Triangle-Free Triangulations, Hyperplane Arrangements and Shifted Tableaux
TTRIANGLE-FREE TRIANGULATIONS, HYPERPLANEARRANGEMENTS AND SHIFTED TABLEAUX
RON M. ADIN AND YUVAL ROICHMAN
Abstract.
Flips of diagonals in colored triangle-free triangulations of aconvex polygon are interpreted as moves between two adjacent chambers ina certain graphic hyperplane arrangement. Properties of geodesics in theassociated flip graph are deduced. In particular, it is shown that: (1) everydiagonal is flipped exactly once in a geodesic between distinguished pairsof antipodes; (2) the number of geodesics between these antipodes is equalto twice the number of standard Young tableaux of a truncated shiftedstaircase shape. Introduction
It was shown in [1] that the diameter of the flip graph on the set of all coloredtriangle-free triangulations of a convex n -gon (to be defined in Subsection 2.1) isexactly n ( n − /
2. Observing that this is the number of diagonals in a convex n -gon, it was conjectured by Richard Stanley that all diagonals are flipped in ageodesic between two antipodes.In this paper Stanley’s conjecture is proved for distinguished pairs of antipodes(Corollary 5.3 below). The proof applies a (cid:101) C n -action on arc permutations, whichyields an embedding of the flip graph in a graphic hyperplane arrangement.Geodesics between distinguished antipodes in the flip graph are then interpretedas minimal galleries from a given chamber c to the negative chamber − c , whilediagonals are interpreted as separating hyperplanes.The set of geodesics between these antipodes is further studied in Section 9.It is shown that the number of these geodesics is equal to twice the number ofYoung tableaux of a truncated shifted staircase shape. Motivated by this result,product formulas for this number, as well as for other truncated shapes, werefound by Greta Panova [9] and Ronald C. King and the authors [2].2. Triangle-Free Triangulations
In this Section we recall basic concepts and main results from [1].
Date : August 6, ’12. a r X i v : . [ m a t h . C O ] A ug RON M. ADIN AND YUVAL ROICHMAN
Basic Concepts.
Label the vertices of a convex n -gon P n ( n >
4) by the elements 0 , . . . , n − Z n . Consider a triangulation (with no extra ver-tices) of the polygon. Each edge of the polygon is called an external edge of thetriangulation; all other edges of the triangulation are called internal edges , or chords . Definition 2.1.
A triangulation of a convex n -gon P n is called internal-triangle-free , or simply triangle-free , if it contains no triangle with internal edges. Theset of all triangle-free triangulations of P n is denoted T F T ( n ) . A chord in P n is called short if it connects the vertices labeled i − i + 1,for some i ∈ Z n . A triangulation is triangle-free if and only if it contains onlytwo short chords [1, Claim 2.3].A proper coloring (or orientation ) of a triangulation T ∈ T F T ( n ) is a labelingof the chords by 0 , . . . , n − i , i + 1.It is easy to see that each T ∈ T F T ( n ) has exactly two proper colorings. Theset of all properly colored triangle-free triangulations is denoted CT F T ( n ).Each chord in a triangulation is a diagonal of a unique quadrangle (the unionof two adjacent triangles). Replacing this chord by the other diagonal of thatquadrangle is a flip of the chord. A flip in a colored triangulation preserves thecolor of the flipped diagonal. Definition 2.2.
The colored flip graph Γ n is defined as follows: the vertices areall the colored triangle-free triangulations in CT F T ( n ) . Two triangulations areconnected in Γ n by an edge labeled i if one is obtained from the other by a flip ofthe chord labeled i . See Figure 2.1 for a drawing of Γ , where the coloring of a triangulation isdisplayed by shading the triangle with the short chord labeled 0 and two externaledges as sides.2.2. A (cid:101) C n − -Action on Triangle-Free Triangulations. Let (cid:101) C n be the affine Weyl group generated by S = { s , s , . . . , s n − , s n } subject to the Coxeter relations(1) s i = 1 ( ∀ i ) , (2) ( s i s j ) = 1 ( | j − i | > , RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 3
Figure 2.1. Γ (3) ( s i s i +1 ) = 1 (1 ≤ i ≤ n − , and(4) ( s i s i +1 ) = 1 ( i = 0 , n − . The group (cid:101) C n − acts naturally on CT F T ( n ) by flips: Each generator s i flipsthe chord labeled i in T ∈ CT F T ( n ), provided that the result still belongs to CT F T ( n ). If this is not the case then T is unchanged by s i . Proposition 2.3. [1, Proposition 3.2]
This operation determines a transitive (cid:101) C n − -action on CT F T ( n ) . This affine Weyl group action on
CT F T ( n ) was used to calculate the diameterof Γ n . Theorem 2.4. [1, Theorem 5.1]
The diameter of Γ n ( n > is n ( n − / . For any colored triangle-free triangulation T , denote by T R the colored triangle-free triangulation obtained by reversing the labeling in T ; namely, the chordlabeled i in T is labeled n − − i in T R (0 ≤ i ≤ n − Theorem 2.5. [1, Proposition 5.6]
For every n > and T ∈ CT F T ( n ) , thedistance between T and T R in Γ n is exactly n ( n − / . RON M. ADIN AND YUVAL ROICHMAN A (cid:101) C n − -Action on Arc Permutations Arc Permutations.
Let S n be the symmetric group on the letters 1 , . . . , n . Denote a permutation π ∈ S n by the sequence [ π (1) , . . . , π ( n )] and transpositions by ( i, j ). Intervals in the cyclic group Z n are subsets of the form { i, i + 1 , . . . , i + k } ,where addition is modulo n . Definition 3.1.
A permutation π ∈ S n is an arc permutation if, for every ≤ k ≤ n , the first k letters in π form an interval in Z n (where n ≡ , namely,the letter n is identified with zero). Example. π = [1 , , , ,
3] is an arc permutation in S , but π = [1 , , , , , S , since { , , } is an interval in Z but not in Z .The following claim is obvious. Claim 3.2.
The number of arc permutations in S n is n n − .Proof. There are n options for π (1) and two options for every other letter exceptthe last one. (cid:3) Denote by U n the set of arc permutations in S n . Definition 3.3.
Define φ : U n → Z n × Z n − as follows: (1) φ ( π ) := π (1) . (2) For every ≤ i ≤ n − , if { π (1) , . . . , π ( i − } is the arc [ k, m ] then π ( i ) is either k − or m + 1 . Let φ ( π ) i := (cid:40) , if π ( i ) = k − , if π ( i ) = m + 1 .φ is clearly a bijection.3.2. A (cid:101) C n − -Action. Let { σ i : 1 ≤ i ≤ n − } be the Coxeter generating set of the symmetric group S n , where σ i is identified with the adjacent transposition ( i, i + 1). Definition 3.4.
For every ≤ i ≤ n − define a map ρ i : U n → U n as follows: ρ i ( π ) = (cid:40) πσ i +1 , if πσ i +1 ∈ U n ; π, otherwise. ( ∀ π ∈ U n )Note that, for π ∈ U n , πσ i +1 ∈ U n iff either i ∈ { , n − } or φ ( π ) i +1 (cid:54) = φ ( π ) i +2 . RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 5
Observation 3.5.
For every π ∈ U n and ≤ j ≤ n − , φ ( ρ ( π )) j = φ ( π ) − n ) , if j = 1 and φ ( π ) = 0; φ ( π ) + 1 (mod n ) , if j = 1 and φ ( π ) = 1; φ ( π ) + 1 (mod 2) , if j = 2; φ ( π ) j , if j (cid:54) = 1 , ,φ ( ρ i ( π )) j = φ ( π ) σ i +1 ( j ) (1 ≤ i ≤ n − , ∀ j ) and φ ( ρ n − ( π )) j = (cid:40) φ ( π ) j , if j (cid:54) = n − φ ( π ) n − + 1 (mod 2) , if j = n − . Proposition 3.6.
The maps ρ i , when extended multiplicatively, determine a welldefined transitive (cid:101) C n − -action on the set U n of arc permutations.Proof. To prove that the operation is a (cid:101) C n − -action, it suffices to show that itis consistent with the Coxeter relations defining (cid:101) C n − when the operator ρ i isinterpreted as an action of the generator s i . All relations may be easily verifiedusing Observation 3.5; we leave the details to the reader.To prove that the action is transitive, notice first that ρ ( π )(1) = π (2) = π (1) ± n ). It thus suffices to prove that, for every 1 ≤ k ≤ n , themaximal parabolic subgroup (cid:104) s , . . . , s n − (cid:105) of (cid:101) C n − acts transitively on the set U ( k ) n := { π ∈ U n : π (1) = k } . Indeed, this parabolic subgroup is isomorphicto the classical Weyl group B n − . By Observation 3.5, the restricted B n − -action on U ( k ) n may be identified with the natural B n − -action on all subsets of { , . . . , n − } , and is thus transitive. (cid:3) A Graphic Hyperplane Arrangement
Real Hyperplane Arrangements.
Let A be an arrangement of finitely many linear hyperplanes in R d that is central and essential , meaning that ∩ H ∈A H = { } . Let L = (cid:116) di =0 L i be the cor-responding graded poset of intersection subspaces, ordered by reverse inclusion. L is a geometric lattice.Let C be the set of chambers of A , namely the connected components of thecomplement R d \ ∪ H ∈A H . Define a graph structure G ( A ) on the set of vertices C , with two chambers c, c (cid:48) ∈ C connected by an edge if they are separated byexactly one hyperplane in A . It is well-known that the diameter of this graph isequal to the number of hyperplanes, | L | = |A| .The reflection arrangement A n − of type A n − , corresponding to the symmet-ric group S n , has as ambient space the ( n − W = { ¯ x = ( x , . . . , x n ) ∈ R n | x + . . . + x n = 0 } RON M. ADIN AND YUVAL ROICHMAN of R n . Its hyperplanes are H ij := { ¯ x ∈ W | x i = x j } for 1 ≤ i < j ≤ n . Thechambers may be identified with permutations in S n , via c π := { ¯ x ∈ W | x π (1) < x π (2) < · · · < x π ( n ) } ( ∀ π ∈ S n ) . The symmetric group S n acts on the chambers via σ i ( c π ) := c πσ i , the uniquechamber which is separated from c π only by the hyperplane H π ( i ) ,π ( i +1) . Then,for every π ∈ S n , c π and − c π = c πw are antipodes in the graph G ( A n − ), where w := [ n, n − , . . . ,
1] is the longest element in S n .A (simple undirected) graph G = ( V, E ) of order n consists of a set V = { v , . . . , v n } of vertices and a set E of edges, which are unordered pairs of distinctvertices. The associated graphic arrangement A ( G ) is the hyperplane arrange-ment in W ∼ = R n − defined by A ( G ) := { H ij | { v i , v j } ∈ E } ⊆ A n − . For example, if K n be the complete graph of order n then the associated graphicarrangement A ( K n ) is the whole reflection arrangement A n − . For more infor-mation see [8].4.2. The Graph of Chambers G ( U (cid:48) n ) .Definition 4.1. Let G be a graph of order n . Two permutations π, τ ∈ S n are G -equivalent if the points ( π (1) , . . . , π ( n )) , ( τ (1) , . . . , τ ( n )) ∈ W lie in the samechamber of the associated graphic arrangement A ( G ) . Index by 1 , . . . , n the vertices of the complete graph K n , and consider thegraph K (cid:48) n obtained by deleting the edges { , } , { , } , . . . , { n − , n } and { n, } from K n . Let A (cid:48) n − := A ( K (cid:48) n ) be the associated graphic arrangement. Twopermutations π, τ ∈ S n are K (cid:48) n -equivalent if and only if there exist permutations π = π , π , . . . , π t = τ such that, for every 0 ≤ r ≤ t −
1, there exists 1 ≤ i ≤ n − π r +1 = π r σ i and π r σ i π − r ∈ { σ j : 1 ≤ j ≤ n − } ∪ { (1 , n ) } . In otherwords, the K (cid:48) n -equivalence is the transitive closure of the the following relation:there exist 1 ≤ j < n such that the letters j and j + 1, or 1 and n , are adjacentin π , and τ is obtained from π by switching their positions. Remark 4.2.
Since K (cid:48) n is invariant under the natural action of the dihedralgroup I ( n ), this group may be embedded in the automorphism group of thegraph G ( A (cid:48) n − ). Indeed, let γ be the cycle (1 , , . . . , n ) ∈ S n and w := [ n, n − , . . . ,
1] the longest element in S n . If A is a K (cid:48) n -equivalence class then for 0 ≤ j < n and (cid:15) ∈ { , } , w γ j A is a also K (cid:48) n -equivalence class. Moreover, edges in G ( A (cid:48) n − ) are indexed by pairs of K (cid:48) n -equivalence classes, where for every sucha pair, ( A, B ) is an edge in G ( A (cid:48) n − ) if and only if w (cid:15) γ j ( A, B ) is an edge in G ( A (cid:48) n − ). RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 7
Definition 4.3. (i)
Define ˜ K (cid:48) n -equivalence on the subset of arc permuta-tions U n ⊂ S n as the transitive closure of the relation: there exist ≤ j < n such that the letters j and j + 1 , or and n , are adjacent in π ,and τ is obtained from π by switching their positions. (ii) Let U (cid:48) n be the set of ˜ K (cid:48) n -equivalence classes in U n . (iii) Let G ( U (cid:48) n ) be the graph whose vertex set is U (cid:48) n ; two ˜ K (cid:48) n -equivalenceclasses in U (cid:48) n are adjacent in G ( U (cid:48) n ) if they have representatives, whosecorresponding chambers in G ( A n − ) lie in adjacent chambers in G ( A (cid:48) n − ) . Observation 4.4.
For n > all ˜ K (cid:48) n -equivalence classes in U n consist of fourpermutations { π, πσ , πσ n − , πσ σ n − } .Proof. For every π ∈ U n and 1 < i < n −
2, if π ( i + 1) = π ( i ) ± πσ i (cid:54)∈ U n .On the other hand, for every π ∈ U n and i ∈ { , n − } , πσ i ∈ U n . (cid:3) Note that, by definition, two ˜ K (cid:48) n -equivalent arc permutations are K (cid:48) n -equivalentin S n ; hence, they lie in same chamber in G ( A (cid:48) n − ). One concludes that G ( U (cid:48) n )contains no loops. Example 4.5. (a) For n = 4 there are four K (cid:48) -equivalence classes in S : = { [1234] , [1324] , [2134] , [1243] , [2143] , [2413] } and its images undercyclic rotations γ j , 0 ≤ j < G ( A (cid:48) ) are all cyclic rotations of ( , ),thus the graph is a 4-cycle. Since, each K (cid:48) -equivalence class contains one˜ K (cid:48) -class in U (cid:48) , the graphs G ( A (cid:48) ) and G ( U (cid:48) ) are identical.(b) For n = 5 there are three types of K (cid:48) -equivalence classes in S : = { [12345] , [13245] , [21345] , [12435] , [21435] , [24135] , [13254] , [21354] } and its ten images under dihedral group action w γ j , 0 ≤ j < (cid:15) ∈ { , } ; = { [13452] , [14352] , [13542] } and its ten images under dihedralgroup action; = { [13524] } and its ten images under dihedral group action.The edges in the graph G ( A (cid:48) ) are { ( , ) , ( , ) , ( , ) , ( , ) , ( , ) , ( , ) , ( , ) } and their images un-der the dihedral group action.There are ten ˜ K (cid:48) -equivalence classes in U , each contained in one ofthe images under the dihedral group action of . Thus the graph G ( U (cid:48) ) is a 10-cycle. RON M. ADIN AND YUVAL ROICHMAN Stanley’s Conjecture
It was conjectured by Richard Stanley [12] that all diagonals are flipped in ageodesic between two antipodes in the flip graph Γ n of colored triangle-free trian-gulations. A bijection between the set of triangle-free triangulations in CT F T ( n )and the subset U (cid:48) n of chambers in the graphic hyperplane arrangement A ( K (cid:48) n ),which preserves the underlying graph structure, is applied to prove Stanley’sconjecture. Theorem 5.1.
The flip graph Γ n (without edge labeling) is isomorphic to thegraph of chambers G ( U (cid:48) n ) . Furthermore,
Theorem 5.2.
There exists an edge-orientation of the flip graph Γ n such that,for any oriented edge of adjacent triangulations ( T, S ) , S is obtained from T byflipping the diagonal [ i, j ] if and only if the corresponding chambers are separatedby the hyperplane x i = x j . An affirmative answer to Stanley’s conjecture follows.
Corollary 5.3.
For every colored triangle-free triangulation T ∈ CT F T ( n ) ,every diagonal is flipped exactly once along the shortest path from T to sametriangulation with reversed coloring T R . Proof of Theorem 5.1 A (cid:101) C n − -Action on U (cid:48) n . For every π ∈ U n denote the K (cid:48) n -class of π in U n by ¯ π . By Observation 4.4,for every π ∈ U n , ¯ π ∈ U (cid:48) n may be represented by a series of n − { π (1) , π (2) } , { π (3) } , . . . , { π ( n − } , { π ( n − , π ( n ) } , where all subsets except ofthe first and the last are singletons.For every simple reflection σ i ∈ S n − , 1 < i < n −
3, let ¯ πσ i be the seriesof subsets obtained from ¯ π by replacing the letters in the i -th and i + 1-st sub-sets. Let ¯ πσ be obtained from ¯ π by replacing letters in the first two subsetsas follows: if { π (1) , π (2) } = { π (3) − , π (3) − } then the first two subsets in¯ πσ are { π (3) , π (3) − } , { π (3) − } ; { π (1) , π (2) } = { π (3) + 1 , π (3) + 2 } thenthe first two subsets in ¯ πσ are { π (3) , π (3) + 1 } , { π (3) + 2 } . Similarly, ¯ πσ n − isobtained from ¯ π by replacing the letter in the n − π ( n − − { π ( n − , π ( n ) } = { π ( n − − , π ( n − − } and with π ( n + 2) otherwise.For every 0 ≤ i ≤ n let θ i : U (cid:48) n (cid:55)→ U (cid:48) n be θ i (¯ π ) = (cid:40) ¯ πσ i +1 , if ¯ πσ i +1 ∈ U (cid:48) n , ¯ π, if ¯ πσ i +1 (cid:54)∈ U (cid:48) n . ( ∀ ¯ π ∈ U (cid:48) n ) Observation 6.1.
The maps θ i , (0 ≤ i ≤ n − , when extended multiplicatively,determine a well defined transitive (cid:101) C n − -action on U (cid:48) n . RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 9
Proof is similar to the proof of Observation 3.6 and is omitted.
Observation 6.2.
Two chambers in ¯ π, ¯ τ ∈ U (cid:48) n are adjacent in G ( U (cid:48) n ) if andonly if there exist ≤ i ≤ n − , such that θ i (¯ π ) = ¯ τ . A Graph Isomorphism.
Let f : CT F T ( n ) (cid:55)→ U (cid:48) n be defined as follows: if [ a, a + 2] is the short chordlabeled 0 then let { π (1) , π (2) } = { a, a + 1 } . For 0 < i < n −
4, assume that thechord labeled i − T is [ a − k, a + m ] for some k, m ≥ k + m = i + 1. Thechord labeled i is then either [ a − k − , a + m ] or [ a − k, a + m + 1]. Let i + 1-stsubset be { a − k − } in the former case and { a + m } in the latter. Finally, letthe last subset consist of the remaining two letters. Claim 6.3.
The map f : CT F T ( n ) (cid:55)→ U (cid:48) n is a bijection.Proof. The map f is invertible. (cid:3) Recall the definition of T R from Section 2. Observation 6.4.
For every T ∈ CT F T ( n ) , f ( T R ) is obtained from f ( T ) byreversing the order of the subsets. Recall from Subsection 2.2 the affine Weyl group (cid:101) C n − -action on CT F T ( n ).To complete the proof of Theorem 5.1 it suffices to show that Proposition 6.5.
For every Coxeter generator s i of (cid:101) C n − (0 ≤ i ≤ n − and T ∈ CT F T ( n ) f ( s i T ) = s i f ( T ) , where s i f ( T ) := θ i ( f ( T )) .Proof. For i = 0, let [ a, a + 2] be the short chord labeled 0 in T . Then the chordlabeled 1 is either [ a, a + 3] or [ a − , a + 2]. In the first case the short chordlabeled 0 in s T is [ a + 1 , a + 3] and all other chords are unchanged, in particular,the chord labeled 1 in s T is [ a, a + 3]. By definition of the map f , the firsttwo subsets in f ( T ) are { a, a + 1 } , { a + 2 } and the first two subsets in f ( s T )are { a + 1 , a + 2 } , { a } and the rest are not changed. On the other hand, bydefinition of θ , the first two subsets in s f ( T ) are { a + 1 , a + 2 } , { a } and the restare unchanged. A similar analysis shows that f ( s T ) = s f ( T ) when the chordlabeled 1 is [ a − , a + 2].For 0 < i < n − i − T be [ a − k, a + m ] forsome k, m ≥ k + m = i + 1. The chords labeled i and i + 1 are then either[ a − k − , a + m ] , [ a − k − , a + m ] respectively, or [ a − k, a + m +1] , [ a − k, a + m +2]or [ a − k − , a + m ] , [ a − k − , a + m + 1] or [ a − k, a + m + 1] , [ a − k − , a + m + 1].In the first two cases s i T = T , so f ( s i T ) = f ( T ). On the other hand, in thesecases f ( T ) σ i (cid:54)∈ U (cid:48) n , so s i f ( T ) = f ( T ).If the chords labeled i and i + 1 in T are [ a − k − , a + m ] , [ a − k − , a + m + 1]respectively, then the chords labeled i and i + 1 in s i T are [ a − k, a + m + 1] , [ a − k − , a + m + 1]. So, the i -th and i + 1-st subsets in f ( T ) are { a − k − } , { a + m } ,and they are switched in s i f ( T ), so same as the corresponding subsets in f ( s i T ).The proof of the forth case is similar.Finally, by Observation 6.4, f ( s T ) = s f ( T ) implies that f ( s n − T ) = s n − f ( T ). (cid:3) Proof of Theorem 5.2
Orienting the Colored Flip Graph.
The goal of this subsection is to equip the colored flip graph Γ n with an edgeorientation that will be used to encode the location of the flipped diagonals. Itwill be proved later that this orientation satisfies the conditions of Theorem 5.2.Our starting point is the edge labeling, mentioned in Section 2, which encodesthe order of the chords.Recall from [1] the bijection ϕ : CT F T ( n ) → Z n × Z n − defined as follows: Let T ∈ CT F T ( n ). If the (short) chord labeled 0 in T is[ a − , a + 1] for a ∈ Z n , let ϕ ( T ) := a . For 1 ≤ i ≤ n −
4, assume that the chordlabeled i − T is [ a − k, a + m ] for some k, m ≥ k + m = i + 1. The chordlabeled i is then either [ a − k − , a + m ] or [ a − k, a + m + 1]. Let ϕ ( T ) i be 0 inthe former case and 1 in the latter.By definition of the map ϕ , Claim 7.1.
For every vector v = ( v , . . . , v n − ) ∈ Z n × Z n − and every ≤ i ≤ n − , the diagonal labeled i in the triangulation T = ϕ − ( v ) is [ k, m ] where k := v − − i + i (cid:88) j =1 v i ∈ Z n and m := v + 1 + i (cid:88) j =1 v i ∈ Z n . Here , ∈ Z are interpreted as , ∈ Z n . It follows that
Corollary 7.2. [1, Lemma 5.7]
For every T ∈ CT F T ( n ) , if ϕ ( T ) = ( v , . . . , v n ) then ϕ ( T R ) = 2 + n − (cid:80) i =0 v i ∈ Z n and ϕ ( T R ) i = 1 − v n − − i ∈ Z (1 ≤ i ≤ n − . RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 11
Observation 7.3. [1, Observation 3.1]
For every T ∈ CT F T ( n ) and a Coxetergenerator s i of (cid:101) C n − ( ϕ ( s T )) j = ϕ ( T ) j , if j (cid:54) = 0 , ,ϕ ( T ) + 1 (mod n ) , if j = 0 and ϕ ( T ) = 0 ,ϕ ( T ) − n ) , if j = 0 and ϕ ( T ) = 1 ,ϕ ( T ) + 1 (mod 2) , if j = 1 and ϕ ( T ) = 0;( ϕ ( s n − T )) j = (cid:40) ϕ ( T ) j , if j (cid:54) = n,ϕ ( T ) n + 1 (mod 2) , if j = n ; and ( ϕ ( s i T )) j = ϕ ( T ) σ i ( j ) (0 < i < n − where σ i := ( i, i + 1) the adjacent transposition. We use this observation to orient the edges in Γ n . Definition 7.4.
Orient the edges in Γ n as follows: If the diagonal labeled n − isflipped orient the corresponding edge from the triangulation encoded by last entry to the one with last entry . If the flip is of the diagonal labeled < i < n − orient the edge from T with ϕ ( T ) i = 0 , ϕ ( T ) i +1 = 1 to the one with these twoentries switched; if it flips the diagonal labeled orient it by the first entry from T with ϕ ( T ) = j to the one with first entry under ϕ being j + 1 ; . See Figure 7.1 for the orientation of Γ , where each colored triangulation T islabeled by the vector ϕ ( T ). Lemma 7.5.
For every T ∈ CT F T ( n ) , the orientation of the edges along anygeodesic from T to T R is coherent with the orientation of Γ n described in Def-inition 7.4; namely, all edges in a geodesic have the same orientation as in theoriented Γ n or all have the opposite orientation.Proof. Consider the dominance order on vectors in Z n × Z n − ; namely,( v , . . . , v n − ) ≤ ( u , . . . , u n − )if and only if k (cid:88) i =0 v i ≤ k (cid:88) i =0 u i (0 ≤ k ≤ n − , where 0 , . . . , n − ∈ Z n are interpreted as 0 , . . . , n − ∈ Z , and similarly for Z .The resulting poset is ranked by (cid:96) ( v , . . . , v n − ) := n − (cid:88) i =0 ( n − − i ) v i . Using Corollary 7.2, the reader can verify that for every T ∈ CT F T ( n ) (cid:96) ( ϕ ( T R )) − (cid:96) ( ϕ ( T )) ≡ n ( n − / n ( n − , Figure 7.1. Γ with orientationwhich is the distance between T and T R (Theorem 2.5).Finally, notice that for every edge e = ( S , S ) in Γ n , the edge e is orientedfrom S to S if and only if (cid:96) ( ϕ ( S )) − (cid:96) ( ϕ ( S )) ≡ n ( n − n ( n −
3) or all steps decrease it by one. Hence the lemma holds. (cid:3)
We note that this proof essentially appears (implicitly) in [1], where an alge-braic interpretation of the rank function as a length function on (cid:101) C n − is given;see, in particular, [1, Sections 3.3 and 5.2].Now color each edge ( S , S ) of Γ n , oriented from S to S , by the chord [ i, j ]which is erased from S . Ignore the edge-orientation and let ˆΓ n be the resultingedge-labeled flip graph. RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 13
Edge-Colored Graph Isomorphism.
Consider an edge-labeled version of the graph G ( U (cid:48) n ), denoted by ˆ G ( U (cid:48) n ),where the edge between two adjacent chambers is labeled by the separating hy-perplane. Theorem 7.6.
The edge-labeled graphs ˆ G ( U (cid:48) n ) and ˆΓ n are isomorphic. Note that this theorem implies Theorem 5.2.
Proof.
By Observation 6.2, two chambers ¯ π, ¯ τ ∈ U (cid:48) n are adjacent in G ( U (cid:48) n ) if andonly if there exist corresponding arc permutations π, τ ∈ U n and 1 < i < n − πσ i = τ . The separating hyperplane is then x k = x m if and only if( k, m ) π = τ , or equivalently πσ i π − = ( k, m ), for the transposition ( k, m ) ∈ S n .Recall the bijection f : CT F T ( n ) (cid:55)→ U (cid:48) n , defined in Subsection 6.2. Since f induces a graph isomorphism, for every 1 < i < n −
1, if π, πσ i are two arcpermutations then f − (¯ π ) , f − ( πσ i )) forms an edge in Γ n . In order to proveTheorem 5.2, it suffices to show that f − (¯ πσ i )) is obtained from f − (¯ π ) byflipping the diagonal [ k, m ], when the edge is oriented from f − (¯ π ) to f − (¯ πσ i ).Indeed, an edge is oriented from f − (¯ π ) to f − ( πσ ) if and only if the lattertriangulation is obtained from the first by flipping the short chord labeled 0[ a − , a + 1]; namely, by replacing the diagonal [ a − , a + 1] by [ a, a + 2], wherethe diagonal labeled 1 is [ a − , a + 2]. By definition of the map f the first twosubsets in ¯ π are { a − , a } , { a + 1 } and in πσ : { a, a + 1 } , { a − } . Letting π = [ a, a − , a + 1 , . . . ] one gets πσ π − = ( a − , a + 1), so the separatinghyperplane is x a − = x a +1 .For 1 < i < n −
3, an edge is oriented from f − (¯ π ) to f − ( πσ i ) if and only ifthe chord labeled i − a − k, a + m ] and the latter triangulation is obtainedfrom the first by flipping a diagonal [ a − k − , a + m ]; namely, by replacing thediagonal [ a − k − , a + m ] by [ a − k, a + m + 1]. Then the i -th and i + 1-st subsetsin ¯ π , which are { a − k − } , { a + m } , are switched in πσ i +1 . So π ( i ) = a − k − π ( i + 1) = a + m + 1, and πσ i +1 π − = ( a − k − , a + m ).Finally, an edge is oriented from f − (¯ π ) to f − ( πσ n − ) if and only if the chordlabeled n − b − , b + 1] and the latter triangulation is obtained from thefirst by flipping the short chord labeled n − b − , b + 1]; namely, by replacingthe diagonal [ b − , b + 1] by [ b, b + 2]. Then the last two subsets in ¯ π are { b − } , { b, b + 1 } and in πσ n − : { b } , { b − , b } . So π = [ . . . , b − , b + 1 , b ] onegets πσ n − π − = [ b − , b + 1]. (cid:3) Proof of Corollary 5.3
Recall that T and T R are antipodes (Theorem 2.5). Proposition 8.1.
For every colored triangle-free triangulation T ∈ CT F T ( n ) ,the corresponding chambers in A ( K (cid:48) n ) satisfy c f ( T R ) = − c f ( T ) . Proof.
Let w := [ n, n − , n − , . . . .
1] be the longest permutation in S n . Itfollows from Observation 6.4 that for every π ∈ U n ( f − (¯ π )) R = f − ( πw ) . Notice that the points π and πw in R n belong to negative chambers c and − c .The proof is completed. (cid:3) Proof of Corollary 5.3.
By Proposition 8.1, the set of hyperplanes whichseparate the chamber c f ( T ) from the chamber c f ( T R ) is the set of all hyperplanesin A ( K (cid:48) n ). By Theorem 7.6 together with Lemma 7.5, one deduces that alldiagonals have to be flipped at least once in a geodesic from T to T R . Finally,by Theorems 2.4 and 2.5, the distance between T and T R in Γ n is equal to thenumber of diagonals in a convex n -gon. Hence, each diagonal is flipped exactlyonce. (cid:3) Geodesics and Shifted Tableaux
Let T be the canonical colored star triangle-free triangulation ; that is the trian-gulation, which consists of the chords [0 , , [0 , , . . . , [0 , n −
2] labeled 0 , . . . , n − Order on the Diagonals.
By Corollary 5.3, every geodesic from the canonical colored star traingle-freetriangulation T to T R determines a linear order on the diagonals. The followingtheorem characterizes these linear orders. Theorem 9.1.
An order on the set of diagonals { [ i, j ] : 1 ≤ i < j − ≤ n − } ofa convex n -gon appears in geodesics in Γ n from T to its reverse T R if and onlyif it is a linear extension of the coordinate-wise order with respect to the naturalorder < < < · · · < n − , or its reverse ≡ n < n − < n − < · · · < . Proof.
Clearly, every geodesic from T to T R starts with either flipping [0 ,
2] or[0 , n − , n -goncorresponding to geodesics from T to T R , which start by flipping [0 , RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 15 linear extension of the coordinate-wise order with respect to the natural order0 < < < · · · < n − c [0 , , ,...,n − to its negative, that start by crossing the hyperplane H , , the hyperplane H k,l is crossed after the hyperplane H i,j , whenever i +1 < j , k + 1 < l , and ( i, j ) < ( k, l ) in point-wise coordinate order. In other words, itsuffices to prove that for every arc permutation π ∈ U n , if ¯ π ∈ U (cid:48) n correspondsto a chamber in such a gallery then π − ( i ) < π − ( j ) = ⇒ π − ( k ) < π − ( l ).Clearly, this holds for the arc permutations which correspond to the identitychamber c [0 , ,...,n − and to its negative − c [0 , ,...,n − = c [ n − ,n − ,..., . Withregard to all other chambers in these galleries, notice first, that all geodesics from T to T R must end by flipping either [ n − , n −
1] or [1 , S ∈ CT F T ( n )be the triangulation, which consists of the chords [1 , , [0 , , [0 , , . . . , [0 , n − , , . . . , n − T R is obtained from S R by flipping[1 ,
3] and S is obtained from T by flipping [0 , S and S R are antipodes,it follows that S does not appear in a geodesic from T to T R which start byflipping [0 , T to T R , which start by flipping [0 , n − , n − π ∈ U n , if ¯ π ∈ U (cid:48) n is a chamber in such a gallery which is not first or last, then π − (2) < π − (0)and π − ( n − < π − ( n − π , π (0), is not 0 or n − π (0) = 1 then, since π − ( n − < π − ( n −
1) and π is an arc permutation,0 = π − (1) < π − (2) < · · · < π − ( n − < π − ( n − π (0) = n − π − (2) < π − (0) and π is an arc permutation,0 = π − ( n − < π − ( n − < · · · < π − (2) < π − (0).Finally, if 2 ≤ π (0) ≤ n − π − ( n − < π − ( n − π − (2) <π − (0) and π is an arc permutation, letting π (0) := i the following holds: 0 = π − ( i ) < π − ( i − · · · < π − (2) < π − (0) and 0 = π − ( i ) < π − ( i + 1) < · · · <π − ( n − < π − ( n − π ∈ U n , such that ¯ π is a chamber in a gallery fromthe identity chamber to its negative that start by flipping [0 , i, j ) < ( k, l ) in point-wise coordinate order, with i + 1 < j and k + 1 < l , suchthat π − ( i ) < π − ( j ) but π − ( k ) > π − ( l ). One concludes that there is no( i, j ) < ( k, l ) in point-wise coordinate order, with i + 1 < j and k + 1 < l , suchthat [ i, j ] is flipped after [ k, l ].It remains to prove the opposite direction, namely, to show that every linearextension of the coordinate-wise order appears as a geodesic. To prove this, first,notice that the lexicographic order does appear. Then observe that if i < j 1, observe that these geodesics may be obtained from geodesicsthat start by flipping [0 , 2] via the reflection which maps every 0 ≤ i ≤ n − n − i . (cid:3) Skew Shifted Young Lattice.Definition 9.2. For a positive integer n let Λ( n ) be the set of all partitions withlargest part ≤ n and with all parts distinct, except possibly the first two partswhen they are equal to n . Namely, Λ( n ) := { λ = ( λ , . . . , λ k ) : k ≥ , n ≥ λ ≥ λ > λ > · · · > λ k > and [ either λ > λ or λ = λ = n ] } . Let (Λ( n ) , ⊆ ) the poset of partitions in Λ( n ) ordered by inclusion of the corre-sponding Young diagrams. Example 9.3. Λ(3) = { (3 , , , , (3 , , , (3 , , , (3 , , (3 , , , (3 , , (3 , , (3) , (2 , , (2) , (1) , () } . Consider the standard tableaux of truncated shifted staircase shape ( n − , n − , n − , n − , . . . , , Y ( n ). Example 9.4. The truncated shifted staircase shape (3 , , , 1) is drawn in thefollowing way: X X X ∗ X X XX XX There are four standard tableaux of this shape1 2 3 ∗ , ∗ , ∗ , ∗ Observation 9.5. The maximal chains in (Λ( n ) , ⊆ ) are parameterizedby the set of standard tableaux of truncated shifted staircase shape ( n − , n − , n − , . . . , . RIANGULATIONS, ARRANGEMENTS AND TABLEAUX 17 The linear extensions of the coordinate-wise order on the set { ( i, j ) : 0 ≤ i + 1 < j ≤ n } \ { (0 , n ) } are parameterized by the set of standard tableaux of truncated shiftedstaircase shape ( n − , n − , n − , . . . , . With any standard tableau of truncated shifted staircase shape T associatetwo words of size (cid:0) n (cid:1) − r ( T ) and c ( T ), where r ( T ) i ( c ( T ) i ), (1 ≤ i ≤ (cid:0) n (cid:1) − i is located. Example 9.6. Let P, Q be the first two tableaux in Example 9.4. Then r ( P ) =(1 , , , , , , , , c ( P ) = (1 , , , , , , , , r ( Q ) = (1 , , , , , , , , c ( Q ) = (1 , , , , , , , , Geodesics and Tableaux. Denote the set of geodesics from T ∈ CT F T ( n ) to T R starting by flipping[0 , 2] by D ( T ) + . Proposition 9.7. There is a bijection from the set of geodesics D ( T ) + to Y ( n − (the set of standard tableaux on truncated shifted staircasepartition ( n − , n − , n − , . . . , ) φ : D ( T ) + → Y ( n − . For every geodesic u ∈ D ( T ) + , the diagonal flipped at the i -th step is [ r ( φ ( u )) i − , c ( φ ( u )) i + 1] . Example 9.8. The bijection φ maps the tableau ∗ to the series of diagonals: [0 , , [0 , , [1 , , [0 , , [1 , , [1 , , [2 , , [2 , , [3 , . Proof. Combining Theorem 9.1 with Observation 9.5(2). (cid:3) Let d n denote the number of geodesics from the canonical star triangulation T of an n -gon to its reverse T R . By Proposition 9.7, d n / n − , n − , , n − , . . . , d n were stated in an early version of this preprint.Subsequently, an explicit multiplicative formula was proved by Greta Panova [9]and Ronald C. King and the authors [2]. Theorem 9.9. The number of geodesics from the canonical star triangulation T of a convex n -gon to its reverse T R is d n = g [ n − · (cid:18) N n − (cid:19) · n − n − N ! · n − n − · ( n − · n − (cid:89) i =0 i !(2 i + 1)! , where g [ n − := g ( n − ,n − ,..., is the number of standard Young tableaux of shiftedstaircase shape ( n − , n − , . . . , and N := n ( n − / . References [1] R. M. Adin, M. Firer and Y. Roichman, Triangle Free Triangulations , Adv. Appl.Math. 45 (2010), 77-95.[2] R. M. Adin, R. C. King and Y. Roichman, Enumeration of standard Young tableaux ofcertain truncated shapes , Electron. J. Combin. 18(2) (2011), The Zeilberger Festschriftvolume, Paper 20, 14 pp.[3] A. Bj¨orner and F. Brenti, Combinatorics of Coxeter groups. Graduate Texts in Math-ematics, 231. Springer, New York, 2005.[4] J. H. Conway and H. S. M. Coxeter, Triangulated polygons and frieze patterns , Math.Gaz 57 (1973), 87–94 and 175–186.[5] P. Dehornoy, Dual presentation of Thompson’s group F and flip distance betweentriangulations , Lecture notes (CIRM, June 2008).[6] P. Dehornoy, The rotation distance between binary trees , Adv. Math. 223 (2010), 1316-1355.[7] S. Elizalde and Y. Roichman, Arc permutations , preprint, 2011.[8] P. Orlik and H. Terao, Arrangements of hyperplanes, Springer-Verlag, Berlin, 1992.[9] G. Panova, Truncated tableaux and plane partitions , preprint 2010, arXiv:1011.0795.[10] B. E. Sagan, Proper partitions of a polygon and k -Catalan numbers , Ars Combin. (2008), 109–124.[11] D. D. Sleator, R. E. Tarjan and W. P. Thurston, Rotation distance, triangulations,and hyperbolic geometry , J. Amer. Math. Soc. (1988), 647–681.[12] R. P. Stanley, personal communication. Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel E-mail address : [email protected] Department of Mathematics, Bar-Ilan University, Ramat-Gan 52900, Israel E-mail address ::