In this section, we shall determine the sign-imbalance of the set Altn(132) and obtain the following theorem.
Theorem 2.2.1. For n ≥ 1, the following identities hold.
(i) I(Alt2n−1(132)) = Fn−1,
(ii) I(Alt2n(132)) = −I(Alt2n−1(132)).
To prove Theorem 2.2.1 we first translate the problem to binary trees. Let Bn be the set of binary trees with n vertices. It is known that |Bn| = Cn. For a binary tree T ∈ Bn, the leftmost path of T is defined by the maximal sequence v1, . . . , vk of vertices of T such that v1 is the root and vi is the left child of vi−1, for 2 ≤ i ≤ k.
Let lmpb(T ) denote the number of vertices in the leftmost path of T . Note that lmpb(T ) = 1 if the root has no left child.
Observe that |Alt2n−1(132)| = Cn (see [22, Theorem 2.2]). Our approach is to establish a bijection Γ between Alt2n−1(132) and Bn such that for a σ ∈ Alt2n−1(132), the sign of σ is opposite to the parity of lmpb(Γ(σ)). Then we can determine the sign-imbalance of Alt2n−1(132) by an involution Ψ on Bn that reverses the parity of lmpb(T ) if T is a binary tree except for the fixed points of Ψ.
Let Γ be a mapping from Alt2n−1(132) to Bn defined inductively as follows. Given a σ = σ1· · · σ2n−1 ∈ Alt2n−1(132), we factorize σ as σ = σ1p1· · · pn−1, where the subword pi = σ2iσ2i+1 consists of an adjacent pair of elements, for 1 ≤ i ≤ n − 1.
For convenience, denote pi = (σ2i, σ2i+1) and let p0 = (0, σ1). We shall associate σ with a binary tree Γ(σ) ∈ Bn with vertices labeled by p0, p1, . . . , pn−1 and write Γ(σ) = Γ(p0p1· · · pn−1). Suppose that pj (0 ≤ j ≤ n − 1) is the subword which con-tains the greatest element of σ. Then take pj as the root of Γ(σ) and put Γ(p0· · · pj−1) and Γ(pj+1· · · pn−1) as the left subtree and the right subtree, respectively (see Exam-ple 2.2.2).
Example 2.2.2. Take a permutation σ = 9 8 10 7 11 4 5 3 6 1 2 ∈ Alt11(132), and fac-torize σ as σ = p0· · · p5 = (0, 9)(8, 10)(7, 11)(4, 5)(3, 6)(1, 2). The corresponding tree Γ(σ), along with the vertex-labeling, is shown in Figure 2.1.
(1,2) (7,11)
(8,10) (3,6)
(0,9) (4,5)
Figure 2.1: The binary tree corresponding to σ = 9 8 10 7 11 4 5 3 6 1 2.
Proposition 2.2.3. Γ is a bijection between Alt2n−1(132) and Bnsuch that a permuta-tion σ = σ1· · · σ2n−1 ∈ Alt2n−1(132) is carried to a tree Γ(σ) ∈ Bn with lmpb(Γ(σ)) = 2n − σ1.
Proof. To show that Γ is a bijection, it remains to find Γ−1. Given a binary tree T ∈ Bn, we shall recover the word Γ−1(T ) by defining inductively a vertex-labeling of T starting with the set S = {1, . . . , 2n − 1}. Let T1 and T2 be the right and left subtrees of the root of T , respectively. If T1 contains k vertices (possibly empty), then we label T1 with S1 which consists of the least 2k elements in S, label the root of T by the pair (x, y), where x is the least element and y is the greatest element in S − S1, and label T2 with S2 = S − S1− {x, y}. Note that |S2| is odd, and inductively the leftmost vertex of T will be labeled by a single element (and then we attach 0 to the left).
Observe that whenever lmpb(T ) = t the elements 2n − 1, 2n − 2, . . . , . . . , 2n − t of σ appear in the leftmost path of T accordingly, and the last vertex of this path is labeled by (0, σ1), where σ1 = 2n − t. The second assertion follows.
With Proposition 2.2.3, the following lemma leads to the fact that the sign of σ is opposite to the parity of lmpb(Γ(σ)).
Lemma 2.2.4. For every σ = σ1· · · σ2n−1 ∈ Alt2n−1(132), the statistic inv(σ) has the opposite parity of σ1.
Proof. For 1 ≤ i ≤ n − 1, let pi = (σ2i, σ2i+1) be the ith ascent of σ. For each pi and 2i + 1 < j ≤ 2n − 1, we observe that σ2i+1 > σj if and only if σ2i > σj since σ is 132-avoiding. It follows that each pi contributes even number of inversions to inv(σ), and the parity of inv(σ) depends on the number of inversions due to σ1, which is equal to σ1− 1.
It is known that the binary trees T ∈ Bn can be transformed into plane trees G with n edges such that the degree of root of G equals lmpb(T ) (similar to leftmost child next right sibling, but here we use rightmost child next left sibling, see Figure 2.2).
Moreover, among many other objects, the nth Fine number Fn counts the number of plane trees with n edges where the root is of even degree (e.g., see [10]). Thus Fn
counts the number of binary trees T ∈ Bn with even lmpb(T ).
a
Figure 2.2: A binary-tree representation for plane trees.
For a vertex x of a rooted tree T , let τ (x) denote the subtree of T rooted at x.
Before proceeding the proof of Theorem 2.2.1, we shall define an involution Ψ on Bn such that it serves the needs of parity-reversing. Given a T ∈ Bn, we construct Ψ(T ) from T according to lmpb(T ) as follows.
(i) lmpb(T ) is even. Let u and v be the left and right children of the root of T , respectively. Let x and y be the left and right children of u, respectively. The tree Ψ(T ) is constructed from T as follows. Separate the subtrees τ (v), τ (x) and τ (y) from T , and then change u to be the right child of the root. Attach τ (y) and τ (v) to u as the left and right subtrees of u, and attach τ (x) to the root as the left subtree of the root (see Figure 2.3).
(ii) lmpb(T ) is odd. If the root has no right child, then let Ψ(T ) = T (i.e., T is a fixed point), otherwise Ψ(T ) is constructed by reversing the operation using in (i).
u Ψ
u
x y v
v y
x
Figure 2.3: An example for the the map Ψ.
Let Fn denote the set of fixed points of the requested involution Ψ, namely, Fn= {T ∈ Bn : lmpb(T ) is odd, and the root has no right child}.
Lemma 2.2.5. For n ≥ 1, we have
|Fn| = Fn−1.
Proof. Given a T ∈ Fn, it is easy to see that T is in one-to-one correspondence to the tree T′ ∈ Bn−1 with even lmpb(T′), where T′ is obtained from T with the root vertex removed. The assertion follows.
We remark that Theorem 2.2.1.(i) is essentially proved by the following bijective result.
Proposition 2.2.6. Ψ is an involution on the set Bn such that (i) Fn is the set of fixed points, and
(ii) lmpb(Ψ(T )) has the opposite parity of lmpb(T ) if T ∈ Bn− Fn.
Proof. It is a routine to check that Ψ is an involution on Bn. The assertion (i) is trivial by the definition of Ψ. For a T ∈ Bn− Fn, we have lmpb(Ψ(T )) = lmpb(T ) − 1 if lmpb(T ) is even, and lmpb(Ψ(T )) = lmpb(T ) + 1 if lmpb(T ) is odd. The proof is completed.
Now we are able to evaluate the sign-imbalance of the set Altn(132).
Proof of Theorem 2.2.1. (i) For the case of odd length, we have I(Alt2n−1(132)) = X
σ∈Alt2n−1(132)
(−1)inv(σ)
= − X
T∈Bn
(−1)lmpb(T ) (by Proposition 2.2.3 and Lemma 2.2.4)
= − X
T∈Fn
(−1)lmpb(T ) (by Proposition 2.2.6)
= Fn−1.
(ii) For the case of even length, given a σ = σ1· · · σ2n ∈ Alt2n(132), observe that σ2n = 1, otherwise there will be a 132-pattern (1, σ2n−1, σ2n) in σ. There is an immediate bijection between Alt2n(132) and Alt2n−1(132) for which σ corresponds to the permutation ω = ω1· · · ω2n−1 ∈ Alt2n−1(132), where ωi = σi−1 for 1 ≤ i ≤ 2n−1.
Note that σ has the opposite parity of ω since inv(σ) = inv(ω) + 2n − 1. Hence I(Alt2n(132)) = −I(Alt2n−1(132)). The proof is completed. ✷