Determining the maximum difference between the number of atoms and number of coatoms of a Bruhat interval of the symmetric group

We determine the largest difference between the number of atoms and number of coatoms of a Bruhat interval of $S_n$.


Introduction
Much work has been done on understanding the structure of Bruhat intervals of the symmetric group (see, e.g., [BW82], [Hul03] and [BB05] along with references therein). Recently, particular interest has arisen in understanding the number of atoms and coatoms of Bruhat intervals of the symmetric group [AR06,Kob11]. There, the maximum number of atoms and coatoms of an interval of a given length is determined. In this note, we determine the largest difference between the number of atoms and the number of coatoms of a Bruhat interval of S n .
Our main results are

Facts about Bruhat Intervals in S n
We will be needing the following definition and two results. Let B at u,v be the partition of [n] whose blocks are the connected components of G at . Similarly, define partition B coat u,v whose blocks are the connected components of G coat . The next result allows us to relate the atoms and coatoms of a Bruhat interval.
The labeled graphs G at and G coat have the same connected components.
The following result gives a sharp upper bound on the number of coatoms an interval of S n can have. .

Largest gap between number of atoms and coatoms of an interval in the Symmetric Group
In this section, we consider the question of how large a gap can there be between the number of atoms and coatoms of a Bruhat interval of the Symmetric group S n . The first result is a simple inequality that will be used later in finding a maximum.
Lemma 3.1. For all k 1 , k 2 ∈ N with k i ≥ 2, We have ⌊k 2 1 /4⌋ + ⌊k 2 2 /4⌋ ≤ ⌊k 2 1 /4 + k 2 2 /4⌋. Therefore it suffices to observe that for k 1 , k 2 ≥ 2, ✷ We now prove the main result of this note, which states that the largest difference between the number of coatoms and atoms of an interval of S n is equal to ⌊n 2 /4⌋ − n + 1. Proof. Let I ∈ I and consider G at and G coat as in definition 2.1. By Proposition 2.2, G at and G coat have the same connected components. Let K i , i = 1, 2, . . . , m, be the connected components of G at and G coat , and let k i ≥ 1 denote their respective number of vertices. Let p be the number of active components and q = m − p. By Proposition 2.3, Also, We maximize (2) over possible K i . Let f (x) = ⌊x 2 /4⌋ − x + 1, so that By Lemma 3.1, if k 1 , k 2 ≥ 2, then

DETERMINING THE MAXIMUM DIFFERENCE BETWEEN THE NUMBER OF ATOMS AND NUMBER OF COATOMS OF A BRUHAT INTER
if n is even n+1 2 − 1 if n is odd, the function f : N → R is monotonically increasing. It follows that for every I ∈ I, Next we show that the value f (n) is attained for some interval I = [u, v]. We consider any v of the form (1). By Proposition 2.4, the interval [1, v] has ⌊n 2 /4⌋ coatoms. The identity permutation has exactly n − 1 elements covering it. ✷ Proof. From the proof of Theorem 3.2, The assumption that n ≥ 4 implies that f (n) > f (n − q) for every q > 0. Moreover, we know that the maximum value of f (n) is attainable. Therefore c(I) − a(I) is maximized only if q = 0. So assume that q = 0. Let K be the single connected component of G at and G coat which contains n vertices. Then Proof. By Proposition 2.4, the number of coatoms is ⌊n 2 /4⌋ only for v of the form (1). ✷ A family of intervals for which the optimal value c(I) − a(I) = ⌊n 2 /4⌋ − n + 1 is attained is given by for v as in (1). There exist other intervals for which this maximum is attained. For example, in S 4 , the intervals for which the maximum is attained are