# On posets and polytopes attached to arbors

Frédéric Chapoton

► To cite this version:

Frédéric Chapoton. On posets and polytopes attached to arbors. Mathematica Scandinavica, 2025, 131 (3), pp.401-449. 10.7146/math.scand.a-159563. hal-04978520v2

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# On posets and polytopes attached to arbors

F. Chapoton

August 22, 2025

À la mémoire des petits bilbis.

## Abstract

Starting from the data of an arbor, which is a rooted tree with vertices decorated by disjoint sets, we introduce a lattice polytope and a partial order on its lattice points. We give recursive formulas for various classical invariants of these polytopes and posets, using the tree structure. For linear arbors, we propose a conjecture exchanging the Ehrhart polynomial of the polytope with the Zeta polynomial of the poset for the reverse arbor. The general motivation comes from a transmutation operator acting on $M$-triangles, which should link the posets considered here with some kind of generalized noncrossing partitions and generalized associahedra. We give some evidence for this relationship in several cases, including notably some polytopes, namely halohedra and Hochschild polytopes.

## Introduction

The combinatorial data that will be our starting point is a rooted tree whose vertices are decorated with a partition of a finite set. Such objects will be called *arbors*. This article introduces two constructions attached to any arbor $t$. The first construction is a lattice polytope $Q_t$ inside some $\mathbb{N}^n$ and the second one is the partial order $P_t$ defined by coordinate-wise comparison on the set of lattice points in this polytope.

Before explaining at length the motivation for these constructions coming from Coxeter groups, cluster theory and representation theory, let us describe our results.

First, it is shown that the polytopes $Q_t$, initially defined by inequalities, are lattice polytopes, which means that they have integral vertices. Moreover, they are simple polytopes and have no interior lattice point. Their vertices are described in a recursive way, using a projection map $\pi$ to a product of smaller polytopes in the same family. The polytopes are also identified as Minkowski sums of simplices. Basic properties of the posets $P_t$ are also stated: they are graded with a unique minimum and the interval between this minimum and any element is a product of total orders. This property allows to view their Hasse diagrams as skeletons of cubical complexes.

Our next set of results consists of recursive formulas to compute several invariants of the polytopes $Q_t$ and posets $P_t$. This uses the projection map $\pi$ and often requires a finer invariant that includes information about the height, which is the sum of coordinates. A recursive formula is obtained for the Ehrhart polynomial of the polytope $Q_t$, which counts lattice points in the dilates, cf. [Bar08].
One considers then the function that measures the volume in the hyperplane sections according to height, and gets recursive formulas for this function and for its Laplace transform. Similarly, recursive formulas are given for the Zeta polynomial, the *M*-triangle and the *f*-vector of the poset *P*<sub>*t*</sub>. The Zeta polynomial is a classical invariant that counts chains in the poset, see [Sta74, §3]. The *M*-triangle is a polynomial in two variables, defined in [Cha04] for any graded poset, as a weighted sum of all Möbius numbers, similar to the more classical characteristic polynomial introduced in [Rot64]. The *f*-vector is defined by counting faces of the underlying cubical complex.

In Section 3, two conjectures are proposed in the case of linear arbors. The first one claims that reversing the arbor upside-down should exchange the Ehrhart polynomial of the polytope *Q*<sub>*t*</sub> and the Zeta polynomial of the poset *P*<sub>*t*</sub>. Here both constructions interact in a rather surprising way. The second conjecture claims that the *h*-vector is unchanged by reversing a linear arbor. Here the *h*-vector of an arbor *t* is the generating polynomial of elements of *P*<sub>*t*</sub> according to the number of non-zero coordinates.

In Section 4, we introduce transmutation as a birational involution acting on the *M*-triangle. Transmutation is central in the original motivation of the article and appears in the results in the next sections. Here the reader may want to read the motivation section below first.

In the five next sections, we gather evidence concerning transmutation for several specific families of arbors. Let us say that two posets are *transmuted partners* if they share the same Zeta polynomial and their *M*-triangles are related by transmutation.

In Section 5, we propose conjectures about unary-binary arbors. In this case, one can define from the arbor *t* a gentle quiver-with-relations whose lattice of support *τ*-tilting objects is finite. Applying the core-label-order construction should lead to transmuted partners of the posets *P*<sub>*t*</sub>. This is also stated in the more combinatorial language of Stokes posets for quadrangulations [Bar01, Cha16].

In Section 6, we consider the case of arbors of type A, meaning linear arbors with no multiple vertex. We introduce a more general family of posets, including the posets *P*<sub>*t*</sub> for arbors of type A. We compute the Zeta polynomials and the *M*-triangles of these more general posets. This implies that noncrossing partitions of Coxeter type A are transmuted partners of the posets *P*<sub>*t*</sub> for arbors of type A.

In Section 7, we consider the case of arbors of type B, meaning arbors with only one vertex. We introduce a more general family of posets, including the poset *P*<sub>*t*</sub> for arbors of type B. We compute the Zeta polynomials and the *M*-triangles of these more general posets. This implies that noncrossing partitions of Coxeter type B are transmuted partners of the posets *P*<sub>*t*</sub> for arbors of type B.

In Section 8, we turn to arbors with one simple vertex and one multiple vertex. The aim is to relate them to the family of simple polytopes named halohedra [DHV11, DFHV12]. We prove that the *h*-vector of the polytope coincides with the *h*-vector of the arbor.

In Section 9, we turn to arbors that are corollas, namely with no multiple vertex and such that all non-root vertices are attached to the root vertex. The aim is to relate these arbors to the family of Hochschild polytopes [San09, RS18] and the associated Hochschild lattices [Com21]. We conjecture that the core-
label-orders of Hochschild lattices are transmuted partners of the posets $P_t$. We prove that the $h$-vector of the Hochschild polytope coincides with the $h$-vector of the arbor.

## 0.1 Motivation

Let us now try to explain, with the help of the following figure, where this work comes from.

<figure><img src="image_2.png" /></figure>

Let us start with the leftmost box. The theory of Cambrian lattices [Rea06] defines, for any finite Coxeter group $W$ and Coxeter element $c \in W$, a finite lattice. For crystallographic simply-laced Coxeter groups, these lattices can also be described inside the theory of cluster algebras, using green mutations, see [KD20] for a survey. There is an alternative representation-theoretic viewpoint using inclusion of torsion classes for finite-dimensional algebras, considered also using support $\tau$-tilting objects [AIR14]. It is known that finite lattices of torsion-classes are congruence-uniform [GM19, DIR$^+$23]. It is also known that Cambrian lattices are congruence-uniform, as quotients of the weak order on the Coxeter group [Rea06].

Let us move to the top box, closely tied to the previous one. It is known that the Hasse diagrams of Cambrian lattices are skeletons of some simple polytopes, sometimes called generalized associahedra [CFZ02]. In this setting, one can define the $F$-triangle, a polynomial in two variables that refines the $f$-vector of the polytope [Cha04] and depends on the choice of a distinguished vertex. Its definition, as recalled in Appendix A, is applied to the polar polytope seen as a spherical simplicial complex with a distinguished facet.

Consider now the bottom box. Starting from a congruence-uniform lattice in the leftmost box, one can define another lattice, using the canonical join representation. This construction was introduced by Reading in [Rea16, §9.7] under the name of *shard-intersection-order*. This is also known as the *core-label-order*, in the terminology of [Müh19]. For Cambrian lattices, the core-label-order only depends on $W$ and recovers the $W$-noncrossing-partitions lattice of Brady-Watt [BW02] and Bessis [Bes15]. As these lattices are graded, one can define their $M$-triangles as recalled in Appendix A.

In the Cambrian context, the $F$-triangle of the generalized associahedra is related to the $M$-triangle of the lattice of $W$-noncrossing-partitions by the explicit
birational transformations (57) and (56) recalled in Appendix A. This equality was conjectured in [Cha04] and proved in [Ath07]. These birational transformations are expected to be more widely applicable in similar contexts, see for example [GM20]. Note that this birational relationship implies that *h*-vectors of generalized associahedra also enumerate non-crossing partitions according to their rank.

All this forms a well-traveled landscape, involving congruence-uniform lattices and directed skeletons of polytopes. Our main claim is that there should exist another aspect of this story, related to the previous landscape by transmutation.

The *core idea* is that some posets from the bottom box (noncrossing partitions and similar posets) should be *transmuted partners* of some posets from the middle box (associated with arbors). By this expression, we mean that two posets share the same Zeta polynomial and that their *M*-triangles are related by an explicit birational involution that we choose to call *transmutation*.

The starting point for the project was in fact in the other direction: starting from the classical lattices of noncrossing partitions, we have found their transmuted partners. This has led us to the present framework of arbors.

We will prove in Section 6 that for linear arbors in which every vertex has only one element, the transmuted *M*-triangle is indeed the *M*-triangle of the noncrossing-partitions lattice of type A. Moreover, we prove that they share the same Zeta polynomial, so they are indeed transmuted partners as claimed.

In a similar vein, it is shown in Section 7 that the noncrossing partitions lattices of type B are transmuted partners of the posets $P_t$ for arbors with one vertex.

<figure><img src="image_3.png" /></figure>

Having introduced a general theory on the arbor side, it is natural to ask if one can use transmutation in the other direction.

The best hope would be the following statement for every arbor $t$:

There should exist a simple polytope $\Phi_t$, some way to orient its skeleton to find the Hasse diagram of a congruence-uniform “Cambrian-like” lattice $\Lambda_t$ and some “noncrossing-like” lattice $NC_t$ built from $\Lambda_t$ by the “core-label-order” construction. The lattice $NC_t$ should be a transmuted partner of $P_t$. If the arbor $t$ has size $n$ and the poset $P_t$ has $N$ elements, then $\Phi_t$ should be a simple polytope of dimension $n$ with $N$ vertices.

There is so far no idea on how to achieve that in general. A proposal for unary-binary arbors is made in Section 5 in terms of gentle algebras.

A much weaker statement would be to produce starting from $t$ the simple polytope $\Phi_t$ together with a distinguished vertex (as in the top box) whose *F*-triangle can be transformed by formulas in Appendix A into an *M*-triangle (as in the bottom box) that would be the transmutation of the *M*-triangle of $P_t$. In particular, Proposition 4.1 says that the *h*-vector of $t$ is then the diagonal of the *M*-triangle of the lattice $NC_t$, hence describes the grading of this poset and also the *h*-vector of the polytope $\Phi_t$.

Again, there is no known idea on how to do that. Two proposals are made in Section 8 and Section 9 that should relate in this way some arbors with known families of polytopes. The evidence is given by a coincidence of *h*-vectors, which would follow from the expected transmutation of *M*-triangles.
Assuming that we can produce the polytope $\Phi_t$ endowed with its distinguished vertex, one could then consider the simplicial complex defined as the dual of the simple polytope $\Phi_t$. One expects that this simplicial complex is flag, and moreover that its Gamma-triangle with respect to the distinguished facet, as defined in [Cha20a], is nonnegative. This holds for all arbors of size at most 13.

Even weaker is the following conjecture, keeping only minimal information.

**Conjecture 0.1.** For every arbor $t$, there exists a simple polytope $\Phi_t$ such that the h-vector of $t$ is equal to the h-vector of $\Phi_t$.

## 0.2 Further remarks

The posets that we attach to arbors of type $\mathbb{A}$ are special cases of the *avalanche posets* introduced by Combe and Giraudo in [CG20, CG22]. Despite their very simple nature and the fact that their cardinality is a Catalan number, they do not seem to have been studied much.

The polytopes that we attach to arbors of type $\mathbb{A}$ have Catalan-many lattice points. They in fact coincide with a special case of the *Pitman-Stanley polytopes* introduced in [SP02], when all parameters in the definition are set to 1.

Our notion of pairs of posets being transmuted partners seems to have some relationship, still to be clarified, with the notion of *Whitney duality* introduced in [GDH21]. Whitney duality relates pairs of posets that exchange their Whitney polynomials of the first and second kind. Some examples involve noncrossing partition lattices. One can in particular note the intriguing fact that the sub- poset of the poset NCDyck$_4$ of [GDH21, Fig. 9] defined by restriction to indices in the order 1, 2, 3, 4 is isomorphic to the poset $P_t$ for the arbor of type $\mathbb{A}$ on 3 elements.

Note that our notion of *arbor* has been considered in the literature under various names, including “multilabelled trees” and “rooted trees of non-empty sets” [MSS99, BOS05].

In our examples of *transmuted partners*, we observe more properties, for example the leading coefficients of the $q$-Zeta polynomials [Cha24] are reciprocal of each other as polynomials in $q$.

**Acknowledgments:** Thanks to Pierre Baumann, Alin Bostan, Marc Mez-zarobba, Philippe Nadeau and Vincent Pilaud for illuminating discussions and constructive suggestions. Thanks to the referee for a careful report and very pertinent information.

## 1 Notations and basic properties

Let us start by describing our main combinatorial objects, using the language of species. For an introduction to the theory of species, see [BLL98].

Let $S$ be the species of sets and $S_{\ge 1}$ be the species of non-empty sets. Let us consider the species $\mathcal{A}$ defined by the implicit equation

$$A = S_{\ge 1} \times S(\mathcal{A}).$$

In words, this means that the $\mathcal{A}$-structures on a finite set $I$ are rooted trees, whose vertices are labeled by pairwise disjoint non-empty subsets of $I$, and the full set of labels is $I$.
Such a rooted tree will be called an *arbor* on the set $I$. For example, here is an arbor on the set $\{a, b, c, d, e, f\}$:

<figure><img src="image_4.png" /></figure>

This example will be used as a running example.

Arbors will either be depicted with their root vertex on the top and growing down as above, or with their root vertex on the left and growing right. The cardinality of a vertex $v$ will be denoted $|v|$. If $|v| \ge 2$, then the vertex $v$ is *multiple*. The root vertex of an arbor $t$ will be denoted $r_t$.

For a vertex $v$ of an arbor $t$, any vertex $w$ such that the unique path in $t$ from $w$ to the root vertex $r_t$ passes through $v$ is called a *descendant* of $v$. This condition is denoted by $v \preceq w$. The set of all descendants of $v$ form the sub-tree rooted at $v$.

To every vertex $v$ of an arbor, one associates the union of all labels that appear in some descendant of $v$. This set is denoted $\mathcal{D}(v)$, standing for “down”. In the running example, the sets $\mathcal{D}(v)$ are $\{a, b, c, d, e, f\}$, $\{b, e\}$, $\{d\}$ and $\{c\}$.

When $|I| = n$, the arbor $t$ has size $n$.

To every arbor $t$ on a set $I$, we will associate both a partial order and a polytope, closely related to each other. Let us start with the polytope.

The ambient space is the vector space over $\mathbb{R}$ with coordinates $(x_i)_{i \in I}$. The polytope $Q_t$ is defined by the inequalities $0 \le x_i$ for every $i \in I$ and

$$ \sum_{i \in \mathcal{D}(v)} x_i \le |\mathcal{D}(v)| \quad (1) $$

for every vertex $v$. In particular, the inequality attached to the root vertex $r_t$ says that the sum of all coordinates is less than or equal to $|I|$, so that this indeed defines a bounded polyhedron.

In the running example, besides all coordinates being nonnegative, the defining inequalities are:

$$ x_c \le 1, \quad x_d \le 1, \quad x_b + x_e \le 2, \quad x_a + x_b + x_c + x_d + x_e + x_f \le 6. $$

The following property is clear from the definition.

**Lemma 1.1.** Points with all coordinates in the interval $[0, 1]$ are contained in $Q_t$. In other words, the polytope $Q_t$ contains the standard hypercube.

The definition of polytopes $Q_t$ by inequalities implies clearly that they are anti-blocking polytopes, in the sense of Fulkerson [Ful71], see also [Sch98, §9.3].

We will often use the height function $ht$ which is the sum of coordinates. Note that the height function is bounded above by $|I|$, which is attained at the point $(1, 1, \dots, 1)$.

Let now $t$ be an arbor on a set of cardinality $n \ge 1$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of
t. Then the projection $\pi$ along the coordinates in the root vertex to the other coordinates defines a map from $Q_t$ to the product $\prod_k Q_{t_k}$. This projection $\pi$ is surjective as the fiber over any point of total height $h$ in $\prod_k Q_{t_k}$ is the intersection of $\mathbb{R}^r_\ge 0$ with the half-space $\sum_{i \in r_t} x_i \le n - h$. Note that the total height on $\prod_k Q_{t_k}$ is less than $n - r$. In particular, the fibers are simplices of strictly positive dimension.

**Lemma 1.2.** *The polytope $Q_t$ is a simple lattice polytope.*

*Proof.* This is true for arbors of size 1, where $Q_t$ is the segment $[0, 1]$.

Assume that this is known for all arbors with size strictly less than $n$. Let $\pi(Q_t)$ denote $\prod_k Q_{t_k}$.

Let $z$ be a vertex of $Q_t$. The point $z$ must be an extremal point in the fiber $\pi^{-1}(\pi(z))$. So its coordinates with indices in the root vertex $r_t$ are either all 0 or exactly one of them is not 0 and equal to $n-h$, where $h$ is the height of $\pi(z)$. If these coordinates are all 0, then $\pi(z)$ must be a vertex of $\prod_k Q_{t_k}$. In this case, we obtain a simple vertex with integer coordinates by induction hypothesis.

In the case of exactly one non-zero coordinate, write $z = (\pi(z), n-h, 0, \dots, 0)$ in the coordinate system where indices in the root come last. Let us show that the point $\pi(z)$ must still be a vertex of $\pi(Q_t)$. Assume by contradiction that $\pi(z)$ is not extremal. Then there exist distinct $\tilde{z}_1$ and $\tilde{z}_2$ in $\pi(Q_t)$ whose middle is $\pi(z)$. Let $h_1, h_2$ be the heights of $\tilde{z}_1$ and $\tilde{z}_2$. Then the height $h$ of $\pi(z)$ is $(h_1 + h_2)/2$. Let $z_1 = (\tilde{z}_1, n - h_1, 0, \dots, 0)$ and $z_2 = (\tilde{z}_2, n - h_2, 0, \dots, 0)$ in the same coordinate system. Then both $z_1$ and $z_2$ are in $Q_t$, distinct, and their middle is $z$. Therefore $z$ is not extremal, which is absurd. Hence $\pi(z)$ is a vertex of $\pi(Q_t)$ and it has integer coordinates and integer height $h$ by induction. Therefore $z$ also has integer coordinates.

It follows that there are $r+1$ vertices of $Q_t$ in the fiber over each vertex of $\pi(Q_t)$. They all have integer coordinates. From the description above, one can also deduce that they are simple. $\square$

One can extract from the proof above a recursive description of the vertices of $Q_t$ themselves. Let us call defect of a point $z$ in $Q_t$ at a vertex $v$ the difference

$$ \delta_v(z) = |\mathcal{D}(v)| - \sum_{i \in \mathcal{D}(v)} x_i(z) \ge 0. \qquad (2) $$

**Lemma 1.3.** For each vertex $v$ of the arbor $t$ and for every vertex $z$ of the polytope $Q_t$,

* either all coordinates $x_i(z)$ with $i \in v$ vanish,

* or exactly one coordinate $x_i(z)$ with $i \in v$ does not vanish.

At every vertex $v$ where all coordinates vanish, the defect $\delta_v(z)$ is at least $|v|$. At every vertex $v$ where one coordinate is non-zero, the defect $\delta_v(z)$ must vanish.

This implies that, assuming all coordinates of a vertex $z$ of $Q_t$ for descendant vertices of a vertex $v$ of $t$ are fixed, the unique possible non-zero value in the vertex $v$ is therefore determined by the zero-defect condition, which means that the defining inequality (1) associated with $v$ must be an equality.

**Lemma 1.4.** *The polytope $Q_t$ has no interior lattice point.*
Proof. Every interior lattice point must have all coordinates $x_i > 0$, but the sum of coordinates must be an integer strictly less than $|I|$. $\square$

Let us now define the poset $P_t$. The elements of $P_t$ are the lattice points of $Q_t$. The partial order is defined by comparison of coordinates, namely $x \le y$ if and only if $x_i \le y_i$ for all $i \in I$. The poset $P_t$ has a unique minimum 0 and is graded by the height function $ht$.

From the definition of $P_t$, one deduces the following property.

**Lemma 1.5.** For every element $a$ in $P_t$, the interval $[0, a]$ is a product of total orders whose lengths are the non-zero coordinates of a plus one.

This property implies that one can consider the Hasse diagram as the skeleton of a cubical complex, by glueing together the cubical complexes for products of total orders.

In the running example, the polytope $Q_t$ is a 6-dimensional simple lattice polytope with $f$-vector $(36, 108, 141, 102, 43, 10, 1)$. The poset $P_t$ has 330 elements, which are the lattice points of $Q_t$. The cubical complex of $P_t$ has $f$-vector $(330, 990, 1152, 654, 186, 24, 1)$. A simpler example is illustrated in Figure 1.

<figure><img src="image_5.png" /><figcaption>Figure 1: Polytope $Q_t$ and poset $P_t$ for the arbor $\{x\} - \{y\}$.</figcaption></figure>

Note that both the polytope $Q_t$ and the poset $P_t$ are species, in the sense that every bijection $I \simeq J$ that induces a relabeling of $t$ also induces an isomorphism of polytopes and posets. We will from now on work with isomorphism classes of arbors.

## 1.1 Minkowski decomposition

Let $t$ be an arbor on the set $I$. Let us show that the polytope $Q_t$ admits a nice decomposition as a Minkowski sum of simplices.

Let us first describe the Minkowski summands. There will be a summand $U_v$ associated with every vertex $v$ of $t$.

For every vertex $v$, define a subset $\mathcal{W}(v)$ of $I$, standing for “up”, as the union of the sets attached to all vertices of $t$ in the unique path in $t$ from the root $r_t$ of $t$ to the vertex $v$. The polytope $U_v$ is the convex hull inside $\mathbb{R}^{\mathcal{W}(v)}$ of the origin and all $|v|e_i$ where $e_i$ are the basis vectors of $\mathbb{R}^{\mathcal{W}(v)}$. Note that it is a Minkowski multiple of the basic simplex whose vertices are the origin and the $e_i$.

In the running example, the subsets $\mathcal{W}(v)$ are $\{a, f\}$, $\{a, f, b, e\}$, $\{a, f, d\}$ and $\{a, f, c\}$.

Let

$$ M_t = \sum_v U_v \qquad (3) $$

be the Minkowski sum of these simplices. We will show that $M_t = Q_t$ by inclusion in both directions.
**Lemma 1.6.** Every point in $M_t$ satisfies all the defining inequalities of $Q_t$.

*Proof.* This is clear for the inequalities $x_i \ge 0$ for $i \in I$, because all Minkowski summands $U_v$ have nonnegative coordinates. So let us consider the inequality attached to a vertex $v$ of $t$. Fix an element $e$ in the Minkowski sum $M_t$, written as

$$e = \sum_{w} e_{w},$$

where $e_w \in U_w$. The inequality to check is

$$\sum_{i \in \mathcal{D}(v)} x_i(e) \le |\mathcal{D}(v)|.$$

So we need to bound the sum

$$\sum_{i \in \mathcal{D}(v)} \sum_{w} x_i(e_w).$$

When $w$ is not a descendant of $v$, the Minkowski summand $U_w$ involves only coordinates in $\mathcal{W}(w)$ which is disjoint from $\mathcal{D}(v)$. Therefore the sum above can be restricted to $w$ in the set of descendants of $v$:

$$\sum_{v \preceq w} \sum_{i \in \mathcal{D}(v) \cap \mathcal{W}(w)} x_i(e_w).$$

By definition, the summand $U_w$ is a dilated basic simplex, hence

$$\sum_{i \in \mathcal{W}(w)} x_i(e_w) \le |w|.$$

Therefore our full sum is bounded above by

$$\sum_{v \preceq w} |w| = |\mathcal{D}(v)|.$$

□

**Lemma 1.7.** Let $t$ be an arbor on the set $I$. Every vertex of the polytope $Q_t$ belongs to $M_t$.

The proof is done by induction on the size of the arbor. This clearly holds for arbors with just one vertex. The proof will use the recursive description of vertices given in Lemma 1.3 and the defects.

*Proof.* Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex $r_t$ of $t$. Let $z$ be a vertex of $Q_t$. By induction on the size of $t$, the statement holds for all sub-trees $t_k$. As the image of $\pi$ is the direct product of the polytopes $Q_{t_k}$, one can take the direct product of the Minkowski decompositions of $Q_{t_k}$. Therefore the image $\pi(z)$ can be written as a sum

$$\pi(z) = \sum_{w \neq r_t} e_w, \qquad (4)$$

where $e_w$ is in the Minkowski summand $U_w$ for some vertex $w$ in some $t_k$.
If all coordinates $x_i(z)$ for $i \in r_t$ vanish, then to each $e_w$ in the sum (4) one can associate a point $f_w$ with the same coordinates plus additional zero coordinates for all vertices $i \in r_t$. This defines a point $f_w$ in the summand $U_w$ of $M_t$, as the sum of coordinates in $\mathcal{W}(w)$ is unchanged. The sum $\sum_{w \neq r_t} f_w$ is therefore in $M_t$ and coincides with $z$ by construction.

Assume now that exactly one coordinate $x_i(z)$ with $i \in r_t$ is non-zero. Because the defect $\delta_{r_t}(z)$ vanishes, one gets

$$x_i(z) + \sum_{j \notin r_t} x_j(z) = n = |r_t| + \sum_k \sum_{w \in t_k} |w|.$$

It follows that

$$x_i(z) - |r_t| = \sum_k \delta_k(z), \qquad (5)$$

where $\delta_k(z)$ is the defect of $z$ with respect to the root vertex of $t_k$.

Consider the set $S$ of vertices that can be reached from the root $r_t$ by repeatedly moving to the root $w$ of a sub-tree with non-zero defect $\delta_w(z)$. Let $w$ be a vertex in $S$. Then from the element $e_w$ in the summand $U_w$ of some $M_{t_k}$, define $f_w$ by adding a new coordinate $x_i$ with value $|w|$ plus other zero coordinates for other vertices $j \in r_t$.

Let us check that $f_w$ is an element in the summand $U_w$ of $M_t$. Certainly all coordinates of $f_w$ are nonnegative. Because the defect $\delta_v(z)$ is not zero for every vertex $v \neq r_t$ on the path between $r_t$ and $w$, all coordinates $x_j(z)$ with $j$ in $\mathcal{W}(w) \setminus r_t$ are zero. The sum of all coordinates of $f_w$ is therefore $|w|$. It follows that $f_w$ is in the summand $U_w$ of $M_t$.

Let us consider $z' = \sum_{w \neq r_t} f_w$. Then for all $j \notin r$, the coordinates $x_j(z)$ and $x_j(z')$ coincide by construction. For $j \in r_t$ not equal to $i$, the coordinates $x_j(z)$ and $x_j(z')$ both vanish. Only the coordinate $x_i$ may differ between $z$ and $z'$. By construction $x_i(z')$ is the sum of $|w|$ over all vertices $w$ in the reachable set $S$.

Let us prove that this sum over $S$ is equal to $x_i(z) - |r_t|$. Let us start with (5) or more precisely its restriction to sub-trees $t_k$ whose root has non-zero defect. For every such $k$, the defect $\delta_k(z)$ is the sum of the size of the root of $t_k$ plus the sum of the defects of the roots of sub-trees of $t_k$. Iterating this process over the set $S$, one obtains the expected equality:

$$x_i(z) - |r_t| = \sum_{w \in S} |w|.$$

Let then $f_{r_t}$ be the element of the summand $U_{r_t}$ with coordinate $i$ equal to $|r_t|$ and other coordinates 0. Then $z' + f_{r_t} = z$, and therefore $z \in M_t$. $\square$

One can use this Minkowski decomposition to define a canonical polynomial whose Newton polytope is $Q_t$, namely

$$\prod_v \left( 1 + \sum_{e \in \mathcal{W}(v)} x_e \right)^{|v|}. \qquad (6)$$

In the case of the arbors of type A, namely linear arbors with no multiple vertex, these polynomials have appeared in the literature in relation with Catalan numbers. They are item 210 in Stanley's book [Sta15]. According to the
errata of this book, these polynomials first appeared in 1982 in a problem by J. Shallit in [Sha82, SW86]. They have been studied recently in [Fri21].

Let us state a corollary of the Minkowski decomposition. Postnikov introduced in [Pos09] the notion of Y-generalized permutohedra, defined as Minkowski sums of scaled standard simplices.

Let $e_0, e_1, \dots, e_d$ denote the standard basis of $\mathbb{Z}^{d+1}$. A scaled standard simplex is the convex hull of a set of vectors $\{k_i e_i\}_{i \in I}$ where $I$ is a non-empty subset of $\{0, 1, \dots, d\}$ and $k_i$ are positive integers.

**Lemma 1.8.** The polytope $Q_t$ is, up to unimodular equivalence, a Y-generalized permutohedra.

*Proof.* Let us consider the finest Minkowski decomposition of $Q_t$, where each summand $U_v$ is further decomposed as a Minkowski sum of equal terms. In this decomposition, each summand is the convex hull in $\mathbb{Z}^d$ of 0 and some vectors $\{e_i\}_{i \in I}$ where $I$ is a non-empty subset of $\{1, \dots, d\}$. This defines a polytope $\Delta_I$ in $\mathbb{Z}^d$, to which one associates a standard simplex $\Delta'_I$ in $\mathbb{Z}^{d+1}$ by replacing the vertex 0 by the basis vector $e_0$. Let us consider the linear map $\rho$ from $\mathbb{Z}^{d+1}$ to $\mathbb{Z}^d$ that forgets the first coordinate. Then $\rho$ maps $\Delta'_I$ to $\Delta_I$ and is moreover a unimodular equivalence from $\Delta'_I$ to $\Delta_I$.

Let us write $Q_t = \sum_{s \in S} \Delta_{I_s}$ and $Q'_t = \sum_{s \in S} \Delta'_{I_s}$ over the appropriate indexing set $S$. Then $\rho$ induces an invertible integral affine map from the hyperplane $H$ in $\mathbb{Z}^{d+1}$ where coordinates sum to $|S|$ to $\mathbb{Z}^d$, that maps $Q'_t$ to $Q_t$. This is the expected unimodular equivalence. $\square$

# 2 Invariants

We will be interested in several invariants of the polytope $Q_t$ and the poset $P_t$. Let us explain how they can be computed by induction.

The basis of the inductions is always the unique isomorphism class of arbor on a set with one element. The polytope $Q_t$ is in this case the line segment $[0, 1]$ in $\mathbb{R}$ and the poset $P_t$ is a total order with 2 elements.

The induction steps will use the following property or some variation of it, that come from the surjective projection map $\pi$ described in Section 1. Let $t$ be an arbor and $t_1, \dots, t_w$ be the sub-trees of the root vertex of $t$. If $e$ is a lattice point in $Q_t$, then one can describe $e$ using one lattice point $e_k$ in $Q_{t_k}$ for every $1 \le k \le w$ by restriction, plus the additional data of the coordinates of $e$ with indices in the root vertex. The only constraint on these root coordinates is their sum, namely the height of $e$ has to be the sum of the heights of all $e_k$ plus the sum of the root coordinates. The same statement holds for all real points.

## 2.1 Ehrhart polynomial

Let us first consider the Ehrhart polynomial $E_t$ of the lattice polytope $Q_t$. Recall that this is the unique polynomial $E_t$ in a variable $u$ such that, for every integer $m \ge 1$, $E_t(m)$ is the number of lattice points in the dilated polytope $mQ_t$. A standard reference on the subject is [Bar08].

In order to compute the Ehrhart polynomial, we will need to refine the count according to height. This is done in a somewhat indirect way, maybe not optimal.
For every integer $m \ge 1$, let us introduce a polynomial $F_{t,m}$ in the variable $X$ that counts lattice points in the dilated polytope $mQ_t$ with weight $X^h$ at height $h$. For an arbor of size $n$, the height on $mQ_t$ is bounded above by $mn$, so $F_{t,m}$ has degree $mn$.

Note that this definition of $F_{t,m}$ could be extended verbatim to any lattice polytope in $\mathbb{R}_{\ge 0}^n$. In this generality, it is clearly multiplicative with respect to direct product of polytopes.

For the arbor $t$ on one element,

$$F_{t,m} = 1 + X + \cdots + X^m. \qquad (7)$$

Let $t$ be an arbor on a set of cardinality $n \ge 2$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of $t$.

**Proposition 2.1.** Let $m \ge 1$ be an integer. Let $F_{W,m}$ be the product $\prod_{k=1}^w F_{t_k,m}$. Write $F_{W,m} = \sum_j W_j X^j$. Then for $0 \le j \le mn$, the coefficient of $X^j$ in $F_{t,m}$ is

$$\sum_{\ell=\max(0,j+m(r-n))}^{j} \binom{r+\ell-1}{\ell} W_{j-\ell}. \qquad (8)$$

*Proof.* Let us fix an integer $m$ and consider the dilated polytope $mQ_t$.

For $0 \le j \le mn$, a point at height $j$ in $mQ_t$ can be described by the coordinates of its image by the projection map $\pi$ together with the remaining $r$ coordinates from the root vertex of $t$. Assume that the image by $\pi$ has height $j-\ell$, with $0 \le \ell \le j$. The possible images are counted by $W_{j-\ell}$, which vanishes if $j-\ell > m(n-r)$ because the degree of $F_{w,m}$ is $m(n-r)$. Then the coordinates from the root vertex describe a point in $\mathbb{N}^r$ with height $\ell$. This set is well-known to be counted by $\binom{r+\ell-1}{\ell}$.

As this is a bijection, we get the expected sum over $\ell$. $\square$

<figure><img src="image_6.png" /></figure>

This allows to compute the usual Ehrhart polynomial by interpolation of the values of $F_{t,m}$ at $X = 1$ with respect to $m$.

For the running example, one finds that $E_t$ is

$$\frac{1}{12} \cdot (2u + 1) \cdot (u + 1)^3 \cdot (83u^2 + 70u + 12),$$

and so the volume is the leading coefficient $83/6$.

**Conjecture 2.2.** For any arbor $t$, all roots of the Ehrhart polynomial of $Q_t$ are negative real numbers in the interval $[-1, 0]$, excluding 0.

Indeed 0 is not a root because the constant term is always 1. This conjecture has been checked for all arbors of size at most 12.

Note that this conjecture would imply that all coefficients of the Ehrhart polynomial are positive. A lattice polytope with this property is said to be *Ehrhart positive*. Postnikov proved in [Pos09] that all Y-generalized permutohedra are Ehrhart positive, see also [Liu19, §3.1] and [FH23, §6.1]. By lemma 1.8, one can conclude that the polytope $Q_t$ is Ehrhart positive.
## 2.2 Volume

Let us now consider the volume of the polytope $Q_t$. This could be computed using the leading coefficient of the Ehrhart polynomial. Instead, we refine here the volume into a volume function $V_t(h)$, that records the volume of every slice of $Q_t$ according to the height function $ht$. This function vanishes outside the segment $[0, n]$ where $n$ is the size of $t$.

This volume-function could be defined verbatim for any polytope in $\mathbb{R}^{n}_{\ge 0}$. In this generality, the volume-function of a direct product of polytopes is the convolution of the individual volume-functions.

For the arbor $t$ on one element,

$$V_t(h) = 1, \qquad (9)$$

for $0 \le h \le 1$ and $0$ otherwise.

Let $t$ be an arbor on a set of cardinality $n \ge 2$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of $t$.

Let $\star$ denote the convolution product of functions on $\mathbb{R}_{\ge 0}$. Let $T_n$ denote the operator that truncates a function to the segment $[0, n]$, extended by zero outside of this segment.

**Proposition 2.3.** *The volume function $V_t$ is given by*

$$T_n(V_{t_1} \star V_{t_2} \star \cdots \star V_{t_w} \star W_{n,r}), \qquad (10)$$

where $W_{n,r}$ is the truncated function $T_n\left(\frac{h^{r-1}}{(r-1)!}\right)$.

*Proof.* Recall that a point in $Q_t$ is determined by a sequence of points $e_k$ in $Q_{t_k}$ by restriction and by a point in $\mathbb{R}^r$ for the coordinates with indices in the root vertex of $t$. The additional constraint is the value of the height. Using multiplicativity, one checks that the volume of a slice in $Q_t$ is the expected convolution, except maybe for heights above $n$. This is fixed by the truncation operator wrapping the convolution products. $\square$

The volume function allows to compute the volume by integration from 0 to $n$. This matches the leading coefficient of the Ehrhart polynomial.

In general, the function $V_t$ is continuous and piecewise-polynomial with possible change of polynomial at integers. In the running example for instance, it is given by $h^5/120$ between 0 and 1, by three complicated polynomials of degree 5 between 1 and 4 and by $2h - 14/3$ between 4 and 6. Integrating from 0 to 6, one computes the volume to be $83/6$ as expected.

## 2.3 Laplace transform of volume

The volume function $V_t$ is in general a rather complicated piecewise-polynomial function of the variable $h$. It is more compactly encoded by its Laplace transform, defined by

$$L_t(v) = \int_0^\infty e^{-vh} V_t(h) dh. \qquad (11)$$

Note that the value at $v = 0$ recovers the volume of $Q_t$.

The Laplace transform sends the convolution product to the product. It therefore remains to understand the truncation operator $T_n$ after the transform.
We will only need the action of $\mathbb{T}_n$ on the functions $e_{k,\ell}(v) = v^{-k}e^{-\ell v}$ with $k > 0$ and $\ell \ge 0$.

**Lemma 2.4.** For integers $k \ge 0$ and $\ell \ge 0$, let $f_{k,\ell}$ be the function on $\mathbb{R}_{\ge 0}$ defined by $\frac{(h-\ell)^k}{k!}$ if $h > \ell$ and 0 otherwise. The Laplace transform of $f_{k,\ell}$ is $v^{-k-1}e^{-\ell v}$.

This statement is easy to prove by standard properties of the Laplace transform, see [DLMF, Table 1.14.4].

**Lemma 2.5.** If $\ell \ge n$, then $\mathbb{T}_n(e_{k,\ell}) = 0$. Otherwise

$$ \mathbb{T}_n(e_{k+1,\ell}) = e_{k+1,\ell} - \sum_{j=0}^{k} \frac{(n-\ell)^{k-j}}{(k-j)!} e_{j+1,n}. \quad (12) $$

*Proof.* Let $g_n(h)$ be the indicatrix function of the interval $[0, n]$. As $g_n = f_{0,0} - f_{0,n}$, its Laplace transform is $\mathcal{L}g_n = \frac{1}{v}(1 - e^{-nv})$. The truncation operator acts on functions of $h$ as the pointwise product with $g_n$. Because $f_{k,\ell}$ has support on $[\ell, \infty]$, this pointwise product $f_{k,\ell}g_n$ is the zero function if $\ell \ge n$. So $\mathbb{T}_n(e_{k,\ell})$ is zero in this case.

Otherwise, assume $\ell < n$. Let us start from

$$ f_{k,\ell}g_n = f_{k,\ell} - f_{0,n}f_{k,\ell} $$

and compute the Laplace transform:

$$ \frac{e^{-\ell v}}{v^{k+1}} - \int_{n}^{\infty} \frac{(h-\ell)^k}{k!} e^{-vh} dh. $$

The second term becomes, first replacing $h-\ell$ by $(h-n)+(n-\ell)$ and expanding,

$$ \sum_{j=0}^{k} \frac{(n-\ell)^{k-j}}{(k-j)!} \frac{e^{-nv}}{v^{j+1}}. $$

This proves the expected formula. $\square$

One can therefore compute the Laplace transform $L_t(v)$ of the volume function $V_t$ by first computing by induction a polynomial $L_t(E, V)$ in variables $E$ and $V$ using the following rules, and then replace $E$ by $e^{-v}$ and $V$ by $1/v$.

The initial value $L_t(E, V)$ for the arbor of size 1 is $-VE + V$.

The truncation formula (12) translates into a similar formula for the action of $\mathbb{T}_n$ on polynomials in $E$ and $V$.

Let $t$ be an arbor on a set of cardinality $n \ge 2$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of $t$.

**Proposition 2.6.** The polynomial $L_t(E, V)$ is $\mathbb{T}_n(\prod_{k=1}^w L_{t_k}(E, V) \cdot \mathbb{T}_n(V^r))$.

*Proof.* This follows from the preceding statements. The function $W_{n,r}$ has Laplace transform equal to $\mathbb{T}_n(v^{-r})$ by Lemma 2.4. $\square$

In the running example, this gives

$$ -V^6 E^4 + 2V^6 E^3 - 2V^5 E^4 + 4V^5 E^3 - 2V^2 E^6 - 2V^6 E - 2V^5 E^2 - \frac{22}{3}VE^6 + V^6. $$

To recover the volume, one needs to take the limit at $v = 0$ of $L_t(e^{-v}, 1/v)$.
## 2.4 Zeta polynomial

Let us now consider the Zeta polynomial of the poset $P_t$. This is a polynomial $Z_t$ in the variable $u$ whose value at every integer $m \ge 2$ is the number of chains $e_1 \le e_2 \le \cdots \le e_{m-1}$ in $P_t$.

We will need a mild refinement of this usual Zeta polynomial. Let $Z_t(u, X)$ be the polynomial in $u$ and $X$ such that for every integer $m \ge 2$,

$$Z_t(m, X) = \sum_{e_1 \le \cdots \le e_{m-1}} X^{\mathrm{ht}(e_{m-1})} \quad (13)$$

where $\mathrm{ht}$ is the height of an element. The existence of this polynomial will follow from the induction below. Setting $X = 1$ forgets about the height and gives the usual Zeta polynomial.

Note that if we consider a product of two graded posets endowed with the sum of height functions, then the definition of $Z$ is naturally multiplicative, as chains are then just pairs of chains of the same length.

For the arbor $t$ on one element, one finds that

$$Z_t(u, X) = 1 + (u-1)X. \quad (14)$$

Let now $t$ be an arbor on a set of cardinality $n \ge 2$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of $t$.

**Proposition 2.7.** Let $Z_W$ be the product of $\prod_{k=1}^W Z_{t_k}$. Write $Z_W = \sum_j W_j X^j$. Then for $0 \le j \le n$, the coefficient of $X^j$ in $Z_t$ is

$$\sum_{\ell=\max(0, j-n+r)}^{j} \binom{r(u-1)+\ell-1}{\ell} W_{j-\ell}. \quad (15)$$

*Proof.* Let us fix an integer $m \ge 2$ and consider a chain $e_1 \le \cdots \le e_{m-1}$ in $P_t$ with $\mathrm{ht}(e_{m-1}) = j$. Because the interval $[0, e_{m-1}]$ is a product of total orders, one can decompose the chain as a product of chains of the same length by separating the coordinates in two disjoint subsets: A for those in the root vertex of $t$ and B for the others. Let $\ell$ be the sum of coordinates of $e_{m-1}$ in A and $j - \ell$ be the sum of coordinates of $e_{m-1}$ in B. Note that $j - \ell \le n - r$ by the defining inequalities for the sub-trees $t_1, \dots, t_w$.

On the set $B$, the restriction of the chain is a product of chains in the sub-trees $t_1, \dots, t_w$ with maximal height $j - \ell$. Such chains are counted by $W_{j-\ell}$ evaluated at $u = m$. On the set $A$, the restriction of the chain is a chain in the poset $\mathbb{N}^r$ ordered coordinate-wise, with maximal height equal to $\ell$. By Theorem 7.2 in Section 7, chains of length $m-1$ in the poset $\mathbb{N}^r$ with maximal height at most $\ell$ are counted by $\binom{r(m-1)+\ell}{\ell}$. Therefore, by taking the difference of values for $\ell$ and $\ell-1$, the possible restricted chains on the set $A$ are counted by $\binom{r(m-1)+\ell-1}{\ell}$.

This decomposition is a bijection, hence the expected formula holds for every $m \ge 2$. Hence it holds as an equality of polynomials in $u$. $\square$
In the running example, one gets

$$
\begin{aligned}
Z_t = \frac{1}{90} \cdot & \left( (802 u^4 - 1839 u^3 + 1598 u^2 - 621 u + 90) (2u-1)(u-1)X^6 \right. \\
& \left. + 6(53 u^2 - 77 u + 30)(4u-3)(2u-1)(u-1)X^5 \right. \\
& \left. + 45(68 u^3 - 156 u^2 + 119 u - 30)(u-1)X^4 \right. \\
& \left. + 60(11u-10)(4u-3)(u-1)X^3 \right. \\
& \left. + 90(17u-15)(u-1)X^2 + 540(u-1)X + 90 \right)
\end{aligned}
$$

and so the usual Zeta polynomial is

$$
\frac{1}{90} \cdot u \cdot (2u - 1) \cdot (802u^4 - 1369u^3 + 893u^2 - 266u + 30).
$$

It seems that the set of arbors $t$ having the property that the Zeta polynomial of $P_t$ factors completely over $\mathbb{Q}$ consists exactly of arbors of type A considered in Section 6, arbors of type $\mathbb{B}$ considered in section Section 7 and one of the two families considered in Section 8.

2.5 M-triangle

Let us now consider the *M*-triangle of the poset $P_t$. This is the polynomial $M_t(X, Y)$ defined by (55) in Appendix A, using values of the Möbius function.

In order to give a recursive formula, we need to compute instead the following
polynomial

$$
K_t(X, Y) = \sum_{a \in P_t} X^{\mathrm{nz}(a)} Y^{\mathrm{ht}(a)}, \quad (16)
$$

where nz(a) is the number of non-zero coordinates in a. Note that the exponent
of X is smaller than or equal to the exponent of Y in any monomial in this
formula.

Note that the polynomial $K_t$ is clearly multiplicative under product for posets of the type that we are considering.

For the arbor *t* on one element,

$$
K_t = 1 + XY. \tag{17}
$$

Let $t$ be an arbor on a set of cardinality $n \ge 2$. Let $(t_1, \dots, t_w)$ be the sub-trees of the root vertex of $t$. Let $r$ be the size of the root vertex of $t$.

**Proposition 2.8.** Let $W = \prod_k K_{t_k}$ and write $W = \sum_{i,j} W_{i,j} X^i Y^j$. Then for $0 \le j \le k \le n$, the coefficient of $X^j Y^k$ in $K_t$ is

$$
\sum_{\ell=\max(0,j-n+r)}^{\min(j,r)} \sum_{m=\max(\ell,k-n+r)}^{k+\ell-j} \binom{r}{\ell} \binom{m-1}{m-\ell} W_{j-\ell,k-m}. \quad (18)
$$

Proof. Any element $a$ of $P_t$ can be described as an element $\pi(a)$ of $\prod_k P_{t_k}$ and $r$ additional coordinates from the root of $t$ for the fiber of $\pi$. Let us assume that $a$ has $j$ non-zero coordinates and height $k$, with $0 \le j \le k \le n$. Assume now that $\pi(a)$ has $j-\ell$ non-zero coordinates and height $k-m$ with $0 \le \ell \le j$ and $0 \le m \le k$. This also requires that $0 \le j-\ell \le k-m \le n-r$.
Then the additional $r$ coordinates must sum to $m$ and among them exactly $\ell$ must be non-zero. This implies that $\ell \le r$ and $\ell \le m$. The choice of these non-zero coordinates is counted by a binomial $\binom{r}{\ell}$. Once this choice is fixed, there remains to choose $\ell$ strictly positive numbers summing to $m$, which amounts to choose $\ell - 1$ cuts among $m - 1$ possible inner cuts in a sequence of length $m$.

As this defines a bijection, one gets the expected double sum, where the projection $\pi(a)$ contributes by the term $W_{j-\ell,k-m}$ and the coordinates from the root contribute the product $\binom{r}{\ell}\binom{m-1}{m-\ell}$. The precise bounds in the double sum follows from the necessary inequalities as previously stated. $\square$

Here is a slightly simplified version of Proposition 2.8 that will be useful.

**Proposition 2.9.** Let $W = \prod_k K_{t_k}$ and write $W = \sum_{i,j} W_{i,j} X^i Y^j$. Then for $0 \le j \le n$, the coefficient of $X^j$ in $K_t(X, 1)$ is

$$ \sum_{\ell=\max(0,j-n+r)}^{\min(j,r)} \sum_{k'=j-\ell}^{n-r} \binom{r}{\ell} \binom{n-k'}{\ell} W_{j-\ell,k'} \quad (19) $$

*Proof.* Starting from Proposition 2.8 and summing over $k$, one gets

$$ \sum_{k, \ell, m} \binom{r}{\ell} \binom{m-1}{m-\ell} W_{j-\ell, k-m} \quad (20) $$

over the appropriate range. By first changing variable to $k' = k - m$, one can perform the inner summation over $m$ from $\ell$ to $n - k'$. $\square$

The polynomial $K_t$ determines the $M$-triangle $M_t$ as follows.

**Proposition 2.10.** For every arbor $t$,

$$ M_t(X, Y) = K_t(1 - 1/X, XY). \quad (21) $$

*Proof.* First, write the $M$-triangle as a double sum

$$ \sum_{b \in P_t} Y^{\text{ht}(b)} \sum_{a \le b} \mu(a, b) X^{\text{ht}(a)}. $$

Recall that the interval $[0, b]$ is a product of total orders whose lengths are the non-zero coordinates of $b$ plus 1. It follows that the inner sum is

$$ \sum_{a \le b} \mu(a, b) X^{\text{ht}(a)} = X^{\text{ht}(b)} (1 - 1/X)^{\text{nz}(b)}. $$

The result then follows from the definition of $K_t$. $\square$

In our running example, the $M$-triangle is

$$ \left( \begin{array}{ccccccc} 1 & -18 & 113 & -334 & 506 & -380 & 112 \\ -6 & 58 & -208 & 352 & -284 & 88 & \\ 15 & -92 & 201 & -188 & 64 & \\ -20 & 78 & -98 & 40 & \\ 15 & -34 & 19 & \\ -6 & 6 & \\ 1 & \end{array} \right) $$

where the south-west to north-east diagonal records the number of elements in $P_t$ according to height.
## 2.6 f-vector and h-vector

As all intervals in the poset $P_t$ are products of total orders, the Hasse diagram of $P_t$ can be viewed as the skeleton of a cubical complex. Let us compute its f-vector, the polynomial in one variable $X$ where the coefficient of $X^k$ is the number of $k$-dimensional cubical faces.

Every cubical face is determined by an interval $[a, b]$ in $P_t$. We will gather them according to their top element $b$. When $b$ has nz($b$) non-zero coordinates, it contributes by the faces of a nz($b$)-dimensional hypercube that are incident to a fixed vertex of this hypercube. The f-vector is therefore

$$f_t(X) = \sum_{b \in P_t} (1 + X)^{nz(b)}. \qquad (22)$$

By (16), this can be computed from the polynomial $K_t$ as

$$f_t(X) = K_t(1 + X, 1). \qquad (23)$$

The h-vector of the arbor $t$ is defined as the generating polynomial

$$h_t(X) = \sum_{b \in P_t} X^{nz(b)}. $$

It satisfies $h_t(X) = f_t(X - 1) = K_t(X, 1)$. By (21), this implies that

$$h_t(X) = M_t \left( \frac{1}{1-X}, 1-X \right). \qquad (24)$$

In our running example, one finds that the f-vector of $P_t$ is

$$f_t(X) = X^6 + 24X^5 + 186X^4 + 654X^3 + 1152X^2 + 990X + 330$$

and its h-vector is

$$h_t(X) = X^6 + 18X^5 + 81X^4 + 130X^3 + 81X^2 + 18X + 1,$$

which is nicely palindromic.

# 3 Ehrhart-Zeta duality for linear arbors

This section proposes a conjectural duality exchanging Ehrhart polynomials and Zeta polynomials.

Let $t$ be an arbor such that each vertex of $t$ has at most one sub-tree. This is called a *linear arbor*. Still the vertices can contain several elements.

For a linear arbor $t$, let $Rev(t)$ be the reverse of $t$, defined as the linear arbor in which the vertices are in the opposite order. This amounts to choose the root vertex of $Rev(t)$ to be the unique leaf of $t$. Reversal is an involution on linear arbors.

For example, here are two arbors $t$ and $Rev(t)$, mutually reverse, with their roots on the left.

<figure><img src="image_7.png" /></figure>
**Conjecture 3.1.** For every linear arbor t, the Ehrhart polynomial $E_t(u)$ of the polytope $Q_t$ is equal to the shifted Zeta polynomial $Z_{Rev(t)}(u + 1)$ of the poset $P_{Rev(t)}$.

By looking at the leading coefficients, Conjecture 3.1 would identify the normalized volume of $Q_t$ to the number of maximal chains of $P_{Rev(t)}$. Recall that in dimension $n$ the normalized volume is $n!$ times the volume.

By the special case of the value at $u = 1$, the posets of the arbors $t$ and $Rev(t)$ should share the same number of elements. Something finer seems to hold.

**Conjecture 3.2.** For every linear arbor $t$, the $h$-vectors of $t$ and $Rev(t)$ are equal.

For example, for the two arbors above, both with 501 elements, one gets

$$E_t = \frac{u+1}{720} \cdot (20167u^5 + 57230u^4 + 62225u^3 + 32170u^2 + 7848u + 720),$$

$$Z_t = \frac{u}{60} \cdot (2u - 1) \cdot (898u^4 - 1611u^3 + 1002u^2 - 249u + 20),$$

$$E_{Rev(t)} = \frac{u+1}{60} \cdot (2u+1) \cdot (898u^4 + 1981u^3 + 1557u^2 + 514u + 60),$$

$$Z_{Rev(t)} = \frac{u}{720} \cdot (20167u^5 - 43605u^4 + 34975u^3 - 12795u^2 + 2098u - 120).$$

The volume of $Q_t$ is 20167/720 and the number of maximal chains of $P_t$ is 21552. Conversely, the volume of $Q_{Rev(t)}$ is 21552/720 and the number of maximal chains of $P_{Rev(t)}$ is 20167. Both $t$ and $Rev(t)$ have $h$-vector

$$x^6 + 23x^5 + 122x^4 + 209x^3 + 122x^2 + 23x + 1.$$

Both conjectures have been checked by computer for all linear arbors of size at most 11. We will not pursue these conjectures further here. One may wonder if the first one can be explained by the existence of unimodular triangulations.

# 4 Transmutation of M-triangles

Recall from Appendix A that the M-triangle of a graded poset is the following polynomial:

$$m = \sum_{a \le b} \mu(a, b) X^{\mathrm{ht}(a)} Y^{\mathrm{ht}(b)},$$

where $\mu$ is the Möbius function of the poset.

Let us introduce the following operation, called transmutation:

$$\bar{m}(X, Y) = m\left(\frac{1-Y}{1-XY}, 1-XY\right). \qquad (25)$$

Transmutation is an involution, as can be checked directly. We will explain later where it comes from.
Note that for the posets $P_t$, the M-triangle $M_t$ can be computed from the simpler polynomial $K_t$ as defined in (16). It follows that the transmuted M-triangle is

$$ \overline{M}_t(X, Y) = K_t \left( \frac{Y(X-1)}{1-Y}, 1-Y \right). \qquad (26) $$

As explained at length in Section 0.1, one of the main motivations for the article is the following hope: there should exist another graded poset $NC_t$ whose M-triangle is $\overline{M}_t$.

For example, consider the arbor t equal to

<figure><img src="image_8.png" /></figure>

Then the poset $P_t$ has 14 elements. Its M-triangle and transmuted M-triangle are

$$ \left( \begin{array}{cccc} -1 & 6 & -10 & 5 \\ 3 & -8 & 5 & 0 \\ -3 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right) \quad \text{and} \quad \left( \begin{array}{cccc} -5 & 10 & -6 & 1 \\ 10 & -16 & 6 & 0 \\ -6 & 6 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$

displayed as matrices of coefficients, with the constant term at the bottom left.

One can check that the transmuted M-triangle is the M-triangle of the classical lattice of noncrossing-partitions of a set of size 4, as defined in [Kre72].

**Proposition 4.1.** The diagonal of the transmuted M-triangle is the h-vector of t.

*Proof.* Formula (26) expresses the transmuted M-triangle as

$$ \sum_{a \in P_t} (1-Y)^{\mathrm{ht}(a)-\mathrm{nz}(a)} (Y(X-1))^{\mathrm{nz}(a)}. $$

In every monomial in this expression, the power of Y is larger than or equal to the power of X. The restriction to monomials with equal powers is

$$ \sum_{a \in P} (YX)^{\mathrm{nz}(a)}, $$

which is equal to $K_t(XY, 1)$ encoding the h-vector of t.

□

Let us now say a few words about the origin of the transmutation operation.

Closely linked to the M-triangle, there is another two-variable polynomial called the H-triangle. They are related by the transformations (58) and (59) recalled in Appendix A.

These triangles and conversion rules between them first appeared in the study of the combinatorics of finite type cluster algebras [Cha04, Cha06]. In this context, the M-triangle was considered for the noncrossing-partitions lattice attached to an ADE-quiver. The H-triangle was a natural refinement of the h-vector of the cluster fan for the same quiver.

One can define transmutation of M-trangles as the conjugation of transposition on H-trangles by the conversion maps between H-trangles and M-trangles. Here transposition just means, viewing H-trangles as triangular matrices in the appropriate way, the transposition of matrices.
Note that the *H*-triangle can be interpreted as a refined decomposition of the homology groups of a smooth toric variety. In this context, the transposition has no clear meaning, as it exchanges the homological grading with the other additional grading.

# 5 Unary-binary arbors and quadrangulations

Let us say that an arbor is unary-binary if

* every vertex has one element and

* every vertex has at most 2 sub-trees.

We will talk about vertices of valency 0, 1 or 2.

For example, here is a unary-binary arbor:

<figure><img src="image_9.png" /></figure>

where the valencies of $a, b, c, d$ are 2, 0, 1, 0.

Given such an arbor $t$, let us choose a planar embedding and assume that the arbor is drawn with the root on the left and growing to the right as above.

Using this picture, one can define a quiver $G_t$ as follows. The vertices of $G_t$ are the vertices of $t$. The edges of $G_t$ are the edges of $t$ suitably oriented. Let $v$ be a vertex $v$ of valency 2. Let $\alpha_v$ and $\beta_v$ the two edges between $v$ and its 2 sub-trees, from top to bottom. We orient $\alpha_v$ towards $v$ and $\beta_v$ from $v$ outwards. Other edges are oriented arbitrarily.

We now consider $G_t$ as a quiver-with-relations for the relations $\alpha_v\beta_v = 0$ for all vertices $v$ of valency 2.

For instance, by choosing to orient the remaining arrow from $d$ to $c$, the previous example gives the quiver-with-relations

<figure><img src="image_10.png" /></figure>

We refer the reader to [ABCJP10, Sch23] for the definition and properties of gentle quivers and algebras. Essentially this means that at every vertex, there are at most two incoming arrows $\alpha, \alpha'$ and at most two outgoing arrows $\beta, \beta'$ and there are relations $\alpha\beta$ and $\alpha'\beta'$ when they make sense.

**Lemma 5.1.** *The quiver-with-relations $G_t$ is a gentle quiver.*

*Proof.* Every vertex is incident to at most 3 edges and there can be no sink and no source vertex. If there are three incident edges, two of them are linked by a relation. $\square$
Let us recall briefly now an important construction starting from the gentle quiver $G_t$. This comes from $\tau$-tilting theory as introduced in [AIR14]. To every finite-dimensional algebra over a field, one can associate a set of support-$\tau$-tilting objects. These objects then naturally appear as vertices in a regular graph called the mutation graph. It happens, for specific classes of finite-dimensional algebras, that the mutation graph of support-$\tau$-tilting objects is finite. That finiteness and its consequences hold notably for the gentle algebras $G_t$ that we consider here.

In this case, the mutation graph is in fact oriented and the Hasse diagram of a partial order which is congruence-uniform. There is also a simple polytope whose skeleton is the mutation graph. We refer to [PPP21] for these properties in the case of gentle algebras.

This finite mutation graph on support-$\tau$-tilting objects for the quiver-with-relations $G_t$ seems to be related to the poset $P_t$ itself by the following conjecture.

**Conjecture 5.2.** *The number of support $\tau$-tilting objects for the gentle quiver-with-relations $G_t$ is equal to the number of elements of the poset $P_t$.*

In fact, there should be a more precise relationship. Let us consider the spherical simplicial complex $SC_t$ which is the polar dual of the simple polytope attached to $G_t$ by $\tau$-tilting theory.

**Conjecture 5.3.** *In the simplicial complex $SC_t$, there exists a facet $C$ such that the $F$-triangle for the pair $(SC_t, C)$ coincides with the $F$-triangle associated with the transmuted $M$-triangle of $P_t$ by the formulas of Appendix A.*

This can been checked concretely in small cases. It has been checked by computer for $n \le 8$.

In our example $t$ as above, the poset $P_t$ has 33 elements. One can check that the gentle quiver $G_t$ also has 33 support $\tau$-tilting objects. The $F$-triangle associated with the transmuted $M$-triangle is

$$ \left( \begin{array}{cccc} 1 & & & \\ 4 & 4 & & \\ 6 & 14 & 8 & \\ 4 & 17 & 23 & 10 \\ 1 & 8 & 22 & 25 & 10 \end{array} \right) \qquad (27) $$

and it appears as expected for some facet of $SC_t$.

## 5.1 Stokes polytopes and quadrangulations

Let us present briefly an alternative description of the same situation, replacing the general $\tau$-tilting theory by the more specific notion of Stokes polytopes and Baryshnikov compatibility [Bar01, Cha16].

Let's consider a regular polygon with $2n+4$ vertices. Fix a quadrangulation $Q$ of this polygon, which means a set of $n$ noncrossing segments joining vertices and cutting the polygon into quadrilateral regions. After Baryshnikov, one can define a compatibility relation between quadrangulations. On the set $R_Q$ of quadrangulations compatible with $Q$, there is a mutation process, that defines a mutation graph. Moreover, this mutation process is oriented and defines a
partial order. The mutation graph is also the skeleton of a polytope, called the Stokes polytope for the quadrangulation [MP19, Man18, BMP17].

There is a simple way to map the planar-embedded arbor $t$ into a quadrangulation. There is also a way to define a gentle quiver from any quadrangulation. The composite map recovers the definition of $G_t$ from $t$. By this correspondence, the $\tau$-tilting theory exactly matches the combinatorial description using mutations of compatible quadrangulations.

It follows that the conjectural statements above could be rephrased entirely using the structures (mutation graph, poset, polytope) coming from this context instead of those coming from $\tau$-tilting theory.

# 6 Arbors of type A

In this section, we consider the case of linear arbors with no multiple vertex, so that their shape is a simple chain. This is the intersection of the set of unary-binary arbors considered before with the set of linear arbors. In order to compute the Zeta polynomial and the M-triangle for the poset $P_t$ in this case, it is convenient to study a more general family of posets.

In this section, we will use the following convention: for a product $\beta_\ell = \prod_{r=2}^\ell \alpha_r$, the value at $\ell = 1$ is 1 and the value at $\ell = 0$ is $1/\alpha_1$.

Let $m \ge 1$ be an integer, used as a slope parameter. Let $R_m$ be the region in the plane of coordinates $(x, y)$ defined by the conditions

$$0 \le x, \quad 0 \le y, \quad my < x. \tag{28}$$

Let now $(x, y)$ be nonnegative integers such that

$$my < x. \tag{29}$$

We consider the set of lattice paths in $R_m$ that go from $(0,0)$ to the fixed point $(x, y)$, by steps $(+1,0)$ (denoted by X) or $(0,+1)$ (denoted by Y). The point $(0,0)$ is the unique allowed point on the line $my = x$ at the boundary of $R_m$.

<figure><img src="image_11.png" /><figcaption>Figure 2: Example of lattice path $X^3YX^2YX^3Y$ with $m = 2$, $x = 8$ and $y = 3$. The word A associated with vertical steps is $(0, 0, 1, 0, 1, 0, 0)$.</figcaption></figure>

To each lattice path $w$ in $R_m$ from $(0,0)$ to $(x, y)$, one associates a word $A = (a_1, \dots, a_{x-1})$, that records the lengths of sequences of consecutive Y between two letters X. The letter $a_i$ is the number of Y between the $i$-th instance of X and the next one. The length of $A$ is $x-1$.
One can characterise the set of such words A by the following inequalities:

$$m \left( \sum_{i \le j} a_i \right) < j, \qquad (30)$$

for $1 \le j \le x - 1$ and the condition

$$\sum_{i} a_i \le y. \qquad (31)$$

In particular, every $a_i$ for $i \le m$ must vanish.

One then defines a poset $\mathcal{P}_{m,x,y}^F$ by term-wise comparison on this set of words. In this notation, the exponent $F$ stands for Fuss.

The poset $\mathcal{P}_{m,x,y}^F$ is graded by the sum of the $a_i$, that we will denote by $ht$. There is a unique minimum $\perp$, with all $a_i$ being zero. Every interval $[\perp, (a_i)_i]$ is isomorphic to the product of total orders of size $a_i + 1$.

The special case where $m = 1$, $x = n + 2$ and $y = n$ recovers the poset $P_t$ attached to the linear arbor of size $n$ with no multiple vertex. Indeed, in terms of words $(a_1, \dots, a_{n+1})$, being strictly below the diagonal $x = y$ amounts to the conditions $a_1 \le 0$, $a_1 + a_2 \le 1$, $\dots$, $a_1 + \dots + a_{n+1} \le n$. Necessarily $a_1 = 0$, so that one finds the defining equations of the poset $P_t$ and the same term-wise comparison.

**Lemma 6.1.** Let $y \ge 1$. The posets $\mathcal{P}_{m,my+1,y}^F$ and $\mathcal{P}_{m,my+1,y-1}^F$ are isomorphic.

*Proof.* For all $m \ge 1$ and for all $y \ge 1$ and $x = my + 1$, lattice paths in the region $R_m$ that ends at $(x, y)$ are in bijection, by removal of the last step $Y$, with lattice paths in the region $R_m$ that ends at $(x, y-1)$. This bijection is the identity on the coordinates used to define the partial order. $\square$

## 6.1 Zeta polynomial

**Theorem 6.2.** The Zeta polynomial of $\mathcal{P}_{m,x,y}^F$ is given by

$$Z_{m,x,y}^{F}(u) = \frac{((x - my - 1)(u - 1) + 1)}{y!} \prod_{j=2}^{y} ((x - 1)(u - 1) + j). \qquad (32)$$

*Proof.* One first notes that if $x = 1$, then $y = 0$ and the right-hand side is the constant polynomial 1. As the poset $\mathcal{P}_{m,1,0}^F$ is just a singleton, the statement holds in that case.

The rest of the proof is by induction based on this initial case and the next four lemmas. $\square$

Let us denote $\mathbf{Zeta}_{m,x,y}^F(u)$ the Zeta polynomial of the poset $\mathcal{P}_{m,x,y}^F$.

**Lemma 6.3.** For all $m \ge 1$ and for all $y \ge 1$ and $x = my + 1$, there is a polynomial equality

$$\mathbf{Zeta}_{m,x,y}^{F}(u) = \mathbf{Zeta}_{m,x,y-1}^{F}(u). \qquad (33)$$

*Proof.* By Lemma 6.1, the posets $\mathcal{P}_{m,x,y}^F$ and $\mathcal{P}_{m,x,y-1}^F$ are isomorphic. $\square$
**Lemma 6.4.** For all $m \ge 1$ and for all $x, y$ satisfying (29) with $x > my + 1$, there is a polynomial equality

$$ \mathrm{Zeta}_{m,x,y}^{F}(u) = \sum_{j=0}^{y} \mathrm{Zeta}_{m,x-1,y-j}^{F}(u) \binom{j-2+u}{j}. \quad (34) $$

*Proof.* By polynomiality of Zeta polynomials with respect to $u$, it suffices to check the formula when $u$ is replaced by any integer $q \ge 2$. We will use a bijection between chains.

Let us assume $q \ge 2$ and consider a chain $A_1 \le A_2 \le \dots \le A_{q-1}$ in $\mathcal{P}_{m,x,y}^F$. We denote by $a_{\ell,i}$ for $1 \le i \le x-1$ the coordinates of the lattice path $A_\ell$. Note that the chain condition implies that $\sum_{i \le k} a_{\ell,i}$ is growing with $\ell$ for every fixed $k$.

Let $j = a_{q-1,x-1}$. The possible values of $j$ are all integers between 0 and $y$.

For $1 \le \ell \le q-1$, we define $\tilde{A}_\ell$ by forgetting the coordinate $a_{\ell,x-1}$ in $A_\ell$. Then every $\tilde{A}_\ell$ defines an element in $\mathcal{P}_{m,x-1,y-j}^F$. Indeed, the conditions (30) remain trivially true and the condition (31) is implied by the definition of $j$ and the growth of $\sum_{i<x-1} a_{\ell,i}$ as a function of $\ell$:

$$ \sum_{i<x-1} a_{\ell,i} + j \le \sum_{i<x-1} a_{q-1,i} + a_{q-1,x-1} \le y. $$

We therefore get a chain $\tilde{A}_1 \le \tilde{A}_1 \le \tilde{A}_2 \le \dots \le \tilde{A}_{q-1}$ in $\mathcal{P}_{m,x-1,y-j}^F$.

To build back the original chain $A_1 \le A_2 \le \dots \le A_{q-1}$ from this truncated chain, one has to choose appropriate values $a_{\ell,x-1}$ for $1 \le \ell \le q-1$.

To recover $A_{q-1}$ in $\mathcal{P}_{m,x,y}^F$, one can only choose $a_{q-1,x-1} = j$. For this choice, the conditions (30) and (31) hold for $A_{q-1}$.

Now choose the other integers $a_{\ell,x-1}$ such that $0 \le a_{1,x-1} \le \dots \le a_{q-1,x-1} = j$. Then the conditions (30) and (31) for $A_\ell$ result from those for $A_{q-1}$. So we have indeed built back a chain in $\mathcal{P}_{m,x,y}^F$.

For each fixed $j$, the number of possible choices for the whole increasing sequence of integers $a_{\ell,x-1}$ ($1 \le \ell < q-1$) is therefore a binomial coefficient $\binom{j+q-2}{j}$. $\square$

**Lemma 6.5.** For all $m \ge 1$ and for all $y \ge 1$ and $x = my + 1$, there is a polynomial equality

$$ Z_{m,x,y}^{F}(u) = Z_{m,x,y-1}^{F}(u). \quad (35) $$

*Proof.* This is an easy computation using (32). $\square$

The following lemma is a computation using hypergeometric functions.

**Lemma 6.6.** For all $m \ge 1$ and for all $x, y$ satisfying (29) with $x \ge my + 1$, there is a polynomial equality

$$ Z_{m,x,y}^{F}(u) = \sum_{j=0}^{y} Z_{m,x-1,y-j}^{F}(u) \binom{j+u-2}{j}. \quad (36) $$

*Proof.* In the right hand side of (36), let us use the formula (32). One finds

$$ \sum_{j=0}^{y} \binom{j+u-2}{j} \frac{((x-m(y-j)-2)(u-1)+1)}{(y-j)!} \prod_{\ell=2}^{y-j} ((x-2)(u-1)+\ell). $$
Using the shorthands $\bar{u} = u - 1$ and $\bar{x} = x - 2$, this becomes

$$ \sum_{j=0}^{y} \binom{j+\bar{u}-1}{j} \frac{((\bar{x}-m(y-j))\bar{u}+1)}{(y-j)!} \prod_{\ell=2}^{y-j} (\bar{x}\bar{u}+\ell). $$

One can separate into two sums:

$$ \sum_{j=0}^{y} \binom{j+\bar{u}-1}{j} \frac{1}{(y-j)!} \prod_{\ell=1}^{y-j} (\bar{x}\bar{u}+\ell) - m\bar{u} \sum_{j=0}^{y} \binom{j+\bar{u}-1}{j} \frac{1}{(y-j-1)!} \prod_{\ell=2}^{y-j} (\bar{x}\bar{u}+\ell). $$

In each of these sums, one can replace the summation upper bound $y$ by $+\infty$ and get an hypergeometric expression:

$$ \frac{1}{y!} \prod_{\ell=1}^{y} (\bar{x}\bar{u} + \ell) \cdot {}_2F_1\left(\begin{array}{c} \bar{u}, -y \\ -y - \bar{x}\bar{u} \end{array}; 1\right) - \frac{m\bar{u}}{(y-1)!} \prod_{\ell=2}^{y} (\bar{x}\bar{u} + \ell) \cdot {}_2F_1\left(\begin{array}{c} \bar{u}, -y+1 \\ -y - \bar{x}\bar{u} \end{array}; 1\right). $$

Then using the Chu-Vandermonde formula for each term, one gets

$$ \frac{1}{y!} \prod_{\ell=1}^{y} (\bar{x}\bar{u} + \ell) \frac{(-y - \bar{x}\bar{u} - \bar{u})_y}{(-y - \bar{x}\bar{u})_y} - \frac{m\bar{u}}{(y-1)!} \prod_{\ell=2}^{y} (\bar{x}\bar{u} + \ell) \frac{(-y - \bar{x}\bar{u} - \bar{u})_{y-1}}{(-y - \bar{x}\bar{u})_{y-1}}, $$

where one has used some Pochhammer symbols. This can be simplified as

$$ \frac{1}{y!} \prod_{\ell=1}^{y} (\bar{x}\bar{u} + \bar{u} + \ell) - \frac{m\bar{u}}{(y-1)!} \prod_{\ell=2}^{y} (\bar{x}\bar{u} + \bar{u} + \ell). $$

which is the expected formula (32). $\square$

<figure><img src="image_12.png" /></figure>

Let $t_n$ be the linear arbor with $n$ vertices and no multiple vertex. It follows from the general Theorem 6.2 at $m = 1, x = n + 2, y = n$ that the Zeta polynomial of the poset $P_{t_n}$ for the arbor $t_n$ is the same as the known Zeta polynomial of the noncrossing-partitions lattice, first computed by Edelman [Ede80], namely

$$ \frac{1}{n!} \prod_{j=0}^{n-1} ((n+1)u-j). $$

## 6.2 M-triangle

Here we will describe the M-triangles of the posets $\mathcal{P}_{m,x,y}^F$.

Because all intervals between the unique minimum and any element are products of total orders, and by the same reasoning as in Section 2.5, the M-triangle is the sum

$$ \sum_{p \in \mathcal{P}^F(m,x,y)} (XY)^{\mathrm{ht}(a)} (1 - 1/X)^{\mathrm{nz}(a)}, \qquad (37) $$

where $\mathrm{ht}(a)$ is the sum of coordinates and $\mathrm{nz}(a)$ is the number of non-zero coordinates.
**Lemma 6.7.** Assume that $y > 0$ and $x > 1 + my$. Then

$$M_{m,x,y} = M_{m,x-1,y} + (1 - 1/X) \sum_{j=1}^{y} (XY)^j M_{m,x-1,y-j}. \quad (38)$$

*Proof.* Let us define a bijection $\varphi$ from $\mathcal{P}_{m,x,y}^F$ to the disjoint union

$$\bigsqcup_{j=0}^{y} \mathcal{P}_{m,x-1,y-j}^{F}.$$

The map $\varphi$ is defined by deleting the subword consisting of the $a_{x-1}$ steps Y just before the last X step, and also removing that X step. Let $j$ denote $a_{x-1}$. If $j = 0$, then one obtains a path in $\mathcal{P}_{m,x-1,y}^F$ because $x > 1 + my$. Otherwise, $j$ can take any value between 1 and $y$, and then the path obtained after deletion belongs to $\mathcal{P}_{m,x-1,y-j}^F$. All this clearly defines a bijection $\varphi$, as one can insert back the removed subword, depending on $j$, just after the last X.

If $j = 0$, the bijection $\varphi$ preserves both the height $ht$ and $nz$. Otherwise, the height $ht$ is decreased by $j$ and the number $nz$ is decreased by 1. This gives the expected formula, using (37). $\square$

Together with the initial conditions

$$M_{m,x,0} = 1 \quad (39)$$

because there is only one path in $\mathcal{P}_{m,x,0}^F$ and

$$M_{m,my+1,y} = M_{m,my+1,y-1} \quad (40)$$

because of the isomorphism of Lemma 6.1, this determines everything by induction.

**Proposition 6.8.** The M-triangle of $\mathcal{P}_{m,x,y}^F$ is the sum of

$$\frac{1}{\ell!} \binom{\ell}{k} (x-1-m\ell) \prod_{r=2}^{\ell} (x-r+k)(-X)^k (-Y)^{\ell}, \quad (41)$$

for $0 \le k \le \ell \le y$.

*Proof.* By induction. The statement holds in the two initial cases by inspection.

Let us expand the right-hand side of (38) as the sum of two sums, with the leftmost term of (38) being incorporated into the first sum:

$$\sum_{j=0}^{y} \sum_{0 \le k \le \ell \le y-j} \frac{1}{\ell!} \binom{\ell}{k} (x-2-m\ell) \prod_{r=2}^{\ell} (x-r-1+k)(-X)^{k+j} (-Y)^{\ell+j}$$

and

$$\sum_{j=1}^{y} \sum_{0 \le k \le \ell \le y-j} \frac{1}{\ell!} \binom{\ell}{k} (x-2-m\ell) \prod_{r=2}^{\ell} (x-r-1+k)(-X)^{k+j-1} (-Y)^{\ell+j}.$$
After a careful manipulation of sums, one obtains

$$ \sum_{0 \le k \le \ell \le y} \sum_{j=0}^{k} \frac{1}{(\ell-j)!} \binom{\ell-j}{k-j} (x-2-m(\ell-j)) \prod_{r=2}^{\ell-j} (x-r-1+k-j) (-X)^k (-Y)^\ell $$

and

$$ \sum_{0 \le k < \ell \le y} \sum_{j=1}^{k+1} \frac{1}{(\ell-j)!} \binom{\ell-j}{k-j+1} (x-2-m(\ell-j)) \prod_{r=2}^{\ell-j} (x-r+k-j) (-X)^k (-Y)^\ell. $$

We will now concentrate on the individual coefficients of $(-X)^k (-Y)^\ell$ in these two sums. Each coefficient is an affine function with respect to $m$. Let us first handle the constant part w.r.t. $m$. Using twice Lemma B.1, one gets the two contributions

$$ \frac{1}{k!(\ell-k)!} \prod_{r=1}^{\ell} (x-1-r+k) \quad \text{and} \quad \frac{1}{k!(\ell-k-1)!} \prod_{r=1}^{\ell-1} (x-1-r+k). $$

Summing both terms, one gets as expected

$$ \frac{1}{k!(\ell-k)!} (x-1) \prod_{r=2}^{\ell} (x-r+k). $$

Let us turn to the linear part w.r.t. $m$. Removing the factor $-m$ and using twice Lemma B.2, one obtains two contributions

$$ \frac{1}{k!(\ell-k)!} \frac{(x-1)\ell-k}{(x-1)(x-2)} \prod_{r=1}^{\ell} (x-1-r+k) $$

and

$$ \frac{1}{k!(\ell-k-1)!} \frac{(x-1)(\ell-1)-k}{(x-1)(x-2)} \prod_{r=1}^{\ell-1} (x-1-r+k). $$

Summing both terms gives as expected

$$ \frac{\ell}{k!(\ell-k)!} \prod_{r=2}^{\ell} (x-r+k). $$

□

The special case $m = 1$ and $y = n$ and $x = n + 2$, is related to the linear arbor $t_n$ of size $n$ with no multiple vertex.

**Proposition 6.9.** The M-triangle of the poset $P_{t_n}$ for the arbor $t_n$ is

$$ M_n^A = \sum_{0 \le k \le \ell \le n} \frac{n+1-\ell}{n+k-\ell+1} \binom{n+k}{n} \binom{n}{\ell-k} (-X)^k (-Y)^\ell. $$
This formula is essentially the same as the F-triangle for the cluster complex of type A_n. Recall that this F-triangle is given by [Cha04, Prop. 11]:

$$F_n^A = \sum_{k=0}^{n} \sum_{\ell=0}^{n} \frac{\ell+1}{k+\ell+1} \binom{n}{k+\ell} \binom{n+k}{n} X^k Y^\ell. \qquad (42)$$

It follows easily that

$$M_n^A(X, Y) = F_n^A(-X, -1/Y) \cdot (-Y)^n. \qquad (43)$$

**Proposition 6.10.** The transmuted M-triangle of $\mathcal{P}_{1,n+2,n}^F$ is the M-triangle of the poset of noncrossing-partitions of type A_n.

*Proof.* The transmutation of $M_n^A$ is

$$M_n^A \left( \frac{1-Y}{1-XY}, 1-XY \right) = F_n^A \left( \frac{1-Y}{XY-1}, \frac{1}{XY-1} \right) \cdot (XY-1)^n.$$

By the proof of the $F = M$ conjecture in [Ath07] and the transformations in Appendix A, the M-triangle for the poset of noncrossing-partitions of type A_n is given by

$$N(X, Y) = F_n^A \left( \frac{Y(X-1)}{1-XY}, \frac{XY}{1-XY} \right) \cdot (1-XY)^n.$$

Because the poset of noncrossing-partitions is self-dual, one has the symmetry

$$N(X, Y) = N(1/Y, 1/X) \cdot (XY)^n.$$

Combining these formulas gives the expected equality. □

An explicit formula for the M-triangle of the poset of noncrossing-partitions of type A_n, not needed here, can be found in [Kra06, Prop. A] with parameter m = 1.

Recall that $t_n$ denotes the arbor of size n with no multiple vertex. It follows from the previous results that the poset $P_{t_n}$ is a transmuted partner of the poset of noncrossing partitions of type A_n. In particular, the h-vector of $t_n$ is equal to the h-vector of the generalized associahedra of type A_n.

# 7 Arbors of type B

Let $n \ge 1$ and $k \ge 0$. Let $\mathcal{P}_{n,k}^B$ be the set of n-tuples $(x_1, \dots, x_n)$ of nonnegative integers with sum less than or equal to k. This is a poset for term-wise comparison.

The poset $\mathcal{P}_{n,n}^B$ is exactly the poset $P_t$ attached by the general construction to the arbor with one vertex containing n elements.

The posets $\mathcal{P}_{n,k}^B$ are graded by the height function $ht$ which is the sum of coordinates. They have a unique minimum element $\perp$. The definition of the poset $\mathcal{P}_{n,k}^B$ implies that its intervals are products of total orders.

**Lemma 7.1.** In a poset $\mathcal{P}_{n,k}^B$, the upper ideal generated by any element of height j is isomorphic to $\mathcal{P}_{n,k-j}^B$.
Proof. This follows directly from the definition. The bijection is given by subtraction of the chosen minimal element. □

**Theorem 7.2.** The Zeta polynomial of $\mathcal{P}_{n,k}^{\mathbb{B}}$ is equal to $\binom{n(u-1)+k}{k}$.

*Proof.* Let us consider a chain $e_1 \le \dots \le e_{u-1}$ in $\mathcal{P}_{n,k}^{\mathbb{B}}$. One can describe the chain by its first element $e_1$ together with all differences $u_{i+1} - u_i$ for $1 \le i \le u-2$. This gives a bijection with $(u-1)$-tuples of $n$ nonnegative integers whose total sum is less than or equal to $k$. The number of these tuples is the expected binomial coefficient. □

It follows from the general Theorem 7.2 at $k=n$ that the Zeta polynomial of the poset $P_t$ for the arbor with just one vertex of cardinality $n$ is the same as the known Zeta polynomial of the noncrossing-partitions lattice of type $\mathbb{B}$, first computed by Montenegro according to Reiner, see [Mon93, Rei97].

Let us now turn to the $M$-triangles of the posets $\mathcal{P}_{n,k}^{\mathbb{B}}$, namely

$$M_{n,k} = \sum_{a \le b} \mu(a, b) X^{|a|} Y^{|b|}.$$

As said above, by definition of $\mathcal{P}_{n,k}^{\mathbb{B}}$, all its intervals are products of total orders. By the same reasoning as in Section 2.5, this leads to the expression

$$M_{n,k} = \sum_{a \in \mathcal{P}_{n,k}^{\mathbb{B}}} (XY)^{\mathrm{ht}(a)} (1 - 1/X)^{\mathrm{nz}(a)}, \quad (44)$$

where $\mathrm{nz}$ is the number of non-zero coordinates.

In this sum, let us write $a$ as a pair $(j, b)$ where $j$ is the first coordinate and $b$ are the remaining coordinates seen as an element in $\mathcal{P}_{n-1,k-j}^{\mathbb{B}}$.

By separating the contribution for $j=0$, this allows to rewrite the sum as

$$M_{n,k} = M_{n-1,k} + (1 - 1/X) \sum_{j=1}^{k} (XY)^j M_{n-1,k-j}. \quad (45)$$

Together with the initial condition

$$M_{0,k} = 1 \quad (46)$$

for the singleton poset, this equation determines the $M$-triangle by induction on $n$.

**Proposition 7.3.** The $M$-triangle for the poset $\mathcal{P}_{n,k}^{\mathbb{B}}$ is

$$\sum_{r=0}^{k} \sum_{\ell=0}^{k-r} \binom{\ell + n - 1}{n - 1} \binom{n}{r} (-X)^{\ell} (-Y)^{\ell + r}. \quad (47)$$

*Proof.* This is done by checking that this formula satisfies the induction (45) and the initial condition (46). The initial condition is clear. For the induction step, the right hand side of (45) is the sum of

$$\sum_{j=0}^{k} \sum_{r=0}^{k-j} \sum_{\ell=0}^{k-j-r} \binom{\ell + n - 2}{n - 2} \binom{n-1}{r} (-X)^{\ell+j} (-Y)^{\ell+r+j}$$
and

$$ \sum_{j=1}^{k} \sum_{r=0}^{k-j} \sum_{\ell=0}^{k-j-r} \binom{\ell+n-2}{n-2} \binom{n-1}{r} (-X)^{\ell+j-1} (-Y)^{\ell+r+j}. $$

After replacing the summation variable $j$ by a new summation variable $\ell' = \ell + j$ and some further careful manipulations of summations, they become

$$ \sum_{r=0}^{k} \sum_{\ell'=0}^{k-r} \binom{\ell' + n - 1}{n - 1} \binom{n - 1}{r} (-X)^{\ell'} (-Y)^{\ell' + r} $$

and

$$ \sum_{r=1}^{k} \sum_{\ell'=0}^{k-r} \binom{\ell' + n - 1}{n - 1} \binom{n - 1}{r - 1} (-X)^{\ell'} (-Y)^{\ell' + r}. $$

Adding these two expressions then gives the expected expression. □

The special case $k = n$ is related to the arbor $t_n$ with one vertex containing $n$ elements.

**Proposition 7.4.** *The M-triangle of the poset $P_{t_n}$ for the arbor $t_n$ is*

$$ M_{n,n}(X,Y) = \sum_{r=0}^{n} \sum_{\ell=0}^{n-r} \binom{\ell+n-1}{n-1} \binom{n}{r} (-X)^{\ell} (-Y)^{\ell+r}. $$

This formula is essentially the same as the F-triangle for the cluster complex of type $\mathbb{B}_n$. Recall that this F-triangle is given by [Cha04, Prop. 13]:

$$ F_n^{\mathbb{B}} = \sum_{k=0}^{n} \sum_{\ell=0}^{n} \binom{n}{k+\ell} \binom{n+k-1}{n-1} X^k Y^\ell. \quad (48) $$

It follows easily that

$$ M_{n,n}(X,Y) = F_n^{\mathbb{B}}(-X, -1/Y) \cdot (-Y)^n. \quad (49) $$

**Proposition 7.5.** *The transmuted M-triangle of $\mathcal{P}_{n,n}^{\mathbb{B}}$ is the M-triangle of the poset of noncrossing-partitions of type $\mathbb{B}_n$.*

*Proof.* The transmutation of $M_{n,n}$ is

$$ M_{n,n} \left( \frac{1-Y}{1-XY}, 1-XY \right) = F_n^{\mathbb{B}} \left( \frac{1-Y}{XY-1}, \frac{1}{XY-1} \right) \cdot (XY-1)^n. $$

By the proof of the $F = M$ conjecture in [Ath07] and the conversion formulas in Appendix A, the M-triangle for the poset of noncrossing-partitions of type $\mathbb{B}_n$ is given by

$$ N(X,Y) = F_n^{\mathbb{B}} \left( \frac{Y(X-1)}{1-XY}, \frac{XY}{1-XY} \right) \cdot (1-XY)^n. $$

Because the poset of noncrossing-partitions is self-dual, one has the symmetry

$$ N(X,Y) = N(1/Y, 1/X) \cdot (XY)^n. $$

Combining these formulas gives the expected equality. □
An explicit formula for the M-triangle of the poset of noncrossing-partitions of type $\mathbb{B}_n$, not needed here, can be found in [Kra06, Prop. B] with parameter $m = 1$.

It follows that the poset $P_{t_n}$ for the arbor $t_n$ with one vertex containing $n$ elements is a transmuted partner of the poset of noncrossing partitions of type $\mathbb{B}_n$. In particular, the $h$-vector of $t_n$ is equal to the $h$-vector of the generalized associahedra of type $\mathbb{B}_n$.

Let us also record here the following value, useful in the next section.

**Proposition 7.6.** For $n \ge 1$, let $t_n$ be the arbor with just one vertex with $n$ elements. Then

$$K_{t_n}(X, Y) = \sum_{0 \le j \le k \le n} \binom{n}{j} \binom{k-1}{k-j} X^j Y^k. \qquad (50)$$

*Proof.* This is a direct application of Proposition 2.8 in the case without sub-tree. The double sum there is restricted to the unique term $\ell = j$ and $m = k$. $\square$

# 8 Arbors and halohedra

The halohedra form a sequence of simple convex polytopes. They have a topological interpretation as a compactification of a configuration space, similar to the case of associahedra and cyclohedra [DHV11, DFHV12]. They also have a combinatorial description in terms of “design tubings” of the cycle graphs, that makes them close relatives of graph-associahedra. The halohedra have also appeared in mathematical physics [SC18, Sal19].

There is one halohedron $H_n$ in dimension $n$ for each dimension $n \ge 1$, starting with a segment and a pentagon. The number of vertices of $H_n$ is

$$\frac{3n-1}{n} \binom{2n-2}{n-1}.$$

For $n \ge 2$, let $h_n$ be the arbor with $n-1$ elements in the root vertex plus one leaf with one element.

**Proposition 8.1.** For $n \ge 2$, the cardinality of $P_{h_n}$ is the number of vertices of $H_n$.

*Proof.* This is done using Proposition 2.1 at $m = 1$, using as starting point the generating polynomial $1 + X$ for the leaf vertex. One finds that the generating polynomial for $P_t$ with respect to height is

$$\sum_{j=0}^{n} \binom{n+j-2}{n-2} + \binom{n+j-3}{n-2} X^j.$$

This can be evaluated at $X = 1$ to

$$\binom{2n-1}{n-1} + \binom{2n-2}{n-1},$$

which is the expected result. $\square$
For $n \ge 1$, let $h'_n$ be the arbor with one element in the root vertex plus one leaf with $n-1$ elements. This is the reverse arbor of $h_n$.

**Proposition 8.2.** For $n \ge 1$, the cardinality of $P_{h'_n}$ is the number of vertices of $H_n$.

*Proof.* This is done using Proposition 2.1 at $m = 1$. For the leaf vertex with $n-1$ elements, the generating polynomial is

$$ \sum_{j=0}^{n-1} \binom{n-2+j}{n-2} X^j. $$

One finds that the generating polynomial for $P_t$ with respect to height is

$$ \sum_{j=0}^{n-1} \binom{n-1+j}{n-1} X^j + \binom{2n-2}{n-1} X^n. $$

At $X = 1$, this simplifies to

$$ \binom{2n-1}{n} + \binom{2n-2}{n-1}, $$

which is the expected result. $\square$

Let us now compute the $h$-vectors for these two families of arbors.

**Proposition 8.3.** As a polynomial, for $n \ge 2$, the $h$-vector of $h_n$ is

$$ \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{n}{j} X^j + \sum_{j=1}^{n} \binom{n-1}{j-1} X^j. \qquad (51) $$

*Proof.* One computes $K_{h_n}(X, 1)$ using Proposition 2.9, with starting point the polynomial $1 + XY$ for the leaf vertex. One gets exactly the expected formula for the coefficient of $X^j$. $\square$

One can easily check that this polynomial is palindromic.

**Proposition 8.4.** As a polynomial, for $n \ge 1$, the $h$-vector of $h'_n$ is

$$ \sum_{j=0}^{n-1} \binom{n-1}{j}^2 X^j + \sum_{j=1}^{n} \binom{n-1}{j-1} \binom{n}{j} X^j. \qquad (52) $$

*Proof.* One computes $K_{h_n}(X, 1)$ using Proposition 2.9, with starting point the polynomial from Proposition 7.6. One gets two terms for the coefficient of $X^j$. The first one is

$$ \sum_{k=j}^{n-1} \binom{n-1}{j} \binom{k-1}{k-j} $$

for $0 \le j \le n-1$, which sums to $\binom{n-1}{j}^2$. The second term is

$$ \sum_{k=j-1}^{n-1} (n-k) \binom{n-1}{j-1} \binom{k-1}{k-j+1} $$

for $1 \le j \le n$, which sums to $\binom{n-1}{j-1} \binom{n}{j}$. $\square$
It follows that the h-vectors for $h_n$ and $h'_n$ coincide, as predicted by Conjecture 3.2.

The h-vectors for the halohedra are known by [Alm22, Theorem 6.1.11], which says, once slightly adapted, that the generating series of h-vectors of halohedra for $n \ge 1$ is

$$h = \frac{1 + (1 + X)s}{2\sqrt{1 - 2(1 + X)s + (X - 1)^2 s^2}} - \frac{1}{2} = (X + 1)s + (X^2 + 3X + 1)s^2 + \dots$$

as a formal power series in $s$. This series therefore satisfies the algebraic equation

$$(1 - 2(1 + X)s + (X - 1)^2 s^2)(2h + 1)^2 = (1 + (1 + X)s)^2,$$

which can also be written as

$$(1 - 2(1 + X)s + (X - 1)^2 s^2)(h + h^2) = Xs^2 + Xs + s. \quad (53)$$

It is proved in Appendix C that the generating series of the common h-vectors for the two families of arbors considered above satisfies the algebraic equation (53). As they start with the same initial terms, this implies that they coincide with the h-vectors of the halohedra.

Let us note that halohedra are related to the Coxeter type $\mathbb{D}_n$: for $n \ge 2$, the h-vector for $H_n$ is equal, as a polynomial in $X$, to the sum of the h-vector for the associahedra of type $\mathbb{D}_n$, with $X$ times the h-vector for the associahedra of type $\mathbb{A}_{n-2}$ (see [FR07, Fig. 5.12] for the necessary formulas).

# 9 Hochschild polytopes

The Hochschild polytopes<fnref n="1" /> were introduced by Saneblidze in the context of algebraic topology of loop spaces [San09, RS18]. They form a sequence of simple polytopes, one in each dimension $n \ge 1$, starting with a segment in dimension 1 and a pentagon in dimension 2. The Hochschild polytope in dimension $n$ has $2^{n-2} \times (n+3)$ vertices and its h-vector is $(X+1)^{n-2}(X^2 + (n+1)X + 1)$ [Com21].

In the study of greedy Tamari lattices in [Cha20b], a lattice structure has been introduced on the vertices of each Hochschild polytope, in such a way that cover relations are edges of the polytope. These Hochschild lattices have been studied in detail in [Com21], who showed that they are congruence-uniform. Some further work can be found in [Pol20, Pol21, Müh22, PP24].

Let us consider the following sequence of arbors $t_n$ for $n \ge 1$. The arbor $t_n$ is made of a root vertex with one element, on which are grafted $n-1$ vertices, all with one element. Here is the example of $t_4$:

<figure><img src="image_13.png" /></figure>

<footnote n="1">¹They are also known as freehedra in anglo-greek.</footnote>
The Hochschild polytope in dimension $n$ is related to the poset $P_{t_n}$ by the following equalities.

**Proposition 9.1.** The number of vertices in the Hochschild polytope in dimension $n$ is the number of elements of the poset $P_{t_n}$. The h-vector of the Hochschild polytope is equal to the h-vector of $t_n$.

*Proof.* Using the recursive description of the Ehrhart polynomial at $m = 1$ in Proposition 2.1 and the initial value $F_{\circ,1} = 1+X$ for the arbor with one vertex, one finds that the number of elements of $P_{t_n}$ is

$$ \sum_{\ell=0}^{n-1} (n+1-\ell) \binom{n-1}{\ell} $$

By breaking this sum into two, one gets $(n+1)2^{n-1} - (n-1)2^{n-2}$. For the finer statement about the h-vectors, one uses the recursive description in Proposition 2.9 and the initial value $K_{\circ} = 1+XY$ for the arbor with one vertex. One first gets that

$$ K_{t_n}(X, 1) = \sum_{j=0}^{n-1} \binom{n-1}{j} X^j + \sum_{1 \le j \le n} (n-j+1) \binom{n-1}{j-1} X^j. $$

These two simple sums can be computed as above and the result follows. $\square$

In fact, there should be a more precise relationship. Let us consider the spherical simplicial complex $\mathcal{S}C_n$ which is the polar dual of the Hochschild polytope in dimension $n$.

**Conjecture 9.2.** In the simplicial complex $\mathcal{S}C_n$, there exists a facet $C$ such that the F-triangle for the pair $(\mathcal{S}C_n, C)$ coincides with the F-triangle associated with the transmuted M-triangle of $P_{t_n}$.

This can been checked concretely in small cases. It has been checked by computer for $n \le 9$. For $n \ge 3$, there seems to be exactly 4 facets with the expected property.

Moreover, it seems that the core-label-order of the Hochschild lattice of dimension $n$ could be a transmuted partner of the poset $P_{t_n}$. The coincidence of the Zeta polynomial with that of $P_{t_n}$ and the transmutation property for M-triangles have been checked by computer up to $n = 9$.

Using FriCAS [Fri24], one can guess the following conjectural formulas for various generating series in $s$ for the arbors $t_n$. For the Zeta polynomials of the posets $P_{t_n}$:

$$ 1 + \sum_{n \ge 1} Z_{t_n}(u, 1) s^n \stackrel{?}{=} \exp \left( \int \frac{u}{(1 - u s)(1 + s - u s)} ds \right). $$

For the M-triangles of the posets $P_{t_n}$:

$$ 1 + \sum_{n \ge 1} M_{t_n}(X, Y) s^n \stackrel{?}{=} \frac{(X Y s - Y s - 1)(X Y s - 1)}{(2 X Y s - Y s - 1)(X Y s - Y s + s - 1)}. $$
For the Ehrhart polynomials of the polytopes $Q_{t_n}$:

$$1 + \sum_{n \ge 1} \mathrm{Ehr}_{t_n}(u)s^n \stackrel{?}{=} \frac{1}{2} \left( 1 - \frac{s}{(us + s - 1)} - \frac{s - 1}{(us + s - 1)^2} \right).$$

For the Laplace transform of volume functions of the polytopes $Q_{t_n}$:

$$\sum_{n \ge 1} L_{t_n}(E, V)s^n \stackrel{?}{=} V \left( \frac{s}{EVs - Vs + 1} + \frac{Es}{Es - 1} \right).$$

# A Conversion formulas

We gather here the conversion formulas that we use between three different kinds of polynomials in two variables, namely F-triangles, M-triangles and H-triangles. These triangles and rational change-of-variables between them were introduced in [Cha04, Cha06] in the context of cluster fans and generalized associahedra. They are called triangles because of the shape of their Newton polygons. Each triangle comes with a size parameter $n$, which is the maximal value of either coordinate in the Newton polygon.

Given a pure simplicial complex of spherical topology SC and a chosen facet C, one can define the F-triangle as the following sum over all faces:

$$F(X, Y) = \sum_{s \in SC} X^{\mathrm{posi}(s)} Y^{\mathrm{nega}(s)}, \qquad (54)$$

where nega(s) (resp. posi(s)) is the number of elements in s that belong (resp. do not belong) to C.

Given a graded poset P, one can define the M-triangle of P as the following sum of Möbius numbers weighted using the rank function ht:

$$M(X, Y) = \sum_{a \le b} \mu(a, b) X^{\mathrm{ht}(a)} Y^{\mathrm{ht}(b)}. \qquad (55)$$

The M-triangles and F-triangles are converted to each other by

$$F(X, Y) = M \left( \frac{Y}{Y-X}, \frac{Y-X}{1+Y} \right) \cdot (1+Y)^n, \qquad (56)$$

where $n$ is the size parameter, and the similar birational formula

$$M(X, Y) = F \left( \frac{Y(X-1)}{1 - XY}, \frac{XY}{1 - XY} \right) \cdot (1 - XY)^n \qquad (57)$$

in the other direction.

We will not define here the H-triangles, that are related to the cohomology ring of the normal fan of the generalized associahedra. We give the following formulas because they motivate the definition of transmutation, as explained at the end of Section 4.

The M-triangles and H-triangles are converted to each other by

$$m(X, Y) = h\left(\left(X - 1\right)Y/(1 - Y), X/(X - 1)\right) \cdot (1 - Y)^n \qquad (58)$$

where $n$ is the size parameter, and the similar birational formula

$$h(X, Y) = m\left(Y/(Y-1), (Y-1)X/(1+(Y-1)X)\right) \cdot (1+(Y-1)X)^n \qquad (59)$$

for the conversion in the other way.
## B Two summation lemmas

These lemmas are used in Section 6.

**Lemma B.1.** *One has*

$$ \sum_{j \ge 0} \frac{1}{(k-j)!} \prod_{r=2}^{\ell-j} (x-r+k-j) = \frac{1}{k!} \frac{1}{x-1} \prod_{r=1}^{\ell} (x-r+k). \quad (60) $$

*Proof.* Write the left hand side as the value at $j = 0$, namely

$$ \frac{1}{k!} \prod_{r=2}^{\ell} (x-r+k), $$

times the hypergeometric series

$$ {}_2F_1 \left( \begin{array}{c} -k, 1 \\ -k+2-x \end{array} ; 1 \right). $$

The later evaluates to $\frac{x+k-1}{x-1}$ by Chu-Vandermonde. $\square$

**Lemma B.2.** *One has*

$$ \sum_{j \ge 0} \frac{1}{(k-j)!} (\ell-j) \prod_{r=2}^{\ell-j} (x-r+k-j) = \frac{1}{k!} \frac{x\ell-k}{x(x-1)} \prod_{r=1}^{\ell} (x-r+k). \quad (61) $$

*Proof.* Write the left hand side as the value at $j = 0$, namely

$$ \frac{1}{k!} \ell \prod_{r=2}^{\ell} (x-r+k), $$

times the hypergeometric series

$$ {}_3F_2 \left( \begin{array}{c} -k, 1, 1-\ell \\ -k+2-x, -\ell \end{array} ; 1 \right). $$

This hypergeometric series can be transformed<fnref n="2" /> into

$$ {}_2F_1 \left( \begin{array}{c} -k, 1 \\ -k+2-x \end{array} ; 1 \right) - \frac{k}{\ell(k+x-2)} {}_2F_1 \left( \begin{array}{c} 1-k, 2 \\ -k-x+3 \end{array} ; 1 \right). $$

The later evaluates to $\frac{(-k+\ell x)(x+k-1)}{\ell x(x-1)}$ by Chu-Vandermonde. $\square$

## C Some computations for halohedra

Following Proposition 8.3, let us define two power series in the variable s:

$$ A = \sum_{n \ge 1} \sum_{j=1}^{n} \binom{n-1}{j-1}^2 X^j s^n \quad \text{and} \quad B = \sum_{n \ge 1} \sum_{j=0}^{n-1} \binom{n-1}{j} \binom{n}{j} X^j s^n. \quad (62) $$

<footnote n="2">²We trust SageMath or Maxima here. This should follow from some contiguity relation.</footnote>
The aim of this section is to prove that $A + B$ coincide with the known generating function for $h$-vectors of halohedra, characterized by (53) and its first few terms. This is done with the help of the software FriCAS [Fri24] and the ore_algebra package for sagemath [KJJ15, The25].

Let us write $P$ for the polynomial $(X - 1)^2 s^2 - 2(X + 1)s + 1$.

First, one can guess an algebraic equation satisfied by $A$:

$$PA^2 = X^2 s^2. \qquad (63)$$

The operator $s\partial_s$ acts on both sides by multiplication by 2. After division by $2A$, one deduces a possible differential equation for $A$:

$$sP\partial_s A + (Xs + s - 1)A = 0,$$

with initial condition $A_0 = 0$. This is equivalent to the following possible recurrence for the coefficients of $A$:

$$(X - 1)^2 n A_n - (X + 1)(2n + 1)A_{n+1} + (n + 1)A_{n+2} = 0,$$

with initial conditions $A_0 = 0$ and $A_1 = X$. One then checks that the formula (62) for $A_n$ indeed satisfies this recurrence with the same initial conditions. So $A$ is the unique power series solution of (63) with zero constant term.

Let us now turn to the case of $B$. First, one can guess an algebraic equation satisfied by $B$:

$$P(B + B^2) = s. \qquad (64)$$

One can also guess a differential equation for $B$:

$$Xs - s + 1 + 2(Xs - s + 1)B + (Xs - s - 1)P\partial_s B = 0$$

with initial condition $B_0 = 0$. Let us consider in the ring $\mathbb{Q}[X, s, B, \partial_s B]$ the ideal generated by (64), its formal derivative with respect to $s$ and this differential equation. Computing the elimination ideal with respect to $\partial_s B$ gives back the principal ideal generated by (64). This proves that the differential equation is compatible with (64). This differential equation is equivalent to the following possible recurrence for the coefficients of $B$:

$$(X - 1)^3 n B_n - (X - 1)(3nX + n + 3X - 1)B_{n+1} + (3nX + n + 6X + 4)B_{n+2} - (n + 3)B_{n+3} = 0,$$

with initial conditions $B_0 = 0$, $B_1 = 1$ and $B_2 = 1 + 2X$.

One then checks that the formula (62) for $B_n$ indeed satisfies this recurrence with the same initial conditions. So one can conclude that $B$ is the unique power series solution of (64) with the given initial terms.

From the algebraic equations (63) and (64), one gets similar equations for $A/s$ and $B/s$. One then obtains by elimination an algebraic equation for the sum $(A + B)/s$, hence

$$P \cdot ((A + B) + (A + B)^2) = Xs^2 + (X + 1)s. \qquad (65)$$

Note that this requires choosing the correct factor among two in the reducible polynomial for $(A + B)/s$ obtained by elimination. This is done by identifying the constant term.

This implies the expected equality between $A + B$ and the generating series of $h$-vectors of halohedra defined by (53), after checking enough initial terms.
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