Theorem 2.3.5. The following identities hold.
(i) For n ≥ 1, I(Alt2n−1(123)) =
(−1)n−12 Cn−1
2 n odd
0 n even.
(ii) For n ≥ 2, I(Alt2n(123)) = −I(Alt2n+1(123)).
Proof. (i) Under the operations of reverse, there is an immediate bijection between Alt2n−1(123) and Alt2n−1(321). Namely, to each ω = ω1· · · ω2n−1 ∈ Alt2n−1(123) there corresponds a σ = σ1· · · σ2n−1 ∈ Alt2n−1(321), where σi = ω2n−i (1 ≤ i ≤ 2n − 1).
Moreover, inv(σ) = 2n−12
− inv(ω) since for i < j, σi > σj if and only if ω2n−j <
ω2n−i. It follows that inv(σ) has the same parity of inv(ω) if n is odd, and has the opposite parity of inv(ω) if n is even. By Theorem 2.3.4, we have I(Alt2n−1(123)) = I(Alt2n−1(321)) = (−1)n−12 Cn−1
2 .
(ii) For n ≥ 2 and for every σ = σ1· · · σ2n+1 ∈ Alt2n+1(123), we observe that σ2n = 1, and that σ2i−1 > σ2i+1, for 2 ≤ i ≤ n (otherwise there will be 123-patterns in σ). Then removing the element σ2nfrom σ and subtracting 1 from the other entries leads to a member σ′ ∈ Alt2n(123) with inv(σ′) = inv(σ) − (2n − 1). This is a parity-reversing bijection between Alt2n+1(123) and Alt2n(123). Hence I(Alt2n(123)) =
−I(Alt2n+1(123)). The proof is completed.
2.4 Sign imbalance of alternating permutations of genus zero
Recall that D(0)n and U(0)n denote the set of alternating and reverse alternating per-mutations of genus zero in Sn, respectively. In this section, we enumerate D(0)n and U(0)n by the method of generating functions for completeness and determine the sign-imbalance of the two sets D(0)n and U(0)n .
We say that an m-cycle of a permutations α ∈ Sn is increasing if its elements are expressible as i < α(i) < α2(i) < · · · < αm−1(i). It is known that a permutation of genus zero can be completely characterized as follows, see [?, ?].
Lemma 2.4.1. Let α ∈ Sn. Then g(α) = 0 if and only if the cycle decomposition of α gives a noncrossing partition of [n], and each cycle of α is increasing.
The following observation is an immediate consequence of Lemma 2.4.1.
Proposition 2.4.2. For n ≥ 1, the following identities hold.
(i) |D(0)2n| = |D(0)2n+1|, (ii) |U(0)2n| = |U(0)2n+2|.
Proof. (i) For every α = α1· · · α2n+1 ∈ D(0)2n+1, we observe that if α2n+1 6= 2n + 1 then by Lemma 2.4.1 and the fact α2n < α2n+1 there will be a crossing in the cycle decomposition of α. Hence α2n+1 = 2n + 1 and the subword α1· · · α2n ∈ D(0)2n. This establishes a bijection between the two sets D(0)2n and D(0)2n+1.
(ii) For every α = α1· · · α2n+2 ∈ U(0)2n+2, by the same argument as in (i) we observe that α2n+2 = 2n+2 and the subword α1· · · α2n+1 ∈ U(0)2n+1. This establishes a bijection between the two sets U(0)2n+1 and U(0)2n+2.
We assume |D(0)0 | = |U(0)1 | = 1, and define the generating functions for |D(0)2n| and
|U(0)2n+1|
S = S(x) =X
i≥0
|D(0)2i |xi, R = R(x) =X
i≥0
|U(0)2i+1|xi. (2.4.1) Proposition 2.4.3. The following relations hold.
(i) R − 1 = 2(S − 1), (ii) R − 1 = 2xRS.
Proof. To prove (i), is suffices to show that |U(0)2n+1| = 2|D(0)2n|, for n ≥ 1. For every permutation α = α1· · · α2n+1 ∈ U(0)2n+1, by Lemma 2.4.1 we observe that either α1 = 1 or α1 = 2. Moreover, the elements 1 and 2 are not adjacent in α. Let U(0)2n+1 be partitioned into two sets A1 and A2, where α ∈ A1 if α1 = 1 and α ∈ A2 if α1 = 2.
There is an immediate bijection between A1 and A2 by interchanging the elements 1 and 2 of the permutation α. For every β = β1· · · β2n ∈ D(0)2n, we associate β with a permutation α = α1· · · α2n+1 ∈ S2n+1, where α1 = 1 and αi = βi−1 + 1 for 2 ≤ i ≤ 2n + 1. We observe that α ∈ U(0)2n+1 and this establishes a bijection between D(0)2n and A1. The assertion (i) follows.
(ii) For every α = α1· · · α2n+1 ∈ A1, we have α1 = 1 and α2k+1 = 2 for some 1 ≤ k ≤ n. We factorize α as α = α1µα2k+1ν, where µ = α2· · · α2k and ν = α2k+2· · · α2n+1. By Lemma 2.4.1, the entries in ν are greater than the entries in µ.
We observe that the word ν is a down-up permutation of length 2n − 2k, and that upon normalized ν is in one-to-one correspondence to the member β = β1· · · β2n−2k ∈ D(0)2n−2k, where βi = α2k+1+i− 2k − 1, for 1 ≤ i ≤ 2(n − k). Moreover, the word µ is a down-up permutation of length length 2k−1, and µ is in one-to-one correspondence to the member γ = γ1· · · γ2k−1 ∈ U(0)2k−1, where γj = 2k + 2 − α2k+1−j, for 1 ≤ j ≤ 2k − 1.
The above argument works well for the case α ∈ A2. This proves the assertion (ii).
Now, we prove the following theorem.
Theorem 2.4.4 (Dulucq-Simion). For n ≥ 1, the following identities hold.
(i) |D(0)2n| = |D(0)2n+1| = Sn,
(ii) |U(0)2n+1| = |U(0)2n+2| = Rn.
Proof. By Proposition 2.4.3, we derive that the generating function R satisfies the equation R = 1 + xR + x2R2. Solving this equation leads to
R = 1 − x −√
1 − 6x + x2
2x ,
which coincides with the generating function for large Schr¨oder numbers. Hence we have |U(0)2n+1| = Rn. Moreover, it follows from the relation R = 2S − 1 that
|D(0)2n| = Sn.
Theorem 2.4.5. For n ≥ 1, we have (i) I(D(0)2n) = I(D(0)2n+1) = (−1)n, (ii) I(U(0)n ) = 0.
Proof. Let γ ∈ D(0)2n be the permutation with cycle decomposition γ = (1 2)(3 4) · · · (2n−
1 2n). Note that sign(γ) = (−1)n. We come up with a sign-reversing involution on the set D(0)2n − {γ}.
For a α = α1· · · α2n ∈ D(0)2n − {γ}, find the least integer j such that (α2j−1 α2j) 6=
(2j 2j−1), say α2k = 2j−1 for some k > j. Then either α2k−1= 2k−1 or α2k−1 = 2k.
Moreover, the elements 2k −1 and 2k are not adjacent in α. The requested involution is obtained by interchanging the elements 2k − 1 and 2k in α.
Chapter 3
Alternating permutations II
3.1 Introduction
Let Sn be the set of all permutations of {1, . . . , n}. For a σ = σ1· · · σn ∈ Sn, the sign of σ, denoted by sign(σ), is defined to be sign(σ) = (−1)inv(σ), where inv(σ) = Card{(σi, σj) : i < j and σi > σj} is the inversion number of σ. Note that sign(σ) = 1 (respectively, −1) if σ is even (respectively, odd). For a subset P ⊆ Sn, the sign-balance I(P) of P is defined by
I(P) =X
σ∈P
sign(σ),
i.e., the difference between the number of even and odd permutations in P.
The problem of sign-balance was first studied by Simion and Schmidt on sets of patten-avoiding permutations in Sn. Let Sn(321) denote the set of 321-avoiding permutations in Sn. (A permutation is called 321-avoiding if it contains no decreasing subsequence of length three.) Simion and Schmidt [29] determined the sign-balance of Sn(321)
I(Sn(321)) =
Cn−1
2 if n odd
0 if n even, (3.1.1)
where Cn= n+11 2nn
is the nth Catalan number. Making use of a multivariate gener-ating function, Adin and Roichman [2] proved a refinement of Eq. (4.1.1) with respect to the statistic ldes(σ) of the permutations σ ∈ Sn(321), where ldes(σ) = max{i : σi > σi+1, 1 ≤ i ≤ n − 1} is the last descent of σ.
Theorem 3.1.1 (Adin-Roichman). The following identities hold.
X
σ∈S2n+1(321)
sign(σ) · qldes(σ) = X
σ∈Sn(321)
q2·ldes(σ) (n ≥ 0), X
σ∈S2n(321)
sign(σ) · qldes(σ) = (1 − q) X
σ∈Sn(321)
q2·ldes(σ) (n ≥ 1).
Shortly after this result had appeared on ArXiv, Reifegerste [25] gave a bijective proof of Theorem 4.1.1 in terms of standard Young tableaux with at most two rows, and proved an analogous refinement respecting the length of the longest increasing subsequence of σ ∈ Sn(321). Mansour [23] found variation of the identities for 132-avoiding permutations.
In this paper, we investigate refined sign-balance properties on 321-avoiding alter-nating permutations. A permutation σ = σ1· · · σn ∈ Sn is alternating (or down-up)
We also prove an analogous result respecting the statistic end(σ), the last entry, of σ.
To prove the above two theorems, we establish a sign-reversing involution on Altn(321), which is defined in terms of plane trees. In fact, the notion of sign-balance has been extended to plane trees. Let Tn be the set of plane trees with n edges.
Regarding the parity of the number of leaves, Eu, Liu and Yeh [15] determined the sign-balance of Tn
Chen, Shapiro and Yang [6] gave an elegant combinatorial proof of this result by establishing a sign-reversing involution on Tn. Their involution turns out to reverse the parity of another statistic hsum(T ) of a plane tree T , the total height of all vertices of T . In this paper, we extend their method to develop a new involution on Tn that reverses the parity of hsum(T ) while preserving the length of the leftmost (respectively, rightmost) path of T , which proves refined sign-balance properties on Tn (Theorems 3.4.2 and 3.4.3) and paves the way to a proof of Theorems 3.1.2 and 3.1.3.
This paper is organized as follows. In section 2, we establish a connection between 321-avoiding alternating permutations of even length and plane trees. In section 3, we review the involution given by Chen, Shapiro and Yang. In section 4, we develop an involution that proves refined sign-balance results on plane trees. In section 5, we prove Theorems 3.1.2 and 3.1.3.