Support Equalities Among Ribbon Schur Functions

In 2007, McNamara proved that two skew shapes can have the same Schur support only if they have the same number of $k\times \ell$ rectangles as subdiagrams. This implies that two ribbons can have the same Schur support only if one is obtained by permuting row lengths of the other. We present substantial progress towards classifying when a permutation $\pi \in S_m$ of row lengths of a ribbon $\alpha$ produces a ribbon $\alpha_{\pi}$ with the same Schur support as $\alpha$; when this occurs for all $\pi \in S_m$, we say that $\alpha$ has"full equivalence class."Our main results include a sufficient condition for a ribbon $\alpha$ to have full equivalence class. Additionally, we prove a separate necessary condition, which we conjecture to be sufficient.


Introduction
The question of when two skew diagrams yield equal skew Schur functions has been studied in detail; for instance, see [1], [6], and [7]. However, the related question of when two skew diagrams have the same Schur support (see Definition .2) has received less attention, with the most substantial progress occurring in [4] (2007) and in [5] (2011).
In [4], P. R. W. McNamara proves that any two skew diagrams with the same Schur support necessarily contain the same number of k × rectangles, for every k, ≥ 1. In [5], P. R. W. McNamara and S. van Willigenburg explicitly determine the Schur support for a special class of skew shapes called equitable ribbons.
In this paper, we expand on the results presented in [4] and [5] by working to classify which connected ribbons have the same Schur support under all permutations of their row lengths; we say these ribbons have full equivalence class (see Definition . 7). Note that in the work that follows, we always assume that each row of a ribbon is at least 2 boxes in length and that each ribbon has at least 3 rows. This is because in the cases where one of these conditions does not hold, it is fairly easy and uninteresting to classify when a connected ribbon has a full equivalence class.
In the next section, we provide preliminary information to aid the understanding of the rest of the paper. In Section I (resp. Section II), we provide a sufficient (resp. necessary) condition for a connected ribbon to have full equivalence class. Finally, we conjecture that the necessary condition from Section II is in fact sufficient.

Preliminaries
We begin by establishing some preliminary information regarding Schur functions and ribbons. Then we will state our main results: Corollary I.4 and Theorem II.1. We will prove these results after stating our final preliminaries: Yamanouchi words, Littlewood-Richardson fillings, and R-matrices, each of which is essential for our proofs.

Schur Functions
The Young diagram corresponding to a partition λ = (λ 1 , λ 2 , . . . , λ m ) of an integer n is a collection of boxes arranged in left-aligned rows, where the i th row from the top has λ i boxes. A filling of a Young diagram with integers is called semistandard if the integers increase weakly across rows and strictly down columns. Such a filled-in Young diagram is called a semistandard Young tableau (SSYT).
Example .1. The following is an example of a semistandard Young tableau: 1 1 1 1 1 2 2 4 2 2 3 3 4 5 5 3 4 5 We use weight or content to refer to multi-set of integers in the filling of a tableau. The weight or content is denoted as a tuple ν = (ν 1 , ν 2 , . . . , ν k ), where ν i is the number of i's in the filling of the tableau. For example, the content of the tableau from Example .1 is ν = (5,4,3,3,3).
Schur functions are often considered to be the most important basis for the ring of symmetric functions. Schur functions are indexed by integer partitions, where the Schur function s λ corresponding to a partition λ is defined as where t i is the number of occurrences of i in T . We can generalize this notion of Schur functions to apply to skew shapes, which are obtained by removing the Young diagram corresponding to the partition µ from the top-left corner of a larger Young diagram corresponding to the partition λ. Here, we require that the diagram for µ is contained in the diagram for λ, and we write the resulting skew shape as λ/µ. When µ is the empty partition, we call λ/µ "straight." Skew Schur functions have an analogous definition to that of straight Schur functions, where the sum in Equation 1 is instead over semistandard Young tableaux of shape λ/µ. Skew Schur functions have the nice property that they are Schur-positive, meaning that for any skew shape λ/µ, where ν denotes a straight partition, and where all coefficients c λ µ,ν ≥ 0. The coefficients c λ µ,ν are called Littlewood-Richardson coefficients, and will play an important role in the Littlewood-Richardson rule (which we introduce in Theorem .10). This relationship between skew Schur functions and straight Schur functions motivates the following definition: Definition .2. The Schur support of a skew shape λ/µ, denoted [λ/µ], is defined as In other words, the support of a skew shape is the set of straight shapes ν such that s ν appears with nonzero coefficient in the expansion of s λ/µ into a linear combination of straight Schur functions.

Ribbons
A ribbon is a skew shape which does not contain a 2 × 2 block as a subdiagram. A skew shape is said to be connected if there exists a path between any two boxes of the diagram using only north, east, south, and west steps such that the path is contained in the diagram.
In this paper, we consider connected ribbons (i.e. skew shapes in which each pair of consecutive rows overlaps in exactly one column). As such, any composition α of an integer n determines a unique connected ribbon. We will use the notation α = (α 1 , α 2 , . . . , α m ) to denote a connected ribbon with m rows, where row 1 is at the top of the ribbon, where row i has length α i , and where α i = 0 for all 1 ≤ i ≤ m. For the remainder of the paper, when we say "ribbon," we mean "connected ribbon." Definition .4. Let α = (α 1 , α 2 , . . . , α m ) and α π = (α π −1 (1) , α π −1 (2) , . . . , α π −1 (m) ) be connected ribbons, where π ∈ S m is a permutation written in cycle notation. We say α π is a permutation of α.
Example .5. Below are all the permutations of the ribbon α = (4, 3, 2): In [4], McNamara proves that two skew shapes have the same Schur support only if they have the same number of k × rectangles for all k, ≥ 1. This result has the following implication for connected ribbons: Proposition .6. Let α and β be connected ribbons such that [α] = [β]. Then β = α π for some permutation π ∈ S m . Proof. By [4], α and β contain the same number of 2 × 1 rectangles as subdiagrams, and therefore have the same number of rows -let's say m rows. Label the row lengths of α, indexing so that the row lengths weakly decrease as the index increases (i.e. α 1 ≥ α 2 ≥ · · · ≥ α m ). Label the row lengths of β in the same way. It suffices to show that α i = β i for all 1 ≤ i ≤ m.
Suppose for the sake of contradiction that there exists an i ∈ {1, 2, . . . , m} for which α i = β i . Choose the minimal such i and assume, without loss of generality, that α i > β i . It follows that α has more 1 × α i rectangles than β, contradicting McNamara's necessary condition for Schur support equality. Therefore, α i = β i for all 1 ≤ i ≤ m, completing the proof.
We define a relation ∼ between connected ribbons α and β, where α ∼ β when [α] = [β]. It is clear that ∼ is an equivalence relation. Then by Proposition .6, α ∼ β only if β is a permutation of α.
For instance, the ribbon α = (4, 3, 2) from Example .5 has full equivalence class, since all of its permutations have support Definition .8. We say integers x ≤ y ≤ z satisfy the strict triangle inequality if z < x + y. In this case, we may also say that {x, y, z} satisfies the strict triangle inequality.
We are now ready to state our main results. We begin with Corollary I.4, which is our sufficient condition for a connected ribbon to have full equivalence class. We now state Theorem II.1, our necessary condition for a connected ribbon to have full equivalence class; we also conjecture this condition to be sufficient in Conjecture II.4. This condition may appear arcane at this point; our constructive proof of the necessity of this condition in Section II will provide some intuition.
Theorem 2.1. Let α = (α 1 ≥ α 2 ≥ . . . ≥ α m ) be a ribbon, with each α i ≥ 2 and m ≥ 3. If α has full equivalence class, then Before we are ready to prove these main results, we need several more preliminaries.

Yamanouchi Words and Tableaux
We now introduce the concepts of Yamanouchi words and Yamanouchi tableaux, which will be essential for using and defining our main tool for proving equality of support -the Littlewood-Richardson rule.
A Yamanouchi word is a word with the property that all of its initial subwords contain no more i + 1's than i's, for all integers i. For our purposes, we are concerned with the reverse reading word of a tableau, which reads right-to-left across rows and top-to-bottom from one row to the next. A Yamanouchi tableau is a tableau whose reverse reading word is Yamanouchi.
Example .9. The following tableau is Yamanouchi because at any point along its reverse reading word, 112213321, the number of 2's is never greater than the number of 1's, and the number of 3's is never greater than the number of 2's.

Littlewood-Richardson Fillings
Littlewood-Richardson fillings (which we often abbreviate as LR-fillings) are fillings which are both semistandard and Yamanouchi. These fillings play an important role in the Littlewood-Richardson rule, which we are now ready to state.

Theorem .10. (Littlewood-Richardson rule) [3] If
then c λ µ,ν is the number of Littlewood-Richardson fillings of λ/µ of content ν. The following corollary follows immediately from Theorem .10 and the definition of Schur support (Definition .2), and will be more directly applicable to the proofs in the remainder of the paper.
Example .9 continued. Notice that the filling in this example is semistandard and has content ν = (4, 3, 2). It follows from Theorem .11 that the straight shape (4,3,2) is in the support of the ribbon with row lengths (2,3,4).
In proofs throughout the remainder of the paper, we frequently utilize Corollary .11 by constructing Littlewood-Richardson fillings of tableaux. In proving a sufficient condition for support equality of two ribbons α and α π (which satisfy the condition), our approach will be to show that if µ is in the support of α, then we can find a LR-filling of α π with content µ given a LR-filling of α with the same content. To prove necessary conditions for support equality, we will show that if α satisfies the conditions ( * ) and α π doesn't, then there exists a content for a LR-filling of α (resp. α π ) that cannot be a content for a LR-filling of α π (resp. α).

R-Matrices
We will now introduce an algorithm which will be instrumental in proving our sufficient condition for a connected ribbon to have full equivalence class. The R-matrix algorithm, described in [2, Section 2.2.3], provides a way to swap two consecutive row lengths in an arbitrary ribbon with a semistandard filling so that the filling within the two rows remains semistandard and has the same content as before. Note, however, that semistandardness of the ribbon as a whole is not necessarily preserved.
Let α = (α 1 , . . . , α m ) be a connected ribbon with filling F, and let α j and α j+1 be the two row lengths we wish to swap. Observe that we can assume α j > α j+1 : if α j = α j+1 , we're done; if α j < α j+1 , we can consider the antipodal rotation of α (by Remark .3). The R-matrix algorithm will utilize this fact and assume that α j > α j+1 . The algorithm proceeds as follows: 1. Convert the filling of rows j and j + 1 to a box-ball system with the boxes corresponding to the j th row on the left and the boxes corresponding to the (j + 1) st row on the right. Steps 1-3 of the R-matrix algorithm as applied to this partial tableau are depicted below. Notice that the only ball movement is two balls in the third box from the top shifting from the left to the right, as these were the only two unconnected balls on the left. Notice that the row lengths have swapped, while the content and semistandardness of the filling has been preserved, as promised.

I A Sufficient Condition
In this section, we prove a sufficient condition for a ribbon to have full equivalence class (Corollary I.4).
We begin with two lemmas, each of which establishes a property about the Rmatrix algorithm (introduced in the Preliminaries), which will be essential for the proof of Theorem I.3 (which implies Corollary I.4). In Lemma I.1, we prove that the R-matrix algorithm preserves the Yamanouchi property of a ribbon with a LR-filling. In Lemma I.2, we prove that the bottom-left entry in the (j+1) st row of a ribbon with a LR-filling is not increased by the R-matrix algorithm -a step towards showing that for certain ribbons, two rows can be swapped while preserving semistandness of the filling.
We use the results of these two lemmas, in addition to some casework, to show in Theorem I.3 that under a certain condition on three adjacent row lengths of a ribbon with an LR-filling, the bottom two of the three adjacent row lengths can be swapped while preserving the Yamanouchi property and semistandardness. By imposing this condition on the entire ribbon, we get as a corollary a sufficient condition for a ribbon to have full equivalence class.
Proof. Since F is Yamanouchi, we only need to show that the initial reverse reading words up to the i th and (i + 1) st rows of F are Yamanouchi. Denote the filling of the i th and (i + 1) st rows of F by R and the filling of the i th and (i + 1) st rows of F by R . Fix any j and assume that j and j + 1 appear in R as follows: · · · j a (j + 1) b · · · · · · j c (j + 1) d · · · Let η j and η j+1 denote the number of j's and (j + 1)'s, respectively, in the reverse reading word of F by the end of the Let x be the number of connected j's on the left when executing the R-matrix algorithm. Similarly, let y be the number of connected (j + 1)'s on the left. Notice that x ≥ min(a, d).
Following the R-matrix algorithm, j and j + 1 occur in R as: · · · j x (j + 1) y · · · · · · j a+c−x (j + 1) b+d−y · · · Define the function r(n) to be the number of (j + 1)'s minus the number of j's which have occurred within the first n elements of the initial reverse reading word of R . (For instance, r(y) = y since the reverse reading word of R begins with (j + 1) y .) Clearly r is maximal after a string of (j + 1)'s, so either after the length y string of (j + 1)'s or after b + d + x elements have been seen. We only have left to show that the function r never exceeds M .
Notice that r(y) = y and r(b On the other hand, if x < d, then since x ≥ min(a, d) (as noted above), we have x ≥ a. Then This completes the proof.
We have just shown that the R-matrix algorithm preserves the Yamanouchi property of a ribbon with an LR-filling. Recall from the Preliminaries that the R-matrix operation as applied to a ribbon with an LR-filling preserves semistandardness of the filling within the two rows that are swapped; however, semistandardness of the filling of the entire ribbon is not necessarily preserved. With the next lemma, we prove another property of the R-matrix algorithm so as to work towards establishing how we might use the R-matrix algorithm while preserving the semistandardness of the entire ribbon. Proof. Suppose x is the entry of the leftmost box in the (i + 1) st row of F. Then by the R-matrix algorithm, x is also in the (i + 1) st row of F . Now the result follows from the fact that the R-matrix operation preserves semistandardness within the two rows.
The remaining way in which α (i i+1) with filling F may not be semistandard is for the number in the rightmost box of the i th row of α (i i+1) to be less than or equal to the leftmost box in the (i − 1) st row (where α and F are as in Lemma I.2). This is the main focus of the following proof.
Proof. Let F be an LR-filling of α with content µ. Since antipodal rotation preserves Schur support (Remark .3), we can assume without loss of generality that α i > α i+1 . Perform the R-matrix algorithm on rows i and i + 1 of α to obtain the ribbon α (i i+1) with filling F . We will refer to the labeling of boxes of α as shown in Figure 4, where the top row shown in the diagram is the (i − 1) st row of α. We denote the corresponding boxes of α (i i+1) as A , B , C , and D . By Lemma I.1, F satisfies the Yamanouchi property. By Lemma I.2, if i = 1 (the top row is swapped), then we have achieved an LR-filling and we are done. So, assume i > 1 and we will now ensure semistandardness in the filling of boxes A and B of α (i i+1) . Let the integers in boxes A and B be a and b, respectively. If a < b, we are done. If b < a, simply swap a and b to obtain an LR-filling of α (i i+1) , and we are done. Now, assume a = b.
Notice that if there is an entry w > b in the (i − 1) st row which is not rightmost, then we may select the minimal such w and swap the b in box B with the leftmost occurrence of w in the (i − 1) st row to obtain an LR-filling with content µ. Thus we may assume that all entries except the rightmost entry of the (i − 1) st row are b. Since F was semistandard, the element in box B of F must have been strictly greater than b. Then since that entry is clearly not in the i th row of F , it must have been taken to the (i + 1) st row by the R-matrix algorithm. Hence, there must be some entry of F in the (i + 1) st row of α (i i+1) which is greater than b. Now, if any entry x (besides the leftmost) of the (i+1) st row is less than b, we can select the maximal such x and swap the b in box A with the rightmost occurrence of x to obtain a valid LR-filling. Therefore we can assume that all entries except the leftmost of the (i + 1) st row are greater than or equal to b.
Additionally, if any entry (besides the leftmost) of the i th row is less than b, we can choose the rightmost such entry and exchange it with the b in box A to obtain a valid LR-filling. Thus we can assume that all entries except the leftmost entry are equal to b. Now, let y denote the leftmost entry of the i th row, and consider the case where y = b. In this case, y < b by the preservation of semistandardness within the two rows swapped. It follows that we can swap y with the b in box A and be done, since we know that the rightmost entry in the (i + 1) st row is strictly greater than b. Therefore we may assume that y = b.
By the above arguments, we may assume the entries shown in Figure 5, where q and z are unknown. (In case the notation is unclear, in the (i + 1) st row, we are attempting to convey that all but possibly the leftmost entry is at least b; additionally, recall that the rightmost entry is strictly greater than b.) We finish the proof by looking at two cases. In order to distinguish between the two cases, we need to set up some additional notation. Let s be value of the leftmost entry in the (i + 1) st row which is greater than b. Let R be the filling of the subdiagram of F consisting exactly of the (i − 1) st , i th and (i + 1) st rows, except excluding the rightmost box of the (i − 1) st row and all boxes to the left of the leftmost entry with values less than or equal to b in the (i + 1) st row. Figure 5: Assumed entries of F Let V z be the initial partial tableau of L which ends immediately after the z. For every integer n, let M n be the number of n's which occur in the initial reverse reading word of V z .
In this case, we argue we can swap the b in box B with an appropriate entry from the row below.
Consider the leftmost box in the (i+1) st row with entry s. If this is the rightmost box of row i + 1, then the i th and (i + 1) st rows together contain at most two entries not equal to b: q and s, where q < b < s. Since s is the lone entry greater than b in rows i and i + 1 of F , in order for F to have been semistandard in rows i − 1 and i, the s must have been in box B of F. Now, looking at the 2 × 1 box overlap between rows i and i + 1, we see q must have been the leftmost entry of the i th row of F.
However, the R-matrix algorithm would have kept q in the i th row; more specifically, having no other numbers smaller than b on the left, a b on the right would be connected to q on the left. This contradicts the filling F . We therefore conclude that the leftmost s cannot be in the rightmost box of the (i+1) st row of F . It follows that swapping the leftmost s with the b in box B of F will produce a semistandard Young tableau. We will call the filling after this swap L.
We now argue that L will also be Yamanouchi. If s = b+1, then the filling clearly retains its Yamanouchi property -this would mean that there are no occurrences of either b + 1 or s − 1 in the i th or (i + 1) st rows and so making s appear earlier and b appear later in this segment will not violate the Yamanouchi property.
Thus we may assume that s = b + 1 (but we will still write s for formatting purposes). Then L has the entries shown in Figure 6, where we indicate that the (i + 1) st row, to the right of its leftmost entry, has a string of b's, followed by a string of numbers at least s.
Let r(n) be the number of b's that occur in the first n elements of the initial reverse reading word of R, minus the number of s's that occur in the same string. Since M b > M s by assumption, in order to show that L is Yamanouchi, it will be sufficient to show that r(n) is never less than −1. Figure 6: Assumed entries of L The reverse reading word of R is b α i−1 −1 sb α i+1 −1 * · · · * s, where the *'s are all at least s. Clearly the function r is minimal when all the *'s are equal to s, so we may assume that we have b α i−1 −1 sb α i+1 −1 s k , where k ≤ α i − 1. To see that r(n) is at least −1 everywhere, we want to show that α i−1 − 1 ≥ 1 (which is clear) and that (α i−1 − 1 + α i+1 − 1) − (k + 1) ≥ −1. It will be helpful to notice that α i < α i−1 + α i+1 can be rewritten as α i + 1 ≤ α i−1 + α i+1 . We then get that This completes Case 1. Proof of Claim 1. Suppose for the sake of contradiction that the last u occurs before the last b in the reverse reading word of V z . Then the initial reverse reading word of V z ending immediately before the last b contains M u u's and M b − 1 = M u − 1 b's, meaning this initial reverse reading word is not Yamanouchi. This contradiction completes the proof.
In particular, this implies that z = b. Now, let t denote the entry immediately above z in F . If t = b, then the b in box B can be swapped with one of z or t to obtain a valid LR-filling. We can therefore assume t = b. Then by Claim 1, we must have z = u. So, we can update our assumed entries of F . Now scan the tableau corresponding to F from the (i − 2) nd row upwards. First suppose we reach a row which contains multiple entries greater than b. Then we can easily swap the leftmost such entry with the b in box B to obtain an LR-filling. As a result, we may assume that we do not encounter a row with more than one entry greater than b. In particular, we may assume that any u appearing above the (i−1) Figure 7: Updated assumed entries of F row must be in the rightmost box of its row, by the semistandardness of F . We will use the following claim to complete the proof of this theorem. Claim 2: There exists a u in the rightmost box of some row above the (i − 1) st such that the entry immediately above the u is less than b.
Proof of Claim 2. First we argue that there exists a u in a row above the (i − 1) st row. Suppose that the u in the (i−1) st row is the topmost u in the tableau. Consider what is immediately to the right of the b in the leftmost box of the (i − 2) nd row. This entry cannot be b by the assumption that M b = M u ; however, it also cannot be greater than u since the tableau is Yamanouchi. Thus we conclude that there exists a u in a row above the (i − 1) st row. Additionally, by the paragraph above the statement of Claim 2, any such u must be in the rightmost box of its row.
Notice that by the Yamanouchi condition, no u can appear in the first row (since u > b), so each u necessarily has a box immediately above it. Assume for the sake of contradiction that every u above the (i−1) st row has a b immediately above it. Then since M b = M u , we have that all of the b's in the first (i − 2) rows are immediately above a u. Now, consider the topmost b. Since it is immediately above a u, it must be the leftmost box in its row. Since all rows are at least 2 boxes in length, there is an entry immediately to the right of this topmost b. This entry must be at least b by semistandardness. However, if it is b, it contradicts the assumption that M b = M u ; if it is greater than b, it contradicts the fact that our tableau is Yamanouchi. In either case, we reach a contradiction, so we conclude that there exists a u in the first (i − 2) rows with an entry not equal to b immediately above it. By semistandardness, this entry must be less than b.
It is easy to see that swapping the u given by Claim 2 with the b in box B produces an LR-filling. This completes the proof of Theorem I.3.
Since adjacent transpositions generate the symmetric group, the above theorem gives the following sufficient condition for a connected ribbon to have full equivalence class.
Proof. Let i ∈ {1, 2, . . . , m − 1} be arbitrary. As noted above, it will be sufficient to show that [α] = [α (i,i+1) ]. If α i = α i+1 , this result follows trivially. Then by Remark .3, we can assume without loss of generality that α i > α i+1 . By assumption, To show containment the other way, consider the antipodal rotation Having proven a sufficient condition for a ribbon to have full equivalence class, we now prove a separate necessary condition.

II A Necessary Condition
Theorem II.1. Let α = (α 1 ≥ α 2 ≥ . . . ≥ α m ) be a ribbon with each α i ≥ 2 and m ≥ 3. If α has full equivalence class, then Although this condition may appear a bit convoluted, the following remark and lemma may help motivate it.
Remark II.2. We will always have α j ≤ N j ≤ α j +m−j −2. In particular, where the second inequality follows from the upper bound on N j given in Remark II.2 and the third inequality follows from the assumption that all rows are at least two boxes long.
Additionally, our proof of Theorem II.1 should make the condition more intuitive.
Proof of Theorem II.1. We prove by contrapositive.
. We will give an LR-filling of α (j j+1) of content µ such that α has no LR-filling of content µ.
Fill α (j j+1) as follows (we'll call this filling F). Fill the i th row entirely with i's for i ≤ j. Put α j+1 (j + 1)'s in the rightmost boxes of the (j + 1) st row and fill the remaining boxes in this row with j's. Note that by Lemma II.3, the leftmost entry of the (j + 1) st row in this filling is a j (meaning this row is longer than α j+1 in length).
We now fill the remaining m − j − 1 rows with as many (j + 1)'s as possible; put (j + 1)'s in all but the leftmost box of the next m − j − 2 rows, as well as in every box in the last row. Now the only empty boxes are the leftmost boxes in rows j + 2, . . . , m − 1. We will call these remaining boxes critical boxes. Fill the critical boxes from top to bottom according to the following algorithm: in each box, put the largest integer ≤ j such that the initial reverse reading word through that box remains Yamanouchi. In practice, we will use exclusively j's until the number of j's in the tableau equals the number of (j − 1)'s. Then, we will alternate between (j − 1)'s and j's until the number of (j − 1)'s equals the number of (j − 2)'s. At this point, we rotate between placing j's, (j − 1)'s, and (j − 2)'s until the number of (j − 2)'s equals the number of (j − 3)'s. We continue in this manner until all boxes have been filled. In order to prove this algorithm gives an LR-filling, we will show that this filling has exactly N j j's.
First we define a "round". Consider the sequence of numbers c 1 , c 2 , . . . , c m−j−2 written into the critical boxes from top to bottom. Let J = {c j : c j = j}. Now partition c 1 , . . . , c m−j−2 into rounds, where each round is a consecutive subsequence of c 1 , . . . , c m−j−2 whose last element is in J but with no other elements in J (i.e. a round ends if and only if a j is encountered).
Claim: If r rounds can be completed before reaching the bottom, then at the end of the r th round, we have filled exactly i≤j: α i <α j +r (α j + r − α i ) critical boxes. In particular, each number i ≤ j such that α i ≤ α j + r has occurred in exactly α j + r − α i critical boxes.
Proof of Claim. We will use induction on r. The claim trivially holds when r = 0. Now consider an arbitrary r > 0 (such that r rounds can be completed before reaching the bottom) and assume the claim holds for r−1. In the r th round, we will write every number that was used in the (r − 1) st round one more time, as well as any number satisfying α = α j + r − 1. Therefore, the latter numbers will each fill exactly 1 critical box after r rounds, as appropriate since, by choice of , α j + r − α = 1. All numbers which appeared in the (r − 1) st round have now occurred in a critical box one more time than before. For a fixed number i, by the induction hypothesis, this is α j + (r − 1) − α i + 1 = α j + r − α i times. This completes the proof of the Claim.
Clearly the number of j's in F is α j plus the number of rounds executed before running out of critical boxes. That is, µ j = α j + max{r | i≤j:α i <α j +r α j + r − α i ≤ m − j − 2} = max{k : In particular, µ j = N j .
By construction, µ j ≤ µ j−1 ≤ · · · ≤ µ 1 . Semistandardness is also clear by construction, so we all that is left to check is that µ j ≥ µ j+1 . Indeed, µ j+1 = m i=j+1 α i − (m − j − 2), so this condition follows by our assumption that N j ≥ m i=j+1 α i − (m − j − 2), which gives that µ j = N j ≥ m i=j+1 α i − (m − j − 2) = µ j+1 . We now show that α does not have an LR-filling of content µ. By the Yamanouchi property and semistandardness, there cannot be any (j + 1)'s above the (j + 1) st row. It follows that the maximum number of (j + 1)'s is m i=j+1 α i − (m − j − 1) < µ j+1 . Therefore there is no LR-filling of α with content µ and by Corollary .11, µ is in the support of α (j j+1) but not of α.
We have proven that the condition in Theorem II.1 is necessary for a ribbon to have full equivalence class. In fact, it can be proven that this condition is both necessary and sufficient for connected ribbons with three or four rows to have full equivalence class. In addition, we have verified by computation that this condition is sufficient for m = 5, m = 6, and m = 7 for certain n. As a result, we conjecture the following: Conjecture II.4. Let α = (α 1 ≥ α 2 ≥ . . . ≥ α m ) be a ribbon with each α i ≥ 2 and m ≥ 3. Then, α has full equivalence class if and only if N j < m i=j+1 α i − (m − j − 2) for all 1 ≤ j ≤ m − 2.