Cyclic Sieving of Matchings

Let Pn denote the poset of matchings on 2n points on a circle, labeled 1, 2, . . . , 2n in cyclic order, in which we define an element to be less than another element if it can be obtained by an uncrossing. Let Pn,k denote the level of the poset with k crossings. Let Pn,k denote the level of the poset with k crossings. We study the cyclic sieving phenomenon of Pn,k. It is known that the triple (Pn,0, g(q), C2n) exhibits cyclic sieving with g(q) = 1 [n+1]q ( 2n n ) q . Motivated by this result, our project focuses on finding similar results for greater number of crossing. By studying to structure of Pn,1, we find that there exists a q-analog polynomial fn(q) such that (Pn,1, fn(q), C2n) exhibits cyclic sieving phenomenon. We have also attained combinatorial interpretations of the polynomial fn(q). 1 Main Result We use the notation in [1], and we let Pn,k = {τ ∈ Pn|c(τ) = k} be the subset of Pn with k crossings (0 6 k 6 n(n+1) 2 ). Consider the operation σ2n acting on the set Pn by τ 7→ σ2n(τ), where σ2n(τ)(i) = { τ(i+ 1) if i 6= 2n τ(1) if i = 2n . (1) The rotation σ2n generates the cyclic group C2n with order 2n. We let C2n act on the poset Pn, and we study the one subset of Pn with one crossing Pn,1. We have proved via a counting argument that |Pn,1| = ( 2n n−2 ) . Let fn,1(q) be the q-analog of |Pn,1|: fn,1(q) = ( 2n n− 2 ) q = [2n]q[2n− 1]q · · · [n+ 3]q [n− 2]q[n− 3]q · · · [1]q , (2) The function fn,1(q) exhibits the cyclic sieving phenomenon for the operation of σ2n on the set Pn,1. Theorem 1. Let ξ2n = e πi n , which is a primitive root of the equation x − 1 = 0, then fn,1(ξ 2n) = #{τ ∈ Pn,1|σ 2n(τ) = τ}. In other words, (Pn,1, fn,1(q), C2n) exhibits the cyclic sieving phenomenon. 2 Proof of the Main Result Lemma 2.1. |Pn,1| = ( 2n n−2 ) . Proof. We first show that |Pn,1| = ( 2n n−2 ) via a bijection between choosing n − 2 element subsets of 2n elements and the matching of 2n points on a circle. We have the labeling of the points as {1, . . . , 2n}, and we choose a subset S of those points of size n− 2. We then must get a matching with one crossing from this subset. Look at each point i in the subset. First, if the point i+ 1( mod n) is not in S, then we connect i and i+ 1. Then for those points in S which are still not connected, we connect i and i+ 3 if we are able. We continue this process, next trying to connect i and i+5, increasing the target point each time by 2. The end result in this process will be that 2(n−2) = 2n−4 of the points will be connected to some other point. Note that by construction there will be no crosses in this set. Then we have exactly