Decomposition of the product of a monomial and a Demazure atom

Abstract. We prove that the product of a monomial and a Demazure atom is a positive sum of Demazure atoms combinatorially. This result proves one particular case in a conjecture which provides an approach to a combinatorial proof of Schubert positivity property. Résumé. Nous montrons combinatoirement que le produit d’un monôme et d’un atome de Demazure est une somme positive d’atomes de Demazure. Ce résultat montre un cas particulier d’une conjecture qui fournit une approche combinatoire de la propriété de positivité de Schubert.


Introduction
Let N (or Z + ) be the set of all positive integers and Z ≥0 be the set of non-negative integers.Also we denote by S k the set of all permutations of the set {1, ..., k}.For n ∈ Z ≥0 and k ∈ N, we say α = (α 1 , α 2 , . . ., α k ) ∈ (Z ≥0 ) k is a weak composition n (denoted as α n) with k parts if k i=1 α i = n and write l(α) = k to denote the length (the number of parts) of α.Furthermore, if α 1 ≥ α 2 ≥ • • • ≥ α k ≥ 0, we call α a partition of n with k parts and write α n (usually we denote l(α) = max{ i | α i > 0} for α being a partition).
Let α be a weak composition.The augmented diagram of shape α is the figure with |α| + l(α) cells (or boxes) where column i has α i + 1 cells.The bottom row is called the basement of the augmented diagram.
For example, if α = (1, 0, 1, 0, 0, 4, 0, 6, 5), then the augmented diagram of α is Also we impose an order, called the reading order, on the cells of the diagram which reads each row from left to right, starting from the top row and ends in the bottom row.So the order of the above diagram is: where the number in each cell represents the order of that cell in reading order.
A filling of an augmented diagram is an assignment of a positive integer to each cell in the diagram.Macdonald [Mac96] defined a family of non-symmetric polynomials, called non-symmetric Macdonald polynomials, {E γ (x 1 , . . ., x n ; q, t)|γ is a weak composition with n parts, n ∈ N} which shares many properties with the family of symmetric Macdonald polynomials [Mac95] {P λ (x 1 , . . ., x n ; q, t)|λ is a partition with n parts, n ∈ N}.
Haglund, Luoto, Mason, Remmel and van Willigenburg [HLMvW11], [HMR13] further studied the combinatorial formulas for Demazure atoms and Demazure characters given by the skyline fillings and obtained results which generalized those for Schurs functions like the Pieri Rule, the Robinson-Schensted-Knuth (RSK) algorithm, and the Littlewood-Richardson (LR) rule.
From now on we only consider fillings whose entries in each column are weakly decreasing from bottom (to top) since those are the fillings related to Demazure atoms and Demazure characters [HHL08].
For any two columns (including the basement cells) i and j with i < j, we pick three cells X, Y and Z, where cell X is immediately above cell Y in the 'taller' column k , where and cell Z from the 'shorter' column to form a triple (X, Y, Z) in the following way: Type A triple: cell Z is in the same row as cell X if α i ≥ α j Type B triple: cell Z is in the same row as cell Y if α i < α j .
We say (X, Y, Z) forms a coinversion triple if the filling F of the diagram assigns each cell in the triple a positive integer, say F (X), F (Y ), F (Z) respectively, in such a way that F (X) ≤ F (Z) ≤ F (Y ).Otherwise we call (X, Y, Z) an inversion triple.
Here are some examples of triples: The first two are type A triples and the last two are type B triples while the second and the third ones are coinversion triples and the first and the last one are inversion triples.
1. Let A α be the Demazure atom of shape α.Then where k ∈ S k is the identity.In particular, if α is a partition, then x αi i is a monomial.
2. Let κ α be the key polynomial (Demazure character) of shape α.Then Note that if α is a partition, κ α = A α which is a monomial by 1.The following are some properties among keys and atoms and their products: 3. A key polynomial is a positive sum of Demazure atoms.In fact,
Let λ, µ be partitions and α, β be weak compositions.Denote + A and + κ as being able to be decomposed into positive sum of atoms and keys respectively.Note that by 3, + κ implies + A .Otherwise, we put a × in the cell.For example, a partition (µ)-shaped atom times a key of any shape (α) is key positive and hence we put + κ in the corresponding box.
Tab. 1: Decomposition of products of atoms and keys into atoms The positive results in the table can be found in [HLMvW11], except for the cells marked 1 ,2 and3 .
This paper proves 1 combinatorially in Section 4 which was open: Theorem 1 The product A µ • A α is atom-positive for any partition µ and weak composition α.
The coefficients in the decomposition into atoms are actually counting the number of ways to insert words arising from SSAF of shape α into a SSAF of shape µ and we will discuss properties of words and how to record different ways of insertion in Section 2 and 3. Also note that the product in the theorem is not key positive.A simple counter example would be just putting µ as the empty partition, that is, with all entries 0.
Conjecture 1 Let α, β be weak compositions.Then the product of the key polynomials of shape α and β can be written as a positive sum of atoms, i.e., It is a classical result in Algebraic Geometry that the product of two Schubert polynomials can be written as a positive sum of Schubert polynomials.A representation theoretic proof is also given recently by using Kráskiewicz-Pragacz modules [Wat14].However a combinatorial proof of the positivity property of Schubert polynomials has long been open.Since every Schubert polynomial is a positive sum of key polynomials [LS89], the product of two Schubert polynomials is a positive sum of product of two key polynomials and hence by Conjecture 1 is a positive sum of atoms.This provides a possible approach to a combinatorial proof of the positivity property of Schubert polynomials by trying to recombine the atoms into keys and hence into Schubert polynomials.
Since words and insertion are the crucial tools used to prove Theorem 13, we will discuss an algorithm on words in Section 2, then we will relate it to SSAF and recording tableaux in Section 3. Finally we will use the results we obtained in Section 2 and 3 to prove the main theorem that the product of A λ • A α is atom-positive for partition λ and weak composition α in Section 4.

Convert a column word to a row word
Definition 1 A word is a sequence of positive integers.
Definition 2 Let a, b, c ∈ N and u, v be some fixed (can be empty) words.Define twisted Knuth relation * by: Then we say two words w and w are twisted Knuth equivalent if w can be transformed to w by repeated use of 1. and 2. and we write w * w .Definition 3 A word w is a column word if it can be broken down into k weakly decreasing subsequences of weakly decreasing lengths Definition 4 A word w is a row word if it can be broken down into k strictly increasing subsequences of weakly decreasing lengths Given a SSAF, one can get its column word by using the algorithm described in [HLMvW11], while one can get its row word by reading the entries of each row in ascending order, starting from the bottom row to the top row.We call a word a column (row resp.)word because when we insert each subsequence of the word using the insertion in [Mas08], a new column (row resp.) will be created.
Example 1 886531|97643|9764|5|6 is a column word whose corresponding SSAF is: where 1 ≤ m < k satisfying The proof is basically induction on k.Lemma 3 shows how to create the first entry of the corresponding row word, leaving the remaining part as a column word .We illustrate the Lemma by an example with a Example 2 Let w 0 be the word in Example 1, hence k = 5, c 1 = 6, c 2 = 5, c 3 = 4, c 4 = c 5 = 1.Set i j = 0 for all j.Then we have  . .b kc k is a column word, we can apply Lemma 4 again and get the second subsequence of the row reading word of F (w), and we can apply Lemma 4 repeatedly on the remaining b ij 's until we get all the subsequences of the row reading word of F (w).As a result, we can convert w into the row reading word of F (w) by applying Lemma 4 repeatedly as described.

Convert a Column Recording Tableau to a Row Recording Tableau
This section gives an interpretation of the twisted Knuth equivalence using recording tableaux.We use the insertion in [Mas08] and the generalized Littlewood-Richardson rule in [HLMvW11].We also use the notation U ← W for a SSAF U and a biword W = x 1 x 2 . . .x n y 1 y 2 . . .y n for some integer n to denote the pair (U , L) where U is the SSAF obtained by U ← y 1 y 2 . . .y n while L is the recording tableau, i.e., by putting x i into the cell created when y i is being inserted.In particular, if y 1 y 2 . . .y n is a column (row resp.)word, then we call L as a column (row resp.)recording tableau.By abuse of notation, we sometimes refer U ← W to either U or L (depending on the context).Note that if we change the basement n into the large basement in [HLMvW11], a column recording tableau is the same as an LRS in [HLMvW11].
Lemma 5 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Lemma 6 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Lemma 6 follows by applying Lemma 5 inductively.
Lemma 7 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Lemma 8 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.
The above result follows by applying Lemma 5 inductively.
Lemma 9 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.
is a column word ).Let L be the recording tableau of U ← W .Let L be the recording tableau of We apply Lemma 10 (which is like the tableaux version of Lemma 3) to get the result above.
Lemma 12 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Let w = a 11 . . .a 1c1 |a 21 . . .a 2c2 | • • • |a k1 . . .a kc k with k ≤ n be a column word.Let V be the SSAF representing w (i.e., inserting w into an empty atom).Let w = x 11 x 12 . . .x r1 • • • x c1,1 . . .x c1,rc 1 (as V has c 1 rows and k columns and so r 1 is k), where k = r 1 ≥ r 2 ≥ • • • ≥ r c1 > 0 are the row lengths of V from bottom to top, be the row reading word of V .Then the recording tableau L of U ← W determines the recording tableau R of U ← W , where By Lemma 11 and the corresponding word in Lemma 4, given L, we know how to enter all the 1's in R, which are those cells marked 1 (1) , 2 (1) , . . ., k (1) after applying Lemma 11 as the b k1 b k−1,1 . . .b 11 = x 11 x 12 . . .x 1r1 (by the argument after Lemma 4).
We remove the cells from L in Lemma 11 to create a new L to apply Lemma 11 on, we can get the second row entries (as described in the paragraphs after Lemma 4 ),and hence we know how to put all the 2's into R.By the same argument, we can fill in all entries in R and hence L determines R. 2 4 Decomposition of the product of a monomial and an atom into positive sum of atoms We prove Theorem 1 mentioned in the Introduction.We rephrase the statement more precisely as follows.
Theorem 13 Let λ be a partition and α be a weak composition.Let A λ and A α be atoms of shape λ and α respectively.Then where c β λα is the number of distinct LRS of shape β/λ created by column words whose corresponding SSAF has shape α.
To prove the theorem, we only need to check that given a LRS L of shape β/λ created by column words whose corresponding SSAF has shape α, if there is some column word such that U λ ← W creates the same L, then the SSAF corresponding to w also has shape α.
First consider the last cell in reading order among all those containing entry k in L, that means it is the very first entry inserted.Since a 11 must be inserted in a cell immediately above the cell (including those in basement) containing a 11 , the position of that cell fixes the value of a 11 .Now consider all the cells with k and also the last cell in reading order among all those containing the entry k − 1, then these cells are created by U λ ← a 11 . . .a 1c1 a 21 .By Lemma 12, we know the corresponding row recording tableau and hence we know which two cells are the first two entries being inserted using the corresponding row word.Note that the first row consists of distinct entries and is inserted in ascending order using the row word, then by Lemma 15 in [HMR13], we know the cells are created in ascending reading order (one after another in reading order) and so they must be the cells immediately above those in U λ , and hence the value inside each of those cells in the SSAF created by inserting the row word into U λ is exactly the value inside the cell just below it.Hence we know what the first two row entries of the corresponding SSAF of a 11 a 12 . . .a 1c1 a 21 are.Since we already know the first row entry, which is the lowest entry of the column corresponding to a 11 . . .a 1c1 is, we now know what the lowest entry of the second column (corresponding to a 11 . . .a 1c1 a 21 and hence the same for a 11 . . .a 1c1 a 21 . . .a 2c2 ) is.
We can repeat the same process until we get all the last entries of the k columns and hence fix the shape of the SSAF corresponding to w.Since we read those entries just by considering L, this shows that L fixes the shape of the corresponding SSAF of w and result follows. 2 the corresponding row word is 13689|589|467|357|46|6.Lemma 2 Let u := a 1 a 2 . . .a n |b and v := a 1 a 2 . . .a n |b|c be two column words, where c > b > a n , n ∈ N. Then 1. u * b a 1 ...a n where b < a 1 , a 1 ≥ a 2 ≥ • • • ≥ a n , and b := a t where t = min{ j | b > a j }, 2. v * b c a 1 a 2 . . .a n where b is defined as in 1., b < c < a 1 and a 1 ≥ a 2 ≥ • • • ≥ a n .Lemma 3 Let k ≥ 2 be an integer and i 1 , . . ., i k be nonnegative integers.Let w 0 = a word.Then there exist a (ir+1) rs where 1 ≤ s ≤ c r , 1 ≤ r < k and a (i k +m) k1 Let w = a 11 . . .a 1c1 • • • a k1 . . .a kc k be a column word.Then there exists a sequence {b ij } 1≤j≤ci,1≤i≤k such that w * b k1 b k−1,1 . . .b 11 b 12 . . .b 1c1 b 22 . . .b 2c2 • • • b k2 . . .b kc k , where b 11 > b 21 > • • • > b k1 and b 11 . . .b 1c1 |b 22 Lemma 3 inductively.Using the notation in Lemma 4, since w * b k1 b k−1,1 . . .b 11 b 12 . . .b 1c1 b 22 . . .b 2c2 • • • b k2 . . .b kc k , where b 11 > b 21 > • • • > b k1 , the SSAF with basement being the identity (i.e., 1 2 • • • n, for some positive integer n) representing the word w is the same as that representing b k1 b k−1,1 . . .b 11 b 12 . . .b 1c1 b 22 . . .b 2c2 • • • b k2 . . .b kc k .Denote F (w) as the SSAF created.Since b k1 b k−1,1 . . .b 11 is strictly increasing, by Lemma 15 in [HMR13], they create new cells in ascending reading order (i.e., one after another) and hence is exactly the first entire row of F (w) as the entries are fixed when inserting b k1 b k−1,1 . . .b 11 into an empty atom, and there are k columns (as w has k subsequences) and so the first row has length k and hence the row reading word has exactly b k1 , b k−1,1 , . . ., b 11 as the first subsequence.That means we can apply Lemma 4 to find the first subsequence of the row reading word of F (w). Since b 12 . . .b 1c1 b 22 . . .b 2c2 • • • b k2 .
ab|c is a column word.).Let L be the recording tableau of U ← W .Let V be the SSAF representing the word abc and a b c be the row reading word of V (so a < b and c ≤ b and abc * a b c ).Consider the biword W = 1 1 2 a b c and let L be the recording tableau of U ← W . Then L determines L.
a 1 a 2 . . .a k |b is a column word ).Let L be the recording tableau of U ← W .If b > a 1 , then L has reading word 22 . . .21, i.e., b creates the last cell in reading order and the insertion of a 2 . . .a k is independent of the insertion of b.
a 1 a 2 . . .a k |b is a column word ).Let L be the recording tableau of U ← W .If i = min{ j | b > a j }, then the cell created by inserting a m for m > i is not affected by the insertion of b.Proof: Since U ← a 1 a 2 . . .a k b = (U ← a 1 a 2 . . .a i−1 ) ← a i . . .a k b.By applying Lemma 6 with the U in the lemma being U ← a 1 . . .a i−1 and result follows. 2

.
one can verify that b a 1 . . .a k is indeed the row reading word of the SSAF representing the word a 1 a 2 ...a k b by 1. in Lemma 2), then L determines L.Proof: If i = k, then we are done by Lemma 8.If i = 1, then by 1. in Lemma 2, we have b a 1 . . .a k = a 1 ba 2 . . .a k and hence by Lemma 6, b would create the last cell in reading order in L when being inserted and hence we know that which two cells are created by a 1 and b in L and hence we know how to label the entries by marking those two cells as 1 and the rest as 2.Suppose i < k then by 1. in Lemma 2, we have b a 1 . . .a k = a i a 1 . . .a i−1 ba i+1 . . .a k .Now by Lemma 7, we know all a j , where j > i, create the same cells in L and L and since a 1 . . .a i−1 a i b * a i a 1 . . .a i−1 b, we just need to first apply Lemma 8 on U ← a 1 . . .a i b as the insertion recorded by L in the lemma in order to find the first (i + 1) cells created in L by the insertion U ← a i a 1 . . .a i−1 b and then label the rest of cells created by (U ← a i a 1 . . .a i−1 b) ← a i+1 . . .a k as 2 and get the entries of L and result follows.2Lemma9 gives a recording tableau interpretation of 1. in Lemma 2 using L and L.Lemma 10 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Let w 0 = a 11 . . .a 1c1 |a 21 . . .a 2c2 | • • • |a k1 . . .a kc k with k ≤ n be a column word and using the notation in Lemma 3 while we assumei j = 0 for 1 ≤ j ≤ c r , 1 ≤ r ≤ k, the recording tableau L of U ← W 0 determines the recording tableau L m of U ← W m for 0 ≤ m ≤ k − 1,where L := L 0 and W m is a biword with the lower word being w m and the upper word has entry k + 1 − t if the lower word entry just below it is a (s) tj for s = i t , i t + 1, and 1 (1) if m > 0 and the entry is a(i k +m) k1Lemma 10 follows by inductively using Lemma 9.Lemma 11 Let U be a SSAF with basement n and shape α for some positive integer n and l(α) ≤ n.Let w = a 11 . . .a 1c1 |a 21 . . .a 2c2 | • • • |a k1 . . .a kc k with k ≤ n be a column word and using the notation in Lemma 4, the recording tableau L of U ← W determines the recording tableau L of U ← W , where