Tiered trees, weights, and Eulerian numbers
Abstract.
Maxmin trees are labeled trees with the property that each vertex is either a local maximum or a local minimum. Such trees were originally introduced by Postnikov [12], who gave a formula to count them and different combinatorial interpretations for their number. In this paper we generalize this construction and define tiered trees by allowing more than two classes of vertices. Tiered trees arise naturally when counting the absolutely indecomposable representations of certain quivers, and also when one enumerates torus orbits on certain homogeneous varieties. We define a notion of weight for tiered trees and prove bijections between various weight 0 tiered trees and other combinatorial objects; in particular order weight 0 maxmin trees are naturally in bijection with permutations on letters. We conclude by using our weight function to define a new analogue of the Eulerian numbers.
Key words and phrases:
Maxmin trees, intransitive trees, Eulerian numbers, Eulerian numbers, nonambiguous trees, permutations2010 Mathematics Subject Classification:
Primary 05C05; Secondary 05A05, 05C311. Introduction
1.1.
Let be a tree with vertices labeled by the ordered set . We say is a maxmin tree if for any vertex , the labels of all its neighbors are either less than or greater than that of . Such trees were introduced by Postnikov [12]; he called them intransitive trees, since they satisfy the property that for any triple , the pairs and cannot both be edges of . These trees first appeared in the study of hypergeometric systems attached to root systems [9], and were later connected to a variety of combinatorial objects:

regions of the Linial hyperplane arrangement (the affine arrangement in defined by the equations , );

local binary search trees (labeled plane binary trees with the property that every left child has a smaller label than its parent and every right child has a larger label than its parent);

and semiacyclic tournaments (directed graphs on the set such that in every directed cycle, there are more edges with than with ).
1.2.
In this paper, we consider a generalization of Postnikov’s trees called tiered trees. Instead of two classes of vertices, maxima and minima, we allow more classes. More precisely, a tiered tree with tiers is a tree on the vertex set with a map from to an ordered set with elements. We require that if is a vertex adjacent to with , then .
Tiered trees naturally arise in two a priori unrelated geometric counting problems [11]:

counting absolutely irreducible representations of the supernova quivers (quivers arising in the study of the moduli spaces of certain irregular meromorphic connections on trivial bundles over [2]), and

counting certain torus orbits on partial flag varieties of type over finite fields, namely those orbits with trivial stabilizers.
These contexts also motivate defining a weight function on tiered trees with values in . We define weights of tiered trees and show that the weights are related to the Tutte polynomials of certain graphs (Theorem 2.6). We show that weight trees of the two extreme types of tiering functions—maxmin trees, which have two tiers, and fully tiered trees, in which every vertex lies on a different tier—are related to other combinatorial objects, namely permutations sorted by their descents and the complete nonambiguous trees of Aval–Boussicault–Bouvel–Silimbani [1]. Finally we use the weights of maxmin trees to define a weight for permutations, which leads to a version of Eulerian numbers different from those studied by Carlitz, Stanley, and Shareshian–Wachs.
1.3.
Here is a guide to the sections of this paper. In §2 we define tiered trees and their weights. We also discuss the connection between the weight and geometry in the case of counting torus orbits. Section 3 discusses the enumeration of tiered trees. The next two sections give combinatorial interpretations of the two extreme cases of weight zero trees: section 4 treats weight zero maxmin trees, and §5 treats weight zero fully tiered trees. Finally in §6 we use weighted maxmin trees to define a notion of weight for permutations and our Eulerian polynomials. We conclude with a discussion of some open questions.
2. Tiered trees and weights
2.1.
For any let be the finite set . Let be a graph with vertices labeled by . We say is tiered with levels if there is a surjective function such that for any adjacent vertices , and such that if are adjacent and , then . We call a tiering function and say is tiered by . Given , we say that the vertices with are at tier .
Any tiered graph determines a composition of into parts by putting . We call the tier type.
2.2 Definition.
A tiered tree with tiering function is a labeled tree tiered by .
We say a tree is fully tiered if , and is a maxmin tree if . For maxmin trees, it is natural to call the vertices in the top tier maxima and those in the bottom tier minima. Figure 1 shows all the maxmin trees on vertices, grouped according to the number of maxima they have. Similarly, Figure 2 shows the fully tiered trees on vertices.
For any tier type let be the corresponding set of tiered trees, and let . Our goal now is to define a weight function .
2.3 Definition.
Let have tiering function . The weight of is defined as follows:

If , i.e. if consists of a single vertex, we put .

If , let be the vertex with the smallest label with respect to , and let ,…, be the connected components of the forest obtained by deleting from . We put
where the integers are defined as follows. For each , let be the vertex that was connected to . We have since was minimal. We define to be the order of the set
(1) Thus records the (zeroindexed) position of the vertex in the ordered list of all those vertices in that could have been connected to .
For example, one can check that all the trees in Figure 1 have weight except for the tree on the bottom right, which has weight . It is easy to check that there is a unique maxmin tree on vertices with maximum and a unique one with maxima, and that the weights of both of these trees are . Similarly, the weights of the trees in Figure 2 are all except for the upper right tree, which has weight . Table 1 gives a short table of the polynomials
for various tier types .
2.4.
Let be a tiered tree with respect to . The pair determines a unique maximal tiered graph containing as a spanning tree. We simply add all edges to between vertices with and . We call the complete tiered graph of order with respect to . By construction is connected. It turns out that the weight of can be interpreted via the Tutte polynomial of . We recall the definition of this polynomial in the form we need (cf. [10, Exercise 15.15]).
Let be a graph with a total ordering of its edge set, let be a spanning tree, and let be an edge of . We say that is internally active with respect to if it is contained in and is the least element in the cut of determined by and . We say that is externally active with respect to if it is not contained in and if it is the least element in the unique cycle determined by . The number of edges internally (respectively, externally) active with respect to is called the internal (resp., external) activity of in .
2.5 Definition.
The Tutte polynomial is defined by
where is the number of spanning trees of of internal activity and external activity .
2.6 Theorem.
Let be a tiered tree, and let be the complete tiered graph containing it. Then the weight is equal to its external activity in , where the edges of are ordered lexicographically.
Proof.
We first prove that, in the algorithm described in Definition 2.3, is equal to the number of externally active edges in that connect to an element of .
To see this, let be a vertex satisfying
and consider the edge of connecting to . Because is a tree and is connected, we have that (for if it was an edge in then would contain a cycle). This unique cycle of contains , , and no other edges containing . Since and are the first two edges in the cycle lexicographically, is externally active if and only if , i.e. lies in the set from (1). Hence the cardinality of is equal to the number of externally active edges connecting to , and our claim is proved.
The full result now follows from our claim, because the sum of the ’s obtained after deleting the vertex will now equal the number of externally active edges that have as the lower of their two vertices. ∎
2.7.
We conclude this section by giving more details about the geometric origin of the weight function. We focus on the connection to counting torus orbits, which is simpler to explain than the connection to representations of quivers. For more details, we refer to [11].
Let be a prime power and let be the finite field with elements. Let be the grassmannian of planes in . The maximal torus acts on via the standard action of on . Let us call a orbit maximal if its stabilizer is trivial.
Let be the number of maximal orbits in , as a function of . Then we have (cf. [11, Theorem 3.15])
(2) 
where the sum is taken over tiered trees of type .
For instance, the variety is just the projective space , which is a toric variety. The unique dense orbit in is maximal and is the only maximal orbit. The equation (2) becomes , and indeed there is a unique maxmin tree with vertices and one maximum (cf. Figure 1 for ).
For a more complicated example, consider the Grassmannian . This is not a toric variety and has a more intricate collection of orbits. See for instance [8, Figure 1], which over shows the images of the different types of orbits under the moment map; the maximal orbits are those whose moment map image is fulldimensional.
To count the orbits over , we can picture planes in a dimensional vector space via lines in . We claim there are two types of maximal orbits, depicted in Figure 3). One, as seen on the left, is given by lines in not passing through any of the four fixed points in (the solid black dots). Any such line determines four points (the grey dots), which are the intersection points of with the fixed planes. The action preserves the crossratio of these points; since the points are distinct, is an element of different from or . Thus there are orbits of this type. The other type of orbit corresponds to points passing through a single fixed line and not meeting a fixed point. Any such line lies in a unique maximal orbit determined by the fixed line it meets. Thus we have such orbits. The expression (2) becomes , which agrees with Figure 1.
We remark that although (2) enumerates maximal orbits, there is not a bijection between maxmin trees and the orbits. We also remark that general tiered trees can be used to count orbits in partial flag varieties, where the partition type of the flag corresponds to the tier type of the tiering function. Table 1 gives a short list of the corresponding orbit counts.
Remark.
One can show, using geometric results from [11], that the polynomials depend only on the partition determined by the tier type , and not the order of its parts. Thus Table 1 gives a complete list of the generating functions for weighted tiered trees with up to vertices. We do not know an elementary proof of this fact.
Tier type  

3. Counting tiered trees
In this section we will show how to enumerate tiered trees; our approach closely follows that of Postnikov [12].
To this end, it will be convenient to extend the notion of tiered trees somewhat. Let be a tree equipped with a tiering function , where . If is surjective, we will say that is properly tiered. Otherwise, if is not surjective, we will say that is improperly tiered if at least two elements of are in the image of . (Thus we never consider tiering functions with one element in their image.) In particular, if is improperly tiered and has the maximum number of empty tiers, then canonically has the structure of a maxmin tree.
Let be the number of properly and improperly tiered trees on vertices with tiers. For fixed let be the exponential generating function of the :
3.1 Proposition.
The generating function satisfies the functional equation
Proof.
It is convenient to consider rooted trees instead. Let us say that a rooted tiered tree is rooted if its root lies in the th tier. Then one can easily show that given and , the total number of rooted trees is
(3) 
Let us put , and let be the exponential generating function for the :
Then using (3), we have
(4) 
In particular, the generating function is independent of .
Now we consider how to create a tiered tree on vertices with tiering function taking values in . We claim that will satisfy the relation
(5) 
This is a consequence of the following. Consider the process of assembling a (possibly improper) tiered tree with root from a rooted tiered forest on vertices. To do this we should first place the root in a given tier, and then should consider all possible ways of joining the root to the roots of the connected components of the forest. It is clear that the number of ways to do this will depend on which tier contains . For instance, if the top tier will contain the root, then we can assemble the forest by joining to any root we like. On the other hand, if root lies in some other tier, then we can only join it to vertices lying in lowerlying tiers.
These considerations, together with the standard exponential formula for generating functions, lead to the expression (5). In particular the term
in (5) counts the contribution of those rooted tiered trees with root in the tier .
Finally, using the fact shown above that all are equal, say to , and applying (4) we find
as desired. ∎
3.2 Theorem.
The total number of tiered trees (proper and improper) with vertices and with tiers is given by
(6) 
We remark that substituting into (6) yields Postnikov’s expression for the number of maxmin trees with vertices:
Proof.
By Proposition 3.1, we have
(7) 
Let . Then , and substituting in (7) we find
We can now use Lagrange inversion to find the coefficients of . Recall that this says that a formal power series , has compositional inverse whose coefficients satisfy
where for any formal power series denotes the coefficient of (cf. [18]). Applying this to our series, we find
But the coefficients of are just those of shifted, which implies
This completes the proof. ∎
Let denote the properly tiered trees on vertices with tiers. We can easily evaluate using the principle of inclusionexclusion:
3.3 Proposition.
We have
(8) 
Proof.
The proof is by induction on the number of tiers . We have by definition. For , the only possible improperly tiered trees are those that take up two tiers, leaving the third empty. Thus we have , and since , we find .
In general, we have
To pass to (8), one substitutes the previously obtained expressions for , and uses elementary properties of binomial coefficients. ∎
We conclude this section by giving a short table (Table 2) of the numbers and . We remark that the can also be recovered from the polynomials in Table 1 by substituting and combining various terms. For instance , and .
3  1  0  0  4  4  306  60  6  1  0  0 
3  2  2  2  5  1  0  0  6  2  246  246 
3  3  11  5  5  2  36  36  6  3  8868  8130 
4  1  0  0  5  3  693  585  6  4  80496  46500 
4  2  7  7  5  4  4304  1748  6  5  400200  83940 
4  3  72  51  5  5  16274  1324  6  6  1414050  46620 
4. Weight zero maxmin trees and Eulerian numbers
Define a polynomial by
where ranges over all maxmin trees with vertices, and for any tree , is its weight and is its number of maxima. Using Table 1, the first few polynomials are
If we set , we find
(9) 
where is the Eulerian number (the number of permutations in with descents). Thus is essentially the Eulerian polynomial.
In fact, it is not difficult to prove the relationship (9) using generating functions. More precisely, the generating function of the Eulerian numbers is well known. Postnikov [12] enumerated all maxmin trees on vertices, which means he computed the specialization . One can modify his result to include the weight parameter , and then by setting finds the generating function for the Eulerian numbers.
Our goal in this section is to give a bijective proof of (9). This has the advantage of revealing a connection between permutations and maxmin trees that was previously unknown. It will also enable us later (§6) to define a analogue of the Eulerian numbers.
4.1 Theorem.
There is a bijection between permutations in with descents and weight maxmin trees on vertices with maxima.
Proof.
We define the bijection recursively. We begin by taking the identity permutation in to the unique maxmin tree on vertices with maximum. Similarly we take the longest element of , the decreasing permutation, to the unique maxmin tree on vertices with maxima.
For the rest of , we regard as embedded in as follows. We take any and represent it as an ordered sequence of . We then adjoin an on the the right and obtain an element of . In other words, we identify with the parabolic subgroup of generated by all simple transpositions other than . We will in fact construct a bijection between permutations in this subgroup of and maxmin trees on vertices.
Thus given represented as above as an ordered sequence on with on the right, we break up into the two subsequences appearing to the left and right of : we write
where the operator denotes concatenation. Note that may be empty, but contains at least the symbol . We further break into a collection of subsequences
as follows. Let be the maximum symbol appearing in . Then consists of the subsequence of from the beginning to . We then let be the maximum of what’s left after deleting , and let be the corresponding subsequence up to . We continue this process until is exhausted.
The result is a collection of ordered sequences and with the following property: the global maximum of each appears in the rightmost position. By induction on the number of vertices, we know how to build a weight maxmin tree with the correct number of maxima and vertices for each of these subsequences (after relabeling them to preserve their permutation patterns). We take the resulting forest of weight maxmin trees and join them together into a maxmin tree for by joining to the smallest maximum available in each connected component. The resulting tree clearly has weight . It is easy to verify that this construction is a bijection with the stated property. ∎
4.2 Example.
Consider the element given by the sequence 8594673201 (we use characters 0,…,9 for for this example). Note that this permutation has five descents, which occur at 8, 9, 7, 3, and 2. We are thus building a maxmin tree with six maxima. We identify with the permutation in obtained by adjoining the character A (hexadecimal ) on the right: . We have
The left permutation becomes broken into four pieces
which means that we must connect vertex to five maxmin trees. The resulting maxmin tree, which has six maxima, is shown in Figure 4.
4.3.
We conclude this section by connecting the decomposition into subsequences in Theorem 4.1 with the Stirling numbers of the first kind.
4.4 Theorem.
The number of permutations in that break into blocks (subsequences) in the decomposition in Theorem 4.1, disregarding the singleton block with the minimum element, equals where is the signless Stirling number of the first kind.
Proof.
The signless Stirling number of the first kind, , counts permutations in with cycles when they are written in disjoint cycle notation. Given such a permutation we construct different permutations in , each of which breaks into blocks as in the proof of Theorem 4.1. Namely, we can insert after any one of the letters in the cycles, or else as a cycle of its own. Once we have done that we write the cycle containing 0 with the 0 first and make that the last cycle in our ordering of the cycles. The remaining cycles we write with their maximum letter last, and order them in decreasing order of their maxima (followed by the cycle containing 0). Erasing the parentheses from the word thus constructed we obtain a permutation in whose blocks are precisely the cycles we started with. It is easy to see that this constructs distinct permutations, and that any permutation in is constructed this way. ∎
4.5 Example.
If in the permutation with cycle decomposition we insert after the we get the following cycles, ordered as in the proof of Theorem 4.4: . Erasing the parentheses gives the permutation , which decomposes into the blocks . Inserting as its own block, on the other hand, gives , which yields .
5. Weight zero fully tiered trees and complete nonambiguous trees
5.1.
A tiered tree is called fully tiered if all its vertices sit on distinct levels, i.e. it has tier type . In this section we describe a bijection between weight fully tiered trees and the complete nonambiguous trees defined by Aval–Boussicault–Bouvel–Silimbani [1]. We begin by defining the latter objects, following [1].
5.2.
Let be the grid of nonnegative lattice points. Given any , let be its coordinates. For later convenience, in figures we represent points in by putting in the upperleft corner, having the coordinate increase from left to right, and having the coordinate increase from top to bottom. A nonambiguous tree is a subset satisfying the following properties:

. This is the root of the tree.

Let be different from the root. Then there exists a unique point with either (i) and or (ii) and . These possibilities are exclusive: the point must satisfy one or the other.

there is no empty line between two given points: if there exists a point such that (resp. ), then for every (resp. ) there exists such that (resp. ).
Any subset satisfying these properties determines a unique binary tree embedded in with edges running along the grid lines. More precisely, if is a point different from the root, then the parent of is the unique point preceding in the same row or column; by condition (ii) above, is uniquely determined. This motivates the name “nonambiguous:” the tree structure on can be recovered from the vertex coordinates even if the edges are missing. Figure 5 shows four different nonambiguous trees with vertices.
5.3.
The trees in Figure 5 have the further property of being complete. A complete nonambiguous tree is one that is complete in the usual sense, that is, every point has or children. Any such tree has vertices, of which are leaves and of which are internal.
Let , be the number of complete nonambiguous trees with internal vertices. The sequence , which begins
(10) 
appears in OEIS [15] as sequence A002190, and is related to the logarithm of the Bessel function:
(11) 
According to [1], complete nonambiguous trees provided the first combinatorial interpretation of the sequence A002190, which was originally defined through (11). The following theorem provides another combinatorial interpretation of this sequence:
5.4 Theorem.
There is a bijection between weight fullytiered trees on vertices and complete nonambiguous trees with internal vertices.
Proof.
We construct the bijection by induction. The unique complete nonambiguous trees with and internal vertices correspond respectively to the unique fully tiered trees with and vertices.
A complete nonambiguous tree with internal vertices sits inside the subsquare of . Each row and each column contains a single leaf of . The tree determines a labeling of the levels of the associated fully tiered tree as follows: if a leaf appears at row (numbering from the bottom of the figure to the top) and column , then at level in we place the vertex (see Figure 6).
Now consider erasing the first column of , along with all the edges connecting vertices in this column to nodes in greater columns. The result is a forest of complete nonambiguous trees, although each is embedded with roots at different points of , and such that their vertex labels are various subsets of . By flattening their vertex labels in the obvious way and shifting the roots, each of these trees corresponds to a complete nonambiguous tree. Thus by induction we know the fully tiered trees to which they correspond. This allows us to add the edges they determine to (using the original vertex labels, not their flattened versions) to obtain a fully tiered forest. To complete to a weight fully tiered tree, we need to connect the vertex to each connected component, and in doing so we choose the minimal vertex we can for each component. The resulting fully tiered tree clearly has weight (see Figure 7 for a complete example).
We claim this is a bijection. To go backwards, we reverse the process. We take a fully tiered tree and delete the vertex . This gives a forest of fully tiered trees with fewer vertices. We can build complete nonambiguous trees from them and can embed them into using the disjoint sets of rows and columns determined by their labels and by the levels on which the labels appear. The final step is to add the root and the necessary edges from the first column heading right to the roots of the smaller trees. Note that these smaller complete nonambiguous trees can be interlaced to form a complete nonambiguous tree exactly because their labels and levels in form disjoint subsets. ∎
6. Weights of permutations and Eulerian numbers
Theorem 4.1 shows that order weight maxmin trees with maxima are in bijection with elements of with descents. What can one say about trees of higher weight? In fact, the proof of Theorem 4.1 shows that any maxmin tree of any weight is built from an underlying permutation. Indeed, in the bijection between trees and permutations, one builds a weight tree from a permutation by extracting subpermutations, building weight trees from them, and then by connecting all the smaller trees to the minimal vertex canonically: one joins the minimal vertex to the smaller trees by picking maxima with the smallest labels.
Thus it is clear how to modify this construction to produce a tree of higher weight. At any stage, one can connect the current minimal vertex to a larger maximum instead of choosing the smallest maximum available. For instance, in Example 4.2, one could connect to instead of to make a tree of weight . Moreover, this is the tree of largest weight one can make from this permutation. This leads to the following definition:
6.1 Definition.
Let . Then the weight of is the largest possible weight of maxmin tree that can be constructed from . More precisely, the function is defined by the following recursion:

If is the identity or the longest element in , then we define .

Otherwise, regard as an element of with the symbol appearing on the right, as in the proof of Theorem 4.1, and let , …, , be the subpermutations extracted there. Then we define
where (respectively, ) is the number of descents in (respectively, ).
6.2 Example.
Let us consider the permutation . We can complete to by putting a new maximal element on the right. Extracting subsequences around the minimal element , we find
Computing proceeds as follows:

flattens to , which is the identity. We have and .

flattens to , which is the completion of the longest word. We have and .

flattens to . A short computation shows (compare with Figure 1 above; the tree on the bottom right corresponds to this permutation). We have .
Thus .
6.3.
With our definition of weight, we can define a analogue of the Eulerian polynomials. Put
Some examples of are given below.
6.4.
We pause to compare the polynomials with other Eulerian polynomials in the literature. Let and for let
Carlitz [4] defined polynomials as follows. First we define by the expansion
and then set . The result is still an integral polynomial in . Then Carlitz’s Eulerian polynomial is defined by
For example,
Stanley [17] defined a different Eulerian polynomial as follows. For any permutation , let be the number of inversions: if is written as an ordered list , then counts the number of pairs with and . Then Stanley defined
For example,
Finally, Shareshian–Wachs [14] generalized Stanley’s definition by considering four different sums of the form
where and are various statistics of permutations; one example yields Stanley’s polynomial.
6.5.
From these examples it is clear that our is quite different. Indeed, there are many interesting problems to investigate about the combinatorial meaning of these polynomials. We indicate some below, and for the present purposes prove the following results:
6.6 Theorem.
There is a bijection between weight 0 permutations on letters with descents and partitions of into subsets. Thus such permutations are counted by , the Stirling numbers of the second kind.
Proof.
We describe an algorithm that constructs a weight 0 permutation given any partition of the set (cf. Example 6.7), and will prove that this provides our bijection.
Let be a partition of the set into subsets. Sort each subset into increasing order, and denote the resulting ordered subsets by . We assemble these subpermutations into a permutation as follows. Let be the sublist containing the minimum element 1, and place it in the rightmost position in . Next, sort the remaining by their respective maxima, and arrange them in order such that their maxima decrease. Finally, adjoin this string of subpermutations to the left of to complete . Because we sorted each and ordered them with their maxima decreasing, the final permutation has descents.
We also claim that has weight 0. To see this, first observe that each in has no descents, so each one has weight 0. Moreover, each tree constructed from the contains only a single maximum, so there is no choice as to how we combine them together to construct the tree corresponding to : the maximum in each subtree must be connected to the global minimal element. Hence the maxmin tree corresponding to has weight 0, and this is the maximum possible weight it could have had.
Finally, notice that the correspondence described is a bijection between partitions of and weight 0 permutations: the steps outlined above demonstrate that each partition determines a unique permutation. Conversely, given the permutation we can recover the original partition of by splitting the permutation at its descent points. This completes the proof. ∎
6.7 Example.
Consider the partition of into parts given by 25, 6130,