Flip posets of Bruhat intervals

In this paper we introduce a way of partitioning the paths of shortest lengths in the Bruhat graph B(u, v) of a Bruhat interval [u, v] into rank posets Pi in a way that each Pi has a unique maximal chain that is rising under a reflection order. In the case where each Pi has rank three, the construction yields a combinatorial description of some terms of the complete cd-index as a sum of ordinary cd-indices of Eulerian posets obtained from each of the Pi. Mathematics Subject Classifications: 20F55, 05E99 , 05E15


Introduction
Given a Coxeter group W with u, v ∈ W , the Bruhat graph associated with a Bruhat interval [u, v] has u-v paths of several lengths (see [9]).The u-v paths of maximum length correspond to maximal chains of [u, v].Dyer [10] defined and proved the existence of a set of total orders on the set of reflections of W , called reflection orders, which were used to prove algebraic and topological properties of [u, v].For instance, [u, v] is Eulerian and Cohen-Macaulay, which was also proved by Björner and Wachs [5] by using a different labeling.One of the advantages of Dyer's reflection orders is that they can be used to label all the paths in the Bruhat graph of [u, v], and not just those of longest length.However, whenever Björner and Wachs's labeling is defined on all paths of a certain length (maximal or not), it has the same descent-set distribution as the any reflection order (see [6,Theorem 2.11]).Utilizing reflection orders, Billera and Brenti [2] defined a polynomial in the noncommutative variables c and d, called the complete cd-index, that encodes the descent-set distribution of any reflection order on the u-v paths.The complete cd-index can be used to express the Kazhdan-Lusztig polynomials and it extends the notion of the cd-index of [u, v] seen as an Eulerian poset.The coefficients of the complete cd-index have been conjectured to be nonnegative, and this has been proven in some cases (see [6,7,11,14]).In this paper we propose an algorithm that separates the u-v paths in the Bruhat graph of [u, v].In the special case of the shortest-length u-v paths, this algorithm yields posets in which every subinterval has at most one rising chain (under any reflection order).Furthermore, when applied to the u-v paths of length two and three, the corresponding terms of the complete cd-index can be obtained as the sum of the cd-index of certain Eulerian posets determined by the algorithm.So not only do we have an alternative proof of the nonnegativity of these terms, but we prove that they arise from a natural decomposition of posets.
The basic definitions are presented in Section 2. The remaining of the paper is organized as follows.The Flip algorithm and some of its properties are presented in Section 3. In Section 4, we apply the Flip algorithm to the maximal chains of the shortest path poset SP (u, v) of [u, v] to obtain certain subposets, which we call flip posets, and show that they satisfy properties similar to SP (u, v).In Section 5 we study the special case of applying the Flip algorithm to B 2 (u, v) and B 3 (u, v), the paths of length two and three, respectively, in the Bruhat graph of [u, v].We furthermore show that certain terms of the complete cd-index, those corresponding to u-v paths of length two or three, can be expressed as a sum of the cd-index of certain Eulerian posets that are obtained from the Flip algorithm, enabling us to conclude nonnegativity in these cases, as well as a connection between the two indices.

Basic definitions
We follow [12, Section 5.1] and let (W, S) denote an arbitrary Coxeter system.Namely, it is well-known that the lexicographic order on all transpositions form a reflection order, and we use this order in our examples.To be more specific, (1 2) ) is a reflection order, and it is customary to refer to these reflections in the order in which they appear.So (1 2) is labeled with a "1," (1 3) is labeled with a "2," and so forth.Any reflection order produces an EL-labeling of a Bruhat interval [u, v] (see [10,Section 4]) and this EL-labeling was used to prove topological properties such as [u, v] being Cohen-Macaulay, which was also proved by Björner and Wachs [5].
The maximal-length u-v paths form the maximal chains in the Bruhat interval [u, v].Furthermore, the set of u-v paths (u . So sometimes we refer to a u-v path in terms of the vertices that are in the path, and sometimes we refer to it in terms of the edges, which are labeled by reflections.Furthermore, we denote by B(u, v) the set of all u-v paths; that is, To every path ∆ = (λ(u 0 , u 1 ), we can associate a monomial w(∆) := x 1 x 2 . . .x k−1 on the variables a and b as follows: The complete ab-index is the polynomial ∆∈B(u,v) w(∆) in the noncommutative variables a and b.Billera and Brenti [2] showed that the complete ab-index can be expressed as a polynomial ψ u,v (c, d) in the noncommutative variables c and d, where c = a + b and d = ab + ba.The polynomial ψ u,v (c, d) is called the complete cd-index of [u, v], where [u, v] denotes the Bruhat interval {x ∈ W | u x v}.Furthermore, it was shown in [2, Proposition 2.5] that the highest-degree terms of ψ u,v (c, d) correspond to the cd-index of [u, v] seen as an Eulerian poset, which was studied by Reading in [15].The cd-index was first defined in [1] as a way of encoding the linear relations on flag vectors of polytopes.Let P be a finite graded poset of rank d + 1 with partial order , rank function r and smallest and largest elements 0 and 1, respectively.For any S = {s 1 , . . ., denotes a maximal chain of P .The vector (f S (P )) S⊆ [d] is called the flag f -vector of P .A closely related object, and sometimes more useful, is the flag h-vector.Each of the components of the flag h-vector is defined from the flag f -vector as follows.
Furthermore, let a and b be non-commutative variables.For S ⊆ [d], let w(S) = The polynomial S⊆[d] h S w(S) is called the ab-index of P .Bayer and Klapper [1] proved that, when P is Eulerian, the ab-index can be rewritten as a homogeneous polynomial in the noncommutative variables c and d, where c = a + b and d = ab + ba.This polynomial is called the cd-index of P , and we denote it by ψ(P ).
Given ∆ = (t 1 , t 2 , . . ., t k ) ∈ B k (u, v), we define the descent set D(∆) of ∆ as Notice that the complete cd-index provides a way of encoding the distribution of the descent sets of all paths in B(u, v).If a given path Γ ∈ B(u, v) has empty descent set, that is, D(Γ) = ∅, then Γ is said to be a rising path.c, d) gives the number of rising (and falling) paths in The shortest path poset of [u, v] was defined in [7].To construct it, one takes the subgraph of B(u, v) formed by the minimal-length u-v paths, and considers this to be the Hasse diagram of a poset, denoted by SP (u, v).The lowest-degree terms of ψ u,v (c, d) correspond to the shortest u-v paths of B(u, v).When SP (u, v) has only one rising chain, SP (u, v) is a Gorenstein* poset; that is, SP (u, v) is Eulerian and Cohen Macaulay (see [7,Theorem 5]).Moreover, under the assumption that SP (u, v) has only one rising chain, SP (u, v) is an EL-labelable poset (the reflection order is an EL-labeling) and therefore the coefficients of the ψ u,v (c, d) corresponding to the maximal chains in SP (u, v) coincide with ψ(SP (u, v)) (see [3,Theorem 2.2]) and are nonnegative (see [13]).

Flip algorithm
In this section we propose an algorithm to divide the elements of B k (u, v) into a graph with [c k−1 ] u,v components so that each component has exactly one u-v path that is rising under < T .
Following [2, Section 6] we now define the flip of Γ ∈ B 2 (u, v) and we use lex to denote the lexicographic order.Let (t 1 , t 2 ) and (r 1 , r 2 ) be in B 2 (u, v).We say that (t 1 , t 2 ) lex (r 1 , r 2 ) if and only if t 1 < T r 1 , or if t 1 = r 1 and t 2 T r 2 .The existence of the complete cd-index implies that there are as many paths with empty descent set in B 2 (u, v) as those with descent set {1}. Order all the paths in B 2 (u, v) lexicographically and let r(Γ) lex , and we denote the flip of a path Γ by flip(Γ).
the electronic journal of combinatorics 25(4) (2018), #P4.16 The following proposition was proved in [2, Proposition 6.2] for affine and finite Coxeter groups, and then proved to hold for general Coxeter groups in [6,Proposition 3.8].The proof of the general case also follows as a consequence of Dyer's proof of Cellini's conjecture [8].
In fact, it turns out that the lexicographically first element in B k (u, v) is rising.
We now define the operation Flip i .In a few words, given a path The pseudocode is given in Algorithm 1 below.
For the remainder of the paper, it will be convenient for us to refer to the elements in B k (u, v) by utilizing the labels of their edges instead of listing the vertices in the path.
We use Flip i to define Flip.Flip(∆) transforms ∆, with D(∆) = ∅, to a path such that D(Flip(∆)) = D(∆) \ {min(D(∆))}, where min(D(∆)) denotes the smallest element in D(∆).In other words, Flip transforms a path ∆ to one that removes the first descent, from bottom to top, of ∆.The pseudocode is written in Algorithm 2.
the electronic journal of combinatorics 25(4) (2018), #P4.16 We write Flip j (∆), with j 1, to denote the path obtained after applying Flip j times; that is, The Flip algorithm constructs a directed graph F G(u, v; k) with vertex set, B k (u, v) and the Flip operator is used to determine the edges.To each path ∆ ∈ B k (u, v) with nonempty descent, we shall assign a unique path Flip(∆) ∈ B k (u, v) obtained by flipping ∆ at its first descent, from bottom to top, and then we add a directed edge from ∆ to Flip(∆).The pseudocode is given in Algorithm 3.
Add (the directed) edge (∆, Flip(∆)) to E. 6: There are two straightforward properties that the Flip algorithm satisfies: it terminates and it creates a graph with [c k−1 ] u,v components.We prove these properties in the following lemma.Lemma 3. The Flip algorithm satisfies the following properties.
(i) The Flip algorithm terminates.
That is, every connected component has a unique vertex v with zero out-degree and every vertex in the component has a unique path to v.
Proof.(i) Notice that if i ∈ D(∆), Flip i (∆) is earlier in the lexicographic order than ∆ by Proposition 1, and since the lexicographically-first path in B k (u, v) is rising by Proposition 2, the Flip algorithm must terminate.
(ii) Every path, seen as a vertex in the flip graph F G(u, v; k) is associated with a unique rising path obtained by iterating Flip until a rising path is reached.Moreover, a vertex representing a rising path has in-degree zero, and is therefore a sink.Hence every rising path must be in a different component, and there are [c k−1 ] u,v of them since there are [c k−1 ] u,v rising paths.Now, if there were two paths between a path ∆ and the rising path associated with it through repeatedly applying Flip, then the electronic journal of combinatorics 25(4) (2018), #P4.16 that would imply that the out-degree of ∆ is at least two, but this is not possible since the only edge having ∆ as a tail is (∆, Flip(∆)).Therefore, there is a unique path between any vertex in the component and the corresponding rising path.
The following corollary is an easy consequence of the Flip algorithm terminating.
Corollary 4. For every ∆ ∈ B k (u, v), there is nonnegative integer i k,∆ so that Then consider the following definitions.
(i) Define F G c (u, v; k) to be the induced subgraph of F G(u, v; k) with vertex set In other words, V c (u, v; k) denotes the set of all elements in V (F G(u, v; k)), the vertex set of F G(u, v; k), that "start" with c and V d (u, v; k) denotes the set of all elements in V (F G(u, v; k)) that "end" with d.
With this definition, we have the following proposition, which we call the Subgraph Property.
as graphs, and In the proof of the Subgraph Property, if ∆ ∈ B k (u, x) and Γ ∈ B (x, v), where x ∈ [u, v], we denote the concatenation of ∆ and Γ by ∆Γ ∈ B k+ (u, v).
Proof.(i) Let (D 1 , D 2 ) be an edge of F G d (u, v; k).We show that there exists an edge (ii) It is enough to show that if (cc 1 , cc 2 ) is and edge in F G c (u, v; k), then (c 1 , c 2 ) is also an edge of F G(x, v; k−p).By definition of the flip graph, cc 2 = Flip(cc 1 ).Then the first descent from bottom to top of cc 1 occurs after c, as otherwise cc 2 ∈ V c (u, v; k).Thus, c 2 = Flip(c 1 ), and the result follows.
Remark 7. We point out that it is not true that We now focus our attention on the case B s(u,v) (u, v), where s (u, v) denotes the length of the shortest u-v path in B(u, v).In this case the Flip algorithm gives a way of partitioning the shortest path poset SP (u, v) into subposets P 1 , . . ., P k so that every subinterval of each P i has at most one rising chain.

Flip(SP (u, v))
In this section we study the Flip algorithm when applied to the shortest u-v paths of B(u, v).Let us denote the number of rising maximal chains in SP (u, v) by r(u, v); that is, r(u, v) := [c s(u,v)−1 ] u,v .Definition 8 (Flip posets).Let F G(u, v) := F G(u, v; s (u, v)), and G 1 , . . ., G r(u,v) be the connected components of the flip graph F G(u, v).For each G i one can form a poset P i whose maximal chains are the vertices of G i .That is, the maximal chains of P i are the elements in the set {C | C ∈ V (G i )}.We call the posets P 1 , . . ., and P r(u,v) the flip posets of [u, v].
To illustrate the Flip algorithm, consider the following example.We discuss this identification in Section 5.
Each P i has properties that resemble those of Bruhat intervals; for instance, consider the following lemma.Lemma 10.Let G i , 1 i r(u, v), be a connected component of F G(u, v) and P i , with 1 i r(u, v), be the corresponding flip poset.Then P i is graded.
Proof.We recall that SP (u, v) is graded (see [7,Proposition 3].)We now prove that the Flip algorithm preserves the rank in SP (u, v).
Let C be a maximal chain of SP (u, v) and w ∈ W be an element in C. Let r C (w) be the length of the u-w path in C. Furthermore, let u 1 and v 1 be elements in C with r C (u 1 ) = r C (w) − 1 and r C (v 1 ) = r C (w) + 1.We show that the Flip algorithm does not the electronic journal of combinatorics 25(4) (2018), #P4.16 change the value r C (w), that is, r Flip j (C) (w) = r C (w) for all j > 1. Suppose otherwise, and let r C (w) be the length of the u-w path C , with Flip j (C) = C for some j > 1.Let u 2 and v 2 be elements in , but in either case there is a contradiction to C and C being maximal chains of SP (u, v).Indeed, if it were the case, there would exist D of the form C w (C ) w or C w C w , where C w (or C w ) and C w (or (C ) w ) denotes the chain C (or C ) of elements that are at most w and the chain C (or C ) of elements that are at least w, respectively.This D would be shorter than C and C , which is not possible since C and C are maximal chains in SP (u, v).So r C (w) = r C (w).Thus the rank function of every flip poset P i of SP (u, v) is the same as the rank function of SP (u, v).
Remark 11.The proof of Lemma 10 holds in more generality.If one selects a maximal chain in a graded poset and there is a suitable definition of the f lip operation, the rank would be preserved after repeated applications of the f lip operation.
Useful notation: We define the following notation.
1. We write x s y to denote that x is smaller than y in SP (u, v).

If
x s y is a cover relation, that is, if there does not exist z with x s z s y, then we use the notation x s y.
3. If P 1 , . . ., P r(u,v) are the flip posets of the Bruhat interval [u, v], we define [x, y] P i := {z ∈ P i : x s z s y}, for x, y ∈ P i and 1 i r(u, v).
As a consequence of the Subgraph Property (Proposition 6), we have the following corollary.
Corollary 12.If x ∈ [u, v] and P 1 , . . ., P r(u,v) are the flip posets of [u, v] then gives rise to a flip poset, which has a unique rising chain, [u, x] P i must have a unique rising chain.
Then the Subgraph Property, Proposition 6(ii), gives the existence of an injection ϕ : Since ϕ is not necessarily a bijection since F G(x, v) need not be a subgraph of the electronic journal of combinatorics 25(4) (2018), #P4.16 F G c (u, v) (cf.Remark 7), there could be an edge (c 1 , c 2 ) of F G(x, v) so that (cc 1 , cc 2 ) is not an edge of F G c (u, v).Hence a connected component of F G c (u, v) might not have a rising chain.
Putting (i) and (ii) together we obtain the following proposition.
Proposition 13.If P 1 , . . ., P r(u,v) are the flip posets of [u, v] then [x, y] P i ⊂ [u, v] P i has at most one rising chain for 1 i r(u, v).
Proof.It is enough to show that the proposition holds for u s x s y or u s y s v, as P i is graded.These cases follow from Corollary 12(i) and 12(ii).
Moreover, Proposition 14.If P 1 , . . ., P r(u,v) are the flip posets of [u, v] and [x, y] P i ⊂ [u, v] P i is an interval of rank 2, then [x, y] P i has at most two atoms for 1 i r(u, v).
Proof.If there were more than two atoms, then either (i) [x, y] P i would have at least two rising chains, or (ii) [x, y] P i would have at least two falling chains.Proposition 13 gives that Case (i) is impossible.On the other hand, if Case (ii) were true, then since the Bruhat interval [x, y] has the same number of rising chains as falling chains, the pigeonhole principle gives that there would exist j with 1 j r(u, v) so that [x, y] P j would have more than one rising chain.Again, this contradicts Proposition 13, and the result follows.
We remark that Corollary 12, Proposition 13, and Proposition 14 hold in the case of Bruhat intervals, for which one can substitute the phrase "at most" with the word "exactly."Indeed, since the Bruhat order is an Eulerian poset, every interval of length two has exactly two atoms (see [4,Lemma 2.7.3]).
Lemma 15.Let P 1 , . . ., P r(u,v) be the flip posets of [u, v] and G 1 , . . ., G r(u,v) be the corresponding connected components in the flip graph.Then each G i , 1 i r(u, v), has an even number of vertices.
5 Flip algorithm applied to B 2 (u, v) and B 3 (u, v) In this section we focus our attention on the case k = 2, 3.In these cases, we have been able to derive connections to the complete cd-index.Indeed, one of the main open questions regarding the complete cd-index is if its coefficients are nonnegative [2,Conjecture 6.1].Partial results exist for some coefficients [6,7,11,14].One approach that has not been explored much is to find a way of partitioning the elements of B k (u, v) of the same length in such a way that each of the parts would produce cd-monomials that when added together would yield the terms of ψ u,v (c, d) of degree k − 1.In this section, we show that the Flip algorithm provides such a procedure for k = 2, 3.

Case
then the maximal chains of SP (u, v) correspond to the elements of B 2 (u, v).In this case, the Flip algorithm will split up the elements of B 2 (u, v) into [c] u,v components, each of them will have two elements, one of which will be falling and the other one will be rising.So in this case, the Flip algorithm will split up the maximal chains of SP (u, v) into [c] u,v pieces, each of them contributing c to ψ u,v(c,d) .Moreover, each of these pieces is isomorphic to a Boolean poset on two elements, which we denote by Boolean(2), and therefore each has ordinary cd-index (seen as an Eulerian poset) of c.In other words, [1] where [k] ψ u,v (c, d) denotes the coefficients of degree k in ψ u,v (c, d).Notice that (1) relates terms in the complete cd-index to terms arising from the ordinary cd-index of Eulerian posets.It turns out we can establish a similar equality for the terms of degree two in the complete cd-index.

Case
We can think of B 3 (u, v) as a poset by considering its elements as maximal chains, even if B 1 (u, v) = ∅.First we need some definitions.
Definition 16 (Order complex, face poset, and face lattice).Following [16, Section 1.1], we recall that for every partially ordered set P , one can define an abstract simplicial complex ∆(P ), called the order complex of P , as follows.The vertices of ∆(P ) are the elements of P and the faces of ∆(P ) are the chains of P .In other words, ∆(P ) = {F ⊆ P | F is totally ordered}.Similarly, if ∆ is a simplicial complex, then its face poset P (∆) is the poset of nonempty faces of ∆ ordered by inclusion.That is, if F 1 and F 2 are two nonempty faces of ∆, then F 1 F 2 if and only if F 1 ⊆ F 2 .Furthermore, the face lattice L(∆) is P (∆) with a smallest element 0 and a largest element 1 attached to it.
Lemma 17.Let P 1 , . . ., P k be the flip poset of B 3 (u, v).Then ∆(P i \{u, v}) is isomorphic to a path or a polygon for 1 i k.
Proof.The result follows if we show that each vertex in G i has degree at most 2.This is clear as any subinterval [x, y] P i ⊂ [u, v] P i of rank two has at most four elements as a consequence of Proposition 13.
To illustrate the previous lemma, for example, the order complexes corresponding to the flip posets of Example 9 are depicted in Figure 2. In this case, they are both paths.

Identification
Let ∆ be the order complex ∆(P i \ {u, v}) for some flip poset P i of B 3 (u, v).Then by Lemma 17, ∆ is either isomorphic to a path or a polygon.We define I(∆) to be the following operator: If ∆ is a polygon, then I(∆) := ∆, and if ∆ is a path, then I(∆) is the unique polygon obtained by identifying the two vertices of degree 1 and preserving the other vertices and edges of ∆.For instance, after identification, the order complexes of the flip posets of Example 9, shown in Figure 2, become a hexagon and a square, respectively.Lemma 15 and Lemma 17 guarantee that, after identification if needed, ∆(P i \ {u, v}) will be an n-gon with an even number of sides.Thus, said order complex is the barycentric subdivision of an n 2 -gon.The face posets of these n 2 -gons shall be utilized to express the terms of degree two in the complete cd-index as sums of the cd-index of Eulerian posets.This is discussed in the next subsection.

Terms of degree two in the complete cd-index
In this sub-section we show how the Flip algorithm separates B 3 (u, v) into Eulerian pieces whose cd-index adds up to the degree-two terms of ψ u,v (c, d).In particular, we establish that the terms corresponding to the paths in B 3 (u, v) are non-negative.Using a different method, the non-negativity of these terms has already been established by Karu [14] as they contain at most one d.Our approach, however, allows us to describe these monomials in the complete cd-index in terms of the (ordinary) cd-index of Eulerian posets, thus providing a nice connection between the two concepts and concluding nonnegativity of certain terms in the complete cd-index.
We now prove that the terms of degree two in the complete cd-index are given by adding up the cd-index of posets coming from the Flip algorithm.
Theorem 18. Suppose that B 3 (u, v) = ∅ and let P i , for 1 i k, be the flip posets of B 3 (u, v).Furthermore, let ∆ i = ∆(P i \ {u, v}), and P i := L(I(∆ i )) (the face lattice of ∆ i after identification).Then the terms of degree two of the complete cd-index are obtained by adding the cd-index of all the P i .In other words, Proof.Lemma 15 gives that I(∆ i ) is a polygon of even length.Thus I(∆ i ) is the first barycentric subdivision of an m i -cycle for some positive integer m i .Therefor, P i (and P i ) has 2m i maximal chains.Notice that the flag f -vector of P i is f ∅ = 1, f {1} = m i = f {2} , and f {1,2} = 2m i and therefore the flag h-vector of P i is h ∅ = 1, h {1} = m i − 1 = h {2} , and h {1,2} = 1.It follows that the ab-index of P i is a 2 + (m i − 1)ab + (m i − 1)ba + b 2 = (a 2 + ab + ba + b 2 ) + (m i − 2)(ab + ba).Thus, ψ(P i ) = c 2 + (m i − 2)d, and the sum of the cd-index of P 1 , . . ., P k is k i=1 Let us now compute the terms of degree two of ψ u,v (c, d).Since there are k flip posets, there are exactly k rising paths and k falling paths in B 3 (u, v).Thus, [2] where n ab and n ba are the number of ab and ba monomials, respectively.Notice that n ab + n ba is the number of paths in B 3 (u, v) that are neither rising nor falling.Thus From Example 9, shown in Figure 1, we know that B 3 (1234, 4312) has two flip posets P 1 and P 2 .The face posets of their order complex, ∆ 1 and ∆ 2 , shown in Figure 2, are that of a triangle and a 2-gon, and they contribute c 2 + d and c 2 , respectively.The sum gives the degree-two terms of ψ 1234,4312 (c, d), 2c 2 + d.

Definition 5 .
∆) is rising, and (iii) The time and space complexity of the Flip algorithm is Ω(|B k (u, v)|).Let x ∈ [u, v], and let c

1 and d 2 d = D 2 .
By definition of the flip graph, we have that D 2 = Flip(D 1 ), so D 1 has a nonempty descent set.If min(D(D 1 )) k − q, then D 2 would not end with d and therefore D 2 ∈ V (F G d (u, v; k)), which contradicts the choice of D 2 .Hence the electronic journal of combinatorics 25(4) (2018), #P4.16 min(D(D 1 )) < k − p, and so min(D(D 1 )) = min(D(d 1 )), and therefore there exists d 2 with Flip(D 1 ) = d 2 d and Flip(d 1 ) = d 2 .Conversely, suppose that (d 1 , d 2 ) is an edge of F G(u, x; k − q) then (d 1 d, d 2 d) must be an edge of F G d (u, v; k), as the first descent from bottom to top of d 1 d occurs in d 1 .

Example 9 .
Consider the 10 elements of B 3 (1234, 4312).Then the output of the Flip algorithm is shown in Figure 1.In the first column we have the two components G 1 and G 2 of the flip graph F G(1234, 4312), and in the right column the corresponding flip posets P 1 and P 2 .The paths of B 3 (1234, 4312) are labeled by reflections, and the same is done to label the edges of the posets.

Figure 1 :
Figure 1: On the left, the two connected components G 1 and G 2 of the flip graph F G(1234, 4312), and on the right, the corresponding flip posets P 1 and P 2 .If 1243 and 1432 are identified, one would obtain the face poset of a triangle and a 2-gon, respectively.We discuss this identification in Section 5.