Bruhat order on fixed-point-free involutions in the symmetric group

We provide a structural description of Bruhat order on the set F2n of fixed-pointfree involutions in the symmetric group S2n which yields a combinatorial proof of a combinatorial identity that is an expansion of its rank-generating function. The decomposition is accomplished via a natural poset congruence, which yields a new interpretation and proof of a combinatorial identity that counts the number of rook placements on the Ferrers boards lying under all Dyck paths of a given length 2n. Additionally, this result extends naturally to prove new combinatorial identities that sum over other Catalan objects: 312-avoiding permutations, plane forests, and binary trees.


Introduction
There is a family of combinatorial identities that express sums over certain Catalan objects in nice closed forms.Perhaps one of the most notable members of this family is Postnikov's hook-length formula, which can be found in [9].This sum is taken over the set B n of binary trees with n nodes, and the value h(v) is the the number of descendants of the node v. Another member of this family, appears in [5], which is a sum over the set F n of plane forests with n nodes.It is wellknown that the Catalan numbers enumerate B n and F n [13,Exercise 6.19].We prove the following identity in this family:  Identity (3) is merely a consequence of our main focus, a combinatorial proof of Theorem 1, which is a decomposition of the rank-generating function of the poset F 2n consisting of the fixed-point-free involutions in S 2n under Bruhat order.
Here, d i (δ) is a statistic on the Dyck path δ ∈ D n , which we describe in Section 3.1.1,and q is a q-analog of the rook number r n (B(δ)).
In order to dissect the poset F 2n , we make use of a bijection ϕ between fixed-point-free involutions and rook placements on Ferrers boards.This bijection is decribed specifically in Section 3.2, and it is well-known as a part of combinatorial folklore among those concerned with perfect matchings.
This realization leads to a decomposition of the poset into disjoint intervals, which in turn yields the expansion of the rank-generating function of F 2n given in Theorem 1.The decomposition of F 2n is accomplished via a poset congruence in the sense of [10].(A poset congruence is the order-theoretic generalization of a lattice congruence.) Figure 2 illustrates the Hasse diagram of Bruhat order on F 6 under the bijection ϕ.
It should be mentioned that (4) appears in [1], in which the authors refer to it as a classical identity.This equation can be derived from [6, Exercise 5.2.9 (b)] using hypergeometric series.Our proof of this result is combinatorial, thus it is very different in nature.Moreover, it describes something fundamental about fixed-point-free involutions.
the electronic journal of combinatorics 21(2) (2014), #P2.20 The connection between these Catalan objects highlights a sense in which binary trees relate to permutations as plane forests relate to fixed-point-free involutions.Specifically, there is a natural map S n → B n (cf.[3], [11]) that is a lattice homomorphism of the weak order on S n , and the natural partial order on the image is the quotient-the Tamari lattice.In particular, the fibers of the map constitute a lattice congruence on the weak order.In this paper, we exhibit a natural map F 2n → F n whose fibers constitute a poset congruence on the Bruhat order on F 2n .The quotient modulo this congruence is a natural partial order on Dyck paths.
We extend this enumerative result to the context of plane forests and binary trees.In particular, under two natural bijections, the rook number statistic of each Dyck path is equivalent to statistics on these objects.This yields additional identities in the family of Catalan sums.

Fixed-point-free involutions
The objects of concern to us are elements of the symmetric group, thus we begin by recalling pertinent notions relating to the symmetric group and Bruhat order in Section 2.1.Section 2.2 contains a characterization of the set F 2n of fixed-point-free involutions, and we describe in Section 2.3 the subposet of these permutations under the order induced by Bruhat order on S 2n .

The symmetric group
For n 1, let S n denote the symmetric group, which consists of all permutations σ : [n] → [n], where [n] denotes the set {1, 2, . . ., n}.Each permutation σ ∈ S n can be written in one-line notation as σ 1 σ 2 • • • σ n , where σ i := σ(i) for all i ∈ [n].Alternatively, σ ∈ S n can be decomposed into its cycle notation in which it is written as a product of unique disjoint cycles of the form We omit 1-cycles from cycle notation.The canonical cycle notation of a permutation is its cycle decomposition with its cycles in ascending order by their minimal elements and the elements within cycles arranged so that the least value appears first.We use the notation appears in the canonical cycle notation of σ.We denote by id n the identity permutation 12 The permutation diagram of σ ∈ S n is the n × n array of squares, which we call cells, with the set  An inversion of σ ∈ S n is a pair (σ i , σ j ) ∈ [n] 2 such that i < j and σ i > σ j .The inversion number inv(σ) of σ is the number of its inversions.A rise of σ ∈ S n is a pair (i, j) ∈ [n] 2 such that i < j and σ i < σ j .A rise is called free if there exists no k ∈ [n] such that i < k < j and σ i < σ k < σ j .Each i ∈ [n] satisfies exactly one of the following the electronic journal of combinatorics 21(2) (2014), #P2.20 relations: σ i < i; σ i > i; or σ i = i.In these cases, i is called a deficiency, an exceedance, or a fixed-point of σ, respectively.
Bruhat order on S n is the partial order relation B which is the transitive closure of the relation → defined by σ → τ if and only if σ(ij) = τ for some transposition (ij) and inv(σ) < inv(τ ).Let B denote the covering relation in Bruhat order.The rank function of S n under Bruhat order is given by inv(σ) for all σ ∈ S n .The following descriptions of Bruhat order will be useful.
Given permutations σ ∈ S n and τ ∈ S m , we say that σ contains τ if there exist  Being composed as a set of ordered pairs lends itself to a more pictorial representation: the arc diagram of x ∈ F 2n is a directed complete matching on the node set [2n] with edge set {(i, j) : (ij) ∈ x}, where (i, j) denotes the directed edge from the initial point i to the terminal point j.Example 6. Figure 4 shows the arc diagram of 73286514 ∈ F 8 .
We adopt from [4] the following definitions relating to the arc diagram of a fixedpoint-free involution.Suppose x ∈ F 2n and (e i d i ), (e j d j ) ∈ x for i < j.The two arcs (e i d i ), (e j d j ) ∈ x are crossing if e i < e j < d i < d j .When no pair of arcs of an arc diagram are crossing, we say the arc diagram is non-crossing.It is well-known that the nth Catalan number C n is the number of non-crossing arc diagrams on 2n vertices [13,Exercise. 6.19(o)].Define the crossing number cross(x) of x ∈ F 2n to be the number of pairs of arcs in its arc diagram that are crossing.Define span(x) := (ed)∈x span(e, d), where span(e, d) := d − e − 1, and define the weight of x ∈ F 2n to be wt(x) := span(x) − cross(x).Proof.We proceed by induction, so first note that the proposition holds in the n = 1 case.Now, consider x ∈ F 2n for some n > 1, and suppose the proposition holds for n − 1.By removing the arc (i, 2n) from the arc diagram of x and subtracting 1 from the entries between i and 2n, we obtain the arc diagram for y ∈ F 2n−2 .By the inductive hypothesis, inv and the proposition holds for n.Thus the proposition holds for all natural numbers.

Bruhat order on F 2n
Let the partial order relation be the restriction of Bruhat order on the elements of F 2n , and let denote the corresponding cover relation.Both the set F 2n and the poset (F 2n , ) will be denoted by F 2n , as no confusion should arise.Because Bruhat order is graded by inversion number, Theorem 4 implies the following lemma.
Lemma 10.Suppose x ∈ F 2n .If (i, j) is a free rise of x and (x i , x j ) is a free rise of x(ij), then inv((ij)x(ij)) = inv(x) + 2.
Proposition 11.The cover relation x y exists in F 2n if and only if y = (ij)x(ij) for some free rise (i, j) of x such that inv(y) = inv(x) + 2.
Proof.Let x, y ∈ F 2n .Suppose inv(y) = inv(x) + 2 and there exists some free rise (i, j) of x such that y = (ij)x(ij).The existence of the cover relations x B x(ij) and x(ij) B y follow from Theorem 4 and the fact that y = x(ij)(x i x j ).Bruhat order is ranked by inversion number, and inv(y) = inv(x) + 2, thus x y.
Conversely, suppose x y.Define ) is a rise of x, there exists at least one element in the set of which j is the minimum.Denote (ij)x(ij) by z.We show that z = y and inv(z) = inv(x) + 2.
If (x i , x j ) is not a free rise of x(ij), then there is some k ∈ [n] such that x i < k < x j and x(ij) x i < x(ij) k < x(ij) x j , i.e., i < x k < j.This contradicts (i, j) being a free rise of x, thus (x i , x j ) is a free rise of x(ij).Moreover, inv(z) = inv(x) + 2 by Lemma 10 and x < z by Theorem 4. Now, note that x k = z k for all k = i, j, x i , x j , and the following inequalities hold by definition: and Theorem 3 implies that z y.By the assumption that x y, we have y = z, thus completing the proof.

The structure of Bruhat order on F 2n
A few structural results concerning F 2n are obtained almost for free.In particular, the following result follows from Propositions 9 and 11.
Proposition 12.The poset F 2n is graded and ranked by the weight function wt.
. It was shown in [4] that the rank-generating function with respect to the weight function is [1] 5   The remainder of this section details a structural decomposition of Bruhat order on F 2n which arises naturally when we realize fixed-point-free involutions as certain rook placements on Ferrers boards.We first proceed by recalling these combinatorial objects.

Dyck boards
A board is a subset of [n] × [n] for some n ∈ N, which we visualize as a subset of the cells on an n × n chessboard.We have already seen an example of a board: the permutation A Dyck word of length 2n is a string δ, consisting of n es and n ds such that no initial segment of δ contains more ds than es.We denote the set of all Dyck words of length 2n by D n .It is well-known that the Catalan numbers count Dyck words [13,Cor. 6.2.3].
For δ ∈ D n , let e i (δ) and d i (δ) denote the positions in δ of the ith e and ith d, respectively, for all i ∈ [n].By letting an e in δ represent a step (0, 1) to the north and a d represent a step (1, 0) to the east, each Dyck word uniquely defines a Dyck path, which is a lattice path from (0, 0) to (n, n), lying above the the main diagonal Proposition 16.Suppose δ ∈ D n .Then (i, j) ∈ B(δ) if and only if e j (δ) < i + j if and only if i + j d i (δ).
Proof.The cell (i, j) is contained in B(δ) if and only if the number of steps north occuring before the ith step east is at least j.Additionally, the cell (i, j) is contained in B(δ) if and only if the number of steps east occuring before the jth step north is less than i.
We impose on D n the following natural partial order: δ D δ if and only if B(δ) ⊂ B(δ ).It is well-known that (D n , D ) is a distributive lattice, which follows from viewing D n as containment order on the set of order ideals of a certain poset.The rank function In terms of the corresponding Dyck boards, the rank of δ can be interpreted as the number of cells of B(δ) lying strictly above the main diagonal, so that δ D δ if and only if B(δ ) is obtained by adding a cell to B(δ).Let D n denote both the set of Dyck words of length 2n and the poset (D n , D ), as no confusion should arise.Figure 8 shows the Hasse diagrams of D 3 and D 4 , with the elements represented by their corresponding Dyck boards.

Rook placements
A (non-attacking) k-rook placement on the board B is a k-subset P ⊂ B such that no two elements of P lie in the same row or column.Let r k (B) denote the number of k-rook placements on the board B. Every σ ∈ S n corresponds to an n-rook placement on the board Example 17. Upon revisiting Figure 7, one finds that for δ = eededded, we have There is an intimate order-theoretic relationship between Dyck paths and rook placements.In fact, Sjöstrand describes in [12,Corollary 5] that the elements in the lower the electronic journal of combinatorics 21(2) (2014), #P2.20 Bruhat interval [id n , σ] correspond to the number of rook placements on the the smallest right-aligned Ferrers matrix covering B(σ).These Ferrers matrices are essentially rotations of our Ferrers boards, thus any Dyck path δ ∈ D n defines a unique order ideal of S n consisting of all permuations σ ∈ S n such that B(δ) contains B(σ).This order ideal is principally generated by the permutation µ(δ), where µ(δ) i := max([d i (δ) − i]\{µ(δ) j : 1 j < i}).One can visually construct µ(δ) on B(δ) by placing a rook as high as possible in each column from left-to-right while maintaining that no two rooks lie in the same row.Proof.Note that if a Ferrers board contains the cell (i, j), then it contains the cells below and to its right.In other words, the Ferrers board contains the cell (i, k) for all 1 k j and the cell (l, j) for all i l n.Now, if τ B σ, then τ (ij) = σ for some free rise (i, j) of τ .Then i < j and τ i = σ j < σ i = τ j , and B(δ) contains the cells (i, τ j ) and (j, τ i ).Thus B(τ ) ⊂ B(δ).
In addition to this fact, the following proposition details why µ is of order-theoretic interest.The result follows from the three lemmas that follow it.
Proposition 22.The map µ is an order-isomorphism from D n to S n (312) under Bruhat order.
Lemma 24.The map µ is a bijection from D n to S n (312) Proof.We already have µ(D n ) ⊆ S n (312) by Lemma 22. Thus it suffices to show µ is injective because D n and S n (312) have the same cardinality.Suppose δ = δ ∈ D n , and let i > 1 be the least index such that δ i = δ i .Without loss of generality, suppose δ i = d and δ i = e.Since it has been assumed that δ j = δ j for all 1 j < i, we have µ(δ) j = µ(δ ) j for all 1 j < i, thus Thus, µ(δ) i = µ(δ ) i , so µ(δ) = µ(δ ), and we have proved that µ is injective.

Hooks and hook lengths
For the Dyck path δ, define the reduced hook H δ (i, j) of cell (i, j) ∈ B(δ) to be the union of the cells in B(δ) in the jth row strictly to the left of cell (i, j) and those in the ith column strictly above cell (i, j).Define the reduced hook length h δ (i, j) of cell (i, j) ∈ B(δ) to be the cardinality of H δ (i, j).(The hook length, usually seen in the context of Young tableaux, counts the cell (i, j).It benefits us here to disregard it, hence the adjective reduced.)The following result shows that one can compute hook lengths on B(δ) directly from the Dyck word δ.
Proposition 26.For all δ ∈ D n , the reduced hook length of the cell Proof.There are d i (δ)−i cells in the ith column of B(δ), and d i (δ)−i−j of them lie above cell (i, j).Similarly, the jth row of B(δ) contains n − (e j (δ) − j) cells, and i + j − e j (δ) − 1 of them lie to the left of cell (i, j).For all σ ∈ S n and δ ∈ D n such that B(σ) ⊂ B(δ), define the hook sum of the pair (δ, σ) to be h(δ, σ) := i∈[n] h δ (i, σ i ).If the board B(δ) contains the rook placements B(σ) and B(τ ) for two permutations σ, τ ∈ S n , then Thus appealing to Proposition 26 and rearranging the summands in the hook sums justifies the following result.
Proposition 28.For any δ ∈ D n , all n-rook placements on the Dyck board B(δ) have equal hook sums.
Because h(δ, σ) does not depend on the choice of σ, we define h(δ) := h(δ, id n ) for all δ ∈ D n .The cells in B(δ) strictly above the main diagonal are counted exactly twice by h(δ), implying that the rank of any δ ∈ D n is h(δ)/2.
Define the crossing number c(δ, σ) to be the number of pairs of hooks that intersect.It is plain to see that if i < j and the two hooks H δ (i, σ i ) and H δ (j, σ j ) intersect, then (i, j) is a rise of σ and (i, σ j ) is the cell in the intersection, provided that it lies in B(δ).
As we see next, the map µ yields the unique rook placement on a given Dyck board with disjoint hooks.
Proof.Suppose δ ∈ D n and σ ∈ S n .We have c(δ, σ) = 0 if and only if there exist values 1 i < j n such that σ i < σ j and H δ (i, σ i ) ∩ H δ (j, σ j ) = {(i, σ j )}, i.e., σ i < σ j d i − i.This occurs if and only if σ = µ(δ).any (ed) ∈ x defines an arc whose terminal point d is paired with the earlier-appearing initial point e.Therefore the arrangement of exceedances and deficiencies, as they occur in the one-line notation of x ∈ F 2n , forms a Dyck word in D n .Definition 32.For x ∈ F 2n , define δ(x) to be the Dyck word δ(x)

F 2n as rook placements on Dyck boards
The remaining characteristic that uniquely determines a fixed-point-free involution is the order in which its n arcs terminate.Definition 33.For x ∈ F 2n , define σ(x) to be the permutation with σ(x) i = j if and only if the arc diagram of x has an arc from e j (δ(x)) to d i (δ(x)) for all i ∈ [n].
For any x ∈ F 2n , the permutation σ(x) can be constructed easily on the arc diagram of x by indexing the es and ds of δ(x) in the order they appear with the set [n] and tracing the arcs backward to realize σ(x) as the bijection from the d-index set to the e-index set.
Example 34.Consider the fixed-point-free involution x = 73286514 in Figure 4. Then δ(x) = eedeeddd and σ(x) = 2413.Figure 12 illustrates the construction detailed in the preceding paragraph.It should be noted that we could equivalently regard σ as a map from the e-index set to the d-index set.This may seem more natural considering the direction of the arrows, and in this case, σ would map to the inverse permutation.However, our definition of σ is more natural in the sense that it corresponds to rook placements under Dyck paths consisting of north and east steps and thus on Ferrers boards that are bottom-and right-justified.
Example 37. The map ϕ identifies the four rook placements on Dyck boards shown in Figure 9 with the four fixed-point-free involutions 35162487, 36154287, 53261487, and 63254187, respectively.
Proposition 38.The map δ : F 2n → D n is order-preserving.
Proof.Suppose x y in F 2n .Then by Proposition 11, there is a free rise (i, j) of x such that y = (ij)x(ij) = (x i x j )x(x i x j ).Thus and y x j = i.
To prove δ(x) D δ(y), it suffices to consider positions i, j, x i , and x j because all other positions have equal values.The definition of a rise guarantees i < j and x i < x j , and we can further require i < x i by swapping i and x i if necessary.There are three cases to consider: (1) i < j < x i < x j , (2) i < x i < x j < j, and (3) i < x i < j < x j .
The third case would imply that x(ij) is not covered by y in Bruhat order on S 2n because (x i , x j ) would not be a free rise of x(ij), contradicting the assumption that x y.
Thus the subsequence of the Dyck word δ(x) affected by moving up in F 2n via a cover relation is either eedd or eded, and in both cases, δ(y) i = δ(y) j = e and δ(y) x i = δ(y) x j = d.In other words, the affected subsequence becomes eedd in δ(y), while the remainder of the two Dyck words are equal.Thus δ(x) D δ(y).

Structural decomposition
Recall that Proposition 12 asserts that the Bruhat order on F 2n has rank function wt(x) = span(x)−cross(x).We have developed the necessary tools to restate this function in terms of our representation of elements of F 2n as rook placements on Dyck boards.
Proof.The first equality is immediate from the definition of span, the definition of hook length, and Proposition 26.Propositions 8, 16, and 29 imply that arcs in the arc diagram of a fixed-point-free involution intersect if and only if the corresponding hooks intersect on its Dyck board, thus justifying the second equality.
In order to prove Theorem 1, we reveal the structural properties of Bruhat order on F 2n with respect to two equivalence relations ∆ and Σ, which we define now.For x, y ∈ F 2n , define x ≡ ∆ y if and only if δ(x) = δ(y), and define x ≡ Σ y if and only if σ(x) = σ(y).We denote the equivalence classes of x under ∆ and Σ by [x] ∆ and [x] Σ , respectively, and we consider these equivalence classes as induced subposets of F 2n .
Propositions 19 and 36 imply that the map σ restricts to a bijection between [x] ∆ and the order ideal of S n generated by the 312-avoiding permutation µ(δ(x)).However, a stronger fact holds: they are isomorphic posets.
Lemma 40.The map σ restricts to an order-isomorphism from [x] ∆ to the principal order ideal of S n generated by µ(δ(x)).
Proof.Consider two fixed-point-free involutions x, y ∈ F 2n such that y ∈ [x] ∆ .The map δ is order-preserving by Proposition 38, and thus [x] ∆ is order-convex.The cover relation x y exists in F 2n if and only if y = (ij)x(ij) for a free rise (i, j) of x and inv(y) = inv(x)+2.
Choose i and j to be deficiencies (by swapping i and j for x i and x j , respectively, if necessary), so that d a := d a (δ(x)) = i and d b := d d (δ(x)) = j for some a, b with a < b.
This cover relation occurs in F 2n if and only if any arc terminating between d a and d b has its initial point between x da and x d b , i.e., between e σ(x)a (δ(x)) and e σ(x) b (δ(x)).In other words, (a, b) is a free rise of σ(x), and σ(y) = σ(x)(ab).Thus σ(x) B σ(y).
Figure 13 illustrates the correspondence described in the proof of Proposition 40.An analogous fact about the map δ on Σ-classes is also true.
Lemma 41.For all x ∈ F 2n , the map δ restricts to an order-isomorphism from [x] Σ to the induced subposet of D n consisting of Dyck paths δ such that B(δ) ⊂ B(σ(x)).
Suppose δ D δ in D n , and we show x y in F 2n .Then δ and δ are identical in all positions except for some two adjacent positions i and i + 1, in which δ has de and δ has ed.Define z := (i i + 1)x(i i + 1), and the following two claims imply z = y.
Claim 1: To conclude that x y in F 2n , we note that the aforementioned de → ed swap from δ to δ implies span(y) = span(x) + 2 and cross(y) = cross(x) + 1.Thus inv(y) = inv(x) + 2, by Proposition 9. Now, because i and i + 1 are adjacent positions, this x i and x i+1 are adjacent in value.Thus it must be the case that (i, i + 1) is a free rise of x, and thus x y.Now, to prove the converse, we suppose x y in F 2n .Then y = (ij)x(ij) for a free rise (i, j) of x and inv(y) = inv(x) + 2. It is either the case, then, that i is a deficiency and j is an exceedance or i is an exceedance and j is a deficiency.Choose i and j so that the former holds (by swapping i and j with x i and x j , respectively, if the latter holds).Note that j = i + k, so that cross(y) = cross(x) + k.Because inv(y) = inv(x) + 2, we must have k = 1, and δ and δ are identical in all positions except i and i + 1 in which δ has de and δ has ed.Thus δ δ .
The remaining structural information about F 2n is how the equivalence classes under ∆ are ordered in F 2n , with respect to one another.We recall from [10] the definition of a poset congruence and the natural partial order on the quotient.An equivalence relation Θ on the elements of a poset P is a poset congruence if (i) every equivalence class [x] Θ of x ∈ P is an interval in the poset P , the electronic journal of combinatorics 21(2) (2014), #P2.20 (ii) the projection Θ ↑ mapping x to the maximal element in its equivalence class is order-preserving, and (iii) the projection Θ ↓ mapping x to the minimal element in its equivalence class is order-preserving.
Define a partial order on the congruence classes of P under Θ by [x] Θ [y] Θ if and only if some x ∈ [x] Θ and y ∈ [y] Θ exist such that x P y .The quotient of P with respect to the poset congruence Θ, denoted P/Θ, consists of the set of congruence classes under Θ under this partial order.The quotient P/Θ is isomorphic to the subposet Θ ↓ (P ) ⊂ P of minimal representatives in each class.We identify these two isomorphic posets.We proceed by showing that the equivalence ∆ satisfies this definition.
Proposition 42.The equivalence ∆ is a poset congruence.
Proof.Each of the three claims below respectively address the three parts of the definition of a poset congruence detailed above.One can verify Proposition 43 for the n = 3 case by comparing the the poset D 3 in Figure 8 to the induced subposet ∆ ↓ (F 6 ) in Figure 2. We now prove our main result.
Recall that the rank-generating function for D n is i∈[n] d i (δ)−2i, justifying the factors of the form q d i (δ)−2i in the expansion.Also recall from Section 3.1.2the rank-generating function of lower Bruhat intervals.Of concern here are the intervals [id n , µ(δ)] ⊆ S n for δ ∈ D n .The rank-generating function for this interval is a fact that follows from Lemma 40.This justifies the remaining factors, and thus proves Theorem 1. Now, setting q = 1 in the expansion proves (3).Interestingly, this identity can be phrased as a sum over 312-avoiding permutations.Let #[id n , σ] denote the cardinality of the lower Bruhat interval [id n , σ] in S n .Then we get the following Catalan sum: In the next and final section, we explore how our main result can be rephrased in terms of other Catalan objects.

Extension to other Catalan objects
Throughout this exposition, we have encountered several Catalan objects: Dyck paths, non-crossing arc diagrams, and 312-avoiding permutations.The ubiquity of the Catalan numbers here and elsewhere [13,Ex. 6.19] incites further investigation into other Catalan objects to which we can relate our previous results.In particular, we express Theorem 1 both as a sum over plane forests in Section 4.1 and as a sum over binary trees in Section 4.2.

Plane trees and plane forests
A tree T is a connected acyclic graph.A rooted tree is one possessing a unique distinguished node called the root of T .We consider only rooted trees and omit the adjective rooted.A node u ∈ T is a descendant of the node v ∈ T if v is in the unique path from u to the root, and v is an ancestor of u in this case.Thus any node v ∈ T is both a descendant and an ancestor of itself.A descendant u of v is a child of v if T contains an edge from u to v, in which case v is the parent of u.A node with no children is called a leaf and an internal node otherwise.A subtree of T is the tree formed by a node and all its descendants.We depict trees with child nodes situated above their parents, thus the root, being the unique ancestor to all nodes of T , is the bottom-most node.
Recall that h(v) is the number of descendants of v, and we dually define α(v) to be the number of ancestors of v.A plane tree is a tree whose subtrees are ordered linearly at each node.Denote by T n the set of plane trees having n nodes.
A plane forest is a linearly-ordered set of plane trees.Let F n denote the set of plane forests having n nodes.There is a natural bijection F n → T n+1 : add a new node and make it the root of all trees in the forest F ∈ F n to obtain T ∈ T n+1 .It appears as an exercise in [13,Ex 6.19(e)] to show that the nth Catalan number enumerates T n+1 .
The bijection F : D n → F n we consider is as follows.We construct the tree T (δ) ∈ T n+1 according to δ as it is read left-to-right.We move around the nodes of T (δ) in a pre-order fashion as we construct it.Add a new edge at the current node if δ i = e or move down the tree to the parent of the current node if δ i = d.When completed, removing the root from T (δ) completes the bijection D n → F n .
Example 44. Figure 14 shows the bijection between T 4 and F 3 , in the order corresponding to the Dyck boards shown in Figure 1.We now show that the rook number statistic of the Dyck board B(δ) is equal to the product of the ancestor statistic on the nodes of the corresponding plane forest F (δ) under the bijection described above.Lemma 45.If v i is the ith node added to F (δ) in the above bijection Proof.The number α(v i ) is one more than the length of the path from v to the root of the tree containing it.This is the number of steps upward less the number of steps downward until v i is reached.In other words, this is the number of es that appear before the ith d minus the number of ds appearing before the ith d in a preorder search.Thus (5) Proof.By the previous lemma the nth rook number r n (B(δ)) of the board B(δ) equals v∈F (δ) α(v), thus we can rephrase Theorem 1 as this sum over plane forests.
In the sum over plane forests given by equation ( 2), the statistic h(v) gives the number of descendants of the node v. Naturally, descendants and ancestors are structurally dual notions, but the fact that the left-hand sides of ( 6) and ( 2) are equal suggests that descendants and ancestors also have an intrinsic duality in a numerical sense.
With regard to Figure 14 in which n = 3, equation ( 6) agrees with equation (3), asserting that 1 + 2 + 2 + 4 + 6 = 5!!.Equation (2) says that 6( 11 + 1 2 + 1 2 + 1 3 + 1 6 ) = 6 + 3 + 3 + 2 + 1 = 5!!.Note that although the sums are equal, the sumands differ!The left-hand side of ( 2) is familiar in the context of binary trees, in which case the sum would be n! rather than (2n−1)!!.This observation inspires an exploration of binary trees, which is the scope of the next section.Before concluding the present section, though, we present two additional nice Catalan sums that follow from (2) by making use of the identity which gives the proportional relationship between double-factorial numbers and the Catalan numbers.For example, applying ( 7) to (2), we obtain With respect to the bijection F n ←→ T n+1 described above, the sum (2) can be rephrased in terms of plane trees as (n + 1)!

Binary trees
We recursively define a plane binary tree as one whose root either has no children or whose root has a left subtree and a right subtree, each of which is also a binary tree.Let PB n denote the set of plane binary trees with n internal nodes.The removal of the n + 1 leaves gives the bijection between plane binary trees and binary trees B n with n nodes.The bijection T : D n → B n is as follows.Read δ ∈ D n left-to-right, and traverse T (δ) in left preorder as we construct it.The es in δ yield nodes of degree 2, and ds yield nodes of degree 1.At δ i , add two children to the current node if δ i = e, and add no new nodes if δ i = d.Continue this way until the end of δ is reached, and removing the leaves completes the bijection D n → B n .To recover δ, add all possible leaves to T to make a binary tree, and traverse T (δ) in preorder and read each internal node as an e and each leaf, disregarding the last, as a d.
Example 48. Figure 15 shows the five binary trees having three internal nodes, listed in the order corresponding to that of the Dyck paths in Figure 1.For the binary tree T ∈ B n , we define a recursive labeling λ on its nodes.Label the root of T with 1, and label a child u of v ∈ T as follows: : u is a right-child of v.
The labeling λ is illustrated by Figure 16.Proof.Consider the binary subtree containing all nodes traversed in the preorder search from the root to the node v i .At each node, we either follow an edge to a left-child or a right-child.These correspond to occurrences of ee and ed, respectively, in δ.Thus λ(v i ) − 1 = d i (δ) − 2i counts the number of times we follow an edge to a left-child.
Proof.By the previous lemma, the nth rook number r n (B(δ)) of the board B(δ) equals v∈T (δ) λ(v), thus we can rephrase Theorem 1 as a sum over binary trees.
Corollary 51.Aside from providing a combinatorial proof of (3), we have provided three natural extensions of this combinatorial identity that are expressed as sums over three other Catalan objects: 312-avoiding permutations, plane forests, and binary trees.Naturally, it would be nice to prove such expressions for other Catalan objects, such as triangulations of (n + 2)-gons into n triangles.

Figure 1 :
Figure 1: The Ferrers boards lying under the five Dyck paths in D 3

Figure 2 :
Figure 2: Bruhat order on F 6 represented as rook placements on Dyck boards

Example 2 .
cells shaded.The cell (i, j) is in the ith column from the left and the jth row from the bottom.The permutation diagram of σ = 416523 ∈ S 6 is pictured in Figure3(a).

2. 2
The set F 2n Let F 2n denote the set of fixed-point-free involutions in S 2n .Note that any x ∈ F 2n has n transpositions in its canonical cycle notation (e 1 d 1 )(e 2 d 2 ) • • • (e n d n ).Additionally, we have e i < d i for all i ∈ n and e i < e i+1 for all i ∈ [n − 1].We use the characters e and d to indicate which values are exceedances and which are deficiencies.The cardinality of F 2n is (2n − 1)!!.
Recall that the edges in the arc diagram of x ∈ F 2n are represented in the canonical cycle notation by disjoint ordered pairs (e, d) ∈ [n] 2 such that x d = e < d = x e , and note that the electronic journal of combinatorics 21(2) (2014), #P2.20

Claim 2 :
z ∈ [x] Σ : Aside from the kth d shifting one position to the right, we have d m (δ(z)) = d m (δ) for all m = k.Therefore the order of the ds in does not change from δ to δ(z), and thus σ(z) = σ(x).
For any i < j, the two arcs (e i , d i ) and (e j , d j ) cross in the arc diagram of x ∈ F 2n if and only if e j < i + j if and only if i + j d i .Proof.By definition, the two arcs cross if and only if e i < e j < d i < d j .It is clear then that the crossing is due solely to the middlemost inequality, meaning that the jth e is preceded by strictly less than i terminal points.Equivalently, the ith d is preceded by at least j initial points.In other words, e j < i + j if and only if i + j d i .
shows the Hasse diagram of F 6 , whose elements are represented in canonical cycle notation.Example 13 describes how F 2n is embedded in S 2n for the n = 2 case.
Ferrers board consists of n adjacent columns of cells that share a bottom edge and whose heights b i are nondecreasing from left to right.
4, as an induced subposet of S 4