Characterizations of L-Bell tableaux

3 4 6 5

7 8 9

Figure 3: A tableau corresponds to the permutation (2, 6, 3, 8, 4, 5, 1, 7, 9).

2.2 Characterizations of L-Bell tableaux

Let φ(LB_n) be the set of permutations under Corteel and Nadeau’s bijection apply-ing on the set LB_n. We use a result from [5] in order to characterize φ(LB_n).

Proposition 1. [5] There exists a bijection between L-Bell tableaux of length n such that the sum of the number of columns and the number of zero rows is k and set partitions of [n] with k blocks.

In the following we prove three lemmas to show that φ(LB_n) ⊆ S_n(3-21) and construct a bijection h between S_n(3-21) and set partitions of [n]. Finally, we apply Proposition 1 to prove the result.

Lemma 1. Let T ∈ LB_n, c₁, c₂, . . . , c_s are the columns of T with c₁ > c₂ > · · · > c_s and r₁, r₂, · · · , r_s be the corresponding rows where cell (r_i, c_i) is a topmost 1, for all 1 ≤ i ≤ s. Then for a column c_l,

1. None of the cells to the right of (r_l, c_l) is a restricted 0, and all of the cells to the left of (r_l, c_l) are 0’s.

2. If r_l is an unrestricted row, then r_t > r_l, for all t < l. That is, the topmost 1 (rt, ct) is in the lower left corner of the topmost 1 (rl, cl), for all t < l

Proof. 1. A restricted 0 to the right of (r_l, c_l) will lead to a exist of a topmost 1 above it in T , which contradicts to the definition of permutation tableaux.

The cells to the left of (r_l, c_l) are all 0’s since T ∈ LB_n.

2. If r_l is an unrestricted row, i.e., none of the cells to the left of (r_l, c_l) is a restricted 0, then the cells in the upper left of it are not topmost 1’s. Thus r_t> r_l, for all t < l.

Lemma 2. Let T ∈ LB_nand follow the notations above. R₀ is the set of unrestricted rows in T and Rp ={r | cell (r, cp) is a rightmost restricted 0 }. Then

1. If r_q∈ R_p for some p < q, then c_q is to the left of c_p in σ = φ(T ).

2. If there exist 1 ≤ k₁, k₂, . . . , k_m ≤ s such that:

(a) k₁ > k₂ > · · · > k_m,

(b) r_kt−1 ∈ R_kt, for all 2 ≤ t ≤ m, and (c) r_km ∈ R₀,

Moreover, if there exists r_w ∈ R₀ such that r_w is to the left of r_km in List₀, then c_k1, c_k2, . . . , c_km in σ are between r_w and max R_km.

Proof. 1. If r_q ∈ R_p for some p < q, then r_q > r_p and we insert R_pc_p into the list before we insert R_qc_q. Since r_q∈ R_p, we insert R_qc_q to the left of r_q. Hence c_q is to the left of c_p in σ.

2. According to (1), the order of ck1, ck2, . . . , ckm in σ is ck1, ck2, . . . , ckm. Suppose that r_w is already to the left of r_km in List₀. Thus after we insert R_kmc_km into the list, Rkmckm are between rw and rkm in Listkm. Moreover, after we insert R_km−1c_km−1 into the list, c_km−1 is between r_w and max R_km(say r_w⁰) in List_km−1 since rkm−1 ∈ Rkm. That is, ck1, ck2, . . . , ckm in σ are between rw and rw⁰, see Figure 4 as an illustion.

According to Lemma 2, we know that if rk is a restricted row, then there exists q < k such that r_k ∈ R_q. Furthermore, if r_q is also a restricted row, then r_q ∈ R_u for some u < q. In other words, for a restricted row rk, we can find rk= rk1, rk2, . . . , rkm

r

r

Figure 4: An illustration of the insertion steps.

satisfy (a), (b), and (c) in Lemma 2. Then we call r_km = r_k. That is, for a restricted row r_k1 = r_k, we can find a zigzag path (r_k1, c_k1), (r_k1, c_k2), (r_k2, c_k2), (r_k2, c_k3), . . . , (r_km−1, c_km), (r_km, c_km) such that the path passes through topmost 1’s and rightmost restricted 0’s alternately, see Figure 5.

1

Figure 5: A zigzag path in LB.

In addition, the order of ck1, ck2, . . . , ckm in σ is ck1, ck2, . . . , ckm, where k1 > k2 >

· · · > k_m. Thus, those descents in σ are in increasing order.

Lemma 3. Let T ∈ LBn, two columns cp, cq with p < q and cp, cq are the corre-sponding columns. Then the behavior of these two paths are either disjoint or they both pass through rightmost restricted 0’s in some column cj where 1 ≤ j ≤ p and then pass through the same cells after that (See Figure 6). Moreover, for the latter

case, we have c_q is to the left of c_p in σ.

1

1

c

c

c

c

c

c

1

c

0 0 or

c

= c

1

Figure 6: The behavior of two zigzag paths in LB.

Proof. Suppose that these two paths are intersecting at some cell and then separate after that. See figure 7. That is, the path starting at c_qpasses cells (r_qt, c_qt), (r_qt, c_qt+1) and the path starting at c_p passes cells (r_pl−1, c_pl), (r_pl, c_pl) where c_qt+1 > c_pl >

c_qt, r_pl < r_qt < r_pl−1. Noth that (r_qt, c_qt), (r_pl, c_pl) are topmost 1’s and (r_qt, c_qt+1), (r_pl−1, c_pl) are rightmost restricted 0’s.

c _q

t+1

c _p

l

c _q

t

r _p

l

r _q

t

r _p

l−1

1 0 0

0 1

c _p c _q

Figure 7: An intersection of two zigzag paths in LB.

Since T ∈ LB, cell (rqt, cqt) is a leftmost 1. Thus the intersection cell (rqt, cpl) is a 0. Furthermore, it is a restricted 0 as cell (r_pl, c_pl) is a topmost 1 above it. This leads to the fact that the cell (rqt, cqt+1) which is to the left of cell (rqt, cpl) is not a rightmost restricted 0, a contradiction.

Since T ∈ LB, all topmost 1’s belong to distinct rows. Thus two paths intersect only if they both pass through rightmost restricted 0’s in some column cj, where 1 ≤ j ≤ p. Then by the behavior of path, they merge together after that. Moreover, if the path starting at cp (respectively cq) passes through the rightmost restricted zero (r_p⁰, c_j) (respectively (r_q⁰, c_j)), then r_q⁰ < rp⁰ and r_q⁰, rp⁰ ∈ R_j. Thus r_q⁰ is to the left of rp⁰ in listj. Hence cq is to the left of cp in σ.

proof of Theorem 2(1). Now we are ready to prove the first part of Theorem 2(1), i.e., φ(LB_n) ⊆ S_n(3-21). For a tableau T ∈ LB_n, we show that the descents in σ = φ(T ) are in increasing order. That is, for any two columns cp, cq with p < q, cq

is to the left to c_p in σ.

We show that for any two columns cp, cq with p < q, cq is to the left of cp in σ.

Hence the descents in σ are in increasing order. we consider four cases:

Case 1. rp, rq are both unrestricted rows. Since rq is unrestricted row, by Lemma 1(2), r_q < r_p. Thus r_q is to the left of r_q in the list List₀. Hence, c_q is to the left of cp in σ.

Case 2. r_p is an unrestricted row, while r_q is a restricted row.

Subcase 1. rq < rp. Then there exists a zigzag path starting from (rq, cq). By Lemma 1(2), the path ends at cell (r_q, c_q), where c_q < c_p and r_q < r_q. Since r_q and rp are both unrestricted rows and rq < rp, we have rq is to the left of rq in list List0. By Lemma 2(2), c_q is to the left of c_q. Hence, it is to the left of c_p in σ.

Subcase 2. rq > rp. By Lemma 1(2), the path starting from (rq, cq) ends in either the upper right or the lower left areas of cell (r_p, c_p). If it ends in the upper right area of (rp, cp), then we have cq is to the left of cp in σ by the similar discussion in Case 1. Suppose it passes cell (r_q⁰, c_p) for some r_q⁰ > r_p, which means (r_q⁰, c_p) is a rightmost restricted 0. Then it ends at (rp, cp). By Lemma 3, cq is to the left of c_p in σ.

Case 3. rp is a restricted row, while rq is an unrestricted row. By Lemma 1(2), r_q < r_p and the path starting from (r_p, c_p) ends at a cell below the row r_q, say (rp, cp). Then rp, rq ∈ R0 and rq < rp. Hence, rq is to the left of rp. By Lemma 2(2), c_p in σ is between r_q and max R_p. Thus we still have the result that c_q is to the left

of c_p in σ.

Case 4. rp, rq are both restricted rows.

Subcase 1. r_p < r_qand these two paths end at the same cell, i.e., r_p = r_q. Since (rp, cp) is a topmost 1, there is no restricted 0 above it in column cp. Thus, these two paths intersect at cells in column c_tfor some t ≤ p according to Lemma 3. We prove that these two path will not intersect at the left area of column cp. Assume that t < p, then we call cell (r_m, c_m) as the last cell, when the path starting at (r_q, c_q) passes before entering the left area of column cp, i.e., p < m < q and rp < rm < rq. Then the cell (r_m, c_p) is a restricted 0 by Lemma 1(1). This implies that these two paths do not intersect at cells in column ct, a contradiction. Hence, t = p. That is, r_m ∈ R_p and r_m is to the left of c_p in List_p. By the similar discussion in Lemma 2(2), cq is to the left of rm in σ, thus cq is to the left of cp in σ.

Subcase 2. r_p < r_q and these two paths are disjoint. By the similar discussion in the previous case, we know that the path starting at cell (rq, cq) do not pass through cells in the lower left area of (r_p, c_p). It neither ends at cells in the lower right area of (rp, cp) according to Lemma 1(2). Thus it passes through the upper right area of (r_p, c_p) and (r_q, c_q) is in the upper right area of (r_p, c_p) in the end. This implies that rq is to the left of rp in List0. By Lemma 2(2), cp is between rq and max R_p. Then we have that c_q is to the left of c_p in σ.

Subcase 3. rp > rq. By the similar discussion, we have (rq, cq) is in the upper right area of (r_p, c_p) if these two paths are disjoint; otherwise, they intersect at cells in column ct for some t < p. For the former case, we have cq is to the left of cp in σ, hence c_q is to the left of c_p in σ. For the latter case, by Lemma 3, let (r_q⁰, c_t) and (rp⁰, ct) are rightmost restricted 0’s, where these two paths pass through column ct, then r_q⁰ < r_p⁰ and r_q⁰, r_p⁰ ∈ R_t. We have that r_q⁰ is to the left of r_p⁰ in List_t. By the similar discussion in Lemma 2(2), cq is to the left of cp in σ.

The second part of the proof is to find a bijection h from S_n(3-21) to set par-titions. For a permutation σ in Sn(3-21), its descents are in increasing order. We obtain the set partition π = h(σ) by cutting σ in front of each left-to-right maximum into blocks. Note that, a left-to-right maximum that is not a descent forms a single

block, while a left-to-right maximum that is a descent forms a block of size greater than 1. For a set partition π, we arrange the blocks in increasing order of their maximum. In each block, we put its maximum in the first place, and then arrange the other elements in increasing order. Finally, we delete the blocks to obtain a permutation which has its descents in increasing order.