Kostka Numbers and Longest Increasing Subsequences Scott Neville Mentor: Arjun Krishnan August 21, 2017 1 Introduction The Kostka numbers appear naturally in several counting problems related to the sym- metric group Sn. Let n be a positive integer, λ a Young diagram with n blocks, and P µ = (µ1; : : : ; µk) be a the content (or weight) vector such that i µi = n. The Kostka number Kλµ is the number of semistandard Young tableaux of shape λ filled with µi copies of the number i. We wish to find the cardinality of the set Ln(k) ⊂ Sn, whose elements have longest increasing subsequences of length at most k. When k = 2, jLn(2)j = Cn, where Cn is the nth Catalan number; we discuss this basic example in some detail below. This paper contains a generalization of this fact and establishes a bijection between certain Kostka numbers and Ln(k). The Robinson-Shensted (RS) correspondence establishes a bijection between permu- tations in Sn and a pairs of Young tableau of the same shape λ. The length of the top row (or width) of λ is the length of a longest increasing subsequence (LIS). Let L(σ) denote the length of an LIS in σ. Consider permutations of length n with L ≤ 2. Via the RS correspondence, they give pairs of tableaux (P; Q) with width at most 2 and size n. There is a standard way to convert these into a single rectangular tableau of shape 2×n (Stanley, 1999) : First replace i with 2n − i + 1 for i 2 [n] in the tableau Q. Then, rotate Q by 180 degrees and glue it to P ; the gluing is done so that the rows with length 1 in P align with the rows of length 1 in Q. Thus we obtain a rectangular diagram with width 2 and height 2n. Example 1. 1 2 ; 1 3 7! 1 2 7! 1 2 3 2 3 2 3 5 3 1 4 6 It's not hard to see that the steps of the transformation can be reversed, and that it is a bijection. The number of such rectangular tableaux is given by the Kostka number 1 K ~ . Since the rectangular tableau is standard, we can use the hook length formula 2×n;1 to count them, and hence (2n!) 1 2n K ~ = = =: Cn: (1) 2×n;1 n!(n + 1)! n + 1 n In this paper we generalize (1). Let ~ak be a vector of k copies of the number a. ~ Theorem 1.1. Let µk = kn ⊕~1n be the vector of n copies of k and n copies of 1. Then, Kw×n,µw−1n = jLn(w)j: The Kostka numbers in Theorem 1.1 count rectangular tableaux with particular weight vectors. The proof establishes a bijection between the rectangular tableaux and LIS with length at most k, by generalizing the argument that proves (1). The generaliza- tion involves a new operation that might be of independent interest. A similar operation on diagrams appears in Briand et al.(2015); our operation extends it to tableaux. 2 Definitions We represent a Young diagram as λ = (λ1; ··· ; λm) with λi being the number of boxes in the ith column. This is nonstandard but it makes the tableaux complement operation P easier to state. A diagram λ gives a partition i λi = n, which we denote λ ` n. Recall that a semistandard Young tableau has each column strictly increasing and each row non decreasing. Let jwj represent the cardinality of a vector w. For 2 vectors w; v, we say w v if jwj > jvj and 8i ≤ jvj; wi ≤ vi. Thus if P is a tableau with th vectors Pi indicating the values contained in the i column of P , we can equivalently say P is a semistandard tableau if all Pi are increasing, and i ≤ j =) Pi Pj. We call a Young tableau a standard Young tableau if each number appears at most once. Finally, we use [n] to denote the set f1; ··· ; ng. 2.1 Tableau operators We have two operations on tableaux that we need to introduce. First, provided λ is a sub-diagram of w×h, we define w×h − λ as the sub-diagram of w×h with a 180 deg rotated λ removed from the bottom right corner of w×h. Definition 2.1 (square diagram/tableau subtraction). γ = w×h−λ is a Young diagram with columns γi that satisfy jγij = h − jλw−i+1j: When R is a tableau, we similarly remove a lambda shaped block from it, with (R − λ)i being composed of the h − jλw−i+1j smallest elements in Ri. 2 For example λ = 3×4 − λ = Definition 2.2 (Column complement). The complement of a column C with distinct n elements with respect to a set [n] is defined as C = [n]−C, where the elements of C are arranged in increasing order. We will omit the superscript n if it's clear from context. Note that nn C = C i.e., the operation is involutive. To produce a new tableau using the complement operation on columns, we have to reverse the column order. Definition 2.3 (Tableau complement). The complement of a tableau P with respect to w;n a width w is denoted by P , where w;n n P i = Pw−i+1 ; We will always set n = maxij Pij, though this is not necessary for the claims below. If we complement all the columns of a tableau and reverse the column order, we obtain a new tableau. This fact will be proved in Lemma 2.4. Example 2. 3;4 P = 1 1 2 P = 1 2 4 2 3 4 3 3 4 w;n Lemma 2.4. If P is a tableau then so is P for all w; n 2 N. We will prove Lemma 2.4 in section3. w;n Lemma 2.5. The map P 7! P is invertible, and is its own inverse. w;n Proof. If C = P then, the complement of the w − i + 1th column of C is equal to the ith column of P . That is, n Cw−i+1 = ([n] − Cw−i+1) = ([n] − ([n] − Pw−(w−i+1)+1)) = ([n] − ([n] − Pi) = Pi: 3 3 Algorithm We provide an invertible algorithm which takes a w×n shaped tableau with weights µw−1n , and returns a pair of tableaux (P; Q) with shape λ and width at most w. This triple in turn corresponds to a permutation with LIS of length at most w by the RS correspondence. Let R be a semistandard tableau of shape w×n with weight µw−1n. In short, we split R into 2 tableaux, Q is taken from the positions of the values [n + 1; 2n] in R, while w;n P is given by (R − λ) . We describe this in detail below. 1. The elements [n + 1; 2n] in R must form a contiguous skew tableau in the bottom left corner of R. If we rotate this skew tableau by 180 degrees and replace the numbers k with 2n − k + 1 for k 2 [n + 1; 2n], we obtain a standard tableau Q of shape λ. Explicitly, if λ = (λ1; : : : ; λk), λi = jfa 2 Rw−i+1 : a > ngj: For all (i; j) in the diagram λ we have Qi;j = 2n − Rw−i;n−j + 1. 2. To get P , we remove the elements [n + 1; 2n] from R, and then take its tableau complement: w;n P = R − λ This forms a standard tableau of shape λ by Lemma 2.4. Example 3. If 1 1 2 R = 2 3 4 3 4 5 6 7 8 First, we split the R into R − λ and a piece that will eventually become Q. 1 1 1 5 2 3 4 6 7 8 3 4 Then, we take the complement of the first tableau to get P . To get Q, we rotate and replace the numbers as specified above. The algorithm returns: 1 2 4 1 2 3 P = Q = 3 4 Lemma 3.1. The P given by the algorithm is a standard tableau, and has the same shape λ as Q. 4 Proof. The proof of the above lemma follows from Claims 3.3, 3.4 and 3.5. We show that the complement operation on R − λ gives the correct shape λ for P , that P has the right entries and that P is standard tableau. Lemma 3.2. The algorithm is a bijection from rectangular tableau of shape w×n and weight µw−1 to pairs of standard tableau of the same shape, with n elements and width at most w. Proof. The algorithm is reversible at every stage: the splitting of R into R − λ and Q is a bijection; the relabeling of the elements in Q is a bijection from f1; : : : ; ng to fn + 1;:::; 2ng and finally the complement operation is a bijection between tableau of shape w×n − λ and weight µw−1 and standard tableau of shape λ (Lemma 2.4). We may invert the algorithm by reversing the steps: given a pair of tableaux (P; Q) of shape λ ` n and width at most w, we can reconstruct a w×n shaped tableau R with weight µw−1. Since each of the stages of the algorithm are bijections, it's easy to see that the algorithm itself is a bijection. Claim 3.3 (Complement gives the right shape). Let λ be a Young diagram. Let A be ~ w;n a tableau with shape w×n − λ and weight (w − 1)n.