Kostka Numbers and Longest Increasing Subsequences

Home , Hook length formula, Kostka number

Scott Neville Mentor: Arjun Krishnan

August 21, 2017

1 Introduction

The Kostka numbers appear naturally in several counting problems related to the symmetric 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, |Ln(2)| = 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 permutations 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 ∈ [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 = |Ln(w)|.

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 generalization 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 |w| represent the cardinality of a vector w. For 2 vectors w, v, we say w v if |w| > |v| and ∀i ≤ |v|, 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 {1, ··· , n}.

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 |γi| = h − |λw−i+1|.

When R is a tableau, we similarly remove a lambda shaped block from it, with (R − λ)i being composed of the h − |λw−i+1| 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 ∈ 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 ∈ [n + 1, 2n], we obtain a standard tableau Q of shape λ. Explicitly, if λ = (λ1, . . . , λk),

λi = |{a ∈ Rw−i+1 : a > n}|.

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 {1, . . . , n} to {n + 1,..., 2n} 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. Then, A has shape λ.

Proof. There are exactly n − λw−i+1 distinct elements in Ai, since A is formed by sub- tracting a rotated λ from the square tableau of height n. The elements are distinct because columns in a tableau must be strictly increasing. Therefore,

|Aw−i+1| = n − |Ai| = λw−i+1.

This shows that all the columns of A have the correct height.

Claim 3.4 (Complement gives the right entries). For all natural numbers n, w, let k be an integer with |k| ≤ n/(w − 1), λ ` n − k(w − 1) a diagram and A a tableau with shape ~ w×n − λ with weight w − 1n−k, then A contains exactly one copy of each element in [n − k].

Proof. Since there are only w columns, and no column can have a duplicate element (if a column did then it wouldn’t be strictly increasing), there can be at most one column that does not contain some element v. Let ρ(v) be the only column of A such that Aρ(v) does not contain v ∈ [n − k]. This means that v ∈ Aw−ρ(v)+1, so all elements [n − k] appear in A. Since |w×n − λ| = (n − k)(w − 1), and n − k = maxi,j Ai,j we have |A| = w(n − k) − (n − k)(w − 1) = n − k, so [n − k] each appear exactly once.

Claim 3.5 (Column complement correctly reverses ≺ order). Let v ≺ w be two column vectors. Then, w ≺ v.

5 Proof. Let U = {a1, . . . , an} be an ordered subset of the natural numbers filling w and v. We write x for the complement of a column vector with respect to U; i.e., x := x ⊂ U. k We assume that both x = (xi)i=1 and x are arranged in increasing order. Let w and v be two columns such that w ≺ v. The proof proceeds by induction on the elements of U. We imagine that we’re growing w and v by examining each element of U sequentially and adding it to either (v) or (w). Define (v)q := {x ∈ U \ v, x ≤ aq} to be the elements of U added to (v) by step q and (w)q similarly. Suppose the first q elements of U have been examined. Then, there are four obvious possibilities for the q + 1th element:

1. aq+1 ∈ v ∩ w, then aq+1 is not added to either (v)q or (w)q.

2. aq+1 ∈ v ∩ w (then aq+1 is added to (v)q).

3. aq+1 ∈ v ∩ w, then aq+1 is added to (w)q.

4. aq+1 ∈ v ∩ w, then aq+1 is added to both (v)q and (w)q.

As the induction hypothesis, suppose |(v)q| ≥ |(w)q| and vi ≤ wi for i = 1,..., |(v)q|. Then, in case1, aq+1 is not added to either set, and hence |(v)q+1| = |(v)q+1| = |(w)q|. In case2, |(v)q+1| ≥ |(w)q+1| and since no element was added to (w)q+1, v|(v)q+1| ≤ w|(v)q+1|. In case3, we observe that we must have |(v)q| > |(w)q|; for if |(v)q| = |(w)q|, we must have seen the same number of elements (r = q − |(w)q|) from both w and v in q steps. Then, we must have aq+1 = vr+1, and hence wr+1 > vr+1. This contradicts w ≺ v. Case 4 is trivial, and hence the induction step is proved. To establish the q = 0 case, we essentially repeat the above argument with empty (v)0 and (w)0. It is easy to see here that case 3 simply cannot occur.

4 Future Work

One can also consider rectangular tableau with slightly different weights. For example, suppose we had n + k entries of w − 1. In these cases we can still apply the above algorithm, but we get tableau of slightly different size and shape. These holes can be filled in via several natural choices to create more bijections. We hope to continue this work and write it up nicely for a short paper.

References

Briand, E., R. Orellana, and M. Rosas (2015). Rectangular symmetries for coeﬃcients of symmetric functions. Electron. J. Combin. 22 (3), Paper 3.15, 18.

Stanley, R. P. (1999). Enumerative combinatorics. Vol. 2, Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge. With a foreword by Gian-Carlo Rota and appendix 1 by Sergey Fomin.