Xk−1
i=0
X
T0∈BBn−1,k
pnin(T0)+k−1−iqinv(T0)+i
= X
T0∈BBn−1,k−1
pnin(T0)qinv(T0)
+ [k]p,q X
T0∈BBn−1,k
pnin(T0)qinv(T0).
3.3 p, q-statistics on L-Bell Tableaux
So far we have Corteel’s bijection f between set partitions and L-Bell tableaux, and a natural bijection g between set partitions and L0-Bell tableaux (see Chapter 3.1 and 3.2). We use these two bijections to find our p, q-statistics on L-Bell Tableaux.
Let us review the p, q-statistics on L-Bell Tableaux. Let LBn,k be the set of L-Bell tableaux of length n with n − k nonzero rows. For T ∈ LBn,k, let a star row
be a row that has a leftmost 1 but no topmost 1’s. Set nin(T ) as the number of pairs x and y, where x is a leftmost 1, and y is an unrestricted 0 that is not in a star row and above x. Let inv(T ) be the cardinality of A ∪ B, where A is the set of cells that is not in a star row and below a topmost 1, and B is the set of pairs u and v in the same column, where u is a cell in A and v is a leftmost 1 below u.
For example, the set partition π = 1, 5, 8, 10/2, 3, 6/4, 9/7 and its corresponding L0-Bell and L-Bell tableaux are shown in Figure 12. The left L0-Bell tableau T0 has nin(T0) = 4 and inv(T0) = 10. For the right L-Bell tableau T , its star rows are 3, 5,
Figure 12: The corresponding L0-Bell and L-Bell tableaux.
proof of Theorem 5(4). We prove this result by using set partitions as the interme-diate step. Let π = B1/B2/ · · · /Bk be a set partition in standard form, and T and T0 be its corresponding L-Bell and L0-Bell tableaux, respectively.
We consider the L0-Bell tableau T0 together with π first. From the bijection g (mapping each column to the row where the topmost 1 belongs), we know that an opener in π represents a row in T0 and an element which is not an opener represents a column in T0. That is, in each block Bt, there are |Bt| − 1 column labels. For a
topmost 1 in cell (i, j) and a cell (i0, j) above it, we have i0 is an opener in π and i0 label j, the number of cells below the topmost 1 cell (i, j) is the number of openers to its right which is smaller than j; that is, rosj(π). Thus
inv(T0) =X
Then we consider the L-Bell tableau T together with π. From the bijection f (mapping each row to the column where the leftmost 1 belongs), we know that the closer of a non-single block in π represents a column in T . An element which is not the closer of a non-single block represents a row in T , so does an element in a single block. The star rows are the elements which are neither openers nor closers of a block. For a pair x and y in the statistic nin(T ), say x = (i, j) and y = (i0, j), let Bt be the block in π containing both i and j, and let Bs be the block containing i0. Since y = (i0, j) is not in a star row, i0 is an opener of Bs. If s > t, then i0 is bigger than the opener i00 of Bt. Thus (i00, j) is a leftmost 1 above y = (i0, j), which contradicts to the assumption that y = (i0, j) is an unrestricted 0. Hence s < t. In other words, the number of y’s which is paired with x is the number of the openers to the left of the non-closer i; that is losi(π) = t − 1. Furthermore, there are |Bt| − 1 non-closers in each block Bt. Hence,
nin(T ) =X
of π which contains i as its opener is to the right of the block containing i0 and j as its opener and closer. For a column label j, the number of cells in A belong it is the number of openers to its right which is smaller than j; that is, rosj(π) where j is a closer of a non-single block. Now we consider a pair u and v in B. Since u is in A, let u be (i, j) and (i0, j) be the topmost 1 above it. Let v be a leftmost 1 below u, say (i00, j). Note that, i0 < i < i00 < j. Thus we have the block containing i as its opener is to the right of the block containing i0, i00, and j, where i0 and j are the opener and closer, respectively , and i00 is a star row. Thus the number of u’s that is paired with v equals to the number of opener i’s which is to the right of the star row i00 and smaller than it; that is, rosi00(π) where i00 is neither an opener nor a closer of a block. Together with the result we get from A,
inv(T ) =X
i is not an opener in π
rosi(π) = X
i∈[n]
rosi(π) = ros(π).
Note that rosi(π) = 0 whenever i is an opener in π.
4 Cardinality of intersections of various subclasses of tableaux
Now we have six subclasses of tableaux counted by the Bell numbers and we can investigate the cardinality of intersections of some of them in this chapter. The results we will prove in this thesis are listed in the table below.
4.1 The Catalan numbers and the Narayana numbers
A Catalan path (or Dyck path) of length 2n is a lattice paths from (0, 0) to (2n, 0) with steps (1, 1) and (1, −1) that never goes below the x-axis.
The Catalan number Cn is the number of the Catalan paths of length 2n. For example, C3 = 5. Figure 13 lists these 5 Catalan paths. It is known that
{Cn}n≥0 := {1, 1, 2, 5, 14, 42, 132, 429, . . . }.
Intersection of The cardinality / The distribution tableaux (w.r.t. some certain statistic) Ln∩ P Tn,
Catalan numbers Cn / Narayana numbers Ln∩ L0Bn,
Ln∩ R0Bn LBn∩ RBn,
Bessel numbers Besn / Number of blocks
*L0Bn∩ R0Bn Ln∩ LBn,
Motzkin numbers Mn/ Number of horizontal steps Ln∩ RBn,
RBn∩ L0Bn, LBn∩ R0Bn BBn∩ LBn,
2n−1 / Binomial numbers BBn∩ RBn,
BBn∩ Ln, BBn∩ L0Bn, BBn∩ R0Bn
LBn∩ L0Bn, Number of matchings in Kn RBn∩ R0Bn / Number of matchings of size k in Kn
Table 4: The intersection of two of these six subclasses
Figure 13: Five Catalan paths of length 6.
Let U and D represent the steps (1, 1) and (1, −1) respectively. We encode a Catalan path with a word over U, D. For example, those 5 paths in Figure 13 are represented by UUUDDD, UUDDUD, UUDUDD, UDUUDD, UDUDUD, respectively.
If δ is a non-empty Catalan path, then δ can be decomposed as δ = αUβD, where α and β are Catalan paths. The Narayana number N(n, k) is the number of the Catalan paths of length 2n with k valleys. By a valley in a path we mean a pair of consecutive DU steps.
Theorem 9. The cardinality of the set of permutation tableaux in Ln is the n-th Catalan numbers.
We give two proofs of this result.
proof I. We give a bijection from the set of permutation tableaux in Ln to the set of rooted plane tree with n edges.
For a permutation tableau where each column contains exactly one 1, say T , we construct a rooted plane tree by the labels of the shape of T as follows: From label 1 to n, a row i represents a leftmost child and a column j represents other children having the common parent with row k where the cell (k, j) is 1. See Figure 14 for an example.
Figure 14: A bijection between permutation tableaux in Ln and rooted plane tree.
For a rooted plane tree with n edges, we traverse it in pre-order and label its edges. Then for a parent, the label of the leftmost child represents a row and the labels of other children represent a column which has a 1 in the row. We need to prove that the resulting tableau T is a permutation tableau where each column contains exactly one 1. From above we know that each column contains exactly one 1 in the row which is the leftmost child of their common parent. Moreover, suppose that there exist i < j < k < l such that cells (i, k) and (j, l) are 1’s and cell (j, k) is a 0. In the corresponding rooted plane tree, i and k have common parent and j and l have common parent, but the parents of i and j are distinct. Thus we have that i < k < j < l if i is not an ancestor of j; otherwise, i < j < l < k. Both of them lead to a contradiction. Hence, the resulting tableau is a permutation tableau in Ln.
proof II. We give another bijection between the set of permutation tableaux in Ln and the set of lattice paths of length 2n. We call a cell is legal in a permutation tableau in Ln if there is no restricted 0 to the right side of it. For a permutation
tableau that each column contains exactly one 1, say T , we recurrently decompose T to construct a lattice path as UD with the following rules:
1. We map label 1 to a consecutive UD.
2. If the next label in tableau T is a row, then we insert a consecutive UD after the rightmost U; if it is a column, and its 1 is located in the ith legal cell, then we insert a consecutive UD behind the ith D counting from the right.
0 1
0 1 0 0
1 2 3 4 5
U U U DDU DU DD 1 2 3 4 5
Figure 15: A bijection between permutation tableaux in Ln and lattice path.
The inverse mapping from lattice path of length 2n to the set of permutation tableaux T in Ln is simple. Besides, in a lattice path we call the (k + 1)-th level of a U if U between the (x, k) and (x + 1, k + 1), for some x, k ∈ N ∪ {0}. For example, there is a Catalan path UUDUDD, the level of the third U is 2. Let the shape of T be defined by the U’s in Catalan path. The k-th U represents a column label k in T if there is a D in front of it; otherwise, it is a row in T . If U is a column, and has level l in the Catalan path, then there is a 1 in the l-th legal cell in column U. See Figure 15 for an example. By the property of the lattice path, the number of permutation tableaux in Ln is the Catalan numbers. Furthermore, we map a column to a pair ‘DU ’ which means a valley in a Catalan path, therefore the number of columns satisfying the Narayana numbers N(n, k) with k columns on Ln∩ P Tn.
As we known in section 3.2 in chapter 3, there is an obvious bijection between Ln,k and L0Bn,k by simply replacing illegal 0’s (0’s which have a 1 to the left of it and
a 1 above it simultaneously) with the superfluous 1’s, we have Ln∩L0Bn= Ln∩P Tn. Since the difference between L0Bn and R0Bn is the position of the superfluous 1’s, we have Ln∩ L0Bn = Ln∩ R0Bn.
Proposition 2. Ln∩ L0Bn = Ln∩ P Tn
Proposition 3. Ln∩ R0Bn= Ln∩ L0Bn