The rest of the thesis is organized as follows. In Section 2, we introduce a bijection φ between non-crossing partitions and Dyck paths. In Section 3, we calculate the generating function of the statistics edes, odes, easc, and oasc in Dyck paths. Then we use it to obtain the generating functions for some interesting quantities. In Section 4, we introduce a bijection ψ between 2-Motzkin paths and Dyck paths, and use these two bijections φ and ψ to get the enumerating polynomials of the statistics emaj, omaj, ecomaj, and ocomaj in Dyck paths. Finally, we summarize our results and list some open questions.
2 Bijection between non-crossing partitions and Dyck paths
Now we introduce a bijection between non-crossing partitions and Dyck paths. De-fine a mapping φ : N C(n) → D(n) by φ(π) = θ(1)θ(2) . . . θ(n), where
θ(i)=
uu, if i ∈ Fs(π) \ Sg(π);
dd, if i ∈ Ls(π) \ Sg(π);
du, if i ∈ Ir(π);
ud, if i ∈ Sg(π).
The mapping φ is a well-known bijection. Since we cannot find a proof in the literature, we take this opportunity to give a proof here.
Theorem 2.1. The mapping φ : N C(n) → D(n) is a bijection.
Proof. For π = B1/B2/ · · · /Bk ∈ N C(n), let Bi = {fi, gi, . . . , li}, where fi <
gi < · · · < li, and bi = |Bi| for all i = 1, 2, . . . , k. (Bi = {fi} for |Bi| = 1.) Then Bi will generate a Dyck path
θ(fi)θ(gi) · · · θ(li) = u(ud)bi−1d ∈ D(bi).
We now claim that φ(π) = θ(1)θ(2) · · · θ(n) is a Dyck path of semilength n. First, let D(1) = θ(f1 = 1)θ(g1) · · · θ(l1) = u(ud)b1−1d ∈ D(b1) be the Dyck path generated by B1. Then, for 1 < j ≤ k, we insert the Dyck path θ(fj)θ(gj) · · · θ(lj) = u(ud)bj−1d generated by Bj into D(j−1) at the position after θ(mj−1) to produce a Dyck path D(j) ∈ D(Pj
i=1bi), where mj−1 = max{t ∈ Sj−1
i=1Bi | t < fj}. Note that we have mj−1+ 1 = fj, and the elements ofSj−1
i=1Bi are ordered increasingly to produce D(j) by the non-crossing property. Hence, we obtain D(k) = θ(1)θ(2) · · · θ(n) which is a Dyck path of semilength n.
For the inverse mapping φ−1 : D(n) → N C(n), given ω = ω1ω2. . . ω2n ∈ D(n), we parenthesize the sequence 1, 2, . . . , n as follows.
(i, put a left parenthesis before i if ω2i−1ω2i=uu, i), put a right parenthesis after i if ω2i−1ω2i=dd,
i , put no parenthesis on i if ω2i−1ω2i=du,
(i), put a pair of parentheses enclosing i if ω2i−1ω2i=ud.
Then create a block of φ−1(ω) for each of the consecutive strings inside lowest level parethesis pairs (i.e. paretheses which pair each other and enclose no others.) Now remove these lowest level parenthesis pairs and all the numbers they enclose, and continue with the remaining parenthesization. Clearly the partition φ−1(ω) is non-crossing.
Let us see an example of the bijection φ.
Example 2.2. (a) π = 146/23/5 ∈ N C(6) is mapping to ω = uuuuddduuddd ∈ D(6). (b) ω = uuuuddduuddd ∈ D(6) is leading to the parenthesization (1(23)4(5)6), then we get π = 146/23/5. Illustrate in Figure 6.
π=146/23/5
Fs(π) \ Sg(π) = {1, 2}, Ls(π) \ Sg(π) = {3, 6}, Ir(π) = {4}, Sg(π) = {5}.
ω1ω2ω3ω4 ω5ω6ω7ω8 ω9ω10ω11ω12
ω = u u u u d d d u u d d d (a) φ
ω=uuuuddduuddd
= u u / u u / d d / d u / u d / d d
(1 (2 3) 4 (5) 6)
π = 146/23/5
(b) φ−1 Figure 6: Example of the bijection φ.
We could describe the bijection φ by the card arrangements as well, which is similar to the technique used in [17]. Let us redefine φ−1 : D(n) → N C(n) for ω = ω1ω1· · · ω2n by φ−1(ω) = λ1λ2. . . λn, where
λi =
use the Uj card if ω2i−1ω2i = uu and the height of ω2i−1 is 2j;
use the Dj card if ω2i−1ω2i= dd and the height of ω2i−1 is 2j;
use the Tj card if ω2i−1ω2i= du and the height of ω2i−1 is 2j;
use the Kj card if ω2i−1ω2i = ud and the height of ω2i−1 is 2j.
The height of ωi is defined to be the y-coordinate of the initial point of ωi, and the cards are described in Figure 7. Then λ1λ2. . . λn would be a graphically represen-tation of a non-crossing partition.
Example 2.3. Let ω = uuuuddduuddd ∈ D(6).
(1 (2 3) 4 (5) 6)
We have φ−1(ω) = 146/23/5 ∈ N C(6) φ−1
λi :
1 U0
2 U1
3 D2
4 T1
5 K1
6 D1
U cards :
U0 U1 U2 Ui
i
i+1
D cards :
D1 D2 D3 Di
i
i-1
T cards :
T1 T2 T3 Ti
i i
K cards :
K0 K1 K2 Ki
i i
Figure 7: Four kinds of cards.
Now we go back to the definition of the bijection φ and discuss the locations of the valleys in the Dyck path ω = ω1ω2· · · ω2n = φ(π) for a non-crossing partition π ∈ N C(n). We consider when the valleys of ω occur at even and odd locations.
An even valley ω2iω2i+1= du occurs in ω if and only if i ∈ Ls(π) and i + 1 ∈ Fs(π) if and only if [i, i + 1] is labelled by l. An odd valley ω2i−1ω2i = du occurs in ω if and only if i ∈ Ir(π). See Figure 8.
even valley: Figure 8: Illustration of the formation of valleys.
Hence we have the following relations between non-crossing partitions and Dyck paths. Thus we have the following theorem.
Theorem 2.4. For π ∈ N C(n), ω = φ(π), we have
maj(ω) = 2 ml(π) + 2 mir(π) − nir(π).
Proof. Using (1) and (2), we have maj(ω) = X
Similarly, we discuss the locations of the peaks in ω. We consider that the peaks of ω occur at even and odd locations. An even peak ω2iω2i+1 = ud occurs in ω if and only if i ∈ (Fs(π) \ Sg(π))S Ir(π) and i + 1 ∈ (Ls(π) \ Sg(π)) S Ir(π) if and only if [i, i + 1] is labelled by r. An odd peak ω2i−1ω2i = ud occurs in ω if and only if i ∈ Sg(π). See Figure 9.
Hence we have the following relations between non-crossing partitions and Dyck paths.
π ∈ N C(n) ω = φ(π) ∈ D(n)
i ∈ R(π) ←→ 2i ∈ eAsc(ω) (4)
i ∈ Sg(π) ←→ 2i − 1 ∈ oAsc(ω) (5)
Also,
easc(ω) = nr(π), oasc(ω) = nsg(π), and asc(ω) = nr(π) + nsg(π). (6)
even peak:
ω2iω2i+1
across two groups
i i + 1 r interval
odd peak:
ω2i−1 ω2i
in one group
i Sg points Figure 9: Illustration of the formation of peaks.
Thus, we have the following theorem.
Theorem 2.6. For π ∈ N C(n), ω = φ(π), we have
comaj(ω) = 2 mr(π) + 2 msg(π) − nsg(π).
Proof. Using (4) and (5), we have comaj(ω) = X
i∈Asc(ω)
i = X
i∈R(π)
2i + X
i∈Sg(π)
(2i − 1)
= X
i∈R(π)
2i + X
i∈Sg(π)
2i − X
i∈Sg(π)
1
= 2 mr(π) + 2 msg(π) − nsg(π).
Note that it is easy to see that maj(ω) = comaj(ω) − n for all ω ∈ D(n).
Example 2.7. Let π = 15/2/34/678 and ω = φ(π), then R(π) = {3, 6, 7} and Sg(π) = {2}. By (4), (5) and Theorem 2.6, we have Asc(ω) = {3, 6, 12, 14} and comaj(ω) = 3 + 6 + 12 + 14 = 35 (maj(ω) = 35 − 8 = 27). See the figure below.
interval: 1
point:
2
Sg
3 r 4 5 6 r 7 r 8
1 2
3 4 5 6 7
8 9
10 11
12 13 14 15 16
Asc(ω) = {3, 6, 12, 14}
3 The generating functions of some statistics in Dyck paths
3.1 The statistics edes, odes, easc, and oasc
It is clear that edes(∅) = odes(∅) = easc(∅) = oasc(∅) = 0. We can easily get the following relations.
Lemma 3.1. Let ω ∈ D(n), n ≥ 1, and ω = uµdν be the first return decomposition of ω, where µ ∈ D(k − 1) and ν ∈ D(n − k) for some 1 ≤ k ≤ n. Then we have
edes(ω) =
( odes(µ) + 1 + edes(ν) if 1 6 k < n;
odes(µ) if k = n; (7)
odes(ω) = edes(µ) + odes(ν) if 1 6 k 6 n; (8)
easc(ω) = oasc(µ) + easc(ν) if 1 6 k 6 n; (9)
oasc(ω) =
( easc(µ) + oasc(ν) if 1 < k 6 n;
1 + oasc(ν) if k = 1. (10)
Proof. Let µ+= uµd ∈ D(k). Since even valleys of µ+ are odd valleys of µ, we have edes(µ+) = odes(µ). And there is an even valley at location 2k if ν 6= ∅. Even valleys of ν remain even in ω, thus we complete the proof of Equation (7). The others can be proved similarly.
Now we encode edes, odes, easc, and oasc by α, β, γ, and δ, respectively, and write down the generating function of the enumerating polynomials according to these statistics by
G(z) = G(α, β, γ, δ; z) :=X
n≥0
X
ω∈D(n)
αedes(ω)βodes(ω)γeasc(ω)δoasc(ω)zn.
Using Lemma 3.1, we can derive the following theorem.
Theorem 3.2. The generating function G(z) = G(α, β, γ, δ; z) has the explicit form of
αodes(µ)+edes(ν)+1βedes(µ)+odes(ν)γoasc(µ)+easc(ν)δeasc(µ)+oasc(ν)
+ X
= 1 + δz + αδz G(α, β, γ, δ; z) − 1 + αz G(β, α, δ, γ; z) − 1
G(α, β, γ, δ; z) − 1 + z(G(β, α, δ, γ; z) − 1)
= 1 + δz + αδzG − αδz + αz
G(β, α, δ, γ; z) · G − G − G(β, α, δ, γ; z) + 1
+ zG(β, α, δ, γ; z) − z.
Hence,
G 1 + αz − αδz − αzG(β, α, δ, γ; z)
= 1 − z + αz + δz − αδz + (z − αz)G(β, α, δ, γ; z). (12) Solving Equation (12) for G, we get
G =G(α, β, γ, δ; z) = 1 − (1 − α − δ + αδ)z + (z − αz)G(β, α, δ, γ; z)
1 + αz − αδz − αzG(β, α, δ, γ; z) . (13) Interchanging variables α and β in Equation (13), we obtain
G(β, α, δ, γ; z) = 1 − (1 − β − γ + βγ)z + (z − βz)G(α, β, γ, δ; z)
1 + βz − βγz − βzG(α, β, γ, δ; z) . (14) Substituting Equation (14) into Equation (12), we have
G(1 + αz − αδz) − αzG 1 − (1 − β − γ + βγ)z + (z − βz)G 1 + βz − βγz − βzG
= 1 − (1 − α − δ + αδ)z + (z − αz) 1 − (1 − β − γ + βγ)z + (z − βz)G 1 + βz − βγz − βzG
.
⇒ G(1 + αz − αδz)(1 + βz − βγz) − G2(1 + αz − αδz)βz
− αzG[1 − (1 − β − γ + βγ)z] − αz(z − βz)G2
= [1 − (1 − α − δ + αδ)z](1 + βz − βγz) − βz[1 − (1 − α − δ + αδ)z]G + (z − αz)[1 − (1 − β − γ + βγ)z] + (z − αz)(z − βz)G.
⇒ G2α(1 − β)z2+ βz(1 + αz − αδz) − Gh
(1 + αz − αδz)(1 + βz − βγz)
− αz[1 − (1 − β − γ + βγ)z] + βz[1 − (1 − α − δ + αδ)z] − (1 − α)(1 − β)z2i +
h
[1 − (1 − α − δ + αδ)z](1 + βz − βγz) + (1 − α)z[1 − (1 − β − γ + βγ)z]
i
= 0. (15)
Solving Equation (15) for G = G(α, β, γ, δ; z), we get the desired result.