A major-index preserving map on fillings

We generalize a map by S. Mason regarding two combinatorial models for key polynomials, in a way that accounts for the major index. We also define similar variants of this map, that regards alternative models for the modified Macdonald polynomials at $t=0$, thus partially answer a question by J. Haglund. These maps imply certain uniqueness property regarding inversion-- and coinversion-free fillings, which allows us to generalize the notion of charge to a non-symmetric setting, thus answering a question by A. Lascoux. The analogous question in the symmetric setting proves a conjecture by K. Nelson.

• In [Mas09], two models for key polynomials are given, with a column-set preserving bijection showing that these are equal. We generalize this map to incorporate a q-parameter, corresponding to major index. It is worth noting that even when q = 0, the bijection between the sets we consider is nontrivial -this case was treated in [Kur16] and the map given in the present paper specializes to the one in [Kur16] when dealing with q = 0.
• We explicitly construct a biword for coinversion-free fillings and extend the biword given for modified Macdonald polynomials in [HHL05]. In the first case, the charge of the biword is shown to be equal to the major index of the filling, while in the second case the cocharge of the biword is equal to the major index of the filling. The extension of the second biword beyond the partition case proves a generalization of a conjecture given in [Nel05].
• We demonstrate a bijective proof thatH σ σλ (x; q, 0) =H λ (x; q, 0), for any fixed basement σ ∈ Sn. In particular, we show that the bijection between the corresponding fillings can be taken to be column-set-preserving, a property that uniquely defines this bijection.
Our proof method amounts to first constructing the maps for fillings with two rows, and then proving that these maps are compatible with a larger filling. The second part of this proof is partially done through computer verification, due to the large (but finite) number of cases that needs to be considered.
The paper is structured as follows: In Section 2, we introduce the necessary terminology regarding the combinatorial model for non-symmetric Macdonald polynomials. In Section 3 and Section 4, we define and prove properties of the maps. Finally, in Section 5, we examine the various applications and consequences of the maps.

Non-symmetric Macdonald polynomials and fillings
In this section, we review the necessary terminology regarding the combinatorial model for non-symmetric Macdonald polynomials, and modified Macdonald polynomials. We use the same notation as in [Ale15], which differs slightly from the one used in [Hag07,HHL08]. In particular, we use English notation, and not the "skyline" way of presenting fillings.
An augmented diagram of shape α is a Young diagram where the length of row i from the top is given by 1 + αi. The leftmost column is considered special and is referred to as the basement. An augmented filling is an assignment of natural numbers to the boxes in the diagram. We specify the entries in the basement by listing them from top to bottom -in most places, the basement is a permutation expressed in the one-line notation. The weight of a filling is the multiset of entries that are not part of the basement. We let x F := u∈F xu, where u ranges over all non-basement entries in F . Example 1. Below is an augmented filling with shape (2, 4, 0, 3, 2), basement given by (4, 5, 1, 3, 2) and x F = x 2 1 x 2 2 x 3 3 x4x 2 5 . 4 1 2 5 5 4 1 3 1 3 3 3 5 2 2 Let F be an augmented filling. Two boxes a, b, are said to be attacking if F (a) = F (b) and the boxes are either in the same column, or they are in adjacent columns with the rightmost box in a row strictly below the other box. A filling is non-attacking if there are no attacking pairs of boxes. The filling in Example 1 is non-attacking.

Inversions, coinversions and descents
A triple of type A is an arrangement of boxes, a, b, c, located such that a is immediately to the left of b, and c is somewhere below b, and the row containing a and b is at least as long as the row containing c. In a similar fashion, the electronic journal of combinatorics 22 (2015), #P00 a triple of type B is an arrangement of boxes, a, b, c, located such that a is immediately to the left of b, and c is somewhere above a, and the row containing a and b is strictly longer than the row containing c.
A type A triple is an inversion triple if the entries ordered increasingly, form a counter-clockwise orientation. Similarly, a type B triple is an inversion triple if the entries ordered increasingly form a clockwise orientation. In the case of equal entries, the one with largest subscript in Eq. (1) is considered to be largest.

a3 b1
(1) A triple which is not an inversion triple is called a coinversion triple.
Let F be an augmented filling and suppose b is a non-basement box, and a is the box immediately to the left of b. We say that b is a descent of F if F (a) < F (b). The set of descents of F is denoted Des(F ).
The leg of a box u, leg(u), in an augmented diagram is the number of boxes to the right of u. The arm, arm(u), of u = (r, c) in an augmented diagram α is the total cardinality of the sets {(r , c) ∈ α : r < r and α r αr} and {(r , c − 1) ∈ α : r < r and α r < αr}.
Given an augmented filling F , the major index, maj(F ), is defined as Let σ ∈ Sn and let α be a composition with n parts, and let NAF(α, σ) denote the set of non-attacking fillings of the augmented diagram of shape α and basement σ, with entries in 1 . . . , n. The non-symmetric permuted basement Macdonald polynomial E σ α (x; q, t) is defined as where u− denotes the box to the left of u, and we consider F (u−) not to be equal to F (u) if u is a box in the basement.
The ordinary non-symmetric Macdonald polynomial Eα(x; q, t) considered in [HHL08] is recovered when taking σ = w0 = (n, . . . , 2, 1), that is, the unique longest permutation in Sn. The basement w0 is often referred to as the key basement -the reason will be evident further down.
It is possible to generalize the modified Macdonald polynomials to permuted basement modified Macdonald Polynomials by considering a big basement σ as the sequence (∞σ 1 , ∞σ 2 , . . . , ∞σ n ). Then definẽ where FIL(α, σ) is the set of fillings of composition shape α with no restriction whatsoever. However, one can show that these polynomials (up to a multiple of a power of t) only depend on the parts of α, see e.g. Eq. (16) further down. This fact is not easy to see from the definition here.

Specializations of Macdonald polynomials
The main topic of this paper is the specialization of (2) and (4) at t = 0. In the first case we have the simplified expression where CoInvFree(α, σ) is the set of coinversion-free fillings of shape α and basement σ. Note that a coinversion-free filling is also automatically non-attacking. We define the key polynomial Kα(x) as the specialization E w 0 α (x; 0, 0)note the use of the key basement w0.
Similarly, we have the simplified expressioñ where InvFree(α, σ) is the set of inversion-free fillings of shape α and basement σ. The standard modified Hall-Littlewood polynomials,Hα(x; q, 0) are recovered by letting σ = w0 and α be a partition.

Brief background on Demazure operators
We only use the following operators briefly in Proposition 3 below, and this subsection is only to give some context.
The non-symmetric Macdonald polynomials and the more general permuted basement Macdonald polynomials can also be defined using Demazure-Lusztig operators, see e.g., [HHL08,Ale15] for details.
Let si act by simple transposition on the indices of the xi, and define It is straightforward to see that ∂i, πi and θi are operators on C[x]. The operators πi and θi are used to define the key polynomials and Demazure atoms, see e.g. [LS90,Mas09]. We define the following t-deformations of the above operators.π Theθi are the Demazure-Lusztig operators, and generators of the affine Hecke algebra. Theπi andθi both satisfy the braid relations, andπiθi =θiπi = t. These operators act on the basement of permuted basement Macdonald polynomials, as well as the indexing composition.
the electronic journal of combinatorics 22 (2015), #P00 Non-symmetric Macdonald, E σ α (x; q, t) Figure 1: Here is an overview of the polynomials. Arrows down indicate the relation specializes to, and plus-arrows down indicate expands positively into. Note that one relation is Conjecture 15, which we discuss further down. Here, λ is a partition, and α is a composition.

A column-set preserving map
By using the operators in Section 2.3 together with the properties proved in [Ale15], it is possible to show the existence of a weight preserving and major index preserving bijection between two sets of coinversion-free fillings: Proposition 3. Let CoInvFree(α, σ) be the set of coinversion-free fillings with shape α and basement σ. Suppose σi = σi+1 + 1 and αi < αi+1. Then there exists a bijection φ : CoInvFree(α, σ) ←→ CoInvFree(siα, σ) CoInvFree(siα, siσ) with the property that φ preserves major index.
Proof. Using relations in [Ale15], we have , where si acts via simple transpositions on the parts of α.
Given a filling, its column-sets is simply the list of (multi)sets of entries in each column. For example, the The purpose of this paper is to explicitly construct a bijection φ with the additional property that it is column-set preserving, that is, F and φ(F ) have the same column sets. Note that it is not clear a priori that we can impose such a strong condition on φ.
In order construct such a φ -which turns out to be unique -we first prove the following statement: Proposition 4. Let CoInvFree(α, σ) be the set of coinversion-free fillings with shape α and basement σ. Suppose σi > σi+1 and αi > αi+1. Then there is an injection with the property that φ preserves column sets and major index.
The first part of this proposition together with Proposition 3 implies that φ is a bijection whenever σi = σi+1 +1.
The proof of Proposition 4 is broken into two major parts: we first construct φ for fillings with two rows in Lemma 5, and then show that the result is compatible with the remainder of the filling in Lemma 6.
Proof. We define a filling rule to transform a filling F in CoInvFree(α, σ) to some F ∈ CoInvFree(s1α, σ) CoInvFree(s1α, s1σ): Start at the end of the first row of F and map that entry to the bottom row (and corresponding column) in F . Label this entry C and consider the adjacent column to the left.
If this adjacent column in F has one entry, map that entry to the bottom row of F and let C now denote this entry.
Otherwise, this column has entries {A, B}. If either entry is greater than C, choose the least element greater than C to be in the bottom row. If neither entry is greater than C, choose the smallest entry to be in the bottom row. Repeat this procedure by letting C be the bottom entry in F in the column to the right of the column being processed. It suffices to show that the resulting F is both coinversion-free and that it has the same major index as F .
The first part is easy -note that A = B since otherwise, F would contain a coinversion. Then exactly one of the arrangements of A and B can produce a coinversion together with C, and it is straightforward to verify that the filling rule gives the coinversion-free choice.
The main difficulty therefore is demonstrating that the filling rule is major index preserving. We proceed by strong induction on the length of the shortest row α2 to demonstrate that the map is major index preserving. For the remainder of the proof, let < indicates the presence of a descent between two entries while indicates the lack of such a descent. For the base case, suppose that the shorter row has length 0.
Suppose F has the form a1 c1 c2 · · · cm b1 where a1 and b1 are the basement entries. There are two separate cases to consider.
Base case a: c 1 > a 1 . Since a1 > b1 it follows that F has the form a1 b1 < c1 c2 · · · cm and therefore both the initial and final fillings have a descent between the first and second columns and the major index is preserved. and in either case, there is no descent between the first and second column. Hence, the major index is preserved and we have proved the base case.
For all the general cases below, we use filling where a1, b1 is the basement with a1 > b1. As before, F is the filling obtained from F via the filling rule. We proceed by casework based on how the filling rule was applied.
The main outline for this casework is as follows: There are a few cases to consider depending on if there is some (smallest) index , 1 < k such that a > b , and if so, how this column appears in F . Note that we can use the induction hypothesis on all descents to the right of such a column, and conclude that these contribute the same amount to the major index in the two fillings.
Case 1: There exist some 3, such that a > b , ai < bi for 2 i − 1 and a appears on By the inductive hypothesis, we have that the major index is preserved after the th column. 3 Since F has no coinversions it follows that bi+1 > ai ai+1 for 2 i − 2 and since bi > ai over the same indices it follows that bi > ai ai+1. Therefore -using the filling rule -it follows that F is of the form σ(a1, b1) b2 · · · b −1 < a · · · σ(a1, b1) a2 · · · a −1 < b · · · c1 · · · cm and the major index is clearly preserved from the second column onwards. To prove that the entries in the first and second column preserves major index, there are two subsubcases to consider. If b1 a2, then b2 > a1 > b1 a2 since F has no coinversions. Thus, Since the descent in both fillings is in the shortest row, it is clear that in both cases it has the same contributes to the major index.
In the second subsubcase, we have a2 > b1, and since F has no coinversions this implies that b2 > a1 a2 > b1.
and major index is preserved again.
Since F is coinversion-free it also implies that bi+1 > ai ai+1 for 2 i − 2. Furthermore, since bi > ai over the same indices, it follows that bi > ai ai+1 and F is of the form Using the same treatment as in Subcase 1a, it follows that maj F = maj F .
The major index is preserved before the ( − 2) nd column by the reasoning in Subcase 1a. Hence, the major index is preserved on the entire filling and statement follows.
In this case, both F and F have two rows which are of the form Continuing left in such a manner, we either reach an index such that bj−1 < bj for 3 j − 1 or there is no such index. If bj−1 < bj is the greatest such j, it follows using logic identical to above that F has the form · · · aj−1 aj aj+1 · · · a −1 < a · · · c1 c2 · · · cm · · · bj−1 < bj bj+1 · · · b −1 b · · · while F has the form · · · bj−1 aj aj+1 · · · a −1 < a · · · · · · aj−1 < bj bj+1 · · · b −1 b · · · c1 c2 · · · cm and the major index is preserved from column j through column . The remaining columns have major index preserved due to Subcase 1a, and thus major index is preserved overall. If there is no such j, note that b2 > a1 a2 since F is coinversion-free with a1 > b1. Therefore, F is of the form and major index is preserved as the sum of leg lengths are the same.
Case 2: There exist some 3, such that a > b , ai < bi for 2 i − 1 and a appears below b in F .
the electronic journal of combinatorics 22 (2015), #P00 Subcase 2a: If a −1 a then it follows that a −1 a > b . Therefore, F is of the form and the major index is preserved using Subcase 1a.
and thus the major index is preserved using the logic of Subcase 1a.
The major index is preserved in this case by following Subcase 1b.ii.
Case 3: Case = 2. This means a2 > b2. By inductive hypothesis, the major index is preserved from the second column onward. We only need to verify that major index is preserved among descents between first and second column. so the major index is preserved, since the descents between first and second columns appear in both fillings.
Subcase 3b: Otherwise a1 b2 and then there are two possible cases.
Subcase 3b.i: a2 appear below b2 in F : • Case b1 a2: It follows that we have the indicated non-descents: a2 . . . and thus the major index is preserved.
• Case b1 < a2: With F being coinversion-free, we have a1 a2 > b2, so where * contributes in the same way to both fillings if it is a descent.
the electronic journal of combinatorics 22 (2015), #P00 Subcase 3b.ii: a2 appear above b2 in F : There are again two possible cases.
• Case b1 b2: This condition together with F being coinversion-free implies that b2 < a2 a1 and we must have b2 . . .
• Case b1 < b2: This gives a1 a2 > b2 > b1 and it follows that In both these subcases, major index is preserved.

Case 4:
The last case is occurs whenever a1 > b1 and ai < bi for 2 i k.
and using the logic in Subcase 1a the result follows.
Using the reasoning from Subcase 1b.ii, there exists a j 1 such that The two marked descents in F has leg lengths that sum to the leg length of the single marked descent in F . Furthermore the major index is preserved before the j th using the logic of Subcase 1a and the therefore the major index is preserved overall.
All cases have now been covered, and this concludes the proof.
The remainder of the proof is the verification that this two row filling rule is compatible with the remainder of the filling -that is, applying the filling rule on two rows in a larger filling, no coinversions are introduced.
Since the presence of a coinversion is a local condition, we can reduce the proof to a finite list of possible configurations. We verify these via computer verification and the procedure is described in the detail in the following proof. Proof. Since a coinversion only depends on elements in two separate rows, it suffices to verify the algorithm for augmented fillings with precisely three rows -the two rows affected by the filling rule, and a fixed third row. We may then assume that these three rows are adjacent. The general strategy is to consider a 2 × 3 sub-grid of the diagram in F , and then conclude that there are no coinversions in F . Furthermore, it suffices to restrict the entries between 1 and 6, since only the relative order among the entries determines the presence of coinversions. Let the two by three grid be as below Since inversions depend on the relative order of the row lengths, we need to take that into consideration, by assigning each row a rank, a number between 1 and 3, indicating the relative order of the row lengths. In case two rows have the same length, the bottom row is assigned a lower rank. The rank together with the up to six entries in the grid allow us to determine which triples in the grid that are coinversions, and we only consider such possible grids in F which are coinversion-free.
Some of the entries in the grid might not be present in the case we are examining the far right of a filling. However, it is straightforward to see that there is no loss of generality to assume that a, b and c are always present, as the other cases can easily be verified by hand.
The filling rule processes the two rows being swapped from right to left, and we need to verify that there is no coinversion in F being produced involving the third fixed row. Note that there might locally be two possible images, F and F given a local grid F , depending on what the filling rule implied in the second column -in this example, the top two rows are swapped: There are exactly three things that can occur locally in the grid. We verify this using the computer.
The diagram is degenerate. One of the rows being swapped has an element missing in the grid. If this is the case, the position of the grid entries in F are uniquely determined by the filling rule and there is only one case in (10). By checking all such local cases, we see that all corresponding grids are coinversion-free.

Both possibilities are valid.
The non-fixed entries in the second column can either be swapped or not by the filling rule, and both these possibilities yield a coinversion-free grid using the filling rule. This is verified by computer. As an example of this situation, we might have F = 3 : 4 2 1 : 1 1 2 : 6 5 F = 1 : 4 2 3 : 1 1 2 : 6 5 F = 1 : 1 1 3 : 4 2 2 : 6 5 and in either case, the filling rule produces valid (coinversion-free) grid.

Only one of the grids is coinversion-free.
This situation requires a more careful analysis, and we need to do a non-local analysis to prove that F is indeed of the form that produce a coinversion-free filling.
Computer check verifies that the event that only one of the two grids are valid occurs only under the conditions in the following claim, which then determines that we are in the case that produces a valid grid: Proof. Suppose that the entries in the second column of F are not as in Eq. (12), that is, we assume they did not "flip".
If the adjacent column to the right of the second column in F also has all three entries present, it follows (via computer verification) that these entries also have the down-increasing condition. This third column in F then also appears identically in F . The down-increasing condition is therefore an invariant, present in all further columns to the right, via induction.
Eventually, we reach the end one of the shortest row, where last complete column satisfies the down-increasing condition and is identical in both F and F . Finally, an exhaustive search on the computer shows that it is impossible for F and F to be of these specified forms and simultaneously be coinversion-free, a contradiction.
Hence, the entries in F must be arranged as in Eq. (12). Computer verification on the local situation in Eq. (12) verifies that the filling rule produces no coinversions.
To conclude the proof, here is a summary of properties that needs to be verified via computer: • Only the three local cases listed above appear among all possible 2×3-grids; degenerate, two valid possibilities, one valid possibility of specified form.
• The filling rule in Claim 7 preserve the down-increasing condition.
• Preserving the down-increasing condition eventually leads to a contradiction -that is, once we reach the end of the shortest row, F and F cannot simultaneously be coinversion-free. This forces F to be of the expected form.
We have done this in Mathematica and Java with two different implementations.

Another column-set preserving map
It is possible to mimic the above proposition in the case of modified Hall-Littlewood polynomials -fillings without inversions.
First note that the above proposition it trivial when ai = ai+1 as we can simply interchange the two basement entries and preserve the remainder of the filling, since this action does not introduce inversions and it clearly preserves the major index. In the case when ai > ai+1, we proceed as before and first establish a bijection in the two row case, followed by proving that this is compatible with a larger filling.
Lemma 9. Let InvFree(α, σ) be the set of inversion-free two-row fillings with shape α = (α1, α2) and big basement σ = (σ1, σ2), with σ1 > σ2 and α1 > α2. Then there is a bijection ϕ ϕ : InvFree(α, σ) ←→ InvFree(s1α, s1σ) with the property that ϕ preserves column sets and major index. Proof. We first consider the case when α2 = 0; in this case we can simply interchange the rows and the result follows trivially. Otherwise for the remainder of the proof, treat the column immediately after the basement to be the first column and note the entries this column are in non-increasing order in order for the filling to have no inversion.
We now define a filling rule to transform a filling F in InvFree(α, σ) to a F ∈ InvFree(s1α, s1σ): Start at the end of the first row and map that entry into to the bottom row in F . Label this entry C and consider the adjacent column to the left.
If this adjacent column in F has one entry, map that entry to the bottom row of F and let C denote this entry.
Otherwise, the adjacent column in F has entries {A, B}. If both entries are greater than or equal to C or less than C, map the larger entry to be in the bottom row of F . Otherwise map the lesser entry to be in the bottom row of F . Repeat this procedure with C being the bottom entry in the adjacent column to the right of the column being processed. It suffices to demonstrate that F is inversion-free and that it has the same major index as F . The first part is note that given a particular {A, B} there is exactly one arrangement of A and B that does not give a coinversion with C, and it follows upon a straightforward verification that the filling rule gives the inversion-free choice.
We now proceed by strong induction on the length of the shorter row α2 to demonstrate that the filling rule is major index preserving. As in the previous proof let < indicate the presence of a descent between two entries while indicates the lack of a descent. For the base case, suppose that the shorter row has length 1. Suppose that F has the form a1 c1 c2 . . . cm b1 where a1 and b1 are the first entries after the basement. Then there are two separate cases to consider.
Base case a: a1 c1. Since a1 b1 it follows that F has the form a1 b1 c1 c2 . . . cm and thus both F and F lack a descent between the first and second columns and the major index is preserved. were a1 and b1 are the first element after the basement and a1 b1. As before, F is the filling obtained from the reverse filling rule. We now proceed by casework on how the filling appears after the filling rule is used.
Case 1: There exists 3 such that a b , ai > bi for 2 i − 1 and a appears on top on b in F . By inductive hypothesis, the major index is preserved after the th column. 4 Subcase 1a: a . Since F is inversion-free, it follows that bi+1 ai < ai+1 for 2 i − 2 and since ai > bi over the same indices it follows that bi < ai < ai+1. Therefore, using the filling rule it follows that F is σ (a1, b1) b2 . . . b −1 a . . .   σ(a1, b1) a2 . . . a −1 b . . . c1 c2 . . . cm and the major index is clearly preserved from the second column onwards. To prove that the major index is preserved between the first and second column, there are two cases. If b1 < a2, then b2 a1 b1 < a2 as F has no inversions. Thus Since the descent is in the longest row in both fillings, it is clear that in both cases it has the same contribution to major index.
In the second subcase, a2 b1 and since F has no inversions it follows that b2 a1 < a2 b1. In this case it follows that we are in the situation F = a1 < a2 . . . b1 b2 . . . and F = b1 b2 . . . a1 < a2 . . . and major index is preserved in this case as well.
This property also implies that bi < ai < ai+1 for 2 i − 2 and thus F has the form Using the same treatment in Subcase 1a, it follows that major index is preserved.
and the major index is preserved up to the ( − 2) nd column by the reasoning in Subcase 1a. Therefore the major index is preserved on the entire filling and statement follows.
Since F is inversion-free it follows that b −1 a −2 < a −1 . In this case, both F and F have two rows which are of the form Continuing left, we either reach an index such that bj+1 bj for 2 j − 2 or there is no such index. If bj+1 bj is the greatest such index it follows using identical reasoning as above that and major index is preserved as the sum of leg lengths are the same.
Case 2: There exist 3 k such that a b , ai > bi for 2 i − 1 and a is below b in F .

Subcase 2a:
If a −1 < a then it follows that b −1 < a −1 < a b . Thus F has the form −1 a . . . b k c1 c2 . . . cm and the major index is preserved using Subcase 1a.
. . . c1 . . . and the major index is preserved using the reasoning in Subcase 1a.
Subcase 2b.ii: Otherwise b −1 < a b , a −1 a . Furthermore a −1 b a as F has no inversions and it follows that the F and F have the form The major index is preserved using the reasoning in Subcase 1b.ii Case 3: Case = 2. In this case a2 b2 and by inductive hypothesis, the major index is preserved from the second column onward. We need therefore simply to verify that the major index is preserved among descent between the first and second column. The major index is preserved between the first and second columns since descents are not present in either possible position.
Subcase 3b: Otherwise b2 > a1 and then there are two possible cases.
Subcase 3b.i: a2 appears below b2 in F : The marked descents in F and F have leg lengths that sum to the same value so the major index is preserved after column j while the major index is preserved from column 1 to column j using Subcase 1a.
We have now verified all possible cases and the result that the filling rule is major index preserving follows via strong induction on the length of the shorter row.
To demonstrate that is is a bijection, we refer to Theorem 5 5.1.1 in [HHL05], with t = 0. It follows that the two sets of fillings ϕ map between are equinumerous. This together with the fact that ϕ is injective implies that ϕ is also bijective.
Analogous to Proposition 4, it remains to show that ϕ is compatible with a larger filling. We proceed as before, and reduce this to a finite set of verifications. The approach is similar to the case when we treated coinversion-free fillings, and the main difference is the details in Claim 11.

Lemma 10. The map ϕ applied to two adjacent rows in a larger filling does not introduce any inversions.
Proof. As before, it suffices to consider the three-row case, where ϕ is applied to two adjacent rows and the third row is fixed. Again, we consider a 2 × 3-grid with entries in 1, . . . , 6, and all three entries in the first column are present. There are again exactly three things that can occur locally in the grid. We verify this using the computer.
The diagram is degenerate. One of the rows being swapped has an element missing in the grid -the entries in F are uniquely determined. By checking all such local cases, we see that all corresponding grids are inversion-free.
Both possibilities are valid.
The non-fixed entries in the second column can either be swapped or not by the filling rule, and both these possibilities yield an inversion-free grid using the filling rule. This is verified by computer. As an example of this situation, we might have

Only one of the grids is inversion-free.
This situation requires a more careful analysis, and we need to do a non-local analysis to prove that F is indeed of the form that produce a inversion-free filling.
Computer check verifies that the event that only one of the two grids are valid occurs only under the conditions in the following claim, which then determines that we are in the case that produces a valid grid: Claim 11. Suppose we swap the longest and shortest row in the 2 × 3 grid, as in Proof. Suppose that the entries in the second column of F are not as in Eq. (14), that is, we assume they did not "flip".
If the adjacent column to the right of the second column in F also has all three entries present, it follows (via computer verification) that these entries also have the up-increasing condition. This third column in F then also appears identically in F . The up-increasing condition is therefore an invariant, present in all further columns to the right, via induction.
Eventually, we reach the end one of the shortest row, where last complete column satisfies the up-increasing condition and is identical in both F and F . Finally, an exhaustive search on the computer shows that it is impossible for F and F to be of these specified forms and simultaneously be inversion-free.
Hence, the entries in F must be arranged as in Eq. (14). Computer verification on the local situation in Eq. (14) verifies that the filling rule produces no inversions.
Thus, ϕ is an injection and together with Theorem 5.1.1 in [HHL05], it is also a bijection. It is also possible to prove that this is a bijection by a computer-style method similar to what is done above.
As a final comment on this section, note that the set InvFree(α, σ) (with big basement) is empty if there are i < j such that αi < αj and σi < σj. This is due to the fact that the first entry in row j, and the two basement entries σi and σj form an inversion triple.

Applications
Despite the relatively lengthy proofs given, the realization the fact that these column set preserving maps exist allow for a multitude of short corollaries.
We begin with a technical result that has been partially proved in a variety of special cases.
Proposition 12. Given fixed column sets, there exists at most one coinversion-free filling with shape α and decreasing basement.
Proof. To prove that there is at most one coinversion free filling with shape α and decreasing basement, simply apply the column set preserving operator φ until the shape becomes a partition.
In the partition case, any list of column sets with sizes compatible with α admits a unique coinversion-free filling. This unique filling can be constructed from the column sets via the following iterative process going column by column, and in each column, top to bottom: Given an entry e in column i, the adjacent entry in column i + 1 is found by taking the largest unused entry in column set i + 1 less or equal to e, and if there is no such element, simply take the largest entry. It is easy to verify that this gives a coinversion-free filling and given E λ (x; 1, 0) = e λ (x) the result follows.
Given a coinversion-free filling F , we construct the biword of F as follows: Let the top row be the non-basement entries of F listed in increasing order and the bottom row be the corresponding columns the entries belong to, listed in increasing order in case of a tie in the first row, see Eq. (15) for an example. We let cw(F ) denote the lower row in this biword, the charge word of F . Finally, define the charge of F as charge(cw(F )), where charge(·) is defined as in e.g., [Hag07], by decomposing cw(F ) into standard subwords, computing charge(w) := maj(rev(w −1 )) of each such subword w (interpreted as a permutation), followed by adding the results. The standard subwords are extracted iteratively by finding the rightmost occurrence of the smallest element, then scanning right to left for the next smallest element, looping around it necessary, and then repeating this process until one has found an occurrence of largest element to the word. These letters form the first subword and this process is repeated on the remaining letters until there are no more letters left. The charge for the permutations is displayed to the right, so the total charge of the word is 3.
Theorem 13. Let F be a coinversion-free filling of shape α with the key basement w0. Then charge(cw(F )) = maj(F ).
Proof. Note that cw(F ) is uniquely determined by the column sets of F by construction. Therefore, after applying φ repeatedly, it suffices to prove the theorem when F is of partition shape, and from hereon, F is assumed to be of partition shape. Under this assumption, F can be constructed from the column sets via the following iterative process given in Proposition 12.
Claim: The first column (the column adjacent to the basement) of F is strictly decreasing. Suppose the largest entry in the basement (and thus in the filling) is n. Note that the i th largest entry in the first column is at the electronic journal of combinatorics 22 (2015), #P00 most n + 1 − i, since all entries in a fixed column are distinct. Furthermore, since the i th largest element in the basement is exactly n + 1 − i, the previous construction rule implies that the first column is strictly decreasing. Now note that the subwords obtained in the word decomposition of cw(F ) naturally correspond to rows of F : The claim ensures that the largest entry in the first column is in the topmost row. It follows that entry is also the first entry of the first subword of cw(F ). Now, by the iterative process on how to recover F from its column sets, it is straightforward to see that the first subword of cw(F ) corresponds to the first row of F . By using the fact that the first column is decreasing, this argument can be repeated for the remaining rows to prove that subword j of cw(F ) corresponds to row j of F .
Finally, since the inverse of the j th subword wj can be seen to have the same relative order as the entries in row j, it follows that maj(rev(w −1 j )) is equal to the contribution of row j to maj(F ). Hence, maj(F ) is equal to charge(cw(F )).
Note that Theorem 13 critically relies on the presence of a decreasing basement, and the equality does not hold for non-decreasing basements.
We are now prepared to give an analog of the famous cocharge formula for partition case of the non-symmetric Macdonald polynomials. Note that the following identity can also be proven using properties of LLT polynomials proved in [HHL05] along with the symmetric cocharge formula, however this proof is far more in the style of the proof of the original cocharge formula given in [HHL05]. This particular version also has the advantage that seems to it generalizes to a non-symmetric setting, see Conjecture 15 below.

Theorem 14. Let λ be a partition. Then
Proof. The partition case of Proposition 12 implies that every possible second row (with strictly decreasing blocks in the second row), in a biword can be obtained from some coinversion-free filling of shape λ.
By performing RSK on such biwords, every possible pair (P, Q) of SSYT's with the insertion tableau P having shape µ and the recording tableau Q having shape µ appears. The proof of this is identical to the one given for the traditional RSK in [Sta01]. Since charge is Knuth-invariant, two coinversion-free fillings F1 and F2 with identical insertion tableau P also have the same charge, and Theorem 13 implies that maj(F1) = maj(F2) = charge(P ). It now follows that as desired.
We believe that the above theorem carries over in a more general setting:
In particular, if SQ is the subset of fillings in CoInvFree(α, w0) with recording tableau Q ∈ SSYT(µ , λ ), then F ∈S Q x F is a key polynomial.
Note that a recent combinatorial proof of the key expansion of Eα(x; q, 0) using weak dual equivalence was given in [Ass17].
The following example illustrates Conjecture 15.
Observe that all coefficients are identical, and only the composition indexing the key polynomials differ by some permutation. Furthermore, several different permutations of the same partition might be present, as with K212 and K221 above. This refines the Kostka-Foulkes polynomials, as different terms in the Kostka-Foulkes polynomials might be associated to different key polynomials in the non-symmetric setting.
The question of finding an appropriate generalization of charge that explains this phenomena was raised by Lascoux in [Las13,, and thus Theorem 13 answers this question.

Inversion-free fillings and Hall-Littlewood polynomials
Similar to Proposition 12, we have the following result: Proposition 17. Given fixed column sets there exists at most one inversion-free filling with shape α and big basement σ.
Proof. Proof is similar to the one in Proposition 12.
A number of analogous results are also possible for the modified Hall-Littlewood polynomials. The following identity is given in [HHL08, Thm. 5.1.1]: Let τ be a permutation and λ be a partition. Theñ H τ w 0 τ λ (x; q, t) =H λ (x; q, t).
The proof of this is rather indirect, and J. Haglund asked for a bijective proof of this identity. With our bijection ϕ, we can prove the t = 0 case: Theorem 18. For every τ ∈ Sn and partition λ, there is a column-set preserving bijection that establish the identityH τ w 0 τ λ (x; q, 0) =H λ (x; q, 0).
Although it might be tedious to carry out the bijections, remember that the resulting bijection is uniquely defined by the column-set preserving property.
We have not been able to prove the more general Eq. (16), but computer experiments suggests the following refinement of the equality: This indicates that one should be able to find a column-set preserving bijection proving Haglund's identity.
Remark 20. One can modify Lemma 5 to show that for every τ , we have where [t top ] indicate the coefficient of the maximal power of q, and the bijection proving this is given by φ. To give an outline of the proof, note first that fillings contributing to both sides of the above identity are exactly coinversion-free fillings. Furthermore, notice in this case with the big basement, is that the first set in the right hand side of Proposition 4 is empty.
We conclude this paper by proving an analogue of Theorem 13, in the case of inversion-free fillings. Let F be an inversion-free filling with a basement. Define the following biword of F (different from the above) as follows: Let the top row be the non-basement entries of F listed in decreasing order and the bottom row be the corresponding columns the entries belong to, listed in increasing order in case of a tie in the first row. Let the cocharge word of F , ccw(F ), be the bottom row in this biword. As an example, 7 1 9 1 4 2 3 8 2 1 3 6 −→ 9 8 7 6 4 3 3 2 2 1 1 1 1 1 1 1 3 4 5 2 4 2 2 3 has subword decomposition 13542, 1423, 12, 1, and cocharge value 4 + 1 + 0 + 0 = 5.
Proof. The case where α is partition is given in [HHL05]. 7 The result then follows by noting that the ccw(F ) is only dependent on the columns sets of F , so by applying the column-set and major-index preserving map ϕ until the partition shape is reached, the statement follows.
This answers a conjecture given in [Nel05] 8