A positive combinatorial formula for symplectic Kostka-Foulkes polynomials I: Rows

We prove a conjecture of Lecouvey, which proposes a closed, positive combinatorial formula for symplectic Kostka-Foulkes polynomials, in the case of rows of arbitrary weight. To show this, we transform the cyclage algorithm in terms of which the conjecture is described into a different, more convenient combinatorial model, free of local constraints. In particular, we show that our model is governed by the situation in type A. We expect our approach to generalize to the general case and lead to a proof of the whole conjecture.


INTRODUCTION
The main motivation for this work is understanding an interplay between combinatorics and representation theory which is highly manifested in the structure of so-called Kostka-Foulkes polynomials. Let g be a complex, simple Lie algebra of rank n. Kostka-Foulkes polynomials K λ,µ (q) are defined for two dominant integral weights as the transition coefficients between two important bases of the ring of symmetric functions in the variables x = (x 1 , ..., x n ) over Q(q): Hall-Littlewood polynomials P λ (x; q) and Weyl characters χ µ (x). They are q-analogues of weight multiplicities [Kat82], affine Kazhdan-Lusztig polynomials [Lus83,Kat82], and appear in various other situations in geometric and combinatorial representation theory (see [NR03] and references therein). We refer the reader to Section 3.1 for a precise definition of Kostka-Foulkes polynomials and recommend [NR03] as a thorough reference. weight λ. In particular, in order to tackle Problem 1.1 and find an appropriate set T (λ, µ), it seems natural to seek for an object which parametrizes the aforementioned µ-weight space of the irreducible, finite dimensional g-module of highest weight λ. This approach turned out to be very succesful in type A n−1 , that is when g = sl(n, C). In this case dominant integral weights are identified with partitions of at most n parts, and a natural candidate for T (λ, µ) is the set SSYT(λ, µ) of semistandard Young tableaux of shape λ and weight µ. In this context, Foulkes conjectured the existence of such a statistic [Fou74], which was explicitly found by Lascoux and Schützenberger [LS78]. They called their statistic charge (which explains our abbreviation ch used also in the general context of arbitrary type) and established the celebrated formula of Problem 1.1 in type A n−1 . Let us briefly describe this statistic. We start by defining the charge statistic ch on standard words in the alphabet A n = {1, ..., n}, that is words where each i ∈ A n appears exactly once. Standard words are naturally identified with permutations by setting w = σ(1) · · · σ(n), where σ ∈ S(n) is a permutation. We define ch(w) -the charge of w -recursively: (1) set c(1) = 0, (2) for r ≥ 2, define c(r) = c(r − 1) if σ −1 (r) < σ −1 (r − 1), and c(r) = c(r − 1) + 1 otherwise, (3) set ch(w) = n i=1 c(i). Let w be a word in the alphabet A n such that the number of occurrences of i + 1 in this word is less or equal to the number of occurrences of i for each i + 1 ∈ A n . For such a word, we can extract its standard subwords w 1 , ..., w m as follows: the first subword w 1 of w is obtained by selecting the rightmost 1 in w, then the rightmost 2 appearing to the left of the selected 1, and so on until there is no k + 1 to the left of the current value k being selected. At this point, we select the rightmost k + 1 in w and continue with the previous process until the largest value appearing in w is reached. The subword w 1 is obtained by erasing all the letters from w that were not selected and we proceed by selecting w 2 by the same procedure performed on the word consisting of the letters that were not selected so far. We continue, until no letters are left. Finally, we will define ch(w) by setting ch(w) = m i=1 ch(w i ). One can show that ch is constant on Knuth equivalent words (see [But94,Proposition 2.4.21]), which allows to define ch as a statistic on semistandard Young tableaux with partition weight. In practice, if T ∈ SSYT(λ, µ) is a semistandard Young tableau of shape λ and weight µ, one may define ch(T ) as ch(w(T )), where w(T ) is its south western row word 1 .
A thorough introduction to Kostka-Foulkes polynomials in type A n−1 and the charge statistic from a purely combinatorial point of view is carried out in [Mac95]. We refer the reader to [But94] for a beautiful exposition and proof of (1.1), which makes use of a recursive formula for computing Kostka-Foulkes polynomials due to Morris [Mor63]. The aforementioned 1 We warn the reader that we will work solely with north eastern column words in the remaining sections of this text. However, to be consistent with the definition of the charge statistic on words [LS78,But94], and to avoid reading words backwards, we prefer to stick to south western row words in our introduction. recursive formula, in turn, is deduced from a formula for Hall-Littlewood polynomials discovered by Littlewood [Lit61].
In this work, we focus on Problem 1.1 for type C n , that is, in case of the symplectic Lie algebra g = sp(2n, C). To the best of our knowledge this is the only case of Problem 1.1 having an explicit conjectural solution, which was formulated by Lecouvey in [Lec05]. In this case, the dominant integral weights λ, µ can again be identified with partitions of at most n parts, however there are several natural combinatorial candidates for the set T (λ, µ) such as King tableaux [Kin76], De Concini tableaux [DC79] or Kashiwara-Nakashima tableaux [KN94] that we also call symplectic tableaux. The last model denoted SympTab n (λ, µ) will be of particular importance in this paper as Lecouvey's conjecture is formulated in terms of symplectic tableaux. These are defined to be semistandard Young tableaux with some additional constraints (see Section 3.2 and [Lec05]) and entries in the ordered alphabet C n = {n < · · · < 1 < 1 < · · · < n}, such that the shape of a tableau is given by λ and its weight by µ. Here, the weight of a tableau with entries in C n is defined slightly differently than the weight of a tableau of type A n−1 and is given by the vector (a n , ..., a 1 ), where a i is the difference between the number of occurrences of i's and i's in T . Lecouvey defined a charge statistic ch n : SympTab n (λ, µ) → Z ≥n by analogy to the situation in type A n−1 . Before we describe Lecouvey's construction, which might seem quite technical, let us first recall this specific situation in type A n−1 to motivate the reader. For a given semistandard Young tableau T , we write its south western row word as ux, where x is the entry of the right-most upper corner and u is a word. Now, it is straightforward to show that there exists a semistandard Young tableau U with word u. It is readily shown that ch(x → U ) = ch(T ) − 1, where x → U denotes the column insertion of x into U , see Section 2.5 for details. We use the notation Cyc A (T ) = x → U since this operation on semistandard Young tableaux is known as cyclage 2 . The reason for introducing cyclage is to compute charge without referring to the standard subwords. It follows directly from the definition of charge that, for any partition λ, the unique semistandard Young tableau T λ of shape and weight λ has charge equal to zero. Moreover, for every semistandard Young tableau T of weight λ, there exists k ∈ N such that Cyc k A (T ) = T λ . It turns out that the minimal such k is equal to the charge ch(T ) of T . Before we describe Lecouvey's conjectural solution to Problem 1.1 involving cyclage it is worth mentioning that a solution to Problem 1.1 in type C n in the weight zero case has been given recently in [LL18, Theorem 6.13], using aforementioned combinatorial model of T (λ, µ) called King tableaux. However, this relies on an interpretation of the Kostka-Foulkes polynomials in terms of generalized exponents which only holds in this special case of weight zero, so that there is little hope to tackle the general weight case with this approach.
1.1. Main result and methodology. In order to define the statistic ch n : SympTab n → N, (with notation as in Section 3.2) and formulate his conjecture, Lecouvey proceeded by analogy to the situation in type A n−1 described above. He used a symplectic version of column insertion, which he introduced in [Lec02], to define a symplectic cyclage operation Cyc C which transforms a symplectic tableau T ∈ SympTab n into a symplectic tableau Cyc C (T ) ∈ SympTab m for m ≥ n. The statistic ch n is defined as follows. Let T ∈ SympTab n be a symplectic tableau. In [Lec05], Lecouvey showed that there exists a 2 In fact, this operation is usually referred to as cocyclage of T in the literature. non-negative integer m such that Cyc m C (T ) is a column C(T ) of weight zero. We denote by m(T ) the smallest non-negative integer with this property. For a symplectic column C of weight zero we set E C = {i ≥ 1|i ∈ C, i + 1 / ∈ C}. The charge of C is defined by and the charge of an arbitrary symplectic tableau T is defined by ch n (T ) = m(T ) + ch n (C(T )). Lecouvey [Lec05] conjectured the following solution of Problem 1.1 in type C n : Conjecture 1.3. Let µ, λ be partitions with at most n parts. Then Our main result reads as follows.
Theorem 1.4. Let λ = (p) and µ = (µ n , . . . , µ 1 ) be an arbitrary partition. Then Conjecture 1.3 holds true: A pivotal point in our methodology, and one which we expect will have impact on the study of the general case of Conjecture 1.3, is a reformulation of Lecouvey's construction in the setting of Theorem 1.4 by providing a new algorithm to compute Cyc k C (T ) for arbitrary integer k. The big advantage of our approach is that in Algorithm 2, which completes this task, we are able to eliminate local constraints which appear in the original construction in two different contexts: • we need to compute Cyc k−1 C (T ) in order to compute Cyc k C (T ); • for each column of Cyc k−1 C (T ), we need to insert boxes recursively into consecutive subcolumns of size 2. In order to free ourselves from the second constraint we give a formula for inserting an entry into a whole column at once, which is given by Proposition 3.3. Although more technical in appearance, our new definition allows us to have a full grasp of the symplectic cyclage procedure. We show in Theorem 4.6 that for a partition λ = (p) which consists of one row and for an arbitrary partition µ the symplectic tableau Cyc k C (T ), where T ∈ SympTab n (λ, µ), is given by Algorithm 2. The main philosophy of Algorithm 2 is that in order to compute Cyc k C (T ), it is enough to only apply Cyc A to certain standard Young tableaux and then apply a very simple function which changes the entries of the outcome.
As an application, we are able to compute ch n (T ) directly from T and, using simple recurrence for Hall-Littlewood polynomials of type C proved by Lecouvey in [Lec05, Theorem 3.2.1.], we deduce Theorem 1.4. We believe that our strategy might lead us to the solution of Conjecture 1.3 in the full generality. Indeed, the restriction λ = (p) is due to the fact that symplectic tableaux of row shape coincide with semistandard tableaux with entries in the alphabet C n (see Proposition-Definition 3.1). In particular, there exists a unique standard tableau of shape (p), and Algorithm 2 consists in applying Cyc A multiple times to this unique tableau. It seems likely that in the more general case there exists a "right" labelling of the boxes of any symplectic tableau T of arbitrary shape, such that a very similar procedure could be followed to compute Cyc k C (T ) and therefore ch n . So far, this question remains open and we will be investigating this question in the future.
1.2. Organization of the paper. In Section 2 we introduce all the necessary combinatorial preliminaries to follow the rest of the paper. In Section 3, we introduce the development related with the combinatorics in type C n , including the original definition of insertion and its non-recursive form given by Proposition 3.3. We also present the cyclage algortithm of Lecouvey, the definition of the charge statistic on symplectic tableaux, and state his conjectural positive formula for symplectic Kostka-Foulkes polynomials. In Section 4 we describe Algorithm 2 producing a certain tableau which we show coincides with the tableau obtained from a row tableau by performing the cyclage operation k times. This section is the most involved, and the reader who wishes to skip the proofs of the results presented in this section may do so and still be able to follow Section 5, where we prove Lecouvey's conjecture for λ = (p) and arbitrary µ.

PRELIMINARIES
2.1. Tableaux. A composition α n of size n ∈ Z ≥0 is a sequence of non-negative integers α = (α 1 , α 2 , . . . ) ∈ Z Z >0 ≥0 such that i α i = n and such that α i = 0 implies that α i+1 = 0 for any i ∈ Z >0 . In particular, there are only finitely many non-zero α i and we denote their number by (α) calling it the length of the composition α. We will also use the notation |α| = n. We denote the set of compositions of size n by Comp n , and we set Comp = n Comp n . For any positive integer i ∈ Z >0 and for any composition α ∈ Comp n we define a new composition α − i as follows: For convenience we denote the unique composition (0, 0, . . . ) of size 0 by 0. To any α ∈ Comp n we associate its diagram defined by: The elements of D α , referred to as boxes, are linearly ordered by the so-called reading order, which is a variant of the lexicographic order: (i 1 , j 1 ) ≤ (i 2 , j 2 ) ⇐⇒ j 1 > j 2 or j 1 = j 2 , i 1 ≤ i 2 . For c ∈ [1, |α|] we denote the c-th box of D α in reading order by c , or by c when it does not lead to a confusion. Let (A, ≺) be a linearly ordered alphabet with minimal element a. For any composition α n we define a tableau T of shape α and entries in A to be a filling of the boxes of the diagram of α by elements from alphabet A. Formally, T is a function The content of a tableau T of shape α is the multiset of its entries. When A is a countable ascending chain (with minimal element a), we say that a tableau has weight β = (β 1 , β 2 , . . . ) when its content is given by the multiset where a+ denotes the successor of a, and a+ k+1 = a + k + . We call a tableau semistandard if for any pair of boxes lying in the same row the content of the left box is smaller than or equal to the content of the right box, and such that for any pair of boxes lying in the same column the content of the upper box is smaller than the content of the lower box, that is such that T (i, j) ≤ T (i + 1, j) and T (i, j) < T (i, j − 1). We call a tableau reading if it is semistandard and if it has the property that for boxes a ≤ b in the reading order T (a) ≤ T (b). We call a tableau standard if it is semistandard of weight β which is a column, that is, β = (1, 1, . . . , 1). We are particularly interested in compositions with some additional properties. We call a composition α unimodal if it is unimodal as a sequence, that is there exists j ∈ Z >0 such that α 1 ≤ · · · ≤ α j ≥ α j+1 ≥ · · · . A partition is a composition with non-increasing elements (in particular, partitions are unimodal). Its diagram is called a Young diagram. A partition λ of size n is denoted by λ n. We denote the set of partitions of size n by Part n and Part = n Part n . Finally we denote the set of tableaux (semistandard and standard tableaux, respectively) of shape α and weight β by Tab A (α, β) (SSTab A (α, β), STab A (α), respectively) and we denote by Tab n (A), SSTab n (A), STab n (A) (YTab n (A), SSYTab n (A), SYTab n (A), respectively) the set of tableaux, semistandard tableaux, standard tableaux (Young tableaux, semistandard Young tableaux, standard Young tableaux, respectively) of size n, that is

Augmented tableaux.
An augmented composition is the data of a composition α and a box b = (i, j) in the diagram of α, called the augmented box. In this case, the augmented composition (α, b) is also called an augmentation of α. The diagram of (α, b) is defined as where b − = (i − 1/2, j) and b + = (i + 1/2, j), and is represented by the diagram of α in which box b is split into two boxes b − and b + . In particular, (α, b) has |α| + 1 boxes, which are again totally ordered by the reading order, and we have b − = c and b + = c+1 for some label c ∈ [1, |α| + 1]. We will call b − and b + the augmented boxes of α.
In turn, an augmented tableau T is the filling of a diagram of an augmented composition by elements of A. Formally, T is a function D (α,b) → A. An augmented tableau T of shape (α, b) induces two regular tableaux T − and T + of shape α defined by Remark 2.2. The augmented tableau T is determined by the tableau T + , the box b and the entry j ∈ A such that T we represent the augmented tableau T by the tableau T − (or equivalently T + ) in which we replace box b by the split box j i . For any (augmented) tableau T , we will denote its shape by shape(T ) ∈ Comp ∪ Comp + . For a composition α n, we denote Tab + α the set of augmented tableaux of shape α + for some augmentation α + of α, and we call n the size of T ∈ Tab + α . As before, we will denote the set of all augmented tableaux of size n by 2.3. Gravity. Reordering the parts of a composition α n gives a partition λ n. Note that λ can be also seen as the result of lifting all the boxes in each column of α so that after the lift, the boxes in the given column are lying in consecutive rows starting from the first row. For this reason, we denote by grav the map Comp n → Part n , α → λ and call it the gravity map. This description induces a map Tab n → YTab n on tableaux, which restricts to a map SSTab(α, β) → SSYTab(grav(α), β) and which we denote by the same symbol. 2.4. Shifting. Let n ∈ Z ≥0 and define shift : Comp n → Comp n as follows where e i = (0, . . . , 0 i − 1 times , 1, 0 . . . ) and i = min{j | α j = max k α k }. Geometrically, it can be interpreted as removing the rightmost upper box from a diagram α and adding a box at the end of the next row. This operator naturally induces a map shift : Tab n → Tab n on tableaux, by setting shape shift(T ) = shift shape(T ) and the i-th entry of shift(T ) to be given by the i-th entry of T in the reading order. Note that shift clearly preserves the subset of unimodal compositions. Input: A partition µ and a composition α.
We extend the domain of the operator shift : where simp(α, µ) i denotes the i-th coordinate of simp(α, µ).
Proof. We define some variation of the lexicographic order ≥ lex on Comp × Part as follows: } is finite, and there exists k ∈ Z ≥0 such that shift(α, µ) = (α, µ). But the only fixpoints of shift are of the form ((1 l ), ν) for some l ≤ |α| and ν 1 = 1, which follows immediately from the definition of shift. The proof is concluded.
We define Corollary 2.7. In the special case α = (p), |µ| ≤ p we have Proof. In order to compute m µ (α), we need to shift the diagram (p) as many times as we need to obtain a column shape, remembering that whenever we obtain a shape β such that µ i = max k β k , we erase the longest row (which we call reduction) and then we apply shift operator to a new shape. In this case, this longest row is the first row of β, which is a direct consequence of the proof of Lemma 2.6. Consider a tableau of shape α filled by numbers in a way that all the entries in i-th row are i − 1. Notice that the difference between the sum of the contents of this tableau of shape shift α and the sum of the contents of this tableau of shape α is equal to 1. In particular, since we were erasing (during reduction) rows of length µ i filled by i − 1, we obtain at the end a column of length p − |µ| filled by consecutive entries starting from (µ) (we performed reduction precisely (µ) times). Therefore Finally, define a local shift operator locshift : Comp + n ∪ Comp n → Comp + n ∪ Comp n+1 by shifting the split box, if it exists, onto the next column if there is a next column (hence preserving the augmented shape), and by replacing the split box by a normal box and putting another box to its right otherwise. For a composition α ∈ Comp n , we define locshift(α) as the augmented composition obtained by removing the rightmost upper box from the diagram of α and by splitting the first box in the next row.
Just as is the case of shift, the map locshift naturally induces a map on augmented reading tableaux.
2.5. Cyclage in type A n−1 . The word w(T ) of a tableau T is obtained from T by reading its entries, column-wise, from right to left and top to bottom. In the rest of this section, fix n ∈ Z ≥0 and consider the type A n−1 alphabet A n = {1, . . . , n}. Following [LS78], we define the cyclage of a semistandard Young tableau T to be the Young tableau Cyc where T is a semistandard Young tableau such that w(T ) ≡ u and w(T ) = xu for a word u and a letter x = 1, and where ≡ is the congruence relation generated by the plactic relations, see [Lot02], and * → U is the column Schensted insertion of the letter * ∈ A into the semistandard Young tableau U .
Proof. Let a , a+1 be consecutive boxes in D α with respect to the reading order, with k = T ( a ), = T ( a+1 ). Let C be the column of T containing a+1 and let C = grav C . Then where D is obtained from C by replacing the entry T ( a+1 ) = by k. Since this property only depends on the relative position of the entries in T , it follows by induction on the number of columns that grav(locshift r+1 (T )) = Cyc A (grav(T )).
where r = α j+1 and j = min{i | α i = max k α k }. By Lemma 2.8 we have which finishes the proof.
We also recall the the definition of the Weyl character: where λ ∈ Λ + is dominant and ρ = 1 2 α∈Φ + α. This is the character of an irreducible gmodule of highest weight λ, where g is the complex semisimple Lie algebra associated with Φ. The Hall-Littlewood polynomial P λ (q) is a one-parameter deformation between Weyl characters and orbit sums m(λ) = |W λ | −1 w∈W e w(λ) , where W λ < W is the stabilizer of λ. Indeed, and P λ (0) = χ(λ) is the Weyl character while P λ (1) = m(λ) is an orbit sum.
The Kostka-Foulkes polynomials K λ,µ (q) ∈ Z[q] for λ, µ ∈ Λ + are then defined as the coefficients in the decomposition of the Weyl characters by the Hall-Littlewood polynomials: , which is the dimension of the µ-weight space of an irreducible g-module of highest weight λ. Moreover, it was conjectured by Lusztig [Lus83] and proven by Kato [Kat82] that Kostka-Foulkes polynomials are appropriately normalized Kazhdan-Lusztig polynomials. This implies that K λ,µ (q) ∈ Z ≥0 [q] has nonnegative integer coefficients, which naturally leads to Problem 1.1.
In the following we are going to investigate Problem 1.1 when Φ is the root system of type C n . We will use the superscript C n to indicate that we work in this case.
We will also use the following notation: with the convention that n = n and For two integers i ≤ j, we will use the following notation: We are interested in the set of symplectic tableaux since these objects give a natural basis of the µ-weight space of an irreducible g-module of highest weight λ in type C, see [KN94]. Therefore We recall the definition of symplectic insertion as introduced in [Lec05]. Given a letter * ∈ C and an admissible column C (again, we do not really need the definition of admissibility in this work, but roughly speaking this is a condition which assures that the insertion * → C described in the following part produces a symplectic tableau, see [Lec05]), the insertion * → C is defined as follows. If * is larger than all the letters of C, then place it in a new box at the bottom of C. This yields a column C and we set * → C = C . Otherwise, if C = a consists of only one box, set * → C := * a .
The insertion of a letter into a column of length at least 2 is defined inductively as follows.
For the base case, assume that C = Note that cases (I1) and (I2) amount to ordinary column bumping. Let C be of length k ≥ 3, and suppose that the insertion of a letter into a column of length k − 1 has been already defined and yields an n-symplectic tableau of shape (2, 1 k−2 ). Write which is a symplectic tableau.
The above definition is not very helpful in practice. Indeed, we would like to understand the global impact of inserting a letter into a column, while the nature of presented definition is local and recursive. The following proposition lets us overcome this difficulty.
Proposition 3.3. Let C be a column, that is, a Young tableau of shape (1, . . . , 1), and let * be an entry not larger than the maximal entry of C. The insertion * → C amounts to performing the operations described below and then applying grav.
• a ≥ 1, b ≥ 0, x < i ≤ y, (in the case a = 1 column C necessarily contains y) • If b > 0, c ≥ 0, x < i ≤ y < z − b (whenever b = 0, c and z are not defined). • Case 2. When * = i is barred and i ∈ C we have the following subcases: with the condition that there is a box between x and i−b+1 • Case 3. When * = i is barred and i / ∈ C we have the following subcases: Proof. By searching the tree presented on Figure 1, we are ensured that we are always in Case 1 -Case 3 and that all the cases are pairwise distinct. We prove the formulas of Case 1 -Case 3 by induction on the length of C. In the case of columns of length at most 2, this description coincides with the original definition. Fix > 2, assume that the claim holds for all columns of length − 1 and let C be a column of length . Let C be a column obtained from C by removing its top box t . By definition, * → C is obtained by first performing * → C = C t and then inserting the top entry of C into Case 1. We have either c > 0 or c = 0. In the case c > 0, performing * → C yields the shape C y+b described by Case 1, by induction hypothesis. Then we have to insert the top entry u of C (which is either the top entry of C in the case c > 1 or is equal to y + b) into the column t y+b . Since we have t < u < y + b, we need to apply the local insertion rule (I1), which yields the shape described by Case 1. In the case c = 0, we have either b > 0 or b = 0.

Figure 1
Suppose first b = 0. Then either y is the top entry of C, the second entry from the top, or the k-th entry from the top with k > 2. In the first case, we have t = y. Therefore, by induction hypothesis, i → C = C t where the top entry of C is i. Thus, it remains to insert i into y t , which, by the local insertion rule (I2), simply bumps out y since i ≤ y. In the second case, by induction hypothesis, after performing i → C we have to insert i to the column x y , which bumps out y by the local insertion rule (I1). In the last case, by induction hypothesis, after performing i → C , we have to insert the top entry u of C , which coincides with the top entry of C and satisfies t < u < y, into the column t y . Here again we apply the local insertion rule (I1), which amounts to bumping out y. In all three configurations, this yields the shape described by Case 1. Finally, suppose that b > 0. By induction hypothesis, after performing i → C , we have to insert the top entry u of C , which coincides with the top entry of C and satisfies y + b − 1 < u < y + b − 1, into the column . Once again, this yields the shape described in Case 1.
Case 2.2. We have either d > 0 or d = 0. In the case d > 0, performing i → C yields the shape described by Case 2.2, by induction hypothesis. We have to insert the top entry u of C , which coincides with the top entry of C and satisfies t < u < i − c + 1, into the column t i-c+1 . By the local insertion rule (I1), this simply bumps out the entry i − c + 1, which yields the shape described in Case 2.2. In the case d = 0, we either have b − c > 1 or b − c = 1. Suppose b − c > 1. By induction hypothesis, after performing i → C , which is described by Case 2.2, we have to insert i − c to the column . Suppose b − c = 1. By induction hypothesis, i → C corresponds to Case 2.1 with c = 0. Therefore after performing i → C , we have to insert . Here again we apply the local insertion rule (I3), which yields . In both cases we obtain Case 2.2 described in the statement.
The proof of the remaining cases is analogous.
We can now define the insertion * → T of a letter * into a symplectic tableau T . This is achieved by the following recursive procedure. Let T denote the result of inserting * into the first column of T according to the previous rule. Denote by T the tableau obtained from T by removing its first column. If T is a column, juxtapose this column with T . Otherwise, T is the juxtaposition of a column and a box b . Then juxtapose this column with (b → T ). It is proved in [Lec05] that this procedure yields a well-defined map between SympTab n and SympTab n+1 . Let α be a unimodal composition and T ∈ Tab Cn (α) such that grav(T ) ∈ SympTab n . We call such a tableau symplectic of shape α. We can use Proposition 3.3 to define the insertion * → T of a letter * ∈ C n . In order to do this, we follow the above definition of the insertion but additionally recording the vertical shift between the columns of T and the vertical shift of the box bumped out. Note that this definition naturally extends the definition of the insertion to tableaux of partition shape to tableaux of unimodal composition shape and grav( * → T ) = * → (grav T ). In particular, the insertion of an entry into an n-symplectic tableau yields an n + 1-symplectic tableau. 3.4. Symplectic cyclage and charge. Before we describe the statistic ch n , we need to introduce the type C analogue of the cyclage presented in Section 2.5. Let T be a symplectic tableau and let w = w(grav T ) be the word of the associated Young tableau. If w = xu where x is a letter, it is readily shown that u is the word of a symplectic tableau U , obtained from T by removing the corresponding box. The cyclage operation on w is η(w) = ux. The cyclage operation may or may not be authorized for a given symplectic tableau T . The following result from [Lec05] characterizes this property.
Proposition-Definition 3.5. The cyclage operation is not authorized on a symplectic tableau T of weight µ if and only if there exists p such that µ p equals the number of columns of T .
In fact, if T is a symplectic tableau for which the cyclage operation is not authorized, we can construct from T a symplectic tableau, called the reduction red(T ) of T , for which the cyclage is authorized. Let t : C → C be the map defined as follows: We define red(T ) of T recursively as follows.
(2) Delete all the n's from P and apply t to all entries x of P such that n < x < n to obtain a new (possibly empty) tableau T . (3) If T is authorized, then set red(T ) = T . Otherwise, set P = T and go back to the previous step.
Remark 3.6. Let T ∈ SympTab n (α, µ). Note that Algorithm 1 was defined in a way that it mimics steps in reduction of T . Therefore it is clear that red(T ) ∈ SympTab n (simp(α, µ)).
By convention, if the cyclage operation is authorized on T we set red(T ) = T . By construction, the cyclage is authorized for red(T ).
Definition 3.7. Let T ∈ SympTab n be a symplectic tableau. If T is a column, we set Cyc C (T ) = red T . Otherwise let w = xu = w(red(T )), where x ∈ C and let U be the symplectic tableau with w(U ) = u. Then we define Cyc C (T ) = red x → U .
and the charge of an arbitrary symplectic tableau T is defined by ch n (T ) = m(T ) + ch n (C(T )).

3.5.
Breaking down the insertion of a letter/box in a tableau. In this section we describe cyclage Cyc C in terms of augmented tableaux introduced in Section 2.2. This description is an important tool to describe an iterated application of cyclage as a simple operation related with an iterated application of cyclage in type A. Let α n − 1 be unimodal, and let T ∈ Tab + α be an augmented tableau of shape (α, b) such that T + has admissible columns and let j = T − (b). Write T + as the concatenation of its columns T = C 1 C 2 . . . C t , and let m be such that b ∈ C m . We define a map locins : Tab n as follows Note that clearly, there exists k ≤ t such that locins k (T ) ∈ Tab n . With this definition, the insertion j → T for a tableau T of shape α can be identified with the following procedure: (1) start with the augmented tableauT of shape (α, b) such thatT + = T , b is the box in the first column of T with the smallest entry j such that j ≤ j , andT − (b) = j (this determines T by Remark 2.2), (2) apply locins recursively until the result is a tableau. In particular, the cyclage of a tableau has the following description in terms of locins.

INSERTION AND SHIFTING
In this section we will construct the new algorithm computing Cyc k C (T ) for arbitrary k > 0 and for T ∈ SympTab((p)), that is T is a symplectic tableau of row shape. Our algorithm does not rely on the particular form of Cyc k−1 C (T ), which allows us to overcome the problem of controlling many local dependencies present in Lecouvey's original algorithm. This will enable us to prove Conjecture 1.3 in Section 5 for λ = (p) and arbitrary µ.
4.1. The content function. Given a composition α and two boxes b and b in α such that b < b in the reading order, their distance in α is defined by where, for a condition C , and row k (s) and col k (s) denote the row and column index of s counted from top to bottom and from left to right, respectively. Let µ = (µ n , . . . , µ 1 ) be a partition, p ∈ Z >0 be an integer and T ∈ SympTab((p), µ). By Proposition-Definition 3.1 we know that there exists (k 1 , . . . , k n ) ∈ Z n ≥0 such that T is the unique tableaux from the set SSYTab Cn ((p), (k n + µ n , k n−1 + µ n−1 , . . . , k 1 + µ 1 , k 1 , . . . , k n )).
As usual, we index the boxes of α and of T α by the integers 1, . . . , |α| in the reading order. We now define a tableau T k of shape α, which we will later show to be equal to Cyc k (T ).
Algorithm 2 Defining the tableau T k .
Example 4.1 (Weight zero). Let T be a tableau of shape (2q) and weight zero (note that all tableaux of weight zero must have an even number of boxes). We may label its boxes by elements in the interval [q, q] ⊂ C. We have α = shift k ((2q), 0) 1 = shift k ((2q)), and the content of a given box in T k is given by Boxes S and S are always partners, and they will hence have opposite contents in T k for each k ≤ m(T ).
Let us first assign labels to all the boxes in α according to the reading order: .
The tableaux T k (respectively T k,s ) have some very useful properties, the most important of which we encompass in the following crucial lemma. For x ∈ C, denote I x = T −1 k ({x}) (respectively I x = T −1 k,s ({x})) and I ≤x = T −1 k ({y ∈ C | y ≤ x}) (respectively I ≤x = T −1 k,s ({y ∈ C | y ≤ x})).
Lemma 4.3. The following properties hold true.
(1) T k and T k,s are reading tableaux, that is for all 1 ≤ t < u ≤ |α| one has T k (t) ≤ T k (u) and T k,s (t) ≤ T k,s (u).
(2) For all 1 ≤ i ≤ n and for all 0 ≤ j < |I i |we have that max I i −j = partner(min I i + j).
(3) For all 1 ≤ i ≤ n, the functions δ α , δ α s are constant on the product of intervals I i ×I i .
Proof. We will prove the statements for T k , since the arguments for T k,s are identical. Let 1 ≤ t < u ≤ |α|. If either t or u is a single then Algorithm 2 gives directly the desired inequality T k (t) ≤ T k (u). Assume that 1 ≤ t < u ≤ |α| are such that T α (t) and T α (u) are unbarred and let t = partner(t), u = partner(u). Note that δ α is bi-increasing, that is for every 1 ≤ x < y < z ≤ |α| we have δ α (x, y) ≤ δ α (x, z) and δ α (y, z) ≤ δ α (x, z). Therefore since the function f α is increasing by definition. This finishes the proof of (1) since for any 1 ≤ d ≤ |α| which is not a single we have Fix i ∈ C. For ∈ {i, i}, let min = min I and max = max I . By monotonicity of δ α , (3) is equivalent to the following statement: Notice first that i max = partner(i min ), and more generally (2) holds true, which is simply a reformulation of the if part of Algorithm 2 for a fixed value of X = i. Therefore, it follows from Algorithm 2 that and by (1) all the inequalities above are equalities. This finishes the proof of (3).
We are ready to prove our main theorem.

(4.2)
Proof. Our proof is by induction on 1 ≤ s < r. Before we start we need to introduce some notation. Let , + 1 denote the labels of the augmented boxes of α s , and let e = T k,s ( ) and f = T k,s ( +1) ≥ e. Therefore, the augmented boxes of α s+1 are labeled by +1, +2. Let C m denote the m-th column of T k,s . For an entry x lying in the column C we denote by C(x) ∈ [1, |α|] \ { } the corresponding label, that is x ∈ C and T k,s (C(x)) = x. We will proceed by going through the cases described in Proposition 3.3. The entry e = T k,s ( ) will play the role of the entry * and from now on we set C = C s which is the column containing the augmented boxes labeled by , + 1.
Case 2.1. We know that e = i for some i ∈ C >0 . Lemma 4.3 (1) implies that + 1 = C(i − b + 1). Since T k,s ( ) = i and T k,s ( + 1) = i − b + 1 we have by Lemma 4.3 (1) that ≤ partner(C(i − b + 1)) < + 1, which is possible only when b = 1. Note that performing Algorithm 2 to obtain T k,s+1 corresponds precisely to performing Algorithm 2 to obtain T k,s . Indeed, in both cases we start from D = + 1, D = and Therefore T k,s+1 (x) = T k,s (x) for all x ∈ [1, |α|], thus T k,s+1 coincides with locins(T k,s ), which is obtained form T k,s by shifting the augmented box as shown in Case 2.1 of Proposition 3.3. This observation concludes the proof in this case.
Case 3.1 We know that e = i for some i ∈ C >0 . Lemma 4.3 (1) implies that + 1 = C(x). Indeed, y < i ≤ x, thus either + 1 = C(x) or + 1 = , where is a box lying directly under C(x) and necessarily x = i. Suppose that + 1 = . If s = 1 then either there exists ∈ I i or µ i−nred = max j α j . The first case contradicts Lemma 4.3 (3) since and the second case contradicts Lemma 4.5. Suppose that s > 1 and that + 1 = . Notice that Lemma 4.3 (1) implies that T k,s ( ) = i for all ∈ [C(x), − 1]. If there exists ∈ I i then again δ α s (C(x), ) > δ α s ( , ) which contradicts Lemma 4.3 (3). If I i = ∅ then by the inductive hypothesis T k,s was obtained as locins(T k,s−1 ), which corresponds to Case 3.1 of Proposition 3.3. In this case T k,s−1 = locshift −1 T k,s . Repeating this argument s − 1 times we get that T k,1 = locshift 1−s T k,s thus T k,1 is not authorized, which is a contradiction with Lemma 4.5. This finishes the proof of our claim that + 1 = C(x). We are going to show that (4.5) T k,s ( ) = T k,s+1 ( ) for every ∈ [1, |α|]. Comparing this with Case 3.1 of Proposition 3.3 we will conclude the proof in this case. First, note that x is barred. Otherwise < partner(C(x)) = partner( + 1) < + 1 by Lemma 4.3 (1), and this is clearly impossible. Note that δ α s ( + 1, ) = δ α s+1 ( + 1, ) for all ∈ I >0 and δ α s ( , ) = δ α s+1 ( , ) for all ∈ I >0 \ C. Up to the step inAlgorithm 2 when D ≤ the construction of T k,s and T k,s+1 coincides. Since m < i < n it is clear that the transition from D > into D ≤ necessarily happens for m < D < n. In particular D ∈ I >0 \ C and δ α s ( , ) = δ α s+1 ( , ).
In particular the construction of T k,s and T k,s+1 coincides at this step of Algorithm 2, and trivially coincides after achieving this step. This finishes the proof.
Since D < C(r) we have that X = δ α s+1 (D , D) + f α (D) = n + b < M + nred and T k,s+1 ( + 1) = n + b, T k,s+1 (C(n + b − 1)) = n + b. At this step of the algorithm D = + 2, D = C(n + b − 1) − µ n+b+1−i − 1 and M + nred = n + b + 1, therefore we have the same parameters of Algorithm 2 as at a certain point of Algorithm 2 performed to construct T k,s . Thus, all the other contents of T k,s+1 are the same as in T k,s . Comparing resulting T k,s+1 with Case 3.2 of Proposition 3.3 we conclude the proof in this case.
Corollary 4.7. Let n, p ≥ 0 be integers and µ = (µ n , µ n−1 , . . . , µ 1 ) a partition. For any T ∈ SympTab n ((p), µ) we have Cyc k C (T ) = red T k . (4.6) Proof. We proceed by induction on k. Proposition-Definition 3.5 implies that T is authorized unless µ n = p, that is, unless µ = (p). If this is the case, then Cyc C (T ) = red(T ) = ∅. From the other hand, applying Algorithm 2 we first compute α = shift((p), µ) 1 = ∅, therefore T 1 = ∅ = Cyc C (T ), as desired. If T is authorized, then Lemma 3.9 implies that Cyc C (T ) = red(locshift(T )) = red(shift(T )) = red(T 1 ), where the last equalities comes from the fact that the shape of T is simply one row and the last entry of T is strictly bigger then the first one. We assume now that Cyc k C (T ) = red T k . Therefore Cyc k+1 C (T ) = Cyc C red T k = red locins r−1 locshift red T k by Lemma 3.9, where r ∈ Z >0 is such that locshift r shape red(T k ) = shift shape red(T k ) . Applying Theorem 4.6 and Lemma 4.5 to the right hand side of the above equalities we have that Cyc k+1 C (T ) = red T k,r which, by the definition and our choice of r, is equal to red T k+1 . This finishes the proof.
We are ready to prove Theorem 1.4.
Therefore E C(T ) consists of all positive entries of C(T ) and due to the construction given by Algorithm 2 we know that nred = (µ), thus E C(T ) = {i + (µ) + 2j : 1 ≤ i ≤ n, l≤i−1 k l ≤ j < l≤i k l }. and comparing this with Proposition 5.1 finishes the proof.