aa r X i v : . [ m a t h . C O ] O c t GENERALIZED DYCK TILINGS
MATTHIEU JOSUAT-VERG`ES AND JANG SOO KIM
Abstract.
Recently, Kenyon and Wilson introduced Dyck tilings, which are certain tilings ofthe region between two Dyck paths. The enumeration of Dyck tilings is related with hookformulas for forests and the combinatorics of Hermite polynomials. The first goal of this workis to give an alternative point of view on Dyck tilings by making use of the weak order and theBruhat order on permutations. Then we introduce two natural generalizations: k -Dyck tilingsand symmetric Dyck tilings. We are led to consider Stirling permutations, and define an analogof the Bruhat order on them. We show that certain families of k -Dyck tilings are in bijectionwith intervals in this order. We also enumerate symmetric Dyck tilings. Introduction
Dyck tilings were recently introduced by Kenyon and Wilson [12] in the study of probabilities ofstatistical physics model called “double-dimer model”, and independently by Shigechi and Zinn-Justin [16] in the study of Kazhdan-Lusztig polynomials. Dyck tilings also have connection withfully packed loop configurations [7] and representations of the symmetric group [6].The main purpose of this paper is to give a new point of view on Dyck tilings in terms of theweak order and the Bruhat order on permutations and to consider two natural generalizations ofDyck tilings.A
Dyck path of length n is a lattice path consisting of up steps (0 ,
1) and down steps (1 , ,
0) to the point ( n, n ) which never goes strictly below the line y = x . We willalso consider a Dyck path λ of length 2 n as the Young diagram whose boundary is determined by λ and the lines x = 0 and y = n .Suppose that λ and µ are Dyck paths of length 2 n with µ weakly above λ . A Dyck tile is aribbon such that the center of the cells form a Dyck path. A
Dyck tiling of λ/µ is a tiling D ofthe region between λ and µ with Dyck tiles satisfying the cover-inclusive property : if η is a tileof D , then the translation of η by (1 , −
1) is either completely below λ or contained in anothertile of D . See Figure 1 for an example. We denote by D ( λ/µ ) the set of Dyck tilings of λ/µ . For D ∈ D ( λ/µ ) we call λ and µ the lower path and the upper path of D , respectively. Then the setof Dyck tilings with fixed upper path λ is denoted by D ( λ/ ∗ ) and similarly, the set of Dyck tilingswith fixed lower path µ is denoted by D ( ∗ /µ ).For D ∈ D ( λ/µ ) we have two natural statistics: the area area( D ) of the region λ/µ and thenumber tiles( D ) of tiles of D . We also consider the statistic art( D ) = (area( D ) + tiles( D )) / X D ∈D ( λ/ ∗ ) q art( D ) = [ n ] q ! Q x ∈ F [ h x ] q , (2) X D ∈D ( ∗ /µ ) q tiles( D ) = Y u ∈ UP( µ ) [ht( u )] q , where F is the plane forest corresponding to λ and, for a vertex x ∈ F , h x denotes the hook lengthof x . The set of up steps of a Dyck path µ is denoted by UP( µ ) and for u ∈ UP( µ ), ht( u ) is the Mathematics Subject Classification.
Figure 1.
An example of Dyck tiling.number of squares between u and the line y = x plus 1. Here we use the standard notation for q -integers and q -factorials: [ n ] q = 1 + q + q + · · · + q n − and [ n ] q ! = [1] q [2] q . . . [ n ] q .Formula (1) was first proved by Kim [13] non-bijectively and then by Kim, M´esz´aros, Panova,and Wilson [14] bijectively. In [14], they find a bijection between D ( λ/ ∗ ) and increasing labelingsof the plane forest corresponding to λ . Kim [13] and Konvalinka independently proved (2) byfinding a bijection between D ( ∗ /µ ) and certain labelings of µ called Hermite histories.Bj¨orner and Wachs showed that the right hand side of (1) is the length generating function forpermutations in an interval in the weak order, see [3, Theorem 6.8] and [2, Theorem 6.1].In this paper we first show that, using the results of Bj¨orner and Wachs [2, 3], (1) can beinterpreted as the length generating function for permutations π ≥ L σ in the left weak orderfor a 312-avoiding permutation σ . We also show that (2) is the length generating function forpermutations π ≥ σ in the Bruhat order for a 132-avoiding permutation σ . We then consider twonatural generalizations of Dyck tilings, namely, k -Dyck tilings and symmetric Dyck tilings.The first generalization is k -Dyck tiling, where we use k -Dyck paths and k -Dyck tiles with thesame cover-inclusive property. We generalize (2) by finding a bijection between k -Dyck tilings and k -Hermite histories. We consider k -Stirling permutations introduced by Gessel and Stanley [9].We define a k -Bruhat order on k -Stirling permutations and show that k -Dyck tilings with fixedupper path are in bijection with an interval in this order. We also consider a connection with k -regular noncrossing partitions. We generalize (1) to k -Dyck tilings with fixed lower path λ when λ is a zigzag path.The second generalization is symmetric Dyck tiling, which is invariant under the reflectionalong a line. We show that symmetric Dyck tilings are in bijection with symmetric matchings and“marked” increasing labelings of symmetric forests.2. Dyck tilings as intervals of the Bruhat order and weak order
As we have seen in the introduction, the two natural points of view for enumerating Dyck tilingsare when we fix the upper path, and when we fix the lower path. We show in this section thatboth can be interpreted in terms of permutations, using respectively the Bruhat order and the(left) weak order, see [1].We begin with the case of a fixed upper path, and the Bruhat order.We denote by S n the set of permutations of [ n ] := { , , . . . , n } . An inversion of π ∈ S n isa pair ( i, j ) of integers 1 ≤ i < j ≤ n such that π ( i ) > π ( j ). The number of inversions of π isdenoted by inv( π ). For a permutation τ , the set of τ -avoiding permutations in S n is denoted by S n ( τ ). For example if τ = 132, σ ∈ S n (132) if there is no i < j < k such that σ i < σ k < σ j .We represent a permutation by a diagram with the “matrix convention”, i.e. there is a dot atthe intersection of the i th line from the top and the j th column from the left if σ ( j ) = i . In thesediagrams, we can represent the inversion of a permutation by putting a cross × in each cell havinga dot to its right and a dot below. See the left part of Figure 2. We need a bijection α between132-avoiding permutations and Dyck paths. It is easy to see that the inversions of a 132-avoidingpermutation are top left justified in its diagram. So we can define a path from the bottom leftcorner to the top right corner by following the boundary of the region filled with × . This turns ENERALIZED DYCK TILINGS 3 b b b b b × × ×× × 7→ × × ×× ×
Figure 2.
The bijection from 132-avoiding permutations to Dyck paths. Thecrosses represent inversions of the permutation 34215.out to be a Dyck path and this defines a bijection (this is an easy exercise). The bijection isillustrated in Figure 2.
Definition 2.1.
Let µ be a Dyck path. • A Hermite history of shape µ is a labelling of the up steps of µ with integers such that astep starting at height h has a label between 0 and h . • A matching of shape µ is a partition of [ n ] in 2-element blocks such that i ∈ [ n ] is theminimum of a block if and only if the i th step of µ is an up step. A crossing of thematching is a pair of blocks { i, j } and { k, ℓ } such that i < j < k < ℓ .The following is well known (see for example [13]). Proposition 2.2.
There is a bijection between Hermite histories of shape µ and matchings ofshape µ . It is such that the sum of weights in the Hermite history is the number of crossings inthe matching. Theorem 2.3.
Let σ ∈ S n (132) and µ = α ( σ ) , then X D ∈D ( ∗ /µ ) q tiles( D ) = X π ≥ σ q inv( π ) − inv( σ ) , where π ≥ σ is the Bruhat order on S n .Proof. From [14], we know that Dyck tilings with a fixed upper path µ are in bijection withHermite histories with shape µ , and the bijection sends the number of tiles to the sum of labels inthe Hermite history. Consequently, Dyck tilings with a fixed upper path µ are in bijection withmatchings of shape µ , and the bijection sends the number of tiles to the number of crossings inthe matching.To show the proposition, we give a bijection between matchings with the same shape µ , andpermutations above σ in the Bruhat order. It is illustrated in Figure 3. The idea is to put dotsin the grid as follows: if there is a pair ( i, j ) in the matching (with i < j ), the i th step in theDyck path is vertical and the j th step is horizontal, so row to the left of the i th step intersectsthe column below the j th step in some cell, and we put a dot in this cell. Then we can read thesedots as a permutation (with the matrix convention). In the example in Figure 3 we get 45321.The crossing in the matchings correspond two inversions of the permutations that lay below theDyck path.The next step is the following: we can prove the set of permutations where all dots are belowthe Dyck path µ is precisely the Bruhat interval { π : π ≥ σ } . First, by construction all the dots of σ are below µ . Suppose all the dots of π are below µ and π ′ ⋗ π in the Bruhat order. It means that π ′ is obtained from π by transforming a pair of dots arranged as bb into a pair of dots arrangedas bb , and the new dots cannot be above µ . So the interval { π : π ≥ σ } is included in the setof permutations where all dots are below the Dyck path µ . Reciprocally, let π be a permutationwhere all dots are below the Dyck path µ . If π = σ , consider an inversion of π wich is as low tothe right as possible. This inversion is in a pattern × bb and the cross is below µ . By transformingthis pattern into bb , we obtain π ′ whith π ′ ⋖ π and has still the property that all dots are below µ . By repeating this operation we must arrive at a permutation whose inversions are exactly thecells above µ , i.e. σ . So π is in the interval { π : π ≥ σ } . (cid:3) MATTHIEU JOSUAT-VERG`ES AND JANG SOO KIM b b b b b b b b b b b b b b b × × Figure 3.
The bijection proving Proposition 2.3. bb b bb bb Figure 4.
The bijection from Dyck paths to plane forests. Here the Dyck pathis rotated clockwise by an angle of 45 ◦ . b b bbb bb b bb bb b
13 3 817 29 10 56 412 11
Figure 5.
An increasing labeling L of a forest of size 13. We have post( L ) =13 , , , , , , , , , , , , L ) = 1 , , , , , , , , , , , , λ with a plane forest: it is obtained from λ by “squeezing” the path as shown in Figure 4, a pair offacing up step and down step corresponds to a vertex.Let F be a plane forest with n vertices. An increasing labeling of F is a way of labeling thevertices of F with 1 , , . . . , n so that the label of a vertex is greater than the label of its parent. Let L be an increasing labeling of F . An inversion of L is a pair i < j such that j is not a descendantof i and j appears to the left of i . For example, if L is the increasing labeling in Figure 5, then L has many inversions including (13 , , (8 , , (2 , L ) (respectively, pre( L )) thepermutation obtained by reading L from left to right using post-order (respectively, pre-order). SeeFigure 5. It is easy to see that there is a unique increasing labeling L of F such that inv(post( L ))is minimal. More specifically, L is the increasing labeling of F with inv( L ) = 0, or equivalently, L is the increasing labeling of F such that pre( L ) = id n , the identity permutation of [ n ]. It isnot difficult to see that the permutation π = post( L ) is 312-avoiding. Theorem 2.4.
Let λ be a Dyck path with corresponding plane forest F . Let L be the increasinglabeling of F such that pre( L ) = id n , and π = post( L ) . Then X D ∈D ( λ/ ∗ ) q art( D ) = X π ≥ L π q inv( π ) − inv( π ) , where ≥ L is the left weak order on S n .Proof. It is shown in [14] that X D ∈D ( λ/ ∗ ) q art( D ) = X L ∈L ( F ) q inv( π ) , where L ( F ) is the set of increasing labelings of F . Thus, it is enough to show that for all k ≥ A k = { L ∈ L ( F ) : inv( L ) = k } , B k = { π ∈ S n : π ≥ L π , inv( π ) − inv( π ) = k } . We will show that the map L post( L ) is a bijection from A k to B k for all k ≥ k . Since A = { L } and B = { π } , it is true when k = 0. Suppose that the claimed statementis true for k ≥
0. We need to show that the map L post( L ) is a bijection from A k +1 to B k +1 .Let L ∈ A k +1 . Since inv( L ) = k + 1 ≥
1, we can find an integer i such that ( i + 1 , i ) is an inversionof L . Since i + 1 is not a descendant of i in L , the labeling L ′ obtained from L by exchanging i and ENERALIZED DYCK TILINGS 5
Figure 6.
An example of 2-Dyck tiling. u
010 4
Figure 7.
The up step u on the left has height ht( u ) = 5 because there are 4squares between u and the line y = x/
2. The diagram on the right is an exampleof 2-Hermite history. i + 1 is also an increasing labeling of F . Since inv( L ′ ) = inv( L ) − k we have L ′ ∈ A k . By theinduction hypothesis, π ′ = post( L ′ ) ∈ B k . One can easily see that the permutation π = post( L )is obtained from π ′ by exchanging i and i + 1. Since i + 1 appears to the left of i in π we haveinv( π ) = inv( π ′ ) + 1 and π ∈ B k +1 . Thus L post( L ) is a map from A k +1 to B k +1 . Similarly, wecan show that, for given π ∈ B k +1 , there is L ∈ B k +1 such that post( L ) = π . Since L is determinedby post( L ) for all L ∈ L ( F ), the map L post( L ) is a bijection from A k +1 to B k +1 . (cid:3) Note that the inversion generating function of increasing labelings of a plane forest is givenby a hook length formula [2]. The fact that some intervals for the weak order have a generatingfunction given by a hook length formula follows from [3].3. k -Dyck tilings For an integer k ≥
1, a k -Dyck path is a lattice path consisting of up steps (0 ,
1) and downsteps (1 ,
0) from the origin (0 ,
0) to the point ( kn, n ) which never goes below the line y = x/k .Let Dyck ( k ) ( n ) denote the set of k -Dyck paths from (0 ,
0) to ( kn, n ). It is well known that thecardinal of Dyck ( k ) ( n ) is the Fuss-Catalan number kn +1 (cid:0) ( k +1) nn (cid:1) (see for example [5]). As in thecase of Dyck path, we denote UP( µ ) the set of up steps of a k -Dyck path µ .A k -Dyck tile is a ribbon in which the centers of the cells form a k -Dyck path. Let λ, µ ∈ Dyck ( k ) ( n ) such that µ is weakly above λ . A (cover-inclusive) k -Dyck tiling is a tiling D of theregion λ/µ between λ, µ ∈ Dyck ( k ) ( n ) with k -Dyck tiles satisfying the cover-inclusive property: if η is a tile in D , then the translation of η by (1 , −
1) is either completely below λ or contained inanother tile of D . See Figure 6 for an example. We denote by D ( k ) ( λ/µ ) the set of k -Dyck tilingsof λ/µ . We also denote by D ( k ) ( λ/ ∗ ) and D ( k ) ( ∗ /µ ) the sets of k -Dyck tilings with fixed lowerpath λ and with fixed upper path µ , respectively.For D ∈ D ( k ) ( λ/µ ), there are two natural statistics: the number tiles( D ) of tiles in D and thearea area( D ) of the region occupied by D . We also defineart k ( D ) = k · area( D ) + tiles( D ) k + 1 . Definition 3.1.
For an up step u of a k -Dyck path, we define the height ht( u ) of u to be thenumber of squares between u and the line y = x/k plus 1. A k -Hermite history is a k -Dyck pathin which every up step u is labeled with an integer in { , , , . . . , ht( u ) − } . See for exampleFigure 7. Given a k -Dyck path µ , we denote by H ( k ) ( µ ) the set of k -Hermite histories on µ . The weight wt( H ) of a k -Hermite history is the sum of the labels in H . MATTHIEU JOSUAT-VERG`ES AND JANG SOO KIM
Figure 8.
An illustration of the bijection between k -Dyck tilings and k -Hermite histories.The following theorem is a generalization of (2). Our proof generalizes the bijective proof in[13]. Theorem 3.2.
For µ ∈ Dyck ( k ) ( n ) , we have (3) X D ∈D ( k ) ( ∗ /µ ) q tiles( D ) = Y u ∈ UP( µ ) [ht( u )] q . Proof.
It suffices to find a bijection f : D ( k ) ( ∗ /µ ) → H ( k ) ( µ ) such that if f ( D ) = H then tiles( D ) =wt( H ). We construct such a bijection as follows. Let D ∈ D ( k ) ( ∗ /µ ). In order to define thecorresponding f ( D ) = H ∈ H ( k ) ( µ ) we only need to define the labels of each up step in µ . For a k -Dyck tile η , the entry of η is the left side of the lowest cell in η and the exit of η is the right sideof the rightmost cell in η . For each up step u in µ , we travel the tiles in D in the following way. If u is the entry of a k -Dyck tile η in D , then enter η at the entry and leave η from the exit. If theexit of η is the entry of another k -Dyck tile of D then travel that tile as well. Continue travelingin this way until we do not reach the entry of any k -Dyck tile in D . Then the label of u is definedto be the number of tiles that we have traveled. See Figure 8 for an example. It is not difficult tosee f ( D ) ∈ H ( k ) ( µ ). Clearly, we have tiles( D ) = wt( H ).It remains to show that f is a bijection. Let H ∈ H ( k ) ( µ ). We will find the inverse image D = f − ( H ) recursively. Suppose that n is the length of µ and m is the number of cells between µ and the line y = x/k . If n = 0 or m = 0, then both D ( k ) ( ∗ /µ ) and H ( k ) ( µ ) have a unique element,and f − ( H ) is the unique element in D ( k ) ( ∗ /µ ) without tiles. Now let n, m ≥ f − ( H ′ ) for every H ′ ∈ H ( k ) ( µ ′ ) if the length of µ ′ is smaller than n or the numberof cells between µ ′ and the line y = x/k is smaller than m . There are two cases.Case 1: H has an up step with label ℓ ≥ H ′ bethe k -Hermite history obtained from H by exchanging the up step and the down step following itand decrease the label ℓ by 1. Then the shape µ ′ of H ′ has one less cells between µ ′ and the line y = x/k . By assumption we can find f − ( µ ′ ). Then f − ( µ ) is the k -Dyck tiling obtained from f − ( µ ′ ) by adding the single square µ \ µ ′ .Case 2: H has no up step with label ℓ ≥ µ is a k -Dyck path ofpositive length, we can find an up step u followed by k down steps. Since u is followed by a downstep, its label is 0. Let P be the point where u starts. Let µ ′ be the k -Dyck path obtained from µ by deleting u and the k down steps following u . By assumption, we can can find f − ( µ ′ ). Then f − ( µ ) is the k -Dyck tiling obtained from f − ( µ ′ ) by cutting it with the line of slope − P and inserting an up step followed by k down steps. For each k -Dyck tile divided by theline, we attach the two divided pieces by connecting the separated points on the border with anup step and k down steps following it. See Figure 9.This gives the inverse map of f . (cid:3) It seems unlikely that there is a hook length formula for D ( k ) ( λ/ ∗ ) when we fix the lower path λ to be arbitrary. If n = 6, k = 2, and λ is the following path, then we have |D ( k ) n ( λ/ ∗ ) | = 607, a ENERALIZED DYCK TILINGS 7
00 20 31 →
00 20 310
Figure 9.
An example of cutting a k -Dyck tiling and inserting an up step and k down steps.prime number.Also if |D ( k ) n ( λ/ ∗ ) | = 71 for the following λ with n = 5 , k = 2.However, when λ is a zigzag path there is a nice generalization of (1). Theorem 3.3.
Let λ be the path u n d kn u n d kn · · · u n ℓ d kn ℓ , where u means an up step and d means a down step. Then we have X D ∈D ( k ) ( λ/ ∗ ) q art k ( D ) = (cid:20) kn + n n (cid:21) q (cid:20) k ( n + n ) + n n (cid:21) q · · · (cid:20) k ( n + · · · + n ℓ − ) + n ℓ n ℓ (cid:21) q , where (cid:2) ab (cid:3) q = [ a ] q ![ b − a ] q ![ b ] q ! .Proof. This can be proved using the same idea in the inductive proof in [13]. We will omit thedetails. (cid:3)
Problem 1.
Find a bijective proof of Theorem 3.3.4. k -Stirling permutations and the k -Bruhat order In this section we consider k -Stirling permutations which were introduced by Gessel and Stanley[9] for k = 2 and studied further for general k by Park [15].A k -Stirling permutation of size n is a permutation of the multiset { k , k , . . . , n k } such that ifan integer j appears between two i ’s then i > j . Let S ( k ) n denote the set of k -Stirling permutationsof size n . We can represent a k -Stirling permutation π = π π . . . π kn as the n × kn matrix M = ( M i,j ) defined by M i,j = 1 if π j = i and M i,j = 0 otherwise. Then the entries of M are 0’sand 1’s such that each column contains exactly one 1, each row contains k (cid:18) (cid:19) . For example, see Figure 10.A k -inversion of π ∈ S ( k ) n is a pair ( i, j ) ∈ [ n ] × [ kn ] such that π j > i and the first i appearsafter π j . Equivalently, we will think of a k -inversion as an entry (or a cell) in the matrix of π which has k k -inversions of π by INV k ( π ), and inv k ( π ) = | INV k ( π ) | . MATTHIEU JOSUAT-VERG`ES AND JANG SOO KIM b b b b b b b b b b b b ×××× × × × × ××
Figure 10.
The matrix corresponding to 422243111334 ∈ S (3)4 , where ones arerepresented as dots. The k -inversions are the cells with crosses.332211332112 322311331122 322113 223311311322 321123 223113113322 311223 221133113223 211233112233 Figure 11.
The Hasse diagram of S (2)3 . Proposition 4.1.
We have X π ∈ S ( k ) n q inv k ( π ) = [ k + 1] q [2 k + 1] q · · · [( n − k + 1] q . Proof.
This is an easy induction. Note that all the 1’s in a Stirling permutation form a block ofconsecutive letters. Starting from σ ∈ S ( k ) n − , we build a Stirling permutation in S ( k ) n by increasingall letters of σ by 1, and inserting a block 1 k at some position. The positions where we can insertthis block give the factor [( n − k + 1] q . (cid:3) As we can see in the following lemma, a k -Stirling permutation is determined by its k -inversions. Lemma 4.2.
For σ, π ∈ S ( k ) n , if INV k ( σ ) = INV k ( π ) , we have σ = π .Proof. Suppose that σ = π . Let r be the smallest index satisfying σ r = π r . We can assume that σ r < π r . Let m = σ r . Note that π j = m for some j > r . Then π does not have the integer m in the first r positions because otherwise π i = m for i < r and we get π i < π r > π j which isforbidden. Then ( m, r ) ∈ INV k ( π ) but ( m, r ) INV k ( σ ), which is a contradiction. Thus we have σ = π . (cid:3) We are now ready to define the k -Bruhat order on k -Stirling permutations. Definition 4.3.
We define the k -Bruhat order on S ( k ) n given by the cover relation σ ⋖ π if π is obtained from σ by exchanging the two numbers in positions a and a k +1 for some integers a < a < · · · < a k +1 satisfying the following conditions:(1) σ a = σ a = · · · = σ a k < σ a k +1 , and(2) for all a k < i < a k +1 we have either σ i < σ a or σ i > σ a k +1 .In fact, if k ≥
2, then for all a < i < a k +1 with i = a j we always have σ i < σ a , see Lemma 4.4.Note that the 1-Bruhat order is the usual Bruhat order. Figure 11 illustrates the k -Bruhatorder. Remark . The elements of S ( k ) n where all occurrences of i are consecutive for any i , form a subsetwhich is in natural bijection with S n . In general ( k > n > ENERALIZED DYCK TILINGS 9
Lemma 4.4.
In this lemma we use the notation in Definition 4.3. If k ≥ , then we have σ i < σ a for all a < i < a k +1 with i = a j .Proof. By the definition of k -Stirling permutation, for all a < i < a k with i = a j we have σ i < σ k ,and for all a < i < a k +1 with i = a j we have π i < π k = σ k . Thus we have σ i = π i < σ k for all a < i < a k +1 with i = a j . (cid:3) Lemma 4.5.
Let σ ⋖ π in S ( k ) n . Then INV k ( π ) is obtained from INV k ( σ ) by adding one cell andmoving some cells (maybe none) to the west or north. Moreover, if a cell is moved to the west(respectively, north), then the new location of the cell is south (respectively, to east) of the newlyadded cell in the same column (respectively, row).Proof. Suppose that π is obtained from σ as described in Definition 4.3. Then INV k ( π ) is obtainedfrom INV k ( σ ) as follows. We add the inversion ( σ a , a ), and change each k -inversion of the form( r, a k +1 ) for some σ a k +1 < r < σ a to ( r, a ). Furthermore if k = 1, we change each k -inversion ofthe form ( σ a k +1 , j ) for some j < a k +1 to ( σ a , j ). (cid:3) We note that in Lemma 4.5, moving cells to north can happen only when k = 1.By Lemma 4.5 we have inv k ( π ) = inv k ( σ ) + 1 if σ ⋖ π . This proves: Proposition 4.6.
Endowed with the k -Bruhat order, S ( k ) n is a graded poset with rank function inv k . The following generalization of 132-avoiding permutations is straightforward and natural.
Definition 4.7. A k -Stirling permutation σ ∈ S ( k ) n is 132-avoiding if there is no 1 ≤ i < j < k ≤ kn such that σ i < σ k < σ j . Let S ( k ) n (132) denote the set of 132-avoiding k -Stirling permutationsin S ( k ) n .In a k -Dyck tilings without tiles, the lower path and upper path are the same. So these tilingsare trivially in bijection with k -Dyck paths. As for the k -Stirling permutations, we have: Proposition 4.8.
The inversions of a 132-avoiding k -Stirling permutation σ are arranged as thecells of a top-left justified Ferrers diagram. Define a path α ( σ ) from the bottom left to the top rightcorner, and following the boundary of this Ferrers diagram with up and right steps. Then α is abijection between 132-avoiding k -Stirling permutations and k -Dyck paths of length n , in particularboth are counted by the Fuss-Catalan numbers kn +1 (cid:0) ( k +1) nn (cid:1) . The proof is simple and similar to the case k = 1. For example, see Figure 12. Proposition 4.9.
Suppose that σ ∈ S ( k ) n is -avoiding. Then, for π ∈ S ( k ) n , we have σ ≤ π ifand only if INV k ( σ ) ⊆ INV k ( π ) .Proof. Since σ is 132-avoiding, INV( σ ) is a Ferrers diagram, say λ . By Lemma 4.5 if τ ⋖ ρ and λ ⊂ INV k ( τ ), then we also have λ ⊂ INV k ( ρ ). This implies the “only if” part.We will prove the “if” part by induction on the number m = inv k ( π ) − inv k ( σ ). If m = 0, it istrue. Suppose m >
0. Let ( i, j ) be an element in INV k ( π ) \ INV k ( σ ) with j as large as possible.Then there are integers a < a < · · · < a k such that π a = π a = · · · = π a k = i and a > j . Thenwe have π t < i for all j < t < a by the maximality of j . Let π ′ be obtained from π by exchangingthe two integers in positions j and a k . Then π ′ ⋖ π . By Lemma 4.5 INV k ( π ′ ) is obtained fromINV k ( π ) by removing the cell ( i, j ) and moving some cell located south or east of ( i, j ). ThusINV k ( π ′ ) still contains INV k ( σ ) which is a Ferrers diagram. By induction we have σ ≤ π ′ , whichcompletes the proof. (cid:3) Proposition 4.10.
Let σ ∈ S ( k ) n (132) , and µ = α ( σ ) the corresponding k -Dyck path. There is abijection between the interval { π : π ≥ σ } in S ( k ) n and k -Hermite histories of shape µ . It is suchthat inv k ( π ) − inv k ( σ ) is sent to the sum of weights in the Hermite history. b b b b b b b b ×× ×× × 7→ Figure 12.
The bijection α in Proposition 4.8. Proof.
Let π ∈ S ( k ) n (132) with π ≥ σ . Note that by Proposition 4.9, the inversions of σ areinversions of π . Then we define a k -Hermite history as follows. For i ≤ n , the label of the upstep in the i th row from the top is the number of × ’s in the matrix of σ that are in the i th row,and below µ . Since inv k ( σ ) is the number of × ’s above the path µ by definition of the bijectionin Figure 12, the number inv k ( π ) − inv k ( σ ) is the sum of weights in the Hermite history. So itremains only to show that these labels define a Hermite history (i.e. they fall in the right range)and that this is a bijection.Let us start with the first row. If the height of the up step in the first row is h , then there are h + k cells to the right of the up step in the first row. Since there are k consecutive dots in thefirst row in π , the × ’s are located in the cells after the up step and before the k consecutive dots.Thus the number of × ’s in the first row is among 0 , , , . . . , h . Now consider the second row. Ifthe height of the up step in the first row is h ′ , then there are h ′ + 2 k cells to the right of the upstep in the second row. By the condition for k -Stirling permtutation, if we delete the columnswhich have a dot in the first row, then the dots in the second row are consecutive, and the × ’sare located in the cells after the up step and before the k consecutive dots. Thus the number of × ’s in the second row is among 0 , , , . . . , h ′ . In this manner, we can see that the number of × ’sin each row is at most the height of the up step in the same row. This argument also shows thatonce the number of × ’s in each row is determined, we can uniquely construct the correspondingpermutation π by putting dots starting from the first row. This implies that the map is a desiredbijection. (cid:3) We now have a generalization of Proposition 2.3. It is a rewriting of Theorem 3.2 using thebijection from Proposition 4.10.
Theorem 4.11. If µ is a k -Dyck path corresponding to σ ∈ S ( k ) n (132) , then X D ∈D ( k ) n ( ∗ /µ ) q tiles( D ) = X π ≥ σ q inv k ( π ) − inv k ( σ ) . k -regular noncrossing partitions In this section, we take another point of view on the k -Stirling permutations studied in theprevious section. Definition 5.1.
We denote by NC ( k ) n the set of k - regular noncrossing partitions of size n , i.e. setpartitions of [ kn ] such that each block contains k elements ( k -regular), and there are no integers a < b < c < d such that a , c are in one block, and b , d in another block (noncrossing). Toeach k -regular noncrossing partition π of [ kn ], we define its nesting poset Nest( π ) as follows: theelements of the poset are the blocks of π , and x ≤ y in the poset when the block x lies betweentwo elements of the block y .There is a natural way to consider a k -Stirling permutation as a linear extension of the nestingposet of a k -regular noncrossing partition. Proposition 5.2.
There is a bijection between S ( k ) n and pairs ( π, E ) where π ∈ NC ( k ) n and E isa linear extension of the poset Nest( π ) .Proof. Let σ ∈ S ( k ) n . We define π by saying that i and j are in the same block if σ i = σ j , and E isdefined by saying that the label of a block of π is σ i where i is any element of the block. We can seethat π is noncrossing from the definition of Stirling permutations, and E is a linear extension by ENERALIZED DYCK TILINGS 11 b b b b b b b b b b b b
12 34
Figure 13.
The bijection proving Proposition 5.2.definition of the nesting poset. In the exemple σ = 422243111334, we get the noncrossing partitionin Figure 13 where the labels define the linear extension of the nesting poset. The inverse bijectionis simple to describe: σ i is equal to j if i is in the block with label j . (cid:3) The poset Nest( π ) is always a forest, so it is possible to consider pairs ( π, E ) as a decreasingforest. Each block of π is a vertex of the forest, and the forest structure is the order Nest( π ).Also, the labelling E naturally gives the decreasing labelling of the forest.Let b be a block of π , the elements of b begin i < · · · < i k . Then the descendants of b in theforest can be distinguished into k − j such that a descendantof b lies between i j and i j +1 .So we arrive at the following definition. Definition 5.3.
Let T ( k − n denote the set of ( k − n vertices defined bythe following conditions: • the descendants of a vertex have a structure of a ( k − • the vertices are labeled with integers from 1 to n and the labels are decreasing from theroots to the leaves.As a result of the above discussion, we obtain that there is a bijection between S ( k ) n and T ( k − n .However we do not insist on this point of view, since the definition of the trees in T ( k − n is notparticularly natural.There is a hook length formula for the number of linear extensions of a forest [3]. If x ∈ Nest( π ),let h x denote the number of elements below x in the nesting poset, then the number of linearextensions of π ∈ NC ( k ) n is n ! Q x ∈ Nest( π ) h x . This gives the following formula for the number of k -Stirling permutations. Proposition 5.4 (Multifactorial hook length formula) . k + 1)(2 k + 1) . . . (( n − k + 1) = X π ∈ NC ( k ) n n ! Q x ∈ Nest( π ) h x . In the case k >
2, it is not clear how to follow the q -statistic, but for k = 2 we can use thebijection between noncrossing matchings and plane forests (which is just π Nest( π )), and usethe q -hook length formula from [2]. We get a hook length formula for [1] q [3] q . . . [2 n − q . Proposition 5.5 ( q -double factorial hook length formula) . We have [1] q [3] q . . . [2 n − q = X F [ n ] q ! Y v ∈ F q h v − [ h v ] q , where the sum is over all plane forests F with n vertices.Proof. We know that the left hand side is the inversion generating function of S (2) n . It remains tounderstand what becomes the number of inversion through the bijection which send a 2-Stirlingpermutation to a noncrossing matching with labeled blocks, or equivalently increasing plane forests.To this end, we distinguish two kinds of inversions. Let i < j and σ ∈ S (2) n . If the four letters i, i, j, j appear in σ in this order, there is no inversion. If they appear in the order j, i, i, j , there isone inversion, and this corresponds to the case where the vertex with label i is below the vertex b b bb bbb bbbbbbb bb bbbb ∗
149 8515 ∗ b bb bb b bbbbbb ∗ ∗ ∗ Figure 14.
A symmetric plane tree of size 18 and a marked increasing labelingof it. The center line is the dashed line.
Figure 15.
An example of symmetric Dyck tiling. It is symmetric with respectto the dashed line.with label j . It means we have to count one inversion for each pair of comparable vertices, andthe number of comparable vertices is clearly P x ∈ π ( h x − j, j, i, i , there are two inversions, and this situationcorresponds to the case where the vertices with labels i, j form an inversion in the forest.So for a particular forest F , we get the term[ n ] q ! Y v ∈ F q h v − [ h v ] q . This completes the proof. (cid:3) Symmetric Dyck tilings and marked increasing forests A symmetric plane forest is a plane forest which is invariant under the reflection about a line,called the center line . A center vertex is a vertex on the center line. The left part of a symmetricplane forest is the subforest consisting of vertices on or to the left of the center line. The size ofa symmetric plane forest is the number of vertices in the left part of it.Let F be a symmetric plane forest of size n . A marked increasing labeling of F is a labelingof the left part of F with [ n ] such that the labels are increasing from roots to leaves, each integerappears exactly once, and each non-center vertex may be marked with ∗ . See Figure 14 for anexample of a marked increasing labeling. We denote the set of marked increasing labelings of F by INC ∗ ( F ).Let L ∈ INC ∗ ( F ). We denote by MARK( L ) the set of labels of the marked vertices in L . An inversion of L is a pair of vertices ( u, v ) such that L ( u ) > L ( v ), u and v are incomparable, and u is to the left of v . Let INV( L ) denote the set of inversions of L .A Dyck path λ of length 2 n is called symmetric if it is invariant under the reflection along theline x + y = n . For two symmetric Dyck paths λ and µ of length 2 n , a Dyck tiling of λ/µ is called symmetric if it is invariant under the reflection along the line x + y = n . See Figure 15 for anexample of symmetric Dyck tiling. We denote by D sym ( λ/µ ) the set of symmetric Dyck tilings ofshape λ/µ .For a symmetric Dyck tiling D , a positive tile is a tile which lies strictly to the left of the centerline, and a zero tile is a tile which intersects with the center line. We denote by tiles + ( D ) and ENERALIZED DYCK TILINGS 13 b b ∗ ↔ b bb ∗ ↔ b bbb b ∗ ∗ ↔ b bbb bb b ∗ ∗ ↔ b bbb bb b bb ∗ ∗ ↔ Figure 16.
An illustration of the bijection between marked increasing labelingof a symmetric forest and symmetric Dyck tilings with fixed lower path. Thenewly added squares in each step are colored gray.tiles ( D ) the number of positive tiles and zero tiles, respectively. Note that the total number oftiles in D is tiles( D ) = 2 · tiles + ( D ) + tiles ( D ). We also define area + ( D ) and area ( D ) to be thetotal area of positive tiles and zero tiles in D , respectively, andart + ( D ) = area + ( D ) + tiles + ( D )2 , art ( D ) = area ( D ) + tiles ( D )2 . For example, if D is the symmetric Dyck tiling in Figure 15, we have tiles + ( D ) = 5, tiles ( D ) = 3,area + ( D ) = 7, area ( D ) = 19, art + ( D ) = (7 + 5) / ( D ) = (19 + 3) / Theorem 6.1.
Let F be a symmetric plane forest of size n and λ the corresponding Dyck path.Then there is a bijection φ : INC ∗ ( F ) → D sym ( λ/ ∗ ) such that if φ ( L ) = D , then tiles ( D ) = | MARK( L ) | and art + ( D ) + art ( D ) = | INV( L ) | + X i ∈ MARK( L ) ( n + 1 − i ) . Proof.
This is a generalization of a bijection in [14] constructed recursively. We “spread” and add“broken strips” to all up steps before the center line and to all down steps after the center line. Ifa vertex is marked, then we also add a square at the center line. See Figure 16. (cid:3)
Corollary 6.2.
Let F be a symmetric plane forest of size n with k center vertices and λ thecorresponding Dyck path. Then |D sym ( λ/ ∗ ) | = 2 n − k n ! Q x ∈ F h x . If k = 0 , we have X D ∈D sym ( λ/ ∗ ) q art + ( D )+art ( D ) t tiles ( D ) = (1 + tq )(1 + tq ) · · · (1 + tq n ) [ n ] q ! Q x ∈ F [ h x ] q . Figure 17.
An example of symmetric Hermite history. The labels of the un-matched up steps are colored blue. The sequence (0 , , , , , , ,
0) of the labelsof unmatched up steps is an involutive sequence.7.
Symmetric Dyck tilings and symmetric Hermite histories
For a symmetric Dyck path µ of length 2 n , let µ + denote the subpath consisting of the first n steps. Note that each up step is matched with a unique down step in a Dyck path. An up step of µ + is called matched if the corresponding down step in µ lies in µ + and unmatched otherwise.A symmetric Hermite history is a symmetric Dyck path µ with a labeling of the up steps of µ + in such a way that every matched up step of height h has label i ∈ { , , . . . , h − } and the labels a , a , . . . , a k of the unmatched up steps form an involutive sequence. Here, a sequence is called involutive if it can be obtained by the following inductive way. • The empty sequence is defined to be involutive. • The sequence (0) is the only involutive sequence of length 1. • For k ≥
2, an involutive sequence of length k is either the sequence obtained from aninvolutive sequence of length k − k − r at the end and inserting a 0before the last r integers, including the newly added integer, for some 1 ≤ r ≤ k − v k of involutive sequences of length k satisfies therecurrence v k = v k − + ( k − v k − with initial conditions v = v = 1. Thus v k is equal to thenumber of involutions in S n . We denote by H sym ( µ ) be the set of symmetric Hermite histories on µ . For H ∈ H sym ( µ ), let k H k be the sum of labels in H and pos( H ) the number of positive labelson unmatched up steps. Proposition 7.1.
There is a bijection ψ : H sym ( µ ) → D sym ( ∗ /µ ) such that if ψ ( H ) = D , then k H k = tiles + ( D ) + tiles ( D ) and pos( H ) = tiles ( D ) . Thus, X D ∈D sym ( ∗ /µ ) q tiles + ( D )+tiles ( D ) t tiles ( D ) = X H ∈H sym ( µ ) q k H k t pos( H ) . Proof.
Given a symmetric Dyck tiling D , considering D as a normal Dyck tiling, we can obtainthe Hermite history corresponding to D . By taking only the labels of up steps before the centerline, we get a symmetric Hermite history. One can check that this gives a desired bijection. (cid:3) Corollary 7.2.
Let µ be a symmetric Dyck path such that µ + has k unmatched up steps. Then X D ∈D sym ( ∗ /µ ) q tiles + ( D )+tiles ( D ) t tiles ( D ) = f k ( q, t ) Y u ∈ UP( µ + ) [ht( u )] q , where UP( µ + ) is the set of matched up steps in µ + and f k ( q, t ) is defined by f ( q, t ) = f ( q, t ) = 1 and f k ( q, t ) = f k − ( q, t ) + tq [ k − q f k − ( q, t ) for k ≥ . A symmetric matching is a matching on [ ± n ] such that if { i, j } is an arc, then {− i, − j } is alsoan arc. We denote by M sym ( n ) the set of symmetric matchings on [ ± n ]. Note that symmetricmatchings are in bijection with fixed-point-free involutions in B n .Let M ∈ M sym ( n ). A symmetric crossing of M is a pair of arcs { a, b } and { c, d } satisfying a < c < b < d and b, d >
0. A symmetric crossing ( { a, b } , { c, d } ) is called self-symmetric if ENERALIZED DYCK TILINGS 15 { c, d } = {− a, − b } . We denote by cr( M ) and sscr( M ) the number of symmetric crossings andself-symmetric crossings of M , respectively.For a symmetric Dyck path µ of length 2 n , let M sym ( µ ) denote the set of symmetric matchings M on [ ± n ] such that the i th smallest vertex of M is a left vertex of an arc if and only if the i thstep of µ is an up step. Proposition 7.3.
We have X H ∈H sym ( µ ) q k H k t pos( H ) = X M ∈M sym ( µ ) q cr( M ) t sscr( M ) . For D ∈ D sym ( ∗ /µ ), let ht( D ) denote the number of unmatched up steps in µ + . Using the resultin [4] and [10] on a generating function for partial matchings we obtain the following formula. Proposition 7.4.
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