A generalised Euler–Poincaré formula for associahedra

We derive a formula for the number of flip‐equivalence classes of tilings of an n ‐gon by collections of tiles of shape dictated by an integer partition λ . The proof uses the Euler–Poincaré formula; and the formula itself generalises the Euler–Poincaré formula for associahedra.


Introduction
In [10][11][12] Fomin and Zelevinsky introduce cluster algebras and in particular find a Lie theoretic manifestation of the associahedron or Stasheff polytope [27]. In [8] Chapoton investigates the combinatorics of the associahedron from the Lie theoretic perspective; while in [22,24], Postnikov and Scott develop these constructions in the direction of total positivity and Grassmannians. (The associahedron also has a wider interconnected network of other uses. See, for example, [7,13,21,28].) Baur et al. [3] develop Fomin-Zelevinsky's and Scott's original construction with emphasis on quiver mutation and cluster categories associated with the Grassmannian. In [4] we generalised Scott's map [24,Section 3] to the whole associahedron. A partial result on counting tilings with prescribed tile sizes up to 'flip equivalence' (cf. [14]) is included in [4]. Based on this, Baur-Schiffler-Weyman (private communication) conjectured a general formula. In this paper, we note that the formula also generalises the Euler-Poincaré formula for the associahedron, and prove the conjecture. The proof uses the polytopal property of the associahedron and the Euler-Poincaré formula itself among other devices.
Let P be a convex polygon with n vertices, labelled 1, 2, . . . , n. Here A n denotes the set of tilings of such an n-gon, or equivalently the set of non-crossing subsets of diagonals [i, j] of P . For example, There are two ways of triangulating a 4-gon. The move between these two is called 'flip'. Two tilings are triangular equivalent, or flip equivalent, if they are related by any sequence of flips (for example, any two triangulations are equivalent [14]). The set of classes of tilings under equivalence is AE n . These are enumerated in [4].
It is convenient to arrange A n into a polytope, called the associahedron [16]. This is ab initio an abstract cell complex: each tiling t is a cell, and the boundary set of a cell is the set of tilings obtained from t by adding one more diagonal. Thus in particular the vertices (0-cells) are the tilings that are triangulations. Removing a diagonal from such a tiling gives a tiling with one quadrilateral tile. Starting now from this tiling, the quadrilateral tile can be triangulated in two ways. The tiling is an edge (1-cell) of the complex and its boundary is the two completions obtained by triangulating the quadrilateral. And so on. Theorem 1.1 (Haiman, Lee [16]; see, for example, [9,Theorem 3.39]). There exists a convex n-dimensional polytope K n+2 called the associahedron whose vertices and edges form the flip graph of a convex (n + 3)-sided polygon. The k-dimensional faces (k-cells) of this polytope are in one-to-one correspondence with the diagonalizations (tilings) of the polygon using exactly n − k diagonals.
Note that shape is well defined on triangular equivalence classes. Let AE n (λ) denote the set, and ae n (λ) denote the number of tilings of P of a given shape up to triangular equivalence.
For μ ∈ P define μ + ∈ P by We extend the combination ∪ of partitions as follows. Let a > 0, λ ∈ P. Define λ ∪ (1 −a ) as (1 m1(λ)−a , 2 m2(λ) , 3 m3(λ) , . . . ). For m 1 (λ) a, λ ∪ (1 −a ) is a partition. For example, (1 3 , 2 2 , 3) ∪ (1 −2 ) = (1, 2 2 , 3). We will consider tilings of shape λ ∪ (1 −a ), for some λ. If m 1 (λ) < a, λ ∪ (1 −a ) is not a partition, and A n (λ ∪ (1 −a )) = ∅. Then Before proving the theorem we focus on one class of special cases: the cases where λ is of form 1 n−2 . In these cases we know ae n (λ) = 1 by Hatcher's theorem (a) in [14]. On the other hand the right-hand side also takes a relatively simple form in these cases. All coefficients Π (1 n−2 ) μ on the right-hand side are 1 since m s (1 n−2 ) = 0 for all s > 1. The right-hand side is thus Noting that μ m a n (1 n−2 ∪ μ + ∪ 1 −|μ + | ) is the number of m-cells of K n−1 , since the signs alternate we actually have the Euler-Poincaré formula. Thus this special case of the theorem asserts: Consider the Schläfli-Poincaré Theorem: Theorem 1.5 (Schläfli, Poincaré; see, for example, [29]). Let f i denote the number of i-cells of the d-dimensional polytope P . Then the Euler-Poincaré formula states that By Theorem 1.1, K n−1 is a n − 3-dimensional polytope, so we can apply Theorem 1.5. We thus see on the one hand that Theorem 1.4 is verified in the case λ = 1 n−2 ; and on the other hand that the special cases of the theorem coincide with the Euler-Poincaré formulae for the various associahedra. In other words, the 'count' of triangulations of P up to flip equivalence computes the Euler-Poincaré characteristic of the corresponding associahedron.
Note that if μ ∈ P n then μ − lies in some P n−m (specifically in P n−l(μ) ). Indeed μ → μ − defines an injective map (See (4), for example, in P 3 .) At the level of i-cells t of associahedron K n−1 then cell dimension Note that this verifies the Euler-Poincaré formula assertion immediately after (3).
(2.2) We define another function from tilings to partitions (cf. (1.3)). A tiling t gives an integer partition f(t) as follows. First consider the partition f + (t) of the number of its triangles into the unordered list of sizes of maximal triangulated regions. Then f(t) = f + (t) − . Again this is well defined on classes. Examples: A further decomposition of A n (λ) is into subsets characterised as the inverse images of f. Thus, for ν in P: and we have AE λ,ν similarly. We write a λ,ν and ae λ,ν for the sizes of these sets. (2. 3) The utility of this decomposition for us lies in the fact that the 'overcount factor' in counting A n instead of AE n is uniform in A λ,ν -since every tiling in this subset belongs to a class of size where C n is the nth Catalan number: C 1 = 1, C 2 = 2 and so on. That is In order to explain the proof it will be helpful to write out the first few terms explicitly. To this end let To obtain shape λ μ we start with λ, but move to a shape with fewer triangular tiles, replacing multiple triangles with larger tiles such that the overall weight is unchanged. Here we note that on the level of tilings, the operation ∪(1 −a ) corresponds to removing a triangles. Then ae n (λ) = a n (λ) − a n (λ (1))Π λ (1) + a n (λ (2))Π λ (2) + a n (λ ( We interpret each of the terms as counting a certain subset of A n (λ). The claim is that the multiplicity of the intersection of this subset with f −1 (ν) is, for each ν, uniform in AE n (λ), in the sense of our discussion of A λ,ν in (2.2) above.
(2.4) The set counted by Π λ μ a n (λ μ) is the subset of A n (λ) where the triangulated regions include regions corresponding to triangulations of the tiles arising from μ -specifically the construction is to count tilings of λ μ but with the tiles with sizes given by μ + marked (hence the multiplicity Π λ μ ). The marked regions can then simply be triangulated. In this way we have tilings of the right shape, but where there are triangulated regions of size at least given by μ + (they might also be bigger). Let us write B λ μ for this subset of A n (λ). (Note that tilings here have only one triangulation in each triangulated polygon from μ -and indeed we do not actually specify one.) That is b λ μ := |B λ μ | = Π λ μ a n (λ μ)

Tilings organised by maximal triangulated regions
As discussed in (2.3), if we organise tilings by taking the fibre of f over some partition ν ∈ P, the overcount factor in counting A n instead of AE n is uniform. So we define the overcount factor By definition, OF ν,ν = 1. We will need the sizes of these factors, but the first thing to note is that they do not depend on λ, since this dependence is removed by the quotient. With this, fibre-wise the claim of Theorem 1.4 becomes Dividing by |AE λ,ν | on both sides, this translates to So we will have proved Theorem 1.4 if we can show that every signed column sum as in (8)

Overcount factor columnwise: small ν
First column: The μth entry in the first column is the overcount factor for the subset of B λ μ (any λ) with no two triangular tiles together. Zero means that in fact the intersection is empty. In the first column we are looking only at tilings in which the triangulation has no clusters of size bigger than 1. Since the equivalence classes are singletons in this case, the count is the same in AE n (λ): All of the remaining combinatorics count sets in which higher tiles are triangulated, so there are larger clusters, and so the intersection in the first column is empty. This means that the theorem counts these classes correctly.  Second column: In the second column, ν = (1), we are intersecting with tilings having a quadrilateral triangulated region and then any other triangles singletons. Such tilings are present in A n (λ) (the first row) and, since the equivalence classes are flip pairs in this case, the count is 2× that in AE n (λ). In the second row (the case of B λ (1) ) we assemble tilings with a marked quadrilateral (which we consider to then be triangulated). However since we only count one tiling here, we count for AE n (λ) correctly. (In this set there are tilings where the triangulated quadrilateral is adjacent to other triangles, but these do not lie in the intersection in this column.) All of the remaining combinatorics count sets in which higher tiles are triangulated, so there are larger clusters, and so the intersection in the second column, for μ all other μ, is empty. This means that the theorem again counts these classes correctly overall.
Column ν = (2): Here we have tilings with a pentagonal triangulated region and any other triangles singletons. Thus the overcount in A n (λ) is 5×. In B λ (1) such pentagons arise when the triangulated quadrilateral is adjacent to one other triangle. Indeed note that a given triangulated pentagonal subregion of the tiling could arise in five different ways in B λ (1)we are counting the number of ways of tiling a pentagon with a quadrilateral and a triangle, which number is a 5 (21). In B λ (2) such pentagons arise when we triangulate a pentagon, with no other adjacent triangles. The construction only counts one such triangulation, so the count is the same as for the class in AE n (λ). The remaining intersections in this column are empty. So again the count in the Theorem is correct.
Column ν = (1 2 ): Here we have tilings with two triangulated quadrilaterals and any other triangles singletons. The overcount factor in A n (λ) is 2 2 . In B λ (1) tilings of the required clustering arise when the triangulated quadrilateral is not adjacent to any triangle, and there are two other triangles that are adjacent (possibly along with further singletons). There are two ways the quadrilateral could be assigned to one of the triangulated quadrilaterals; and there are two triangulations of the other triangulated quadrilateral, so we have a 2×2× overcount here. In B λ (2) there is a triangulated pentagon, and hence no intersection. In B λ (1 2 ) there are two marked quads. Each is given a triangulation, so the count is the same as for classes. There are no more non-empty intersections in this column. Again the theorem is verified.
Column ν = (3): Here we have tilings with one triangulated hexagon. Such tilings are overcounted in A n (λ) by the appropriate Catalan number, 14. In B λ (1) we have the triangulated hexagon formed by a quadrilateral and two triangles. The number of ways of doing this is a 6 (21 2 ).
Proof of Lemma 2.7. Observe that in each product l(ν) i=1 a νi+3 (γ i ) each factor addresses one maximal triangulated region (for these regions their sizes, as polygons, are the ν i + 3). We have to take the maximal subregions prescribed by ν (note that we can, and this does, ignore those that are already simple triangles) and populate them according to the polygons available from μ. The contribution to the overcount factor is a νi+3 (γ i ) since this counts the number of ways of tiling the subregion as indicated by γ i (after this, in our construction, recall from (2.4), we simply triangulate to any one triangulation). Note that the sum over γ l(ν) |= μ constructively exhausts all possibilities. Indeed it does not check against internal details of ν, so also formally includes some impossibilities, in general -the cases where |γ i | > ν i + 1 for some i and the factor a νi+3 (γ i ) = a νi+3 (γ i ) is zero. (Since the extra terms are zero this does not affect the identity.) Our next step in proving (8) is Lemma 2.13. We will need a couple of preliminaries. (2.9) Note that for any m and γ ∈ P Meanwhile γ → γ − is defined and injective on P r for r > 0 by (5), and then Lemma 2.10. For m ∈ N the map γ → γ + m is an injection P → P m ; and the image includes P m .
Proof. First note that γ + m ∈ P m by construction. Then note that injectivity follows from (11). Finally the image property follows from (12).
Proof. We write the maps from 2.10 as γ → γ + → γ + m taking P → P + αm → P m . By (9) we have μ∈P |= ν μ = (P + ) l(ν) . Thus the map here is the Cartesian product × Thus it is injective. Since the image of a Cartesian product is the product of images, the image bound of Lemma 2.10 also implies the image bound here.
Proof. Observe from (3) that F r is the Euler-Poincaré characteristic. Now use Theorems 1.1 and 1.5.

On combinatorial isometries and other discussion points
This section takes the form of some extended remarks, on open questions and possible connections.
Here we have drawn attention to the classification of cells of the associahedron by mechanisms such as 'shapes' of tilings as defined in (1.3). One question thus raised is how (literal) shapes of cells in the associahedron are related to shapes of tilings.
Consider the associahedron as a combinatorial complex (see, for example, [2]). We call a selfmap of the (combinatorial) complex a combinatorial isometry. This should be distinguished from an isometry of one of the concrete polytopal realisations, since these depend on the realisation. A typical realisation has no true isometries. On the other hand at least some nontrivial self-maps of the complex exist. That is, the ones induced by rotating the polygon P . We illustrate by considering K 5 and K 6 .
3.1. Associahedron K 5 . Consider Figure 4 (interpolating between the polytopal and tiling complex realisations of K 5 , by an extension of Brown's figure in [7]). Here we see that there are various subclassifications of i-cells of K n available. Firstly there is the evident and well-known intrinsic classification by products of smaller associahedra. This essentially coincides with our tiling shape classification via λ × i K λ + i . This corresponds to ignoring all triangles in tilings, since K 2 is trivial. What about extrinsic properties?
Firstly consider the case of 0-cells. In terms of their intrinsic physical shape these are obviously all the same -points. (Compare this with their equivalence under flip in the tiling realisation.) From another perspective, 0-cells are 'tristate' points (points belonging to three cells). They can be sorted into black circles ('tristate' points for three 5-gonal faces); white circles (tristate points for two 5-gonal and one 4-gonal face that are adjacent to 5 3 tristate points); the rest (tristate points of 5 2 4 type that are not adjacent to 5 3 tristate points). It will be clear from the figure that these classes coincide (in this case) with the classes induced under rotation of P .
What about 1-cells? Again these all have the same shape. (They might have different lengths in a concrete realisation, but this is not canonical.) We leave it as an exercise to consider extrinsic properties.
In preparation for looking briefly at K 6 (where our 4d sketching skills fail) we include a tabulation of cells in Figure 4 in a format that does lift to K 6 . The table is organised by dimension of cell. The μ label runs through the integer partitions of the dimension and λ = μ + (let us generally exclude those μs that give an impossible λ). The final component gives representative tilings, up to polygon isometries.