A Bijection Between K -Kohnert Diagrams and Reverse Set-Valued Tableaux

Lascoux polynomials are K -theoretic analogues of the key polynomials. They both have combinatorial formulas involving tableaux: reverse set-valued tableaux ( RSVT ) rule for Lascoux polynomials and reverse semistandard Young tableaux ( RSSYT ) rule for key polynomials. Furthermore, key polynomials have a simple algorithmic model in terms of Kohnert diagrams, which are in bijection with RSSYT . Ross and Yong introduced K -Kohnert diagrams, which are analogues of Kohnert diagrams. They conjectured a K -Kohnert diagram rule for Lascoux polynomials. We establish this conjecture by constructing a weight-preserving bijection between RSVT and K -Kohnert diagrams.


Introduction
Fix a positive integer n throughout this paper. A weak composition of length n is a sequence of n non-negative integers. If α is a weak composition, we use α i to denote its i th entry.
The paper is organized as follows. In section 2, we review related combinatorial rules for κ α and L (β) α . In section 3, we define two maps Ψ α and Φ α on KKD(α) and RSVT(α) respectively. The following sections will prove Theorem 1. 3. In section 4, we introduce a partial order on all weak compositions. We call it the Bruhat order and show it is equivalent to the left swap order in [AS18]. In section 5, we describe the sets KKD(α) and RSVT(α) recursively using the Bruhat order. In section 6, we introduce two auxiliary operators g and e on KD(α) and discuss their properties. In section 7, we give recursive descriptions of maps Ψ α and Φ α in terms of g and e . Then we show Ψ α (resp. Φ α ) is a well-defined map to RSVT(α) (resp. KKD(α)) using the recursive definitions developed in section 5. Finally we prove Theorem 1.3.

Background
2. 1. RSSYT(α) and RSVT(α). Given a partition λ = (λ 1 λ 2 · · · λ 0), a Young diagram of shape λ is a finite collection of boxes, aligned at the left, in which the i th row has λ i boxes. We use English convention for our Young diagrams and tableaux, so the first row is the highest row.
A reverse semistandard Young tableau of shape λ is a filling of the Young diagram λ with positive number such that (i) each box contains exactly one number, (ii) the entries in each row weakly decrease from left to right, and (iii) the entries of each column strictly decrease from top to bottom. Let RSSYT λ be the set of all the reverse semistandard Young tableaux of shape λ.
Following [SB02], we introduce another set of tableaux where a box might have more than one number. A reverse set-valued tableau of shape λ is a filling of the Young diagram λ with positive numbers such that (i) each box contains a finite and non-empty set of positive integers, (ii) if a set A is to the left of a set B in the same row, then min(A) max(B), and (iii) if a set C is below a set A in the same column, then min(A) > max(C). Let RSVT λ be all the reverse set-valued tableaux of shape λ.
Let the weight vector for T be the weak composition whose i th component is the the total number of appearance of i in T , denoted by wt (T ). Given any weak composition α, let |α| = i 1 α i . Given T ∈ RSVT λ , define L(T ) to be an element in RSSYT λ constructed by only keeping the largest number in each box of T . We call these numbers the leading numbers of T . Any number in T that is not a leading number is called a extra number . Let the excess of T be the number of extra numbers in T , so we can denote it by ex(T ) = |wt(T )| − |λ|.
Next we give the definition of left key of T , denoted by K − (T ), where T is a RSSYT. It was first given in [Wil13,Section 5]. We give the description as in [SY21,Definition 3.11].

2.2.
Viewing RSVT(α) as a pair of diagrams. In this subsection, we introduce another perspective on RSVT(α). A diagram is a finite subset of N × N. We may represent a diagram by putting a box at row r and column c for each (c, r) in the diagram. We adopt the convention where columns begin at 1 from the left and rows begin at 1 from the bottom. The weight of a diagram D, denoted as wt(D), is a weak composition whose i th entry is the number of boxes in its i th row. A diagram pair is an ordered pair D = (D 1 , D 2 ) such that D 1 and D 2 are disjoint diagrams. We may represent D by putting a box at (c, r) for each (c, r) ∈ D 1 and putting a box with label X at (c, r) for each (c, r) ∈ D 2 . Cells in D 1 are called Kohnert cells. Cells in D 2 are called ghost cells. The weight of D, denoted as wt(D), is a weak composition whose i th entry is the number of Kohnert cells and ghost cells in its i th row. Let the excess of D, denoted by ex(D), be |D 2 |. Now we embed the set of RSVT into the set of diagram pairs. Given an RSVT T , we send it to (L, E). The set L (resp. E) consists of all (r, c) such that r is a leading (resp. extra) number in column c of T . This map is injective. If we know (L, E) is the image of some RSVT T , we can uniquely recover T : First, for each c, build a column that consists of r such that (c, r) ∈ L. The column should be decreasing from top to bottom. Then for each (c, r) ∈ E, put r in the lowest cell whose largest number is larger than r. This will be column c of T . Now we may view each RSV T as a diagram pair. We write T = (L, E) to denote this correspondence. It is clear that this correspondence preserves wt(·) and ex(·).

KD(α)
and KKD(α). We give another combinatorial definition of key polynomials due to Kohnert [Koh90]. A diagram pair is called a key diagram pair if its Kohnert cells are left-justified and has no ghost cells. Given a weak composition α, we let D α the key diagram pair associated to α: On its row i, there are α i left-justified Kohnert cells and no ghost cells.
Next, we define a Kohnert move on a diagram pair with no ghost cells: Select the rightmost box in any row and move it downward to the first position available, possibly jumping over other cells as needed. Let KD(α) be the closure of {D α } under all possible Kohnert moves. Theorem 2.6. [Koh90] The key polynomials indexed by α, is given by Remark 2.7. There is a natural identification between KD(α) and RSSYT(α) which yields KD(α) = RSSYT(α). Take T ∈ RSSYT(α). By our convention in the previous subsection, T is viewed as a diagram pair (L, ∅). This result is well-known to experts. For example, it follows from work done in [AS18]. For completeness, we will recover this result in section 5.
Ross and Yong [RY15, Section 1.2] generalized Kohnert moves. We state their construction below.
A K-Kohnert move is an operation on a diagram pair. It selects the rightmost cell in a row. The selected cell cannot be a ghost cell. Then move this cell downward to the first position available. It can jump over other Kohnert cells, but cannot jump over any ghost cells. After the move, it may or may not leave a ghost cell at the original position. When a K-Kohnert move leaves a ghost cell, we also refer it as a ghost move. Let KKD(α) be the closure of {D α } under all possible K-Kohnert moves. We make the following observations. Remark 2.9. Let α be a weak composition. We have Remark 2. 10. Usually, an element of KD(α) is viewed as a diagram. We defined KD(α) as a set of diagram pairs so we can work with KD(α) and KKD(α) using the same technique. In particular, with our convention, KD(α) is viewed as a subset of KKD(α).
(2) If a number i appears in row r, then i r.
(3) If a number i appears in (c, r) and (c + 1, r ), then r r .
(4) Let i, j appear in column c with j > i and j is lower than i. Then there is an i in column c + 1 that is strictly above the j in column c.
Let KT(α) be the set of all Kohnert tableaux with content α.
Assaf and Searles constructed bijections between KT(α) and KD(α) in [AS18]. For each T ∈ KT(α), we may ignore its numbers and view each cell as a Kohnert cell. By [AS18, Lemma 2.4], the resulting diagram pair is in KD(α).
The inverse of the map above is called Kohnert Labeling with respect to α, denoted as Label α (·). We may describe it as the following algorithm on certain diagram pairs.
Let D be an arbitrary diagram pair such that column c of D has |{i : α i c}| Kohnert cells and no ghost cells. Initialize sets S 1 , S 2 , . . . as S c = {i : α i c}. Iterate through boxes of D from right to left, and from bottom to top within each column. For the box (c, r), find the smallest i ∈ S c such that i does not appear at (c + 1, r ) for all r > r. We remove i from S c and fill i in (c, r) of D. If no such i exists or i < r, terminate the algorithm. After all boxes are filled, output the final tableau.
By [AS18, Lemma 2.6, Lemma 2.7, Theorem 2.8], the labeling algorithm on D produces an output if and only if D ∈ KD(α). Moreover, if we restrict the algorithm on KD(α), then this is a bijection from KD(α) to KT(α) whose inverse is described above.
Example 2.14. Let α = (0, 2, 1), we have where the relative order in the sets corresponds to KD(α) from Example 2.8 under the above labeling algorithm.

3.1.
An informal description of Ψ α . In this section, we describe a map Ψ α from KKD(α) to the set of all diagram pairs. First, we describe an operator on KD(α). Let G be a diagram. Then G (·) acts on KD(α) in the following way: Take D ∈ KD(α). Iterate through cells of G from right to left. Within each column, go from bottom to top. For (c, r) ∈ G, search for the largest r r such that (c, r ) is a Kohnert cell in D. Moreover, if we raise the cell (c, r ) to (c, r), the resulting diagram is still in KD(α). After finding such r , we move cell (c, r ) to (c, r). After iterating over all cells in G, we denote the final Kohnert diagram by G (D). If we cannot find such an r during an iteration, then G (D) is undefined.   .
We say this is an informal description of Ψ α since the map is not obviously well-defined. It seems possible that G ((K, ∅)) is undefined. In section 7, we will provide an alternative description of the map Ψ α and check the following.

3.2.
An informal description of Φ α . The map Φ α can be described similarly on RSVT(α). First, we need an analogue of the G (·) operator. Let E be a diagram. Then E (·) acts on KD(α) in the following way: Take D ∈ KD(α). Iterate through cells of E from left to right. Within each column, go from top to bottom. For (c, r) ∈ G, search for the smallest r r such that (c, r ) is a Kohnert cell in D. Moreover, if we drop the cell (c, r ) to (c, r), the resulting diagram is still in KD(α). After finding such r , we move cell (c, r ) to (c, r). After iterating over all cells in G, we denote the final Kohnert diagram by G (D). If we cannot find such an r during an iteration, then G (D) is undefined.
Example 3.4. Let D be the first Kohnert diagram in Example 2. 8. Let E be the diagram {(1, 1), (2, 1)}. We may compute E (D) as follows. We label (c, r) and (c, r ) involved in each step above and below the arrows.

· ·
Notice that this is an element of KKD(α).
Again, Φ α is not obviously well-defined. In section 7, we will provide an alternative description of the map Φ α and check the following.
Lemma 3. 6. The map Φ α is a well-defined map from RSVT(α) to KKD(α). Now we restate our main result. The proof is also in section 7.

Bruhat Order on Weak Compositions
Partial order on weak compositions has been studied in [AS18, AQ19, FG21, FGPS20]. In this section, we give a definition via the key tableau associated to the weak composition. In subsection 4.1, we also study m(α, S) (resp. M (α, S)) which is the unique minimum (resp. maximum) weak composition in a certain set of weak compositions. In subsection 4.2, we used properties of M (α, S) to show that our Bruhat order is equivalent to the left swap order defined in [AS18,AQ19], which implies that the Bruhat order is equivalent to the inclusion order on KD(α), KKD(α), RSSYT(α) and RSVT(α).

Bruhat order.
We may define a partial order on all weak compositions.
Definition 4.1. Let α, γ be two weak compositions. We define α γ if key(α) and key(γ) have the same shape and key(α) key(γ) entry-wise. This order is called the Bruhat order.   Proof. We first prove statement 1 and 2 are equivalent. By definition S S if and only if key(1 S ) key(1 S ). The number on row j column 1 of key(1 S ) (resp. key(1 S )) is the j th largest number in S (resp. S ). Thus, key(1 S ) key(1 S ) is equivalent to statement 2.
Next, we show the statement 2 and 3 are equivalent. Assume the statement 2 is true. Let s ∈ S be the j th largest number in S. Then |[s, n] ∩ S| = j. Since the j th largest number in S is at least s, |[s, n] ∩ S | j. Now assume statement 3 is true. Let s be the j th largest number in S. We know there are at least j numbers in [s, n] ∩ S , so the j th largest number in S is at least s. Take S ⊆ [n] and a weak composition α. Consider the set {γ : γ α, supp(γ) ⊆ S}. Let m(α, S) be the unique minimum element in the set, if it exists. Later, we will show m(α, S) exists as long as the set is non-empty. First, we introduce an algorithm to compute m(α, S) or assert it does not exist. Initialize list to be an empty list and initialize σ to be the weak composition with all 0s. Iterate over i = 1, . . . , n. Perform the following two processes in each iteration: • (Adding process): If α i > 0, then add α i to list.
• (Removing process): If i ∈ S and list is non-empty, then remove max(list) from list and assign it to σ i . After all iterations, if list is empty, then m(α, S) is σ. Otherwise, such m(α, S) does not exist.
We take steps to show this algorithm is correct. We start with the following observation, which connects this algorithm with the operator in Definition 2.1.
Lemma 4.6. Assume the algorithm outputs m(α, S) = σ. Let T c (resp. A c ) be the set consisting of numbers in column c of key(σ) (resp. key(α)). Then T c = S A c . Proof. We know i ∈ A c if and only if during the adding process of the iteration i, a number at least c is added to list. Similarly, i ∈ T c if and only if during the removing process of the iteration i, a number at least c is removed from list.
During the adding process of iteration a 1 , the algorithm puts α a 1 into list. This is the first time that list gains a number at least c. Thus, t 1 a 1 . Moreover, assume there exists t ∈ S such that a 1 t < t 1 . During the removing process of iteration t, list has a number at least c. The algorithm will remove a number at least c from list, contradicting to t / ∈ T c . Thus, t 1 is the smallest in S with t 1 a 1 . Now consider t j with j > 1. During the removing process of t j , we remove a number at least c for the j th time. Thus, we have added at least j such numbers to list, so a j t j . Now assume there is t < t j such that t ∈ S, t a j , and t > t j−1 . During the removing process of iteration t, there is a number at least c in list, so such a number will be removed. We have a contradiction since t / ∈ T c . Thus, t j is the smallest in S such that t j a j and t j > t j−1 .
Next, we investigate the condition for the algorithm to assert m(α, S) does not exist.
(2) Consider the i th iteration. First, the size of list is increased by one if i ∈ supp(α). Next, the size of list is fixed or decreased by one if i ∈ S. Throughout this algorithm, list gains |supp(α)| numbers and loses at most |S| numbers. It is not empty when the algorithm ends.
(2) By Lemma 4.4, there exists j ∈ supp(α) such that Between the j th iteration and the last iteration inclusively, list gains |[j, n] ∩ supp(α)| numbers and loses at most |[j, n] ∩ S| numbers. It is not empty when the algorithm ends.
(2) Assume when the algorithm ends, list is not empty. Find the largest j such that list is not empty since the j th iteration. First, we know list is empty right before the j th iteration. Second, we know a number is added to list during the j th iteration, so j ∈ supp(α). By Between the j th iteration and the last iteration inclusively, list gains |[j, n] ∩ supp(α)| numbers. Since list is not empty since iteration j, list loses |[j, n] ∩ S| numbers. Thus, list is empty after the last iteration.
(3) From the previous statement, we know the algorithm will produce a weak composition σ.
Just need to check supp(σ) ∈ S and σ α. It is apparent that supp(σ) ⊆ S. Since all positive numbers in α are assigned into σ, key(σ) and key(α) are of the same shape. Next we show that σ α by comparing key(σ) and key(α) column-by-column: Let T c (resp. A c ) be the set consisting of numbers in column c of key(σ) (resp. key(α)). By Lemma 4.6, Now we can prove the correctness of our algorithm. In other words, consider the set {γ : γ α, supp(γ) ⊆ S}.
• If list is empty after the iterations, then the output σ is the unique minimum in the set.
• Otherwise, the set is empty. Moreover, the second case happens only when the set is empty. Proof. If list is not empty when the algorithm ends, {γ : γ α, supp(γ) ⊆ S} = ∅ by the previous lemma.
Now assume list is empty when the algorithm ends. Then the set is non-empty: By the proof of the previous lemma, the output σ is in this set. We check σ is the least element. Assume there is a γ in the set with γ σ. Let T c consists of numbers in column c of key(γ). Define T c and A c similarly for key(σ) and key(α) respectively. Then we can find c such that T c T c . Let t j be the j th smallest number in T c . Define t j and a j similarly for T c and A c . Then we can find smallest j such that t j < t j . Notice that t j a j and t j ∈ S. Moreover, t j > t j−1 t j−1 if j > 1. We have a contradiction to the fact T c = S A c from Lemma 4.6. Proof. Follows from the previous two lemmas.
Analogously, we may also look at the set {γ : γ α, supp(γ) ⊆ S}. Similarly, if it is non-empty, it will contain a unique maximum element. Let M (α, S) be this element. To compute it, we only need to slightly change our algorithm above: Let i goes from n to 1, instead of 1 to n. Similar to Corollary 4.9, we have the following for M (α, S). Proof. The proof is similar to the proof of Corollary 4.9.
We end this subsection by a property that connects m(γ, S) and M (α, S). Let γ be a weak composition. We use γ to denote the weak composition obtained by decreasing each positive entry of γ by 1.
Then we get the first statement since α M (α, S).

4.2.
Left swap order. Assaf and Searles [AS18] also defined a partial order on weak compositions called the left swap order . In this subsection, we introduce this order and show that it is equivalent to the Bruhat order. When γ is obtain from α by exchanging the i th and j th parts of α, we write γ = (i j)α. To show the equivalence between the left swap order and the Bruhat order, we need a few lemmas. We start with the following, which summarizes how M (α, S) is changed when we changes S in a nice way.
Lemma 4.14. Let α be a weak composition and S ⊆ [n] with S supp(α). Take g / ∈ S such that |[g, n] ∩ supp(α)| > |[g, n] ∩ S|. Then there exists s ∈ S such that s < g and Proof. Run the algorithm that computes M (α, S). After initialization, the algorithm iterates from i = n to i = 1. Right after the iteration with i = g, the algorithm has put |[g, n] ∩ supp(α)| numbers to list and has removed at most |[g, n] ∩ S| numbers from list. Since |[g, n] ∩ supp(α)| > |[g, n] ∩ S|, the list is non-empty. Let x be the largest number in the current list. Then this x will be picked sometime in the future. Let s ∈ S ∩ [1, g) be the largest such that in the iteration i = s, the algorithm assigns x to σ i . We may run the algorithm to compute We know from definition that M (α, S) α. The next lemma will describe their relationship in the left swap order.
Finally, we need the following intuitive lemma, which says both partial orders are preserved by the operator α → α.  Proof. First we show if γ α, then γ α. It suffices to show when γ is a left swap of α. Say γ = (i j)α where i < j and α i < α j . Let T c (resp. T c ) consists of numbers in column c of key(α) (resp. key(γ)). When c α i or c > α j , we have T c = T c . When α i < c α j , T c is obtained from T c by replacing j with i. Therefore key(γ) key(α) and γ α.

Recursive descriptions of KKD(α) and RSVT(α)
To check our maps Φ α and Ψ α are well-defined, we need to study their domains. In this section, we give necessary and sufficient criteria on when a diagram pair is in KKD(α) (resp. RSVT(α)). Our criteria will be recursive.
For a diagram pair D = (K, G), we can send it to a triple (K 1 , G 1 , d) where K 1 , G 1 are disjoint subsets of [n] and d is a diagram pair. They are defined as follows: • K 1 := {r : (1, r) ∈ K} • G 1 := {r : (1, r) ∈ G} • d is the diagram pair with a Kohnert cell (resp. ghost cell) at (c, r) if D has a Kohnert cell (resp. ghost cell) at (c + 1, r) with c 1. The map D → (K 1 , G 1 , d) is invertible: given disjoint K 1 , G 1 ⊆ [n] and a diagram pair d, we can uniquely recover D. Thus, we may identify a diagram pair with its image and write D = (K 1 , G 1 , d).
Fix a weak composition α in this section. An element of KKD(α) or RSVT(α) can be written as (K 1 , G 1 , d). We will find conditions on this triple to determine when (K 1 , G 1 , d) ∈ KKD(α) and RSVT(α).
Notice D ∈ KKD(γ) ⊆ KKD(α) by Corollary 4.18. Thus, this lemma implies the reverse direction of Theorem 5.1. Proof. We describe an algorithm that turns D into a key diagram pair of some weak composition γ via reverse K-Kohnert moves. First, we consider d. Since d ∈ KKD(M (α, K 1 )), we may do reverse K-Kohnert moves on d to obtain the key diagram pair of M (α, K 1 ). Now, Kohnert cells in D form the key diagram pair of M (α, K 1 ). If G 1 is empty, we are done. Otherwise, let g = min(G 1 ). By Lemma 4.14, we can find k ∈ K 1 ∩ [1, g) such that M (α, (K 1 {g}) − {k}) = (g k)M (α, K 1 ). We perform reverse K-Kohnert moves to lift the entire row k of D to row g, and remove the ghost at (1, g). Next we assign (K 1 {g}) − {k} to K 1 and assign G 1 − {g} to G 1 . Now, Kohnert cells in D still form the key diagram pair of M (α, K 1 ). We may repeat the steps above until G 1 is empty. The resulting diagram pair is a key diagram pair for some weak composition γ. Clearly, γ i = 0 if (1, i) was not a cell in D.
Proof of Theorem 5.1. The reverse direction is already shown. For each D = (K 1 , G 1 , d) ∈ KKD(α), we need to show the four conditions are satisfied. We prove by induction on max(α). When max(α) = 0, the claim is immediate.
Assume the statement is true for any weak composition whose maximum entry is less than max(α). For any D ∈ KKD(α), we want to check it satisfies the four conditions. Condition (1) is immediate. We prove the other conditions by an induction on K-Kohnert moves. If D is the key diagram pair of α, then the last three conditions are immediate. Now assume (K 1 , G 1 , d) ∈ KKD(α) satisfies the last three conditions. Perform one K-Kohnert move and obtain (K 1 , G 1 , d ). We want to show (K 1 , G 1 , d ) also satisfies the last three conditions. If the K-Kohnert move is not on column 1, then K 1 = K 1 and G 1 = G 1 , which gives us the second and the third condition. Notice that d is obtained from d by one K-Kohnert move, so the last condition is also satisfied. Now suppose the K-Kohnert move is on column 1. K 1 = (K 1 − {i}) {j} with i > j, and G 1 is either G 1 or G 1 {i}. We check the last three conditions below.
Let T = key(M (α, K 1 )). For each column of T that contains i, we replace i by the largest i such that i < i and i is not in this column. Then we sort the column into strictly decreasing order. Let T be the resulting tableau. It is clear that T is a key. Let σ = wt(T ). We make two observations about T and σ: • Column 1 of T consists of K 1 , so supp(σ) = K 1 .
It remains to check γ σ. Let g 1 > · · · > g s be the numbers in column c of key(γ). By γ i = 0, none of these numbers is i. Let t 1 > · · · > t s (resp. t 1 > · · · > t s ) be numbers in column c + 1 of T (resp. T ). By γ M (α, K 1 ), we know g k t k for 1 k s. If i is not in column c + 1 of T , then we are done. Otherwise, assume t a = i. We know t 1 , . . . , t s are obtained from t 1 , . . . , t s by changing t a , · · · , t b into t a − 1, . . . , t b − 1. Since g a i and g a = i, we have g a i − 1 = t a . Then g a+1 i − 2 = t a+1 . Following this argument, we have g k t k for 1 k s. Thus, γ σ M (α, K 1 ).

5.2.
Describing RSVT(α). Take T ∈ RSVT(α) and write T as (L 1 , E 1 , t). First, we notice the following relation between K − (T ), K − (t) and L 1 : Proof. Let S = key(1 L 1 + m(γ, L 1 )). View S and T as tableaux. Let T c be the set consisting of leading numbers in column c of T , so T 1 = L 1 . It suffices to show that K − (T ) = S. We compare these two tableaux column by column. Apparently, column 1 of K − (T ) and column 1 of S both consist of L 1 .
(3) View T as a tableau. For e ∈ E 1 , assume it is on row j column 1 of T . Then |L 1 ∩(e, n]| = j.
Thus t ∈ RSVT(M (α, L 1 )). Now, we check if we have T = (L 1 , E 1 , t) satisfying these four conditions, then T ∈ RSVT(α). We may construct T as a tableau. First, we build column 1 of T . We arrange numbers in L 1 into a strictly decreasing column. For each e ∈ E 1 , by condition 2, |(e, n] ∩ L 1 | > 0. Then we put e on row |(e, n] ∩ L 1 |. Clearly, this column is an RSVT with leading numbers from L 1 and extra numbers from E 1 .
Next, we may build the tableau corresponding to t recursively. Let L 1 be the set of leading numbers in column 1 of t. Put the column we just constructed on the left of t. Only need to check all numbers in row j of our new column are weakly larger than the leading number in row j column 1 of t, which is the j th largest number in L 1 .
• Let e be an extra number on row j of our new column, By condition 3, |[e, n] ∩ L 1 | < j. Thus, the j th largest number in L 1 is at most e. • By condition 4, if we let γ = wt(K − (t)), then γ M (α, L 1 ). By Lemma 4.11, m(γ, L 1 ) exists. Then by Corollary 4.9, the j th largest number in supp(γ) = L 1 is weakly less than the j th largest number in L 1 , which is the leading number in row j of our new column. Now we have constructed a tableau T which if viewed as a diagram pair, corresponds to (L 1 , E 1 , t). It remains to check K − (T ) key(α). Notice that K − (T ) = key(1 L 1 + m(γ, L 1 )). By Lemma 4.11 and γ M (α, L 1 ), we have K − (T ) key(α). Thus, (L 1 , E 1 , t) ∈ RSVT(α).
The recursive descriptions of KKD(α) and RSVT(α) share many similarities. They only differ at the third condition, which is the condition on the positions of ghost cells. From this observation, we get the following result which was discussed in Remark 2.7.

Two operators on Kohnert diagrams
In order to prove the well-definedness and bijectivity of Ψ α and Φ α defined in Section 3, we introduce two auxiliary operators g and e on KD(α) and study their properties. Later in Section 7, we will use these two operators to give alternative descriptions of Ψ α and Φ α . 6.1. Introducing the g operator. We define an operator g on KD(α) for each g ∈ [n].
We would like to determine when g (D) is well-defined. This is partially answered by the following lemma: The proof involves Kohnert tableaux from subsection 2.4. Proof. Assume g (D) = (D , k). Then in column 1 of D , there are |(g, n] ∩ K 1 | + 1 cells weakly above row g. In Label α (D ), these cells are filled by distinct number in [g, n] ∩ supp(α). Thus, |[g, n] ∩ supp(α)| > |(g, n] ∩ K 1 |. Next, we will show the converse of this lemma. First, we introduce an algorithm called sharp algorithm. Its input would be a number g and D = (K 1 , ∅, d) ∈ KD(α) such that |[g, n] ∩ supp(α)| > |(g, n] ∩ K 1 |. It will output a diagram pair D with only Kohnert cells. It will also output a filling of D . Later, we will check the filling is a Kohnert tableau with content α, which implies D ∈ KD(α). Finally, we will check D is the first component of g (D).
The sharp algorithm consists of five steps: • Step 1: Compute Label α (D). • Step 2: Since |[g, n] ∩ supp(α)| > |(g, n] ∩ K 1 |, there is a number m such that m g but m is weakly below row g in column 1. Find the highest such m. Let k be the row index of this m.  Lemma 6.4. The filling produced by the sharp algorithm is a Kohnert tableau with content α. Consequently, the filling is Label α (D ) and D ∈ KD(α).

Proof. We claim that after
Step 4 and after each iteration of Step 5, the filling satisfies the first three conditions of Definition 2.13. Moreover, if i < j violates the last condition in column c, then j = m and c = 1. After Step 4, the filling clearly satisfies the first three conditions. Now assume i < j violates the last condition in column c. Clearly, c = 1 and m is i or j. Assume m is i, then j is below row g in column 1. If j were below row k, then m, j would have violated condition 4 before this move. On the other hand, if j were above row k, then we would have picked j instead of m. In either case, we reach a contradiction. Thus, j must be m.
If there is a u such that u < m violates condition 4, we pick the smallest such u. Assume our claim holds now. We need to show our claim is still true after we swap u and m. We check the first three conditions: (1) Condition 1 clearly holds.
(2) Only need to check condition 2 for m. Recall u < m. Since u satisfies condition 2 before the move, so does m after the move.
(3) Only need to check condition 3 for u. Since u, m violate condition 4 before this swap, and that u satisfies condition 3 before moving m to (1, g), there is no u in column 2 strictly above the m in column 1. Thus, after the move, u satisfies condition 3. Now assume i < j violates condition 4 in column c. Clearly, c = 1 and one of i, j is u or m. We just need to check u cannot be i or j and m cannot be i: • Assume i = u. Then u, j would have violated condition 4 before the move, contradicting to our claim. • Assume j = u. Then i < u < m. Before the move, i, m violates condition 4. Then we would have picked i and swapped it with m, rather than u. Contradiction. • Assume i = m. If j were below m before the move, then m, j would have violated condition 4 before the move. Now assume j were between u and m before the move. Then u, j would have violated condition 4 before the move. Now our claim holds after each move. When the sharp algorithm terminates, there is no violation of condition 4 with the form i, m. Thus, the filling satisfies condition 4, so it is in KT(α).
Lemma 6.5. The D yielded by the sharp algorithm is the first component of g (D). Proof. The lemma is trivial if g ∈ K 1 . Thus, we may assume g / ∈ K 1 . We already showed D ∈ KD(α) by constructing Label α (D ). Take r ∈ K 1 with k < r < g. It suffices to show (K 1 − {r} {g}, ∅, d) / ∈ KD(α).
In column 1 of Label α (D), assume s 1 , . . . , s p are the numbers below row g and weakly above row r. By how we picked m, we know s 1 , . . . , s p < g. We may run the labeling algorithm on ((K 1 − {r}) {g}, ∅, d). It behaves the same as on D on cells prior to (1, r). After filling all these cells before (1, r), we know s 1 , . . . , s p are still in the set S 1 . However, there remains only p − 1 empty cells below row g. Thus, at least one number of s 1 , . . . , s p will be placed weakly above row g. Since this number is less than g, the labeling algorithm will terminate and produce no output. Thus, this diagram is not in KD(α).
Thus, we know the sharp algorithm outputs the first component of g (D), together with its Kohnert Labeling. Now we can tell when g (D) is well-defined. • c > 1; • r < g and r = k; • r > m. Proof. By the behavior of the sharp algorithm, Label α (D ) is obtained from Label α (D) by moving m from (1, k) to (1, g) and repeatedly swapping m with a number above it. The number m will not go above row m, so only (1, k) and cells between row g and row m in column 1 are affected. 6.2. Commutativity of g operators. Next, we observe that two g operators might "commute" under certain conditions. Consider the following example: Example 6.8. Let α = (0, 0, 2, 1). Let D be the following element in KD(α). We first apply 3 and get (D 1 , 2). Then apply 4 on D 1 and get (D f inal , 1).
We can try to swap the order of these two operators. We first apply 4 on D and get (D 2 , 1). Then we apply 3 on D 2 and get (D f inal , 2).
Observe that changing the order of these two operators will not affect the final Kohnert diagram.
This phenomenon is captured by the following two lemmas.
Lemma 6.9. Take D ∈ KD(α). Take g 1 , g 2 ∈ [n] with g 1 < g 2 . Assume g 1 (D) = (D 1 , k 1 ) and g 2 (D 1 ) = (D f inal , k 2 ). If k 1 > k 2 , then the two operators "commute". That is: • g 2 (D) = (D 2 , k 2 ) for some D 2 ∈ KD(α), and Proof. Let C be the first column of Label α (D). Define C 1 and C f inal similarly. In C, let m 1 (resp. m 2 ) be the number at row k 1 (resp. k 2 ). First, we claim m 1 is below row g 2 in C 1 . If not, then we may find a number u that is weakly above row g 2 in C but below row g 2 in C 1 . By Corollary 6.7, the u is above row g 1 in C 1 , so u is higher than m 2 . By u g 2 , we should pick u rather than m 2 when computing g 2 (D 1 ). Contradiction. Now consider C 1 . By Corollary 6.7, m 2 is still at row k 2 . All numbers between m 2 and row g 2 will be less than g 2 . Thus, m 1 < g 2 . Since C and C 1 only differ between row g 1 and row m 1 , all numbers between m 2 and row g 2 will be less than g 2 in C. Thus, g 2 will also pick m 2 when acting on D. g 2 (D) = (D 2 , k 2 ). Column 1 of Label α (D 2 ) agrees with C between row k 1 and g 1 . Thus, g 1 will pick m 1 when acting on D 2 . Lemma 6. 10. Take D ∈ KD(α). Take g 1 , g 2 ∈ [n] with g 1 < g 2 . Assume g 2 (D) = (D 2 , k 2 ) and g 1 (D 2 ) = (D f inal , k 1 ). If k 1 > k 2 , then the two operators "commute". That is: • g 1 (D) = (D 1 , k 1 ) for some D 1 ∈ KD(α), and Proof. Let C be the first column of Label α (D). Define C 2 and C f inal similarly. In C, let m 1 (resp. m 2 ) be the number at row k 1 (resp. k 2 ). First, C and C 2 agree between row k 2 and row g 2 . Thus, g 1 would also pick m 1 when acting on D, so g 1 (D) = (D 1 , g 2 ).
Since g 2 picks m 2 when acting on D, all numbers between row k 2 and row g 2 in C are less than g 2 . In particular, m 1 < g 2 . We know column 1 of D 1 is obtained from C by changing cells between row k 1 and row m 1 . Thus, in column 1 of D 1 , all numbers between row k 2 and row g 2 are still less than g 2 . When acting on D 1 , g 2 would still pick m 2 . 6.3. Introducing the k operator. Next, we define the operator k , which can be viewed as the (partial) inverse of g . Lemma 6.12. Take D ∈ KD(α).
We would like to determine when k (D) is well-defined. This is answered by the following lemma: Lemma 6.13. Assume D = (K 1 , ∅, d) ∈ KD(α). Let K 2 be the set of row indices for cells in column 2 of D. Then k (D) is well-defined if and only if k ∈ K 1 or |K 1 ∩ (k, n]| > |K 2 ∩ (k, n]|. Proof. First, assume the condition fails. We show k (D) is undefined. Assume by contradiction that k (D) = (D , g). Define K 1 and K 2 similarly for D . Then K 1 = (K 1 − {g}) {k} and K 2 = K 2 . Thus, |K 1 ∩ (k, n]| < |K 1 ∩ (k, n]| |K 2 ∩ (k, n]| = |K 2 ∩ (k, n]|. Then consider Label α (D ). There are |K 2 ∩ (k, n]| distinct numbers above row k in column 2. They all must appear above row k in column 1, but there are not enough cells for them. Contradiction. Now assume the condition holds, we show k (D) is well-defined. Clearly, we are done if k ∈ K 1 . Now assume |K 1 ∩ (k, n]| > |K 2 ∩ (k, n]| and consider Label α (D). By our assumption, we can find m above row k in column 1 such that there is no m above row k in column 2. Pick the lowest such m and move it to (1, k). Then the resulting filling is in KT(α): (1) Condition 1 of KT(α) is clear.
(2) Since we moved a cell down, condition 2 is clear. Remark 6.14. In the previous proof of well-definedness, we choose the cell containing m and move it down to row k. The resulting diagram is still in KD(α). Notice that this might not be the lowest cell that can do this job. See the following example.
Example 6.15. Following Example 6.3. We would like to compute 1 (D ). In D , there are 4 cells in column 1 above row 1 and there are 2 cells in column 2 above row 1. Thus, the condition in Lemma 6.13. is satisfied. We want to check 1 (D ) is well-defined. The proof of well-definedness gives m = 6. After moving the 6 to row 1, the resulting filling is in KT(α), which implies the underlying diagram is in KD(α). However, moving the cell (1, 3) to (1, 1) in D will also make the resulting diagram in KD(α).
Proof. It follows directly from Lemma 6.10 and Lemma 6.12. 6.4. Relations between g and e . In this section, we investigate the relationship between the two operators introduced above. We already know the effect of g can be reversed by the e operator, and vice versa. Next, we show that a sequence of g can also be reversed by a sequence of e , and vice versa.
If we compute Ψ α (D) using the description in subsection 3.1, we would go through the following iterations.
It is clear that this recursive description agrees with the original description of Ψ α . To prove Lemma 3.3, we need to show t, D 0 , . . . , D |G 1 | exist and satisfy our assumptions. Besides, we need to check the final output is a diagram pair in RSVT(α).
Then clearly if we ignore ghost cells in D 0 , it is in KD(α). Moreover, D 0 has no ghost cells in column 1. Next, we need to show the diagram pairs D i are well-defined. By Theorem 5.1, we know for each g i , |[g i , n] ∩ supp(α)| > |[g i , n] ∩ K 1 |. Notice that the first i − 1 iterations will not move any cells above row g i−1 . Thus, [g i , n] ∩ K 1 = [g i , n] ∩ K i−1 1 . By Lemma 6.6, D i exists. Next, we need to check the image is in RSVT(α). In other words, should satisfy all four conditions in Theorem 5.4. Let L 1 be the set of row indices of Kohnert cells in column 1 of t.
(1) The first condition is immediate.

7.2.
Recursive description of Φ α . Let E be an arbitrary diagram. We can recursively describe the operator E (·). If D is empty, then E (D) is also empty if E = ∅, or undefined otherwise. If  Proof. Recall that E (L) iterates over cells of E from left to right. Within each column, it goes from top to bottom. For a cell (c, r) ∈ E, it picks the lowest cell weakly above (c, r) such that once this cell is lowered to (c, r), the diagram is still in KD(α). Then it moves the chosen cell to (c, r).
Let D = (L 1 , ∅, t) be the Kohnert diagram at the beginning of the iteration of (c, r) ∈ E. The iterations of (1, e) ∈ E can be characterized by the e operator. Assume c 2. By the recursive description of KD(α), the following two statements are equivalent for any r < r: • (c, r ) is a cell in D such that if we move it to (c, r), the diagram is still in KD(α).
• (c − 1, r ) is a cell in t such that if we move it to (c − 1, r), the diagram is still in KD(M (α, L |E 1 | 1 )).
It is clear that this recursive description agrees with the original description of Φ α . To prove Lemma 3.6, we need to show d, D 0 , . . . , D |E 1 | exist and satisfy our assumptions. Moreover, we need to check the final output is a diagram pair in KKD(α).
Clearly D 0 is a diagram pair whose Kohnert cells form a diagram in KD(α) and has no ghost cells in column 1. Next, we need to show D i is well-defined for an arbitrary 1 i |E 1 |. Notice that the first i − 1 iterations will not move any cells weakly below row e i−1 . Let (1) The first condition is immediate.
Now we have the desired weight-preserving bijection between KKD(α) and RSVT(α). We can claim the Ross-Yong conjecture is correct.

Acknowledgments
The authors thank Sami Assaf, Brendon Rhoades and Mark Shimozono for helpful conversations.