translation. id) either to the starting vertex of the long inner edge j , (op) or to the ending vertex of the long inner edge j .Let us orient the flips from the case of adjacent starting vertices (id) to the case of far-away startingvertices (op) .This is illustrated in Figure 3, with the (id) case on the left and the (op) case on the right.For an oriented flip from Q (cid:48) to Q (cid:48)(cid:48) , one says that it is an out-flip for Q (cid:48) and an in-flip for Q (cid:48)(cid:48) . For every Q -compatible quadrangulation Q (cid:48) , the number of in-flips plus the number of out-flips is n −
1, as every inneredge corresponds either to an in-flip or to an out-flip.This allows to define an orientation of the flip graph γ Q . Let us call this oriented graph Γ Q .To study properties of the graph Γ Q , one will proceed by induction on the size n of the quadrangulation Q , which is its number of squares.In any quadrangulation with n ≥ e of Q that boundsa square s with exactly one neighbour square . Up to switching black and white, one can assume that theedge e goes from white to black in counterclockwise order. Let s be the square bordering the chosen edge e and let Q \ s be the quadrangulation obtained from Q by removing s .Let Q (cid:48) be any Q -compatible quadrangulation. The underlying red polygon has two white vertices i and i + 2 (modulo 2 n + 2) that lie just before and just after the chosen blue edge e . Recall that the vertices arenumbered in clockwise order.One can associate with Q (cid:48) a Q \ s -compatible quadrangulation θ s ( Q (cid:48) ) as follows. Pictorially, one collapsesthe boundary section of the red polygon between white vertices i and i + 2 to a single white vertex in asmaller red polygon. Edges of Q (cid:48) then get identified if they happen to coincide after the collapsing.Let us describe now the fibres of the map θ s . If there are k inner edges in θ s ( Q (cid:48) ) starting at the collapsedvertex, then there must be k + 1 inner edges in Q (cid:48) starting either at i or i + 2. There are k + 2 choices forthe distribution of these k + 1 starting points between i and i + 2. Therefore the fibre has k + 2 elements.Moreover, these k +2 elements are naturally totally ordered by oriented flips, as there is a unique sequenceof k + 1 oriented flips going from the case where all k + 1 edges start at i to the case where all k + 1 edgesstart at i + 2. This gives a natural total order on every fibre of θ s .Oriented flips between Q -compatible quadrangulations are of two kinds. Proposition 1.3
Let Q (cid:48) −→ Q (cid:48)(cid:48) be a flip of Q -compatible quadrangulations. Then exactly one of thefollowing holds: (inFib) θ s ( Q (cid:48) ) = θ s ( Q (cid:48)(cid:48) ) and the flip goes down one step in this fibre of θ s , (outFib) θ s ( Q (cid:48) ) −→ θ s ( Q (cid:48)(cid:48) ) is a flip between Q \ s -compatible quadrangulations. Proof.
Let H the red hexagon that contains and defines the flip. Let f be the inner blue edge boundingthe square s .Assume first that f is a long inner edge in H . Then H must has two opposite sides made of an edgestarting from i and an edge starting from i + 2. There is at most one such hexagon where an out-flip ispossible, and this flip does not change the image by θ s , but moves down by one step in the total order insidethe fibber. This is the situation (inFib) .Otherwise, f is not a long inner edge in H . In this case, the hexagon H remains an hexagon in thereduced quadrangulation Q \ s , hence the flip induces a flip inside Q \ s . This is the situation (outFib) .We will say that these are in-fibre flips and out-fibre flips . Lemma 1.4
For every sequence of two flips Q (cid:48) (inFib) −→ Q (cid:48)(cid:48) (outFib) −→ Q (cid:48)(cid:48)(cid:48) , one can find a quadrangulation Q (cid:93) and flips Q (cid:48) (outFib) −→ Q (cid:93) (inFib) −→ . . . (inFib) −→ Q (cid:48)(cid:48)(cid:48) . (2) This is because a quadrangulation is a tree of squares, hence has at least one leaf. roof. If the two hexagons involved in the two consecutive flips do not overlap, the flips do just commute.One can simply exchange them.Otherwise everything happens inside an octagon. There are essentially just two cases to consider: therecan be either 4 or 5 ways to fill the octagon. In both cases, one can find the expected sequence of flips byinspection. It will involve 1 or 2 final in-fibre flips.
Proposition 1.5
For every pair of quadrangulations Q (cid:48) to Q (cid:48)(cid:48) related by a sequence of flips, one can find asequence of flips from Q (cid:48) to Q (cid:48)(cid:48) starting with out-fibre flips and ending with in-fibre flips. Proof.
This follows from the previous lemma 1.4, by repeated application at the first possible place in thesequence. This rewriting preserves the number of out-fibre flips. This process is finite, because at everyreduction step, the first out-fibre flip not already packed at the beginning get strictly closer to the beginningof the sequence.Using these tools, one can now proceed to proofs by induction on n . Proposition 1.6
Every sequence of flip-moves is finite.
Proof.
By induction on the number of squares n . This is obvious for the case n = 1, where there is no flipat all.Otherwise, let us consider an infinite sequence of flips. One can map it to a sequence made either of flipsin the smaller quadrangulation Q \ s or a down-step in the total order of a fibre of θ s . Because these fibresare finite, one can obtain an infinite sequence of flips in Q \ s . This is absurd by induction.It follows that there cannot be any oriented cycles in Γ Q . Theorem 1.7
The flip graph Γ Q is the Hasse diagram of a partial order. Proof.
The proof is by induction on n . One has to show that the graph is transitively reduced.This means that for a single flip Q (cid:48) −→ Q (cid:48)(cid:48) , there should be no longer sequence of flips starting at Q (cid:48) and ending at Q (cid:48)(cid:48) . Here longer means having at least 2 steps.Assume first that the flip Q (cid:48) −→ Q (cid:48)(cid:48) is inside a fibre of θ s . Then any flip exiting this fibre would nevercome back to this fibre by induction. So it is only possible to look for the longer sequence inside the fibre.But there cannot be any longer path between two consecutive elements of a total order.Assume now that the flip Q (cid:48) −→ Q (cid:48)(cid:48) induces a flip θ s ( Q (cid:48) ) −→ θ s ( Q (cid:48)(cid:48) ). Assume moreover that thereis some longer sequence of flips from Q (cid:48) −→ Q (cid:48)(cid:48) . By proposition 1.5, one can suppose that it starts without-fibre flips and ends with in-fibre flips.If the longer sequence starts with an out-fibre flip, this must be the flip Q (cid:48) −→ Q (cid:48)(cid:48) , otherwise one cannotreach Q (cid:48)(cid:48) , by induction hypothesis. It must then stop, otherwise it will just stop at some lower point in thefibre. So in fact, it was not longer, which is absurd.If the longer sequence starts with an in-fibre flip, it can only contain in-fibre flips, and therefore can neverreach Q (cid:48)(cid:48) which is in a different fibre. This is a contradiction.By convention, the partial order will be decreasing along the edges of the oriented graph Γ Q . Proposition 1.8
The map θ s is a morphism of posets. Proof.
This is clear by the description of flips using θ s in proposition 1.3.Recall the definition of two special Q -compatible quadrangulations Q and τ ( Q ) at the end of subsec-tion 1.1. Proposition 1.9
The unique Q -compatible quadrangulation with only out-flips is Q . The unique Q -compatiblequadrangulation with only in-flips is τ ( Q ) . roof. Let us first note that Q has only out-flips. This follows from its definition by a small negativerotation. Similarly, τ ( Q ) has only in-flips, because it is defined by a small positive rotation.It remains to show the converse statements. Let us prove that for Q and out-flips, the case of τ ( Q ) andin-flips being similar. The proof is illustrated in Figure 4.Let Q (cid:48) be a Q -compatible quadrangulation with only out-flips.Let s = ( i, j, k, (cid:96) ) be a square of Q that is a leaf in the tree structure of Q , where the boundary edges are i − j, j − k and k − (cid:96) . One can assume without restriction that j is a black vertex and k a white vertex.Let ˆ j , ˆ k, ˆ (cid:96) be the boundary edges of Q (cid:48) corresponding in the obvious way to the vertices j, k, (cid:96) . If theunique square of Q (cid:48) containing ˆ j and ˆ k also contains ˆ (cid:96) , then one can remove the boundary edges i − j, j − k and k − (cid:96) from Q and the boundary edges ˆ j , ˆ k, ˆ l from Q (cid:48) and proceed by induction. It then follows that Q (cid:48) is equal to Q .Otherwise, there are inner edges of Q (cid:48) starting at the common vertex of ˆ k and ˆ (cid:96) . One of them is boundingthe square of Q (cid:48) containing ˆ j and ˆ k . Let H be the hexagon containing this edge in its interior. One cancheck that the inner edge of H is an in-flip for Q (cid:48) , which is absurd. i j k (cid:96) i j k (cid:96) m Figure 4: Induction using a leaf of Q Proposition 1.10
The oriented flip graph Γ Q is the Hasse diagram of a connected poset. The undirectedflip graph γ Q is connected. Proof.
This follows from the existence of a unique minimal element τ ( Q ) (and a unique maximal element Q ) in the poset, proved in proposition 1.9.From now on, Γ Q will denote both the poset and its Hasse diagram. Note that these posets are not graded,as one can see already with the pentagon that one can obtain starting with some of the quadrangulations ofan octagon. Let us now consider another potential way to define the directed graph Γ Q . This is only conjectural for themoment.Fix an initial quadrangulation Q . Consider the directed graph D with vertices all quadrangulations,and edges defined by counter-clockwise rotation inside hexagons (where the edge extremities are moved toadjacent vertices). This large directed graph D obviously contains Γ Q .One would like to find another way to recover Γ Q , by keeping only some vertices and edges of D , withouthaving to use the compatibility relation.The idea is that the set of vertices of Γ Q should have the following equivalent description. Let us saythat a path in D is repetition-free if no quadrangulation occurs twice. A quadrangulation Q (cid:48) is said to be8 eachable from Q if no repetition-free path in D from Q to Q (cid:48) contains two consecutive flips inside the samehexagon. Then quadrangulations reachable from Q should be exactly the same as elements of Γ Q .If this holds, then one can build the set of reachable quadrangulations step by step, using for examplethe following algorithm. • Start from the singleton set S = { Q } . • Assume that some set S of reachable quadrangulations have been found. Then either1. there exists a quadrangulation Q (cid:48) not in S , obtained from an element of S by a flip, reachable,and such that all quadrangulations on all repetition-free paths in D from Q to Q (cid:48) are already in S ,2. or there is no such Q (cid:48) and the algorithm stops.To see what this algorithm does inside Γ Q , let us fix a linear extension of Γ Q , starting with Q and endingwith τ Q . Then the algorithm would succeed by adding the elements of Γ Q according to this linear extension.This would allow to define the graph Γ Q without using the compatibility condition introduced by Barysh-nikov, at the prize of a somewhat greater complexity. This could be useful in more general contexts whereno analogue of compatibility is known.This description of Γ Q would be summarised by the following sentence: do not rotate twice in the samehexagon. Let us first consider the special case of the quadrangulations C n with inner edges(0 , , (0 , , . . . , (0 , n −
1) (3)inside the polygon with 2 n + 2 vertices. They have n squares. An example is depicted in Figure 5. * Figure 5: The quadrangulation C of the Catalan familyLet us choose the border edge of the red polygon corresponding to the vertex 0 of the blue polygon asa root. Recall that choosing such a root allows to identify quadrangulations with rooted ternary trees, byplanar duality. The root is marked by ∗ in the figure. Proposition 2.1
The C n -compatible quadrangulations are in bijection with rooted ternary trees with nomiddle branch, and therefore also with rooted binary trees. The flips correspond to the left-to-right rotationmoves of rooted binary trees. In-flips correspond to right-branches and out-flips to left branches. Proof.
This description holds at the starting quadrangulation, which has only left branches. It remains toshow that this correspondence is preserved under flips. Details are left to the reader.Therefore, in this case, the number of Q -compatible quadrangulations is the Catalan number1 n + 1 (cid:18) nn (cid:19) , (4)9nd the poset Γ Q is the Tamari lattice (see for example [24] for its usual definition using rotation of binarytrees).More generally, for ribbon quadrangulations (defined as quadrangulations where no square has neighboursquares on opposite sides), the posets Γ Q are expected to match with the Cambrian lattices of type A .Figure 6 illustrates the directed flip graph Γ Q for a quadrangulation with 4 squares, having 12 compatiblequadrangulations. This is the simplest example of a Stokes poset that is really new, as the quadrangulationis not a ribbon one.Figure 6: A quadrangulation Q and the flip graph of its 12 Q -compatible quadrangulationsSee Figure 7 for the numbers of Q -compatible quadrangulations for some (connected) quadrangulationswith up to 6 squares.Figure 7: Quadrangulations with up to 6 squares and number of compatible quadrangulations. Let us now describe a factorisation property, that allows to restrict the attention to a special class ofquadrangulations.A square s in a quadrangulation is called a bridge if it has exactly two neighbour squares and thesesquares are attached to opposite edges of s . A quadrangulation is called connected if it does not contain anybridge. See Figure 8 for a quadrangulation with several bridges.Let Q be a quadrangulation. Assume that s is a bridge in Q . Let Q (resp. Q ) be the quadrangulationobtained from Q by removing all squares from one side of s (resp. from the other side). Equivalently, Q and Q are obtained by cutting along one edge of s and keeping the part that contains s . Proposition 2.2
The directed graph Γ Q is the direct product of the directed graphs Γ Q and Γ Q . Proof.
The hypothesis on Q says that in the polygon there are four vertices i, i + 1 and j, j + 1 that formthe given square s of Q . Because the edges ( i, j + 1) and ( j, i + 1) are oriented in opposite directions, no Q -compatible edge can cross them both. This implies that Q -compatible quadrangulations can be describedas pairs made of one Q -compatible quadrangulation and one Q -compatible quadrangulation. One cancheck that the flips occur in each factor independently. F -triangle Let Q be a fixed quadrangulation. Let G Q be the simplicial complex whose simplices are the sets of non-crossing Q -compatible inner edges. Every Q -compatible quadrangulation correspond to a maximal simplexin G Q .This simplicial complex is pure, because every partial Q -compatible quadrangulation can always becompleted into a Q -compatible quadrangulation. To see this, one can observe that a set of Q -compatibleedges cuts the polygon into pieces, and each piece behave with respect to compatibility as a smaller polygon,with an underlying quadrangulation inherited from Q .The simplicial complex G Q is the flag complex, on the ground set of Q -compatible inner edges, for thecompatibility given by being non-crossing.Let I − Q be the set of initial inner edges in Q (seen as Q -compatible inner edges), and let Φ + Q be the setof all other Q -compatible inner edges.Let us define the F -triangle as the polynomial in two variables F Q ( x, y ) = (cid:88) f ∈ G Q x f ∩ Φ + Q y f ∩ I − Q . (5)When evaluated at ( x, x ), this reduces to the usual f -vector of the simplicial complex G Q .There is some kind of parabolic structure in quadrangulations, akin to the classical theory of roots systemsor Coxeter groups. The role of simple roots is played by the inner edges. Proposition 2.3
There holds y∂ y F Q = y (cid:88) e ∈ Q F Q − e F Q + e , (6) where the sum runs over inner edges of Q , and Q − e , Q + e are the quadrangulations obtained by cutting Q along e . Proof.
The left hand side is counting simplices in the simplicial complex G Q , with a marked initial inneredge. These are clearly in bijection with the disjoint union, over inner edges of Q , of pairs of simplices inthe simplicial complexes G Q + e and G Q − e .Let now s be a bridge in a quadrangulation Q , as defined in subsection 2.2.Let Q (resp. Q ) be the quadrangulation obtained from Q by removing all squares from one side of s (resp. from the other side). Proposition 2.4
The simplicial complex Γ Q is the direct product of the simplicial complexes Γ Q and Γ Q .The F -triangle of Q is the product of the F -triangles of Q and Q . Proof.
This follows from the same basic properties of the compatibility relation as prop. 2.2.11n the work of Baryshnikov [4], all the simplicial complexes G Q are described as the dual of simplepolytopes. In particular, they are all spherical, and we will use here this result. It would certainly beinteresting to study the fans and the polytopes that are involved.The F -triangle has a nice symmetry. Proposition 2.5
There holds F Q ( − − x, − − y ) = ( − n F Q ( x, y ) . (7) Proof.
This is a simple consequence of the fact that G Q are spherical simplicial complexes, see [7, Prop. 5]for the proof of this equation in a very similar context. Let Q be a quadrangulation, and e be a inner edge of Q . Let Q (cid:48) e and Q (cid:48)(cid:48) e be the two quadrangulations definedby cutting Q along e . Let Q (cid:48)(cid:48) e be the image of Q (cid:48)(cid:48) e by a reflection in the plane (the mirror image of Q (cid:48)(cid:48) e ).The twist of Q along e (on the Q (cid:48)(cid:48) e side) is the quadrangulation defined by gluing back Q (cid:48) e and Q (cid:48)(cid:48) e alongtheir boundary edges corresponding to e .Twisting twice on the same side gives back Q . Twisting successively on opposite sides give the mirrorimage of Q .Let Q and Q (cid:48) be quadrangulations related by twisting along one inner edge. Conjecture 2.6
The undirected flip graphs γ Q and γ Q (cid:48) are isomorphic. The simplicial complexes G Q and G Q (cid:48) are isomorphic. It should be noted that the oriented flip-graphs are not the same under this hypothesis.In particular, the number of Q -compatible quadrangulations should be the same as the number of Q (cid:48) -compatibles ones.The statement of conjecture 2.6 was already proposed in [4] for the dual polytopes.Moreover, one also expects some stronger enumerative invariance. Conjecture 2.7
The F -triangles of G Q and G Q (cid:48) are equal. In this section, which does not contain much details, some generalisations of the flip graphs are proposed,using exceptional sequences on Dynkin diagrams or equivalent objects for Coxeter groups. Beware that thecorrectness of this proposal depends on the unproven alternative description of Γ Q in subsection 1.4.Let us start by a simple combinatorial reformulation of the flip graphs of quadrangulations in terms ofother combinatorial objects.A noncrossing tree in the regular polygon with n + 1 vertices is a set of edges between vertices of thepolygon, with the following properties • edges do not cross pairwise, • any two vertices are connected by a sequence of edges, • there is no loop made of edges.Note that boundary edges are allowed in the set, and a typical noncrossing tree will contain only some ofthem.There is a simple bijection between quadrangulations of the 2 n + 2-polygon and noncrossing trees in the n + 1-polygon. Assume that vertices of the 2 n + 2-polygon have been coloured black and white alternating.Then every quadrangle contains a unique black-black diagonal (one of its two diagonals). The collection ofthese diagonal edges can be seen to form a noncrossing tree. Conversely, by drawing a noncrossing tree usingthe black vertices of the 2 n + 2-polygon, one can consider all black-white edges that do not cross edges ofthe noncrossing tree. This gives back the quadrangulation.12igure 9: Bijection between quadrangulations and noncrossing treesPassing through this bijection, one can translate the flips of quadrangulations into a simple operation onnoncrossing trees. It is in fact enough to consider what happens for the flip inside an hexagon, as every flipwill behave locally the same.The result of as follows. A flip of noncrossing trees will change just one edge of the noncrossing tree.Assume that there are edges i − j and k − j in the noncrossing tree T , with i < j < k in the clockwise cyclicorder, and that there is no other edge incident to j between them inside the ambient polygon. Then the flipconsists in replacing i − j by i − k .Flipping twice in the same hexagon of a quadrangulation translates into flipping again along two othersides of the same triangle ( i, j, k ) of a noncrossing tree.Using the conjectural alternative description of subsection 1.4, one would therefore be able to restate thedigraph Γ Q in terms of flips of noncrossing trees, not flipping twice consecutively along two sides of the sametriangle. For short, let A n denote in this section the equi-oriented quiver of type A n . The noncrossing trees in the n + 1-polygon have been related in [1] to exceptional sequences (up to shifts and reordering) in the derivedcategory D mod A n .The correspondence goes as follows. By numbering the vertices of the n +1-polygon from 0 to n clockwise,every inner edge gets a label ( i, j ) with i < j . Let the inner edge ( i, j ) correspond to the indecomposablemodule over A n with support [ i + 1 , j ].Then Araya proved that a noncrossing tree is sent by this map to a collection of indecomposable modulesthat can be ordered into an exceptional sequence and, conversely all the underlying sets of indecomposablemodules coming from exceptional sequences are obtained in this way. Let us call such a collection ofindecomposable modules an exceptional set .Given the description above of the flips acting on noncrossing trees, one can readily check that theycorrespond, after the bijection to exceptional sets, to the action of the braid group on exceptional sequences(see for example [11] for the braid group action). More precisely, given an exceptional set, pick an orderinginto an exceptional sequence, act by one generator s i of the braid group to get another exceptional sequence,and then forget about the ordering. This describe what the flips are.Then moving twice consecutively in a same triangle gets translated to acting twice by the same s i at thesame place.This would allow to generalise the definition of the flip graphs to any finite Dynkin diagram, providedthat the conjectural description of subsection 1.4 holds. In fact, it is even possible to get from here to the more general setting of finite Coxeter groups.It has been known since the article [18] (see also [17]) that exceptional sequences in derived categories ofrepresentations of Dynkin quivers are closely related to the corresponding lattices of noncrossing partitions.More precisely, let W be a finite crystallographic Coxeter group, and c be a Coxeter element in W . Asis well-known, this data can be translated into a quiver Q with underlying graph the Dynkin diagram. It could also be called an exceptional collection, but there is no universal agreement on the meaning of that. c , andexceptional sequences of modules over Q . In fact, maximal chains in the noncrossing partition lattices arethe same as factorisations of the Coxeter element c as a product of reflections .In this new language, the action of the braid group on exceptional sequences get translated to a similaraction of the braid group on maximal chains of the noncrossing partition lattice . When thinking to maximalchains as factorisations of the Coxeter element into reflections, this is given by a simple conjugation action.This interpretation would allow to extend the definition of the flip graphs to all finite Coxeter groups,where quadrangulations are replaced by maximal chains in the noncrossing partition lattice, up to reordering.In these generalisations, some properties are lost. In particular, the flip graphs are no longer regular, asone can see already for some examples in type D . In this section, we turn to another aspect of quadrangulations, that should in fact be related to the previousone.Let us start by some combinatorial definitions.A serpent in a quadrangulation Q is a set of two distinct squares s, s (cid:48) such that the unique path from s to s (cid:48) in Q is made of a sequence of right-angle turns. This means that following this path, one neverenters and leaves a square through opposite sides. Abusing notation, one will also consider that the path isthe serpent. By convention, the path starts and ends at the centre of the squares. See the leftmost part ofFigure 11 for a visual example.A serpent nest is a set of serpents inside a quadrangulation, satisfying an extra condition and modulo anequivalence relation: • The extra condition is the following: two serpent extremities s and s can share the same square ifand only if the serpents do not leave this square by the same side. • The equivalence relation is the following: two sets of serpents are considered equivalent if the patternof paths inside every square is the same in both.Let us explain in more details the meaning of this equivalence relation. Inside every square, there canbe 8 kinds of path segments: 4 kinds entering by an edge and stopping at the centre (these can appear atmost once), and 4 kinds entering by an edge and leaving by an adjacent edge (these can appear any numberof times). The pattern of paths just remembers how many paths segments of each kind there are.Figure 10: An example of the pattern of paths in a squareAnother way to describe this equivalence relation is the following. Consider two serpents that cross thesame edge. Cut both of them into two pieces along this edge, and glue them back after swapping them. Thisgives another set of serpents that also satisfies the extra condition. The equivalence relation is the closureof this kind of move.The extra condition implies that every square can contain at most 4 serpent extremities. Therefore, thereare only a finite number of different serpent nests in each quadrangulation.Let SN Q be the set of serpent nests in a quadrangulation Q .Let us define the rank rk( σ ) of a serpent nest σ as the number of serpents in it. By a well-known principle,this is half the number of serpent extremities. All reflections, not only simple ones. Sometimes named the Hurwitz action. The word “serpent” is a synonym for “snake”, which is already used for something else in the context of cluster algebras. Q be a quadrangulation. Assume that s is a bridge in Q (as defined in subsection 2.2). Let Q (resp. Q ) be the quadrangulation obtained from Q by removing all squares from one side of s (resp. from theother side). Theorem 4.1
There is a bijection between serpent nests in Q and pairs of serpent nests in Q and Q . Proof.
In fact, no serpent can cross the bridge, by definition, because they must make right-angle turnsat every step. Therefore every serpent in Q is either in Q or in Q . The extra condition and equivalencerelation are also compatible with this decomposition.Let now Q and Q (cid:48) be quadrangulations related by twisting along some inner edge e , as defined in 2.4.Every serpent in Q can be mapped to a serpent in Q (cid:48) by just twisting it. This is clearly a bijection. Theorem 4.2
This bijection induces a bijection between serpent nests in Q and in Q (cid:48) , that preserves therank. Proof.
The extra condition in the definition of serpent nests is obviously preserved. Compatibility with theequivalence relation is easy.For a quadrangulation Q , let h Q be the generating polynomial for the rank of serpent nests: h Q ( x ) = (cid:88) σ ∈ SN Q x rk( σ ) . (8)Abusing notation, this is called the h -vector.Let us now define a duality on serpent nests.For this we need the following definition. A short serpent is a serpent whose extremities are adjacentsquares. We say that a serpent nest σ contains a short serpent at the edge e if one set of serpents in theequivalence class of σ contains a short serpent at this edge. Equivalently, there are serpent extremities ineach square adjacent to e that both leave their respective squares through e .Let σ be a serpent nest. The dual σ is another serpent nest, defined as follows.Consider an edge e between two squares s and s (cid:48) of Q . If both s and s (cid:48) contains a serpent extremityleaving through e , remove them both. If neither s nor s (cid:48) contains a serpent extremity leaving through e ,add both. Otherwise (exactly one of s or s (cid:48) has a serpent extremity leaving through e ), do nothing.This change has to be performed for every inner edge of Q .This is clearly an involution on serpent nests. This amounts to remove all existing short serpents, andintroduce some wherever possible. Proposition 4.3
The rank of the dual of a serpent nest σ is n − − rk( σ ) , where n − is the number ofinner edges of the quadrangulation. roof. The quadrangulation has n squares and n − σ containing k serpents among which (cid:96) short serpents. There are k − (cid:96) longserpents, and therefore 2 k − (cid:96) inner edges where the duality does nothing (near the extremities of the longserpents). The duality changes something around n − − k + 2 (cid:96) inner edges. So in the dual σ , there are n − − k + (cid:96) short serpents. Therefore σ contains n − − k serpents.This implies that the h -vector is palindromic: h Q ( x ) = x n − h Q (1 /x ) . (9)For the four examples of size 5 in Figure 7, one finds the following h -vectors x + 8 x + 15 x + 8 x + 1 , (10) x + 8 x + 16 x + 8 x + 1 , (11) x + 9 x + 17 x + 9 x + 1 , (12) x + 10 x + 20 x + 10 x + 1 . (13)From the description of the duality, fixed points are serpent nests where each inner edge is adjacent toexactly one serpent extremity that crosses this edge.It seems that the number of fixed points is (up to sign) the value of the h -vector at x = −
1. This couldbe related to the Charney-Davis conjecture about h -vectors of simplicial spheres, see [27, 10, 23].Let us now define a refinement of the h -vector.A serpent in a serpent nest σ is called a simple serpent if two conditions hold. The first condition is thatits extremities are adjacent squares. The second condition is that there is no other serpent crossing the sameedge. This is a stronger restriction than just being a short serpent.The h -triangle of a quadrangulation Q is defined as the following polynomial in two variables: H Q ( x, y ) = (cid:88) σ ∈ SN Q x rk( σ ) y s ( σ ) , (14)where s ( σ ) is the number of simple serpents in the serpent nest σ .Given an inner edge e in a quadrangulation Q , one can define two quadrangulations by cutting Q alongthis edge. Let us call them Q (cid:48) e and Q (cid:48)(cid:48) e . This will be called a parabolic decomposition .The H -triangle has a simple property with respect to parabolic decomposition. Proposition 4.4
There holds y∂ y H Q = xy (cid:88) e ∈ Q H Q (cid:48) e H Q (cid:48)(cid:48) e . (15) Proof.
The left hand side is counting pairs made of a serpent nest in Q with a marked simple serpent. Eachterm of the right hand side is counting pairs of serpent nests in Q (cid:48) e and Q (cid:48)(cid:48) e . The equality is given by thebijection that cuts along the unique edge crossed by the simple serpent.There seems to be a close relation with the F -triangle defined in subsection 2.3. Let n be the number ofinner edges. Conjecture 4.5
There holds the equation H Q ( x, y ) = ( x − n F Q (cid:18) x − , y − xx − (cid:19) . (16)This would imply in particular a relation between the f -vector and the h -vector, that is exactly the usualrelation defining the h -vector from the f -vector for simplicial complexes.Given prop. 2.5, this would imply the following symmetry. Conjecture 4.6
There holds H Q ( x, y ) = x n H Q (1 /x, − x + xy ) . (17)16he very same relationship (16) between F -triangles and H -triangles has first been conjectured in [7] inthe context of Catalan combinatorics of finite Weyl groups and Coxeter groups. In this case, the F -triangleis related to cluster algebras and the H -triangle to nonnesting partitions. A similar formula was conjecturedin [8] to relate the H -triangle to a third triangle, called the M -triangle, defined using the M¨obius numbersof the lattices of noncrossing partitions.These conjectures have later been generalised in [2, Ch. 5] to the Fuss-Catalan combinatorics of Coxetergroups, where one more parameter m appears. The interested reader can find a lot of material on thissubject in this reference.This story is now completely proved, but maybe not fully understood, after the works of several authors,see [3, 22, 21, 32] and [31].In fact, all the statements proved about these triangles in the Catalan setting seems to extend to ourcurrent context, where the initial data is a quadrangulation instead of a Coxeter type. The M -triangle ismissing, as there is no known analogue of the noncrossing partitions here. Note that the symmetry abovegets a nice explication in the Catalan world by the self-duality of the lattices of noncrossing partitions. Recall the quadrangulations C n introduced in subsection 2.1, with inner edges(0 , , (0 , , . . . , (0 , n −
1) (18)inside the polygon with 2 n + 2 vertices, see Figure 5.In this case, every pair of squares define a serpent and every serpent nest is uniquely determined by theset of serpent extremities. In one square, there can be at most two serpent extremities.Recall that a nonnesting partition of the set { , . . . , n − } is a set of segments ( i, j ) with 1 ≤ i ≤ j ≤ n ,such that no segment contains another one.By looking at serpent extremities as opening or closing parentheses, one can find a simple bijectionbetween serpent nests and nonnesting partitions of { , . . . , n − } , that maps the number of serpents to thenumber of segments.It follows that the number of serpent nests inside C n is the Catalan number and the h -vector of C n isgiven by the Narayana numbers.Another nice family of examples is given by the quadrangulations of the general shape displayed inFigure 12, namely a linear string of n + 1 squares, with one square attached below each of them exceptedthe two extremes. Let us denote by L n this quadrangulation, which has 2 n squares. We call them the Lucasfamily, for reasons explained below. Let us also denote by K n the quadrangulation obtained from L n byremoving the rightmost square in the linear string of n + 1 squares.Figure 12: The quadrangulation L of the Lucas family and the quadrangulation K Let us denote by (cid:96) n and k n the h -vectors h L n and h K n . Recall that they are polynomials in x countingthe serpent nests according to their rank. Proposition 4.7
The sequence (cid:96) n satisfies the recurrence (cid:96) n +2 = (1 + 4 x + x ) (cid:96) n +1 + x (1 + x + x ) (cid:96) n , (19) with initial conditions (cid:96) = 0 and (cid:96) = 1 + x . roof. One can obtain the following coupled recursions for (cid:96) n and k n by a combinatorial reasoning onserpent nests: (cid:96) n = (1 + x ) k n + x(cid:96) n − and k n = (1 + x ) (cid:96) n − + xk n − + x (1 + x ) (cid:96) n − . (20)By elimination, one finds the required formula.With more care, a similar recursion can be found for the H -triangles.It follows that the (cid:96) n form a Lucas sequence, and therefore that there exist polynomials ( Z n ) n ≥ suchthat (cid:96) n = (cid:89) d | n Z d , (21)for all n ≥
1. Apparently, the Z n have positive coefficients. There is no combinatorial explanation of thisfactorisation so far. One may speculate that this could be explained by some generalisations of noncrossingpartitions lattices.The first few numbers of serpent nests in L n are2 , , , , , , , , , . . . (22)(twice sequence A090018 of oeis.org ) and the first few values of Z n at x = 1 are2 , , , , , , , , , , , , , . . . (23) A natural question to ask is, given a quadrangulation, how many serpent nests does it afford ? Even better,one would like to know the h -vector that counts serpent nests according to their rank.This does not seem to be easy. There is nevertheless an idea that allows to compute these numbers orpolynomials for some large quadrangulations. It consists of1. cutting a quadrangulation along one edge,2. counting what could be called open serpent nests in both half-quadrangulations3. and gluing the results back.Let us call open quadrangulation a quadrangulation together with a distinguished boundary edge, thatwill be considered to be open. An open serpent nest inside an open quadrangulation is defined in the sameway, except that now some of the serpents can end on the open edge. They are considered up to the sameequivalence relation as serpent nests. Clearly, there is a finite number of such configurations.The rank rk( σ ) of an open serpent nest σ is the number of (closed) serpents. Let op( σ ) be the numberof open serpents in σ , namely those that end on the open edge.Given an open quadrangulation Q , define its open h -vector as h Q ( x, t ) = (cid:88) σ ∈ SN Q x rk( σ ) t op( σ ) . (24)One will glue back two open h -vectors to get an h -vector. The gluing is defined by the following rule.Let A and B be two polynomials in x and t . Define a polynomial A B in x by A B = (cid:88) k ≥ A k B k x k , (25)where A k and B k are the coefficients of t k in A and B (when considered as polynomials in t with coefficientsin Z [ x ]).Let Q be a quadrangulation, e be an inner edge of Q . Let Q (cid:48) e and Q (cid:48)(cid:48) e be the open quadrangulationsobtained by cutting Q along the edge e . 18 roposition 4.8 The h -vector of Q is given by h Q = h Q (cid:48) e h Q (cid:48)(cid:48) e . Proof.
Cutting along e gives a bijection between serpent nests in Q and pairs of open serpent nests in Q (cid:48) e and Q (cid:48)(cid:48) e with the same number of open serpents. This implies the formula.To give just a very simple example, consider the quadrangulation of an hexagon. Its h -vector is 1 + x .On the other hand, for the open quadrangulation of a square, one find that the open h -vector is 1 + t . Onecan recover 1 + x as (1 + t ) t ).This method allows for example to compute the number of serpents nests for quadrangulations madeby gluing a large open Catalan-type quadrangulation (or a large open Lucas-type one) to any small openquadrangulation. Consider the free commutative algebra F generated by all quadrangulations (considered up to rotation). Let A be its quotient by the relations Q = Q Q (26)when there is a bridge s in a quadrangulation Q defining two quadrangulations Q and Q as in subsection 2.2.Then A is a graded commutative algebra, where the degree is the number of inner edges.On this algebra, let us define an operator ∂ : A → A . It is given on a quadrangulation Q by ∂Q = (cid:88) e ∈ Q Q (cid:48) e Q (cid:48)(cid:48) e (27)where the right hand side is the product of the two pieces obtained by cutting along the edge e . This extendsuniquely as a derivation of the free commutative algebra F , which then descends to the quotient algebra A .The operator ∂ is a derivation with respect to the product, and homogeneous of degree − ∂ ). This is a well-defined automorphism of A . It does not preserve thegrading, but does preserve an increasing filtration by the number of inner edges.Recall the F -triangle of a quadrangulation was defined in 2.3 as a polynomial in x, y . Proposition 5.1
The F -triangle is a morphism of commutative algebra from A to Z [ x, y ] . It sends thederivation ∂ to ∂ y and ∆ to the substitution operator y (cid:55)→ y + 1 . Proof.
The fact that it is a morphism follows from Prop. 2.4. The derivation part follows from Prop. 2.3.The statement about ∆ is a consequence of the statement about ∂ .Let us call a cross-tree the equivalence class of a quadrangulation under the moves of twisting along anedge as defined in subsection 2.4.Cross trees can also be considered as abstract sets of squares, glued together along their sides so as toform a tree-like shape, but where the two possible gluings along an edge are not distinguished.The notion of bridge introduced in subsection 2.2 still make sense for cross-trees, as well as the associatednotion of connectedness.By the theorem 4.2, the set of serpent nests inside a quadrangulation really only depends on the cross-tree.One can define an algebra similar to A using cross-trees instead of quadrangulations. This is a quotientalgebra of A .If conjecture 2.7 holds, the F -triangle only depends on the cross tree, and therefore is defined on thisquotient algebra. We shall compute here the number of connected quadrangulations up to rotation and the number of connectedcross-trees, up to isomorphism.This is done using the classical language of species, see [6] for the basics of this fundamental theory.Let X be the species of singletons. Let E be the species of unordered pairs. Let C be the species oforiented cycles of length 4. 19 .1 Connected quadrangulations up to rotation Let T be the species of all connected quadrangulations, T (cid:3) the species of those pointed in a square, T i thespecies of those pointed along an inner edge, and T (cid:3) ,i the species of those pointed along a pair (square,adjacent inner edge).Let T b be the species of quadrangulations that can be obtained from a connected quadrangulation bycutting along an inner edge, keeping one of the two halves and marking the cut edge. The species T b isan auxiliary one, in which it is not allowed that the square next to the marked edge has exactly one otherneighbour opposite to the edge, but where it is allowed that this square has only two opposite neighbours.One has the following combinatorial relations: T (cid:3) + T i = T + T (cid:3) ,i ,T i = E ( T b ) ,T (cid:3) ,i = ( T b ) ,T (cid:3) = X (1 + T b + ( T b ) + ( T b ) + C ( T b ))) ,T b = X (1 + 2 T b + 3( T b ) + ( T b ) ) . The first equation is an instance of the dissymetry equation for trees, see [6, Chap. 4] for the general principlebehind it. Other equations are just obvious recursive descriptions. The last equation determines T b and theother ones in turn determines the other species.From these equations, one can compute the first few terms of the associated generating series: T b : 1 , , , , , , , . . .T : 1 , , , , , , , , , . . .T i : 0 , , , , , , , . . .T (cid:3) : 1 , , , , , , , . . .T (cid:3) ,i : 0 , , , , , , , . . . It is also interesting to consider the sub-species T B of T b where the square near the marked boundaryedge does not have exactly two opposite neighbours. In some sense, this condition is a stronger form ofconnectedness. By removing one term in the equation defining T b , one gets that T B = X (1 + 2 T b + 2( T b ) + ( T b ) ) , (28)which gives the first few terms T B : 1 , , , , , , , , . . . Let us compute here the number of connected cross-trees.Let T be the species of all connected cross-trees, T (cid:3) the species of those pointed in a square, T i thespecies of those pointed along an inner edge, and T (cid:3) ,i the species of those pointed along a pair (square,adjacent inner edge).Let T b be the species of cross-trees that can be obtained from a connected cross-tree by cutting alongan inner edge, keeping one of the two halves and marking the cut edge. The species T b is an auxiliary one,in which it is not allowed that the square next to the marked edge has exactly one other neighbour oppositeto the edge, but where it is allowed that this square has only two opposite neighbours.One has the following combinatorial relations: T (cid:3) + T i = T + T (cid:3) ,i , T i = E ( T b ) , T (cid:3) ,i = ( T b ) , T (cid:3) = X (1 + T b + E ( T b ) + T b E ( T b ) + E ( E ( T b ))) , T b = X (1 + T b + ( T b ) + E ( T b ) + T b E ( T b )) . T b and the other ones in turn determines theother species.From these equations, one can compute the first few terms of the associated generating series for thenumber of isomorphism classes: T b : 1 , , , , , , , , . . . T : 1 , , , , , , , , , . . . T i : 0 , , , , , , , , . . . T (cid:3) : 1 , , , , , , , . . . T (cid:3) ,i : 0 , , , , , , , . . . Just as for quadrangulations, it is also interesting to consider the sub-species T B of T b where the squarenear the marked boundary edge does not have exactly two opposite neighbours. By removing one term inthe equation defining T b , one gets that T B = X (1 + T b + ( T b ) + T b E ( T b )) , (29)which gives the first few terms T B : 1 , , , , , , , , . . . The first few terms of sequence T are illustrated in Figure 7. For reasons related to fans (that are not considered in this article), it is also interesting to count quadrangu-lations and cross-trees avoiding the cross (which means that no square has four neighbours). We will abusenotation by keeping the same names for these new species.Let us start with quadrangulations. Using similar notations for various species of connected quadrangu-lations avoiding the cross, one has the following combinatorial relations: T (cid:3) + T i = T + T (cid:3) ,i ,T i = E ( T b ) ,T (cid:3) ,i = ( T b ) ,T (cid:3) = X (1 + T b + ( T b ) + ( T b ) ) ,T b = X (1 + 2 T b + 3( T b ) ) . The first equation is an instance of the dissymetry equation. Other equations are just obvious recursivedescriptions. The last equation determines T b and the other ones in turn determines the other species.From these equations, one can compute the first few terms of the associated generating series: T b : 1 , , , , , , , . . .T : 1 , , , , , , , . . .T i : 0 , , , , , , , . . .T (cid:3) : 1 , , , , , , , . . .T (cid:3) ,i : 0 , , , , , , , . . . .4 Connected cross-trees with no cross Using the same notations as before, one has the following combinatorial relations: T (cid:3) + T i = T + T (cid:3) ,i , T i = E ( T b ) , T (cid:3) ,i = ( T b ) , T (cid:3) = X (1 + T b + E ( T b ) + T b E ( T b )) , T b = X (1 + T b + ( T b ) + E ( T b )) . The first equation is an instance of the dissymetry equation for trees. Other equations are just obviousrecursive descriptions. 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