A combinatorial Schur expansion of triangle-free horizontal-strip LLT polynomials

In recent years, Alexandersson and others proved combinatorial formulas for the Schur function expansion of the horizontal-strip LLT polynomial $G_\lambda(x;q)$ in some special cases. We associate a weighted graph $\Pi$ to $\lambda$ and we use it to express a linear relation among LLT polynomials. We apply this relation to prove an explicit combinatorial Schur-positive expansion of $G_\lambda(x;q)$ whenever $\Pi$ is triangle-free. We also prove that the largest power of $q$ in the LLT polynomial is the total edge weight of our graph.


Introduction
LLT polynomials are remarkable symmetric functions with many connections in algebraic combinatorics. Lascoux, Leclerc, and Thibon [11] originally defined LLT polynomials in terms of ribbon tableaux in order to study Fock space representations of the quantum affine algebra. Haglund, Haiman, Loehr, Remmel, and Ulyanov [9] redefined them in terms of tuples of skew shapes in their study of diagonal coinvariants. Haglund, Haiman, and Loehr [8] found a combinatorial formula for Macdonald polynomials, which implies a positive expansion in terms of these LLT polynomials G λ (x; q). LLT polynomials are also closely connected to chromatic quasisymmetric functions and to the Frobenius series of the space of diagonal harmonics [4]. Grojnowski and Haiman [7] proved that LLT polynomials, and therefore Macdonald polynomials, are Schur-positive using Kazhdan-Lusztig theory, but it remains a major open problem to find an explicit combinatorial Schur-positive expansion. We give a brief account of some recent results in this direction.
In the unicellular case, meaning that every skew shape of λ consists of a single cell, we can associate a unit interval graph to λ. Huh, Nam, and Yoo [10] found an explicit Schur-positive expansion whenever this graph is a melting lollipop, namely q wt a (T ) s shape(T ) .
Moreover, they proved that for arbitrary unit interval graphs, this formula gives the correct coefficient of s µ whenever the partition µ is a hook.
More generally, we focus on the horizontal-strip case, meaning that every skew shape of λ is a row. Grojnowski and Haiman [7] showed that if the rows of λ are nested, then G λ (x; q) is a transformed modified Hall-Littlewood polynomial and so its Schur expansion is given by the celebrated Lascoux-Schützenberger cocharge formula [13], namely q cocharge(T ) s shape(T ) .
Alexandersson and Uhlin [3] found a generalization of cocharge to prove an analogous formula when the rows of λ come from a skew shape σ/τ with no column having more than two cells. They formulated it for vertical-strips but we can equivalently state it as q cocharge τ (T ) s shape(T ) .
In this paper, we define a weighted graph Π associated to λ. In Section 3, we use our weighted graph to express linear recurrences of horizontal-strip LLT polynomials. We further generalize cocharge and apply our recurrences in Section 4 to prove the explicit combinatorial Schur-positive formula whenever the weighted graph Π is triangle-free. We also prove that the largest power of q in the LLT polynomial G λ (x; q) is the total edge weight of Π.

Background
A partition σ is a finite sequence of nonincreasing positive integers σ = σ 1 · · · σ ℓ . By convention, we set σ i = 0 if i > ℓ. A skew diagram λ is a subset of Z × Z of the form (2.1) λ = σ/τ = {(i, j) : i ≥ 1, τ i + 1 ≤ j ≤ σ i } for some partitions σ and τ with σ i ≥ τ i for every i. When τ is empty, we write σ instead of σ/∅. The elements of λ are called cells and the content of a cell u = (i, j) ∈ λ is the integer c(u) = j − i. We will focus heavily on rows, which are skew diagrams of the form for some a ≥ b ≥ 0. We denote by c(R) = {b, b + 1, . . . , a − 1} the set of contents of cells in R and by ℓ(R) = b and r(R) = a − 1 the smallest and largest contents of c(R) respectively. A semistandard Young tableau (SSYT) of shape λ is a function T : λ → {1, 2, 3, . . .} that satisfies where we write T i,j to mean T ((i, j)). The weight of T is the sequence w(T ) = (w 1 , w 2 , . . .), where w i = |T −1 (i)| is the number of times the integer i appears. We denote by SSYT λ the set of SSYT of shape λ and by SSYT(α) the set of SSYT of weight α. We define the skew Schur function of shape λ = σ/τ to be where x T is the monomial x w 1 1 x w 2 2 · · · . When τ is empty, we call s λ a Schur function.
A multiskew partition is a finite sequence of skew diagrams λ = (λ (1) , . . . , λ (n) ). If each λ (i) is a row, then we call λ a horizontal-strip. We denote by (2.5) SSYT λ = {T = (T (1) , . . . , T (n) ) : the set of semistandard multiskew tableaux of shape λ. Cells u ∈ λ (i) and v ∈ λ (j) with i < j attack each other if c(u) = c(v) or c(u) = c(v) + 1. The skew shapes λ (i) and λ (j) attack each other if some cells u ∈ λ (i) and v ∈ λ (j) attack each other. Entries T (i) (u) and We denote by inv(T ) the number of inversions of T . Now we define the LLT polynomial Example 2.1. Let λ = (4/0, 5/2, 2/0). When λ is a horizontal-strip we draw it so that cells of the same content are aligned vertically as on the left. We have written the content in each cell using our convention that content increases from left to right. We have also drawn two tableaux T , U ∈ SSYT λ with dotted red lines indicating the inversions. The tableau T contributes q 5 x 3 1 x 2 2 x 4 3 to (2.6) and the tableau U contributes q 3 x 2 1 x 2 2 x 3 3 x 2 4 . We can expand the LLT polynomial G λ (x; q) in the basis of Schur functions as G λ (x; q) = q 5 s 432 + q 5 s 441 + q 5 s 522 + (q 5 + q 4 )s 531 + 2q 4 s 54 + 2q 4 s 621 (2.7) We cite some helpful properties of LLT polynomials. The first three are immediate from the definition. (1) , . . . , λ (n) ) be a multiskew partition.
The special case where λ is a horizontal-strip was proven in [9, Theorem 3.1.3] using some results introduced in [14]. Both this special case and Theorem 2.3 were proven using Kazhdan-Lusztig theory. It is a major open problem to find an explicit combinatorial Schurpositive expansion of LLT polynomials. We conclude this section with a discussion of a successful solution in a special case. We first introduce the jeu de taquin algorithm. Definition 2.4. Let λ = σ/τ be a skew shape and T ∈ SSYT λ . An inside corner of λ is a cell u ∈ τ such that λ ∪ {u} is a skew shape. The jeu de taquin slide of T into an inside corner u is obtained as follows. There is a cell v directly above or directly right of u. If both, let v be the one with a smaller entry, and if they have the same entry, let v be the cell above u. We move the entry in v to u. We continue by considering the cells directly above and directly to the right of v and we stop when we vacate a cell on the outer boundary of λ, so that the result is a skew tableau. The rectification of T is the tableau obtained by successive jeu de taquin slides until the result is of partition shape. The following definition of cocharge is not the classical one but it is an equivalent characterization and it has the advantage that the weight of T is not required to be a partition.
Definition 2.6. Let T ∈ SSYT λ . The cocharge of T is the integer defined by the following three properties.
(3) Suppose that the shape of T is disconnected so that T = X ∪ Y with every entry of X above and left of every entry of Y . Let i be the smallest entry of T and suppose that no entry of X is equal to i. Let S be a tableau obtained by swapping X and Y so that every entry of Y is above and left of every entry of X. Then cocharge(T ) = cocharge(S) + |X|.
The classical definition of cocharge satisfies these three properties [12, Lemma 6.6.6]. Conversely, these properties suffice to calculate cocharge(T ) by using the following process called catabolism, which was introduced in [12, Problem 6.6.1]. If T is a single row, then cocharge(T ) = 0, otherwise by applying jeu de taquin slides, we can slide the top row of T to the left to disconnect it, swap the pieces, and rectify to produce a new tableau S of smaller cocharge. Repeated catabolism will terminate with a single row of cocharge zero.
Example 2.7. Given the tableau T below left, repeated catabolism produces the following sequence of tableaux and cocharge(T ) = 1 + 2 + 1 = 4. Example 2.8. If T is a tableau of partition shape and every entry equal to i or j with i < j, then T must be of the form T = j · · · j i · · · i i · · · i j · · · j and a single catabolism will produce a tableau with one row, so the cocharge of T is the number of entries in the second row. We can think of cocharge(T ) as measuring the extent to which there are entries above others.
We now present a case in which a combinatorial formula for the Schur-positivity of a horizontal-strip LLT polynomial is known. Theorem 2.9. [7, Theorem 7.15] Let λ = (R 1 , . . . , R n ) be a horizontal-strip such that ℓ(R 1 ) ≤ · · · ≤ ℓ(R n ) and r(R 1 ) ≥ · · · ≥ r(R n ) and let λ i = |R i |. Then the LLT polynomial G λ (x; q) is a transformed modified Hall-Littlewood polynomial, whose Schur expansion is known [13] to be The central problem of this paper is to generalize cocharge in order to prove an analogous combinatorial formula for the Schur expansion of any horizontal-strip LLT polynomial. Our main result, Theorem 4.6, is such a combinatorial formula in the case where no three rows of λ pairwise attack each other. Our strategy is to define a weighted graph Π(λ) associated to λ and to use it to express linear recurrences of LLT polynomials.

A weighted graph description of horizontal-strip LLT polynomials
We begin by defining our weighted graph Π(λ). Definition 3.1. Let R and R ′ be rows. We define the integer where as before, Definition 3.2. Let λ = (R 1 , . . . , R n ) be a horizontal-strip. Consider the cells of λ that are the rightmost cells in their row and label these cells 1, . . . , n in content reading order, meaning in order of increasing content and from bottom to top along constant content lines. We define the weighted graph Π(λ) with vertices v 1 , . . . , v n as follows. The weight of a vertex v i , denoted |v i |, is the size of the row R i ′ whose rightmost cell is labelled i. Vertices v i and v j corresponding to rows R i ′ and R j ′ with i ′ < j ′ are joined by an edge if R i ′ and R j ′ attack each other and the weight of the edge Example 3.3. Let λ = (R 1 , R 2 , R 3 ) = (4/0, 5/2, 2/0) as in Example 2.1. We have drawn λ with the rightmost cells in each row labelled in content reading order, and we have drawn Π(λ) below right. We have M(R 1 , R 2 ) = 2, M(R 1 , R 3 ) = 2, and M(R 2 , R 3 ) = 1.
Remark 3.4. It follows immediately from the definition that M(R, R ′ ) = M(R ′+ , R) and therefore the map κ from Proposition 2.2, Part 3 preserves our weighted graph. We can think of the integers M(R i , R j ) as measuring the extent to which the rows R i and R j attack each other. We now present a result to make this idea precise.
Remark 3.6. Theorem 3.5 tells us that M(λ) is the largest power of q in the LLT polynomial G λ (x; q). In particular, if G λ (x; q) = G µ (x; q), then M(λ) = M(µ). We also remark that by Proposition 2.2, Part 5, the smallest power of q is I(λ) − M(ω(λ)). If λ is a horizontalstrip, the largest power of q in the LLT polynomial is the total edge weight of Π(λ). In the Hall-Littlewood case of Theorem 2.9, we have M(λ) = n(λ) = i (i − 1)λ i .
Proof. We first observe that an entry x ∈ T makes at most one inversion with the entries in each row. If x made an inversion with y ∈ T and z ∈ T in the same row with y to the left of z, then we would require y > x > z, but y ≤ z because the rows of T are weakly increasing.
To prove the first statement, we show that for rows R of λ (i) and R ′ of λ (j) with i < j, the number of inversions inv R,R ′ (T ) of T between cells in these rows is at most M(R, R ′ ). Suppose that ℓ(R) ≤ ℓ(R ′ ) and let v 1 be the leftmost cell of row R ′ . If there is no cell u 1 of R with c(u 1 ) = c(v 1 ), then inv R,R ′ (T ) = 0, so suppose otherwise. Let v 2 , . . . , v m be the cells in R ′ to the right of v 1 and let u 2 , . . . , u m ′ be the cells in R to the right of u 1 as pictured below left. By our observation above, . . , v m be the cells in R ′ to the right of v 1 and let u 2 , . . . , u m ′ be the cells in R to the right of u 1 as pictured below right. Again, by our observation above, we have inv R, To prove the second statement, we construct a tableau T ∈ SSYT λ with inv(T ) = M(λ). Consider the cells of λ that are the leftmost cells in their row. We label these cells 1, 2, 3, . . . in reverse content reading order, meaning in order of decreasing content and from top to bottom along constant content lines. We then fill each cell with the label of the leftmost cell in its row. As an illustration, when λ = (4/0, 5/2, 2/0), the tableau we have constructed is the tableau T in Example 2.1.
By construction, the rows of T are constant and therefore weakly increasing, and if a cell u is directly below a cell v in the same skew diagram, then the leftmost cell in the row of u must precede that of v in reverse content reading order, so T (u) < T (v) and indeed T ∈ SSYT λ . Finally, we show that for rows R of λ (i) and , then every cell in R has a strictly larger entry than every cell in R ′ , so we have an inversion , then every cell in R ′ has a strictly larger entry than every cell in R, so we have an inversion We continue to use the integers M(R, R ′ ) to describe relationships between rows. The following definition will be justified by Lemma 3.15.
Definition 3.7. We say that two rows R and R ′ commute if M(R, R ′ ) = M(R ′ , R). We write R ↔ R ′ if R and R ′ commute and R R ′ otherwise.
Proof. We compute directly from the definition. If r(R) < ℓ(R ′ )−1, then This completes the proof.
Remark 3.9. For a more visual description, we have that two rows R and R ′ commute if and only if they are disjoint and separated by at least one cell, or if one is contained in the other.
In the examples below, only the pairs of rows on the left and on the right commute.
Definition 3.10. Multiskew partitions λ and µ are LLT-equivalent, denoted λ ∼ = µ, if for every multiskew partition ν we have the equality of LLT polynomials More generally, finite formal Q(q)-combinations of multiskew partitions i a i (q)λ i and Remark 3.11. By repeatedly applying the map κ from Proposition 2.2, Part 3, we have that if λ ∼ = µ, then G ν·λ·ν ′ (x; q) = G ν·µ·ν ′ (x; q) for every multiskew partitions ν and ν ′ . We can think of LLT-equivalence as a local linear relation because we can locally replace λ with µ while preserving the LLT polynomial.
We can prove an LLT-equivalence by rearranging to an LLT-equivalence of N[q]-linear combinations of multiskew partitions and finding an appropriate bijection of tableaux. f : and for every c ∈ Z, the multiset of entries in cells of content c are preserved, that is We now use Theorem 3.12 to establish some valuable LLT-equivalence relations. These relations appear in [5,Lemma 5.2] and [2, Theorem 2.1] in terms of operators on Dyck paths and Schröder paths.
Lemma 3.13. Let R and R ′ be rows such that ℓ(R ′ ) = r(R) + 1. We have the LLTequivalence Proof. By Theorem 3.12, it suffices to find an appropriate bijection For a tableau T ∈ SSYT (R,R ′ ) , let x and y denote the entries in the cells of content r(R) and ℓ(R ′ ) respectively. We partition depending on whether x ≤ y or x > y, and we similarly partition SSYT (R ′ ,R) . We will now assemble our bijection f as the union of the unique bijections Example 3.14. These bijections are illustrated in an example below. We have written q's to indicate how the numbers of inversions are supposed to change.
Our next LLT-equivalence relation justifies the terminology of commuting rows.
Lemma 3.15. Let R and R ′ be rows such that R ↔ R ′ . We have the LLT-equivalence Remark 3.16. Lemma 3.15 tells us that if λ = (R 1 , . . . , R n ) and R i ↔ R i+1 , then we can switch rows R i and R i+1 so that G λ (x; q) = G µ (x; q), where µ = (R 1 , . . . , R i+1 , R i , . . . , R n ). Also note that Theorem 3.5 implies that if (R, Proof. By Theorem 3.12, it suffices to find an appropriate bijection are not inversions, and otherwise if i ≤ c and c ≤ j the cells of content c are unchanged so any inversions are preserved. Therefore, f also satisfies (3.11). We can recover the contents i and j from U by noting that By combining Lemma 3.13 and Lemma 3.15, we obtain the following recurrence relation.
Lemma 3.17. Let R and R ′ be rows such that R R ′ and ℓ(R ′ ) < ℓ(R). We have the LLT-equivalence Proof. Note that by Proposition 3.8, we must have that ℓ( The following Proposition will help us visualize the relation (3.21). Proof. Recall that by Proposition 3.8, we must have ℓ(R 1 ) − 1 ≤ r(R 2 ). We now consider several cases. If ℓ(R) ≥ ℓ(R 1 ), then Similarly, if r(R) < r(R 2 ) − 1, then At this point, if r(R) ≥ r(R 1 ), then |R 1 ∩ R| = M 1 and again a b x We conclude this section by characterizing the triangle-free weighted graphs Π(λ) that can arise from a horizontal-strip λ.
Definition 3.20. A graph G is a caterpillar if it is a tree and its vertices can be partitioned V = P ⊔ L so that the induced subgraph G[P ] is a path and every v ∈ L has degree one. Proposition 3.21. Let λ be a horizontal-strip such that Π = Π(λ) is triangle-free. Recall that for a vertex v ∈ Π, we denote by |v| the size of the corresponding row of λ.
(1) If i < j < k and v i is adjacent to v k , then M j,k = |v j |.
(2) Every vertex v i is adjacent to at most one vertex v j for which i < j.
(3) Every connected component of Π is a caterpillar C = (P ⊔ L, E) and if v j ∈ L is adjacent to v k , then M j,k = |v j |. (5) Let Π ′ be a graph whose vertices {v 1 , . . . , v n } have positive integer weights |v i | and whose edges (v i , v j ) have positive integer weights M i,j ≤ min{|v i |, |v j |}. If Π ′ satisfies the above four conditions, then Π ′ = Π(µ) for some horizontal-strip µ.
Remark 3.22. In particular, Part 5 tells us that the property in Part 1 precisely characterizes the labellings of Π that can arise from a horizontal-strip.
Example 3.23. For the horizontal-strip λ below left, we have drawn the caterpillar Π(λ) below right. We have P = {v 1 , v 4 , v 6 } and note that 8 + 1 ≥ 3 + 2 + 2 + 2 and 4 + 1 ≥ 2 + 3. (1) Let R i ′ , R j ′ , and R k ′ be the rows of λ corresponding to the vertices v i , v j , and v k respectively. By Proposition 2.2, Part 3, we may assume without loss of generality that and v j is adjacent to v k . Now using that Π is triangle-free, v i is not adjacent to v j , so r(R i ′ ) < ℓ(R j ′ ), R j ′ ⊆ R k ′ , and M j,k = |v j |.
(2) If v i is adjacent to v j and v k with i < j and i < k, then v j and v k are adjacent by Part 1, creating a triangle in Π.
(3) We first note that Π must be acyclic because if the vertices {v it } r t=1 with i 1 < · · · < i r form a cycle, then the vertex v i 1 is adjacent to two vertices v it and v i t ′ with i 1 < i t , i t ′ , contradicting Part 2. Therefore, the connected components of Π must be trees. Let C = (V, E) be a connected component of Π and note that by Part 1, we must have V = {v i : i 1 ≤ i ≤ i r } for some i 1 , i r . Because C is connected, there must be a path P = (v i 1 , v i 2 , . . . , v i r−1 , v ir ) and it follows from Part 2 that i 1 < i 2 < · · · < i r−1 < i r . By Part 1, if i t < j < i t+1 , then v j must be adjacent to v i t+1 . Because C is a tree, this accounts for all of the edges of C so indeed C is a caterpillar. Moreover, by Part 1, if v i ∈ L = C \ P is adjacent to v j , then M i,j = |v i |.
(4) Let R i ′ and R j ′ t be the rows of λ corresponding to vertices v i and v jt respectively, and again by Proposition 2.2, Part 3, we may assume that i ′ = 1. Because Π is triangle-free, we have M jt,j t ′ = 0 for t = t ′ so assuming without loss of gnerality that and summing we have (5) It suffices to construct a horizontal-strip µ for each connected component C = (P ⊔ L, E) because in general we can translate each µ to avoid attacking the others. Let P = {v i 1 , . . . , v ir } with i 1 < · · · < i r and define the integers

Now our desired horizontal-strip is
As an illustration, the horizontal-strip in Example 3.23 is the one constructed here.

The combinatorial formula
In this section, we generalize cocharge in order to prove a combinatorial formula for the LLT polynomial G λ (x; q) whenever the weighted graph Π(λ) is triangle-free.
Definition 4.1. Let µ be a partition and let T ∈ SSYT µ be a tableau of shape µ with smallest entry i. We define the integer where as before, w i (T ) is the number of i's in T .  Let T be an SSYT and let i < j be integers. We denote by T | i,j the rectification of the skew tableau obtained by restricting T to the entries x with i ≤ x ≤ j, and we define the integer We now state our main Theorem.
Theorem 4.6. Let λ be a horizontal-strip such that the weighted graph Π = Π(λ) is triangle-free and let α i = |v i |. Then the LLT polynomial of λ is where cocharge Π (T ) = i<j min{M i,j , cocharge i,j (T )}.
Before we prove Theorem 4.6, we will present some examples and special cases to familiarize ourselves with this generalization of cocharge.
To calculate the coefficient of s 733 , we consider the three tableaux of weight α = 4153 and shape 733 as follows. The values of cocharge Π are calculated below. Therefore, the coefficient of s 733 is (q 6 + 2q 5 ).
Corollary 4.8. Let λ be a horizontal-strip whose weighted graph Π(λ) is the path below.
Then the LLT polynomial of λ is where cocharge Π (T ) = n−1 i=1 min{M i , number of entries in the second row of T | i,i+1 }. Remark 4.9. Corollary 4.8 generalizes [3, Theorem 1], which gives a combinatorial formula for G λ (x; q) in certain cases where Π(λ) is a path as above, with the additional constraint that M i−1 + M i ≤ α i for 2 ≤ i ≤ n − 1. Note that by Proposition 3.21, Part 4, the weaker inequality M i−1 + M i ≤ α i + 1 holds for 2 ≤ i ≤ n − 1.
Example 4.10. Let λ be a horizontal-strip with exactly two rows, so that Π(λ) is respectively. In either case, for each 0 ≤ k ≤ b, there is a unique tableau T k with content α and shape (a + b − k)k. Therefore, by Corollary 4.8, the LLT polynomial is q min{M,k} s (a+b−k)k = s (a+b) + · · · + q M s (a+b−M )M + · · · + q M s ab .
Note that in this example, the formula (4.3) does not depend on the labelling of Π.
We now illustrate the idea of the proof in the case of Example 4.7.
Proof of Theorem 4.6 in the case of Example 4.7. We will use induction on M(λ). By applying Lemma 3.17 to rows R 3 and R 4 and by using Proposition 3.18, we can write where the weighted graphs Π = Π(λ), Π ′ = Π(λ ′ ), and Π ′′ = Π(λ ′′ ) are given below. By our induction hypothesis, our task is now to prove that Let us consider the coefficient of s 733 . The first sum on the right hand side corresponds to the three tableaux from Example 4.7 with the values of cocharge Π ′ calculated below. The factor of q tells us to increase these values by one, and this corresponds to increasing M 1,3 from 2 to 3. Indeed, for the tableaux T 1 and T 2 , because cocharge 1,3 (T i ) = 3, the contribution to cocharge Π (T i ) is now min{3, 3} = 3 instead of min{2, 3} = 2. However, the tableau T has cocharge 1,3 (T ) = 2, so in this case we do not want to increase the cocharge by one. The second sum allows us to make this correction. It corresponds to the tableau below.
S = 4 4 4 3 3 3 1 1 2 3 3 3 3 2 + 1 + 2 = 5 The factor of (1 − q) tells us to change the term q 6 s 733 corresponding to T back into the term q 5 s 733 . In general, this second sum precisely corrects for those tableaux for which we do not want to increase the cocharge by one when we increase M 1,3 by one. To be specific, we will define a bijection q cocharge Π (T ) s shape(T ) , as desired.
We will now make this argument precise and check the details. We first recall the Littlewood-Richardson rule, which gives a combinatorial formula for the product of two Schur functions.
Theorem 4.11. [6, Section 5.1, Corollary 2 and Corollary 3] (The Littlewood-Richardson rule) Let λ, µ, and ν be partitions and fix a tableau S ∈ SSYT µ . Denote by c λ µ,ν the number of tableaux T ∈ SSYT λ/ν whose rectification is S. Then the product of Schur functions s µ and s ν expands in the basis of Schur functions as We now prove Theorem 4.6.
Proof of Theorem 4.6. Let λ = (R 1 , . . . , R n ) and let R i ′ be the row corresponding to the vertex v 1 . We use induction on n. If n = 1, then both sides of (4.3) are s α 1 , so we may assume that n ≥ 2.
We first consider the case where the vertex v 1 has no neighbour. Letα = (0, α 2 , . . . , α n ) and note that we can associate a tableauT ∈ SSYT λ/(α 1 ) (α) with a tableau T ∈ SSYT λ (α) by placing α 1 1's underneathT . Because cocharge Π is defined by restricting to the appropriate entries and rectifying, we have cocharge Π (T ) = cocharge Π (T | 2,n ). Now using Proposition 2.2, Part 2, our induction hypothesis, and the Littlewood-Richardson rule, we have as desired. So we now suppose that the vertex v 1 has a neighbour v j corresponding to some row R j ′ and note that by Proposition 3.21, Part 2, this neighbour is unique. We also use induction on M(λ). If M(λ) = 0, then the vertex v 1 has no neighbour, so we may assume that M(λ) ≥ 1. Using Proposition 3.21, Part 2 again, the vertex v j has at most one neighbour v k for which j < k. By Proposition 2.2, Part 2, we may assume that i ′ > j ′ . We claim that we may further assume that i ′ = j ′ + 1 and that R i ′ R j ′ .
For 1 ≤ t < i ′ with t = j ′ , because M(R t , R i ′ ) = 0 and r(R i ′ ) ≤ r(R t ), we must in fact have r(R i ′ ) < ℓ(R t ) − 1 and R i ′ ↔ R t by Proposition 3.8. Therefore, by Lemma 3.15, we may move row R i ′ down to assume without loss of generality that i ′ = j ′ + 1. If R i ′ R j ′ , then we have established our claim. Otherwise, if R i ′ ↔ R j ′ , then we continue to use Lemma 3.15 to move the row R i ′ down, then we use Proposition 2.2, Part 2, to move R i ′ back to the top and decrease ℓ(R i ′ ) by one, and we use Lemma 3.15 to again assume that i ′ = j ′ + 1.
If R i ′ R j ′ then we are done, otherwise we continue this process, decreasing ℓ(R i ′ ) by one each time. Because R i ′ and R j ′ will not commute when ℓ(R i ′ ) < ℓ(R j ′ ), this process will eventually terminate and we may assume that i ′ = j ′ + 1 and R i ′ R j ′ . Now we can apply Lemma 3.17 in order to write . . , R n ). By Proposition 3.18, the graphs Π = Π(λ), Π ′ = Π(λ ′ ), and Π ′′ = Π(λ ′′ ) are as below, where M = M 1,j and c = α 1 + α j − (M − 1).
. . , α j−1 , c, α j+1 , . . . , α n ) and let t = α 1 − (M − 1). We define a map (4.14) ϕ : as follows. For S ∈ SSYT(β), let ϕ(S) = T be the tableau of the same shape given by where j 1 is the column of the rightmost j in S. As an illustration, for t = 2 and the tableaux S 1 and S 2 below left, we have T 1 = ϕ(S 1 ) and T 2 = ϕ(S 2 ) below right. Informally, we change the two red 3's on the bottom row into 1's. We first show that the map ϕ is well-defined and is a bijection. Let n j denote the number of j's in S that are not on the bottom row. Because the columns of S are strictly increasing, n j is at most the number of entries in S less than j, that is n j ≤ (M − 1) + α 2 + · · · + α j−1 . By Proposition 3.21, Part 3 and Part 4, this means that n j ≤ α j , so S has at least c − α j = t j's on the bottom row and indeed t j's have been replaced by t 1's and ϕ(T ) ∈ SSYT(α). Furthermore, by construction, f (T ) = f (S) + t ≥ t, so cocharge 1,j (T ) = cocharge 1,j (S) ≤ α 1 − t = M − 1. Also, given T ∈ SSYT(β) such that cocharge 1,j (T ) ≤ M − 1, then f (T ) ≥ t and we can define S = ϕ −1 (T ) ∈ SSYT(α) by We now claim that cocharge Π ′′ (S) = cocharge Π ′ (T ). We have already shown that f (T ) = f (S) + t and therefore cocharge 1,j (S) = cocharge 1,j (T ). For 2 ≤ i ≤ j − 1, the tableau S| i,j is simply T | i,j with t j's appended to the right of the first row, and therefore f (S| i,j ) = f (T | i,j ) and cocharge i,j (S) = cocharge i,j (T ). It remains to consider cocharge j,k . In fact, it could happen that cocharge j,k (S) = cocharge j,k (T ). However, we claim that this is only possible when both integers are at least M j,k , so that (4.17) min{M j,k , cocharge j,k (S)} = M j,k = min{M j,k , cocharge j,k (T )}.
We restrict S and T to entries x with j ≤ x ≤ k and we consider how the entries move when we rectify these tableaux. By Theorem 2.5, the rectification does not depend on the order of choices of inside corners, so let us begin by performing jeu de taquin slides into all inside corners that are not on the first row to produce tableaux S ′ and T ′ . Because S and T differ only in their first row, that is S i ′ ,j ′ = T i ′ ,j ′ for all i ′ ≥ 2, we also have S ′ i ′ ,j ′ = T ′ i ′ ,j ′ for all i ′ ≥ 2. Also note that the n j j's of S ′ and T ′ that are not on the first row must now be on the second row, or in other words S ′ 2,j ′ = T ′ 2,j ′ = j for 1 ≤ j ′ ≤ n j . Now repeatedly perform jeu de taquin slides on S ′ until we obtain a skew tableau S ′′ of shape σ/(n j ) for some σ and define T ′′ similarly. Let t 0 be the number of jeu de taquin slides performed on S ′ to produce S ′′ , so that t + t 0 is the number of slides performed on T ′ to produce T ′′ . We have two cases to consider.
In particular, we have f (S| j,k ) = f (T | j,k ) + t and cocharge j,k (S) = cocharge j,k (T ). As an illustration, for j = 3 and k = 5, these stages in the rectification of the tableaux S 1 and T 1 are given below.
When we perform the final n j jeu de taquin slides, the entries of the first rows of S ′′ and T ′′ do not move, the n j j's in the second rows of S ′′ and T ′′ each move one cell down, and the remaining entries in the second rows move at most n j cells to the left. Therefore, we now have f (T | j,k ) ≤ n j and f (S| j,k ) ≤ n j + t. By Proposition 3.21, Part 3 and Part 4, we have n j ≤ (M −1)+α 2 +· · ·+α j−1 ≤ α j −M j,k , and therefore cocharge j,k (S), cocharge j,k (T ) ≥ M j,k .
As an illustration, for j = 3 and k = 5, these stages in the rectification of S 2 and T 2 are given below. When we rectify T ′ 2 , a 5 moves down from the second row so the second rows of S ′′ 2 and T ′′ 2 will be different. However, when we rectify S ′′ 2 and T ′′ 2 , these 5's move at most three cells to the left, so f (S 2 | 3,5 ) ≤ 5 and f (T 2 | 3,5 ) ≤ 3, which means that cocharge 3,5 (S), cocharge 3,5 (T ) ≥ 4. Informally, the only way that cocharge j,k (S) could differ from cocharge j,k (T ) is if a cell in the second row of T moves down prematurely, but if this happens, then f (T | j,k ) will be small enough to make cocharge j,k (T ) ≥ M j,k . Theorem 4.6 expresses the LLT polynomial G λ (x; q) in terms of the weighted graph Π(λ) but not in terms of λ itself. In other words, if Π(λ) and Π(µ) are equal and triangle-free, then G λ (x; q) = G µ (x; q). We conjecture that the formula (4.3) does not depend on the labelling of the vertices of Π(λ), provided that whenever i < j < k and v i is adjacent to v k , then M j,k = |v j |. By Proposition 3.21, Part 5, this is equivalent to the following statement.
We further conjecture that in general a horizontal-strip LLT polynomial G λ (x; q) is determined by its unlabelled weighted graph.