Equidistributed statistics on Fishburn matrices and permutations

Recently, Jel\'inek conjectured that there exists a bijection between certain restricted permutations and Fishburn matrices such that the bijection verifies the equidistribution of several statistics. The main objective of this paper is to establish such a bijection.


Introduction
Given a sequence of integers x = x 1 x 2 · · · x n , we say that the sequence x has an ascent at position i if x i < x i+1 . Let ASC(x) denote the set of the ascent positions of x and let asc(x) denote the number of ascent of x. A sequence x = x 1 x 2 · · · x n is said to be an ascent sequence of length n if it satisfies x 1 = 0 and 0 ≤ x i ≤ asc(x 1 x 2 · · · x i−1 ) + 1 for all 2 ≤ i ≤ n. Let A n be the set of ascent sequences of length n. For example, by the Fishburn number F n (sequence A022493 in OEIS [10] ) for memory of Fishburn's pioneering work on the interval orders [4,5,6]. More examples of Fishburn objects are constantly being discovered. Levande [7] introduced the notion of Fishburn diagrams and proved that Fishburn diagrams are counted by Fishburn numbers, confirming a conjecture posed by Claesson and Linusson [2]. Jelínek [8] showed that some Fishburn triples are enumerated by Fishburn numbers.
Zagier [14] and Bousquet-Mélou et al. [1] obtained the generating function of F n , that is Kitaev and Remmel [9] extended the work and found the generating function for (2 + 2)-free posets when four statistics are taken into account. Levande [7] and Yan [13] independently presented a combinatorial proof of a conjecture of Kitaev and Remmel [9] concerning the generating function for the number of (2 + 2)-free posets.
Let us recall the notions of pattern avoiding permutations and Fishburn matrices before we state our main results. Let S n be the symmetric group on n elements and π = π 1 π 2 · · · π n be a permutation of S n . We say that π contains the pattern if there is a subsequence π i π i+1 π j of π satisfying that π i + 1 = π j < π i+1 , otherwise we say that π avoids the pattern . For example, the permutation 42513 contains the pattern while the permutation 52314 avoids it.
The pattern can be defined similarly. Let S n ( ) be the set of ( )-avoiding permutations of [n] and S n ( ) be the set of ( )-avoiding permutations of [n], respectively. These two sets are both enumerated by Fishburn numbers [1,11]. In a permutation π, we say π i is a left-to-right maximum (or LR-maximum) if π i is larger than any element among π 1 , π 2 , . . . , π i−1 . Let LRMAX(π) denote the set of LR-maxima of π and let LRmax(π) denote the number of LR-maxima of π. Analogously, we can define LR-minima, RL-maxima, RL-minima of a permutation π. Denote by LRMIN(π), RLMAX(π) and RLMIN(π) the set of LR-minima, RL-maxima and RL-minima of π, their cardinalities being denoted by LRmin(π), RLmax(π) and RLmin(π), respectively. Fishburn matrices were introduced by Fishburn [6] to represent interval orders. A Fishburn matrix is an upper triangular matrix with nonnegative integers whose every row and every column contain at least one non-zero entry. The weight of a matrix is the sum of its entries. Similarly, the weight of a row (or a column) of a matrix is the sum of the entries in this row (or column). Denote by M n the set of Fishburn matrices of weight n. For example, Given a matrix A, we use the term cell (i, j) of A to refer to the the entry in the i-th row and j-th column of A, and we let A i,j denote its value. We assume that the rows of a matrix are numbered from top to bottom and the columns of a matrix are numbered from left to right in which the topmost row is numbered by 1 and the leftmost column is numbered by 1. A cell (i, j) of a matrix A is said to be zero if A i,j = 0. Otherwise, it is said to be nonzero. A row ( or column) is said be zero if it contains no nonzero cells. Otherwise, it is said to be nonzero row ( or column).
A cell (i, j) of a matrix A is a weakly north-east cell (or wNE-cell) if it is a nonzero cell and any other cell weakly north-east form c is a zero cell. More precisely, a cell (i, j) of a matrix A is a wNE-cell if A s,t = 0 for all s ≤ i and t ≥ j.
By using generating functions, Jelínek [8] proved the following symmetric joint distribution on M n . Jelínek [8] also posed the following weaker conjecture which can be followed directly from Theorem 1.1 and Conjecture 1.1. The main objective of this paper is to establish a bijection between S n ( ) and M n which satisfies the former four items of Conjecture 1.1, thereby confirming Conjecture 1.2.

Bijection between permutations and ascent sequences
In this section, we shall construct a bijection θ between S n ( ) and A n , and show that the map θ proves the equidistribution of two 4-tuples of statistics.
Let π be a permutation in S n ( ) and let τ be the permutation obtained by deleting n from π. Then we have that τ is also a permutation in S n ( ). If not, we assume that τ i τ i+1 τ j is a pattern in τ . Since π is ( )-avoiding, we have π i+1 = n. Then π i π i+1 π j+1 forms a pattern in π, a contradiction. This property allows us to construct the permutation of S n ( ) inductively, starting from the empty permutation and adding a new maximal value at each step.
Let τ be a permutation in S n−1 ( ). The positions where we can insert the element n into τ to obtain a -avoiding permutation are called active sites. The site after the maximal entry n in π is always an active site. We label the active sites in π from right to left with 0, 1, 2 and so on.
The bijection θ between S n ( ) and A n can be defined recursively. Set θ(1) = 0. Suppose that π is a permutation in S n ( ) which is obtained from τ by inserting the element n into the x n -th active site of τ . Then we set θ(π) = x 1 x 2 · · · x n−1 x n , where θ(τ ) = x 1 x 2 . . . x n−1 .

Example 2.1
The permutation 85231647 corresponds to the sequence 01102103 since it is obtained by the following insertion, where the subscripts indicate the labels of the active sites.
Lemma 2.1 Let π = π 1 π 2 · · · π n be a permutation in S n ( ) and θ(π) = x = x 1 x 2 · · · x n . Then we have that s(π) = 2 + asc(x) and a(π) = x n , where s(π) denotes the number of active sites of π and a(π) denotes the label of the site located just after the entry n of π.
Proof. Suppose that π is obtained from τ by inserting the element n into the x n -th active site of τ . Then we have For any entry i which is to the right of n, i is followed by an active site in π if and only if i is followed by an active site in τ . Since the site after n in π is always active, we obtain a(π) = x n Now let us focus on the equation s(π) = 2 + asc(x). We will prove it by induction on n. It obviously hold for n = 1. Assume that it holds for n − 1. For any entry i < n − 1, i is followed by an active site in π if and only if i is followed by an active site in τ . The site after n in π is always an active site. Thus, to determine s(π), the only question is whether the site after n − 1 is active. We need consider two cases.
Case 1: If 0 ≤ x n ≤ a(τ ) = x n−1 , then the entry n in π is to the right of n − 1. It follows that the site after n−1 is not an active cite in π. Since the site after n−1 is an active cite in τ , we have that s(π) = s(τ ). By the induction hypothesis, s(τ ) = 2+asc(x ′ ) = 2+asc(x). Hence we deduce that s(π) = 2 + asc(x).

Theorem 2.2
The map θ is a bijection between S n ( ) and A n .
Proof. We prove this conclusion by induction on n. It obviously holds for n = 1. Assume that θ is a bijection between S n−1 ( ) and A n−1 .
We first show that θ is a map from S n ( ) to A n . Let π = π 1 π 2 · · · π n be a permutation in S n ( ) which is obtained from τ by inserting a maximal entry n in the active site labeled by x n in τ . Then To prove that x ∈ A n , it suffices to show that x n ≤ asc(x ′ ) + 1. Recall that the rightmost active site is labeled 0. Hence the leftmost active site in τ is labeled s(τ ) − 1. By the recursive description of the map θ, we have that x n ≤ s(τ ) − 1. From Lemma 2.1 we see that s(τ ) = 2 + asc(x ′ ). Thus we have x n ≤ asc(x ′ ) + 1. Since x encodes the construction of π, θ is an injective map from S n ( ) to A n .
It remains to show that θ is surjection. Let y = y 1 y 2 · · · y n be an ascent sequence and p = p 1 p 2 · · · p n−1 = θ −1 (y ′ ), where y ′ = y 1 y 2 · · · y n−1 . From the definition of ascent sequence and Lemma 2.1, we have that y n ≤ asc(y ′ ) + 1 = s(p) − 1. Let q be the permutation obtained from p by inserting the maximal entry n into the active site labeled y n in p. By the construction of the map θ, it can be easily seen that θ(q) = y. This concludes the proof. Let x = x 1 x 2 · · · x n be an ascent sequence in A n . The modified ascent sequence of x, denoted byx, is defined by the following procedure: Modified ascent sequence were introduced by Bousquet-Mélou et al., see more details in [1].
For an ascent sequence x = x 1 x 2 · · · x n , let zero(x) denote the number of zeros in x and let max(x) denote the number of elements x i satisfying x i = asc(x 1 x 2 · · · x i−1 ) + 1.
If x n > x n−1 , then n is to the left of n − 1 in π. In this case, τ i is a LR-maximum in π if and only if τ i is a LR-maximum in τ and l(τ i ) ≥ x n . After the inserting n into τ , l(τ i ) is increased by 1 if τ i is also a LR-maximum in π. Hence we have that where the last equality follows from the fact that This completes the proof.
Combining Theorems 2.2 and 2.3, we are led to the following result.

Bijection between ascent sequences and Fishburn matrices
The main objective of this section is to establish a bijection φ between A n and M n . To this end, we will define a removal operation and an addition operation on the matrices of M n .  The following lemma shows that the removal operation on a Fishburn matrix of M n will yield a Fishburn matrix in M n−1 . Proof. It is easily seen that for any removal operation applied on the matrix A, the weight of f (A) is one less than the weight of A. It is trivial to check that there exists no zero columns or rows in f (A). Moreover, the removal operation also preserves the property of being upper-triangular. Thus, f (A) ∈ M n−1 . This completes the proof. Lemma 3.1 tells us that for any A ∈ M n , after n applications of the removal operation f to A, we will get a sequence of Fishburn matrices, say

Given a matrix
We now define an addition operation g on a Fishburn matrix which is shown to be the inverse of the removal operation later. Given a matrix A ∈ M n and i ∈ [0, dim(A)], We construct a matrix g(A, i) in the following manner.  By similar arguments as in the proof of Lemma 3.1, one can easily verify that the addition operation will also yield a Fishburn matrix. We now define a map φ from A n to M n recursively as follows. Given an ascent sequence x = x 1 x 2 . . . , x n , we define A (1) = (1) and Next we aim to show that the map φ is well defined and has the following desired properties.
Proof. We will prove by induction on n. It is trivial to check that the statement holds for n = 1. Assume that it also holds for n − 1, that is, From the construction of the addition operation, one can easily verify that index(φ(x)) = x n + 1 and The result follows.
Denote by tr(A) the number of nonzero cells belonging to the main diagonal of A. Lemma 3.4 For any x = x 1 x 2 · · · x n ∈ A n and A ∈ M n with A = φ(x), we have the following relations.
(1) zero(x) = rsum 1 (A); (2) max(x) = tr(A);  Proof. Point (5) follows directly from point (3). Similarly, point (6) is an immediate consequence of the proof of point (4) with q = 1. Now we verify points (1)-(4) by induction on n. Clearly, the statement holds for n = 1. Assume that it also holds for any some n − 1 with n ≥ 2. Let x ′ = x 1 x 2 · · · x n−1 and B = φ(x ′ ). Recall that A = g(B, x n ). From the definition of the addition operation g and the induction hypothesis, it is not difficult to verify that and For point (3), from the construction of the addition operation g, we see that the cell (x n + 1, dim(A)) is always a wNE cell. Moreover, there is a wNE-cell in row i of A if and only if there is a wNE-cell in row i of B and i < x n + 1. This yields that For point (4), we have two cases.
If x n > x n−1 = index(B) − 1, then either rule (Add2) or rule (Add3) applies. It is not difficult to verify that where the last equality follows from the fact that This completes the proof. Lemma 3.5 For any x = x 1 x 2 . . . x n ∈ A n , we have ψ(φ(x)) = x.
Proof. Suppose that we get a sequence of matrices A (1) , A (2) , . . . , A (n) when we apply the map φ to x, where A (1) = (1) and A (k) = g(A (k−1) , x k ) for all 1 < k ≤ n. Similarly, suppose that when we apply the map ψ to φ(x), we get a sequence y = y 1 y 2 . . . y n and a sequence of matrices B (1) , B (2) , . . . , B (n) , where B (n) = φ(x), B (k) = f (B (k+1) ) for all 1 ≤ k < n, and y k = index(B (k) ) − 1. Lemma 3.3 ensures that index(A (k) ) = x k + 1. In order to prove x = y, it suffices to show that A (k) = B (k) for all 1 ≤ k ≤ n. We proceed to prove this assertion by induction on n. Clearly, we have B (n) = φ(x) = A (n) . Assume that we have A (j) = B (j) for all j ≥ k + 1. In the following we aim to show that A (k) = B (k) . By the induction hypothesis, it suffices to show that f (A (k+1) ) = A (k) . We have three cases.
If index(A (k) ) ≤ x i+1 < dim(A (k) ), then rule (Add3) applies and A (k+1) is obtained from A (k) in the following way. First we insert a new (empty) row between rows x i+1 and x i+1 + 1, and insert a new (empty) column between columns x i+1 and x i+1 + 1. Let the new row be filled with all zeros except for the rightmost cell which is filled with a 1. Denote by A ′ the resulting matrix. Let T be the set of indices j such that j ≥ x i+1 + 1 and column j contains at least one nonzero cell above row x i+1 + 1. Suppose that T = {c 1 , c 2 , . . . , c ℓ } with c 1 < c 2 < . . . < c ℓ . Let c 0 = x i+1 + 1. For all 1 ≤ a ≤ x i+1 and 1 ≤ b ≤ ℓ, move all the entries in the cell (a, c b ) to the cell (a, c b−1 ), and fill all the cells in column dim(A ′ ) and above row x i+1 + 1 with zeros. It is easy to check that dim(A (k+1) ) = dim(A (k) ) + 1, index(A (k+1) ) = x i+1 + 1 and rsum x i+1 +1 (A (k+1) ) = 1. So rule (Rem3) applies and f (A (k+1) ) is obtained from A (k+1) by the following procedure. Let S be the set of indices j such that j ≥ x i+1 + 1 and column j contains at least one nonzero entry above row x i+1 +1. It is not difficult to check that S = {c 0 , c 1 , c 2 , . . . , c ℓ−1 }. Let c ℓ = dim(A (k+1) ). For all 1 ≤ a < x i+1 − 1 and 1 ≤ b ≤ ℓ − 1, move all the entries in the cell (a, c b ) to the cell (a, c b+1 ). Simultaneously delete row x i+1 + 1 and column x i+1 + 1. These operations simply reverse the construction of A (k+1) from A (k) , and therefore f (A (k+1) ) = A (k) . This completes the proof.
Theorem 3.6 The map φ is a bijection between A n and M n . Moreover, for any x ∈ A n and A ∈ M n with φ(x) = A, we have (zero, max, Rmin)x = (rsum 1 , tr, ne)A