Symplectic keys and Demazure atoms in type C

We compute, mimicking the Lascoux-Sch\"utzenberger type A combinatorial procedure, left and right keys for a Kashiwara-Nakashima tableau in type C. These symplectic keys have a similar role as the keys for semistandard Young tableaux. More precisely, our symplectic keys give a tableau criterion for the Bruhat order on the hyperoctahedral group and cosets, and describe Demazure atoms and characters in type C. The right and the left symplectic keys are related through the Lusztig involution. A type C Sch\"utzenberger evacuation is defined to realize that involution.


Introduction
The irreducible characters of the general linear group over C, the Schur functions, are combinatorially expressed as sums on semistandard Young tableaux [27]. When restricting to the symplectic group, two different types of symplectic tableaux have been proposed. King showed that the irreducible symplectic characters, the symplectic Schur polynomials, can be seen as a sum on a family of tableaux that are known as King tableaux [13], and De Concini has proposed the ones known as De Concini tableaux [6]. Kashiwara and Nakashima [11] described symplectic tableaux, which are just a variation of De Concini tableaux, with a crystal graph structure. That crystal structure allows a plactic monoid compatible with insertion and sliding algorithms, and Robinson-Schensted type correspondence, studied by Lecouvey in terms of crystal isomorphisms [17,18]. The generalization of the notion of plactic monoid for finite Cartan types was first introduced by Littelmann using his path model [22]. Symplectic Kashiwara-Nakashima tableaux are the ones that we work with, in the corresponding ambient plactic monoid. We however note that very recently Lee has endowed King tableaux with a crystal structure [19].
Kashiwara [10] and Littelmann [21] have shown that Demazure characters [7], for any Weyl group, can be lifted to certain subsets of the crystal B λ of Kashiwara-Nakashima tableaux of a given shape λ, called Demazure crystals. That is, a Demazure character (key polynomials) is the generating function of the weights over Demazure crystals. In type C, they are indexed by integer vectors and are certain non symmetric Laurent polynomials, with respect to the action of the Weyl group, which can be seen as "partial" symplectic characters, i.e., sums of a certain portion of monomials in a symplectic Schur polynomial.
In type A, the Demazure crystals are subsets of the crystal B λ , the crystal of all semistandard Young tableaux of shape λ. Lascoux and Schützenberger [16] identified the tableaux with nested columns as keys tableaux, and defined the right key map that sends tableaux to key tableaux. Their right key map gives a decomposition of B λ into non intersecting subsets , in bijecttion with elements v in the orbit of λ, under the action of the Weyl group [16,Theorem 3.8]. They have called standard bases to the sum of sum monomial weight over U(v), which, after Mason [23], are coined Demazure atoms.
The decomposition describes what tableaux contribute to the Demazure crystal B v , as a union of Demazure atoms, over an interval in the Bruhat order, on the classes modulo the stabilizer of λ. This order, induced on the orbit of λ, gives B v = λ≤u≤v U(u).
Our work has been motivated by the questions raised in a presentation by Azenhas [2], in The 69th Séminaire Lotharingien de Combinatoire. In those questions, Azenhas identified some Kashiwara-Nakashima tableaux as key tableaux, which match our identification, but it lacks a construction of the right key map, thus lacking a proof of the combinatorial description of type C Demazure characters. Note also that, during the preparation of this paper, Jacon and Lecouvey informed us about their paper [9], where they find the same key in type C, but their approach is different from ours.
Inspired by the Lascoux-Schützenberger's construction of the left and right keys of a semistandard Young tableau, we give a similar construction in type C. Our construction of the left and right keys of a Kashiwara-Nakashima tableau in type C, based on frank words in type C, that we introduce in Section 4, and Sheats symplectic jeu de taquin. Our Theorem 18 is the type C analogue of [16,Theorem 3.8].
The paper is organized as follows. In Section 2, we discuss the Weyl group of type C, the signed permutation group B n , the Bruhat order on B n and on its cosets, modulo the stabilizer of λ, the Kashiwara-Nakashima tableaux and the symplectic key tableaux.
Those key tableaux are used in Proposition 6 to explicitly construct the minimal lenth coset representatives. We recall some results from Brenti and Bjorner [4] and Proctor [24], that lead to a tableau criterion for the Bruhat order on B n and on its cosets, theorems 5 and 7 using the symplectic key tableaux. In Section 3, we recall the Baker-Lecouvey insertion, the Sheats symplectic jeu de taquin and use them to discuss the plactic and coplactic monoids and the Robinson-Schensted type C correspondence. These monoids have a natural interpretation in terms of type C Kashiwara crystal, for a U q (sp 2n )-module, in terms of connected components and crystal isomorphic connected components. In Section 4, we extend the concept of frank word, in type A, to type C and, with the help of symplectic jeu de taquin, we present our right and left key maps, Theorem 15. Using the right key map, we describe the tableau that contribute to a Demazure atoms and to a Demazure crystal, which is our main result, Theorem 18. In Section 5, we develop a type C evacuation, an analogue of the J-operation discussed by Schützenberger for semistandard Young tableaux in [25]. Proposition 22 shows that the evacuation of the right key of a Kashiwara-Nakashima tableau is the left key of the evacuation of the same tableau.
2. Weyl group of type C, Bruhat order and symplectic key tableau In the second set we will consider the following order of its elements: 1 < · · · < n < n < · · · < 1 instead of the usual order.
Consider the group B n , with generators s i , 1 ≤ i ≤ n, having the following presentation, regarding the relations among the generators, B n := s 1 , . . . , s n |s 2 i = 1, 1 ≤ i ≤ n; (s i s i+1 ) 3 = 1, 1 ≤ i ≤ n − 2; (s n−1 s n ) 4 = 1; (s i s j ) 2 = 1, 1 ≤ i < j ≤ n, |i − j| > 1 , known as hyperoctahedral group or signed symmetric group. This group is a Coxeter group and we consider the (strong) Bruhat order on its elements [4]. The elements of B n can be seen as odd bijective maps from [±n] to itself, i.e., for all σ ∈ B n we have σ(i) = −σ(−i), i ∈ [±n]. The subgroup with the generators s 1 , . . . , s n−1 is the symmetric group S n and its elements can be seen as bijections from [n] to itself. Both groups can also be seen as groups of n × n matrices. The elements of the symmetric group can be identified with the permutation matrices, and if we allow the non-zero entries to be either 1 or −1, we have the elements of B n . Hence B n has 2 n n! elements. The groups S n and B n are the Weyl groups for the root systems of types A n−1 and C n , respectively.
Let σ ∈ B n . We call to [a 1 a 2 . . . a n ], where a i = σ(i) for i ∈ [n], the window notation of σ, and write σ = [a 1 a 2 . . . a n ]. Given a vector v ∈ Z n , we have that s i , with i ∈ [n − 1], acts on v swapping the i-th and the (i + 1)-th entries, and s n acts on v changing the sign of the last entry. The window notation of s i σ is obtained after applying s i to the window notation of σ, if we see it as a vector. Ignoring signs, vσ The i-th letter of vσ changes its sign if and only if i appears in σ.
2.1. Bruhat order on B n . The length of σ ∈ B n , ℓ(σ), is the least number of generators of B n needed to go from [1 2 . . . n], the identity map, to σ. Any expression of σ as a product of ℓ(σ) generators of B n is called reduced. We say that two letters of the window notation of σ form an inversion if the bigger letter appears first. Next proposition gives a way to compute ℓ(σ) that only requires to look at the window notation of σ. This is a variation of the length formula presented on [4, Proposition 8.1.1], where the authors consider the usual ordering of the alphabet [±n] and the generator that changes the sign of an entry of the window notation acts on the first entry instead of the last one.

Remark.
• If i does not appear in the window presentation of σ, for all i ∈ [n], we may identify σ, in one-line notation, with σ(1) . . . σ(n) ∈ S n and ℓ(σ) = #{inversions of σ} The Bruhat order on the set of the elements of B n can be defined in the following way [4]: , where σ i are generators of B n , and u be two elements in B n . Then u ≤ w in the Bruhat order if By definition, if u ≤ w then ℓ(u) ≤ ℓ(w), but the reverse is not true. If σ(n) is positive and i = n, or, σ(i) < σ(i + 1) and i = n, we can also say that s i σ > σ.
The combinatorics of crystal graphs in type C and the Bruhat order combinatorics on B n and cosets are strongly related. In Subsection 2.3, we give a tableau criterion for the Bruhat order on B n . For this aim, we recall Kashiwara-Nakashima (KN) tableaux in type C and define symplectic key tableau.
2.2. Kashiwara-Nakashima tableau in type C. This subsection focuses on the notion of symplectic tableaux introduced by Kashiwara and Nakashima to label the vertices of the type C crystal graphs [12], which are a variation of the De Concini tableaux [6]. (See [26] for more details.) The Young diagram of shape λ is an array of boxes, left justified, in which the i-th row, from top to bottom, has λ i boxes. We identify a partition with its Young diagram. For example, the Young diagram of shape λ = (2, 2, 1) is .
Given µ and ν two partitions with ν ≤ µ entrywise, we write ν ⊆ µ. The Young diagram of shape µ/ν is obtained after removing the boxes of the Young diagram of ν from the Young diagram of µ. For example, the Young diagram of shape µ/ν = (2, 2, 1)/(1, 0, 0) is . Definition 2. Let ν ⊆ µ be two partitions and A a completely ordered alphabet. A semistandard skew tableau of shape µ/ν on the alphabet A is a filling of the diagram µ/ν with letters from A, such that the entries are strictly increasing in each column and weakly increasing in each row. When |ν| = 0 then we obtain a semistandard Young tableau of shape µ.
Denote by SSY T (µ/ν, A) the set of all semistandard Young skew tableaux T of shape µ/ν, with entries in A. When A = [n] we write SSY T (µ/ν, n).
When considering tableaux with entries in [±n], it is usual to have some extra conditions besides being semistandard. We will use a family of tableaux known as Kashiwara-Nakashima tableaux. From now on we consider tableaux on the alphabet [±n].
A column is a strictly increasing sequence of numbers in [±n] and it is usually displayed vertically. A column is said to be admissible if the following one column condition (1CC) holds for that column: Definition 3 (1CC). Let C be a column. The 1CC holds for C if for all pairs i and i in C, where i is in the a-th row counting from the top of the column, and i in the b-th row counting from the bottom, we have a + b ≤ i.
If a column C satisfies the 1CC then C has at most n letters. If 1CC doesn't hold for C we say that C breaks the 1CC at z, where z is the minimal positive integer such that z and z exist in C and there are more than z numbers in C with absolute value less or equal than z.

Example 2. The column
breaks the 1CC at 1.
The following definition states conditions to when C can be split: Definition 4. Let C be a column and let I = {z 1 > · · · > z r } be the set of unbarred letters z such that the pair (z, z) occurs in C. The column C can be split when there exists a set of r unbarred letters J = {t 1 > · · · > t r } ⊆ [n] such that: (1) t 1 is the greatest letter of [n] satisfying t 1 < z 1 , t 1 ∈ C, and t 1 ∈ C, (2) for i = 2, . . . , r, we have that t i is the greatest letter of [n] satisfying t i < min(t i−1 , z i ), t i ∈ C, and t i ∈ C.
The 1CC holds for a column C if and only if C can be split [26,Lemma 3.1]. If C can be split then we define right column of C, rC, and the left column of C, ℓC, as follows: (1) rC is the column obtained by changing in C, z i into t i for each letter z i ∈ I and by reordering if necessary, (2) ℓC is the column obtained by changing in C, z i into t i for each letter z i ∈ I and by reordering if necessary. If C is admissible then ℓC ≤ C ≤ rC by entrywise comparison. If C doesn't have symmetric entries, then C is admissible and ℓC = C = rC. In the next definition we give conditions for a column C to be coadmissible.

Definition 5.
We say that a column C is coadmissible if for every pair i and i on C, where i is on the a-th row counting from the top of the column, and i on the b-th row counting from the top, then b − a ≤ n − i.
Note that, unlike in Definition 3, in the last definition b is counted from the top of the column.
Given an admissible column C, consider the function Φ that sends C to the column of the same size in which the unbarred entries are taken from ℓC and the barred entries are taken from rC. The column Φ(C) is a coadmissible column and the algorithm to form Φ(C) from C is reversible [17, Section 2.2]. In particular, every column on the alphabet [n] is simultaneously admissible and coadmissible. Let T be a skew tableau with all of its columns admissible. The split form of a skew tableau T , spl(T ), is the skew tableau obtained after replacing each column C of T by the two columns ℓC rC. The tableau spl(T ) has double the amount of columns of T .
If T is a tableau without symmetric entries in any of its columns, i.e., for all i ∈ [n] and for all columns C in T , i and i do not appear simultaneously in the entries of C, then in order to check whether T is a KN tableau it is enough to check whether T is semistandard in the alphabet [±n]. In particular SSY T (µ/ν, n) ⊆ KN (µ/ν, n).
The weight of a word w on the alphabet [±n] is defined to be the vector wt(w) ∈ Z n where the entry i is obtained by adding the multiplicity of the letter i and subtracting the multiplicity of the letter i, for i ∈ [n]. If T is a skew tableau, the column reading of T , cr(T ), is the word obtained after concatenating all columns of T from right to left. The length of w is the total number o letters in w. The weigth of a KN tableau T is the vector wt T := wt(cr(T )) = ( In Section 3.2, we recall a way to go from a word in the alphabet [±n] to a KN tableau, the Baker-Lecouvey insertion.
2.3. Key tableaux in type C and the Bruhat order on B n .

Definition 7.
A key tableau in type C is a KN tableau in type C in which the set of elements of each column is contained in the set of elements of the previous column and the letters i and i do not appear simultaneously as entries, for any i ∈ [n]. The set of key tableaux in type A is the subset of the key tableaux in type C consisting of the tableaux having only positive entries, hence they are SSYT for the alphabet [n].
Every vector v of Z n is in the B n -orbit of exactly one partition, the one obtained by sorting the absolute values of all entries of v. Consider a partition λ ∈ Z n . The B n -orbit of λ is the set λB n := {λσ | σ ∈ B n }. For instance, the vector v = (1, 3, 0, 3, 2) is in the B 5 -orbit of λ = (3, 3, 2, 1, 0).

Proposition 2.
Let v ∈ Z n in the B n -orbit of the partition λ. There is exactly one key tableau K(v) whose weight is v. In addition, the shape of the key tableau K(v) is λ v . When v = λ, K(λ) is the only KN tableau of weight and shape λ, also called Yamanouchi tableau of shape λ.
Proof. Existence: Given v = (v 1 , . . . , v n ) ∈ Z n there exists a key tableau K of weight v by putting in the first |v i | columns the letter i if v i ≥ 0 or i if v i ≤ 0, and then sorting the columns properly. Clearly the columns of K are nested and it is a KN tableau without symmetric entries, hence it is a key tableau. Also, its shape is λ v .
Unicity: Since the key tableau doesn't have symmetric entries then, for all i ∈ [n], we have that in K the letter sgn(v i )i appears |v i | times in its entries. In order to the columns of K be nested we have that these |v i | entries appear in the first |v i | columns, hence we have determined exactly which letters appear in which column of K and now we just have to order them correctly. So the key tableau K with weight v is unique. When v = λ, K(λ) has only i's in the row i, for i ∈ [n]. Hence there is a bijection between Z n and key tableaux in type C on the alphabet [±n], given by the map v → K(v). If σ ∈ B n we put K(σ) := K((n, n − 1, . . . , 1)σ). The shape of K(σ) is the stair partition Λ = (n, n − 1, . . . , 1). One has a natural bijection between B n and the B n -orbit of Λ. Proposition 3. If σ ∈ B n has the letter α in the j-th position then α appears in the first n + 1 − j columns of the corresponding key tableau, K(σ).
We now append 0 to the alphabet [±n], obtaining [±n] ∪ {0}, where n < 0 < n, and, for all σ ∈ B n , we put σ(0) := 0. Given an element σ ∈ B n consider the map This map, originally defined in [4], produces a table which is related to key tableaux in type C. See example below: Example 8. Let σ = 3 124. Then (4, 3, 2, 1)σ = (3, 2, 4, 1) and We then have the following result: Remark. In [4, Chapter 8.1] the authors use the same alphabet as here, but with the usual ordering on the integers. So, to translate the results from there to here, it is needed to apply the ordering isomorphism defined by

The Bruhat order on cosets of
be the set of right cosets of B n determined by the subgroup W λ . Each coset W λ σ has a unique minimal length element σ v , such that v = λσ v . Reciprocally, given a vector v ∈ λB n , there is a unique minimal length element σ v ∈ B n such that v = λσ v . We have then a bijection between the elements in the B n -orbit of λ, and the right cosets of B n determined by the subgroup W λ , given by The set J c detects the minimal length coset representatives of W λ \ B n : σ is a minimal coset representatives of W λ \ B n if and only if all its reduced decompositions starts with a generator s i ∈ J c [4]. However key tableaux, K(v), v ∈ λB n , may be used to explicitly construct the minimal length coset representatives of W λ \ B n . Given a vector v ∈ λB n , we show that there is a unique minimal length element σ v ∈ B n such that v = λσ v and we show how to obtain σ v explicitly. The next proposition is a generalization of what Lascoux does in [14] for vectors in N n (hence σ v ∈ S n ).

Proposition 6.
Let v ∈ λB n . Consider T the tableau obtained after adding the column obtained after reading T where entries with the same absolute value are read just once.
Proof. Consider λ = (λ 1 , . . . , λ n ). In this proof we will write λ as (λ Let σ = [α 1 . . . α n ] ∈ B n read from T . Let's prove that α j appears λ j times in K(v): If j = 1 then α 1 appears in all columns of K(v), because it was the first letter read and the columns are nested. Hence it appears λ 1 times. Also, the |α 1 |-th entry of λσ is sgn(α 1 )λ 1 which is the weight of |α 1 | in K(v). For j > 1, proceeding inductively, we have that α j appears in all columns of K(v) not fully occupied by α i , with i < j, hence it appears λ j times. Also, the |α j |-th entry of λσ is sgn(α j )λ j , which is the weight of |α j | in K(v). This makes sense even if λ j = 0. So we have that λσ = v.
We only have to see that σ is the minimal length element of the set {ρ ∈ B n | λρ = v}. The subset of elements B n that applied to λ returns v is the coset W λ σ. Looking at σ, this allows us to swap α i and α j in σ if λ i = λ j and to change the sign of α i if λ i = 0. Since for each column the reading to obtain σ is ordered from the least to the biggest, we have that between these elements of B n , σ has minimal number of inversions and the letter α j is unbarred if λ j = 0 because α j will only be added to σ when reading the column C. Hence, by Proposition 1, σ is the minimal length element of W λ σ.
Given a partition λ ∈ Z n we identify each coset W λ σ with its minimal length representative σ v , where v = λσ ∈ λB n . Under this identification, we now induce the Bruhat order in the B n -orbit of λ and in the coset space of W λ B n .
Let v ∈ λB n . If K := K(v) is the key tableau with weight v, consider the tableau K obtained from K after erasing the minimal number of columns in order to have a tableau with no duplicated columns. Call v and λ to the weight and the shape of K, respectively. If K and K ′ are two key tableaux with shape λ, we have that K ≥ K ′ (by entrywise comparison) if and only if K ≥ K ′ . Note that to recover K from K we just have to know λ, and that K = K( v).
It is also possible to obtain v from v without having to look at the key tableau. If i is posistive, i and i do not appear in v but i + 1 or i + 1 appears then change all appearances of i + 1 and i + 1 to i and i, respectively, and repeat this as many times as possible, obtaining the vector v. The set of the absolute values of its entries is a set of consecutive integers starting either in 0 or 1. Hence the key tableau associated to it doesn't have repeated columns.
Due to Proposition 6 we have that σṽ Recall J and J c defined above. Note that the set J is the same for λ and λ. If i ∈ J c and i = n then all entries of λ are different from 0, which implies K(v) (and K(v)) having columns of length n; if i ∈ J c and i < n then λ i > λ i+1 , hence K(v) will have exactly i rows with length greater then λ i+1 , hence K(v) (and K(v)) will have columns of length i. Since K(v) doesn't have repeated columns, J c have exactly the information of what columns length exist in K(v). Theorem 3BC of Proctor's Ph.D. thesis [24] states that given a partition λ there is a poset isomorphism between the poset formed by the key tableaux of shape λ (ordered by entrywise comparison) and the poset formed by the Bruhat order in the vectors of the orbitλB n = {λσ : σ ∈ B n }.
The following theorem gives a tableau criterion for the Bruhat order on vector the same B n -orbit and for the corresponding B n -coset space.
Proof. Let v, u ∈ λB n two vectors in the same B n − orbit.We have that where (1) holds by Definition 8. Note that in (2) we also need to record λ, because it is needed in (4) to recover the shape of K(v) from the shape K(v). Finally the equivalence (3) is an application of Theorem 3BC of Proctor's Ph.D. thesis [24].

Crystal graphs in type C and symplectic plactic monoid
We recall two equivalence relations of words in the alphabet [±n], the Knuth type C equivalence, or (symplectic) plactic equivalence, and the (symplectic) coplactic equivalence. On the basis of these two equivalence relations is the Robinson-Schensted (RS) type C correspondence, in which each word is uniquely parametrized by a KN tableau and an oscillating tableau of the same final shape and whose length is of the word. This bijection has a natural interpretation in terms of crystal connectivity and crystal isomorphic connected components in Kashiwara theory of crystal graphs [5,11,17,18]. For this aim and reader convenience, we begin to recall the symplectic jeu de de taquin and Baker-Lecouvey insertion.
3.1. Sheats symplectic jeu de taquin. The symplectic jeu de taquin [17,26] is a procedure that allows us to change the shape of a KN skew tableau and eventually rectify it.
To explain how the symplectic jeu de taquin behaves, we need to look to how it works on 2-column KN skew tableaux. Let T be a 2-column KN skew tableau with splittable columns C 1 and C 2 such that C 1 has an empty cell.
Consider the tableau spl(T ) such that the columns ℓC 1 and rC 1 have an empty cell in the same row as C 1 . Call α to the entry under the empty cell of rC 1 and β to the entry right of the empty cell of rC 1 .
If α ≤ β or β does not exist, then both empty cells will change their position with the cells under them. This is a vertical slide.
If the slide is not vertical, then it is horizontal. So we have α > β or α does not exist. Call C ′ 1 and C ′ 2 to the columns after the slide. In this case we have two subcases, depending on the sign of β: (1) If β is barred we are moving a barred letter from ℓC 2 to rC 1 . Remember that ℓC 2 has the same barred part as C 2 and that rC 1 has the same barred part as Φ(C 1 ). So, looking at T , we will have an horizontal slide of the empty cell, . In a sense, β went from C 2 to Φ(C 1 ). (2) If β is unbarred we have a similar story, but this time β will go from Φ( Although in this case it may happen that C ′ 1 is no longer admissible. In this case, if the 1CC breaks at i, we erase both i and i from the column and remove a cell from the bottom and from the top column, and place all the remaining cells orderly. Eventually the empty cell will be a cell such that α and β do not exist. In this case we redefine the shape to not include this cell and the jeu de taquin ends. An entry of the tableau without cells under it or to the right of it is called an inner corner.
Given a KN skew tableau T of shape µ/ν, the rectification of T consists in playing the jeu de taquin until we get a tableau of shape λ, for some partition λ. The rectification is independent of the order in which the inner corners of ν are filled [ Remark. If the columns C 1 and C 2 do not have negative entries then the symplectic jeu de taquin coincides with the jeu de taquin known for SSYT.

Baker-Lecouvey insertion.
The Baker-Lecouvey insertion [3,17] is an algorithm that given a word in the alphabet [±n] returns a KN tableau. Let w be a word in the alphabet [±n], we call P (w) to the tableau obtained after inserting w. This insertion is similar to the usual column insertion for SSYT tableau. In fact both have the same behavior unless one the following three cases happens: Suppose that we are inserting the letter α in the column C of the KN tableau and (1) y ∈ C is the smallest letter bigger or equal then α and y ∈ C, for some y ∈ [n]: there is in C a maximal string of consecutive decreasing integers y, y − 1, . . . , u + 1 starting in the entry y in the column C. Then the bump consists of replacing the entry y with α and subtracting 1 to the entries y, y − 1, . . . , u + 1. The entry u is then inserted in the next column to the right. This is known as the Type I special bump. was. The entry u is then bumped into the next column. This is known as the Type IIb special bump. (3) after adding α in the bottom of the column C, the 1CC breaks at i: then we will slide out the cells that contain i and i via symplectic jeu de taquin. In the case 3 of the Baker-Lecouvey insertion we will be removing a cell from the tableau instead of adding. Despite the length of cr(P (w)) might be less than the length of w, the weight is preserved during Baker-Lecouvey insertion, wt(w) = wt(P (w)). The following result is corollary of the proposition. In particular we have that if we insert cr(T ) we obtain T again. This implies that during the insertion of cr(T ) the case 3 of the Baker-Lecouvey insertion cannot happen. In Example 11, we may conclude that P (23231) = P (cr(P (23231))) = P (11133).

Remark. The Baker-Lecouvey insertion is
3.3. Robinson-Schensted type C correspondence, plactic and coplactic equivalence. Let [±n] * be the free monoid on the alphabet [±n]. The Robinson-Schensted (RS) type C correspondence [17, Theorem 5.2.2] is a combinatorial bijection between words w ∈ [±n] * and pairs (T, Q) where T is a KN tableau and Q is an oscillating tableau, a sequence of Young diagrams that record, by order, the shapes of the tableaux obtained while inserting w, whose final shape is the same as T . Every two consecutive shapes of the oscillating tableau differ in exactly one cell and its length is the same of w. Since both the symplectic jeu de taquin and the Baker-Lecouvey insertion are reversible [3,17], we have that every pair (T, Q), with the same final shape, is originated by exactly one word. The Robinson-Schensted type C correspondence is the following map: where the union is over all partitions λ with at most n parts, and O(λ, n) is the set of all oscillating tableau with final shape λ and all shapes of the sequence have at most n rows. Given w 1 , w 2 ∈ [±n] * , the relation w 1 ∼ w 2 ⇔ P (w 1 ) = P (w 2 ) defines an equivalence relation on [±n] * known as Knuth equivalence. The type C plactic monoid is the quotient [±n] * / ∼ where each Knuth (plactic) class is uniquely identified with a KN tableau [15,17] 3.4. Crystal graphs in type C and coplactic monoid. Crystals were originally defined in quantum groups. Here we define them axiomatically associated to a root system Φ and a weight lattice Λ [5]. Let V be an Euclidian space with inner product ·, · . Fix a root system Φ with simple roots {α i | i ∈ I} where I is an index set and a weight lattice Λ ⊇ Z-span{α i | i ∈ I}. A Kashiwara crystal of type Φ is a nonempty set B together with maps [5]: where i ∈ I and 0 / ∈ B is an auxiliary element, satisfying the following conditions: In this case, we also have wt(b) = wt(a)+α i , ε i (b) = ε i (a) − 1 and ϕ i (b) = ϕ i (a) + 1; (2) for all a ∈ B, we have ϕ i (a) = wt(a), 2α i The crystals we deal with are the ones of a U q (sp 2n )-module. They are seminormal [5], and satisfy ϕ i (a) = max{k ∈ Z ≥ 0 | f k i (a) = 0} and ε i (a) = max{k ∈ Z ≥ 0 | e k i (a) = 0}. An element u ∈ B such that e i (u) = 0 for all i ∈ I is called a highest weight element. A lowest weight element is an element u ∈ B such that f i (u) = 0 for all i ∈ I. We associate with B a coloured oriented graph with vertices in B and edges labelled by This is the crystal graph of B.
If B and C are two seminormal crystals associated to the same root system, the tensor product B ⊗ C is also a seminormal crystal. As a set, we will have the Cartesian product B × C, where its elements are denoted by If B and C are finite, In type C n , we consider {e i } n i=1 the canonical basis of R n . The root system is Φ C = {±e i ± e j | i < j} ∪ {±2e i } and the simple roots are α i = e i − e i+1 , if i ∈ {1, . . . , n − 1}, α n = 2e n . The weight lattice Z n has dominant weights λ = (λ 1 ≥ · · · ≥ λ n ≥ 0). In type C n , the standard crystal is seminormal and has the following crystal graph: −e i . The highest weight element is the word 1, and the highest weight e 1 . We denote the crystal by B e 1 . The crystal B e 1 is the crystal on the words of [±n] * of a sole letter. The tensor product of crystals allows us to define the crystal B ⊗k of words of length k ≥ 1 on the alphabet [±n], where the vertex w 1 ⊗ · · · ⊗ w k is identified with the word w 1 . . . w k ∈ [±n] * . Given i ∈ [n], the action of the operators e i and f i on w is easily given by the signature rule [12,17,5]. For later convenience, (this becomes clear in the next section) the crystal operators will act, from now on, on the right. We substitute each letter w j by + if w j ∈ {i, i + 1} or by − if w j ∈ {i + 1, i}, and erase it in any other case. Then successively erase any pair +− until all the remaining letters form a word that looks like − a + b . Then ϕ i (w) = b and ε i (w) = a, e i acts on the letter associated to the rightmost unbracketed − (i.e., not erased), whereas f i acts on the letter w j associated to the leftmost unbracketed

Proposition 10. Let w be a word in the alphabet [±n]. Then w is a highest weight word if and only if the weight of all its prefixes (including itself) is a partition.
In this case, one has that P (w) = K(λ) the Yamanouchi tableau of shape λ, the weight of w Proof. Part "if": We will prove the contrapositive of the statement. There is a i such that (w)e i = 0. Let k be the position of the leftmost − of the signature rule of w, and consider the prefix w k with the first k letters. Since the k-th letter of w had an unbracketed − in the signature rule then the last letter of w k will also be an unbracketed −. Hence there are more − than + in the signature rule of w k . Call t α to the number of α in w k . We have that Part "only if": We will once again prove the contrapositive of the statement. Let w k be a prefix such that its weight is not a partition. Hence there is i ∈ [n] such that hence for this i there will be more − than + in the signature rule of w k . So in the first k letters of w there will be more − than +, so there is an unbracketed − in w, hence (w)e i = 0. Note that the argument works even if i = n. In this case we need to assume t n+1 = t n+i = 0.
It follows from [17, Proposition 3.2.6] that the insertion of the highest word w of weight λ is K(λ).
Let G n = k≥1 B ⊗k be the crystal of all words in [±n] * . Given k ≥ 1, the crystal B ⊗k , as a graph, is the union of connected components where each component has a unique highest weight word. Two connected components are isomorphic if and only if they have the same highest weight [11]. Two words w 1 , w 2 ∈ [±n] * are said to be crystal connected or coplactic equivalent if and only if they belong to the same connected component of G n . This means that both words are obtained from the same highest weight word, through a sequence of crystal operators f i , or one is obtained from another by some sequence of crystal operators f i and e j , i, j ∈ [n]. The connected components of G n are the coplactic classes in the RS correspondence that identify words with the same oscillating tableau [17,Proposition 5.2.1]. Also, two words w 1 , w 2 ∈ [±n] * are Knuth equivalent if and only if they occur in the same place in two isomorphic connected components of G n , that is, they are obtained from two highest words with the same weight through a same sequence of crystal operators [17]. This means if w 1 ∼ w 2 then (w 1 )e i ∼ (w 2 )e i and ( . Choose a word w ∈ [±n] * such that the shape of P (w) is λ. If we replace every word of its coplactic class with its insertion tableau we obtain the crystal of tableaux B λ that has all KN tableaux of shape λ. The crystal B λ does not depend on the initial choice of word w in the plactic class of w [17, Theorem 6.3.8].

Right and Left Keys and Demazure atoms in type C
In this section, we define type C frank words on the alphabet [±n] and use them to create the right and left key maps, that send KN tableaux to key tableaux in type C. The main result of this section is the type C version [16,Theorem 3.8], due to Lascoux and Schützenberger, which, using the right key map, gives a combinatorial description of Demazure atoms in type C.

Frank words in type C. Frank words were introduced in type A by Lascoux and
Schützenberger in [16]. We start by defining frank words in the alphabet [±n].

Definition 9.
Let w be word on the alphabet [±n]. We say that w is a type C frank word if the length of its columns form a multiset equal to the multiset formed by the length of the columns of the tableau P (w). Proof. Suppose that the statement is false. So there is a factor of w that is a nonadmissible column with all of its proper factors admissible. Hence we can apply the Knuth relation K5, meaning that w is Knuth related to a smaller word w ′ . But in this case, the number of letters of w ′ is less then the number of cells of P (w) = P (w ′ ), which is a contradiction.
The following proposition is an extension of [8, Proposition 7] on SSYT to KN tableaux.

Proposition 12. Let T be a KN tableau of shape λ. Let µ/ν be a skew diagram with same number of columns of each length as T . Then there is a unique KN skew tableau S with shape µ/ν that rectifies to T and cr(S) is a frank word.
Proof. If T is a Yamanouchi tableau K(λ) and S ∈ KN (µ/ν, n) rectifies to K(λ), then, since S and K(λ) have the same number of cells, all entries of S are unbarred, hence S is a semistandard skew tableau. So, it follows from [8,Proposition 7] that S exists and is unique. If T is not a Yamanouchi tableau, note that T is crystal connected to K(λ) and from [17,Theorem 6.3.8] we have that the symplectic jeu de taquin slides commutes with the action of the crystal operators. Consider Y ′ λ the only tableau on the skew-shape µ/ν that rectifies to Y λ , which exists due to [8,Proposition 7]. Since S rectifies to T , which is crystal connected to K(λ), and Y ′ λ rectifies to K(λ), S is crystal connected to Y ′ λ and the path has same sequence of colours as the one from T to K(λ). Hence S exists and is uniquely defined. Proof. All other skew tableaux with given last column length can be found from a given one by playing the symplectic jeu de taquin or its reverse in all columns except the last one. Note that S has the same number of cells of the tableau obtained after rectifying, hence we can't lose cells when applying the symplectic jeu de taquin or its reverse.
Fixed a KN tableau T , consider the set of all possible last columns taken from skewtableaux with same number of columns of each length as T . Corollary 13 implies that this set has one element for each distinct column length of T . For every column C in this set, consider the columns rC, its right column. The next proposition implies that this set of right columns is nested, if we see each column as the set formed by its elements.

Proposition 14.
Consider T a two-column KN skew tableau C 1 C 2 with an empty cell in the first column. Slide that cell once via symplectic jeu de taquin, obtaining a two-column KN skew tableau C ′ 1 C ′ 2 with an empty cell. Then rC ′ 2 ⊆ rC 2 . Proof. If the sliding was vertical then C ′ 2 = C 2 , hence rC ′ 2 = rC 2 . If the sliding was horizontal, call β to the number on the cell right of the empty cell on spl(T ). Call Φ to the function that takes an admissible column to the associated coadmissible column.
If β = b is unbarred then C ′ 2 = Φ −1 (Φ(C 2 ) \ {b}). In this case Φ(C ′ 2 ) = Φ(C 2 ) \ {b}, hence rC 2 and rC ′ 2 have the same barred part. Consider z 1 < · · · < z ℓ the unbarred letters that appear on C 2 and not on Φ(C 2 ). When we take b from Φ(C 2 ), if b ∈ Φ(C 2 ) our set of letters z 1 < · · · < z ℓ will lose an element, giving the inclusion of the unbarred part of C ′ 2 in C 2 ; if b ∈ Φ(C 2 ), then b ∈ C 2 and in C ′ 2 the least z i > b may reduce to b, and subsequent z j may reduce to z j−1 . Hence we have the inclusion of the unbarred part of C ′ 2 in C 2 . If β = b is barred then C ′ 2 = C 2 \ {b}. In this case rC 2 and rC ′ 2 have the same unbarred part. Consider t 1 > · · · > t ℓ the barred letters that appear on Φ(C 2 ) and not on C 2 . When we take b from C 2 , if b ∈ C 2 our set of t 1 > · · · > t ℓ letters will lose an element, giving the inclusion of the barred part of rC ′ 2 in rC 2 ; if b ∈ C 2 , then b ∈ Φ(C 2 ) and in C ′ 2 the least z i > b may reduce to b, and subsequent bigger z j 's may reduce to z j+1 . Hence we have the inclusion of the barred part of Φ(C ′ 2 ) in Φ(C 2 ). This proposition defines a map that sends a KN tableau to key tableau in type C, identified as the (symplectic) right key of a KN tableau. (Right key map). Given a KN tableau T , we can replace each column with a column of the same size taken from the right columns of the last columns of all skew tableaux associated to it. We call this tableau the right key tableau of T and denote it by K + (T ).

Theorem 15
Proof. The previous proposition implies that the columns of K + (T ) are nested and do not have symmetric entries. So, it is indeed a KN key tableau.

Remark.
• Recall the set up of Proposition 12. If the shape of S, µ/ν, is such that every two consecutive columns have at least one cell in the same row, then each column of S is a column of the word cr(S), hence cr(S) is a frank word. Moreover, the columns of S appear in reverse order in cr(S). Therefore, given a KN tableau T , the columns of K + (T ) can be also found as the right columns of the first columns of frank words associated to T . • If T is a SSYT then this right key map coincides with the one defined by Lascoux and Schützenberger in [16].

Example 14.
The tableau T = 1 3 1 3 3 3 is associated to 6 KN skew tableaux with same number of columns of each length as T , each one corresponding to a permutation of its column lengths, and each one is associated to its column reading, which is a frank word. The right key tableau associated to T has as columns r 3 3 1 , r 3 1 and r 1 . Hence In the same spirit of the right key, we define the left key of a KN tableau. Just like in Proposition 14, we can prove that the slides of the symplectic jeu de taquin are effectively adding an entry to ℓC 1 , i.e. ℓC 1 ⊆ ℓC ′ 1 , hence the left columns of the first columns of all skew tableaux with the same number of columns of each length as T will be nested.
So, if we replace each column of T with a column of the same size taken from the left columns of the first columns of all skew tableaux associated to it we obtain the left key K − (T ).

Demaure crystals and right key tableaux.
Let λ ∈ Z n be a partition and v ∈ λB n . We define U(v) = {T ∈ KN (λ, n) | K + (T ) = K(v)} the set of KN tableaux of B λ with right key K(v). Given a subset X of B λ , consider the operator D i on X, with i ∈ [n] defined by XD i = {x ∈ B λ | (x)e k i ∈ X for some k ≥ 0} [5]. If v = λσ where σ = s i 1 . . . s i ℓ(σ) ∈ B n is a reduced word, we define the Demazure crystal to be This definition is independent of the reduced word for σ [5,Theorem 13.5]. In particular, when σ is the longest element of B n we recover B λ . Also this definiton is independent of the coset representative of W λ σ, that is, where the two rightmost identities follow from Theorem 7.

Lemma 1. Let σ = s i be a generator of B n and C an admissible column such that
Proof. Let i = n. We can apply f i to C if and only n ∈ C and n ∈ C. In this case n ∈ rC and after applying f i we have n ∈ C and n ∈ C, hence n ∈ rC. So wt(rC) = wt(r((C)f n ))s n .
Let i < n. We can apply f i to C, so we have 6 cases to study: (1) i ∈ C, i + 1, i + 1, i ∈ C: In this case we have that i + 1 ∈ (C)f i , i, i + 1, i ∈ (C)f i . Note that i / ∈ rC and i + 1 / ∈ r((C)f i ). If i + 1 ∈ rC then i ∈ r((C)f i ), hence f i swaps the weight of i and i + 1 from (1, 0) to (0, 1), respectively. If i + 1 ∈ rC then i ∈ r((C)f i ), hence f i swaps the weight of i and i + 1 from (1, −1) to (−1, 1).
(2) i, i + 1 ∈ C, i + 1, i ∈ C: In this case we have that i Note that i, i + 1 ∈ rC, i+1, , i ∈ rC and that i+1, i ∈ r((C)f i ), i, i + 1 ∈ r((C)f i ), and all other appearances in rC are intact. Hence f i swaps the weight of i and i + 1 from (1, −1) to (−1, 1).
and all other appearances in rC are intact. Hence f i did nothing to weight of rC.
and all other appearances in rC are intact. Hence f i did nothing to weight of rC. (5) i, i + 1, i ∈ C, i + 1 ∈ C: In this case we have that i Note that i, i + 1 ∈ rC, i+1, i ∈ rC and that i+1, i ∈ r((C)f i ), i, i + 1 ∈ r((C)f i ), and all other appearances in rC are intact. Hence f i swaps the weight of i and i + 1 from (1, −1) to (−1, 1). (6) i + 1 ∈ C, i, i + 1, i ∈ C: In this case we have that i ∈ (C)f i , i, i + 1, i + 1 ∈ (C)f i .
Note that i, i + 1 ∈ rC and i + 1 ∈ rC. If i ∈ rC then we have i, i + 1 ∈ r((C)f i ) and i + 1, i ∈ r((C)f i ), so f i did nothing to weight of rC. If i ∈ rC then i + 1 ∈ r((C)f i ) and i ∈ r((C)f i ), hence f i swaps the weight of i and i + 1 from (0, −1) to (−1, 0).

Remark.
All the cases where the weight is preserved happen to have equal weight for i or i + 1 in rC or we are in a column C in which we can also apply e i .
Hence we have the following corollaries: Proof. Consider a multiset of frank words F such that the multiset of length of their first columns is the same of the multiset of lengths of columns of T .
If K + ((T )f i ) = K + (T ) then we are done. Else there are two cases: 1 ≤ i < n and i = n.
Consider 1 ≤ i < n. There is a column of T whose weight of i is bigger than its weight for i + 1. Since T is a key tableau, this implies that in all columns of T weight of i is bigger or equal than the weight of i + 1.
Let A be the subset of F such that the weight of i and i + 1 in the right column of its first column is different and does not swap when we apply f i to the frank word.

Lemma 2. Let i ∈ [n]
and C be an admissible column such that one of the following happens (1) i < n and the weight of i in rC is less than the weight of i + 1 in rC; (2) i = n and weight of i is negative in rC, then we can apply e i to C (in the sense (C)e i = 0).
Proof. If i = n then −n appears on rC and n does not. Since n is the biggest unbarred letter of the alphabet we have that −n also appears in C and n does not. Hence we can apply e n to C.
If i < n and the weight of i in rC is less than the weight of i + 1 in rC then the weight of both can be one of the following three options: (0, 1), (−1, 1), (−1, 0). Note that rC does not have symmetric entries. So in the first two cases we have that i + 1 exists in rC and i does not, hence i + 1 exists in C and i does not, so we can apply e i to C. In the last case, we have that i exists in rC and i + 1 and i + 1 does not. Hence we have that i exists in C and i or i + 1 does not, so we can apply e i to C.
The next theorem is the main theorem of this paper. It gives a description of a Demazure crystal atom in type C using the right key map Theorem 15. Lascoux and Schützenberger, in [16,Theorem 3.8], proved the type A version of this theorem, which consists in considering the case when v ∈ N n and, consequently, σ v ∈ S n . For inductive reasoning, used in what follows, we recall the chain property on the set of minimal length coset representatives modulo W λ [4,Theorem 2.5.5].
Let ρ ≥ 0. Consider σ = s i a generator of B n such that ρσ > ρ and λρσ = λρ = v, i.e., ρσρ −1 / ∈ W λ . Recall e i , ε i , f i and φ i from the definition of the crystal B λ . If T ∈B λρσ then T is obtained after applying f i (maybe more than once) to a tableau inB λρ , which by inductive hypothesis exists in U(v). By Corollary 16, if (T )f i / ∈ U(v) then (T )f i ∈ U(vσ). So it is enough to prove that given a tableau T ∈ U(v) ∪ U(vσ) then (T )e We have two different cases to consider: i = n and i < n. If T ∈ U(vσ) then, if i < n, there exists a frank word of T such that, if V 1 is its first column then rV 1 has less weight for i than for i + 1 (less in the usual ordering of real numbers); if i = n, there exists a frank word of T such that, if V 1 is its first column then rV 1 has negative weight for i. Since we are in the column rV 1 , if i < n, i and i + 1 can have weights (0, 1), (−1, 1) ou (−1, 0) and if i = n then i has weight −1. Note that these are the exact conditions of Lemma 2. In either case, due to Lemma 2, we can applying e i enough times to the frank word associated until this no longer happens. This is true because we only need to look to V 1 to see if it changes after applying e i enough times to the frank word. In the signature rule we have that successive applications of e i changes the letters of a word from the end to the beginning, so, from the remark after Lemma 1, the number of times that we need to apply e i , in order to conditions of Lemma 2 do not hold for the first column, is ε i (T ). So K + (T )e ε(T ) i = K(vσ), hence, from Corollary 17, will be in a Demazure crystal associated to ρ ′ ∈ B n , with ρ ′ < ρ such that ρ ′ σ = ρ. This cannot happen because in this case ρ ′ = ρσ < ρ, which is a contradiction. The crystal is split into several parts. Each one of those parts is a Demazure atom and contains exactly one symplectic key tableau, so we can identify each part with the weight of that key tableau, which is a vector in the B 2 -orbit of (2, 1). From the previous theorem we have that all tableaux in the same part have the same right key.

Combinatorial description of type C Demazure characters and atoms.
Given v ∈ λB n define the Demazure character (or key polynomial), κ v , and the Demazure atom in type C, κ v , as the generating functions of the KN tableaux weights in B v and B v , respectively:

Lusztig Involution in types
For type A we have that i ′ = n − i and ω 0 is the reverse permutation, and for type C we have i ′ = i and ω 0 = −Id, where Id is the identity map. In both cases the involution can be seen as flipping the crystal upside down. Definition 11. [5] Let C be a connected component in the crystal G n of type C, the crystal of words in [±n] * . The dual crystal C ∨ is the crystal obtained from C after reversing the direction of all arrows. Also, the if x ∈ C, then for its correspondent in C ∨ , x ∨ , we have wt(x) = −wt(x ∨ ).
In type C, since i ′ = i and ω 0 = −Id, it follows from the definition that C and C ∨ , as crystals in G n , have the same highest weight. Therefore, they are isomorphic. In the case of B λ , the crystal of KN tableaux of shape λ, the Lusztig involution is a realization of the dual crystal of B λ . Hence the crystal B λ is self-dual. We shall see other realizations of the dual.

Evacuation algorithms.
In type A, the Lusztig involution on the crystal of SSYTs with same shape is known as Schützenberger involution or evacuation, and takes T , a tableau with weight wt T , to T Ev . The tableau T Ev has weight (wt T )ω 0 , where ω 0 is the longest permutation of S n , in the Bruhat order. Note that (wt T )ω 0 is the vector wt T in reverse order, i.e., (v 1 , . . . , v n )ω 0 = (v n , . . . , v 1 ). In type C we will work with KN tableaux instead of SSYTs. Consider T ∈ KN (λ, n). In this case T Ev is a tableau with the same shape of T , such that wt T = −wt T Ev = (wt T Ev )ω C 0 , where ω C 0 is the longest permutation of B n . The complement of a tableau or a word in types A and C consists in applying ω 0 and ω C 0 , respectively, to all its entries. In type A, it sends i to n + 1 − i for all i ∈ [n], i.e., w 0 (i) = n + 1 − i and in type C we have w 0 (i) = −i. Given a SSYT, there are several algorithms to obtain a SSYT with the same shape whose weight is its reverse. We recall some versions of them for which one is able to find analogues for KN tableaux.
(2) Define w ⋆ the word obtained by complementing its letters and writing it backwards.
In both Cartan types we have that algorithms 1 and 2 produce the same tableau since the column reading of T 0 is w ⋆ , assuming that, in type C, T 0 is admissible. This can be concluded using the following lemma. Lemma 3. For type C, the split of a column C, (ℓC, rC) is the rotation and complement of the split of the column C 0 = Complement(π-rotate(C)), (ℓC 0 , rC 0 ).
Proof. Let's say that (ℓC, rC) = A ′ A BB ′ where C = A B , ℓC = A ′ B and rC = A B ′ , where A and A ′ are the unbarred letters of the columns C and ℓC, respectively, and B and rB are the barred letters of C and rC, respectively. Note that ℓC and C share the barred part and C and rC share the unbarred part.
Consider a KN tableau T with column reading w. The column reading of the tableau obtained after applying Algorithm 1 to T is Knuth-related to w ⋆ , because both give the same tableau if inserted. Since ⋆ is an involution ((w ⋆ ) ⋆ = w), if we apply the algorithm again we will get a tableau whose column reading, by the last theorem, is Knuth equivalent to (w ⋆ ) ⋆ = w, hence we will have T again. So Algorithm 1 is an involution. Next we conclude that Algorithm 1 is a realization of the Lusztig involution for type C. Theorem 21. Let w ∈ [±n] * . The connected component of the crystal G n that contains the word w is isomorphic to the one that contains the word w ⋆ . Therefore P (w) and P (w ⋆ ) have the same shape and weights of opposite sign. Moreover, the two crystals are dual of each other and the ⋆ map is a realization of the dual crystal.
Proof. Remember the crystal operators e i and f i from the definition of crystal. Note that ((w)f i ) ⋆ = (w ⋆ )e i , because in the signature rule applied to w and w ⋆ , the distance of the leftmost unbracketed + of w to the beginning of the word is equal to the distance of the rightmost unbracketed − of w ⋆ to the end of this word. Hence, the letter that changes when applying f i to w is the complement of the letter that changes when applying e i to w ⋆ , and the letter obtained on their position after applying the crystal operators are also complement of each other. Hence the crystal that contains the word w ⋆ is the dual to the one that contains w. But the crystal that contains w is self-dual, hence the crystals that contains any of the words are isomorphic. From [17, Theorem 3.2.8] P (w) and P (w ⋆ ) have the same shape.

Right and left keys and Lusztig involution.
The next result shows that the right and left key maps defined for KN tableaux anticommutes with the Lusztig involution. The evacuation of the right key of a tableaux if the left key of the evacuation of the same tableau.

Proposition 22. Let T be a KN tableau and Ev the type C Lusztig involution. Then
Proof. Since the tableaux K + (T ) and K − (T Ev ) are a key tableaux, they are completely determined by their weights. Then we just need to prove that their weights are symmetric.
Fix a column C of K + (T ). There is a frank word w, Knuth related to cr(T ), such that C is the right column of the first column of w. Let's say the w k is the first column of w. From Propostion 20, w ⋆ is Knuth related to cr(T ) ⋆ , hence P (w ⋆ ) = T Ev . Also note that the w ⋆ has the same number of columns of each length as w, hence it is a frank word, and its last column is w ⋆ k . Note that Lemma 3 implies that if v is an admissible column, then l(v ⋆ ) = (rv) ⋆ . So we have that l(w ⋆ k ) = (rw k ) ⋆ is a column of K − (T Ev ). Therefore, for each column C of K + (T ) there is a column of K − (T Ev ) whose weight is ω 0 (C), hence K + (T ) and K − (T Ev ) have symmetric weights.

Final Remarks
In [23], Mason showed that atoms are specializations of non-symmetric Macdonald polynomials of type A with q = t = 0. This allowed to use the shape of semi-skyline augmented fillings, in bijection with SSYT, to detect the right key tableaux. It would be interesting to obtain a similar object for a KN tableau in type C. For example, semiskyline augmented fillings have been instrumental to obtain a RSK type bijective proof [1] for the Lascoux non-symmetric Cauchy identity in type A [14].
In [20], using the model of alcove paths, Lenart defined an initial key and a final key that have a similar behaviour to the left and right keys defined here. Jacon and Lecouvey [9] have shown that the right key map can be computed as the last direction for paths in that model.

Acknowledgements
This work was partially supported by the Center for Mathematics of the University of Coimbra -UID/MAT/00324/2019, funded by the Portuguese Government through FCT/MEC and co-funded by the European Regional Development Fund through the CMUC, Department of Mathematics, University of Coimbra, Apartado 3008, 3001-454 Coimbra, Portugal E-mail address: jmsantos@mat.uc.pt