Uncrowding algorithm for hook-valued tableaux

Whereas set-valued tableaux are the combinatorial objects associated to stable Grothendieck polynomials, hook-valued tableaux are associated to stable canonical Grothendieck polynomials. In this paper, we define a novel uncrowding algorithm for hook-valued tableaux. The algorithm"uncrowds"the entries in the arm of the hooks and yields a set-valued tableau and a column-flagged increasing tableau. We prove that our uncrowding algorithm intertwines with crystal operators. An alternative uncrowding algorithm that"uncrowds"the entries in the leg instead of the arm of the hooks is also given. As an application of uncrowding, we obtain various expansions of the canonical Grothendieck polynomials.


Introduction
Set-valued tableaux play an important role in the K-theory of the Grassmannian. They form a generalization of semi-standard Young tableaux, where boxes may contain sets of integers rather than just integers [Buc02]. In particular, the stable symmetric Grothendieck polynomial indexed by the partition λ is the generating function of set-valued tableaux where SVT(λ) is the set of set-valued tableaux of shape λ and weight(T ) is the vector with i-th entry being the number of i in T . Here |T | is the number of entries in T and |λ| is the size of λ. Stable symmetric Grothendieck polynomials G λ can be viewed as a K-theory analogue of the Schur functions s λ (while the Grothendieck polynomial is an analog of the Schubert polynomial [LS83]). Buch [Buc02] also described the structure coefficients c ν λµ , which is the coefficient of G ν in the expansion of G λ G µ in terms of set-valued tableaux, generalizing the Littlewood-Richardson rule for Schur functions.
The Grassmannian Gr(k, C n ) of k-planes in C n has a fundamental duality isomorphism Gr(k, C n ) ∼ = Gr(n − k, C n ).
This implies that the structure constants have the symmetry c ν λµ = c ν ′ λ ′ µ ′ , where λ ′ denotes the conjugate of the partition λ (see for example [Ful97,Example 9.20]). Hence one expects a ring homomorphism on the completion of the ring of symmetric function defined on the basis of stable symmetric Grothendieck polynomials τ (G λ ) = G λ ′ . The standard involutive ring automorphism ω defined on the Schur basis by ω(s λ ) = s λ ′ does not have this property [LP07] where J λ is the weak symmetric Grothendieck polynomial.
2.1. Hook-valued tableaux. A semistandard Young tableau U of hook shape is a tableau of the form where the integer entries weakly increase from left to right and strictly increase from bottom to top. Note that we use French notation for Young diagrams and tableaux throughout the paper. In this case, H(U ) = x is called the hook entry of U , L(U ) = (ℓ 1 , ℓ 2 , . . . , ℓ p ) is the leg of U , and A(U ) = (a 1 , a 2 , . . . , a q ) is the arm of U . Both the arm and the leg of U are allowed to be empty. Additionally, the extended leg of U is defined as L + (U ) = (x, ℓ 1 , ℓ 2 , . . . , ℓ p ). We denote by max(U ) (resp. min(U )) the maximal (resp. minimal) entry in U .
Definition 2.1. [Yel17] Fix a partition λ. A semistandard hook-valued tableau (or hook-valued tableau for short) T of shape λ is a filling of the Young diagram for λ with (nonempty) semistandard Young tableaux of hook shape such that: (i) max(A) min(B) whenever the cell containing A is in the same row, but left of the cell containing B; (ii) max(A) < min(C) whenever the cell containing A is in the same column, but below the cell containing C. The set of all hook-valued tableaux of shape λ (respectively, with entries at most m) is denoted by HVT(λ) (respectively, HVT m (λ)).
Given a hook-valued tableau T , its arm excess is the total number of integers in the arms of all cells of T , while its leg excess is the total number of integers in the legs of all cells of T .
Remark 2.2. In the special case when a hook-valued tableau has arm excess 0, it is also called a set-valued tableau. Similarly, a multiset-valued tableau is a hook-valued tableau with leg excess 0. We use the notation SVT(λ) (resp. SVT m (λ)) and MVT(λ) (resp. MVT m (λ)) for the set of all set-valued tableaux of shape λ (resp. with entries at most m) and the set of all multiset-valued tableaux of shape λ (resp. with entries at most m), respectively.

2.2.
Crystal structure on hook-valued tableaux. Hawkes and Scrimshaw [HS20] defined a crystal structure on hook-valued tableaux. We review their definition here.
Definition 2.3 ([HS20], Definition 4.1). Let C be a hook-valued tableau of column shape. The column reading word R(C) is obtained by reading the extended leg in each cell from top to bottom, followed by reading all of the remaining entries, arranged in a weakly increasing order.
For a hook-valued tableau T , its column reading word is formed by concatenating the column reading words of all of its columns, read from left to right, that is, where ℓ is the number of columns of T and C i is the ith column of T . i < m, we employ the following pairing rules. Assign − to every i in R(T ) and assign + to every i + 1 in R(T ). Then, successively pair each + that is adjacent and to the left of a −, removing all paired signs until nothing can be paired.
The operator f i acts on T according to the following rules in the given order. If there is no unpaired −, then f i annihilates T . Otherwise, locate the cell c with entry the hook-valued tableau B = T (c) containing the i corresponding to the rightmost unpaired −. We remark that the definition of crystal operators on HVT specializes to the definition on SVT in [MPS21] or the one on MVT in [HS20] when the arm excess or leg excess of the tableaux is set to 0, respectively. . For a given cell (r, c) in row r and column c in a hook-valued tableau T , let L T (r, c) be the leg of T (r, c), let A T (r, c) be arm of T (r, c), let H T (r, c) be the hook entry of T (r, c), and let L + T (r, c) be the extended leg of T (r, c).

Uncrowding map on hook-valued tableaux
In Section 3.1, we first review the uncrowding map on set-valued tableaux. In Section 3.2, we give a new uncrowding map on hook-valued tableaux and prove some of its properties in Section 3.3. The relation to the uncrowding map on multiset-valued tableaux is given in Section 3.4. In Section 3.5, we give the inverse of the uncrowding map on hook-valued tableaux, called the crowding map. In Section 3.6, an alternative definition of the uncrowding map on hook-valued tableaux is provided.
3.1. Uncrowding map on set-valued tableaux. For set-valued tableaux, there exists an uncrowding operator, which maps a set-valued tableau to a pair of tableaux, one being a semistandard Young tableau and the other a flagged increasing tableau (see for example [Len00,Buc02,BM12,RTY18]). In this setting, the uncrowding operator intertwines with the crystal operators on setvalued tableaux and semistandard Young tableaux, respectively [MPS21].
Consider partitions λ, µ with λ ⊆ µ and λ 1 = µ 1 . A flagged increasing tableau (introduced in [Len00] and called (strict) elegant fillings by various authors [LP07,BM12,Pat16]) is a row and column strict filling of the skew shape µ/λ such that the positive integer entries in the i-th row of the tableau are at most i − 1 for all 1 i ℓ(µ), where ℓ(µ) is the length of partition µ. In particular, the bottom row is empty. The set of all flagged increasing tableaux is denoted by F. The set of all flagged increasing tableaux of shape µ/λ with λ 1 = µ 1 is denoted by F(µ/λ).
We now review the uncrowding operation on set-valued tableaux. We call a cell in a set-valued tableau a multicell if it contains more than one letter.
Definition 3.1. Define the uncrowding operation on T ∈ SVT(λ) as follows. First identify the topmost row r in T with a multicell. Let x be the largest letter in row r that lies in a multicell; remove x from the cell and perform RSK row bumping with x into the rows above. The resulting tableau, whose shape differs from λ by the addition of one cell, is the output of this operation.
The uncrowding map on set-valued tableaux is defined as follows. Let T ∈ SVT(λ) with leg excess ℓ.
(1) Initialize P 0 = T and Q 0 = F 0 , where F 0 is the unique flagged increasing tableau of shape λ/λ. (2) For each 1 i ℓ, P i is obtained from P i−1 by applying the uncrowding operation. Let C be the cell in shape(P i )/shape(P i−1 ). If C is in row r ′ , then F i is obtained from F i−1 by adding cell C with entry r ′ − r.
It was proved in [Buc02, Section 6] that U SVT in (3.1) is a bijection. Monical, Pechenik and Scrimshaw [MPS21] proved that U SVT intertwines with the crystal operators on set-valued tableaux (see also [MPPS20]). A similar uncrowding algorithm for multiset-valued tableaux was given in [HS20, Section 3.2].
3.2. Uncrowding map on hook-valued tableaux. In [HS20], the authors ask for an uncrowding map for hook-valued tableaux which intertwines with the crystal operators. Here we provide such an uncrowding map by uncrowding the arm excess in a hook-valued tableaux to obtain a set-valued tableaux. An alternative obtained by uncrowding the leg excess first is given in Section 3.4.
Definition 3.2. The uncrowding bumping V b : HVT → HVT is defined by the following algorithm: (1) Initialize T as the input.
(2) If the arm excess of T equals zero, return T.
(3) Else, find the rightmost column that contains a cell with nonzero arm excess. Within this column, find the cell with the largest value in its arm. (In French notation this is the topmost cell with nonzero arm excess in the specified column.) Denote the row index and column index of this cell by r and c, respectively. Denote the cell as (r, c), its rightmost arm entry by a, and its largest leg entry by ℓ. (4) Look at the column to the right of (r, c) (i.e. column c + 1) and find the smallest number that is greater than or equal to a.
• If no such number exists, attach an empty cell to the top of column c + 1 and label the cell as (r, c + 1), wherer is its row index. Let k be the empty character.
• If such a number exists, label the value as k and the cell containing k as (r, c + 1) wherẽ r is the cell's row index. We now break into cases: (a) Ifr = r, then remove a from A T (r, c), replace k with a, and attach k to the arm of A T (r, c + 1).
remove a from A T (r, c), insert (a, ℓ] ∩ L T (r, c) into L T (r, c + 1), replace the hook entry of (r, c + 1) with a, and attach k to A T (r, c + 1). (5) Output the resulting tableau. See Figures 1 and 2   Proof. It suffices to check that V b preserves the semistandardness condition of both the entire hookvalued tableau and the filling within each cell. We break into two cases depending on whether Step (4)a or (4)b in Definition 3.2 is applied. Case 1: Assume Step (4)a is applied. To verify semistandardness within each cell, it suffices to check cells (r, c) and (r, c+ 1). The semistandardness within cell (r, c) is clearly preserved as the only change to the hook-shaped tableau in cell (r, c) is that an entry was removed from A T (r, c). We now check the semistandardness condition within cell (r, c + 1). We have that V b either created the cell (r, c+ 1) and inserted the number a in it or V b replaced k with a and appended k to the arm of cell (r, c + 1). In both cases, the tableau in cell (r, c + 1) is a semistandard hook-shaped tableau. In the second case this is true since k is weakly greater than H T (r, c + 1) and k is the smallest number weakly greater than a in column c + 1. We now check the semistandardness of the entire tableau. Note that it suffices to check the semistandardness in rowr and column c + 1. Sincer < r, the semistandardness in row r is preserved as a is larger than every number in (r, c) and k remains in the same cell. Also, the semistandardness in column c + 1 is preserved as k is chosen to be the smallest number in column c + 1 that is weakly greater than a. Case 2: Assume Step (4)b is applied. The semistandardness within cell (r, c) is clearly preserved as the only change to (r, c) is that entries from L T (r, c) and A T (r, c) are removed. We now check the semistandardness condition within cell (r, c + 1). If (a, ℓ] ∩ L T (r, c) = ∅, then a is weakly larger than all elements of (r, c). In this case, the semistandardness within cell (r, c + 1) follows from the argument in Case 1. If (a, ℓ] ∩ L T (r, c) = ∅, then a is not weakly larger than all elements of (r, c). After applying V b the semistandardness condition in the leg of (r, c + 1) will still hold as a < x < z for all x ∈ (a, ℓ] ∩ L T (r, c), where z is the smallest value in L T (r, c + 1). Similarly, the semistandardness condition in the arm of (r, c + 1) holds as a < k or k is the empty character. Thus, the semistandardness condition in each cell is preserved. The semistandardness of row r is preserved as all numbers strictly greater than a in (r, c) are moved to (r, c + 1) along with a. The semistandardness condition within column c + 1 is preserved as every number in (r + 1, c + 1) is strictly greater than ℓ and every number in (r − 1, c + 1) is strictly less than a.
Definition 3.4. The uncrowding insertion V : HVT → HVT is defined as A column-flagged increasing tableau is a tableau whose transpose is a flagged increasing tableau. LetF denote the set of all column-flagged increasing tableaux. LetF(µ/λ) denote the set of all column-flagged increasing tableaux of shape µ/λ. (1) Let P 0 = T and let Q 0 be the column-flagged increasing tableau of shape λ/λ.
(2) For 1 i α, let P i+1 = V(P i ). Let c be the index of the rightmost column of P i containing a cell with nonzero arm excess and letc be the column index of the cell shape(P i+1 )/shape(P i ).
Then Q i+1 is obtained from Q i by appending the cell shape(P i+1 )/shape(P i ) to Q i and filling this cell withc − c.
Define U (T ) = (P (T ), Q(T )) := (P α , Q α ). Then, we obtain the following sequence of tableaux V i b (T ) for 0 i 2 = d when computing the first uncrowding insertion:   Proof. By Lemma 3.3 and Definition 3.4, we have that V(T ) is a hook-valued tableau. Note that if the arm excess of T is nonzero, then the arm excess of V(T ) is one less than that of T . Since P (T ) = V α (T ), where α is the arm excess of T , we have that the arm excess of P (T ) is zero. Thus, P (T ) is a set-valued tableau.   j . This implies that the a when applying V b to V j b (V i (T )) is weakly smaller than the a when applying V b to V j b (V i−1 (T )). Thus, we must have r i+1 j+1 r i j+1 .
3.3. Properties of the uncrowding map. Let T be a hook-valued tableau. Define R i (T ) as the induced subword of R(T ) consisting only of the letters i and i + 1. In the next lemma, we use the same notation as in Definition 3.2. Furthermore, two words are Knuth equivalent if one can be transformed to the other by a sequence of Knuth equivalences on three consecutive letters xzy ≡ zxy for x y < z, yxz ≡ yzx for x < y z.
Lemma 3.10. For T ∈ HVT, R i (T ) = R i (V b (T )) unless T satisfies one of the following three conditions: (a) a = i or a = i + 1 and column c + 1 contains both an i and an i + 1, (b)r = r, i ∈ (a, ℓ] ∩ L T (r, c), k = i, and column c + 1 contains an i + 1, (c)r = r, a = i, i + 1 ∈ (a, ℓ] ∩ L T (r, c), and (r, c) contains another i besides a.
Proof. Let R i (T ) = r 1 r 2 . . . r m . We break into cases based on the value of a.
Case 1: Assume a = i, i + 1. Assume Step (4)a is applied by V b . If k = i, i + 1, then R i (T ) = R i (V b (T )) as the position of all letters i and i + 1 remains the same. Let k = i. We have that k is the only i in column c + 1. Hence, when k gets bumped from L T (r, c + 1) and appended to A T (r, c + 1), the relative position of k to the other letters i and i + 1 in R i (T ) does not change. Thus, R i (T ) = R i (V b (T )). Let k = i + 1. Note that column c + 1 cannot have a cell containing an i as k is the smallest number weakly greater than a. Hence, moving k from L T (r, c + 1) to A T (r, c + 1) will not change R i (T ). Therefore, we once again have that R i (T ) = R i (V b (T )). Assume Step (4)b is applied by V b . Consider the subcase when (a, ℓ] ∩ L T (r, c) = ∅. By a similar argument to the previous paragraph, we have that R i (T ) = R i (V b (T )). Next, consider the subcase when i + 1 ∈ (a, ℓ] ∩ L T (r, c). This implies that a < i and the only time i + 1 occurs in column c is in L T (r, c). Note that if an i exists in column c, it must be contained in L T (r, c). We also have that k i + 1 or k is the empty character and no cell in column c + 1 contains an i. Thus, removing (a, ℓ] ∩ L T (r, c) from L T (r, c), replacing k with (a, ℓ] ∩ L T (r, c) in L T (r, c + 1), and appending k to A T (r, c + 1) does not change R i (T ). Therefore R i (T ) = R i (V b (T )). Let i ∈ (a, ℓ] ∩ L T (r, c) and i + 1 ∈ (a, ℓ] ∩ L T (r, c). Note that the only place i + 1 can occur in column c is as H T (r + 1, c) and the only place i can occur in column c is in L T (r, c). This implies that removing (a, ℓ] ∩ L T (r, c) from L T (r, c), replacing k with (a, ℓ] ∩ L T (r, c) in L T (r, c + 1) and appending k to A T (r, c + 1) will not change R i (T ) unless both i + 1 and i show up in column c + 1. This can only occur when k = i which implies that R i (T ) = r 1 . . . i i + 1 k . . . r m and R i (V b (T )) = r 1 . . . i + 1 i k . . . r m . We see that R i (T ) and R i (V b (T )) only differ by a Knuth relation implying they are Knuth equivalent. Assume that i, i + 1 ∈ (a, ℓ] ∩ L T (r, c) = ∅. If a > i + 1 the positions of all letters i and i + 1 remain the same after V b is applied. If a < i, then the positions of all letters i and i + 1 also remain the same unless k = i or k = i + 1. In both of these special subcases, it can be checked that still Case 2: Assume a = i. Assume Step (4)a is applied by V b . If column c + 1 does not contain both an i and an i + 1, then we have R i (T ) = R i (V b (T )). However, if both an i and an i + 1 are in column c + 1, then Assume Step (4)b is applied by V b . Consider the subcase when (a, ℓ] ∩ L T (r, c) = ∅. By a similar argument to the previous paragraph, we have that R i (T ) = R i (V b (T )) unless both an i and an i + 1 are in column c + 1 in which case R i (T ) and R i (V b (T )) are only Knuth equivalent. Consider the subcase given by i + 1 ∈ (a, ℓ] ∩ L T (r, c). Note that no cell in column c + 1 can contain an i, the only cell that could contain an i + 1 in column c + 1 is (r, c + 1), and the only cell containing letters i or i + 1 in column c is (r, c). This implies that it suffices to look at the changes to (r, c) and (r, c + 1). We see that R i (T ) = r 1 .
where γ 1 is the number of letters i in cell (r, c) including a. We see that R i (T ) and R i (V b (T )) are Knuth equivalent. Consider the subcase when i + 1 ∈ (a, ℓ] ∩ L T (r, c) = ∅. We have that both i and i + 1 cannot be in a cell in column c + 1 and an i + 1 cannot be in column c. Thus applying Case 3: Assume a = i + 1. Assume Step (4)a is applied by V b . If column c + 1 does not contain both i and i + 1, then we have that R i (T ) = R i (V b (T )). However, if both i and i + 1 occur in column c + 1, then ) unless both i and i + 1 occur in column c + 1. In this exceptional case, we have that R i (T ) and R i (V b (T )) are only Knuth equivalent by a similar argument to the previous paragraph. If (a, ℓ] ∩ L T (r, c) = ∅, then k > i + 1 or k is the empty character and no cell in column c + 1 contains an i + 1. Thus Remark 3.11. In general, the full reading words are not Knuth equivalent under the uncrowding map. For example, take the following hook-valued tableau T , which uncrowds to a set-valued tableau S: The reading word changed from 4321254 to 2143254, which are not Knuth equivalent.
Proof. Since P (T ) = V s b (T ) for some s ∈ N and Knuth equivalence is transitive, we have that R i (T ) is Knuth equivalent to R i (P (T )) by the previous lemma. As f i (T ) = 0, we have that every i in R i (T ) is i-paired with an i + 1 to its left. This property is preserved under Knuth equivalence giving us that f i (P (T )) = 0. The same reasoning implies (2).
Lemma 3.13. Let T ∈ HVT. ( Proof. We are going to prove (1). Part (2) follows since e i and f i are partial inverses. Let a, ℓ, k, r, c, andr be defined as in Definition 3.
. . r ′ m be the corresponding reading words. Let (r,ĉ) denote the cell containing the rightmost unpaired i in T , wherer andĉ are its row and column index respectively. We break into cases based on the position of (r,ĉ) to (r, c).
Case 1: Assume (r,ĉ) = (r, c). We break into subcases based on how f i acts on T .
• Assume that (r + 1, c) contains an i + 1. As every entry in (r, c) must be strictly smaller than the values in (r + 1, c) and (r, c) must contain an i, we have that ℓ = i or a = i. If ℓ = i, then ℓ is i-paired with the i + 1 in (r + 1, c). Hence a is always equal to i and a must correspond to the rightmost unpaired i of T . Thus, f i acts on T by removing a from (r, c) and appending an i + 1 to A T (r + 1, c). Note that (a, ℓ] ∩ L T (r, c) = ∅ implying V b acts on T by removing a from A T (r, c), replacing k in (r, c + 1) with a, and appending k to A T (r, c + 1) wherer r. We break into subcases based upon where the values of i and i + 1 are in column c + 1 utilizing the fact that column c + 1 cannot contain an i without an i + 1 (since the arm excess of cell (r + 1, c) is zero and cell (r, c) contains the rightmost unpaired i).
Assume that column c + 1 does not contain an i. Since a corresponds to the rightmost unpaired i in T and column c + 1 does not contain an i, we have that the rightmost unpaired i in V b (T ) is precisely a in the cell (r, c + 1). Note that (r + 1, c + 1) does not contain an i + 1 in V b (T ) as k i + 1 or k is the empty character. Similarly, we have that (r, c + 2) does not contain an i. Thus, f i acts on V b (T ) by changing a to an i + 1 in (r, c + 1). We now consider V b (f i (T )). When applying V b to f i (T ), a ′ is precisely the i + 1 appended to A T (r + 1, c) and k ′ is the same as k.
Assume that column c + 1 contains both an i and an i + 1 in the same cell. Note that this implies that k = i. Since a is the rightmost unpaired i in T and the only cell in column c + 1 that contained an i + 1 or an i is (r, c + 1), we have that the rightmost unpaired i in V b (T ) is the i appended to A T (r, c + 1). Since (r, c + 1) contains an i + 1, we have that (r + 1, c + 1) cannot contain an i + 1 and (r, c + 2) cannot contain an Assume that column c + 1 contains both an i and an i + 1 in different cells. Note that this implies that k = i. Since a corresponds to the rightmost unpaired i in R i (T ) and the only i + 1 and i in column c + 1 are in cells (r + 1, c + 1) and (r, c + 1) respectively, we have that the rightmost unpaired i in R i (V b (T )) corresponds to the i appended to A T (r, c + 1). By assumption, we have that (r + 1, c + 1) contains an i + 1. Thus, f i acts on V b (T ) by removing the i from A V b (T ) (r, c + 1) and appending an i + 1 to , a ′ is precisely the i + 1 appended to A T (r + 1, c) and k ′ is the i + 1 in cell (r + 1, c + 1). Ifr ′ = r + 1, then i + 1 is weakly larger than every value in (r + 1, c). Thus, either • Assume that (r + 1, c) does not contain an i + 1 and (r, c + 1) contains an i. Under these assumptions, we have that no cell in column c can contain an i + 1. This implies that column c + 1 must contain an i + 1. The cell (r + 1, c + 1) cannot have an i + 1 as this would force (r + 1, c) to also have an i + 1. Thus, (r, c + 1) must contain an i + 1 in its leg. By our assumption we have that f i acts on T by removing the i from (r, c + 1) and appending an i + 1 to L T (r, c). We break into subcases according to where the rightmost unpaired i sits inside the cell (r, c). If the rightmost unpaired i is in H T (r, c), then a i which would either contradict the hook entry being the rightmost unpaired i or cell (r, c + 1) containing an i. Thus, we only need to consider the subcases where the rightmost unpaired i is either in the leg or arm of (r, c).
Assume that the rightmost unpaired i is in L T (r, c) for this entire paragraph. This implies that ℓ = i. Since (r, c + 1) contains an i, we have that a < i. Ifr < r, then V b acts on T by removing a from (r, c), replacing k with a in (r, c + 1), and appending k to A T (r, c + 1). Since a, k < i, we have that V b does not change position of the rightmost unpaired i. Note that (r + 1, c) still does not contain an i + 1 while (r, c + 1) still contains an i. Thus, f i acts on V b (T ) by removing the i from (r, c + 1) and appending an i + 1 to L V b (T ) (r, c). We now consider V b (f i (T )). Note that (r ′ , c ′ ), a ′ , and k ′ are the same as (r, c), a, and k respectively. Thus, V b acts in the same way as before. This gives us that . Note that (r + 1, c + 1) does not contain an i + 1 and (r, c + 2) does not contain an i because (r, c + 1) contains an i + 1. Thus, Assume that the rightmost unpaired i is in A T (r, c). This implies that a = i and forces a to correspond to the rightmost unpaired i. We also have that k is the i in (r, c + 1). Since i is weakly greater than all values in (r, c), we have that (a, ℓ]∩L T (r, c) = ∅. Thus, V b acts on T by removing a from (r, c), replacing k with a in (r, c + 1), and appending k to A T (r, c + 1). Since a was the rightmost unpaired i in T and cell (r, c + 1) contains an i+1 in its leg, we have that the rightmost unpaired As i+1 is in (r, c+1), we have that (r+1, c+1) cannot contain an i+1 and (r, c+2) cannot contain an i. This implies that f i acts on V b (T ) by changing the i in A V b (T ) (r, c + 1) to an i + 1. We now consider V b (f i (T )). We have that a ′ is the same as a and k ′ is equal to the i + 1 in (r, c + 1). Note that (a ′ , ℓ ′ ] ∩ L T (r, c) = {i + 1}. This implies that V b acts on f i (T ) by removing i + 1 from L f i (T ) (r, c) and a from A f i (T ) (r, c), replacing the i + 1 in (r, c + 1) with {i + 1, a}, and appending an i + 1 to A f i (T ) (r, c + 1). We see • Assume that (r + 1, c) does not contain an i + 1 and (r, c + 1) does not contain an i.
We break into subcases based on where the rightmost unpaired i sits inside (r, c).
Assume that the rightmost unpaired i is in the hook entry of (r, c) for the remainder of this paragraph. Note that this implies that a > i and the rightmost unpaired i in V b (T ) is still the hook entry of (r, c). We see that V b does not insert an i + 1 into (r + 1, c) nor an i into (r, c + 1). This implies that f i acts on T and V b (T ) in the same way by changing the hook entry of (r, c) into an i + 1. Next, we note that (r ′ , c ′ ), a ′ , k ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are the same as (r, c), a, k, and (a, ℓ] ∩ L T (r, c) respectively. Thus, V b acts on T and f i (T ) in the same manner without affecting the hook entry of (r, c). Therefore, we have that the actions of f i and V b on T are independent and Assume that the rightmost unpaired i is in the leg of (r, c) for the remainder of this paragraph. This implies that a = i. First, we assume that a > i orr < r. Under this extra assumption, we observe that the action of V b does not change the position of the rightmost unpaired i. Also, V b does not insert an i + 1 into (r + 1, c) nor an i into (r, c + 1). We see that f i acts on T and V b (T ) in the same way by changing the i in the leg of (r, c) into an i + 1. Next, we note that (r ′ , c ′ ), a ′ , and k ′ are the same as (r, c), a, and k respectively. If a > i, we have that a i + 1 implying that orr < r. This implies that V b acts on T and f i (T ) in the same manner and does not affect the i or i + 1 in the leg of (r, c). Therefore, we have that the actions of f i and V b on T are independent and ). Next, assume thatr = r and a < i. + 1), and appending k to A T (r, c + 1). By assumption, there was no i in (r, c + 1) to begin with. Thus, we have that the rightmost unpaired i of V b (T ) is the i in (r, c+1) that replaced k. Since k i+1 or k is the empty character, we have that the cell (r + 1, c + 1) does not contain an i + 1 and the cell (r, c + 2) does not contain an i. Hence, f i acts on V b (T ) by replacing the i in L V b (T ) (r, c + 1) with an i + 1. We now consider V b (f i (T )). We have that f i acts on T by changing the i in L T (r, c) to an i + 1. We see that a ′ and k ′ are the same as a and k respectively. Since i > a, we have that i + 1 > a or in other words Assume that the rightmost unpaired i is in A T (r, c) andr < r or (a, ℓ] ∩ L T (r, c) = ∅ for this entire paragraph. Under this assumption, f i acts on T by changing the rightmost i in the arm of (r, c) to an i+1. Also, V b acts on T by removing a from A T (r, c), replacing k in (r, c + 1) with a, and appending k to A T (r, c + 1). First, we make the additional assumption that i < a. Since we assume the rightmost unpaired i is in the arm of (r, c) and i < a, we have the rightmost unpaired i in V b (T ) is in the same position as in T . Note that the cell (r + 1, c) still does not contain an i + 1 and the cell (r, c + 1) still does not contain an i. Thus, we have that f i acts on V b (T ) by changing the rightmost i in A V b (r, c) into an i + 1. We now consider V b (f i (T )). We see that a ′ and k ′ are the same as a and k respectively. This implies that V b acts on f i (T ) by removing a from (r, c), replacing k with a in (r, c), and appending k to A f i (T ) (r, c + 1). We see ). Next, we make the assumption that a = i and column c + 1 does not contain both an i and an i + 1. We have that the rightmost unpaired i in V b (T ) is precisely the i that replaced k in (r, c + 1). We also have that k i + 1 or k is the empty character implying that the cell (r + 1, c + 1) does not contain an i + 1 and the cell (r, c + 2) does not contain an i. This implies that f i acts on V b (T ) by changing the i in L + V b (T ) (r, c + 1) to an i + 1. We now consider V b (f i (T )). We see that a ′ is the i + 1 in (r, c) created by appying f i and k ′ is the same as k. Thus, V b acts on f i (T ) by removing the i + 1 from (r, c), replacing k with an i + 1 in (r, c), and appending k to A f i (T ) (r, c + 1). We see that f i (V b (T )) = V b (f i (T )). Next, we assume that a = i and column c + 1 contains both an i and an i + 1 in the same cell. Note that this implies that k = i. Since a corresponded to the rightmost unpaired i in T and the only cell in column c + 1 that contains an i + 1 or an i is (r, c + 1), we have that the rightmost unpaired i in V b (T ) corresponds to the i appended to A T (r, c + 1). Since (r, c + 1) contains an i + 1 in V b (T ), we have that (r + 1, c + 1) cannot contain an i + 1 and (r, c + 2) cannot contain an i. Thus, f i acts on V b (T ) by changing the i in A V b (T ) (r, c + 1) to an i + 1. We now consider V b (f i (T )). We see that a ′ is the i + 1 in (r, c) obtained after applying f i and k ′ is the i + 1 in cell (r, c + 1). This implies that V b acts on f i (T ) by removing the i + 1 from (r, c), replacing k ′ with an i + 1 in (r, c + 1), and appending k ′ to A f i (T ) (r, c + 1). We see that Finally, we make the assumption that a = i and column c + 1 contains both an i and an i + 1 but in different cells. We once again have that k = i, but now we have that (r + 1, c + 1) contains an i + 1. We have that the rightmost unpaired i in V b (T ) is the i that was appended to A T (r, c + 1). Since (r + 1, c + 1) contains an i + 1, we have that f i acts on V b (T ) by removing the i from A V b (T ) (r, c + 1) and appending an i + 1 to A V b (T ) (r + 1, c + 1). We now consider V b (f i (T )). We see that a ′ is the i + 1 in (r, c) obtained after applying f i and k ′ the i + 1 in cell (r + 1, c + 1). This implies that V b acts on f i (T ) by removing the i + 1 from (r, c), replacing k ′ with an i + 1 in (r+1, c+1), and appending Assume that the rightmost unpaired i is in the arm of (r, c),r = r, and (a, ℓ]∩L T (r, c) = ∅ for this entire paragraph. First, we make the additional assumption that i < a. This gives us that V b (T ) is attained from T by removing (a, ℓ] ∩ L T (r, c) from L T (r, c) and a from A T (r, c), replacing k in cell (r, c + 1) with ((a, ℓ] ∩ L T (r, c)) ∪ {a}, and appending k to A T (r, c + 1). Since k, a > i, we have that the rightmost unpaired i in V b (T ) remains the same as in T . We also have that the cell (r + 1, c) does not contain an i + 1 and the cell (r, c + 1) does not contain an i. Thus, f i acts on V b (T ) by changing the rightmost i in A V b (T ) (r, c) to an i + 1. We now consider V b (f i (T )). We have that f i acts on T by changing the rightmost i in A T (r, c) to an i + 1. We see that a ′ , k ′ , and (a ′ , l ′ ] ∩ L f i (T ) (r ′ , c ′ ) are the same as a, k, and (a, ℓ] ∩ L T (r, c) respectively. This implies that V b acts on f i (T ) by removing (a, ℓ] ∩ L T (r, c) from L f i (T ) (r, c) and a from A f i (T ) (r, c), replacing k in cell (r, c + 1) with ((a, l] ∩ L T (r, c)) ∪ {a}, and appending k to A f i (T ) (r, c + 1). We see that f i (V b (T )) = V b (f i (T )). Next, we assume that a = i and (r, c) contains an i + 1. Since a = i, the i + 1 in (r, c) must be in its leg. Also as a is the rightmost unpaired i of T , we must have that (r, c) contains another i besides a. This gives us that V b (T ) is attained from T by removing (a, ℓ] ∩ L T (r, c) from L T (r, c) and a from A T (r, c), replacing k in cell (r, c + 1) with ((a, ℓ] ∩ L T (r, c)) ∪ {a}, and appending k to A T (r, c + 1). Note that the i inserted into (r, c + 1) becomes i-paired while an i in (r, c) becomes unpaired. This implies that the rightmost unpaired i in V b (T ) still sits in the cell (r, c). We see that the cell (r + 1, c) still does not contain an i + 1; however, the cell (r, c + 1) now contains an i. This implies that f i acts on V b (T ) by removing the i from the cell (r, c + 1) and appending an i + 1 to L V b (T ) (r, c). We now consider V b (f i (T )). We have that f i acts on T by changing a into an i+1. We have that a ′ is the i+1 obtained from applying f i and k ′ is the same as k. We see that (a ′ , ℓ ′ ]∩L f i (T ) (r ′ , c ′ ) is the same as (a, ℓ] ∩ L T (r, c) excluding the i + 1. We have that V b acts on f i (T ) by removing (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) from L f i (T ) (r, c) and i + 1 from A f i (T ) (r, c), leaving the i + 1 in L f i (T ) (r, c), replacing k in (r, c + 1) with ((a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ )) ∪ {a ′ }, and appending k to A f i (T ) (r, c + 1). We see that f i (V b (T )) = V b (f i (T )). Finally, we assume that a = i and i + 1 is not in the cell (r, c). This gives us that V b (T ) is attained from T by removing (a, ℓ] ∩ L T (r, c) from L T (r, c) and a from A T (r, c), replacing k in cell (r, c + 1) with ((a, ℓ] ∩ L T (r, c)) ∪ {a}, and appending k to A T (r, c + 1). Since k j > i + 1 for all j ∈ (a, ℓ] ∩ L T (r, c), we have that the i inserted into the cell (r, c + 1) is the rightmost unpaired i in V b (T ). Note that the cell (r + 1, c + 1) does not contain an i + 1 and the cell (r, c + 2) does not contain an i. Thus, f i acts on V b (T ) by changing the i in (r, c + 1) to an i + 1. We now consider V b (f i (T )). We have that f i acts on T by changing a into an i + 1. We have that a ′ is the i + 1 obtained from applying f i and k ′ is the same as k. We see that (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) = (a, ℓ] ∩ L T (r, c).
We have that V b acts on f i (T ) by removing (a, ℓ] ∩ L T (r, c) from L f i (T ) (r, c) and i + 1 from A f i (T ) (r, c), replacing k in (r, c + 1) with ((a, ℓ] ∩ L T (r, c)) ∪ {a ′ }, and appending k to A f i (T ) (r, c + 1). We see that f i (V b (T )) = V b (f i (T )). Case 2: Assume thatr < r andĉ = c.
Note that a > i. By Lemma 3.10 we have that R i (T ) = R i (V b (T )) unless a = i + 1 and column c + 1 contains both an i and an i + 1. However, even in this special case, we see that the rightmost unpaired i of V b (T ) is in the same position as the rightmost unpaired i of T . We also see that V b (T ) does not change whether or not cell (r + 1, c) contains an i + 1 and whether or not cell (r, c + 1) contains an i. Thus, f i acts on the same i and in the same way for both T and V b (T ). Since a > i, we have that k ′ is the same as k. Note that the only way for f i to affect the cell (r, c) in T is ifr = r − 1 and (r, c) contains an i + 1. However, even in this special case, we see that (r ′ , c ′ ), a ′ , l ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are the same as (r, c), a, ℓ, and (a, ℓ] ∩ L T (r, c). Thus, V b acts on T and f i (T ) in the same way. Therefore, we have that the actions of f i and V b on T are independent and f i (V b (T )) = V b (f i (T )). Case 3: Assume thatĉ < c.
Letĩ denote the rightmost unpaired i of T . From the proof of Lemma 3.10, we have that V b does not change whether or not the i's to the right ofĩ in R i (T ) are i-paired. Thus, the rightmost unpaired i in R i (T ) and R i (V b (T )) are in the same position. As V b does not affect any column to the left of column c, we have that the rightmost unpaired i for V b (T ) is in the same position as the rightmost unpaired i for T . Note that V b also does not affect whether or not cell (r + 1,ĉ) contains an i + 1 and whether or not cell (r,ĉ + 1) contains an i. Thus, f i acts on the rightmost unpaired i in T and V b (T ) in exactly the same way. Next, we note that (r ′ , c ′ ), a ′ , k ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are the same as (r, c), a, k, and (a, ℓ] ∩ L T (r, c) respectively. Thus, V b acts on T and f i (T ) in the same way. Therefore, we have that the actions of f i and V b on T are independent and f i (V b (T )) = V b (f i (T )). Case 4: Assume thatr r andĉ = c + 1.
Under this assumption, we have that column c + 1 does not contain an i + 1 and a = i + 1 since the cells in column c + 1 do not contain any arms. We break into subcases.
• Assume that k = i. This implies that the rightmost unpaired i in V b (T ) is in the same position as the rightmost unpaired i in T . We see that V b does not change whether or not cell (r + 1, c + 1) contains an i + 1 and whether or not cell (r, c + 2) contains an i. Thus, f i acts on the rightmost unpaired i in T and V b (T ) in exactly the same way. We also observe that (r ′ , c ′ ), a ′ , ℓ ′ , k ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are the same as a, ℓ, k, and (a, ℓ] ∩ L f i (T ) (r, c) respectively. Thus, V b acts on T and f i (T ) in the same way. Therefore, we have that the actions of f i and V b on T are independent and ). • Assume that k = i. We see that the rightmost unpaired i in V b (T ) is the i that was appended to A T (r, c+ 1). Note that V b does not change whether or not cell (r + 1, c+ 1) contains an i+1 and whether or not cell (r, c+2) contains an i. We first make the extra assumption that (r, c+2) in T contains an i. This implies that f i acts on V b (T ) and T in the same way by removing the i from the hook entry of (r, c+2) and appending an i+1 to the leg of (r, c + 1). We also have that (r ′ , c ′ ), a ′ , ℓ ′ , k ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are equal to (r, c), a, ℓ, k, and (a, ℓ] ∩ L f i (T ) (r, c) respectively. Thus, V b acts on T and f i (T ) in the same way. Therefore, we have that the actions of f i and V b on T are independent and f i (V b (T )) = V b (f i (T )). We now assume that (r, c + 2) does not contain an i. This implies that f i acts on V b (T ) by changing the i in A V b (T ) (r, c + 1) to an i + 1 and acts on T similarly by changing the i in L V b (T ) (r, c + 1) to an i + 1. Note that (r ′ , c ′ ), a ′ , ℓ ′ , and (a ′ , ℓ ′ ] ∩ L f i (T ) (r ′ , c ′ ) are equal to (r, c), a, ℓ, and (a, ℓ] ∩ L f i (T ) (r, c) respectively while k ′ is the i + 1 in L f i (T ) (r, c + 1). Thus, besides the value of the number that is bumped from the leg of (r, c + 1) to its arm, we have V b acts on T and f i (T ) in the same way. Looking at f i (V b (T )) and V b (f i (T )), we see that f i (V b (T )) = V b (f i (T )). Case 5: Assume thatr > r andĉ = c or c + 1.
Under this assumption, we have that V b does not change the cells (r,ĉ), (r + 1,ĉ), and (r,ĉ + 1). We also have that R i (T ) = R i (V b (T )) implying that the rightmost unpaired i in V b (T ) is in the same position as the rightmost unpaired i in T . Thus, f i acts on the rightmost unpaired i in V b (T ) and T in the same way. Note that i + 1 cannot be in column c implying that f i can only make changes to the legs and hook entries of (r,ĉ) and (r,ĉ + 1).
Since these changes only affect the legs and hook entries of cells outside of the possible cells that V b can change, we have that V b acts on T and f i (T ) in the same way. Therefore, we have that the actions of f i and V b on T are independent and f i (V b (T )) = V b (f i (T )). Case 6: Assume thatĉ c + 2.
Letĩ denote the rightmost unpaired i of T . From the proof of Lemma 3.10, we have that V b does not change whether or not the i + 1's to the left ofĩ are i-paired. Thus, the rightmost unpaired i in R i (T ) and R i (V b (T )) are in the same position. As V b does not affect any column to the right of column c + 1, we have that the rightmost unpaired i for V b (T ) is in the same position as the rightmost unpaired i for T . Note that V b also does not affect whether or not cell (r + 1,ĉ) contains an i + 1 and whether or not cell (r,ĉ + 1) contains an i. Since the cells that f i and V b could change are different and the rightmost unpaired i does not change, we have that the actions of f i and V b on T are independent and Theorem 3.14. Let T ∈ HVT.
Proof. Part (2) follows from part (1) since e i and f i are partial inverse. We prove part (1) here. Let T ∈ HVT with arm excess α such that f i (T ) = 0 for some i. Then f i (P (T )) = P (f i (T )) follows from Lemma 3.13, as P (T ) is obtained by successive applications of V on T and each application of V is a string of applications of V b .
Since crystal operators do not change arm excess, we may employ the notation in Definition 3.5 and denote the pair of insertion and recording tableaux produced at the j-th step for 0 j α of the uncrowding map U for T and f i (T ) as (P j (T ), Q j (T )) and (P j (f i (T )), Q j (f i (T ))), respectively. As crystal operators do not change the shape of T , we have shape(P j (f i T )) = shape(f i (P j (T ))) = shape(P j (T )) for all 0 j α. Hence (3.2) shape(P j+1 (T ))/shape(P j (T )) = shape(P j+1 (f i (T )))/shape(P j (f i (T ))) for all 0 j α − 1.
Suppose Q j (T ) = Q j (f i (T )) for a given j 0. It suffices to show that the cells shape(Q j+1 (T ))/shape(Q j (T )) = shape(P j+1 (T ))/shape(P j (T )) and shape(Q j+1 (f i (T )))/shape(Q j (f i (T ))) = shape(P j+1 (f i (T )))/shape(P j (f i (T ))) in Q j+1 (T ) and Q j+1 (f i (T )) are at the same position with the same entry. By (3.2), the cells are in the same position, say in columnc. By Definition 2.5, f i does not move elements in the arm to a different column, so the columns in which we start the uncrowding insertion V on P j (T ) and P j (f i (T )) are the same, say c, by Definition 3.5. Hence the cells shape(Q j+1 (T ))/shape(Q j (T )) and shape(Q j+1 (f i (T )))/shape(Q j (f i (T ))) are at the same position with entryc − c. The theorem follows.
Hawkes and Scrimshaw [HS20, Theorem 4.6] proved that HVT m (λ) is a Stembridge crystal by checking the Stembridge axioms. This also follows directly from our analysis above.
Corollary 3.15. The crystal HVT m (λ) of Definition 2.5 is a Stembridge crystal of type A m−1 .
Proof. According to [MPS21], SVT m (µ) is a Stembridge crystal of type A m−1 . By Theorem 3.14, the map is a strict crystal morphism (see for example [BS17, Chapter 2]). The statement follows.

3.4.
Uncrowding map on multiset-valued tableaux. The uncrowding map on hook-valued tableaux described above turns out to be a generalization of the uncrowding map on multisetvalued tableaux by Hawkes and Scrimshaw [HS20, Section 3.2]. We will prove that this is indeed the case in this section. Let us recall the definition of the uncrowding map in [HS20, Section 3.2]. (1) Set U λ 1 +1 = ∅ and F λ 1 +1 be the unique column-flagged increasing tableau of shape ∅/∅.
(2) Let 1 k λ 1 and assume that the pair (U k+1 , F k+1 ) is defined. The pair (U k , F k ) is defined recursively from (U k+1 , F k+1 ) using the following two steps: (a) Define U k as the RSK row insertion tableau from the word where C j is the j-th column of T for every 1 j λ 1 . In other words, if we denote by T k the tableau formed by the columns weakly to the right of the k-th column of T , U k is obtained by performing the RSK row insertion using the column reading word of T k . (b) Form the tableau F k of shape shape(U k )/shape(T k ) as follows. Shift F k+1 by one column to the right and fill the boxes in the same positions into F k ; for every unfilled box in the shape shape(U k )/shape(U k+1 ), label each box in column i with entry i − 1.
Recall that as long as f i T = 0 for T ∈ MVT(λ), we have U (f i T ) = f i U (T ) by [HS20, Theorem 3.17] and P (f i T ) = f i P (T ) by Theorem 3.14. Now let T ∈ MVT(λ) be arbitrary. Then T = f i 1 · · · f i k (S) for some sequence of i 1 , . . . , i k and S highest weight. Hence, It remains to show that Q(T ) = F (T ) for all T ∈ MVT(λ). To do this, we show that the newly created boxes of the uncrowding map up to a specified column in Definition 3.16 are in the same positions as those for the uncrowding insertion in Definition 3.5. For every Y ∈ MVT(µ) and for every 1 j µ 1 , denote by Y j the tableau formed by the rightmost j columns of Y ; here Y µ 1 +1 is the empty tableau.
Let T ∈ MVT(λ) be arbitrary. For 1 k λ 1 +1, let P (k) be the tableau obtained by performing the uncrowding map U on T on the columns from right to left up to and including the k-th column of T ; here P (λ 1 +1) = T . In other words, P (k) = V α k (T ) as in Definition 3.4, where α k is the arm excess of T k . As the entries to the left of column k of T are untouched by the uncrowding insertion in Definition 3.4, for every 1 k λ 1 +1, we have (P (k) ) k = P (T k ) = U (T k ). It follows that for every 1 k λ 1 , up to horizontal shifts, the newly formed boxes in shape(P (k) )/shape(P (k+1) ) = shape[(P (k) ) k+1 ]/shape[(P (k+1) ) k+1 ] and shape([U (T k )] k+1 )/shape([U (T k+1 )] k+1 ) are in the same positions. Since the entries in these boxes both record the difference in column indices relative to the k-th column for each 1 k λ 1 and since the recording tableaux for both maps are formed from the union of these boxes, we conclude that Q(T ) = F (T ), completing the proof.
3.5. Crowding map. In this section, we give a description of the "inverse" of the uncrowding map.
We begin by introducing some notation. Let F ∈F with e entries. For each cell (r, c) in F with entry F (r, c), define the corresponding destination column to be d(r, c) = c − F (r, c). Define the crowding order on F by ordering all the cells in F with a filling, first determined by their destination column (smallest to largest) and then by column index (largest to smallest). Denote the order by (r 1 , c 1 ), (r 2 , c 2 ), . . . , (r e , c e ). Set α(F ) = (α 1 , α 2 , . . . , α e ), where α i = F (r i , c i ). Let the arm excess for a column of a hook-valued tableau be the sum of arm excesses of all its cells.
Definition 3.19. Let h ∈ HVT and let (r, c) be a cell in h with c > 1 and with at most one element in A h (r, c). If A h (r, c) is empty, we also require that the cell (r, c) is a corner cell in h. Then we define the crowding bumping C b on the pair [h, (r, c)] by the following algorithm: (1) If A h (r, c) is nonempty, set m to be the only element in A h (r, c) and b = max{x ∈ L + h (r, c) | x m}. Otherwise, set m = H h (r, c) and b = max(L + h (r, c)).
Step 2 equals r, perform either of the following: and keep it to be strictly increasing. Remove cell (r, c) from h. (b) Otherwise, r ′ = r and perform either of the following: c) is empty, then remove cell (r, c) from h. Denote the resulting (not necessarily semistandard) hook-valued tableau by h ′ . We write C b ([h, (r, c) We also define the projections p 1 and p 2 by Figures 3 and 4 for illustration.   • if r ′ = r, then h ′ is always semistandard and has the same weight as h; • if r ′ = r and A h (r, c) is empty, then h ′ might have fewer letters than h. In Example 3.20, S contains 5 letters while S ′ only contains 4. This happens precisely when L h (r, c) is nonempty.
In principle, the arm in cell (r ′ , c − 1) could be greater than the q that is to be inserted. However, we only consider the cases as defined in the order described by the next paragraph. We refer to Proposition 3.27 which states that all tableaux we deal with in this section are indeed semistandard hook-valued tableaux.
• Since F ∈F, we always have c > 1.
• The case that A h (r, c) is empty can only occur in T (0) j−1 for some j > 0. In this case, (r, c) = (r j , c j ), which is a corner cell.
• Consider the α j steps in T (0) j−1 j . We first delete cell (r j , c j ), which has no arm. Then at every step after that, we move leftward one column at a time. Before we reach column d(r j , c j ), there is exactly one column with arm excess being 1 and the rest has zero arm excess among columns to the right of d(r j , c j ) since recall that the cells (r j , c j ) are ordered from smallest to largest destination column. Once we reach column d(r j , c j ), the cell there may contain more than one arm element, but we then go to (r j+1 , c j+1 ), which is a corner cell instead. Thus there is at most one element in A h (r, c).
Definition 3.23. With the same notation as above, define the insertion path of T We obtain the sequence from the algorithm: j , we consider the following cases. q (r j+1 , c j+1 )) and Figure 5 for illustration.
. This completes the proof. Proof. In any path j , consider the arm excess of its columns. Those with column index c such that d(r j , c j ) < c < c j started with arm excess 0, then changed to arm excess 1 when the insertion path passed through that column, and immediately decreased to 0.
Thus the q (s) j that is being moved to cell (r ′ , c − 1) is always at the rightmost column containing nonzero arm excess. When c−1 > d(r j , c j ), the arm excess of the column c−1 is exactly 1, (r ′ , c−1) is also the topmost cell containing an arm. For c − 1 = d(r j , c j ), the path path j has reached its destination. At that point, any column to the right of d(r j , c j ) has 0 arm excess. It follows from Lemma 3.25 that the cell (r is also the topmost cell containing an arm. Proposition 3.27. The tableau T (s+1) j is a semistandard hook-valued tableau for all 0 j e − 1 and for all 0 s < α j+1 .
Proof. We only need to check that the q in Step 2 of Definition 3.19 is greater or equal to the hook entry and arm of the cell q is to be inserted into. When q is the only arm element, it is obvious that q is greater or equal to the hook entry.
The case when q is not the only arm element can only happen when we reach the destination column of the path. By the proof of Lemma 3.25, we have that for q (s) j+1 q (s+z j ) j for s 1 and for j such that d(r j , c j ) = d(r j+1 , c j+1 ). Hence the statement follows by setting k = α j+1 .
Proof. We show that ifh = V b (h), where h ∈ HVT, V b is as defined in Definition 3.2 andh is obtained by moving some letter(s) from the cell (r, c) to (r, c + 1) (potentially adding a box), then We follow the notation used in Definitions 3.2 and 3.19. Thus a = max (A h (r, c)). We have that H h (r, c) a. If cell (r + 1, c) is in h, then H h (r + 1, c) > a. Case (1):r = r. Case (1A): If cell (r, c + 1) is not in h, then h ′ is obtained by adding cell (r, c + 1) and moving a from A h (r, c) to H h (r, c + 1). Under the action of C b , by Step 1, b = a and r ′ = r. C b appends a to Ah(r, c) and removes cell (r, c + 1), which recovers h.
Case (1B): If cell (r, c + 1) is in h, then k ∈ L + h (r, c + 1) is the smallest number that is greater than or equal to a in column c + 1. h ′ is obtained by removing a from A h (r, c), replacing k with a, and attaching k to A h (r, c + 1). Under the action of C b , by Step 1, we can see that and r ′ = r. By Step 3(b)i, q = b = a, and a is appended to Ah(r, c) and q = a in Lh(r, c + 1) is replaced with m = k. In the end, m is removed from Ah(r, c + 1). We recover h.
Case (2):r = r. Let ℓ = max(L + h (r, c)). Case (2A): If cell (r, c + 1) is not in h, V b adds cell (r, c + 1), removes the part of L h (r, c) that is greater than a to L h (r, c + 1) and moves a from A h (r, c) to H h (r, c + 1). Under the action of C b , by Step 1, m = a and b = ℓ. Thus r ′ = r. By Step 3(a)ii, we move Lh(r, c + 1) into Lh(r, c) and we recover h. Case (2B): If cell (r, c + 1) is in h,h is obtained by moving the part of L h (r, c) that is greater than a to L h (r, c + 1), moving a from A h (r, c) to H h (r, c + 1), and appending k to A h (r, c + 1). Under the action of C b , by Step 1, m = k and b = ℓ. Then r ′ = r and q = a. By Step 3(a)i, we move the set {x ∈ Lh(r, c) | a < x k} from Lh(r, c + 1) into Lh(r, c), which is the set that was moved from cell (r, c) by V b . Removing k from Ah(r, c + 1) and setting Hh(r, c + 1) = k, we recover h. is semistandard and has the same weight as S for all 0 j e − 1, for all 0 s α j+1 . Thus image(U ) ⊂ K λ and C • U = 1 HVT(λ) .
Proposition 3.31. K λ is a subset of the image of U : HVT(λ) → ⊔ µ⊇λ SVT(µ)×F (µ/λ). Moreover, Proof. Let (S, F ) ∈ K λ , then for all 0 j < e and for all 0 for all 0 j < e and for all 0 s < α j+1 . Following the notation in Definition 3.2, we first locate the rightmost column that contains nonzero arm excess, then determine the topmost cell in rowr in that column with nonzero arm excess. We denote by a the largest arm element in that cell.
By Lemma 3.26, in T  Corollary 3.32. The uncrowding map U is a bijection between HVT(λ) and K λ with inverse C.
3.6. Alternative uncrowding on hook-valued tableaux. In Section 3.2, we defined an uncrowding map sending hook-valued tableaux to pairs of tableaux with one being set-valued and the other being column-flagged increasing. As hook-valued tableaux were introduced as a generalization of both set-valued tableaux and multiset-valued tableaux, it is natural to ask if there is an uncrowding map taking hook-valued tableaux to pairs of tableaux with one being multiset-valued. In this section we provide such a map.
Definition 3.33. The multiset uncrowding bumpingṼ b : HVT → HVT is defined by the following algorithm: (1) Initialize T as the input.
(2) If the leg excess of T equals zero, return T.
(3) Find the topmost row that contains a cell with nonzero leg excess. Within this column, find the cell with the largest value in its leg. (This is the rightmost cell with nonzero leg excess in the specified row.) Denote the row index and column index of this cell by r and c, respectively. Denote the cell as (r, c), its largest leg entry by ℓ, and its rightmost arm entry by a. (4) Look at the row above (r, c) (i.e. row r + 1) and find the leftmost number that is strictly greater than ℓ.
• If no such number exists, attach an empty cell to the end of row r + 1 and label the cell as (r + 1,c), wherec is its column index. Let k be the empty character. • If such a number exists, label the value as k and the cell containing k as (r + 1,c) wherẽ c is the cell's column index. We now break into cases: (a) Ifc = c, then remove ℓ from L T (r, c), replace k with ℓ, and attach k to the leg of L T (r + 1,c). + 1,c), replace the hook entry of (r + 1,c) with ℓ, and attach k to L T (r + 1,c). (5) Output the resulting tableau.
(2) For 1 i α, letP i+1 =Ṽ(P i ). Let r be the index of the topmost row ofP i containing a cell with nonzero leg excess and letr be the row index of the cell shape(P i+1 )/shape(P i ). ThenQ i+1 is obtained fromQ i by appending the cell shape(P i+1 )/shape(P i ) toQ i and filling this cell withr − r. DefineŨ (T ) = (P (T ),Q(T )) := (P α ,Q α ).

Applications
In this section, we provide the expansion of the canonical Grothendieck polynomials G λ (x; α, β) in terms of the stable symmetric Grothendieck polynomials G µ (x; β = −1) and in terms of the dual stable symmetric Grothendieck polynomials g µ (x; β = 1) using techniques developed in [BM12]. We first review the basic definitions and Schur expansions of the two polynomials.
Definition 4.1. The reading word word(S) = w 1 w 2 · · · w n of a set-valued tableau S ∈ SVT(µ) is obtained by reading the elements in the rows of S from the top row to the bottom row in the following way. In each row, first ignore the smallest element of each cell and read all remaining elements in descending order. Then read the smallest elements of each cell in ascending order.
Example 4.2. The reading word of P (T ) in Example 3.6 is word(P (T )) = 8675423362111567. The dual stable symmetric Grothendieck polynomials g µ (x; 1) are dual to G µ (x; −1) under the Hall inner product on the ring of symmetric functions.
Definition 4.4. A reverse plane partition of shape µ is a filling of the cells in the Ferrers diagram of µ with positive integers, such that the entries are weakly increasing in rows and columns. We denote the collection of all reverse plane partitions of shape µ by RPP(µ) and the set of all reverse plane partitions by RPP.
The evaluation ev(R) of a reverse plane partition R ∈ RPP is a composition α = (α i ) i 1 , where α i is the total number of columns in which i appears. The reading word word(R) is obtained by first circling the bottommost occurrence of each letter in each column, and then reading the circled letters row-by-row from top to bottom and left to right within each row.  ((3, 2)).
Lam and Pylyavskyy [LP07] showed that the dual stable symmetric Grothendieck polynomials g µ (x; 1) are generating functions of reverse plane partitions of shape µ g µ (x; 1) =
They also provided the Schur expansion of the dual stable symmetric Grothendieck polynomials [LP07, Theorem 9.8] where the sum is over all flagged increasing tableaux whose outer shape is µ.
According to [BM12], a symmetric function f α over the ring R is said to have a tableaux Schur expansion if there is a set of tableaux T(α) and a weight function wt α : T(α) → R so that wt α (T )s shape(T ) .
Furthermore, any symmetric function with such a property has the following expansion in terms of G µ (x; −1) and g µ (x; 1).
Applying [HS20, Theorem 4.6] and the above correspondence, we obtain Note that Proposition 4.8 in particular implies that the canonical Grothendieck polynomials are Schur positive. This was known from [HS20], but here an explicit tableaux formula is given.