A crystal on decreasing factorizations in the 0-hecke monoid

. We introduce a type A crystal structure on decreasing factorizations of fully-commu-tative elements in the 0-Hecke monoid which we call ⋆ -crystal. This crystal is a K -theoretic generalization of the crystal on decreasing factorizations in the symmetric group of the ﬁrst and last author. We prove that under the residue map the ⋆ -crystal intertwines with the crystal on set-valued tableaux recently introduced by Monical, Pechenik and Scrimshaw. We also deﬁne a new insertion from decreasing factorization to pairs of semistandard Young tableaux and prove several properties, such as its relation to the Hecke insertion and the uncrowding algorithm. The new insertion also intertwines with the crystal operators.


Introduction
The symmetric Grothendieck polynomials [LS82,LS83,FK94] represent Schubert classes in the K-theory of the Grassmannian. They can be expressed [Buc02] as generating functions of semistandard set-valued tableaux where M µ λ is the number of highest weight set-valued tableaux of weight µ in the crystal SVT m (λ). A different combinatorial formula for the multiplicities M µ λ was given by Lenart [Len00, Theorem 2.2] in terms of flagged increasing tableaux.
A generalization [LS82,LS83,FK94] of the symmetric Grothendieck polynomials in (1.1) are the stable Grothendieck polynomials labeled by permutations w ∈ S n G w (x 1 , . . . , x m ; β) = (k,h) where the sum is over decreasing factorizations [k, h] t of w in the 0-Hecke algebra. When β = 0, G w specializes to the Stanley symmetric function F w [Sta84].
In this paper, we define a new crystal, which we call the ⋆-crystal, on decreasing factorizations of w in the 0-Hecke algebra, when w is 321-avoiding. This crystal is local in the sense that the crystal operators f i and e i only act on the i-th and (i + 1)-st factor of the decreasing factorization. It generalizes the crystal of Morse and Schilling [MS16] for Stanley symmetric functions (or equivalently reduced decreasing factorizations of w) in the 321-avoiding case. We show that the ⋆-crystal and the crystal on set-valued tableaux intertwine under the residue map (see Theorem 2.16). We also show that the residue map and the Hecke insertion [BKS + 08] are related (see Theorem 3.5), thereby resolving [MPS18,Open Problem 5.8] in the 321-avoiding case. In addition, we provide a new insertion algorithm, which we call ⋆-insertion, from decreasing factorizations on 321-avoiding elements in the 0-Hecke monoid to pairs of (transposes of) semistandard Young tableaux (see Definition 3.8 and Theorem 3.16), which intertwines with crystal operators (see Theorem 4.22). This recovers the Schur expansion of G w of [FG98] when w is 321-avoiding, stating that if G w = µ β |µ|−ℓ(w) g µ w s µ , then g µ w = |{T ∈ SSYT n (µ ′ ) | w C (T ) ≡ w}|, where w C (T ) is the column reading word of T (see Remark 4.23). We also show that the composition of the residue map with the ⋆-insertion is related to the uncrowding algorithm [Buc02] (see Theorem 4.29). Other insertion algorithms have recently been studied in [CP19].
The paper is organized as follows. In Section 2, we introduce the ⋆-crystal on decreasing factorizations in the 0-Hecke monoid and show that it intertwines with the crystal on semistandard set-valued tableaux [MPS18] under the residue map. In Section 3, we discuss two insertion algorithms for decreasing factorizations. The first is the Hecke insertion introduced by Buch et al. [BKS + 08] and the second is the new ⋆-insertion. In Section 4, properties of the ⋆-insertion are discussed. In particular, we prove that it intertwines with the crystal operators and that it relates to the uncrowding algorithm. We conclude in Section 5 with some discussions about the non 321-avoiding case.
A reduced expression for an element w ∈ S n is a word a 1 a 2 . . . a ℓ with a i ∈ [n−1] := {1, 2, . . . , n−1} such that (2.1) w = s a 1 · · · s a ℓ and ℓ is minimal among all words satisfying (2.1). In this case, ℓ is called the length of w.
Definition 2.1. The 0-Hecke monoid H 0 (n), where n 1 is integer, is a monoid of finite words generated by positive integers in the alphabet [n − 1] subject to the relations (2.2) pq ≡ qp if |p − q| > 1, pqp ≡ qpq for all p, q, pp ≡ p for all p.
We may form an equivalence relation ≡ H 0 on such words based on the relations (2.2). A word a = a 1 a 2 . . . a ℓ representing a ∈ H 0 (n) is reduced if ℓ 0 is the smallest among all words equivalent to a. Note that the set of all reduced words a in H 0 (n) is in bijection with the set of all reduced words in S n by considering a = a 1 a 2 . . . a ℓ as a word for the element s a 1 s a 2 · · · s a ℓ in S n .
is either empty (meaning ℓ i = 0) or strictly decreasing (meaning h i 1 > h i 2 > · · · > h i ℓ i ) for each 1 i m and h ≡ H 0 w in the 0-Hecke monoid H 0 (n).
The set of all possible decreasing factorizations into m factors shall be denoted as H m or H m (n) if we want to indicate the value of n. We call ex(h) = len(h) − ℓ the excess of h, where len(h) is the length of h as a word and ℓ is the length of the underlying reduced Hecke word of h as an element in H 0 (n).
We say that an element w ∈ H 0 (n) (resp. w ∈ S n ) is 321-avoiding if none of the reduced expressions for w contain a consecutive subword of the form i i + 1 i for any i ∈ [n − 1].

2.2.
The ⋆-crystal. In this section, we introduce a new crystal structure of type A m−1 on the set of 321-avoiding Hecke factorizations, which we call the ⋆-crystal H m,⋆ . As a set, H m,⋆ consists of all 321-avoiding Hecke factorizations in H m . (1) If x + 1 ∈ h i ∩ h i+1 , then remove x + 1 from h i , add x to h i+1 .
(2) Otherwise, remove x from h i and add x to h i+1 . If all letters in h i+1 have been paired, then e ⋆ i annihilates h. Let y be the smallest unpaired letter in h i+1 . The crystal operator e ⋆ i acts on h in either of the following ways: (1) If y − 1 ∈ h i ∩ h i+1 , then remove y − 1 from h i+1 , add y to h i .
(2) Otherwise, remove y from h i+1 and add y to h i .
It is not hard to see that e ⋆ i and f ⋆ i are partial inverses of each other. (1) Note that the ⋆-crystal of Definition 2.3 is local in the sense that the crystal operators f ⋆ i and e ⋆ i only depend on and change the i-th and (i + 1)-st factors h i and h i+1 (unless they annihilate the decreasing factorization).
(2) Compared to [MS16], one pairs a letter b in h i+1 with the smallest letter a b in h i rather than a > b. To see that the underlying word for h andh is the same, it suffices to show that h i+1 h i ≡ H 0 h i+1hi . Let x be the largest unpaired letter in h i . By the bracketing procedure this implies that x / ∈ h i+1 . We can write h i+1 as w 1 w 2 , where w 1 is a word containing only letters greater than x, and w 2 is a word containing only letters smaller than x. We can write h i as w 3 xw 4 , where w 3 contains only letters greater than x and w 4 contains only letters smaller than x.
The pairing process will result in one of the two following cases: (1) If x + 1 ∈ h i ∩ h i+1 , then obtainh i by removing x + 1 from h i , andh i+1 by adding x to h i+1 . (2) Otherwise, obtainh i by removing x from h i and obtainh i+1 by adding x to h i+1 .
We first argue that in either case we must have x − 1 / ∈ w 2 . Assume x − 1 ∈ w 2 and let k be the largest number such that the interval [x − k, x − 1] ⊆ w 2 . By assumption k 1. In order for x to be the largest unpaired letter in h i , [x − k, x − 1] must be contained in w 4 . We can write where all letters in w ′ 2 are smaller than x − k − 1. When k = 1, we have the following subword , which contains a braid (x − 1)x(x − 1). When k > 1, we also have the following subword , which also contains a braid.
Case (1): Let k be the largest letter such that [x + 1, x + k] ⊆ w 3 . Clearly k 1. Suppose k > 1, then we can write w 3 = w ′ 3 (x + k) . . . (x + 1). Since x is the largest unpaired letter in h i , everything in [x + 1, x + k] ⊆ w 3 must be paired. The letter x + 1 in w 3 is paired with x + 1 ∈ w 1 , which implies that x + i in w 3 is paired with x + i ∈ w 1 for all 1 i k. This implies that [x + 1, x + k] ⊆ w 1 . Then we have the following subword which contains a braid. Thus we must have k = 1, which implies that x + 2 / ∈ w 3 . Write w 1 = w ′ 1 (x + 1). Then by direct computation Case (2): We claim that if x + 1 / ∈ h i+1 , then x + 1 / ∈ h i . Otherwise the x + 1 ∈ h i must be paired with some z ∈ h i+1 , so we have z x + 1. But x is unpaired, which implies z > x, that gives us a contradiction. Hence x + 1 / ∈ w 3 . Recall that x − 1 / ∈ w 2 . Therefore, by a straightforward computation The above arguments show that the underlying word and the total length of the Hecke factorization are unchanged under f ⋆ i . Furthermore, the excess remains unchanged under f ⋆ i . Similar arguments hold for e ⋆ i . Remark 2.7. Here we summarize several results from the proof that will be needed later. Namely, if x is the largest unpaired letter in h i , then • One and only one of the three statements hold: It will be shown in Section 2.4 that H m,⋆ is indeed a Stembridge crystal of type A m−1 (for an introduction to crystal and terminology, see [BS17]).

2.3.
The crystal on set-valued tableaux. In this section, we review the type A crystal structure on set-valued tableaux introduced in [MPS18]. In fact, in [MPS18] the authors only considered the crystal structure on straight-shaped set-valued tableaux. Here we consider the crystal on skew shapes as well, see Theorem 2.10.
We use French notation for partitions λ = (λ 1 , λ 2 , . . .) with λ 1 λ 2 · · · 0, that is, in the Ferrers diagram for λ, the largest part λ 1 is at the bottom. Definition 2.8. A semistandard set-valued tableau T of skew shape λ/µ is a filling of the Ferrers diagram of λ/µ with nonempty subsets of positive integers such that: • for all adjacent cells A, B in the same row with A to the left of B, we have max(A) min(B), • for all adjacent cells A, C in the same column with A in the row below B, we have max(A) < min(C). The weight of T , denoted by wt(T ), is the integer vector whose i-th component counts the number of i's that occur in T . The excess of T is defined as ex(T ) = | wt(T )| − |λ|. We denote the set of all semistandard set-valued tableaux of shape λ/µ by SVT(λ/µ). Similarly, if the maximum entry is restricted to m, the set is denoted by SVT m (λ/µ).
We define a crystal structure on semistandard set-valued tableaux of skew shape. Definition 2.9. Let T ∈ SVT m (λ/µ). We employ the following pairing rule for letters i and i + 1. Assign − to every column of T containing an i but not an i + 1. Similarly, assign + to every column of T containing an i + 1 but not an i. Then, successively pair each + that is adjacent to a −, removing all paired signs until nothing can be paired.
The operator f i changes the i in the rightmost column with an unpaired − (if this exists) to i + 1, except if the cell b containing that i has a cell to its right, denoted b → , that contains both i and i + 1. In that case, f i removes i from b → and adds i + 1 to b. Finally, if no unpaired − exists, then f i annihilates T .
Similarly, the operator e i changes the i + 1 in the leftmost column with an unpaired + (if this exists) to i, except if the cell b containing that i + 1 has a cell to its left, denoted b ← , that contains both i and i + 1. In that case, e i removes i + 1 from b ← and adds i to b. Finally, if no unpaired + exists, then e i annihilates T .
Based on the pairing procedure above, ϕ i (T ) is the number of unpaired − while ε i (T ) is the number of unpaired +.
One can easily show that the crystal on SVT m (λ/µ) of Definition 2.9 defines a seminormal crystal (for definitions see [BS17]). It was proved in [MPS18, Theorem 3.9] that the above described operators e i and f i define a type A m−1 Stembridge crystal structure on SVT m (λ). We claim that their proof goes through also for skew shapes.
Theorem 2.10. The crystal SVT m (λ/µ) of Definition 2.9 is a Stembridge crystal of type A m−1 .
Proof. Since the proof is exactly the same as in [MPS18,Theorem 3.9], we just state the outline and give a brief description. For details we refer to [MPS18].
First note that the signature rule given by column-reading is compatible with the signature rule given by row-reading (top to bottom, left to right, and arrange the letters in the same cell by descending order) by semistandardness. Hence we may consider the crystal to live inside the tensor product of its rows. A single-row semistandard set-valued tableaux of a fixed shape is isomorphic to a Stembridge crystal, as shown in [MPS18, Proposition 3.5]: where Λ k are the fundamental weights of type A m−1 .
2.4. The residue map. In this section, we define the residue map from set-valued tableaux of skew shape to 321-avoiding decreasing factorizations in the 0-Hecke monoid. We then show in Theorem 2.16 that the residue map intertwines with the crystal operators, proving that H m,⋆ is indeed a crystal of type A m−1 (see Corollary 2.17).
Definition 2.11. Given T ∈ SVT m (λ/µ), we define the residue map res : SVT m (λ/µ) → H m as follows. Label all cells (i, j), where i, j 1, with ℓ(λ) + j − i. Produce a decreasing factorization h = h m h m−1 . . . h 2 h 1 by declaring h i to be the (possibly empty) sequence formed by taking the labels of all cells in T containing i and then arranging these labels in decreasing order. This defines res(T ) := h.
Recall from Definition 2.2 that the underlying element w ∈ H 0 (n) associated to res(T ) may be recovered by reading the letters that appear in res(T ) from left to right. We label the cells of T as follows: For instance, to read off the third factor, we search for all cells containing 3; these cells come with labels 1 and 2, so we have 21 in the third factor. Altogether, we obtain res(T ) = (21)(31)(3) ∈ H 3 .
We now discuss that the image of the residue map res is in fact H m,⋆ , the set of 321-avoiding decreasing factorizations into m factors.
For this purpose, let us describe the inverse of the residue map. Let h = h m h m−1 . . . h 2 h 1 ∈ H m . Begin by filling m along the diagonals labeled by the letters that appear in h m . As the result T is supposed to be of skew shape, the cells containing m along increasing diagonals need to go weakly down from left to right. If these diagonals are consecutive, then the cells have to be in the same row of T since T is semistandard. Continue the procedure above by filling i along the diagonals specified by h i for all i = m − 1, m − 2, . . . , 1, applying the condition that the resulting filling should be semistandard. Proof. We shall show more generally that at any given stage in the algorithm above, the tableau produced is of skew shape if and only if the corresponding Hecke word is 321-avoiding.
Assume that the resulting tableau is not of skew shape. Consider the earliest stage in the algorithm when the produced tableau is not of skew shape. Then, either one of the following cases must have occurred for the first time.
Case 1: There are adjacent cells with nonempty sets A and B (where max(A) min(B)) in the same row on diagonals i and i + 1 respectively with no cells appearing directly below these cells, as illustrated on the left side of Figure 1. Moreover, by minimality, we have an integer x with the following properties: (1) i + 1 ∈ h x and x < min(A), (2) there does not exist a y with x y < min(B) and i + 2 ∈ h y . By applying semistandardness, a cell containing x is created directly below the cell containing the set A as in the right side of Figure 1. Furthermore, by (2), for all x y < min(B), we have that every letter in h y is either at most i + 1 or at least i + 3. It follows that, after possibly applying commutativity (i + 1 with letters at most i − 1 or at least i + 3) and the idempotent relation, the Hecke word h min(B) . . . h x+1 h x is equivalent to one containing the braid subword i + 1 i i + 1. This implies that h is equivalent to a Hecke word containing the same braid subword.
Case 2: There are adjacent cells with nonempty sets A and B in the same column on diagonals i+ 1 and i respectively with no cells appearing directly to the left of these cells, as illustrated on the left side of Figure 2. Moreover, by minimality, we have an integer x with the following properties: (1) i ∈ h x and x min(A), (2) there does not exist a y with x < y min(B) and i − 1 ∈ h y . By applying semistandardness, a cell containing x is created directly to the left of the cell containing the set A as in the right side of Figure 2. Furthermore, by (2), for all x < y min(B), we have that every letter in h y is either at most i − 2 or at least i. Similar to the argument in Case 1, the Hecke word h min(B) . . . h x+1 h x is equivalent to one containing the braid subword i i + 1 i. This implies that h is equivalent to a Hecke word containing the same braid subword.
The above arguments imply that the image of res is contained in H m,⋆ . Conversely, if h is 321avoiding, then the algorithm for res −1 does not produce Case 1 or Case 2 above and hence the x i+1 Figure 1. A forbidden case while inverting the residue map. resulting tableau T is of skew shape and hence the algorithm is well-defined (up to shifts along the diagonal if a gap of size at least 3 occurs in the labels).
If the skew shape λ/µ of the tableau T is known, then one may simplify the procedure above noting that the filling of i specified by letters in h i must occur along a horizontal strip for all i = m, m − 1, . . . , 1. In this case, the recovered tableau T is unique and there is no shift ambiguity if a gap of size at least 3 occurs in the labels. In the procedure to determine a suitable skew tableau whose residue map is h, after filling 4's along the diagonals with labels 1 and 6 respectively, due to semistandardness, the 3 in diagonal 2 is below the 4 in diagonal 1, while the 3's in diagonals 5 and 7 are respectively to the left and below the 4 in diagonal 6. Continuing with the remaining fillings, we have two possibilities: where T 1 ∈ SVT 4 ((4, 4, 1, 1)/(2, 2)) and T 2 ∈ SVT 4 ((3, 3, 1, 1, 1)/(1, 1, 1)).
Proof. Let T ∈ SVT m (λ/µ), h = res(T ) and ℓ = ℓ(λ). We prove the following three statements associated to f k (T ) and f ⋆ k (h). (1) We claim that if there is no unpaired k in T , then f ⋆ k annihilates h. Furthermore, if the rightmost unpaired k in cell b of T has content x, then x is also the largest unpaired letter in h k .
For the proof of (1) it suffices to notice that the signature rule on tableaux is equivalent to the pairing process in decreasing Hecke factorizations. We rephrase the pairing procedure in decreasing Hecke factorizations on tableaux: • At the beginning, no letter is paired.
• Then start with the rightmost column and work westward.
• Successively, for each k+1, compute its content a, then pair it with the k of smallest content weakly greater than a that is yet unpaired. Next, we argue that the signature rule yields the same result on the rightmost unpaired letter. Assume we are looking at cell b containing the current k + 1 with content a. Case (a): Suppose there is no unpaired k with content a but at least one unpaired k with strictly greater content(s). Then pair it with the current k + 1. This is the direct signature rule. Case (b): Suppose there is no unpaired k with content weakly greater than a, then this k + 1 is unpaired. This is also the direct signature rule. Case (c): Suppose there is an unpaired k with content a. Then it must be either in the same cell b, or one row below and one column to the left of b on the diagonal labelled a. If they are in the same cell, then the pairing is the direct signature rule.
Otherwise, there must be cells to the left and below b since the shape is skew. Suppose cell b is in row r. Consider the rightmost entry in cell (r, j) in row r containing a k + 1, and the leftmost entry in cell (r − 1, q) in row r − 1 containing a k. Considering this as the first of a consecutive occurrence, cell b is cell (r, j), so we have ℓ + j − r = a. By semistandardness and the condition that the shape is skew, we can partially fill out the involved subtableau of T for rows r − 1, r from column q to j: All the cells (s, t) with q < t < j and s ∈ {r, r − 1} and the cells (r, q) and (r − 1, j) are single-valued by semistandardness as shown in the above figure.
From the k + 1 in (r, j), we start the pairing process. First, we claim that the k in cell (r − 1, j) must be unpaired at this point. Suppose that there is a k + 1 to the east of cell (r, j) with content smaller or equal to ℓ + j − r + 1, then it must be cell (r, j + 1), which violates that (r, j) is the rightmost cell in row r containing a k + 1. Then the pairing says the k + 1 in cell (r, t) pairs with the k in cell (r − 1, t − 1) for q < t j. Lastly, the k + 1 in cell (r, q) has to pair with the previously unpaired k in cell (r − 1, j) since there are no unpaired k with label greater or equal to ℓ + q − r and smaller than ℓ + j − r + 1.
Although the pairing is different than the usual signature rule pairing, which pairs k + 1, k in the same column, the 2(j − q + 1) letters end up being paired. Since it will not influence which one will be the rightmost unpaired letter, it is still equivalent to the signature rule.
So in any case, the pairing is equivalent to the signature rule. Thus the rightmost unpaired k in T corresponds to the largest unpaired letter in h k .
(2) We claim that if f k changes the rightmost unpaired k in T to a k + 1 (with content x) without moving it, then f ⋆ k moves a letter x from h k to h k+1 . Since f k does not need to move any letter, it means the cell to the right of b, denoted by b → , does not contain a k. It is the only cell with content x + 1 that could contain a k. This implies that (3) We claim the following. If f k changes a k from b → into a k + 1 and moves to cell b, then f ⋆ k removes an x + 1 from h k and changes it to an x in h k+1 . That f k needs to move a number means that k and k + 1 are in b → , which implies that x + 1 ∈ h k ∩ h k+1 . By Definition 2.3, f ⋆ k removes the x + 1 from h k and adds an x to h k+1 . We have proved the three statements and they complete the proof that f k and f ⋆ k intertwine under the residue map. The proof is similar for e k and e ⋆ k . Proof. By Theorem 2.16 and the fact that the residue map is invertible, this follows from the fact that SVT m (λ/µ) is a Stembridge crystal proven in [MPS18, Theorem 3.9] (see also Theorem 2.10).
Example 2.18. Consider the tableau T (with labels in red) given by For the crystal operators on set-valued tableaux we obtain with res (f 1 (T )) = (31)(32) (2). Then it can be easily checked that the following diagram commutes: .

Insertion algorithms
In this section, we discuss two insertion algorithms for decreasing factorizations in H m (resp. H m,⋆ ). The first is Hecke insertion introduced by Buch et al. [BKS + 08], which we review in Section 3.1. We prove a relationship between Hecke insertion and the residue map (see Theorem 3.5).
In particular, this proves [MPS18, Open Problem 5.8] for 321-avoiding permutations. The second insertion is a new insertion, which we call ⋆-insertion, introduced in Section 3.2. It goes from 321avoiding decreasing factorizations in the 0-Hecke monoid to pairs of (transposes of) semistandard tableaux and is well-behaved with respect to the crystal operators.
3.1. Hecke insertion. Hecke insertion was first introduced in [BKS + 08] as column insertion. Here we state the row insertion version as in [PP16]. In this section, we represent a decreasing factor- We describe how to insert x into P i , denoted P i ← x, by describing how to insert x into a row R. The insertion may modify the row and may produce an output integer, which will be inserted into the next row. First, we insert x into the first row R of P i following the rules below: (1) If x z for all z ∈ R, the insertion terminates in either of the following ways: (a) If we can append x to the right of R and obtain an increasing tableau, the result P i+1 is obtained by doing so; form Q i+1 by adding a box with y in the same position where x is added to P i . (b) Otherwise row R remains unchanged. Form Q i+1 by adding y to the existing corner of Q i whose column contains the rightmost box of row R. (2) Otherwise, there exists a smallest z in R such that z > x.
(a) If replacing z with x results in an increasing tableau, then do so. Let z be the output integer to be inserted into the next row. (b) Otherwise, row R remains unchanged. Let z be the output integer to be inserted into the next row. The entire Hecke insertion terminates at (P n , Q n ) after we have inserted every letter from the Hecke biword. The resulting insertion tableau P n is an increasing tableau. If k = (n, n − 1, . . . , 1), the recording tableau Q n is a standard set-valued tableau. Example 3.3. Take [k, h] t from Example 3.1. Following the Hecke row insertion, we compute its insertion tableau and recording tableau: However, in Theorem 3.5 below we will see that in certain cases it is. Note, however, that unlike in [MPS18] we use row Hecke insertion from right to left rather than column insertion from left to right (in analogy to [MS16] for Edelman-Greene insertion).
Since k ∈ T (i, j) if and only if ℓ + j − i ∈ h k under the residue map, where ℓ = ℓ(λ) and h k is the k-th factor of h, the statement of Theorem 3.5 is equivalent to applying Hecke insertion on the entries of T sorted first by ascending order of entries, followed by ascending diagonal content. .

Cell
(1,1) (2,1) (1,2) (1,2) (2,2) Content  2  1  3  3  2  Entry  1  2  2  3  4 The insertion sequence by entry is listed in the table below: We will prove Theorem 3.5 by induction by considering all subtableaux of T , obtained by adding the entries in T one by one in the order above: In addition, the corresponding sequence of insertion tableaux and recording tableaux is listed here: Proof of Theorem 3.5. We prove the theorem by proving the following more specific statement. For a given step in the insertion process, suppose that the entries of T that are involved so far form a nonempty subtableau T ′ of T with shape µ containing cell (1, 1), and the insertion tableau and recording tableau at the corresponding step are P (T ′ ) and Q(T ′ ). Then, they both have shape µ, and the entry of where µ ′ is the transpose of the partition µ and ℓ := λ ′ 1 = ℓ(λ). We prove this by induction on subtableaux of T . Base step: Suppose T ′ only contains a single cell (1, 1) and T ′ (1, 1) = S, where S is a subset of T (1, 1) with cardinality d. Then P (T ′ ) is obtained by inserting d times the number ℓ. So we have P (T ′ ) = ℓ and Q(T ′ ) = T ′ . Here µ = (1), so for (i, j) = (1, 1), we have ℓ + j − µ ′ j + i − 1 = ℓ. Inductive step: Suppose that the statements hold for some subtableau T ′ of shape µ. Assume the next insertion step involves adding the entry k in cell (p, q) of T to T ′ to obtain T ′′ . There are two cases: (1) the cell (p, q) is already in T ′ , or (2) the cell (p, q) is not in T ′ . Case (1): We must have (p, q) to be an inner corner of T ′ (no cell is to its right or above it), so p = µ ′ q and p > µ ′ q+1 . In this case, k is recorded in Q(T ′ ). Then by the induction on T ′ , every cell To determine the insertion path of P (T ′ ) ← ℓ + q − p, we compute the columns q and q + 1 of P (T ′ ) as follows: Following Case 2(b) of Hecke insertion, the insertion path is vertically up column q + 1. At the top of the column, ℓ + q is inserted into row µ ′ q+1 + 1. Furthermore, ℓ + q is greater than . By Hecke insertion Case 1(b), the insertion ends in row µ ′ q+1 +1. Also P (T ′ ) is unchanged, and k is recorded in cell (p, q) of Q(T ′ ) since it is the corner whose column contains the rightmost box of row µ ′ q+1 + 1. In this case, we get Q(T ′′ ) = T ′′ . Since the shape µ is unchanged, we have that P (T ′′ ) = P (T ′ ) also satisfies the statement. Case (2): If cell (p, q) is not in T ′ , then it must be an outer corner of T ′ , so µ ′ q = p − 1 and µ ′ q−1 > p − 1. Specifically two cases can happen: (a) p = 1 and (1, it is appended to the end of the first row which is the cell (1, q). The letter k is recorded in the same new cell of Q(T ′ ). In this case, the only entry in P that is changed is (1, q), and its entry ℓ + q − 1 satisfies the statement. Also Q(T ′′ ) equals T ′′ .
the number q − p + ℓ is in-between the two when i = 1. So the insertion starts by bumping (1, q). To get the insertion path, we compute columns q − 1 and q as follows: By Hecke insertion Case 2(a), ℓ + q − p is placed in cell (1, q) and the original column q is shifted one position higher. By Hecke insertion Case 1(a), the insertion terminates at row p and the original entry in cell (p − 1, q) is appended at the rightmost box of row p. Thus µ ′ q increases by 1. The updated entries in column q still satisfy the statement. Since the entries in other columns of P (T ′ ) are unchanged and µ ′ j is unchanged for j = q, they also satisfy the statement. So we have P (T ′′ ) satisfies the statement. The letter k is inserted into the new cell (p, q) of Q(T ′ ), which makes Q(T ′′ ) = T ′′ .
Thus, the statement holds, proving the theorem.
3.2. The ⋆-insertion. We define a new insertion algorithm, which we call ⋆-insertion, from a 321avoiding decreasing Hecke biword [k, h] t to a pair of tableaux P and Q, denoted by ⋆([k, h] t ) = (P, Q).
Definition 3.8. Fix a 321-avoiding decreasing Hecke biword [k, h] t . The insertion is done by reading the columns of this biword from right to left. Begin with (P 0 , Q 0 ) being a pair of empty tableaux. For every integer i 0, we recursively Case 1: If R is empty or x > max(R), then form P i+1 by appending x to row R and form Q i+1 by adding q in the corresponding position to Q i . Terminate and return (P i+1 , Q i+1 ). Case 2: Otherwise, if x / ∈ R, locate the smallest y in R with y > x. Bump y with x and insert y into the next row of P i . Case 3: Otherwise, x ∈ R, then locate the smallest y in R with y x and interval [y, x] contained in R. Row R remains unchanged and y is to be inserted into the next row of P i .
Furthermore, denote by P ← x the tableau obtained by inserting x into P . The collection of all cells in P ← x, where insertion or bumping has occurred is called the insertion path for P ← x. In particular, in Case 1 the newly added cell is in the insertion path, in Case 2 the cell containing the bumped letter y is in the insertion path, and in Case 3 the cell containing the same entry as the inserted letter is in the insertion path. The corresponding sequence of insertion tableaux and recording tableaux under the ⋆-insertion is listed here: Then we have ⋆([k, h] t ) = (P, Q), and the cells in the insertion paths at each step are highlighted in yellow.
. Then, the following statements hold: (1) P t is semistandard and Q has the same shape as P .
(2) Let x be an integer such that the Hecke word x · h is 321-avoiding. Then, the insertion path for P ← x goes weakly to the left.
Proof. We will prove (1) by induction on the number of cells of P . Statement (2) will follow by some results in the proof of statement (1). Consider the leftmost column [q, x] t of [k, h] t and let [k ′ , h ′ ] t be the Hecke biword formed by taking the remaining columns in the same order. If the ⋆-insertion of [k ′ , h ′ ] t yields (P ′ , Q ′ ), note that we have P = P ′ ← x. For all integers j 1, denote by R j the (possibly empty) j-th row of P ′ . Denote by u the entry to be inserted into R j and B j as the cell in the insertion path at R j , where 1 j k. Additionally, if bumping occurs at R j , denote the entry bumped out as y.
(1) We will prove that if (P ′ ) t is semistandard, then the transpose of the updated tableau is semistandard.
Case (a): Suppose that the insertion terminates at R 1 . Then Case 1 of the ⋆-insertion has occurred, with a cell containing x appended at the end of the row. If R 1 is nonempty, then x > max(R 1 ). Additionally, as (P ′ ) t is semistandard, integers strictly increase along R 1 but weakly increase along the column containing B 1 . Hence, the transpose of the resulting tableau P is semistandard. Case (b): Suppose that insertion terminates at R k , where k > 1. We will show that for all 1 j k, the changes introduced at row R j of P ′ maintain the property that the transpose of the updated tableau is semistandard. Case (b)(i): Suppose that j = k. In this case, a new cell containing u is appended at the end of R k and u > max(R k ) if the row is nonempty, proving that the integers increase strictly along R k . If Case 2 occurs at R k−1 , then u is the entry bumped out of R k−1 with the property that when u ′ is inserted into R k−1 , u ∈ R k−1 is the smallest entry with u > u ′ . Let z be the entry below cell B k . We claim that z u. If we assume instead that z > u, then the cell containing z is strictly to the right of B k−1 . However, the cell above B k−1 has value greater than u since (P ′ ) t is semistandard and u / ∈ R k . This contradicts the minimality of u ′ , as u ′ is greater than this value, hence proving the claim.
If Case 3 occurs at R k−1 , then u is bumped out of R k−1 with the property that when u ′ is inserted into R k−1 , u ∈ R k−1 is the smallest entry with [u, u ′ ] ⊆ R k−1 . Let z be the entry below cell B k . Then, similar to the argument immediately before, z u ′ . Hence, we have established that the integers weakly increase along the column containing B k after u is appended at the end of R k . Case (b)(ii): Suppose that 1 j < k and Case 2 occurs at R j . Then y is the entry bumped out of R j with the property that when u is inserted into R j , y ∈ R j is the smallest entry with y > u. Thus, as u / ∈ R j , for all entries z and z ′ respectively to the left and to the right of B j , we have z < u < y < z ′ .
If Case 2 occurs at R j−1 , then u is bumped out of R j−1 with the property that when u ′ is inserted into R j−1 , u ∈ R j−1 is the smallest entry with u > u ′ . Let z be the entry below cell B j . Then by repeating the same argument as in the first subcase of in Case (b)(i), we obtain z u.
If Case 3 occurs at R j−1 , then u was bumped out of R j−1 with the property that when u ′ is inserted into R j−1 , u ∈ R j−1 is the smallest entry with [u, u ′ ] ⊆ R j−1 . Let z be the entry below cell B j . Then by repeating the same argument as in the second subcase of in Case (b)(i), we obtain z u ′ .
Hence, we have established that integers increase weakly along the column containing B j but increase strictly along R j after u bumps out y. Case (b)(iii): Suppose that 1 j < k and Case 3 occurs at R j . In this case, there are no changes to row R j after inserting u and bumping y. Hence, it is trivial that integers increase weakly along the column containing B j but increase strictly along R j after u bumps out y.
In all cases, we have shown that if (P ′ ) t is semistandard, then the transpose of the updated tableau remains semistandard. Therefore, by induction on the number of added cells, we have proved that the insertion tableau P under ⋆-insertion satisfies the property that P t is semistandard. Finally, note that the shape of the recording tableau is modified only when Case 1 of the ⋆insertion has occurred. In this case, a cell is added to form Q at the same position as the cell added to form P . Since we always begin with a pair of empty tableaux, by inducting on the number of added cells, the shapes of P and Q are the same.
(2) Suppose that the insertion terminates at R k , where k 1. We shall prove that B j is weakly to the left of B j−1 for all 1 < j k by revisiting the cases explored in the proof of part (1) (note that P should replace the role of P ′ ).
If Case 2 occurs at R j−1 , then u is the entry bumped out of R j−1 with the property that when u ′ is inserted into R j−1 , u ∈ R j−1 is the smallest entry with u > u ′ . As in the proof of the first subcase of Case (b)(i) in part (1), we conclude that the entry z of the cell below B k satisfies z u, showing that B j is weakly to the left of B j−1 .
If Case 3 occurs at R j−1 , then u was bumped out of R j−1 with the property that when u ′ is inserted into R j−1 , u ∈ R j−1 is the smallest entry with [u, u ′ ] ⊆ R j−1 . As in the proof of the second subcase of Case (b)(i) in part (1), we conclude that the entry z of the cell below B j satisfies z u ′ , B j is weakly to the left of B j−1 .
This completes the proof.
For the following results, given a tableau P with positive integer entries, row(P ) denotes its row reading word, obtained by reading these entries row-by-row starting from the final row, reading from left to right. We will consider row(P ) as an element in a fixed 0-Hecke monoid. Lemma 3.11. Let P be a tableau such that P t is semistandard and row(P ) is 321-avoiding. Let x be an integer such that row(P ) · x is 321-avoiding. Then, Proof. To prove (3.1), let us first prove the following statements for all row tableaux P : • With the assumptions in lemma, if insertion terminates at row P while computing P ← x, then row(P ← x) ≡ H 0 row(P ) · x.
• With the assumptions in lemma, if y is bumped from row P and P changes to P ′ while computing P ← x, then Assume that insertion terminates at row P while computing P ← x. Then, Case 1 must have occurred and P changes to P ′ , where P ′ is P appended by a cell containing x. Hence, we have Assume that y is bumped from row P and P changes to P ′ while computing P ← x. Then, either Case 2 or Case 3 must have occurred.
If Case 2 occurs at P , then x ∈ P and there is a y ∈ P with y > x; furthermore, y is the smallest value with such property. Write P as AyB, where A and B are the row subtableaux of P formed by entries to the left and to the right of y, respectively. Then, P ← x is the tableau with row Axb followed by row y. As x / ∈ P , we have max(A) < x < y < min(B). Hence by commutativity relations, for all z ∈ B, we have z · x ≡ H 0 x · z and for all z ∈ A, we have z · y ≡ H 0 y · z, so that regarding A and B as Hecke words, we obtain It follows that If Case 3 occurs at P , then x, y ∈ P with y being the smallest value such that [y, x] ⊆ P . Write P as ABC, where B = [y, x], A and C are respectively the row subtableaux of P formed by entries to the left and to the right of B. Then, P ← x is the tableau with row ABC followed by row y. As row(P ) · x was assumed to be 321-avoiding, x + 1 / ∈ P . Furthermore, by minimality of y, y > max(A) + 1. Hence, by commutativity relations, for all z ∈ A, we have z · y ≡ H 0 y · z and for all z ∈ C, we have x · z ≡ H 0 z · x, so that Moreover, by using the relations p − 1 p p = p − 1 p − 1 p, we have y · B ≡ H 0 B · x. It follows that Hence, the two statements above hold for all row tableaux P .
We are now ready to prove (3.1) in full generality. The result follows once we prove by induction on the number of rows of P , with the given setup above, that the following statements hold: • If the insertion terminates within tableau P while computing P ← x, then • If y is bumped from tableau P and P changes to P ′ while computing P ← x, then row(P ← x) ≡ H 0 y · row(P ′ ).
Indeed, if P is a (possibly empty) row tableau, then we are done by the two previous statements that have been proved. Let k 1 be an arbitrary integer. Assume that both statements mentioned above hold for all such tableaux P with k rows.
Let P be a tableau with k+1 rows with the setup as above. Then, we may consider the subtableau P * formed from its first k rows and denote the final row as R. Note that row(P ) = row(R) · row(P * ) and row(R) is a 321-avoiding Hecke word.
Assume that the changes from P to P ← x involve at most the first k rows of P . Then P ← x is the same tableau as P * ← x with an extra row R, so that by the inductive hypothesis, Now assume that the changes from P to P ← x involves all k + 1 rows of P . Let P ′ be the resulting tableau after performing these changes on P * and let y be the entry bumped from the final row of P * . Then, P ← x is the tableau obtained by concatenating tableau R ← y after P ′ .
If the insertion terminates at row R, then by the previous statements for all row tableaux and the inductive hypothesis, we obtain Otherwise, if the insertion bumps z from R and R changes to R ′ while computing R ← y, then it holds that the insertion bumps z from P while computing P ← x. In this case, if we denote P ′′ as the tableau P ′ concatenated by row R ′ , then This completes the induction.
Remark 3.12. Observe that the assumption that row(P ) is 321-avoiding implies that row(R) is 321-avoiding for each row R of P . Moreover, in the proof of Lemma 3.11, if x is to be inserted into row R of P when computing P ← y and x ∈ R, then the extra assumption that row(P ) · x is 321-avoiding implies that R does not contain x + 1. Lemma 3.13. Let P be a tableau such that P t is semistandard and row(P ) is 321-avoiding. Let x, x ′ be integers such that the Hecke words row(P ) · x and row(P ) · xx ′ are 321-avoiding.
Denote the insertion paths of P ← x and (P ← x) ← x ′ as π and π ′ respectively. Also, suppose that P ← x and (P ← x) ← x ′ introduce boxes B and B ′ respectively. Then the following statements about ⋆-insertion are true: (1) If x < x ′ , then π ′ is strictly to the right of π. Moreover, B ′ is strictly to the right of and weakly below B. (2) If x x ′ , then π ′ is weakly to the left of π. Moreover, B ′ is weakly to the left of and strictly above B.
Proof. Similar to Fulton's proof [Ful96] of the Row Bumping Lemma, we will keep track of the entries as they are bumped from a row. Consider a row R of tableau P and suppose that u and u ′ are to be inserted into R when computing P ← x and (P ← x) ← x ′ respectively, where u < u ′ . Denote by C (similarly C ′ ) the box in π (similarly π ′ ) that is also in R.
Case 1: x < x ′ . We will prove that the following assertions hold for R: (a) If the insertion terminates at R while computing P ← x, then the insertion terminates at R while computing (P ← x) ← x ′ . (b) C ′ is strictly to the right of C. Note that the insertion terminates at R when computing P ← x precisely when Case 1 of the ⋆-insertion occurs at R. Box C containing u is appended at the end of R. As u ′ > u, Case 1 occurs again at R with box C ′ containing u ′ appended to the right of C, so bumping does not occur at R when computing (P ← x) ← x ′ . This proves (a) and simultaneously, (b) for this case.
Let us assume that bumping occurs at R with y bumped out when computing P ← x. Case A: If y is bumped from R because Case 2 occurs, the insertion at row R introduced to box C ′ occurs strictly to the right of C (containing u) because: (i) If u ′ > max(R), then box C ′ containing u ′ is appended to the end of R by Case 1. In particular, C ′ appears strictly to the right of C. (ii) Otherwise, since u ′ > u, the letter u ′ is inserted into a box C ′ strictly to the right of C with y ′ bumped out. If u ′ / ∈ R, Case 2 occurs and y ′ > y because C ′ and C originally contained y ′ and y respectively. Else, u ′ ∈ R and Case 3 occurs. Suppose that [y ′ , u ′ ] is the longest interval of consecutive integers contained in R. Since box C that originally contained y is strictly to the left of C ′ , we have u < y < u ′ . Therefore, [u, u ′ ] cannot be contained in R, so y < y ′ . Case B: Otherwise, y is bumped from R because Case 3 occurs when computing P ← x and [y, u] is the longest interval of consecutive integers contained in R by Remark 3.12. The insertion at row R introduced to box C ′ occurs strictly to the right of C (containing u) because: (i) If either u ′ > max(R) or u ′ / ∈ R, then by similar arguments as in Case A(i) and Case A(ii), C ′ appears to the right of C. Furthermore, in the latter situation, by a similar argument in Case A(ii), we have y < y ′ . (ii) Otherwise, u ′ ∈ R and Case 3 occurs. As u ′ > u, u ′ is inserted into box C ′ strictly to the right of C with y ′ bumped out. In addition, [y ′ , u ′ ] is the longest interval of consecutive integers contained in R. As row(R) is 321-avoiding before computing P ← x, u+1 / ∈ R. Hence [u, u ′ ] cannot be contained in R. It follows that y u < u + 1 < y ′ . Note that in the arguments above, we have also shown that if y and y ′ are bumped from R when computing P ← x and (P ← x) ← x ′ respectively, then y < y ′ . It follows that we may apply similar arguments in the rows following R. Since assertion (b) now holds for all rows, we conclude that π ′ is strictly to the right of π. In addition, π ′ cannot continue after π ends because of assertion (a). Considering that π ′ goes weakly left by Lemma 3.10, we conclude that box B ′ is strictly to the right of and weakly below B.
Case 2: x x ′ . We will prove that the following assertions hold for R: (1) If the insertion terminates at R while computing P ← x, then bumping occurs at R while computing (P ← x) ← x ′ . (2) C ′ is weakly to the left of C. If the insertion terminates at row R when computing P ← x, then Case 1 occurs and box C containing u is appended at the end of R. If u ′ ∈ R, Case 3 occurs at R with y ′ u ′ u bumped out. Furthermore, box C ′ containing u ′ is weakly to the left of C. If u ′ / ∈ R, Case 2 occurs at R with y ′ > u ′ bumped out and u ′ < u. We have y ′ u by minimality of y ′ , so that box C ′ is weakly to the left of C. In either of the subcases, bumping occurs at R when computing (P ← x) ← x ′ . This proves (a) and simultaneously, (b) for this case.
Let us assume that bumping occurs at R with y bumped out when computing P ← x. Case A: If y is bumped from R because Case 2 occurs when computing P ← x, the insertion at row R introduced to box C ′ occurs weakly to the left of C (containing u) because: (i) If u ′ / ∈ R, then u ′ is inserted into box C ′ containing y ′ by Case 2, while bumping out this y ′ . As u ′ < u, we have y ′ u < y and that C ′ appears weakly to the left of C.
(ii) Otherwise, u ′ ∈ R and Case 3 occurs. The letter u ′ is inserted into box C ′ weakly to the left of C as u ′ u. In addition, if [y ′ , u ′ ] is the longest interval of consecutive integers in R, then y ′ is bumped out. Furthermore, we have y ′ < y as C, which originally contained y before computing P ← x, is to the right of the box containing y ′ .
Case B: Otherwise, y is bumped from R because Case 3 occurs when computing P ← x. Let [y, u] be the longest interval of consecutive integers that is contained in R. The insertion at row R introduced to box C ′ occurs weakly to the left of C (containing u) because: (i) If u ′ / ∈ R, then u ′ < u, u ′ is inserted into box C ′ containing y ′ and y ′ is bumped out by Case 2. As row(P ) · x is 321-avoiding, in particular row(R) is 321-avoiding. Hence u ′ < y, so that C ′ is weakly to the left of box containing y (hence also weakly to the left of C). Furthermore, we have y ′ y by the minimality of y ′ . (ii) If u ′ ∈ R, then either u ′ = u or u < u ′ . The former case is easy as Case 3 occurs again with u ′ inserted into C ′ = C and y ′ = y is bumped out. If u < u ′ , then as row(P ) · x is 321-avoiding, row(R) is 321-avoiding, so that u ′ < y − 1. It follows that C ′ is strictly to the left of box containing y (hence also strictly to the left of C). Furthermore, we have y ′ u ′ < y − 1 < y.
Note that in the arguments above, we have also shown that if y and y ′ are bumped from R when computing P ← x and (P ← x) ← x ′ respectively, then y y ′ . It follows that we may apply similar arguments in the rows following R. Since assertion (b) now holds for all rows, we conclude that π ′ is weakly to the left of π. In addition, π ′ must continue after π ends because of assertion (a). Considering that π ′ goes weakly left by Lemma 3.10, we conclude that box B ′ is weakly to the left of and strictly above B.
Let U be a tableau such that U t is semistandard and the Hecke word row(U ) is 321-avoiding. We describe the reverse row bumping for ⋆-insertion of U as follows. Locate an inner corner of U and remove entry y from that row. Perform the following operations until an entry is bumped out of the bottommost row. Suppose that we are reverse bumping y into a row R. If y / ∈ R, find the largest x ∈ R with x < y; insert y and bump out x. Otherwise, y ∈ R, so find the largest x ∈ R such that [y, x] is the longest interval of consecutive integers. In this case, row R remains unchanged but x is bumped out. Then reverse bump x into the next row below unless there is no further row below. In this case, terminate and return the resulting tableau as T along with the bumped entry x. It is straightforward to see that reverse row bumping specified above reverses the bumping process specified by the ⋆-insertion.  Let x 1 < x 2 < . . . < x k (similarly x k . . . x 2 x 1 ) be integers such that row(T ) · x 1 x 2 . . . x i is 321-avoiding for all 1 i k. Then, the collection of boxes added to T to form the tableau U = ((T ← x 1 ) ← x 2 ) · · · ← x k has the property that no two boxes are in the same column (similarly row).
Conversely, if U is a tableau of shape µ such that λ ⊆ µ and µ/λ consists of k boxes with no two boxes in the same column, i.e, a horizontal strip of size k (similarly row, i.e., a vertical strip of size  k), then there is a unique tableau T of shape λ and unique integers x 1 < x 2 < · · · < x k (similarly x k · · · x 2 x 1 ) such that Proof. Assume that x 1 < x 2 < · · · < x k . By statement (1) of Lemma 3.13, the sequence of added boxes in U = ((T ← x 1 ) ← x 2 ) · · · ← x k moves weakly below and strictly to the right when computing U . In particular, no two of the added boxes can be in the same column.
To recover the required tableau T and integers x 1 < x 2 < · · · < x k , perform reverse row bumping on the boxes specified by the shape µ/λ within U starting from the rightmost box, working from right to left. The tableau T and the integers x 1 , x 2 , . . . , x k are uniquely determined by the operations. Moreover, by Lemma 3.13, the integers x k , x k−1 , . . . , x 1 obtained in the given order of operations satisfy x 1 < x 2 < · · · < x k .
Now assume x k · · · x 2 x 1 . By statement (2) of Lemma 3.13, the sequence of added boxes moves strictly above and weakly to the right when computing U . In particular, no two of the added boxes can be in the same row.
Similarly, one may perform reverse row bumping on the boxes specified by the shape µ/λ within U starting from the topmost box, working from top to bottom. Again, the operations uniquely determine the tableau T and the integers x 1 , x 2 , . . . , x k . Moreover, by Lemma 3.13, the integers x k , x k−1 , . . . , x 1 obtained in the given order of operations satisfy x k · · · x 2 x 1 .
Finally, note that in a decreasing Hecke biword [k, h] t , where h = h m . . . h 2 h 1 , entries within a fixed a i are inserted in increasing order. It follows that the collection of all boxes with label i form a horizontal strip within the tableau Q. Collecting all these horizontal strips with values i from m to i in order by using the converse recovers Q, implying that Q is semistandard. Proof. By successive applications of Lemma 3.11, if (P, Q) = ⋆([k, h] t ), then as h is a 321-avoiding Hecke word, row(P ) is also 321-avoiding. Hence, using Lemma 3.10 and Corollary 3.15, ⋆-insertion is a well-defined map from the set of all 321-avoiding, decreasing Hecke biwords to the set of all pairs of tableaux (P, Q) of the same shape with both P t , Q semistandard and row(P ) being a 321-avoiding Hecke word.
It remains to show that the ⋆-insertion is an invertible map. Assume that P and Q are tableaux of the same shape with both P t , Q semistandard and row(P ) being a 321-avoiding Hecke word. Since Q is semistandard, the collection of boxes with the same entry form a horizontal strip. Starting with the largest such entry m, perform reverse row bumping with the boxes in the strip from right to left. By Lemma 3.13, this recovers the entries in h m in decreasing order. Repeating this procedure in decreasing order of entries recovers h = h m . . . h 2 h 1 , which automatically yields a decreasing Hecke biword [k, h] t . Furthermore, by repeated applications of Lemma 3.11, since row(P ) was 321-avoiding, then the reverse word of h is 321-avoiding, so that h is 321-avoiding too.
Finally, by repeated applications of the converse stated in Corollary 3.15, the recovered decreasing Hecke biword [k, h] t is unique.

Properties of the ⋆-insertion
In this section, we show that the ⋆-insertion intertwines with the crystal operators. More precisely, the insertion tableau remains invariant on connected crystal components under the ⋆insertion as shown in Section 4.1 by introducing certain micro-moves. In Section 4.2, it is shown that the ⋆-crystal on H m,⋆ intertwines with the usual crystal operators on semistandard tableaux on the recording tableaux under the ⋆-insertion. In Section 4.3, we relate the ⋆-insertion to the uncrowding operation.
4.1. Micro-moves and invariance of the insertion tableaux. In this section, we introduce certain equivalence relations of the ⋆-insertion in order to establish its relation with the ⋆-crystal. From now on we are focusing on the sequence in the insertion order. Since the decreasing Hecke factorizations is inserted from right to left, we look at the Hecke word read from right to left. (1) Knuth moves, for x < z < y: (I1) xyz ∼ yxz (I2) zxy ∼ zyx (2) Weak Knuth moves, for y > x + 1: (II1) xyy ∼ yxy (II2) xxy ∼ xyx (3) Hecke move, for y = x + 1: (III) xxy ∼ xyy Note that the micro-moves preserve the underlying Hecke words. Next, we use the following notation on ⋆-insertion tableaux. For a single-row increasing tableau R, let R x denote the first row of the tableau R ← x and let R(x) denote the output of the ⋆-insertion from the first row. If the ⋆-insertion outputs a letter, then denote it by R(x); if x is appended to the end of the row R, then the output R(x) is 0, which can be ignored. We always have x · 0 ∼ x ∼ 0 · x. and R(7) = 6. Furthermore, the first row of R 7 ← 9 is R 7,9 = 1 3 4 6 7 8 9 and R 8 (9) = 0.
Lemma 4.4. Let R be a single-row increasing tableau, and x, y, z be letters such that row(R)·x·y ·z is 321-avoiding. Let x ′ , y ′ , z ′ be letters such that xyz ∼ x ′ y ′ z ′ . Following the above notation, we have Proof. Let R be a single-row increasing tableau and M be the largest letter in R. First note that if a ∈ R and row(R) · a is 321-avoiding, then a + 1 / ∈ R, see also Remark 3.12. There are five types of equivalence triples, so we discuss them in 3 groups.
Case (1A): M < x < z y. In this case, the first resulting tableau is R xyz = R x z and the outputs are R(x) = R x (y) = 0 and R xy (z) = y. The second resulting tableau is R yxz = R x z and the outputs are R(y) = 0 = R yx (z) and R y (x) = y. So we have R xyz = R yxz and also 0 · 0 · y ∼ 0 · y · 0.
Case (1B): x M < z y. In this case, we have R xy = R yx and R(x) = R y (x) since y is just appended to the end of R and does not influence how x is inserted. This gives R xyz = R yxz . The related outputs are R x (y) = R(y) = 0, R xy (z) = R yx (z) = y. Thus R(x) · 0 · y ∼ 0 · R(x) · y.
Case (1C): x < z M < y. In this case, we also have that R xy = R yx and R(x) = R y (x), for the same reason as case (1B). Thus we have R xyz = R yxz and R xy (z) = R yx (z). Since we have Case (1D): x < z y M . If x is the maximal letter in R x , then it follows as case (1B). Otherwise, this case needs further separation into subcases.
(2) If R(x) > y, let the letter to the right of R(x) in R be R(x) → . Then both R xyz and R yxz are obtained by replacing R(x) with x and R(x) → with z. For the output, we have R(x) = R(y), R x (y) > R(y), R y (x) = y, R yx (z) = R x (y) and R xy (z) = y. Since y < R(x) < R x (y), we have that Case 1D-(ii): x ∈ R, y / ∈ R. Then R(x) x, R(y) > y and x + 1 / ∈ R. In this case, we have R x (y) = R(y) and R y (x) = R(x), thus R xy = R yx , R xy (z) = R yx (z) and R xyz = R yxz . Since x + 1 / ∈ R, we have R xy (z) > x + 1. This implies R(x) x < R xy (z) y < R(y), thus R(x)R x (y)R xy (z) ∼ R(y)R y (x)R yx (z) as it is a type (I1) move.
(1) If R(x) = y, denote the box to the right of y in y as y → . Note that y → > y + 1. Then R x (y) = y → , R xy (z) = y, R y (x) = y and R yx (z) = y → . Note y − 1 / ∈ R, otherwise R(x) y − 1. Thus R(y) = y. Both R xyz and R yxz are obtained by replacing y ∈ R with x and y → with z, so R xyz = R yxz . The outputs R(x)R x (y)R xy (z) = yy → y ∼ yyy → = R(y)R y (x)R yx (z) as it is a type (II2) move.
(2) Suppose R(x) < y and R(x) = R(y). Then [R(x), y] ⊂ R and R x (y) = R(x) + 1. Since R y = R and R xy = R x , we have that both R xy and R yx equal R x and furthermore R y (x) = R(x). Note that z can either be equal to y or z < R x (y), otherwise z ∈ R xy and z + 1 ∈ R xy , which will give us a braid from row(R xy ) · z. Thus we have R xy (z) = R yx (z) = R(x) + 1. In either case, the outputs are R(x)R x (y)R xy (z) = R(x)(R(x) + 1)(R(x) + 1) ∼ R(x)R(x)(R(x) + 1) = R(y)R y (x)R yx (z) as they are type (III) moves.
(3) Suppose R(x) < y and R(x) < R(y). Then R(y) > R(x) + 1 and R x (y) = R(y). Similar to the previous case, both R xy and R yx are equal to R x , and z is either y or z < R(y). In either case, R xy (z) R(y).
Then the outputs are R(x)R x (y)R xy (z) = R(x)R(y)R x (z) ∼ R(y)R(x)R x (z) = R(y)R y (x)R yx (z) as they are type (I1) or (II1) moves.
Then R x (y) = R(y), R y (x) = R(x) and R xy = R yx = R. Since z > x and x + 1 / ∈ R, we have R(z) > x + 1 R(x) + 1. By similar reasons to the previous two subcases of Case 1D-(iii), z can either be y or z < R(y) in order to avoid a braid in row(R xy )z. So we have R xy (z) R(y). Then the outputs are R(x)R x (y)R xy (z) = R(x)R(y)R(z) ∼ R(y)R(x)R(z) = R(y)R y (x)R yx (z) as they are type (I1) moves.

Cases (I2) and (II2):
We have z < x < y, or z = x < y and y > z + 1. In both cases Case (2A): M < x < y, then R(x) = R x (y) = 0. R xy = R x y is obtained by appending x and y to the end of R. Since x ∈ R x and z x < y, we have R xy (z) = R x (z). Moreover, R xzy is obtained by appending y to the end of R xz and hence R xyz = R xzy . The outputs are Case (2C): z x < y M , then we have R x (z) x. We discuss the following subcases.
Case 2C-(i): x, y / ∈ R, then we have R(x) > x and R x (y) > y. Since y > x and x replaces R(x) in R, we have R x (y) > R(x) from row strictness. Since R x (y) > R(x) and R x (z) x, we have R xy (z) = R x (z) and R xz (y) = R x (y). Furthermore, R xyz = R xzy . Moreover, we have Case 2C-(iii): x / ∈ R, y ∈ R. Then R xy = R x , R(x) > x and R x (y) y. Let the letter to the right of R(x) in R be R(x) → . Then R(x) → > R(x) > x implies R(x) → > x + 1. This also shows that x + 1 / ∈ R x and thus R x (y) > R(x). Since R xy = R x and R xzy = R xz , we have R xyz = R xz = R xzy . Since R x (z) x < R(x) < R x (y), we have R x (y) > R x (z) + 1. Since z x, we also have that R x (y) = R xz (y). Thus the outputs are
Case (3A): x > M . Then R x is obtained by appending x to the end of R and R(x) = 0. Also R xx = R x with output R x (x). Note R x,x+1 (x + 1) = R x (x). Both R xx,x+1 and R x,x+1,x+1 are obtained by appending x + 1 to the end of R x , thus they are the same. The outputs are Case (3B): x M, x + 1 > M . Both R xx,x+1 and R x,x+1,x+1 are obtained by appending x + 1 to the end of R x , so they are equal. Since x ∈ R x , we have R x,x+1 (x + 1) = R x (x). Thus the outputs are R(x)R x (x)R xx (x + 1) = R(x)R x (x)0 ∼ R(x)0R x,x+1 (x + 1).

Case (3C):
x + 1 M . It is clear that x ∈ R x . If x is the maximal letter in R x , then the rest follows as case (3B).
Otherwise, let x → be the letter to the right of x in R x . Since x ∈ R x , we must have x + 1 / ∈ R x , thus x → > x + 1. Moreover, we have R xx = R x , R x (x + 1) = R xx (x + 1) = x → . Since R x,x+1 is obtained from R x by replacing x → with x + 1 and x, x + 1 ∈ R x,x+1 , we have R x,x+1 (x + 1) = R x (x). Both R xx,x+1 and R x,x+1,x+1 are obtained from R x by replacing x → with x + 1, thus they are the same. Furthermore, since R x (x) x and x → > x + 1, we have that Proof. Let P be a ⋆-insertion tableau. By Lemma 3.10, P t is a semistandard tableau. Let the rows of P be R 1 , . . . , R ℓ . Then each row is strictly increasing. The row R j is considered to be empty for j > ℓ.
Let x 1 , y 1 , z 1 and x ′ 1 , y ′ 1 , z ′ 1 be letters such that x 1 y 1 z 1 ∼ x ′ 1 y ′ 1 z ′ 1 and row(P ) · x 1 · y 1 · z 1 is 321avoiding. Let the output of the ⋆-insertion algorithm of P ←  In the next couple of lemmas, we prove that the crystal operators f ⋆ k act by a composition of micro-moves as given in Definition 4.1. More precisely, for a 321-avoiding decreasing factorization h, we have h rev ∼ f ⋆ k (h) rev as long as f ⋆ k (h) = 0, where h rev is the reverse of h. Remark 4.7. By Definition 2.3 and Remark 2.7, there are two cases for the k-th and (k + 1)-st factors under the crystal operator f ⋆ k , where x is the largest unpaired letter in the k-th factor, w i , v i > x and u i , b i < x: where v s = x + 1.
where v s = w p = x + 1. In both cases, u i < x − 1 since if u 1 = x − 1 then b 1 = x − 1 due to the fact that x is unbracketed; but this would mean that the word is not 321-avoiding. We also notice that since all u i are paired with some b j , we have that t q and b i u i . Similarly, all v i are paired with some w j , so we have that p s and v i w p−s+i . Let u denote the sequence u 1 . . . u q and let b denote the sequence b 1 . . . b t .
(1): When b i−1 > b i + 1 or u i < b i , the result follows directly. Consider the case that u i = b i = a and b i−1 = b i + 1 = a + 1 for some letter a. Since a = u i < u i−1 b i−1 = a + 1, we must have u i−1 = a + 1. Let c be the largest letter such that [a, c] ⊆ b. Then c a + 1 and c + 1 / ∈ b. Moreover, since all u i are paired, u i b i and u j−1 > u j , it is not hard to see that [a, c] ⊆ u and c, c − 1 ∈ u. Since c + 1 / ∈ b, we can use commutativity to move c ∈ b to the left and obtain a subword c(c − 1)c, which contradicts that the original Hecke word is 321-avoiding.
(2): The proof is almost identical to the first part. When w p−s+i > w p−s+i+1 + 1 or v i > w p−s+i , the result follows.
Consider the case w p−s+i = w p−s+i+1 + 1 = a + 1 and v i = w p−s+i = a + 1 for some letter a. Since a = w p−s+i+1 v i+1 < v i = a + 1, we must that v i+1 = a. Let c be the smallest letter such that [c, a + 1] ⊆ w. Then c a and c − 1 / ∈ w. Moreover, since all v i are paired, v j w p−s+j and v j+1 < v j , we can see that [c, a + 1] ⊆ v and c, c + 1 ∈ v. Since c − 1 / ∈ w, we can use commutativity to move c ∈ w to the right and form a subword c(c + 1)c, which contradicts that the original Hecke word is 321-avoiding.
We now summarize several observations that will be used later. (1) For 1 i q, and v i > w p−s+i+1 + 1 by Lemma 4.8.
Remark 4.10. When v s = x + 1, we have the following equivalence relations: (1) 1 i s, xv i w p ∼ v i xw p , since x < w p v s and v s > x + 1.
Proof. With the equivalence relations from Remark 4.9 (1)-(5), we can make the sequences of equivalence moves as follows: Proof. With the equivalence relations from Remark 4.9 (6)-(7), we can make the following equivalence moves: Lemma 4.13. We have Proof. With the equivalence relations from Remark 4.9 (6), (7) and (11), we can make the following equivalent moves: Lemma 4.14. When v s = x + 1, we have Proof. With the equivalence relations from Remark 4.9 (6)-(7) and Remark 4.10 (1), we can make the following equivalence moves: Proof. With the equivalence relations from Remark 4.9 (4), (5), (8)-(9) and Remark 4.10 (2), we can make the following equivalence moves: Proof. With the equivalence relations from Remark 4.9 (1), (3), (5) and (10) we have the following equivalence moves:  . . b t ), where x is the largest unpaired letter in h k . Then by Lemmas 4.11 and 4.12, we have the following sequence of equivalence moves: . By Lemmas 4.14 and 4.15, we have Case (2): Then by Lemmas 4.13 and 4.16, we have Therefore, we have shown that in both cases, f ⋆ k keeps the underlying reverse Hecke words equivalent.  Proof. Without loss of generality, we may assume that r = 1. We prove the statement by induction on m. The case m = 1 is trivial. Let m 1 be arbitrary and suppose that the statement holds for this m. We prove the statement for m + 1. We need to insert P ⋆ (h) ← h m+1  Proof. The proof is done by induction on subtableaux of T similarly to the proof of Theorem 3.5.
For a given step in the insertion process, suppose that the entries of T that are involved so far form a nonempty subtableau T ′ of T with shape µ containing cell (1, 1). Furthermore, assume that the insertion and recording tableau at the corresponding step are P (T ′ ) and Q(T ′ ). Then they both have shape µ, and the entry of cell (i, j) of P (T ′ ) is ℓ + j − µ ′ j + i − 1. In addition, Q(T ′ ) = T ′ , where µ ′ is the conjugate of the partition µ and ℓ := λ ′ 1 = ℓ(λ). Note that we do not encounter Case (1) in the proof of Theorem 3.5. All other arguments still hold since for every insertion the letter is not contained in the row it is inserted into, that is, the insertion always bumps the smallest letter that is greater than itself. Thus we omit the detail of the proof.

4.2.
The ⋆-insertion and crystal operators. In this section, we prove that the ⋆-insertion and the crystal operators on 321-avoiding decreasing Hecke factorizations and semistandard Young tableaux intertwine.
. In other words, the following diagram commutes: The crystal operator f ⋆ i acts only on factors h i+1 and h i . Hence it suffices to prove the statement for h = h i+1 h i . . . h 1 with i + 1 factors.

4.3.
Uncrowding skew set-valued tableaux. Buch [Buc02] introduced a bijection from a setvalued tableau of straight shape to a pair (P, Q), where P is a semi-standard tableau and Q is a flagged increasing tableaux. The map involves the use of a dilation operation [BM12,RTY18] which can be defined equally to act on skew set-valued tableaux. Chan and Pflueger [CP19] recently studied the operation in this more general context. We review here the results needed for our purposes.
Let λ and µ be partitions such that λ ⊆ µ and λ 1 = µ 1 . A flagged increasing tableau (introduced in [Len00] and called elegant fillings by various authors [Len00, LP07, BM12, Pat16]) is a row and column strict filling of the skew shape µ/λ such that the positive integers entries in the i-th row of the tableau are at most i − 1 for all 1 i ℓ(µ). Denote the set of all flagged increasing tableaux of shape µ/λ by F µ/λ .
We use multicell to refer to a cell in a set-valued tableau with more than one letter.
Definition 4.24. For a skew shape λ/µ, the uncrowding operation is defined on T ∈ SVT(λ/µ) as follows: identify the topmost row r in T containing a multicell. Let x be the largest letter in row r which lies in a multicell; delete this x and perform RSK row bumping with x into the rows above. The resulting tableau is the output of this operation. Note that its shape differs from λ/µ by the addition of one cell. The uncrowding map, denoted uncrowd, is defined as follows. Let T ∈ SVT(λ/µ) with ex(T ) = ℓ.
• Start with P 0 = T and Q 0 = F , where F is the unique flagged increasing tableau of shape λ/λ. • For each 1 i ℓ, P i is obtained from P i−1 by successively applying the uncrowding operation until no multicells remain. Each operation involves the addition of cell C to form P i by first deleting an entry in cell B of P i−1 ; this is recorded by adding a cell with entry k to Q i−1 at the same position as C, where k is the difference in the row indices of cells B and C.
Proof. Chan and Pflueger [CP19] prove that the image of T ∈ SVT(λ/µ) under the uncrowding map is a pair (P, Q), where P is a semi-standard tableau of shape ν/µ and Q is a flagged increasing tableau of shape ν/λ. Monical, Pechenik and Scrimshaw in [MPS18, Theorem 3.12] proved that the crystal operators on SVT m (λ) intertwine with those on SSYT m (ν) under uncrowd. Since uncrowd is defined equally on skew shapes, the result follows.

4.4.
Compatibility of ⋆-insertion with uncrowding. For a partition µ, let T µ be the unique tableau of shape µ with µ i letters i in each row i. Note that uncrowd(T µ ) = (T µ , ∅) since ex(T µ ) = 0. Proof. For T ∈ SVT m (λ/µ), let T * be the set-valued tableau of shape λ obtained from T by adding ℓ(µ) to each entry and filling in the cells of µ with T µ . By Proposition 4.21, we have (4.2) ⋆ •res(T µ ) = (P µ , T µ ) , where P µ is the semistandard tableau specified in the proof of Proposition 4.21. The claim follows by noting that res(T * ) = res(T )res(T µ ).
Definition 4.28. A modification of ⋆-insertion is defined on H * ,m as follows: for h ∈ H * ,m , let λ/µ be the shape of res −1 (h) (which is well-defined up to a shift by Proposition 2.13). For h ′ = res(T µ ), let (P * , Q * ) = ⋆(hh ′ ). Define⋆(h) = (P, Q) where P is obtained from P * by deleting all entries in cells of µ and Q is defined from Q * by deleting T µ from it and decreasing all other letters by ℓ(µ).
Note that this is well-defined by Lemma 4.27 and the fact that each h ∈ H * ,m can be associated to a skew shape λ/µ which is the shape of res −1 (h) by Proposition 2.13. Also note that⋆(h) = ⋆(h) if µ = ∅. Proof. We start by addressing the straight-shape case; for T * ∈ SVT m (λ), consider the following compositions of maps: (P ,Q) T * h (P, Q) (f k (P ),Q) f k (T * ) f ⋆ k (h) (P, f k (Q)).  Figure 3. The crystal graph for H 3 (2) restricted to Hecke factorizations with four letters.
It is not difficult to check that the operators f i and e i defined above preserve the underlying word of the Hecke factorizations whenever they do not annihilate it. Furthermore, the structure above defines an abstract, seminormal A m−1 crystal on H m (2).
We note that one may also verify that the crystal is a Stembridge crystal by checking that the axioms formulated in [Ste03] are satisfied. Figure 3 displays the crystal graph on H 3 (2) restricted to decreasing factorizations that use exactly 4 letters.

5.2.
Nonlocality. In this subsection, we show that it is impossible to construct a crystal on H m with the following properties for f i : (1) f i only changes the i-th and (i + 1)-st decreasing factors; (2) f i is determined by the first (i + 1) factors; (3) f i does not change the underlying Hecke word and its excess. The above assumptions imply that the f i operator does not change the underlying Hecke subword formed by the first (i + 1) factors. Obviously the crystal on H m (2) defined in Section 5.1 satisfies these assumptions.
(Note that s 22111 is zero in four variables and hence could be omitted). The linear term in β implies that there are two connected components with highest weight (2, 2, 2, 0) (lowest weight (0, 2, 2, 2)) for the crystal H 4 (3) with one excess. All decreasing factorizations mentioned below share the same underlying Hecke word 12132 with excess 1 and 4 factors.
There are two decreasing factorizations of weight (2, 2, 2, 0): ( )(21)(21)(32) and ( )(21)(32)(32). Focus on the connected component with highest weight ( )(21)(32)(32) and try to complete the crystal graph from top to bottom. Since the only decreasing factorization of weight (2, 2, 1, 1) with the first and second factors both being (32) is (2)(1)(32)(32), we can compute the f 3 action on this highest weight element. By some similar arguments we can fill in part of the crystal graph as indicated in Figure 4 with the above assumption. The dashed spaces are undetermined.
Yet note that the red f 2 highlighted in the graph changed the first factor from (3) to (2). Hence Condition (1) is violated, providing a counterexample that crystals with the above conditions always exist on H m (n) for n > 2.