Bijective enumeration of some colored permutations given by the product of two long cycles
Abstract
Let be the permutation on symbols defined by . We are interested in an enumerative problem on colored permutations, that is permutations of in which the numbers from to are colored with colors such that two elements in a same cycle have the same color. We show that the proportion of colored permutations such that is a long cycle is given by the very simple ratio . Our proof is bijective and uses combinatorial objects such as partitioned hypermaps and thorn trees. This formula is actually equivalent to the proportionality of the number of long cycles such that has cycles and Stirling numbers of size , an unexpected connection previously found by several authors by means of algebraic methods. Moreover, our bijection allows us to refine the latter result with the cycle type of the permutations.
keywords:
Colored Permutations, Bipartite Maps, Long Cycle Factorization1 Introduction
The question of the number of factorizations of the long cycle into two permutations with given number of cycles has already been studied via algebraic or combinatorial
Unfortunately, even though generating series have nice compact forms,
the formulae for one single coefficient are much more complicated
(see for example [GS98]).
The case where one factor has to be also a long cycle is particularly interesting.
Indeed, the number of permutations of with cycles, such that is a long cycle, is known to be the coefficient of some linear monomial in Kerov’s and Stanley’s character polynomials (see ([Bia03], Theorem 6.1) and [Sta03]; [Fér10]). These polynomials express the character value of the irreducible representation of the symmetric group indexed by a Young diagram on a cycle of fixed length in terms of some coordinates of .
The numbers admit a very compact formula in terms of Stirling numbers.
Theorem 1.1 ([Kl93]).
Let be two positive integers with the same parity. Then
(1) 
where is the unsigned Stirling number of the first kind, that is the number of permutations of with cycles.
This formula has been found independently by several authors: J.H. Kwak and J. Lee ([KL93], Theorem 3), then D. Zagier ([Zag95], Application 3) and finally R. Stanley ([Sta11], Corollary 3.4). Very recently, a combinatorial proof of this statement has been given by R. Cori, M. Marcus and G. Schaeffer [CMS10]. This paper is focused on an equivalent statement in terms of colored (or partitioned) permutations.
Definition 1.2.
A colored permutation of with colors is a couple where:

is a permutation of ;

is a surjective map from to a set of colors of cardinality . We require that two elements belonging to the same cycle of have the same color.
In what follows, we consider that two colored permutations differing only by a bijection on the set of colors are the same object. As such, coloration can be seen as a set partition of the set of cycles of , or as a set partition of coarser than the set partition into cycles of (in other words, if and lie in the same cycle of , they must be in the same part of ). The set of colored permutations of with colors is denoted .
According to the last remark of definition 1.2, we rather denote colored permutations
where is a set partition coarser than the set partition into cycles of .
These objects play an important role in the combinatorial study of the factorizations in the symmetric group,
as it is much easier to find direct bijections for colored factorizations than it is for classical ones
(see [GN05]; [GS10]; [Ber10]; [SV08]; [MV09]).
Generating series of colored and classical factorizations are linked through simple formulae
(Lemma 6.1).
We consider here an analogue of Theorem 1.1 for colored permutations, that is the problem of enumerating colored permutations such that is a long cycle. We obtain the following elegant result:
Theorem 1.3.
Let be two positive integers. Choose randomly (with uniform probability) a colored permutation in . Then the probability for to be a long cycle is exactly .
Given a colored permutation in , the (unordered) sequence of the numbers of elements having the same color defines an integer partition of with parts, which we call the type of . For any integer partition of , we note the set of all colored permutations of type . Our main result is the following refinement of Theorem 1.3:
Theorem 1.4 (Main result).
Let be two positive integers. Fix an integer partition of size and length . Choose randomly (with uniform probability) a colored permutation in . Then the probability for to be a long cycle is exactly .
In fact, counting colored permutations and counting permutations without additional structure are two equivalent problems. Therefore, one can deduce from Theorem 1.4 a refinement of Theorem 1.1.
To state this new theorem, we need to introduce a few notations. Recall that the type of a permutation is defined as the sequence of the lengths of its cycles, sorted in increasing order. With this notion, it is natural to refine the numbers and : if (i.e. is a partition of ), let (resp. ) be the number of permutations of type (resp. with the additional condition that is a long cycle). Of course, is given by the simple formula , where is the number of parts in and .
Then, as Theorem 1.1 deals with permutations of and , we need operators on partitions which modify their size, but not their length. If (resp. ) has at least one part (resp. ), let (resp. ) be the partition obtained from (resp. ) by erasing a part (resp. ) and adding a part (resp. ). For instance, using exponential notations (see ([Mac95], chapter 1, section 1)), and .
Theorem 1.5 (Corollary).
Let be two positive integers with the same parity. For each partition of length , one has:
(2) 
From this result, one can immediately recover Theorem 1.1 by summing over all partitions of length and size . Indeed,
To be comprehensive on the subject, we mention that G. Boccara has found an integral formula for (see [Boc80]), but there does not seem to be any direct link with our result.
Remark 1.6.
Note that the statement of Theorem 1.4 is much nicer than Theorem 1.5 (in particular, the fact that the ratio depends only on and is quite surprising). This suggests that it is interesting to work with colored permutations rather than with permutations without additional structure (as it is done in [CMS10] for example).
Outline of the paper. Thanks to an interpretation of colored permutations in terms of partitioned hypermaps (Section 2), we prove bijectively Theorem 1.4 in Sections 3, 4 and 5. Finally, in Section 6, we use algebraic computations in the ring of symmetric functions to show the equivalence with Theorem 1.5.
2 Combinatorial formulation of Theorem 1.4
2.1 Blackpartitioned maps
By definition, a map is a graph drawn on a twodimensional oriented closed compact surface (up to deformation), i.e. a graph with a cyclic order on the incident edges to each vertex.
The faces of a map are the connected components of the surface without the graph (we require that these components are isomorphic to open discs).
As usual [Jac68]; [Cor75],
a couple of permutations in can be represented as a bipartite map (or hypermap)
with edges labeled with integers from to .
In this identification, (resp. ) is the edge following
when turning around its white (resp. black) extremity.
White (resp. black) vertices correspond to cycles of (resp. ).
In this setting, faces of the map correspond to cycles of the product .
Hence, the condition (which we will assume from now on) means that the map is unicellular (i.e. has only one face) and that the positions of the labels are determined by the choice of the edge labeled by (which can be seen as a root). In this case, the couple of permutations is entirely determined by .
Therefore, if , the quantity is the number of
rooted unicellular maps with black vertices’ degree distribution
(there are no conditions on white vertices).
The condition that the product is a long cycle is equivalent
to the fact that the corresponding rooted bipartite map has only one white vertex
(we call such maps star maps).
Thus is the number of star
rooted unicellular maps with black vertices’ degree distribution .
As in the papers [SV08] and [MV09], our combinatorial construction deals with maps with additional structure:
Definition 2.1.
A blackpartitioned (rooted unicellular) map is a rooted unicellular map with a set partition of its black vertices. We call degree of a part (block) of the sum of the degrees of the vertices in . The type of a blackpartitioned map is its blocks’ degree distribution.
In terms of permutations, a blackpartitioned map consists of a couple in with the condition and a set partition of coarser than the set partition in orbits under the action of . Note that couples with are in bijection with permutations . Therefore, a blackpartitioned map is the same object as a colored permutation (see Definition 1.2). The number of colors corresponds to the number of blocks in the set partition .
Example 2.2.
Let , , and be the partition . Here, the type of is . Associating the triangle, circle and square shape to the blocks, is the blackpartitioned star map pictured on figure 1.
If , we denote by (resp. ) the number of blackpartitioned maps (resp. blackpartitioned star maps) of type . Equivalently, (resp. ) is the number of couples as above such that is a partition of type (resp. and is a long cycle).
With this notations, Theorem 1.4 can be rewritten as:
(3) 
2.2 Permuted star thorn trees and Morales’Vassilieva’s bijection
The main tool of this article is to encode blackpartitioned maps into star thorn trees, which have a very simple combinatorial structure. Note that they are a particular case of the notion of thorn trees, introduced by A. Morales and the second author in [MV09].
Definition 2.3 (star thorn tree).
An (ordered rooted bipartite) star thorn tree of size is a planar tree with a white root vertex, black vertices and thorns connected to the white vertex and thorns connected to the black vertices. A thorn is an edge connected to only one vertex. “Planar“ means that the sons of a given vertex are ordered (here, a thorn should be considered as a son of its extremity).
We call type of a star thorn tree its black vertices’ degree distribution (taking the thorns into account). If is an integer partition, we denote by the number of star thorn trees of type .
Two examples are given on Figure 2 (for the moment, please do not pay attention to the labels). The interest of this object lies in the following theorem.
Theorem 2.4 ([Mv09]).
Let be a partition of length . One has:
(4) 
This theorem corresponds to the case of ([MV09], Theorem 2) (note that the proof is entirely bijective).
The righthand side of (4) is the number of couples where:

is a star thorn tree of type .

is a bijection between thorns with a white extremity and thorns with a black extremity (by definition, has exactly thorns of white extremity and thorns of black extremity).
We call such a couple a permuted (star) thorn tree. By definition, the type of is the type of . Examples of graphical representations are given on Figure 2: we put symbols on edges and thorns with the following rule.
Two thorns get the same symbol if they are associated by and,
except from that rule, all symbols are different (the chosen symbols and their order do not matter, we call that a symbolic labeling).
Using this result, one obtains another equivalent formulation for Theorem 1.4:
(5) 
Sections 3, 4 and 5 are devoted to the proof of equation (5). We proceed in a three step fashion. Firstly, we define a mapping from the set of blackpartitioned star maps of type (counted by ) into the set of permuted star thorn trees of the same type. Secondly, we show it is injective. As a final step, we compute the cardinality of the image set of and show it is exactly .
Remark. Although there are some related ideas, is not the restriction of the bijection of paper [MV09].
3 Mapping blackpartitioned star maps to permuted thorn trees
3.1 Labeled thorn tree
Let be a blackpartitioned star map. First we construct a labeled star thorn tree :

Let be the integer list such that and such that the long cycle is equal to . The root of is a white vertex with descending edges labeled from right to left with ( is the rightmost descending edge and the leftmost).

Let be the maximum element of the block . For , if for some , we draw a black vertex at the other end of the descending edge labeled with . Otherwise the descending edge is a thorn.
Remark 3.1.
As the leftmost descending edge is never a thorn and is labeled with .

For , let be the cycles included in block such that is the maximum element of cycle . (We have ). We also order these cycles according to their maximum, i.e. we assume that . As a direct consequence, .
We connect thorns to the black vertex linked to the root by the edge . Moving around the vertex clockwise and starting right after edge , we label its thorns with the integersin this order. Note that the last one is as is the label of the edge. Then is the resulting thorn tree.
Remark 3.2.
Moving around a black vertex clockwise starting with the thorn right after the edge, a new cycle of begins whenever we meet a lefttoright maximum of the labels.
The idea behind this construction is to add a root to the map , select one edge per block, cut all other edges into two thorns and merge the vertices corresponding to the same black block together. Step (i) tells us where to place the root, step (ii) which edges we select and step (iii) how to merge vertices (in maps unlike in graphs, one has several ways to merge given vertices).
Example 3.3.
Let us take the blackpartitioned star map of example 2.2. Following construction rules (i) and (ii), one has , , and the descending edges indexed by , and connect a black vertex to the white root. Other descending edges from the root are thorns. Using (iii), we add labeled thorns to the black vertices to get the labeled thorn tree depicted on Figure 3. Focusing on the one connected to the root through the edge , we have . Reading the labels clockwise around this vertex, we get . The three cycles can be simply recovered looking at the lefttoright maxima , and .
Remark 3.4.
Let us fix a labeled thorn tree coming from a blackpartitioned star map . Then can be found from by reading the labels around the root in counterclockwise order and is the following setpartition: for each black vertex of , the block of is the set of the labels of the edge and of the thorns linked to . Hence, a labeled thorn tree corresponds at most to one blackpartitioned star map .
3.2 Permuted thorn tree
We call the star thorn tree obtained from by removing labels and the permutation that associates to a white thorn in the black thorn with the same label in .
Finally, we define:
4 Injectivity and reverse mapping
Assume for some black partitioned star map . We show that is actually uniquely determined by .
As a first step, we recover the labeled thorn tree . Let us draw the permuted thorn tree as explained in paragraph 2.2. We show by induction that there is at most one possible integer value for each symbolic label.

By construction, the label of the rightmost edge or thorn descending from the root is necessarily .

Assume that for , we have identified the symbols of values . We look at the edge or thorn with label connected to a black vertex . In this step, we determine which symbol corresponds to .
Recall that, when we move around clockwise finishing with the edge (in this step, we will always turn in this sense), a new cycle begins whenever we meet a lefttoright maximum (Remark 3.2). So, to find , one has to know whether is a lefttoright maximum or not.
If all values of symbols of thorns before have not already been retrieved, then is not a lefttoright maximum. Indeed, the remaining label values are , …, and at least one thorn’s label on the left of lies in this interval. According to our construction necessarily corresponds to the symbolic label of the thorn right at the left of (case a)
If all the symbol values of thorns before have already been retrieved (or there are no thorns at all), then is a lefttoright maximum. According to the construction of , corresponds necessarily to the symbolic label of the thorn preceding the next lefttoright maximum. But one can determine which thorn (or edge) corresponds to the next lefttoright maximum: it is the first thorn (or edge) whose value has not been retrieved so far (again moving around the black vertex from left to right). Indeed, all the values retrieved so far are less than and those not retrieved greater than . Therefore is the thorn right at the left of (case b).If all the values of the labels of the thorns connected to have already been retrieved then is the maximum element of the corresponding block and corresponds to the symbolic label of the edge connecting this black vertex to the root (we can see this as a special case of case b).

Consider the element (thorn of edge) of white extremity with the symbolic label corresponding to . The next element (turning around the root in counterclockwise order) has necessarily label .
As a result, the knowledge of the thorn or edge with label uniquely determines the edge or thorn with label .
Applying the previous procedure up to we see that is uniquely determined by and so is (see Remark 3.4).
Example 4.1.
Take as an example the permuted thorn tree drawn on the lefthand side of Figure 2, the procedure goes as described on Figure 5. First, we identify . Then, as there is a non (value) labeled thorn on the left of the thorn connected to a black vertex with label value , necessarily is not a lefttoright maximum and is the label of the thorn immediately to the left of . Then as follows around the white root, we have .
We apply the procedure up to the full retrieval of the edges’ and thorns’ labels. We find , , . Finally, we have , , as shown on figure 5.
5 Characterisation and size of the image set
5.1 A necessary and sufficient condition to belong to
Why is not surjective?
Let us fix a permuted star thorn tree . We can try to apply to it the procedure of section 4 and we distinguish two cases:

it can happen, for some , when one wants to give the label to the edge following (step (iii)), that this edge has already a label (). If so, the procedure fails and is not in .

if this never happens, the procedure ends with a labeled thorn tree . In this case, one can find the unique blackpartitioned star map corresponding to and by construction .
For instance, take the couple on the right of Figure 2, the procedure gives successively
and then we should choose , but this is impossible because we already have .
Lemma 5.1.
If the procedure fails, the label of the edge that should get a second label is always .
Proof.
Assume . As the reconstruction procedure did not fail for , there are two distinct pairs of thorns with labels and . We will prove that the reconstruction provides labels and to two distinct elements.
We assume that the labels and have been given to the same element. In particular, and must belong to the same black vertex. Let us consider the different possible cases in the reconstruction step (ii):

If is obtained via case b (the lefttoright maximum case), the label must be just to the right of and not a lefttoright maximum. But this is impossible because all thorns to the left of (including ) have labels smaller than .

If is obtained via case a (the not lefttoright maximum case) and is a lefttoright maximum. The label is just to the right of the thorn/edge labeled by both and . Then is before the next lefttoright maximum. So the edge to the right of has a label greater than and can not be .

If is obtained via case a (the not lefttoright maximum case) and is not a lefttoright maximum. The label is still just to the right of the thorn/edge labeled by both and . Label must be as well just to the right of . It is not possible as and are the labels of two distinct thorns or edge since the procedure has not failed at step .
Finally and correspond to two different symbolic labels and hence and also (they are respectively the symbolic label of the elements right at the left of and when turning around the root). Hence, the procedure can not fail for a value of . ∎
An auxiliary oriented graph
Remark 3.1 gives a necessary condition for to be in : its leftmost edge attached to the root must be a real edge and not a thorn. From now on, we call this property : note that, among all permuted thorn trees of a given type of length , exactly over have this property. Whenever is satisfied, we denote the leftmost edge leaving the root and its black extremity. The lemma above shows that the procedure fails if and only if is chosen as for some . But this can not happen at any time. Indeed, the following lemma is a direct consequence from step (ii) of the reconstruction procedure:
Lemma 5.2.
A real edge (i.e. which is not a thorn) can be chosen as only if the edge and all thorns attached to the corresponding black vertex have labels smaller or equal to . If this happens, we say that the black vertex is completed at step .
Corollary 5.3.
Let be a real edge of black extremity . Let us denote the element (edge or thorn) immediately to the left of around the white vertex. Let be the black extremity of the element associated to (i.e. itself if it is an edgeand its image by otherwise). Then can not be completed before .
Proof.
If is completed at step , by Lemma 5.2, the element has a label . As has the same label, this implies that has label or in other words, that is completed at time . ∎
When applied for every black vertex , this corollary gives some partial information on the order in which the black vertices can be completed. We will summarize this in an oriented graph : its vertices are the black vertices of and its edges are , where and are in the situation of the corollary above. This graph has one edge attached to each of its vertices except . As examples, we draw the graphs corresponding to and to (see Figures 2 and 4) on Figure 6.
The graph gives all the information we need!
Can we decide, using only , whether belongs to or not? There are two cases, in which the answer is obviously yes:

Let us suppose that is an oriented tree of root (all edges are oriented towards the root). In this case, we say that has property . Then, the vertex can be completed only when all other vertices have been completed, i.e. when all edges and thorns have already a label. That means that can be chosen as only for . Therefore, in this case, the procedure always succeeds and belongs to . This is the case of .

Let us suppose that contains an oriented cycle (eventually a loop). Then all the vertices of this cycle can never be completed. Therefore in this situation the procedure always fails and does not belong to . This is the case of .
In fact, we are always in one of these two cases:
Lemma 5.4.
Let be an oriented graph whose vertices have outdegree , except one vertex which has outdegree . Then is either an oriented tree with root or contains an oriented cycle.
Proof.
We consider two different cases:

either, there exists a vertex with no paths from to . In this case, we denote the vertices such that is the successor of and is the successor of . As the number of vertices is finite, there are at least two indices and such that . The chain is an oriented loop.

or there is a path from each vertex to . So contains an oriented tree of root . As the number of edges is exactly one less than the number of vertices, is an oriented tree.∎
Finally, one has the following result:
Proposition 5.5.
The mapping defines a bijection:
(6) 
5.2 Proportion of permuted thorn trees in
To finish the proof of equation (5), one has justto compute the size of the righthand side of (6). We do it via a quite technical (but pretty easy) induction, it would be nice to find a more elegant argument.
Proposition 5.6.
Let be a partition of of length . Denote by the proportion of couples with properties and among all the permuted thorn trees of type . Then, one has:
Proof.
In fact, we will rather work with the proportion of couples verifying among the permuted thorn trees of type verifying . As the proportion of couples with property among couples of type is , one has: . We will prove by induction over that:
The case is easy: as has only one vertex and no edges, it is always a tree. Therefore, for any , one has .
Suppose that the result is true for any of length and fix a partition of length .
Let (resp. ) be the set of permuted thorn trees of type , verifying (resp. verifying and ). With these notations, is defined as the quotient
It will be convenient to consider marked permuted thorn trees, i.e. permuted thorn trees with a marked black vertex different from . The marked vertex will be denoted and the corresponding edge . We denote (resp. ) the set of marked permuted thorn trees of type , verifying (resp. verifying and ). To each permuted thorn tree of type corresponds exactly marked permuted thorn trees, so:
Let us now split these sets depending on the degree of the marked vertex:
where denote the subset of of trees with a marked vertex of degree . By Lemma 5.7 (see next paragraph), one has:
Let us consider an element of . We distinguish two cases:

either the end of the edge leaving in the graph is itself. In this case, the graph contains a loop and the element is not in .

or it is another vertex of the tree. We call such marked permuted thorn trees good. We will prove below (Lemma 5.9) with the induction hypothesis that, in this case, exactly elements over are in .
By Lemma 5.8, the second case concerns exactly elements over . Therefore:
and we can compute as follows
5.3 Technical lemmas
Let be a partition of size and length .
Lemma 5.7.
For all ,
Proof.
Consider the action of on consisting in permuting the black vertices (with their thorns). In each orbit and hence in the whole set , the proportion of elements for which the leftmost black vertex has degree is . To each element in correspond exactly elements in obtained by choosing a marked vertex among the black vertices different from . Therefore the probability that has degree is also , which is what we wanted to prove.
Note that this is not true if we consider elements with property as the action of does not preserve this property. ∎
We denote by the set of good marked permuted thorn trees of type , for which is a vertex of degree .
Lemma 5.8.
Proof.
Consider the action of on consisting in changing the cyclic order of the edges and thorns incident to the root without moving the leftmost edge. In each orbit of this action, the edge or thorn just after is uniformly distributed among the edges and thorns incident to the root and different from . Among these edges and thorns, there are thorns which are associated by to a thorn incident to the black vertex . By definition, an element in is good if and only if is not one of these thorns, therefore, in each orbit, the proportion of good elements is . ∎
Recall that any marked permuted thorn tree verifying property is good. In other terms, is a subset of .
Lemma 5.9.
We assume that, for of size and length , the proportion of permuted star thorn trees of type verifying among those which verify does not depend on . We denote this proportion . Then one has:
Proof.
Consider the following application
where is obtained as follows. Consider the edge or thorn immediately to the left of and denote the black extremity of the element with the same symbolic label. Then, starting from , erase the marked black vertex with its edge and move its thorns to the black vertex (at the right of its own thorns). For example,
This application has nice properties:

it preserves property . Indeed, if , then is obtained form by contracting its edge attached to the vertex .

the number of preimages of a given permuted star thorn tree depends only on its type . Indeed, there are no preimages if is not of the form for some (from now on, we use the notation ). Otherwise, the preimages are obtained as follows: choose a vertex of of degree (there are possible choices), choose the edge or a thorn of white extremity associated to one of its leftmost thorns ( choices per vertex ), add a new black vertex just at the right of this element and attach the last thorns of to this new vertex. With this description, it is clear that the cardinality of preimage is .
Recall that we assumed the number ( dependent only on and , but not on ) to be the proportion of permuted star thorn trees of type verifying among those which verify . With the two above properties, we can compute the proportion of elements verifying in . Indeed,
which is exactly what we wanted to prove. ∎
6 Link between Theorems 1.4 and 1.5
The goal of this section is to prove the equivalence between Theorem 1.4 and Theorem 1.5. This will be done using differential calculus in the symmetric function ring : we present this algebra in paragraph 6.1. Then, in paragraph 6.2, we explain how the generating series of blackpartitioned maps and maps are related. Finally, after a small lemma on thorn trees (paragraph 6.3), we use all these tools to prove the equivalence of Theorems 1.4 and 1.5 in paragraph 6.4.
6.1 Symmetric functions
Let us begin by some definitions and notations on symmetric functions. As much as possible we use the notations of I.G. Macdonald’s book [Mac95].
We consider the ring of symmetric polynomials in variables . The sequence admits a projective limit , called ring of symmetric functions. This ring has several classical linear bases indexed by partitions.

monomial symmetric functions: for monomials we use the short notation . Then, we define
where the sum runs over all vectors which are permutations of (without multiplicities).
Remark. We use upper case for the monomial symmetric functions instead of the usual lower case because a lot of formulae in this paper involve multiplicities of some parts and monomial symmetric functions at the same time.

power sum symmetric functions: by definition
Besides, we consider the differential operator given by:
Let us compute the image by this operator of the symmetric polynomials and . If , we denote the set (without multiplicities) of all vectors obtained by a permutation of the vector of size .
where is the vector of length , whose components are all equal to , except for its th component which is equal to . It is clear that, if is a permutation of a partition , then is a permutation of some for .
We will group together terms with the same exponent. So the question is: given a vector , which is a permutation of , in how many ways can it be written as