Necessary conditions for Schur-maximality

McNamara and Pylyavskyy conjectured precisely which connected skew shapes are maximal in the Schur-positivity order, which says that $B\leq _s A$ if $s_A-s_B$ is Schur-positive. Towards this, McNamara and van Willigenburg proved that it suffices to study equitable ribbons, namely ribbons whose row lengths are all of length $a$ or $(a+1)$ for $a\geq 2$. In this paper we confirm the conjecture of McNamara and Pylyavskyy in all cases where the comparable equitable ribbons form a chain. We also confirm a conjecture of McNamara and van Willigenburg regarding which equitable ribbons in general are minimal. Additionally, we establish two sufficient conditions for the difference of two ribbons to be Schur-positive, which manifest as diagrammatic operations on ribbons. We also deduce two necessary conditions for the difference of two equitable ribbons to be Schur-positive that rely on rows of length $a$ being at the end, or on rows of length $(a+1)$ being evenly distributed.

Within the algebra of symmetric functions, perhaps the most acclaimed basis is that consisting of Schur functions due to their ubiquitous nature: arising in enumerative combinatorics as generating functions for tableaux, in the representation theory of the symmetric and general linear groups, and in algebraic geometry when studying the cohomology ring of the Grassmannian, in addition to other areas such as quantum physics. One vibrant research avenue concerning them is that of determining when a symmetric function is Schur-positive, that is, when a symmetric function expanded as a linear combination of Schur functions has nonnegative coefficients. Schur-positive functions have two particular representationtheoretic interpretations. The first is that if a homogeneous symmetric function f of degree N is Schur-positive, then it arises as the Frobenius image of some representation of the symmetric group S N . The second is that f (x 1 , . . . , x n ) is the character of a polynomial representation of the general linear group GL(n, C). Consequently, knowing when a symmetric function is Schur-positive is very desirable. However, this phenomenon is incredibly rare, as has been calculated by F. Bergeron, Patrias and Reiner [3] who proved that the probability of a monomial-positive symmetric function, which is homogeneous of degree N , to be Schur-positive is where λ and µ are partitions of N and K λµ is the Kostka coefficient. To help explain this phenomenon, in addition to dealing with Schur functions directly, tools have been developed to determine Schur-positivity such as dual equivalence graphs [1,2,6], or the theory of crystal bases that has also been applied as a means to determine Schur-positivity [7]. Despite their rarity, examples of Schur-positive functions arise in a variety of contexts from graph theory and the study of chromatic symmetric functions [10] and chromatic quasisymmetric functions [27] to enumerative combinatorics where sets of partitions determining a quasisymmetric function have been shown to determine a function that is in fact symmetric and Schur-positive [8,11,12]. However, the most classic rendition of the question of Schurpositivity is to determine when the difference of two skew Schur functions is Schur-positive. This question is still considered to be intractable in full generality, however partial results and conjectures do exist [4,9,15,16,17,18,20,22,24] and more progress has been made when the difference is zero [5,13,21,25,29]. One particularly intriguing conjecture is due to McNamara and Pylyavskyy, who gave a construction for connected skew shapes A i such that for any other connected skew shape B the difference of skew Schur functions s A i − s B is Schur-positive for some A i with the same number of cells as B. These A i were ribbons, that is, connected skew shapes with no 2 × 2 subdiagram, and in [23] McNamara and van Willigenburg showed that such a connected skew shape must indeed be a ribbon whose row lengths are all a or (a + 1). We extend these results by establishing additional necessary conditions on such ribbons and by proving the conjecture of McNamara and Pylyavskyy in the case where the set of ribbons whose row lengths are a fixed number of a's and (a + 1)'s is totally ordered by A ≥ s B if s A − s B is Schur-positive. That is, while not confirming the conjecture in full generality we do confirm it in the important case of chains.
This article is structured as follows. In the next section we introduce necessary definitions and concepts, and prove that the ribbon conjectured to be maximal is indeed an equitable ribbon in Lemma 14 and Corollary 15. In Section 3 we introduce two operations on ribbons that guarantee that the difference of two ribbon Schur functions is Schur-positive. More precisely, in Theorems 22 and 25 we create a larger ribbon using an operation that moves cells from the top row of a ribbon. Meanwhile in Theorem 33 we create a larger ribbon using an operation that exchanges two rows of a ribbon. We then apply these tools in Section 4 when we identify all chains of equitable ribbons in Theorem 35, thereby confirming McNamara and Pylyavskyy's conjecture in these cases. In Section 5 we confirm another case of this conjecture in Corollary 53, which is reliant on Theorem 40 that a larger equitable ribbon must have short rows as its first and last rows. In Section 6 we introduce the notions of profile and quasi-profile, the latter of which is particularly crucial as it yields another condition for the Schur-positivity of a difference of equitable ribbons in Theorem 60, which implies that a larger equitable ribbon must have its longer rows being more evenly distributed, confirms that the chains identified in Theorem 35 are the only ones in Corollary 83, and confirms the conjecture of McNamara and van Willigenburg on minimal equitable ribbons in Theorem 84.
Any composition whose parts when read from left to right are weakly decreasing is called a partition. In our above example β = 31 is a partition. Given partitions λ and µ such that = max{ (λ), (µ)}, append zeroes to λ or µ to artificially increase the length so that (λ) = (µ). If λ i = µ i for all 1 ≤ i < j and λ j = µ j , then if λ j > µ j we say that λ is lexicographically greater than µ, denoted by λ > lex µ. Also note that every composition α determines a partition λ(α) by reordering the parts of α into weakly decreasing order. Lastly, for convenience, we denote by ∅ the empty composition or partition of size and length 0.
2.2. Skew shapes and ribbons. Given a partition λ, its diagram, also denoted by λ, is the array of left-justified cells with λ i cells in row i from the top. If we refer to cell (i, j) then this refers to the cell in the i-th row from the top and j-th column from the left. The only exception to this is Subsection 2.4 where for ease of exposition this will refer to the cell in the i-th row from the bottom and j-th column from the left. Let λ and µ be partitions such that (λ) ≥ (µ) and λ i ≥ µ i for all 1 ≤ i ≤ (µ). Then we define the skew shape λ/µ to be the array of cells We call λ the outer shape and µ the inner shape. Also by row (respectively, column) length we refer to the number of cells in a given row (respectively, column) of λ/µ. We define the transpose of λ/µ to be the array of cells Skew shapes can be either disconnected or connected, where a skew shape is said to be disconnected if it can be partitioned into nonempty skew shapes D 1 and D 2 so no cell in D 1 has a row or column in common with any cell in D 2 . Otherwise λ/µ is said to be connected. Connected skew shapes that will be our focus later are ribbons, which are connected skew shapes not containing the subdiagram 22 = . Observe that, therefore, a ribbon is completely determined by its row lengths read from top to bottom. This yields a composition, and henceforth we abuse notation by identifying a ribbon with its corresponding composition.
Example 2. Below is the skew shape λ/µ = 4431/32. Note that it is also the ribbon 1231.
Given a ribbon, we refer to the top row and the bottom row as the end rows and every other row as an intermediate row. Also note that given a ribbon with R rows, S columns and N cells in total that R + S = N + 1 and knowing any two of these statistics will determine the third.
Narrowing our focus even further, we now turn our attention to equitable ribbons. We describe a ribbon as row-equitable if all its row lengths are a or (a + 1) for some a ≥ 1 and column-equitable if all its column lengths are b or (b + 1) for some b ≥ 1. A ribbon is said to be equitable if it is both row-equitable and column-equitable. Given such an equitable ribbon we call the rows of length a short rows and of length (a + 1) long rows. Furthermore, given an equitable ribbon α we denote by SE(α) the number of end rows of α that are short. We also note that if a ribbon is row-equitable and a ≥ 2, then we are guaranteed that it is equitable, and similarly if we are given that a ribbon is column-equitable and b ≥ 2, then again we are guaranteed that it is equitable.
3. Skew Schur functions and ribbon Schur functions. Now that we have introduced skew shapes and ribbons, we can use them to define our algebraic focus, namely skew Schur functions. But before this we need to introduce tableaux. Let λ/µ be a skew shape. Then a semistandard Young tableaux (SSYT) T of shape λ/µ, denoted by sh(T ) = λ/µ, is a filling of the cells of λ/µ with positive integers such that the entries in each row weakly increase when read from left to right, and the entries in each column strictly increase when read from top to bottom. The entry of cell (i, j) in an SSYT T is denoted by T i,j and the content of T is where c i (T ) is the number of i's appearing in T . The skew Schur function s λ/µ in variables {x 1 , x 2 , . . .} is then defined to be where the sum is over all SSYTs T with sh(T ) = λ/µ and For example, x c(T ) = x 4 1 x 2 2 x 3 for the above T . Two types of skew Schur functions will be of particular interest to us. The first is when λ/µ is a ribbon corresponding to some composition α, then we define the ribbon Schur function r α to be the skew Schur function s λ/µ . The second of these is when λ/µ has µ = ∅, then we define the Schur function s λ to be the skew Schur function s λ/µ . Schur functions are of particular interest since they form a basis for the algebra of symmetric functions to which skew Schur functions belong, and hence every skew Schur function can be written as a linear combination of Schur functions. To see exactly how we require Littlewood-Richardson tableaux. We say that a tableau is a Littlewood-Richardson (LR) tableau if, as we read the entries from right to left along each row, taking the rows from top to bottom, the number of i's we have read is always weakly greater than the number of (i + 1)'s we have read, for all i ≥ 1. This is known as the lattice word condition. See Example 4 for an example of an LR tableau. Note that the lattice word condition guarantees that every cell in the top row of an LR tableau is filled with a 1. This seemingly innocuous consequence will play a powerful role later.
Theorem 5 (Littlewood-Richardson rule). [19,26,28] Let λ/µ be a skew shape. Then where c λ µν is the number of LR tableaux T satisfying sh(T ) = λ/µ and c(T ) = ν. It is well-known that the LR tableau obtained by filling the i-th cell in each column with i yields the lexicographically largest ν such that c λ µν = 0, and moreover c λ µν = 1. The Littlewood-Richardson rule also guarantees us that skew Schur functions are Schur-positive, that is, when expanded as a linear combination of Schur functions all the coefficients are nonnegative, and this leads us to our key definition. Definition 6. Given symmetric functions F and G we say that We will now use this to define a poset on skew shapes. Given a skew shape λ/µ let [λ/µ] be the equivalence class consisting of all skew shapes whose skew Schur function is equal to s λ/µ . Then say [λ/µ] ≥ s [ν/ρ] if s λ/µ ≥ s s ν/ρ . Since, by definition, s λ/µ is homogeneous of degree N , where N is the number of cells in λ/µ it follows that [λ/µ] and [ν/ρ] will be incomparable unless λ/µ and ν/ρ have the same number of cells, N . Hence now consider all the equivalence classes [λ/µ] where λ/µ has N cells, and let P N be the poset whose elements are these equivalence classes, with order relation ≤ s .
As noted, r α = r α * , and furthermore if α is an equitable ribbon then the following corollary confirms that this is all.
Corollary 8. Let α be an equitable ribbon. Then the equivalence class of α only contains α and α * .
Proof. We show that α = α 1 · · · α (α) does not factorize as a composition β • γ of two nonsymmetric compositions β and γ. Suppose that it does. We know that the row lengths of α are all either a or (a + 1) for some a. Now because α is a concatenation of nearconcatenations of γ, we must have γ 1 = α 1 ∈ {a, a + 1} and γ (γ) = α (α) ∈ {a, a + 1}. Also, since β is nonsymmetric and hence not all 1's there will be some near-concatenation, so γ 1 + γ (γ) ∈ {a, a + 1}; the only possibility is if In particular, the first row of α has length 1. Now by applying the ω involution ω(s λ/µ ) = s (λ/µ) t , we can use the same argument to show that the first column of α has length 1. However, now α is just a single cell, so can not factorize as described.
Returning to P N it is straightforward to verify certain properties of it, such as by the Littlewod-Richardson rule P N has a unique maximal element consisting of N disconnected components each containing a single cell. Therefore, the more interesting question is what is a maximal element among connected skew shapes? If we restrict our attention to connected skew shapes then it was shown [23, Corollary 3.6] that the skew shape must be an equitable ribbon, but the question still remains, which one? It is this question that will serve as the impetus of our paper, and hence before we continue we will make two reductions in order to simplify our study.
Reduction 1: This reduction is notational and we will henceforth identify the equivalence class [α] with its representative α. This is because by Corollary 8 the equivalence class of an equitable ribbon α only contains α and α * .
Reduction 2: This reduction fixes the number of rows of any ribbon we consider to some integer R. This is because by [23, Lemma 3.8] two ribbons, and hence two equitable ribbons, are incomparable in P N if they have a different number of rows.
Hence we will focus our attention on the convex subposet of P N , which we will denote by R((a + 1) n a m ), which consists of all equitable ribbons with n rows of length (a + 1) and m rows of length a. By the involution ω(s λ/µ ) = s (λ/µ) t it additionally suffices to consider only a ≥ 2. However, many of our results intriguingly hold when a = 1, and hence we include these results for completeness. With these reductions in place we now define the box diagonal diagram, which is crucial to the question of maximal connected skew shapes in P N . Definition 9. The box diagonal diagram P R,S is the skew shape consisting of R rows, S columns and N = R + S − 1 cells constructed as follows. Draw a grid R cells high and S cells wide, and a line L given by y = R S x from the bottom left corner to the top right corner. Then P R,S consists of the cells whose interior or top left corner point is intercepted by L.
Example 10. The box diagonal diagrams P 5,7 and P 3,6 are indicated below by the shaded blue and green cells. Note that P 3,6 illustrates the "top left corner point" phrase of the definition indicated by the green cells.
We can now conjecture which connected skew shapes are maximal in P N . This conjecture is due to McNamara and Pylyavskyy and is stated in [23,Conjecture 1.3].
Conjecture 11. [23, Conjecture 1.3] In the subposet of P N consisting of connected skew shapes, there are exactly N maximal elements given by P R,N −R+1 for R = 1, . . . , N .
As we will show in Corollary 15, P R,S is an equitable ribbon and hence we obtain the following conjecture for R((a + 1) n a m ), which we prove in certain cases in Theorem 35 and Corollary 53, and confirm up to uniqueness in Corollary 80.

Conjecture 12.
In the poset R((a + 1) n a m ) there is exactly one maximal element given by P n+m,(n+m)a−m+1 .
In [23, Section 5.3] they also conjectured a unique minimal element, as follows. We confirm this element is indeed minimal in Theorem 84.

2.4.
Box diagonal diagrams. In order to better understand Conjecture 12, we present a new procedure to determine the box diagonal diagram. This will also verify that the box diagonal diagram is indeed an equitable ribbon. Because this construction relies on the Cartesian coordinate system, we adopt the convention of counting rows from the bottom for this subsection only. In this way, the top right corner of cell (x, y) is at the point (x, y).

Lemma 14.
(1) Suppose that S ≥ R and write S R = a − , where a ∈ N and ∈ [0, 1). Then the box diagonal diagram P R,S is a ribbon and: • The R-th row has length a.
• For 1 ≤ i ≤ R − 1, the i-th row has length a if i / ∈ { t
• For 1 ≤ i ≤ R − 1, the i-th row has length (a + 1) if i ∈ { t 1− : t ∈ Z}. In particular, these long rows occur precisely at (2) Suppose that R ≥ S and write R S = b − , where b ∈ N and ∈ [0, 1). Then the box diagonal diagram P R,S is a ribbon and: • The first column has length b.
In particular, these long columns occur precisely at Proof. Because the proofs of both parts are so similar, we only prove the first part. Consider the point (x, y) on the line L given in Definition 9.
Let 1 ≤ i ≤ R. We will determine which cells are in the i-th row of P R,S .
that is, L intersects the right side of cell ( (i − 1) S R + 1, i) at a point above the lower right corner. Since L has positive slope, it therefore also intersects the interior of this cell. So the leftmost cell in the i-th row of P R,S is cell The rightmost cell of the R-th row is cell (S, R).
that is, L intersects the right side of cell ( i S R , i), which is the left side of cell ( i S R + 1, i). Either L intersects this cell in the top left corner, or since L has positive slope, in the interior of this cell. So the rightmost cell in the i-th row of P R,S is cell Note that for 1 ≤ i ≤ R − 1, the rightmost cell of the i-th row is directly below the leftmost cell of the (i + 1)-th row, so adjacent rows overlap in exactly one column and indeed P R,S is a ribbon.
Corollary 15. For any R and S, the box diagonal diagram P R,S is an equitable ribbon.

Proof.
• Corollary 16. The box diagonal diagrams for the following posets are the following equitable ribbons.
a(a + 1)a(a + 1)a (5) R((a + 1) n a 2 ) : a(a + 1) n a (6) Remark 17. Later we will prove in each case that these box diagonal diagrams are indeed maximal elements of their respective posets, thus confirming Conjecture 12 in these cases. We prove the first four in Theorem 35 of Section 4, the fifth in Corollary 53 of Section 5, and the sixth in Corollary 80 of Section 6.
Proof. We prove only (5) because it is the most representative of how each part can be proved. We use Lemma 14 to calculate P R,S . Because the cases of a ≥ 2 and a = 1 are qualitatively different, we handle them separately.
If a ≥ 2, then • so by Part 1 of Lemma 14 the long rows of P R,S are at So P R,S = a(a + 1)a(a + 1)a.
If a = 1, then • so by Part 2 of Lemma 14, the long columns of P R,S are at So P R,S has column lengths 232, and hence P R,S has row lengths 12121.

Ribbon operations for Schur-positivity
In the next two subsections, we establish two ribbon Schur function inequalities, which will be pivotal for calculations in Section 4.

3.1.
Moving a cell from the first row. Our first inequality compares a ribbon α to a ribbon β that is obtained from α by moving a cell from the first row to a different row.
Definition 18. Let α = α 1 · · · α R be a composition with α 1 ≥ 2 and 2 ≤ i ≤ R. Then define that is, M i (α) is the composition formed by decrementing the first part of α by one and incrementing the i-th part of α by one. Definition 20. Let α = α 1 · · · α R be a composition, 2 ≤ i ≤ R, and ν be a partition. Then we define C α,i,ν to be the set of LR tableaux T such that where the first cell in the i-th row of T is in column i 1 . We refer to such T as row-i-restricted LR tableaux, or when the context is clear, simply as restricted LR tableaux. Also set Informally, we can illustrate a row-i-restricted LR tableau in the following way. If the first cell of row i denoted by the asterisk is a 1 then the bullet to the right must be at least the bullet below. * • • Example 21. The following LR tableau T is of shape α = 8554 and has content ν = 11 74.
We illustrate this restriction as follows, where if the asterisk is a 1 then the bullet to the right must be at least the bullet below. * • • Proof. Let A be the set of LR tableaux of shape α and let B be the set of LR tableaux of shape β. We will prove the desired inequality by finding an injective content-preserving function and we will calculate the difference r β − r α by determining which LR tableaux of B are not in the image of A.
For an LR tableau T ∈ A, set f i (T ) to be the tableau of shape β where the top row is filled with all 1's, the i-th row is filled with a 1 followed by the entries of the i-th row in T , and all other rows are filled as in T . Informally, we can think of f i as moving a 1 from the top row to the i-th row.
By construction, f i (T ) is an SSYT of shape β with the same content as T . Also note that f i is injective because we can reconstruct T from f i (T ). It remains to verify the lattice word condition for f i (T ). Because the reading word of f i (T ) differs from that of T , which is a lattice word, only by reading a 1 after the other entries of the i-th row, it suffices to check that there have not been too many 2's read before this point. The tableau f i (T ) has β 1 1's in the first row. Then in the subtableau U of f i (T ) supported on the following (i − 1) rows, with the leftmost cell of row i removed, since we know that cell will be filled with a 1, we have that the number of 2's in U ≤ the number of columns of U = the number of cells of U − the number of rows of U + 1 by hypothesis. Therefore the lattice word condition is preserved and indeed f i (T ) ∈ B.
Finally, consider an LR tableau f i (T ) in the image of f i . Let i 1 be the column of the first cell in the i-th row of f i (T ). By construction, we know that because T is column strict. Therefore, the LR tableaux of B that are not in the image of f i are precisely the row-i-restricted LR tableaux of shape β and so as desired.
Example 24. If T is the tableau below, then f 4 (T ) is as indicated.
We now make a slight generalization of Theorem 22.
Proof. To prove this, we will use three applications of Theorem 22. First, we can check that and so by Theorem 22 the first two terms of Equation (8) give us Second, because the first part of M i (α) is one larger than the first part of β, and otherwise the two compositions agree to at least the first i parts, the hypothesis of Theorem 22 is satisfied for M i (α) and so the latter two terms of Equation (8) give us Third, letting A be the set of row-i-restricted LR tableaux of shape M i (α) and B be the set of row-i-restricted LR tableaux of shape β, then we can apply the same injection f j : A → B as in the proof of Theorem 22, noting that it preserves the restriction on row i. The restricted LR tableaux of B not in the image of f j will now have the restriction on both row i and row j, and so

3.2.
Exchanging adjacent rows. Our second inequality compares a ribbon α to a ribbon β that is obtained from α by exchanging adjacent rows of α. We will need to develop some machinery before we make the comparison.
Definition 29. For two partitions λ = λ 1 λ 2 · · · λ (λ) and µ = µ 1 µ 2 · · · µ (µ) where (λ) ≥ (µ) without loss of generality, define the union of λ and µ to be the partition and the intersection of λ and µ to be the partition These operations are indeed the set-theoretic union and intersection of the corresponding diagrams. For two skew shapes λ/µ and ν/ρ, define the union and intersection to be the skew shapes We warn that these are the skew shapes arising from the set-theoretic union and intersection of the outer and inner shapes separately, and may not be the set-theoretic union and intersection of the overall diagram.
We can now state the main result of this subsection.
By Theorem 31, it remains to express the ribbons 554 and 444 on the right side of Equation (9) as skew shapes λ/µ and ν/ρ such that ribbons on the left side of Equation (9) will be their union and intersection.
The skew shapes will be as follows.
Informally, an empty column was added to ν/ρ to align the starred cells. Then the dominance condition tells us that the λ/µ shape, corresponding to a δ * , is more to the left, which causes the union to be γ with the longer row a at the top, and the intersection to be δ * with the shorter row a at the top.
By Theorem 31, it remains to express the ribbons 554 and 44 on the right side of Equation (10) as skew shapes λ/µ and ν/ρ such that the ribbons on the left side of Equation (10) will be their union and intersection.
The skew shapes will be as follows.
Informally, four empty columns were added to ν/ρ to align the starred cells. The strict dominance condition also allowed us to add an empty row to ν/ρ to match the λ/µ shape. This causes the bottom row of the λ/µ shape, corresponding to a δ * , to disappear in the union, producing γ with the longer row a at the top; and the bottom row of the ν/ρ shape to appear in the intersection, producing δ * with the shorter row a at the top.
By Theorem 31, it remains to express the ribbons a δ * and aγ on the right side of Equation (11) as skew shapes λ/µ and ν/ρ so that the ribbons on the left side of Equation (11) will be their union and intersection.

Totally ordered equitable ribbons
Our goal for this section is to prove the following theorem, which identifies cases in which R((a + 1) n a m ) is a chain. We will see in Corollary 83 of Section 6 that these are in fact the only cases in which R((a + 1) n a m ) is a chain.
Theorem 35. The partially ordered set R((a + 1) n a m ) of ribbons with n rows of length (a + 1) and m rows of length a is a chain in the following cases.
We first establish the strictness of the inequalities above. Proof. First note that if a ≥ 2, then all of the ribbons above are equitable, so by Corollary 8 could only be equal to their reverse, but since none of the ribbons are reverses of each other the result follows. Therefore it remains to consider the a = 1 case. We will show that in each chain, any element α can not factor as a composition β • γ of two nonsymmetric factors, and therefore again the equivalence class of α will only contain itself and its reverse.
(1) Chain (12): If β is nonsymmetric then it can not be all 1's, so there must be a nearconcatenation of γ within α, giving rise to a part at least 2. But this accounts for the single 2 in α, leaving γ to be all 1's and not nonsymmetric.
(3) Chain (14): By Equation (1), the total number of cells would have to factor in a way such that both factors admit nonsymmetric compositions, but neither 6 = 6 · 1 nor 6 = 3 · 2 satisfy this.
(4) Chain (15): By Equation (1), the total number of cells would have to factor in a way such that both factors admit nonsymmetric compositions, but 7 = 7 · 1 does not satisfy this.
We now establish the order relations in the next two lemmas, from which Theorem 35 will follow immediately.
Lemma 38. The following relations hold.
There are now three remaining relations to establish. We will use a combination of Lemma 28 and Theorem 22, along with its generalization Theorem 25 to simplify differences. Then we use the Littlewood-Richardson rule to calculate the Schur function expansions.
Lemma 39. The following relations hold.

By Theorem 22,
is Schur-positive and is illustrated by the following diagram.  is Schur-positive and is illustrated by the following diagram.
Here if * = 2 there would be too many 2's in the second row, so * = 1 and the right bullet is at least the bullet below, so must both be 2's. Then the third row can not have any more 2's so must have (a − 1) 1's. The fourth row will then have some number of 1's, some number 0 ≤ x ≤ a − 1 of 2's, and some number 0 ≤ y ≤ a + 1 of 3's. We can not have x = y = 0 otherwise there would be a 1 below a 1. So It remains to find restricted LR tableaux of shape (2a + 1)a(a + 1)a, as illustrated in Diagram (16), with content (4a + 1 − x − y)(a + 1 + x)y for each (x, y) ∈ A.
If y = 0, then since 1 ≤ x ≤ a − 1, we can form the restricted LR tableaux as follows, with x 2's in total in the fourth row. If x = 0 and 2 ≤ y ≤ a + 1, then form the restricted LR tableaux as follows, with (y − 1) 3's in total preceded by 1's in the fourth row.  Otherwise, x = 0 and 1 ≤ y ≤ a, and we form the restricted LR tableaux as follows, with x 2's in total preceded by 1's in the second row, and y 3's in total preceded by 1's, in the fourth row. This completes the proof of this part.
Because the left sides are equal by Theorem 7, we have that (2a+1) ) .
We will now use Theorems 22 and 25 to analyze the three differences in order to show that is Schur-positive.
By Theorem 25, the sum is Schur-positive and is illustrated by the diagram below. If an asterisk is a 1, then the bullet to its right must be at least the bullet below that asterisk.
Because the left sides are equal by Theorem 7, we find that Therefore, we have that by Theorem 25, G 2 is Schur-positive and is illustrated by the following diagram.
If the top asterisk is a 1, then both bullets must be 2's and the third row is all 2's. If the top asterisk is a 2, then the second row is all 2's and the third row is some number 0 ≤ x ≤ a of 2's followed by (a + 1 − x) 3's. Therefore we have that

Now let us consider
By Theorem 22, the first difference is Schur-positive and is illustrated by the following diagram.
If * = 1 then both bullets are 2's and third row is some number 0 ≤ x ≤ a + 1 additional 2's preceded by 1's (or 0 ≤ x ≤ a if a = 1). If * = 2 then the second row is all 2's and the third row is some number of 1's followed by some number 0 ≤ x ≤ a of 2's, some number 0 ≤ y ≤ a of 3's, and one more 3 for the bottom bullet. So we have (18) where by convention if a = 1, then s (3a)(2a+2) = s 34 = 0.
Similarly, by Theorem 22, the second difference is Schur-positive and is illustrated by the following diagram.
If * = 1 then both bullets are 2's and the third row is some number 0 ≤ x ≤ a additional 2's preceded by (2a − x) 1's. If * = 2 then the second row is all 2's and the third row is some number of 1's followed by some number 0 ≤ x ≤ a − 1 of 2's, some number 0 ≤ y ≤ a of 3's, and one more 3 for the bottom bullet. So we have Subtracting Equation (19) from Equation (18), we find that Therefore, we must show that where B is the multiset union It remains to find restricted LR tableaux of shape (2a)(a+1)(a+1)a, as illustrated in Diagram (17), with content (4a + 1 − x − y)(a + 1 + x)y for each (x, y) ∈ B. We must find two fillings where (x, y) = (a, 1), and we can ignore the pair (x, y) = (a + 1, 0) if a = 1.
If x = a + 1, y = 0, and a ≥ 2, then form the restricted LR tableau as follows. If x = a and 2 ≤ y ≤ a then form the restricted LR tableaux as follows, with y 3's in total preceded by 1's in the fourth row. Finally, if x + y = a + 1 with 1 ≤ x ≤ a − 1 then form the restricted LR tableaux as follows, with (x − 1) 2's and y 3's in total in the third row. This completes the proof of this part, and therefore, the proof of this lemma.

Large ribbons and short end rows
In the previous section, Chain (13) from Theorem 35 suggests that a large ribbon in R((a + 1) n a m ) tends to have short end rows. This is also corroborated by Theorem 22, which can produce a larger ribbon in P N by shortening the first row by one cell. Additionally, Lemma 14 tells us that the conjectured maximal element of P N has a short last row. In this section, we prove the following necessary condition for an order relation in R((a + 1) n a m ) along these lines.
Example 41. Chain (14) with a = 4 gives the following. r 4554 > s r 5454 > s r 5544 > s r 5445 The numbers of short ends of the ribbons are respectively 2, 1, 1, and 0, weakly decreasing as the ribbon Schur functions decrease.
Remark 42. It follows immediately from Theorem 40 that the class of ribbons in R((a + 1) n a m ) with zero (similarly one or two) short ends forms a convex subposet of R((a+1) n a m ).
We now prove Theorem 40. Because the cases where a = 1 and a ≥ 2 are qualitatively different, we handle them separately. Proof. We may assume that m, n ≥ 1, otherwise R(2 n 1 m ) only consists of one element so α = β. Suppose that SE(α) < SE(β). We will identify a positive term of r β − r α , contradicting r α ≥ s r β .
Let δ ∈ R(2 n 1 m ) and let σ be the content of the lexicographically largest LR tableau filling of shape δ. We then have Because by assumption δ has at least two rows and all rows are of length 1 or 2, we can verify that a single-cell column occurs exactly when δ has a 2 as its first part or as its last part. Therefore σ 2 = (n + 1) − (2 − SE(δ)) = n − 1 + SE(δ). Now let ν be the content of the lexicographically largest LR tableau filling of shape β and let ρ be the content of that of α. By the above, we know that ν 1 = n + 1 = ρ 1 , and because SE(α) < SE(β), that ρ 2 < ν 2 . Therefore ρ < lex ν, and the lexicographically largest term of r β − r α is s ν , a positive term. This contradicts r α ≥ s r β . Before proving the a ≥ 2 case of Theorem 40, we make a quick definition.
Definition 45. An LR tableau T of equitable ribbon shape is comfortable if every long row has a 1 and every long intermediate row has two 1's. Otherwise, T is uncomfortable.
Example 46. Below are two LR tableaux of shape 5544. The left tableau is comfortable; note that the last row is short so it need not have a 1. The right tableau is uncomfortable because the second row, which is long, has only one 1. Proof. We may assume that m, n ≥ 1, otherwise R((a + 1) n a m ) only consists of one element so α = β. Suppose that SE(α) < SE(β). We will identify a positive term of r β − r α , contradicting r α ≥ s r β .
Let ν be the two-part partition (aR + n − (a + R − 2))(a + R − 2); we will count LR tableaux fillings of shapes α and β of content ν. First let us describe such fillings.
The lattice word condition tells us that the first row must be all 1's and that the (a+R−2) 2's must each be preceded by their own 1. Because we only have 1's and 2's, each of the (R − 1) two-cell columns must be filled with a 1 above a 2. Then there are (a − 1) remaining 2's left to place. Now because the first row has a or (a + 1) 1's, only one of which was in a two-cell column, there are enough 1's in the first row to satisfy the lattice word condition. In other words, any SSYT of shape α or β with content ν and with all 1's in the top row is automatically an LR tableau.
It is useful to count separately the comfortable and uncomfortable LR tableaux. Let A c and A u be respectively the sets of comfortable and uncomfortable LR tableaux of shape α and content ν, and let B c and B u be respectively the sets of comfortable and uncomfortable LR tableaux of shape β and content ν.
First we handle the comfortable tableaux by defining a bijection Given an LR tableau T ∈ A c , set g(T ) to be the tableau of shape β where the k-th row is filled as follows: • If α k = β k , fill the k-th row of g(T ) in the same way as the k-th row of T .
• If α k = a and β k = a + 1, fill the k-th row of g(T ) with a 1 followed by the entries of the k-th row of T . • If α k = a + 1 and β k = a, fill the k-th row of g(T ) with the second through (a + 1)-th entries of the k-th row of T . Because T is comfortable, all long rows of T have a 1, so the content is preserved. Also, because all long intermediate rows of T have two 1's, the resulting tableau g(T ) is still an SSYT. By construction, g(T ) has all 1's in the top row, is of shape β, and is comfortable, so indeed g(T ) ∈ B c . Finally, g is a bijection because the inverse can be constructed in exactly the same way. Therefore, |A c | = |B c |.
We now handle the uncomfortable tableaux. Let us characterize T ∈ A u . We know that the first row of T is all 1's, the (R − 1) two-cell columns of T have a 1 followed by a 2, and that as long as T is an SSYT, the lattice word condition will be satisfied, no matter how the remaining (a − 1) 2's are distributed among the remaining rows. The bottom row of T , if it happens to be long, will have a remaining cells, so it must contain a 1. Therefore, for T to be uncomfortable, there must be a long intermediate row i that fails to have two 1's. Since row i has (a + 1) cells, the first of which is a 1 and the last of which is a 2, the (a − 1) remaining cells must be filled with the (a − 1) remaining 2's, thus determining the entire LR tableau. Hence |A u | is the number of long intermediate rows of α; similarly |B u | is the number of long intermediate rows of β. Therefore |B u | − |A u | = SE(β) − SE(α). Now the difference r β − r α contains the positive term Example 48. Below are the two uncomfortable tableaux of shape 5554 and content 13 6. For each long intermediate row of the ribbon, there is a unique such tableau where that row is filled with a single 1 followed by 2's. It is tempting to assert the converse statement that for α, β ∈ R((a + 1) n a m ), if SE(α) > SE(β), then r α ≥ s r β as well; however, we will see counterexamples in Corollary 83 of Section 6. We can however prove the following partial converse, which resembles [23, Proposition 3.5].
Remark 51. If m = 1, then R((a + 1) n a) is Chain (13) and by Theorem 35 the full converse to Theorem 40 does in fact hold. This is why we assume that m ≥ 2 in Proposition 50.
Proof. If SE(β) = 2 we simply take α = β. Otherwise, reversing if necessary, we can write β in the form β = δ(a + 1)aγ, where δ is a possibly empty string of all (a + 1)'s and because m ≥ 2, γ has a part of size a.
This allows us to iteratively find successively larger ribbons until we obtain a ribbon with short first row. If necessary, we reverse and continue iterating to obtain a still larger ribbon α with two short ends.
Corollary 53. The ribbon α = a(a + 1) n a is the unique maximal element of R((a + 1) n a 2 ).
Remark 54. According to (6) of Corollary 16, Corollary 53 confirms that for R((a + 1) n a 2 ) the box diagonal diagram is indeed the unique maximal element, confirming Conjecture 12 in this case.
Remark 55. It is tempting to try to extend Theorem 40 to the rows next to the end rows and continuing to work inward. For example, we may think that if r α ≥ s r β and that SE(α) = SE(β), it would be necessary that the number of short rows of α next to the end rows must exceed that of β. However, the inequality from Chain (15) shows that this is not the case. In Section 6 we explain what is going on here.

Large ribbons and even distribution of long rows
In this section we present our second necessary condition for an order relation in R((a + 1) n a m ). We begin with an example to motivate our result.
According to (7) of Corollary 16, the conjectured maximal element of Conjecture 12 for R(5 4 4 15 ) is β = 444 5 444 5 444 5 444 5 444, suggesting that the long rows, namely the 5's, are evenly distributed, far from each other and far from the ends of the ribbon. Our goal for this section is to show that indeed for α, β ∈ R((a + 1) n a m ) to satisfy r α ≥ s r β , the long rows of α must be more evenly distributed than those of β. We now make this notion precise.
Definition 56. Let us write a composition α ∈ R((a + 1) n a m ) as α = a p 1 (a + 1)a p 2 (a + 1) · · · (a + 1)a p n+1 , where p i ≥ 0 for 1 ≤ i ≤ n + 1. Then we define the profile of α to be the tuple p(α) = p 1 p 2 · · · p n+1 and we define the quasi-profile of α to be the tuple q(α) = q 0 q 1 · · · , where q j = |{i : p i = j}|; that is, q j is the number of j's in p(α).
Remark 59. Given a fixed value of a, one can recover a composition α from its profile p(α). However, one can not in general recover α from its quasi-profile q(α) because compositions whose profiles are permuted will have the same quasi-profile.
We are now able to state our result regarding the distribution of long rows. The proof of this result will require much preparation, namely Corollary 67, Theorem 70, and Theorem 71.
Remark 62. We observe that the quasi-profile of an α ∈ R((a+1) n a m ) represents how evenly distributed its long rows are. Now Theorem 60 allows us to understand the inequality r a(a+1)a(a+1)a ≥ s r a(a+1)(a+1)aa presented in Remark 55 of the previous section. After maximizing the number of short ends, the next priority in looking for a large ribbon in R((a + 1) n a m ) is not to place more short rows toward the ends of the ribbon, but rather to evenly distribute the long rows.
r 4554 > s r 5454 > s r 5544 > s r 5445 The quasi-profiles of the ribbons are respectively 12, 12, 201, and 201, weakly increasing lexicographically as the ribbon Schur functions decrease.
Remark 64. From Examples 41 and 63, we see that if one is initially only told that R(5 2 4 2 ) is a chain, then Theorems 40 and 60 are sufficient to put the ribbons in order.
Remark 65. Similar to Remark 42 from Section 5, it immediately follows from Theorem 60 that the parameter q with lexicographic order partitions R((a+1) n a m ) into convex subposets according to its possible values.
We will now prepare to prove Theorem 60. We first use the Jacobi-Trudi identity to obtain leading terms in the Schur function expansion of a ribbon Schur function. We then recast this identity in a combinatorial light and perform an extensive enumeration.
Recall that the Jacobi-Trudi identity states that for partitions λ and µ, we have (20) s Lemma 66.
(1) The lexicographically least term of the complete homogeneous symmetric function expansion of s λ is h λ . In other words, we can write for some coefficients c µ .
(2) The lexicographically least term of the Schur function expansion of h λ is s λ . In other words, we can write Proof.
(1) We use induction on the number of parts of λ. If λ has zero parts, then s ∅ = 1 = h ∅ . Otherwise, expanding the determinant of Equation (20) along the first row, we have that for λ = λ 1 λ 2 · · · λ (λ) where A j is the minor obtained by removing the first row and j -th column from the original matrix. The partitions appearing in the sum have a part λ 1 − 1 + j > λ 1 so are lexicographically larger than λ, while by the induction hypothesis the lexicographically least term of det(h λ i+1 −i+j ) is h λ 2 ···λ (λ) , so the lexicographically least term of s λ is h λ 1 h λ 2 ···λ (λ) = h λ , as desired.
(2) By the first part, the transition matrix from the Schur function basis to the complete homogeneous symmetric function basis, when the indices are ordered using reverse lexicographic order, is lower triangular with 1's along the main diagonal. Therefore, so is the inverse matrix.
Corollary 67. Let f be a nonzero symmetric function. Expand f in the complete homogeneous symmetric function basis and separate the lexicographically least term to write Proof. By Part 2 of Lemma 66, a term of the Schur function expansion of f is now c λ s λ .
We make use of the following convenient reformulation of the Jacobi-Trudi identity for ribbon Schur functions, after a definition.
Let us summarize where we are. By Corollary 67, we have a necessary condition for a difference r α − r β to be Schur-positive, based on the lexicographically least term appearing in its complete homogeneous symmetric function expansion. By Theorem 70, the complete homogeneous symmetric function expansion of r α can be determined by finding the multiplicities of partitions in M(α). Therefore, our next task is to count the multiplicities of the lexicographically least partitions in M(α) and M(β). For α ∈ R((a + 1) n a m ), the lexicographically least partitions that can appear in M(α) are of the form We now find that the multiplicities of these partitions are intimately connected to the quasiprofile q(α) of α. This will be exactly what we need to prove Theorem 60.
Theorem 71. Let α ∈ R((a+1) n a m ) and k ≥ 0. Let λ k be the partition (2a) k (a+1) n a m−2k . There is an integer C k depending only on m, n, and q j (α) for 0 ≤ j ≤ k − 2 such that Before we embark on the proof of Theorem 71, we will make some remarks, work through examples, and develop some tools.
Remark 72. We emphasize that the integer C k does not depend on the profile p(α). In other words, to count the multiplicity m α (λ k ) one does not require the profile p(α), which is equivalent to knowing α, but only the quasi-profile q(α). In addition, one specifically requires the values q j (α) for 0 ≤ j ≤ k − 1, and the dependence on q k−1 (α) is known.
We now prepare to prove Theorem 71.
Note 75. Let x be an integer and k be a nonnegative integer. Set choices for the remaining pairs. Summing over j, we have as desired.
We will also need a helpful identity of Jensen. The following formulation, specialized at z = −1, is sufficient for our purposes.
Theorem 77. [14] Let x and y be integers and v be a nonnegative integer.
Now we are ready to prove Theorem 71.
Proof of Theorem 71. We use induction on k. When k = 0 we simply have m α (λ 0 ) = 1, so assume the result for k < k. The partition λ k arises from coarsening by adding k disjoint pairs of adjacent a's from among the (n + 1) strings of a's in α, whose lengths are given by p(α). Note that even if a = 1, this is the only way to form λ k . We sum over the different ways of choosing how these k pairs are distributed, indexed by nonnegative tuples η = η 1 · · · η n+1 , and by Lemma 76 there are p i −η i η i ways to choose η i pairs from a string of p i a's. We have From here there are two steps. First we replace the binomial coefficient , using the induction hypothesis to show how the correction is related to the q j (α) for small values of j. Then we use Theorem 77 to show that what remains depends only on m and n.
For our first step, we write where the D j (α, k) represent the correction incurred by terms where p i = j and η i > j. We will show that the D j (α, k) depend only on m, n, and q j (α) for j ≤ k − 2, with the exception of D k−1 = (−1) k q k−1 (α).
As a short practice calculation to aid with the full calculation, suppose that p 1 = 0. Whenever η 1 = r > 0, then in order to compensate for adding a p 1 = 0 term, we will now Now it remains to show that the sum depends only on m and n. Moreover, it suffices to show that f (p) = f (p ), where This is because f (p) is symmetric in the p i and any tuple with fixed size m and fixed length n + 1 can be achieved by a sequence of these steps. We separate the first two parts of η, summing over different values of u = η 1 + η 2 , and apply Theorem 77 to conclude that This concludes the proof.
Remark 78. From the induction in the proof of Theorem 71, we can show that C k is an integer-valued polynomial in m, n, and q j (α) for j ≤ k − 1 with rational coefficients. Moreover, the exponents of any monomial j q e j j appearing here satisfy j (j + 1)e j ≤ k.
Now that we have proven Theorem 71, it is a straightforward consequence to prove the goal of this section, namely Theorem 60, which states that for α, β ∈ R((a + 1) n a m ), if r α ≥ s r β , then q(α) ≤ lex q(β).
Remark 79. In the proof of Theorem 40 we determined the lexicographically largest term that appears in a difference. Now, in the proof of Theorem 60, we determined the lexicographically smallest term.
Corollary 80. Suppose that m = d(n + 1) for some d. Then α = a d (a + 1)a d · · · a d (a + 1)a d is a maximal element of R((a + 1) n a m ).
Proof. The ribbon α is the unique element of R((a + 1) n a m ) that minimizes q(α).
Remark 81. According to (7) of Corollary 16, Corollary 80 confirms that the box diagonal diagram for R((a + 1) n a d(n+1) ) is indeed a maximal element, corroborating Conjecture 12 in this case. Corollary 83.
6.1. Minimal equitable ribbons. Having developed this machinery, we may now answer Conjecture 13 by proving that indeed the conjectured minimal element of R((a + 1) n a m ) is minimal. As independent motivation, recall that Theorem 40 tells us that a large element of R((a + 1) n a m ) tends to have short rows near the ends. Also, Theorem 60 tells us that a large element of R((a + 1) n a m ) tends to have its long rows evenly spaced out. In view of these, it could be expected that the minimal element should be the one that is farthest from achieving either of these, and it is with this result that we conclude.
Proof. If n = 0 or m = 0 then α is the only element of R((a + 1) n a m ) so the statement is trivially true, and if n = 1 or m = 1 then R((a + 1) n a m ) is a chain so the result follows from Theorem 35. So suppose that m, n ≥ 2. Let β ∈ R((a + 1) n a m ) be a ribbon with β < s α. By Theorem 60, q(β) ≥ lex q(α). However, q(α) = n0 m−2 1 achieves the lexicographically maximum possible value of q(α), so q(β) = q(α) = n0 m−2 1 and in β all of the a's are together, so we have β = (a + 1) n−t a m (a + 1) t for some 0 ≤ t < n 2 . By Theorem 40, t ≥ 1.
Because q(α) = q(β), we have by Theorem 71 that the multiplicities in M(α) and M(β) agree at every λ k = (2a) k (a + 1) n a m−2k . So we now move to count the multiplicities at the next smallest partitions.
Let us consider the partitions µ k = (2a + 2) k (a + 1) n−2k a m , which arise from joining k pairs of (a + 1)'s. Note that simply by switching the roles of the a's and the (a + 1)'s, we have m (a+1) n−t a m (a+1) t ((2a + 2) k (a + 1) n−2k a m ) = m a n−t (a+1) m a t ((2a) k (a + 1) m a n−2k ), because the right hand side now counts the number of ways to join k pairs of a's. Therefore, the problem of counting the multiplicities of the µ k reduces to the problem of counting those of the λ k from Theorem 71. It follows from Theorem 71 that m α (µ k ) = m β (µ k ) for k ≤ t and m α (µ t+1 ) − m β (µ t+1 ) = (−1) t+1 .
We should note that the multiplicities in M(α) and M(β) not only agree at every µ k for k ≤ t, but they also agree at any partition that is obtained from further coarsening by adding any a's together in any way. This is because both α and β have all m a's together, and they are not affected by the coarsenings that form the (2a + 2)'s. So we need not worry about any of these partitions, even though when a = 1 they may be smaller lexicographically than some of the µ k .
Again, because in α and β all the a's are together, we need not consider partitions that arise from µ k for k ≤ t − 1 by further coarsening by adding a's in any way.
Therefore, by Theorem 70 the difference r α − r β has as the lexicographically least term so by Corollary 67 can not be Schur-positive, a contradiction.