A Decomposition of Parking Functions by Undesired Spaces

There is a well-known bijection between parking functions of a fixed length and maximal chains of the noncrossing partition lattice which we can use to associate to each set of parking functions a poset whose Hasse diagram is the union of the corresponding maximal chains. We introduce a decomposition of parking functions based on the largest number omitted and prove several theorems about the corresponding posets. In particular, they share properties with the noncrossing partition lattice such as local self-duality, a nice characterization of intervals, a readily computable M\"obius function, and a symmetric chain decomposition. We also explore connections with order complexes, labeled Dyck paths, and rooted forests.


Introduction
An n-tuple of integers is called a parking function if it can be rearranged so that its ith entry is at most i. We can then record which of the missing numbers (if any) is the largest, which gives a natural yet little-studied decomposition of parking functions into n parts. These sets can be constructed recursively by noticing that any parking function for which k < n is the largest missing number can be obtained from a parking function of length n − 1 by adding an n, so we are particularly interested in the set of all parking functions of length k which omit k.
By utilizing the well-known correspondence between parking functions and noncrossing partitions [Sta97], we can view elements from our decomposition as maximal chains in the noncrossing partition lattice NC n+1 . Focusing on chains which come from parking functions where k is the largest missing number yields a poset Poset(PF n,k ) with elements from NC n+1 (although this is not an induced subposet). From this viewpoint, we can describe the observation above by noting that Poset(PF n,k ) is the direct product of Poset(PF k,k ) with a Boolean lattice of rank n − k. Further investigation of these posets yields a surprising number of nice results.
Following the example set by NC n+1 , we present several theorems which underscore the appeal of these posets, the first of which depicts the stucture of their intervals.

Theorem A. Each interval in Poset(PF n,k ) is either of the form
Similar to the story for NC n+1 , we can leverage this result to see the rich symmetry exhibited by these posets.
This result is pleasing in its own right, but we can also use it to prove several other facts about the structure of these posets. In particular, we find that each Poset(PF n,n ) is irreducible under direct product and use this to prove that there are no non-trivial order-preserving automorphisms of Poset(PF n,n ). Perhaps most interestingly, we use our findings to analyze the Möbius function of our posets.
Theorem C. The Möbius function of Poset(PF n,k ) is zero.
If we consider the topology of these posets, this computation becomes circumstantial evidence for the homotopy type of the corresponding simplicial complex.
Conjecture. The order complex of Poset(PF n,k ) without its bounding elements is contractible.
In addition to these results, we can follow the decomposition of parking functions through to other common Catalan-type objects such as labeled Dyck paths and labeled rooted forests. In each of these settings, we illustrate the sets which correspond to our decomposition and find a natural way to view the recursion mentioned above, lending weight to the notion that this method illustrates some interesting information.
In the first section of this article, we review the basic definitions for parking functions and introduce the proposed decomposition. Section 2 is dedicated to the posets associated to our decomposition and our proofs of their properties. We discuss the implications regarding order complexes that our results appear to make in Section 3, and we illustrate our decomposition in the areas of labeled Dyck paths and labeled rooted forests in Section 4.

Background
We begin by reviewing the basic definitions and introducing our proposed decomposition.
Definition 1.1. An n-tuple of integers (a 1 , . . . , a n ) is a parking function if there is a permutation σ ∈ Sym n such that a σ(1) ≤ . . . ≤ a σ(n) and a σ(i) ≤ i for all i ∈ [n], where [n] = {1, . . . , n}. A parking function is called primitive if it is already in weakly increasing order -these form the canonical representatives of the equivalence classes formed under the relation of permutation. A clever argument [Sta99] shows that there are (n + 1) n−1 parking functions of length n and C n = 1 n+1 2n n primitive parking functions, where C n is the nth Catalan number. We denote the set of all parking functions of length n by PF n .
With these basic properties out of the way, we can define our decomposition of PF n into n subsets.
Definition 1.2. Let n, k ∈ N with 1 < k ≤ n and define PF n,k to be the set of all parking functions of length n for which k is the largest missing number. That is, PF n,k = {a ∈ PF n | k ∈ a but k + 1, . . . , n ∈ a}.
We can then identify the remaining parking functions with permutations of [n] by thinking of each (a 1 , . . . , a n ) as the element in Sym n which sends i to a i . Then PF n is the disjoint union PF n = Sym n ⊔ PF n,2 ⊔ . . . ⊔ PF n,n .
In many of our proofs for properties of PF n,k , it will suffice to demonstrate them for PF n,n . Proposition 1.3. The number of parking functions of length n for which k is the largest missing number is Proof. By the remark above, we know that |PF n,k | = n|PF n−1,k |. Then so it suffices to compute |PF k,k |. We know that |PF k | = (k + 1) k−1 , and since every parking function with a k can be obtained uniquely from PF k−1 by inserting a k, there are k|PF k−1 | = k · k k−2 parking functions in PF k with a k. Hence, |PF k,k | = (k + 1) k−1 − k k−1 and the claim follows. In addition to deserving study in their own right, parking functions have an intimate relationship with the lattice of noncrossing partitions. This setting proves an interesting one to consider our decomposition of PF n .
Definition 2.1. A partition σ of [n] is said to be noncrossing if, when we consider the elements of [n] arranged in clockwise ascending order around a circle, the convex hulls of the blocks of σ are disjoint (Figure 1). The poset of all such partitions (a subposet of the partition lattice Π n ) is called the noncrossing partition lattice and is denoted NC n (Figure 2). It is wellknown that |NC n | = C n , the nth Catalan number. In addition, the number of maximal chains in NC n+1 (paths from bottom to top in its Hasse diagram) is (n + 1) n−1 .
The reappearance of (n+1) n−1 is not a coincidence -the connection which proves this is essential to our results.
Theorem 2.2. There is a one-to-one correspondence between parking functions of length n and maximal chains of NC n+1 .
Given a maximal chain in NC n+1 , we produce a parking function by labeling its covering relations, i.e. the edges in the Hasse diagram. For each covering relation σ < τ , the larger partition τ is obtained by joining two blocks B 1 and B 2 in σ to form one block B in τ . Without loss of generality, suppose B 1 contains min B; we will then label this edge by the largest number in B 1 which is less than all the elements in B 2 -see Figure 3 for an example. Performing this process on each edge and reading the labels on a maximal chain from bottom to top creates a parking function of length n, and this map is a bijection. [Sta97] Definition 2.3. By the correspondence above, each PF n,k corresponds to a collection of maximal chains in NC n+1 ; define Poset(PF n,k ) to be the poset whose Hasse diagram is the union of these chains in NC n+1 .
Notice that although Poset(PF n,k ) is a subset of NC n+1 , it is not an induced subposet in general since there are missing relations in the smaller poset. In addition, Poset(PF n,k ) is not usually a lattice, but there is still a nice structure here which deserves exploration.  Since our definition for Poset(PF n,k ) is based on the maximum chains of NC n+1 , it is not immediately clear which noncrossing partitions will appear. Focusing for a moment on when n = k, we can see that the elements of NC n+1 which do not lie in Poset(PF n,n ) are precisely those for which every maximal chain containing them has an n-label on one of its edges.
A covering relation σ < τ in NC n+1 is labeled n if and only if a block in τ is the union of the block containing n in σ and {n + 1} ∈ σ. There are exactly two ways to guarantee that this will happen in every maximal chain passing through some π ∈ NC n+1 : either {n, n + 1} is a block in π or 1 and n are in the same block in π (written 1 ∼ n in π or 1 ∼ π n) and {n + 1} is a block in π. This can be summarized in a proposition as follows: i.e those whose labels are a permutation of (1, 2, 3) Then the elements in Poset(PF n,n ) are precisely those in NC n+1 −(L 1 ∪L 2 ).
Notice that L 1 ∼ = L 2 ∼ = NC n−1 since any π ∈ NC n−1 can be sent to an element of L 1 by adding {n, n + 1} as a block or to L 2 by adding {n + 1} as a block and inserting n into the block containing 1, and removing n and n + 1 reverses these inclusions. In particular, the proposition above tells us that |Poset(PF n,n )| = C n+1 − 2C n−1 since L 1 and L 2 are disjoint copies of NC n−1 . Similar to our result for PF n,k and PF n−1,k , we can see that there is a recursive structure to these posets. In this setting, the corresponding result is that Poset(PF n,k ) is the direct product of Poset(PF n−1,k ) and a two-element chain. Applying this repeatedly, we obtain a useful result. an explicit decomposition of Poset(PF n,k ) into two isomorphic copies of Poset(PF n−1,k ) which are related appropriately in the poset structure. Define two elements of Poset(PF n,k ) as follows: 1 . . , n}, {n + 1}} Let0 and1 denote the minimum and maximum elements of NC n+1 , respectively, and consider the intervals [σ,1] and [0, τ ] in Poset(PF n,k )these consist of elements in which n and n + 1 share a block and in which n + 1 forms a singleton block, respectively. Notice that their (disjoint) union is Poset(PF n,k ) since each element in this poset must be part of a maximal chain in which an n-label is created. Now, construct a map from Poset(PF n,k ) to Poset(PF n−1,k ) which "forgets" n + 1 by removing it from whichever block it was in. This map respects the poset structure and is a bijection when restricted to either of the above intervals, hence each is isomorphic to Poset(PF n−1,k ). Following these bijections gives us an isomorphism Φ : [0, τ ] → [σ,1], and we can see that for each π ∈ [0, τ ], π ≤ Φ(π). Therefore, we have found two copies of Poset(PF n−1,k ) which satisfy the structure of a direct product with a 2-element chain. That is, Theorem 2.5 tells us that we may focus our investigation on Poset(PF n,n ) since many desirable poset properties respect the operation of direct product. Interestingly, this poset cannot be factored any further via the direct product -it is irreducible in this sense. Proof. Notice that if P and Q are posets of height m and n, respectively, then there is a 1-to-m+n m correspondence between pairs of maximal chains from P and Q and the set of maximal chains in P × Q. We can see this by realizing that the maximal chains in the direct product of P and Q are formed by merging a maximal chain from each, and there are m+n m ways to do this.
If Poset(PF n,n ) were reducible, then the number of maximal chains in Poset(PF n,n ) could be expressed as n k for some k ∈ {1, . . . , n − 1}. However, we know that the number of maximal chains in Poset(PF n,n ) is |PF n,n | = (n + 1) n−1 − n n−1 , which is coprime to n, whereas n k shares a nontrivial factor with n since gcd( n i , n j ) > 1 whenever 0 < i, j < n. [ES78] Therefore, Poset(PF n,n ) is irreducible. In addition to the recursive structure depicted above, the symmetry exhibited by these posets convinces us that they are a worthwhile setting for investigation.
Proof. There are 2n + 2 order-reversing automorphisms of NC n+1 [Bia97], so we will describe one and show that it fixes Poset(PF n,n ). Self-duality of Poset(PF n,k ) then follows from Theorem 2.5 since the product of self-dual posets is self-dual.
Let π ∈ NC n+1 and suppose i ≤ j for some i, j ∈ [n + 1]. Define We then define a map ρ : NC n+1 → NC n+1 by declaring that i ∼ j in ρ(π) if and only if x ∼ y in π for all x ∈ A i,j and y ∈ B i,j .
Put another way, we can obtain ρ(π) by first drawing the circular representation of π and interspersing n + 1 extra white points between the preexisting black points so that, proceeding clockwise, the black point labeled i is just before the white point labeled n − i, modulo n + 1. Then, to obtain the noncrossing partition ρ(π), we take the convex hulls of white points which do not cross the original partition -see Figure 9.
It is easy to see that this forms an order-reversing involution on NC n+1 and is thus an isomorphism. We need to check that this map fixes the elements Poset(PF n,n ) and is order-reversing in that poset's relation.
To see that ρ is order-reversing with respect to Poset(PF n,n ), it suffices to show that any covering relation in NC n+1 which produces an n-label is sent to another n-labeled covering relation. Thankfully, we know that such a covering relation σ < τ occurs if and only if {n + 1} ∈ σ and n ∼ n + 1 in τ . Examining the definition of ρ, we see this is equivalent to knowing that n ∼ n + 1 in ρ(σ) and {n + 1} ∈ ρ(τ ), so the new covering relation is also labeled by an n. Hence, ρ is an order-reversing automorphism of NC n+1 which fixes Poset(PF n,n ) and respects its poset structure, so ρ is an order-reversing automorphism of Poset(PF n,n ).
The rest of our results make use of the convenient structure of intervals in Poset(PF n,n ), on which we must first prove a somewhat technical theorem, mirroring the fact that each interval in NC n+1 is a direct product of smaller noncrossing partition lattices.
Proof. Let σ, τ ∈ Poset(PF n,n ) with σ < τ . If σ =0 and τ =1, then [σ, τ ] = Poset(PF n,n ) and we're done. For now, suppose that τ =1. Recall that the only way to produce an n-label in a chain between two elements in NC n+1 is if n ∼ n + 1 in the coarser partition and n + 1 is a singleton in the finer partition. If one of these conditions is not satisfied, i.e. n ∼ τ n + 1 or and since we know that intervals in NC n+1 are products of smaller noncrossing partition lattices [NS97], the result follows for this case. Now, suppose that n ∼ τ n + 1 and {n + 1} ∈ σ. Then there are elements between σ and τ in NC n+1 which do not appear in Poset(PF n,n ), so we need to try something different.
Let B ∈ τ be the block containing n and n + 1 and let B 1 , . . . B k ∈ σ be the blocks whose union is B. If we write B = {b 1 , . . . , b l+1 } such that b 1 < . . . < b l+1 , then we can "factor out" an interval in Poset(PF l,l ) from [σ, τ ] in the following way: define f : B → [l + 1] by f (b i ) = i and notice that In fact, σ ′ ∈ Poset(PF l,l ) by Proposition 2.4, which we will use momentarily. Next, define Notice that σ and τ ′ have blocks which agree on the elements of B. The benefit of making this modification is that the interval [σ, τ ′ ] Poset(PFn,n) is equal to [σ, τ ′ ] NC n+1 since we have "removed" the blocks which could create n-labels. Similarly, the interval [σ ′ ,1] Poset(PF l,l ) contains (an isomorphic copy of) only those elements. More concretely, we have decomposed our interval to obtain Applying our order-reversing map ρ, we can see that in Poset(PF l,l ). Notice that since {n + 1} ∈ σ, then {l + 1} ∈ σ ′ , so we have l ∼ l + 1 in ρ(σ ′ ). Then the interval [0, ρ(σ ′ )] decomposes into the direct product of Poset(PF r,r ) (where r is the size of the block containing l and l + 1 in ρ(σ ′ )) and j i=1 NC m i (where the m i 's are the sizes of the other blocks in ρ(σ ′ )). Hence, we have Since we know the form of intervals in NC n+1 , we're done.
With this result in hand, we can prove a few more properties about the structure of these posets. In particular, we have proven a stronger version of Theorem 2.7.
We know that Poset(PF n,n ) is self-dual via our map ρ, but it is worth wondering whether there are other maps which would work just as well. Interestingly, this symmetry is unique -put another way, there is only one order-preserving automorphism.
Theorem 2.8. The identity map is the unique order-preserving automorphism on Poset(PF n,n ).
Proof. Let f be an order-preserving automorphism of Poset(PF n,n ) and consider the two elements characterized as an edge from 1 to n + 1 and the triangle between 1, n, and n + 1: In fact, we can conclude something stronger -notice that [E,1] is isomorphic to N C n by merging 1 with n + 1. Since the automorphisms of N C n are the 2n natural dihedral symmetries [Bia97], f must restrict to one of All that remains is to show that elements outside these two intervals are fixed by f . Let σ ∈ Poset(PF n,n ) be such an element -then σ covers σ ∧ ρ(E) by splitting off {n} and σ ∨ E covers σ by joining the blocks containing 1 and n + 1. Since σ ∧ ρ(E) and σ ∨ E are each fixed by f , the interval [σ ∧ ρ(E), σ ∨ E] is fixed setwise. If we can show that σ is the unique element in this interval which does not also lie in [E,1] ∪ [0, ρ(E)], then we can conclude that σ is fixed and we're done.
Suppose first that n ∼ n + 1 in σ. Then 1, n, and n + 1 lie in separate blocks in σ ∧ ρ(E) and share the same block in σ ∨ E. The only way to obtain an intermediate element is to combine two of the blocks in σ ∧ ρ(E); we obtain σ by combining the blocks with n and n + 1. If we were to merge the 1 and n + 1 blocks (leaving n a singleton) we would obtain an element in [E,1]∩[0, ρ(E)], and if we merge the 1 block with the n block, the result is an element not in Poset(PF n,n ). Hence, in this case the interval [σ ∧ ρ(E), σ ∨ E] has two elements of height 1, each of which must be fixed. Now suppose n ∼ n + 1 in σ. Then 1, n, and n + 1 live in distinct blocks in σ ∧ ρ(E) as above, but 1 and n + 1 share a block without n in σ ∨ E. Then there are only two possibilities for intermediate elements in Therefore, we can see that f must fix each element in Poset(PF n,n ), and we're done.
In the spirit of analyzing the symmetries and recursive structure of these posets, we also compute the Möbius function of Poset(PF n,k ).
In the case of Poset(PF n,k ), we find a similarly interesting result.
Theorem C. The Möbius function of Poset(PF n,k ) is zero.
Proof. Since the Möbius function respects direct products, it suffices to prove this for Poset(PF n,n ); the first case is straightforward since Poset(PF 2,2 ) is simply a 3-element chain.
Definition 2.11. A poset P with height n is said to have a symmetric chain decomposition (SCD) if the elements of P can be partitioned into saturated chains so that ranks of the minimum and maximum in each chain sum to n.
It follows quickly from this definition that if posets P and Q have a symmetric chain decomposition, then so does P × Q.
Proof. By the observation above, it suffices to show that Poset(PF n,n ) has an SCD, and we will do so by mirroring Simion and Ullman's proof for NC n . [SU91] First, notice that we can easily find SCDs when n is 1,2, or 3 by glancing at the Hasse diagram. Now, suppose that this result holds for values less than some fixed n > 3 and decompose Poset(PF n,n ) as follows: That is, R i is the subposet consisting of elements for which i is the secondsmallest number in the block containing 1. We claim that Poset(PF n,n ) is the disjoint union of these subposets. It is straightforward to see from the definition that each R i is disjoint from R 1 , R ′ 1 , and each other R j . Observing that R 1 ∩ R ′ 1 = ∅ amounts to noticing that no element of R 1 could be less than the maximum element in R ′ 1 . Hence, this is a disjoint collection of subposets for Poset(PF n,n ). As for their union, notice that for any π ∈ Poset(PF n,n ), there are three cases. If {1} ∈ π, then π ∈ R i for some i = 1. If not, then we either have 2 ∼ n in π, in which case π ∈ R ′ 1 , or 2 ∼ n and thus π ∈ R 1 . So Poset(PF n,n ) is the disjoint union of these subposets.
Observing each R i as an interval for i ∈ {3, . . . , n−1}, we can see that each is isomorphic to a direct product of Poset(PF l,l ) and/or NC m for values of l and m less than n, so each has a symmetric chain decomposition by our inductive hypothesis and the analogous result for NC n . [SU91] Additionally, each R i ranges from height 1 to height n − 1 in Poset(PF n,n ), so it is "symmetrically embedded". Additionally, R 1 and R 2 are each isomorphic to Poset(PF n−1,n−1 ) and embedded so that R 2 covers R 1 -that is, R 1 ∪ R 2 is the direct product of a 2-element chain and Poset(PF n−1,n−1 ). Hence it admits an SCD which is also symmetrically embedded from height 0 to height n.
It then remains to show that R ′ 1 , R n , and R n+1 together admit a symmetrically embedded SCD. To this end, define two subposets of R n+1 as follows: Then we can see that A ∼ = R ′ 1 , B ∼ = R n , and each is isomorphic to NC n−2 . Moreover, A coverse R ′ 1 and R n covers B in such a way that A ∪ R ′ 1 and B ∪ R n are each isomorphic to the direct product B 2 × NC n−2 . Thus, each has an SCD and ranges from height 1 to n − 1, so they are symmetrically embedded.
All that remains is R n+1 − (A ∪ B), which can be expressed as Similar to how we began, we can partition this subposet into the intervals each of which is isomorphic to a direct product of noncrossing partition lattices. Therefore, each has an SCD which ranges from height 2 to heignt n − 2.
Since we have accounted for all subposets in our decomposition, we can conclude that Poset(PF n,n ) admits a symmetric chain decomposition.

Order Complexes
The study of order complexes gives us a topological way to understand Poset(PF n,k ), possibly leading to greater insight for this decomposition.
Definition 3.1. Let P be a poset and define the order complex ∆(P ) to be the simplicial complex constructed by associating a k-simplex to each finite chain x 0 < x 1 < · · · < x k in P in the natural way.
Notice that maximal chains in P correspond to top-dimensional simplices in ∆(P ), which thus can be labeled by parking functions in the case when P is NC n+1 . In particular, our decomposition of PF n produces a decomposition of the facets of ∆(NC n+1 ) and we can view our order-reversing map ρ as a symmetry of this topological space.
In general, when P is bounded we refer to the edge in the order complex which corresponds to the chain0 <1 by the diagonal. Notice that then ∆(P ) is contractible since each of0 and1 is a cone point, so we may deformation retract the complex to either one. In order to see the combinatorial data of P topologically, we disregard0 and1 and examine ∆(P ), where P = P −{0,1}.
It turns out that ∆(P ) has another topological name -it is the link of the diagonal in ∆(P ). As such, we refer to ∆(P ) as the link complex or simply link of P . There is an intimate connection between the topology of this complex and the combinatorics of P . Hence, by Theorem 2.10, we know that the link of Poset(PF n,k ) has reduced Euler characteristic 0. This, in combination with low-dimensional computations, gives compelling evidence for the following conjecture. Conjecture. The link of Poset(PF n,k ) is contractible.

Other Settings
While NC n+1 is an exceptionally fruitful setting for exploring our decomposition, there are other objects in bijection with PF n with matching decompositions. Specifically, there are geometric ways to visualize PF n,k with labeled Dyck paths and labeled rooted forests. If nothing else, these examples speak for the natural definition of PF n,k .
Definition 4.1. A Dyck path of length 2n is a lattice path in Z ⊕ Z from (0, 0) to (n, n) which stays weakly above the diagonal y = x and consists only of "up" and "right" steps. A labeled Dyck path is a Dyck path for which the n "up" steps are distinctly labeled by the {1, . . . , n}.
Proposition 4.2. There is a one-to-one correspondence between labeled Dyck paths of length 2n and PF n .
Given (a 1 , . . . , a n ) ∈ PF n , let b i be the number of times i appears in this parking function and draw a path from (0, 0) to (n, n) with b 1 steps up followed by one step to the right, then b 2 steps up followed by one step to the right, and so on. The restrictions on parking functions ensure that this path will stay weakly above the diagonal. Label this Dyck path as follows: label the bottom-most unlabeled "up" step in column a 1 with 1. Label the same such step in column a 2 with a 2, and so on. This results in a labeled Dyck path, and this map is a bijection. It follows immediately from the bijection that if k is the largest number missing from a parking function f , then the corresponding labeled Dyck path has no up-step in the kth column, followed by n−k columns each with exactly one up-step. In other words, we have characterized this decomposition with the following theorem.
Theorem 4.4. D n,k is the set of labeled Dyck paths of length 2n which end in a string of exactly n − k "up-right" pairs.
There is a similar way to see the PF n,k decomposition in the setting of rooted forests.
Definition 4.5. A rooted forest is a collection of trees in which each tree has one vertex designated as the root. A labeled rooted forest is one in which each vertex is labeled with a distinct element of {1, 2, . . . , n}, where n is the total number of vertices. We denote the set of labeled rooted forests on n vertices as RF n .
There is a bijection from RF n to PF n that extends our decomposition to this setting in a natural way. [Pak09] Definition 4.6. Let v and w be vertices in a rooted forest f .
(1) h(v) is the length of the shortest path between v and a root.
(2) If v and w are adjacent and h(w) = h(v) + 1, then v is the parent of w and w is a child of v. Definition 4.7. Let f be a labeled rooted forest. Define φ : RF n → PF n as follows: (1) First, we construct a (labeled) binary rooted tree b on n vertices using the information from f . For each vertex v in f , we will define a corresponding vertex v ′ in b with the same label.
(a) If v is the (unique) vertex in f such that h(v) = 0 and η(v) = 1, then define v ′ to be the root of b. (b) If v has a sibling w such that η(w) = η(v) + 1, then v ′ will have a left-edge connecting v ′ to w ′ such that w ′ is a child of v ′ . (c) If v has any children, let x be the unique child such that such that η(x) = 1. Then, v ′ has a right-edge connecting v ′ to x ′ such that x ′ is a child of v ′ .
(2) We now use b to construct a labeled Dyck path. Start at the root of b and proceed counterclockwise around the outside of the tree, visiting each vertex twice. Each time we encounter a vertex that we have not already seen, we construct an up-step in our lattice path with the same label. If we first encounter a vertex immediately before we encounter a left-edge, we construct a right-step after encountering the vertex for the second time. Otherwise, we construct a right-step immediately after recording the up-step corresponding to the vertex. (3) We can now use the bijection described in Proposition 4.2 to find the corresponding parking function φ(f ).
We now define a way of partitioning these rooted forests that behaves nicely with respect to φ.
Definition 4.8. Let f 1 , f 2 be labeled rooted forests and define an equivalence relation by f 1 ∼ f 2 if and only if there exists a graph isomorphism ψ : f 1 → f 2 such that for all vertices v in f 1 , h(v) = h(ψ(v)) and η(v) = η(ψ(v)).
Proposition 4.9. φ induces a one-to-one correspondence between equivalence classes of RF n and equivalence classes of PF n under permutation.
Proof. Let f 1 and f 2 be labeled rooted forests such that f 1 ∼ f 2 and let ψ : f 1 → f 2 be the associated graph isomorphism. Since h and η are preserved by the isomorphism, the corresponding binary trees b 1 and b 2 are identical up to relabeling. Hence, the corresponding labeled Dyck paths have the same shape and thus correspond to parking functions which are rearragements of one another.
We can now realize our decomposition of parking functions in the setting of labeled rooted forests.
Definition 4.10. Let RF n,k be the set of labeled rooted forests which correspond to the parking functions in PF n,k via the bijection above, and let RF n,1 be those which correspond to parking functions identified with Sym n .
First, we notice that the recurrence relation between PF n,k and PF n+1,k has an analogue in this setting.
Definition 4.11. Each labeled rooted forest has a unique path from the root to a leaf such that every vertex v i on this path satisfies η(v i ) = 1. We denote this path as P f and its leaf as v f .
Proposition 4.12. The rooted forest equivalence classes in RF n+1,k can be obtained from the equivalence classes in RF n,k by adding a single edge and leaf to v f for each f ∈ RF n,k .
Proof. Fix an equivalence class in RF n+1,k and select a representative f so that the vertex v f is labeled n + 1. Notice that in the binary tree b corresponding to f , P f becomes the maximal path of right-edges from the root. Removing v f produces a labeled rooted forest f ′ ∈ RF n with the property that the associated binary tree b ′ differs only from b in that its path of right-edges from the root has one fewer vertex. In other words, the Dyck paths associated to f and f ′ are the identical except for an additional "up-right" pair at the end of the former, so by the bijection in Definition 4.7, we can see that f ′ ∈ RF n,k . Rearranging the labels for f would change nothing except the labels on f ′ , so any equivalence class in RF n+1,k can be obtained from one in RF n,k by the procedure above.
We can use similar reasoning to determine which equivalence classes of rooted forests are contained in each RF n,k .
Theorem 4.13. For each f ∈ RF n , consider the longest sub-path of P f containing v f and not containing a root vertex such that none of the vertices on the path have siblings. Then f ∈ RF n,k if and only if the number of vertices on this sub-path is n − k.
Proof. By Proposition 4.12, it suffices to show this for RF n,n . To this end, suppose that f ∈ RF n and that v f has at least one sibling. Then, under RF 3,1 :  Figure 13. The equivalence classes of RF 3 split according to the decomposition the bijection given in Definition 4.7, the vertex associated to v f in the corresponding binary tree has a child via a left-edge, but no right-edge. It follows that the labeled Dyck path for this tree ends with (at least) two right-steps, and thus the parking function φ(f ) does not contain n, hence f ∈ RF n,n . By reversing our steps through the bijection we can see that v f has a sibling if and only if f ∈ RF n,n and our claim is proven.
It seems that these are simply a few of many fruitful ways of studying our decomposition of PF n . We expect that there are several other settings or generalizations for these ideas which could give an interesting perspective for the structure of parking functions.