# An algebra over the operad of posets and structural binomial identities

José Antonio Arciniega-Nevárez1<EMAIL_ADDRESS>1División de Ingenierías,
Universidad de Guanajuato, Guanajuato, México  ,  Marko Berghoff2
<EMAIL_ADDRESS>2Mathematical Institute, University of Oxford, Oxford, UK
and  Eric Rubiel Dolores-Cuenca3<EMAIL_ADDRESS>3Department of
mathematics, Yonsei University, Seoul, Korea

###### Abstract.

We study the generating functions of strict and non-strict order polynomials
of series-parallel posets. These generating functions are closely related to
Ehrhart series and $h^{*}$-polynomials of the associated order polytopes. We
argue that working with power series instead of polynomials highlights the
structure of an underlying operad of posets. The action of this operad gives
rise to an algorithm that finds for a given power series $f(x)$, if possible,
a poset $X$ such that $f(x)$ is the generating function of the order
polynomial of $X$. Furthermore, we derive a new proof of Stanley’s reciprocity
theorem and we show that $h^{*}_{|X|}=0$ for series-parallel posets. As an
application we find new identities for binomial coefficients and describe
properties of the negative hypergeometric distribution.

##  1\. Introduction

Given a partially ordered set (poset) $X$, Stanley [18] considered the problem
of counting the number $\Omega^{\circ}(X,n)$ and $\Omega(X,n)$ of strict and
non-strict order preserving maps from $X$ to the poset $\langle
n\rangle=1<\ldots<n$. He showed that these are given by polynomials in $n$.
Moreover, he showed that one determines the other by a special instance of
what later became known as _combinatorial reciprocity_ [4]. Wagner [22]
described how order polynomials of (series-parallel) posets behave under the
operations of concatenation and disjoint union.

In this paper we take a different point of view. Instead of the order
polynomial we consider the generating functions of the sequences
$(\Omega^{\circ}(X,n))_{n\in\mathbb{N}}$ and $(\Omega(X,n))_{n\in\mathbb{N}}$,

$\mathfrak{Z}\left(X\right)=\sum_{n=1}^{\infty}\Omega^{\circ}(X,n)x^{n}\quad\text{
and }\quad\mathfrak{Z}^{+}\left(X\right)=\sum_{n=1}^{\infty}\Omega(X,n)x^{n}.$

We call them the _strict_ and _non-strict order series_ of the poset $X$. In
the case of chains $\langle n\rangle$ of length $n$ we find

$\mathfrak{Z}\left(n\right)=\frac{x^{n}}{(1-x)^{n+1}}\quad\text{ and
}\quad\mathfrak{Z}^{+}\left(n\right)=\frac{x}{(1-x)^{n+1}}.$

Let $P(X)$ denote the order polytope of a poset $X$ and $E_{P(X)}$ its Ehrhart
polynomial. Using results of Stanley [19] and Macdonald [17] we find

$\displaystyle\mathfrak{Z}^{+}\left(X\right)$
$\displaystyle=\sum_{n=1}^{\infty}\Omega_{X,n}x^{n}=x+\sum_{n=2}^{\infty}E_{P(X)}(n-1)x^{n}=x\frac{h^{*}(x)}{(1-x)^{|X|+1}}.$

In other words, up to a degree shift we work with the Ehrhart series of
$P(X)$, or, its $h^{*}$-polynomial (aka $h^{*}$-vector) if we use the basis
$\\{\frac{x^{i}}{(1-x)^{|X|+1}}\\}_{0\leq i\leq|X|}$ on the Ehrhart series of
$P(X)$. If we write the (strict) order series (or rather its analytic
continuation) in the basis $\\{\frac{x^{i}}{(1-x)^{i+1}}\\}_{1\leq i\leq|X|}$,
then

$\mathfrak{Z}\left(X\right)=\sum_{i=1}^{|X|}c_{i}\frac{x^{i}}{(1-x)^{i+1}}.$

The coefficients $c_{i}$ are closely related to the $f^{*}$-vector of $P(X)$
[6]. The $f^{*}$ vector correspond to the choice of basis
$\\{\frac{x^{i}}{(1-x)^{i}}\\}_{1\leq i\leq|X|+1}$ on the Ehrhart series of
$P(X)$.

Ehrhart series, $h^{*}$\- and $f^{*}$-vectors are well-studied objects in
combinatorics. They encode information about polytopes (or, more generally,
polytopal complexes). For example , the $f^{*}$-vector counts the number of
atomic lattice points of a polytope [6], the $h*$-vectors relates, among other
things, to its volume. When specialized to order polytopes they admit also
interpretations directly in terms of the poset. For instance, the
$h^{*}$-vector of $P(X)$ counts the number of linear extensions of $X$,
refined by the number of descents [19]. See [1, 3] for an overview.

We argue here that using order series has its own advantages. Firstly, their
coefficients have an obvious combinatorial interpretation (as the values of
order polynomials). Secondly, working in the basis
$\\{\frac{x^{i}}{(1-x)^{i+1}}\\}_{1\leq i\leq|X|}$ we can describe (strict)
order series as finite sums $\sum c_{i}\frac{x^{i}}{(1-x)^{i+1}}$ and show
that the coefficients $c_{i}$ encode topological properties of the underlying
poset (cf. see Remark 3.4). Thirdly, by working with power series we clarify
the structure of an algebra over an operad that is hidden by the definition of
the $h^{*}$-polynomials. The language of operads also simplifies the study of
Doppelgänger posets, that is, posets having the same order polynomial (cf.
Proposition 3.2).

A precise list of result is given in the next section. We now introduce the
operad structure under consideration.

Consider $\mathbf{P}(n)$ the set of posets with $n$ elements. The family
$\\{\mathbf{P}(n)\\}_{n\in\mathbb{N}}$ has an operad structure: Given
$\\{P_{i}\in\mathbf{P}(n_{i})\\}_{i=1,\cdots,N}$ and a poset
$P_{0}\in\mathbf{P}(N)$, we define
$P_{0}(P_{1},\cdots,P_{N})\in\mathbf{P}(\sum n_{i})$ as the result of
replacing the $i$ vertex of $P_{0}$ with $P_{i}$ for all $i$. Here, the new
order is the dictionary order; the first order is the order of $P_{0}$ and the
second is the order of each $P_{i}$. In other words, the image of every
element of $P_{i}$ is smaller than the image of every element of $P_{j}$ if
$i<_{P_{0}}j$, and if $x,y\in P_{i}$ then their images preserve the order of
$P_{i}$ (cf. [7]). The action of $S_{n}$ on $P(n)$ permutes the vertices of
the poset. In this setting, the disjoint union $\sqcup$ of two posets is given
by the operadic composition with $P_{0}=\\{x,y\\}$, which is commutative and
associative, while taking $P_{0}=\\{x<y\\}$ corresponds to the associative
concatenation of posets. We denote the latter operation in the following by
$\mu$.

In this paper we consider the suboperad $\mathcal{SP}$ generated by $\sqcup$
and $\mu$. Series-parallel posets and their order series are examples of
algebras over this operad. An operadic morphism between this algebras allow us
to compute the order series of an SP poset. First, we find the decomposition
of the poset in terms of $\mu,\sqcup$ via [21]. Then we apply the operadic
morphism between posets and power series, similarly to [10].

There is a closely related operad which has particular nice properties.
Consider the unary operation

$D(X)=\mu\left(\langle 1\rangle,\mu\big{(}\langle 1\rangle\bigsqcup X,\langle
1\rangle\big{)}\right).$

This add a new maximum $x_{1}$ and a new minimum $x_{0}$ to $X$, and a third
element $y$ with $x_{0}<y<x_{1}$. Geometrically we are attaching a handle
(with a point) to the Hasse diagram of $X$. There is a corresponding operad
$\mathcal{P}$ generated by $\mu$ and $D$ acting on a $\mathcal{P}$-algebra of
posets and power series.

###  1.1. Outline of results and structure of the paper

Note that most of the literature on order polynomials focuses on the _non-
strict_ order polynomial. However we derive and phrase our results in terms of
the _strict_ order series because this makes the formulae slightly simpler
(compare propositions 2.10 and 2.11 for instance). Moreover, from our basis we
find topological properties of the (Hasse diagram of the) underlying poset,
see for example Remark 3.4. After all, it does not matter much as both objects
are related by the respective versions of reciprocity.

In Section 2, we introduce our operadic point of view. We explain how posets
and power series can be viewed as algebras over the operads $\mathcal{SP}$ and
$\mathcal{P}$. In particular, we define a $\mathcal{P}$–algebra, called
Wixárika posets, that is obtained from the one-element poset $\langle
1\rangle$ by iterated applications of concatenation and attaching handles. An
example is given in Figure 2 (in this paper we depict Hasse diagrams ordered
from left to right).

We proceed then to derive expressions for the operations of $\mu$, $D$ and
$\sqcup$ on the level of power series. In our setting, the strict or non-
strict order series of a disjoint union $X\sqcup Y$ is given by the Hadamard
product of the order series of $X$ and $Y$, and the order series of the
concatenation of $X$ and $Y$ is given by the product of their order series,
times a factor $1-x$ in the strict and $\frac{1-x}{x}$ in the non-strict case.

In Section 2.2, we show that the analytic continuations of strict and non-
strict order series of a series-parallel poset $X$ are related by the
inversion $i\colon x\mapsto x^{-1}$,

$\mathfrak{Z}^{+}\left(X\right)=(-1)^{|X|+1}\mathfrak{Z}\left(X\right)\circ
i.$

Note that this is equivalent to Ehrhart/Macdonald reciprocity as formulated in
[19, Theorem 1.5], but we derive it solely in terms of the algebra structure
on the order series of $X$. We then proceed to show how this statement is
equivalent to Stanley’s well-known reciprocity theorem.

In Section 3, we discuss order series of Wixárika posets. To a poset, we
assign a sum $\sum c_{i}\mathfrak{Z}\left(i\right)$ indexed over a finite
number of chains. We demonstrate that the coefficients
$\\{c_{i}\\}_{i\in\mathbb{N}}$ encode topological information such as the
Betti numbers of the poset’s Hasse diagram. From the point of view of Ehrhart
theory, our results imply that by writing the Ehrhart series of a polytope on
the basis $\\{\frac{x^{i}}{(1-x)^{i+1}}\\}$, and if the polytope is the order
polytope of a Wixárika (or series-parallel) poset, the coefficients encode
topological information about the poset (see Proposition 3.1, Proposition
3.3). Conversely, given the order polytope of a series-parallel poset, then we
prove that $h^{*}_{|X|}=0$, as well as other properties of the $h^{*}$ vector
(see Lemma 2.23 and Corollary 2.26). These results reduce the search space of
algorithms to decide whether a given power series of the form $f(x)=\sum
a_{i}\frac{x^{i}}{(1-x)^{i+1}}$ is the order series of a series-parallel
poset, as we explain in Proposition 3.2.

In Section 4 we use our framework to derive identities for binomial
coefficients. The operadic action implies relations which we call structural
identities.

Given $p,q,s\leq p+q$ then for $t\leq s$:

${p+q+t\choose s}{s\choose t}=\sum_{\begin{subarray}{c}a+c=s\\\ n\leq a\\\
r\leq c\\\ n+r=t\end{subarray}}{p+n\choose a}{a\choose n}{q+r\choose
c}{c\choose r}-\sum_{\begin{subarray}{c}a+c=s-1\\\ n\leq a\\\ r\leq c\\\
n+r=t-1\end{subarray}}{p+n\choose a}{a\choose n}{q+r\choose c}{c\choose r}.$

Let $n$ be a non-negative integer and $k\in\mathbb{N}$. For any $k$-partition
$n=n_{1}+\cdots+n_{k}$, with $n_{i}\geq 0$, and for $m\geq n+k-1$ we have

${m\choose n}={k-1\choose 0}\sum_{\begin{subarray}{c}\sum m_{i}=m\\\ m_{i}\geq
n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose n_{i}}-{k-1\choose
1}\sum_{\begin{subarray}{c}\sum m_{i}=m-1\\\ m_{i}\geq
n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose n_{i}}\pm\cdots\\\
+(-1)^{k-2}{k-1\choose k-2}\sum_{\begin{subarray}{c}\sum m_{i}=m-(k-2)\\\
m_{i}\geq n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose
n_{i}}+(-1)^{k-1}{k-1\choose k-1}\sum_{\begin{subarray}{c}\sum
m_{i}=m-(k-1)\\\ m_{i}\geq n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose
n_{i}}.$

As far as we are aware, those identities are new. For example they cannot be
found in [12].

We also give a new proof that the _negative hypergeometric distribution_ is a
distribution and compute its expectation value.

We conclude with a section of open problems.

###  1.2. Related work

We give a (short) account of closely related work.

Consider the category of finite sets, and let B be the skeleton of the
category with objects finite sets but the only morphisms are isomorphisms. The
theory of B-species, see [5, 24], studies functors from B to finite sets and
power series associated to them. In comparison, we study a operadic
homomorphism from the algebra of series-parallel posets to power series. For
example, the poset $D(\langle 1\rangle)$ has the associated power series

$\frac{x^{3}}{(1-x)^{4}}+2\frac{x^{4}}{(1-x)^{5}}=x^{3}+6x^{4}+\cdots.$

There is no B-species with the corresponding generating function, because a
species only depends on the number of elements of the input set. We believe
our series are a topological version of species.

In this article, a $n$-labeling of a poset $X$ is a map to $\langle n\rangle$
that preserves the order. This is a different notion to [15] in which a
$n$-labeling on a poset is defined as a poset $X$ and a map of sets.

In regard to labelings of Hasse diagrams, there are many different variations
of labelings of (directed) graphs. See for instance [11]. However, we could
not find a notion that is related to ours.

The papers [2, 14, 20] study conditions on the coefficients of $h^{*}$\-
polynomials, and the paper [6] describes conditions on $f^{*}$-vectors. As far
as we are aware, our results are independent.

Note that if we let $t=0$ on Proposition 4.1 we recover the Vandermonde
identity. Most identities in [12] fix the top part of the combinatorial
expression and vary the bottom part. For example, consider the generalized
Vandermonde identity,

(1) ${m_{1}+\cdots+m_{p}\choose n}=\sum_{n_{1}+\cdots
n_{p}=n}\prod_{i=1}^{p}{m_{i}\choose n_{i}}.$

Fixing a partition $m_{1}+\cdots+m_{p}$, the formula relates the binomial
coefficient ${m_{1}+\cdots+m_{p}\choose n}$ with the product of binomial
coefficients ${m_{i}\choose n_{i}}$ over all possible $p$-partitions $n=\sum
n_{i}$. In comparison, in Corollary 4.2, we fix a partition
$n_{1}+\cdots+n_{p}$ and relate the binomial coefficient ${m\choose
n_{1}+\cdots+n_{p}}$ with the product of binomial coefficients ${m_{i}\choose
n_{i}}$ where the terms $m_{i}$ depend on $m$ and the length $p$ of the
partition.

While the multivariate negative hypergeometric distribution is a well known
topic in probability, as far as we are aware, the proofs in the section of
probability are new.

##  2\. Operad Algebras

In this section we work with algebras over the operad $\mathcal{P}$ generated
by $\mu,D$. We then explain how our results generalize for algebras over the
operad $\mathcal{SP}$ generated by $\mu,\sqcup$. We are interested in posets
and their order series as algebras over these operads. For an introduction to
operads see [16].

Any element of $\mathcal{P}$ can be represented by a binary rooted tree with
markings on the vertices of the tree. The binary operation $\mu$ is
represented by a branching point with two inputs and one output, the operation
$D$ is represented by a branching point with one input and one output, labeled
by a black dot. See Figure 1.

Figure 1. Left: The identity operation, the operation $D$ and the operation
$\mu$. Right: A tree representing an element of $\mathcal{P}$.

Given a set $A$, $End_{A}$ is the operad of endomorphisms of $A$. The set $A$
is an $\mathcal{P}$-algebra if there is a morphism $\mathcal{P}\rightarrow
End_{A}$, that is, the operations of $\mathcal{P}$ are defined on $A$.
Evaluation of a $\mathcal{P}$-operation is obtained by replacing the leaves of
the tree associated to the operation by elements of the algebra. We imagine
that elements of the algebra are drops of water. Whenever the elements of the
algebra reach an operation, we replace the input drops by the result of the
operation. We repeat until we reach the root.

We are interested in the $\mathcal{P}$–algebra generated by the poset with a
single point where $\mu$ is concatenation and $D$ is attaching a handle. We
call this the $\mathcal{P}$–algebra of _Wixárika posets_ and denote it by
$W_{Po}$. For an example see Figure 2, which depicts the Hasse diagram of the
Wixárika poset given by the $\mathcal{P}$ element in Figure 1. The name is due
to the resemblance of the Hasse diagrams of such posets to the intricate
colorful necklaces crafted by the Wixárika community in North America.

Figure 2. Hasse diagram of the right tree of Figure 1.

######  Remark 2.1.

An example of an algebra over $\mathcal{SP}$ are series-parallel posets (see
Figure 3). This algebra is generated by the poset with a single point.

Figure 3. Hasse diagram of a series-parallel poset
$X=(\\{x,y\\}\bigsqcup\\{z\\})*(\\{x,y\\}\bigsqcup\\{z\\}\bigsqcup\\{x,y\\})$.

In the remainder of this section we prove a couple of statements that are
valid for finite posets. They will be used to define other algebras over
$\mathcal{P}$ (and $\mathcal{SP}$).

######  Definition 2.2.

Let $X$ be a finite poset. Define $O_{X,n,+}$ as the set of _(weak or non-
strict) order preserving maps_ from $X$ to $\langle n\rangle$ and $O_{X,n}$ as
the set of _strict order preserving maps_ from $X$ to $\langle n\rangle$.

For a finite poset $Y$, the inclusions of $Y$ and $W$ in $\mu(Y,W)$ allow us
to see $Y$ and $W$ as a subposets of the concatenation. There is thus a
function

(2) $G_{n}:\bigcup_{l=1}^{n-1}O_{Y,l}\times O_{W,n-l}\to O_{\mu(Y,W),n}.$

On the term $O_{Y,l}\times O_{W,n-l}$ the function is given by
$G_{n}(h,k)(v)=h(v)$, if $v\in Y$, and $G_{n}(h,k)(v)=k(v)+l$, if $v\in W$.
This function is surjective. Furthermore, for any element
$f:\mu(Y,W)\rightarrow\langle n\rangle$ let $M$ be the maximum of $f|_{Y}$ and
$m$ the minimum of $f|_{W}$ (actually $m=M+i$ for some integer $i\geq 0$). If
$j\in\\{0,1,\cdots,i-1\\}$, we can partition $\langle n\rangle$ in $\langle
M+j\rangle$ and $\langle n-M-j\rangle$ in order to define
$h_{j}=f|_{Y}:Y\to\langle M+j\rangle$, and $k_{j}:W\to\langle n-M-j\rangle$,
given by $k_{j}(v)=f|_{W}(v)-M-j$. See Figure 4 for an example.

Figure 4. The function from $\langle 2\rangle$ to $\langle 4\rangle$ that
sends $1\mapsto 1,$ and $2\mapsto 4$ can be split in 3 different ways:
$\\{\langle 1\rangle\rightarrow\langle 1\rangle\\}\times\\{\langle
1\rangle\rightarrow\langle 3\rangle\\},\\{\langle 1\rangle\rightarrow\langle
2\rangle\\}\times\\{\langle 1\rangle\rightarrow\langle 2\rangle\\}$ and
$\\{\langle 1\rangle\rightarrow\langle 3\rangle\\}\times\\{\langle
1\rangle\rightarrow\langle 1\rangle$}.

We will need the following auxiliary lemma.

######  Lemma 2.3.

Assume that there is a sequence of sets

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{i}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$

where

  * •

$i:A\rightarrow B$ is injective,

  * •

$p:B\rightarrow C$ is surjective,

  * •

there is a split $s:C\to B$, such that $p\circ s=id$ and $s(C)\cap
i(A)=\varnothing$, and

  * •

for every point $c\in C,s(c)\bigsqcup(i(A)\cap p^{-1}(c))=p^{-1}(c)$.

Then $s(C)\bigsqcup i(A)=B$.

######  Remark 2.4.

A sequence satisfying the previous lemma is called a _splitting sequence of
sets._

######  Proposition 2.5.

Let $Y$ and $W$ be finite posets. There is a bijection between

$O_{\mu(Y,W),n}\cup\left(\bigcup_{a+b+1=n}O_{Y,a}\times O_{W,b}\right)$

and

$\bigcup_{c+d=n}O_{Y,c}\times O_{W,d}.$

###### Proof.

For $k\in O_{W,n-1}$ define $(k+1)\in O_{W,n+1}$ by $(k+1)(v)=k(v)+1$. The
shifting function

$Sh:\bigcup_{a+b=n-1}O_{Y,a}\times O_{W,b}\to\bigcup_{a+(b+1)=n}O_{Y,a}\times
O_{W,b+1}$

given by sending $(h,k)$ into $(h,k+1)$ is injective. Consider the function
defined in (2),

$G_{n}:\bigcup_{c+d=n}O_{Y,c}\times O_{W,d}\to O_{\mu(Y,W),n},$

and preimage $G_{n}^{-1}(f)$ of $f\in O_{\mu(Y,W),n}$. Then

$G_{n}^{-1}(f)-Sh\left(\bigcup_{a+b=n-1}O_{Y,a}\times O_{W,b}\right)$

has only one element, namely the pair $(h,k)$ where
$h=f|_{Y}:Y\rightarrow\langle m\rangle$ and $k:W\rightarrow\langle
n-m+1\rangle$ is given by $k(v)=f|_{W}(v)-m+1$ (again, $m$ is the minimum of
$f|_{W}$).

Choosing this unique point we define a section

$s:O_{\mu(Y,W),n}\to\bigcup_{c+d=n}O_{Y,c}\times O_{W,d}$

whose image is disjoint from the image of the shifting function $Sh$. This
holds because $k(m)=1$ but no function of the form $g+1$ has $1$ in the
codomain. Then the hypothesis of Lemma 2.3 holds and we conclude that there is
a set bijection

$O_{\mu(Y,W),n}\cup\left(\bigcup_{a+b+1=n}O_{Y,a}\times
O_{W,b}\right)\to\bigcup_{c+d=n}O_{Y,c}\times O_{W,d}.$

∎

For an integer $n>1$ define

$\triangle\langle n\rangle=\bigsqcup_{i=1}^{n}(\langle i\rangle,\langle
n-i\rangle).$

Also define ${O}_{(X,Y),\triangle\langle n\rangle}$ as the set of pairs of
strict order preserving maps $(k:X\to\langle i\rangle,h:Y\to\langle
n-i\rangle)$. In this notation, the implicit split sequence in Proposition 2.5
is written as

(3) $\textstyle{{O}_{(X,Y),\triangle\langle
n-1\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{O}_{(X,Y),\triangle\langle
n\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{O}_{\mu(X,Y),n}.}$

Recall that $O_{Y,n,+}$ is the set of order preserving maps. Given a pair
$(i,j)$, if $n=i+j$, then there is a function

$O_{(X,Y),(\langle i\rangle,\langle j\rangle),+}\to O_{\mu(X,Y),n-1,+}$

that takes the pair of order preserving maps $h:X\mapsto\langle i\rangle$ and
$k:Y\mapsto\langle j\rangle$ and defines $F:\mu(X,Y)\to\langle n-1\rangle$ by
$F(v)=h(v)$ if $v\in X$ and $F(v)=k(v)+i-1$ if $v\in Y$.

Note that the result of Proposition 2.5 is also valid for the split sequence

(4) $\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
27.88733pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{O_{(X,Y),\triangle\langle
n-1\rangle,+}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
51.88733pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
51.88733pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{O_{(X,Y),\triangle\langle
n\rangle,+}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
126.99533pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
126.99533pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{O_{\mu(X,Y),n-1,+}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

We will use these last two exact sequences in the following section.

###  2.1. Algebra of power series

We emphasize the fact that the order polynomials $\Omega^{\circ}(X,n)$ and
$\Omega(X,n)$ count the cardinality of $O_{X,n}$ and $O_{X,n,+}$,
respectively.

######  Definition 2.6.

For a poset $Y$ we define a formal power series, called its _order series_ ,
by

$\mathfrak{Z}\left(Y,x\right)=\sum_{n\geq 1}\Omega^{\circ}(Y,n)x^{n}.$

We use the auxiliary variable $x$ to keep track of the degree, but from now on
we omit it for ease of notation. For a chain $\langle n\rangle$ we simplify
the notation by $\mathfrak{Z}\left(\langle
n\rangle\right)=\mathfrak{Z}\left(n\right)$.

Recall the definition of multiset coefficients by
$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)\kern-3.00003pt\right)={n+k+1\choose
k}$. We frequently use the following identities,

(5) $\displaystyle\mathfrak{Z}\left(m\right)=\sum_{n=m}^{\infty}{n\choose
m}x^{n}=\frac{x^{m}}{(1-x)^{m+1}},$ (6)
$\displaystyle\frac{x}{(1-x)^{m+1}}=\sum_{n=1}^{\infty}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\kern-3.00003pt\right)x^{n}=\sum_{n=1}^{\infty}{n+m-1\choose
m}x^{n},$ (7)
$\displaystyle\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{m}\right)\kern-3.00003pt\right)=\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{m+1}{n-1}\right)\kern-3.00003pt\right).$

For a proof of Equation (5) see [23, Equation (1.5.5)] or [12, Equation
(1.3)]. Equation (6) follows from Equation (5) and Equation (7) can be checked
by direct inspection.

An alternative description for the power series of the chain $\langle
m\rangle$ can be read from Equation (5):

$\displaystyle\mathfrak{Z}\left(m\right)=\frac{x^{m}}{(1-x)^{m+1}}$
$\displaystyle=\sum_{n}{n\choose m}x^{n}$ (8)
$\displaystyle=\sum_{n}\frac{x^{m}}{m!}\frac{d^{m}}{dx^{m}}x^{n}$
$\displaystyle=\frac{x^{m}}{m!}\frac{d^{m}}{dx^{m}}\frac{1}{(1-x)}.$

If $x$ is the common variable for the power series
$\mathfrak{Z}\left(Y\right)$ and $\mathfrak{Z}\left(W\right)$, we define

$\mathfrak{Z}\left(Y\right)*\mathfrak{Z}\left(W\right)=\mathfrak{Z}\left(Y\right)(1-x)\mathfrak{Z}\left(W\right).$

Here we are using the Cauchy product of power series.

Suppose that there is a short splitting sequence of finite sets.

(9) $\textstyle{O_{(X,Y),\triangle\langle
n-i_{2}\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{O_{(X,Y),\triangle\langle
n-i_{1}\rangle}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{O_{\mu(X,Y),n}}$

with $i_{1}$ and $i_{2}$ non-negative integer and $n\in\mathbb{N}$ (see for
instance Equation (3)). Then

$\Omega^{\circ}(\mu(X,Y),n)=\Omega^{\circ}((X,Y),\triangle\langle
n-i_{1}\rangle)-\Omega^{\circ}((X,Y),\triangle\langle n-i_{2}\rangle).$

For $n\leq 1$ we request that $\Omega^{\circ}((X,Y),\triangle\langle
n-i_{2}\rangle)=0$. The Cauchy product satisfies

(10)
$\sum_{i=1}\Omega^{\circ}(X,i)x^{i}\sum_{j=1}\Omega^{\circ}(Y,j)x^{j}=\sum_{n=2}\Omega^{\circ}((X,Y),\triangle\langle
n\rangle)x^{n}.$

This property together with Equation (9) gives

(11) $\displaystyle\sum_{n=2}\Omega^{\circ}(\mu(Y,W),n)x^{n}=$ $\displaystyle
x^{i_{1}}\sum_{n=2}\Omega^{\circ}((Y,W),\triangle{\langle
n-i_{1}\rangle})x^{n-i_{1}}-$
$\displaystyle-x^{i_{2}}\sum_{n=2}\Omega^{\circ}((Y,W),\triangle{\langle
n-i_{2}\rangle})x^{n-i_{2}}$ $\displaystyle=$ $\displaystyle
x^{i_{1}}\sum_{j=1}\Omega^{\circ}(Y,j)x^{j}\sum_{k=1}\Omega^{\circ}(W,k)x^{k}$
$\displaystyle-x^{i_{2}}\sum_{j=1}\Omega^{\circ}(Y,j)x^{j}\sum_{k=1}\Omega^{\circ}(W,k)x^{k}.$

Then we have the following

######  Proposition 2.7.

Let $Y$ and $W$ be posets. Then

$\mathfrak{Z}\left(\mu(Y,W)\right)=\mathfrak{Z}\left(Y\right)*\mathfrak{Z}\left(W\right).$

###### Proof.

The sequence in Equation (3) is exact, so by Equation (11) with $i_{1}=0$ and
$i_{2}=1$ we have

$\mathfrak{Z}\left(\mu(Y,W)\right)=\sum_{n=2}\Omega^{\circ}(\mu(Y,W),n)x^{n}=\mathfrak{Z}\left(Y\right)(1-x)\mathfrak{Z}\left(W\right).$

∎

Note that

(12) $\displaystyle\mathfrak{Z}\left(n\right)*\mathfrak{Z}\left(m\right)$
$\displaystyle=\frac{x^{n}}{(1-x)^{n+1}}*\frac{x^{m}}{(1-x)^{m+1}}$
$\displaystyle=\frac{x^{n+m}}{(1-x)^{n+m+1}}$
$\displaystyle=\mathfrak{Z}\left(n+m\right).$

We can repeat the above discussion for non-strict order preserving maps.
Define

$\mathfrak{Z}^{+}\left(Y\right)=\sum_{n=1}^{\infty}\Omega(Y,n)x^{n},$

We define

(13)
$\mathfrak{Z}^{+}\left(Y\right)*^{+}\mathfrak{Z}^{+}\left(W\right)=\mathfrak{Z}^{+}\left(Y\right)\frac{(1-x)}{x}\mathfrak{Z}^{+}\left(W\right).$

Since the sequence in Equation (4) is exact, Equation (11) with $i_{1}=-1$ and
$i_{2}=0$ also proves

######  Proposition 2.8.

$\mathfrak{Z}^{+}\left(Y\right)*^{+}\mathfrak{Z}^{+}\left(W\right))=\mathfrak{Z}^{+}\left(\mu(Y,W)\right).$

In particular, $\mathfrak{Z}^{+}\left(n\right)=\frac{x}{(1-x)^{n+1}}$.

We denote the _Hadamard product_ of power series by

$\sum_{n=1}^{\infty}a_{n}x^{n}\odot\sum_{n=1}^{\infty}b_{n}x^{n}=\sum_{n=1}^{\infty}a_{n}b_{n}x^{n}.$

Since the number of $n$-labeling maps on a parallel composition $X\bigsqcup Y$
is the product of the number of $n$-labeling maps on each poset, we obtain:

(14) $\mathfrak{Z}\left(Y\bigsqcup
W\right)=\mathfrak{Z}\left(Y\right)\odot\mathfrak{Z}\left(W\right),\quad\mathfrak{Z}^{+}\left(Y\bigsqcup
W\right)=\mathfrak{Z}^{+}\left(Y\right)\odot\mathfrak{Z}^{+}\left(W\right)$

######  Proposition 2.9.

Let $Y=\bigsqcup_{i}Y_{i}$ and $W=\bigsqcup_{j}W_{j}$ be finite posets. Then

$\mathfrak{Z}\left(\mu(Y,W)\right)=\bigodot_{i}\mathfrak{Z}\left(Y_{i}\right)*\bigodot_{j}\mathfrak{Z}\left(W_{j}\right).$

###### Proof.

By Proposition 2.7 and (14)
$\mathfrak{Z}\left(\mu(Y,W)\right)=\mathfrak{Z}\left(Y\right)*\mathfrak{Z}\left(W\right)=\odot_{i}\mathfrak{Z}\left(Y_{i}\right)*\odot_{j}\mathfrak{Z}\left(W_{j}\right).$
∎

######  Proposition 2.10.

For $0\leq m,k:$

$\mathfrak{Z}\left(k\right)\odot\mathfrak{Z}\left(m\right)=\frac{x^{k}}{k!}\frac{d^{k}}{dx^{k}}\mathfrak{Z}\left(m\right)=\sum_{n=0}^{k}{m+n\choose
k}{k\choose n}\mathfrak{Z}\left(m+n\right).$

###### Proof.

From (8) we obtain

$\displaystyle\mathfrak{Z}\left(k\right)\odot\mathfrak{Z}\left(m\right)$
$\displaystyle=\sum_{n}{n\choose k}{n\choose m}x^{n}$
$\displaystyle=\sum_{n}\frac{x^{k}}{k!}\frac{d^{k}}{dx^{k}}{n\choose m}x^{n}$
$\displaystyle=\frac{x^{k}}{k!}\frac{d^{k}}{dx^{k}}\mathfrak{Z}\left(m\right)$

The proposition follows from the chain of equalities below.

$\displaystyle\mathfrak{Z}\left(k\right)\odot\mathfrak{Z}\left(m\right)$
$\displaystyle=\frac{x^{k}}{k!}\frac{d^{k}}{dx^{k}}\mathfrak{Z}\left(m\right)$
$\displaystyle=\frac{x^{k}}{k!}\frac{d^{k}}{dx^{k}}\frac{x^{m}}{(1-x)^{m+1}}$
$\displaystyle=\frac{x^{k}}{k!}\sum_{n=0}^{k}{k\choose
n}\left(\frac{d^{k-n}}{dx^{k-n}}x^{m}\right)\left(\frac{d^{n}}{dx^{n}}(1-x)^{-m-1}\right)$
$\displaystyle=\sum_{n=0}^{k}{k\choose
n}\frac{m!}{(m+n-k)!k!}\frac{(m+n)!}{m!}x^{m+n}(1-x)^{-m-n-1}$
$\displaystyle=\sum_{n=0}^{k}{m+n\choose k}{k\choose
n}\mathfrak{Z}\left(m+n\right).$

∎

######  Proposition 2.11.

For $0\leq m,k$

$\mathfrak{Z}^{+}\left(k\right)\odot\mathfrak{Z}^{+}\left(m\right)=\frac{x}{k!}\frac{d^{k}}{dx^{k}}\left(x^{k-1}\cdot\mathfrak{Z}^{+}\left(m\right)\right)$

is a linear combination of chains.

###### Proof.

Since disjoint union is expressed by the Hadamard product we have

$\displaystyle\mathfrak{Z}^{+}\left(k\right)\odot\mathfrak{Z}^{+}\left(m\right)$
$\displaystyle=\sum_{i\geq 1}\binom{i+k-1}{k}\binom{i+m-1}{m}x^{i}$
$\displaystyle=\sum_{i\geq
1}\binom{i+m-1}{m}\frac{x}{k!}\frac{d^{k}}{dx^{k}}x^{i+k-1}$
$\displaystyle=\frac{x}{k!}\frac{d^{k}}{dx^{k}}\left(x^{k-1}\cdot\mathfrak{Z}^{+}\left(m\right)\right).$

For the second statement we need two auxiliary results. The first is
$\frac{x^{m}}{(1-x)^{n+1}}=\frac{x^{m-1}}{(1-x)^{n+1}}-\frac{x^{m-1}}{(1-x)^{n}}$
which implies that for $m<n$, the term $\frac{x^{m}}{(1-x)^{n}}$ is a linear
combination of terms of the form $\frac{x}{(1-x)^{i}}$. The second auxiliary
result is

$\displaystyle\frac{d^{k}}{dx^{k}}\frac{x}{(1-x)^{n}}$ $\displaystyle=$
$\displaystyle\sum_{j}{k\choose
j}\frac{d^{k-j}}{dx^{k-j}}x\frac{d^{j}}{dx^{j}}\frac{1}{(1-x)^{n}}$
$\displaystyle=$ $\displaystyle
x\frac{n+k-1!}{n-1!}\frac{1}{(1-x)^{n+k}}+{k\choose
k-1}\frac{n+k-2!}{n-1!}\frac{1}{(1-x)^{n+k-1}}.$

Now we compute

$\displaystyle\mathfrak{Z}^{+}\left(k\right)\odot\mathfrak{Z}^{+}\left(m\right)$
$\displaystyle=$
$\displaystyle\frac{x}{k!}\frac{d^{k}}{dx^{k}}\left(x^{k-1}\cdot\mathfrak{Z}^{+}\left(m\right)\right)$
$\displaystyle=$ $\displaystyle\frac{x}{k!}\sum_{i}{k\choose
i}\frac{d^{k-i}x^{k-1}}{dx^{k-i}}\frac{d^{i}\mathfrak{Z}^{+}\left(m\right)}{dx^{i}}$
$\displaystyle=$ $\displaystyle\frac{x}{k!}\sum_{i}{k\choose
i}\frac{x^{i-1}(k-1)!}{(i-1)!}\frac{d^{i}\mathfrak{Z}^{+}\left(m\right)}{dx^{i}}$
$\displaystyle\overset{\eqref{aux}}{=}$
$\displaystyle\frac{x}{k!}\sum_{i}{k\choose
i}\frac{x^{i-1}(k-1)!}{(i-1)!}\frac{(m+i-1)!}{(m-1)!}\frac{x}{(1-x)^{m+i}}$
$\displaystyle+$ $\displaystyle\frac{x}{k!}\sum_{i}{k\choose
i}\frac{x^{i-1}(k-1)!}{(i-1)!}{i\choose
i-1}\frac{(m+k-2)!}{(m-1)!}\frac{1}{(1-x)^{m+i-1}}$ $\displaystyle=$
$\displaystyle\sum_{i}{k\choose i}\frac{m+i-1}{k}{m+i-2\choose
m-1}\frac{x^{i+1}}{(1-x)^{m+i}}$ $\displaystyle+$
$\displaystyle\sum_{i}{k\choose i}\frac{1}{k}{i\choose i-1}{m+i-2\choose
m-1}\frac{x^{i}}{(1-x)^{m+i-1}}.$

By the first auxiliary result this is a linear combination of non-strict order
series of chains. The exact coefficients can be obtained from Proposition 2.10
and Theorem 2.19. ∎

The fact that ${k\choose m}=0$ for $m>k$ implies

$\mathfrak{Z}\left(k\right)\odot\mathfrak{Z}\left(m\right)=\mathfrak{Z}\left(m\right)\odot\mathfrak{Z}\left(k\right)\text{
and
}\mathfrak{Z}^{+}\left(k\right)\odot\mathfrak{Z}^{+}\left(m\right)=\mathfrak{Z}^{+}\left(m\right)\odot\mathfrak{Z}^{+}\left(k\right).$

The previous discussion shows that the Hadamard product is closed in
$\mathbb{N}\left[\left\\{\mathfrak{Z}\left(j\right)\right\\}_{j}\right]$ and
$\mathbb{N}\left[\left\\{\mathfrak{Z}^{+}\left(j\right)\right\\}_{j}\right]$,
respectively.

######  Corollary 2.12.

The strict or non-strict order series of any series-parallel poset is a linear
combination of chains.

###### Proof.

The order series are generated by the chain $\frac{x}{(1-x)^{2}}$ under the
operations $*$ (or $*^{+}$) and $\odot$. These operations are linear on power
series and send chains into linear combinations of chains. ∎

The explicit description $\mathfrak{Z}\left(X\right)=\sum
a_{i}\mathfrak{Z}\left(i\right)$ can be found by evaluating on the tree of
operations as in [10], with the difference that we work with power series
instead of polynomials. Once we find the series, the $n$-order polynomial can
be obtained as $\sum a_{i}{i\choose n}$.

Finally we turn to the operation $D$ that we used to define Wixárika posets.
Abusing notation we set
$D(\mathfrak{Z}\left(X\right))=\mathfrak{Z}\left(D(X)\right)$ and
$D^{+}(\mathfrak{Z}^{+}\left(X\right))=\mathfrak{Z}^{+}\left(D(X)\right)$.
Note that $D(X)$ can also defined by $\langle 1\rangle*(\langle
1\rangle\bigsqcup X)*\langle 1\rangle$, so

(16)
$D(\mathfrak{Z}\left(n\right))=n\mathfrak{Z}\left(n+2\right)+(n+1)\mathfrak{Z}\left(n+3\right)$

and similar for $D^{+}$ (this case is discussed in the next section).

Summarizing the above, we have related the concatenation of posets (series
composition) and disjoint union (parallel composition) to concatenation of
power series $*$ and $*^{+}$, and Hadamard product, respectively. The order
series of the one-element poset is $\frac{x}{(1-x)^{2}}$. This fact allows us
to compute the strict and non-strict order series for any series-parallel
poset.

In the language of operads, we have an action of the operad $\mathcal{SP}$
generated by $\mu$ and $\sqcup$ on posets and power series. Restricting to
Wixárika posets, we have an action of $\mathcal{P}$ on them and on power
series. We call the $\mathcal{P}$–algebra generated by $\frac{x}{(1-x)^{2}}$
the $\mathcal{P}$–algebra of _(Wixárika) power series_ and denote it by
$W_{PS}$.

######  Proposition 2.13.

The morphisms $\mathfrak{Z}\left(\cdot\right)\colon W_{Po}\to W_{PS}$ from
Wixárika posets to Wixárika power series is a $\mathcal{P}$–algebra
homomorphism.

###### Proof.

This is a consequence of Proposition 2.7 and the definition of
$D(\mathfrak{Z}\left(X\right))$. ∎

Similarly, we may use Proposition 2.8 to derive the existence of another
$\mathcal{P}$-morphism $\mathfrak{Z}^{+}\left(\cdot\right)\colon W_{Po}\to
W_{PS}$ (by mapping $\mu$ to $*^{+}$ and $D$ to $D^{+}$).

######  Remark 2.14.

The operation $*$ is written in terms of the usual product of power series.
This means that it is distributive with respect to addition of power series.
Since $\frac{d}{dx}$ is also distributive with respect to addition, the
operation $D$ is also distributive. It follows that any iteration of $D$ and
$*$ is a finite sum of the form $\sum a_{j}\mathfrak{Z}\left(j\right)$. Due to
the explicit action of the operad, the $a_{j}$ are non-negative integer
coefficients.

######  Remark 2.15.

The $\mathcal{P}$–algebra morphism $\mathfrak{Z}\left(\cdot\right)$ from
$W_{Po}$ to $W_{PS}$ is surjective but not injective, since $*$ is commutative
on power series but $\mu$ is not commutative on posets. For example, consider
$\mu(\langle 1\rangle,D(\langle 3\rangle))$ and $\mu(D(\langle
3\rangle),\langle 1\rangle)$. They are different posets, but they have the
same associated power series. Posets with this property are called
Doppelgängers [13].

We conclude this section, with the following result explaining the interaction
between $*$ and $\odot$.

######  Lemma 2.16.

Let $s>0$. The following relationship holds for $f,g$ power series

(17)
$\mathfrak{Z}\left(s\right)\odot(f*g)=\sum_{a+c=s}(\mathfrak{Z}\left(a\right)\odot
f)*(\mathfrak{Z}\left(c\right)\odot g)-\\\
\left(\sum_{a+c=s-1}(\mathfrak{Z}\left(a\right)\odot
f)*(\mathfrak{Z}\left(c\right)\odot g)\right)*\mathfrak{Z}\left(1\right)$

###### Proof.

We use the notation ${s\choose a,b,c}=\frac{s!}{a!b!c!}$ for the multinomial
coefficient.

$\displaystyle\mathfrak{Z}\left(s\right)\odot(f*g)$ $\displaystyle=$
$\displaystyle\frac{x^{s}}{s!}\frac{d^{s}}{dx^{s}}(f(1-x)g)$ $\displaystyle=$
$\displaystyle\frac{x^{s}}{s!}\sum_{a+b+c=s}{s\choose
a,b,c}\frac{d^{a}}{dx^{a}}f\frac{d^{b}}{dx^{b}}(1-x)\frac{d^{c}}{dx^{c}}g$
$\displaystyle=$ $\displaystyle\frac{x^{s}}{s!}(\sum_{a+c=s}{s\choose
a,0,c}\frac{d^{a}}{dx^{a}}f(1-x)\frac{d^{c}}{dx^{c}}g$
$\displaystyle+\sum_{a+c=s-1}{s\choose
a,1,c}\frac{d^{a}}{dx^{a}}f\frac{d}{dx}(1-x)\frac{d^{c}}{dx^{c}}g)$
$\displaystyle=$ $\displaystyle\sum_{a+c=s}(\mathfrak{Z}\left(a\right)\odot
f)*(\mathfrak{Z}\left(c\right)\odot g)$
$\displaystyle-\frac{x^{s-1}}{s-1!}\sum_{a+c=s-1}{s-1\choose
a,c}\frac{d^{a}}{dx^{a}}f\frac{d^{c}}{dx^{c}}g(1-x)^{2}\frac{x}{(1-x)^{2}}$
$\displaystyle=$ $\displaystyle\sum_{a+c=s}(\mathfrak{Z}\left(a\right)\odot
f)*(\mathfrak{Z}\left(c\right)\odot g)$
$\displaystyle-\left(\sum_{a+c=s-1}(\mathfrak{Z}\left(a\right)\odot
f)*(\mathfrak{Z}\left(c\right)\odot g)\right)*\mathfrak{Z}\left(1\right)$

∎

###  2.2. A reciprocity theorem

A natural question to ask at this point is whether a relation exists between
$\mathfrak{Z}\left(X\right)$ and $\mathfrak{Z}^{+}\left(X\right)$ as in
Stanley’s reciprocity theorem [18, Theorem 3]:
$\Omega(X,n)=(-1)^{|X|}\Omega^{+}(X,-n)$. This is indeed the case. It follows
from Ehrhart/Macdonald reciprocity [19, Theorem 1.5] (see also [4]), but we
give here an independent proof.

######  Lemma 2.17.

The strict or non-strict order series of any series-parallel poset has an
analytic continuation to $\mathbb{P}^{1}-\\{1\\}$.

###### Proof.

In the previous section we have seen that the order series of any series-
parallel poset is a linear combination of order series of chains. Since both
$\frac{x}{(1-x)^{n+1}}$ and $\frac{x^{n}}{(1-x)^{n+1}}$ are analytic on
$\mathbb{P}^{1}-\\{1\\}$ for every $n>0$, the result follows. ∎

Abusing notation we denote these analytic continuations also by
$\mathfrak{Z}\left(X\right)$ and $\mathfrak{Z}^{+}\left(X\right)$.

######  Remark 2.18.

The singularities of $\mathfrak{Z}\left(X\right)$ and
$\mathfrak{Z}^{+}\left(X\right)$ at 1 are simple poles. It follows that the
Moebius transformation $x\mapsto\frac{1}{1-x}$, sending 1 to $\infty$, turns
the analytic continuations of order series into polynomials. This is very
reminiscent of the relation between Ehrhart seris and $h*$-polynomials. For
example, for chains we obtain

$\mathfrak{Z}\left(n\right)|_{y=\frac{1}{1-x}}=\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}y^{n-k+1},\quad\mathfrak{Z}^{+}\left(n\right)|_{y=\frac{1}{1-x}}=y^{n+1}-y^{n}.$

Now let $i$ denote the map $x\mapsto x^{-1}$ and define a map $\iota$ as
follows. If $f$ is the analytic continuation of a strict or non-strict order
series of a poset $X$, then

$\iota(f):=(-1)^{|X|+1}\big{(}f\circ i\big{)}.$

######  Theorem 2.19.

For any series-parallel poset $X$ we have

$\iota(\mathfrak{Z}\left(X\right))=\mathfrak{Z}^{+}\left(X\right)\text{ and
}\iota^{2}=\mathrm{id}.$

In other words, $\mathfrak{Z}\left(X\right)$ is the expansion of its analytic
continuation at 0, while $\mathfrak{Z}^{+}\left(X\right)$ is the expansion at
$\infty$ of the _same_ function (up to a sign).

To prove the theorem, we will show that $\iota$ commutes with concatenation
and disjoint union of chains, that is, it is a morphism between two
$\mathcal{SP}$-algebras. Since these operations generate every series-parallel
poset, the claim will follow. For the case of disjoint union we need a
preparatory calculation.

######  Lemma 2.20.

Let $f$ be $n$-times differentiable and $x\neq 0$. Then

$\frac{x^{n}}{n!}\frac{d^{n}}{dx^{n}}(f\circ
i)=(-1)^{n}\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{x^{-k}}{k!}\frac{d^{k}f}{dx^{k}}({x}^{-1}).$

###### Proof.

Let us abbreviate $\frac{d^{n}}{dx^{n}}$ by $d_{n}$. Applying _Faà di Bruno
formula_ for the chain rule of higher derivatives we get

$d_{n}(f\circ
i)=\sum_{k=1}^{n}d_{k}f({x}^{-1})B_{nk}(d_{1}i,d_{2}i,\ldots,d_{n-k+1}i).$

Here the _(incomplete) Bell polynomials_ $B_{nk}$ are defined by

$B_{nk}(x_{1},\ldots,x_{n-k+1})=\sum\frac{n!}{j_{1}!j_{2}!\cdots
j_{n-k+1}!}\left(\frac{x_{1}}{1!}\right)^{j_{1}}\cdots\left(\frac{x_{n-k+1}}{(n-k+1)!}\right)^{j_{n-k+1}},$

where the sum is taken over all sequences $j_{1},\ldots,j_{n-k+1}$ of non-
negative integers such that

$j_{1}+j_{2}+\cdots+j_{n-k+1}=k,\quad
j_{1}+2j_{2}+3j_{3}+\cdots+(n-k+1)j_{n-k+1}=n.$

Now we compute $B_{nk}(d_{1}i,d_{2}i,\ldots,d_{n-k+1}i)$ to be given by

$\displaystyle\sum\frac{n!}{j_{1}!j_{2}!\cdots
j_{n-k+1}!}\left(\frac{-x^{-2}}{1!}\right)^{j_{1}}\cdots\left(\frac{(-1)^{n-k+1}(n-k+1)!x^{-n+k-2}}{(n-k+1)!}\right)^{j_{n-k+1}}$
$\displaystyle=$ $\displaystyle\sum\frac{n!}{j_{1}!j_{2}!\cdots
j_{n-k+1}!}(-1)^{\sum_{i=1}^{n-k+1}ij_{i}}x^{-\sum_{i=1}^{n-k+1}ij_{i}}x^{-\sum_{i=1}^{n-k+1}j_{i}}$
$\displaystyle=$ $\displaystyle\sum\frac{n!}{j_{1}!j_{2}!\cdots
j_{n-k+1}!}(-1)^{n}x^{-n}x^{-k}$ $\displaystyle=$
$\displaystyle(-1)^{n}x^{-n}x^{-k}B_{nk}(1!,\ldots,(n-k+1)!)$ $\displaystyle=$
$\displaystyle(-1)^{n}x^{-n}x^{-k}\binom{n-1}{k-1}\frac{n!}{k!}$

where the last identity is a well-known property of the incomplete Bell
polynomials [8]. ∎

###### Proof of Theorem 2.19.

Since $(1-x^{-1})=-\frac{(1-x)}{x}$, we see that $\iota$ commutes with
concatenation,

$\iota(\mathfrak{Z}\left(n\right)*\mathfrak{Z}\left(m\right))=(-1)^{n+m+1}\left(\mathfrak{Z}^{+}\left(n\right)\frac{-(1-x)}{x}\mathfrak{Z}^{+}\left(m\right)\right)=\iota\mathfrak{Z}\left(n\right)*^{+}\iota\mathfrak{Z}\left(m\right).$

We claim

$\iota\mathfrak{Z}^{+}\left(n\right)\odot\iota\mathfrak{Z}^{+}\left(m\right)=\iota\big{(}\mathfrak{Z}^{+}\left(n\right)\odot\mathfrak{Z}^{+}\left(m\right)\big{)}.$

Since $\iota^{2}=\mathrm{id}$ this readily also implies

$\iota\big{(}\mathfrak{Z}\left(n\right)\odot\mathfrak{Z}\left(m\right)\big{)}=\iota^{2}\left(\mathfrak{Z}^{+}\left(n\right)\odot\mathfrak{Z}^{+}\left(m\right)\right)=\iota\mathfrak{Z}\left(n\right)\odot\iota\mathfrak{Z}\left(m\right).$

Abbreviate $\mathfrak{Z}^{+}\left(m\right)$ by $f$. Using Lemma 2.20 we
calculate

$\displaystyle\iota\mathfrak{Z}^{+}\left(n\right)\odot\iota(f)$
$\displaystyle=\mathfrak{Z}\left(n\right)\odot\iota(f)=(-1)^{m+1}\frac{x^{n}}{n!}d_{n}(f\circ
i)$
$\displaystyle=(-1)^{m+n+1}\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{x^{-k}}{k!}d_{k}f({x}^{-1})=\iota\left(\mathfrak{Z}\left(n\right)\odot
f\right).$

∎

From the calculations above it follows that for non-strict order series the
operation of attaching a handle is represented by

$D^{+}(f)=\mathfrak{Z}^{+}\left(1\right)*^{+}\big{(}f\odot\mathfrak{Z}^{+}\left(1\right)\big{)}*^{+}\mathfrak{Z}^{+}\left(1\right)=\frac{x}{(1-x)^{2}}\frac{df}{dx}.$

In fact, $\iota$ also commutes with $D$,

$\displaystyle\iota D(\mathfrak{Z}\left(X\right))$
$\displaystyle=\iota\left(\frac{x^{3}}{(1-x)^{2}}\frac{d\mathfrak{Z}\left(X\right)}{dx}\right)$
$\displaystyle=\frac{(-1)^{|X|+4}}{x(1-x)^{2}}\
\frac{d\mathfrak{Z}\left(X\right)}{dx}(x^{-1})$
$\displaystyle=(-1)^{|X|+5}\frac{x}{(1-x)^{2}}\frac{d(\mathfrak{Z}\left(X\right)\circ
i)}{dx}$ $\displaystyle=D^{+}\iota\big{(}\mathfrak{Z}\left(X\right)\big{)}.$

######  Example 2.21.

We compute the non-strict order series of the poset $X=D\big{(}\langle
2\rangle\big{)}$. Its strict order series is

$\mathfrak{Z}\left(X\right)=D\big{(}\mathfrak{Z}\left(2\right)\big{)}=\frac{x^{3}}{(1-x)^{2}}\frac{d}{dx}\left(\frac{x^{2}}{(1-x)^{3}}\right)=\frac{x^{3}(2x+x^{2})}{(1-x)^{6}},$

and therefore

$\mathfrak{Z}^{+}\left(X\right)=\iota\Big{(}D\big{(}\mathfrak{Z}\left(2\right)\big{)}\Big{)}=\frac{x^{-3}(2x^{-1}+x^{-2})}{(1-x^{-1})^{6}}=\frac{x(2x+1)}{(1-x)^{6}}$

which indeed is equal to

$D^{+}\big{(}\mathfrak{Z}^{+}\left(2\right)\big{)}=\frac{x}{(1-x)^{2}}\frac{d}{dx}\left(\frac{x}{(1-x)^{3}}\right).$

It is important to note that the map $\iota$ is not linear. For instance,
applying it term by term to the representation
$\mathfrak{Z}\left(X\right)=2\mathfrak{Z}\left(4\right)+3\mathfrak{Z}\left(5\right)$
does not give the desired answer; the sign rule is determined by the number of
elements of $X$, not of the components in a factorization.

We now show that our reciprocity theorem implies Stanley’s version, and vice
versa.

######  Proposition 2.22.

For series-parallel posets Theorem 2.19 is equivalent to Stanley’s reciprocity
theorem.

###### Proof.

Corollary 2.12 states that for a series-parallel poset $X$ its order series
$\mathfrak{Z}\left(X\right)$ is a linear combination of chains
$\mathfrak{Z}\left(X\right)=\sum_{i=1}^{|X|}a_{i}\mathfrak{Z}\left(i\right)$.
This implies $\Omega^{\circ}(X,n)=\sum_{i=1}^{|X|}a_{i}{n\choose i}$ because
the term of degree $n$ in $\mathfrak{Z}\left(i\right)$ is ${n\choose i}$.
Assuming Stanley reciprocity we obtain

$\Omega(X,n)=(-1)^{|X|}\sum_{i}a_{i}{-n\choose
i}=(-1)^{|X|}\sum_{i}(-1)^{i}a_{i}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{i}\right)\kern-3.00003pt\right).$

Then

$\displaystyle\mathfrak{Z}^{+}\left(X\right)$ $\displaystyle=$
$\displaystyle\sum_{n}\Omega(X,n)x^{n}$ $\displaystyle=$
$\displaystyle(-1)^{|X|}\sum_{i}(-1)^{i}\sum_{n}a_{i}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{i}\right)\kern-3.00003pt\right)x^{i}$
$\displaystyle=$
$\displaystyle(-1)^{|X|}\sum_{i}(-1)^{i}a_{i}\frac{x}{(1-x)^{i+1}}$
$\displaystyle=$
$\displaystyle(-1)^{|X|+1}\sum_{i}a_{i}\frac{(\frac{1}{x})^{i}}{(1-\frac{1}{x})^{i+1}}$
$\displaystyle=$ $\displaystyle\iota(\mathfrak{Z}\left(X\right)).$

On the other hand, from Corollary 2.12 we obtain
$\mathfrak{Z}\left(X\right)=\sum a_{i}\mathfrak{Z}\left(i\right)$. By
comparing coefficients of $\mathfrak{Z}^{+}\left(X\right)$ and
$\iota(\mathfrak{Z}\left(X\right))$ we recover the Stanley reciprocity
theorem. ∎

###  2.3. $h*$-polynomials

We conclude by relating the $h^{*}$-polynomial
$h^{*}(x)=\sum_{i=0}^{|X|}h^{*}_{i}x^{i}$ of the order polytope of $X$ to its
strict order series $\mathfrak{Z}\left(X\right)$. From Corollary 2.12, we know
that there exist $c_{i}$ such that
$\mathfrak{Z}\left(X\right)=\sum_{i=1}^{|X|}c_{i}\frac{x^{i}}{(1-x)^{i+1}}.$

######  Lemma 2.23.

For a series-paralell poset $X$ we have $h^{*}_{|X|}=0$

###### Proof.

By definition we have
${\mathfrak{Z}^{+}\left(X\right)}{}=x\frac{h^{*}(x)}{(1-x)^{|X|+1}}.$ But note
that $\frac{x^{|X|+1}h^{*}_{|X|}}{(1-x)^{|x|+1}}$ can not be in the algebra
$\\{\frac{x}{(1-x)^{i+1}}\\}_{1\leq i\leq|X|}$ unless $h^{*}_{|X|}=0$. ∎

######  Remark 2.24.

Note that for a series-parallel poset $c_{|X|}\neq 0$ either by the explicit
definition of $*$ and $\odot$ or by Ehrhart theory since this expression is
the volume of the polytope.

######  Proposition 2.25.

If $X$ is a series-parallel poset, then

(18) $h_{j}^{*}=\sum_{i=1}^{|X|-j}(-1)^{|X|-i+j}c_{i}{|X|-i\choose j}$

and

(19) $c_{|X|}={\sum_{{i=0}}^{|X|}h^{*}_{i}},\quad
c_{j}=(-1)^{j}\sum_{\begin{subarray}{c}i=j\end{subarray}}^{|X|}h^{*}_{i}{i\choose
j},\quad 0<j<|X|.$

###### Proof.

Follows from Theorem 2.19 by comparing coefficients. ∎

######  Corollary 2.26.

For a chain $\langle n\rangle$ the $h^{*}$-polynomial satisfies
$\sum_{\begin{subarray}{c}i=j\end{subarray}}^{|X|}(-1)^{j}h^{*}_{i}{i\choose
j}=0$ for every $0<j<|X|$.

###### Proof.

This follows because for a chain $\langle n\rangle$ the coefficients $c_{i}=0$
for $i<|n|$. ∎

Note that given a poset $X$, we can construct the order polytope, construct
the normal fan of this polytope, and associate a toric variety [9]. In this
situation the Ehrhart series coincides with the Hilbert series and our results
apply as well.

##  3\. Representability of Wixárika posets

Recall that elements of the operad $\mathcal{P}$ are rooted labeled trees. We
can describe them by words in the characters $\mu$ and $D$. Let $w$ be such an
element. By abuse of notation, we write $w(\langle 1\rangle)$ to represent the
substitution of the poset $\langle 1\rangle$ on all leaves of the tree
describing the word $w$. In [10] an algorithm is described to compute the
order polynomials. The same algorithm can be adapted to our setting to compute
order series, and in fact, from the point of view of operads, the algorithm
amounts to evaluate the operations on the tree from top to bottom. Similarly,
we write $w(\mathfrak{Z}\left(1\right))$ to represent the substitution of the
corresponding elements in power series. For example, the tree of Figure 1
corresponds to $w=D(\mu(,\mu(,\mu(,D(D(\mu(D(),\mu(,D()))))))))$. Thus
$w(\mathfrak{Z}\left(1\right))=882\mathfrak{Z}\left(16\right)+7995\mathfrak{Z}\left(17\right)+27232\mathfrak{Z}\left(18\right)+143792\mathfrak{Z}\left(19\right)+33552\mathfrak{Z}\left(20\right)+9880\mathfrak{Z}\left(21\right)$.

We will say that a power series $f(x)$ is _represented_ if there is a series-
parallel poset $X$ such that $f(x)=\mathfrak{Z}\left(X\right)$. In order to
determine which power series can be represented, we shall determine
topological invariants of a poset $X$ from its associated power series
$\mathfrak{Z}\left(X\right)$, such as the Betti number of $X$ (which is short
for _the first Betti number of the Hasse diagram of $X$_). With that
information, we describe an algorithm which determines if $f(x)$ can be
represented and if so, it explicitly constructs all possible series-parallel
posets $X$ (Doppelgänger) so that $f(x)=\mathfrak{Z}\left(X\right)$.

######  Proposition 3.1.

Consider $w\in\mathcal{P}$, a word in the characters $\mu$ and $D$. Let
$f(x)=w(\mathfrak{Z}\left(1\right))$ and $X=w(\langle 1\rangle)$.

  1. (1)

The series $f(x)$ admits a unique description

$f(x)=a_{i}\mathfrak{Z}\left(i\right)+a_{i+1}\mathfrak{Z}\left(i+1\right)+\cdots+a_{k}\mathfrak{Z}\left(k\right),$

where the coefficients $a_{j}$ are non-negative integers.

  2. (2)

The number $i$ is the number of elements in a maximal chain in $X$.

  3. (3)

The number $k$ is the number of elements in $X$.

  4. (4)

The difference $d=k-i$ is the Betti number of $X$, equal to the number of
times the character $D$ occurs in $w$.

  5. (5)

The difference $m=i-2d-1$ is the number of times the character $\mu$ occurs in
$w$.

  6. (6)

The number $m+1$ is the number of leaves in the tree $w$.

  7. (7)

$\sum_{u=1}^{k}(-1)^{k-u}a_{u}=1$

  8. (8)

The first coefficient $a_{i}$ admits a factorization
$a_{i}=\prod_{j=1}^{d}t_{j}$ where for any $t_{j}\geq 1$ there are
$r_{t_{j}},s_{t_{j}},$ and $u_{t_{j}}$, so that
$(r_{t_{j}},t_{j},s_{t_{j}},u_{t_{j}})$ is an input of $D$ in the evaluation
$w(1,1,1,1)$.

###### Proof.

We shall give a proof of each item.

  1. (1)

This was proved in Remark 2.14.

  2. (2)

If $\mu$ appears in $w$, it concatenates subposets of any maximal chain. If
$D$ appears in $w$, it adds a maximum and a minimum point to the maximal chain
in the input poset, which is a subposet of any maximal chain.

We conclude from Equation (16) and Equation (12) that when we evaluate $w$,
the lowest term counts the number of points in a maximal chain.

  3. (3)

We prove this in Lemma 3.3 in a more general setting.

  4. (4)

From Equation (16) it follows that we can track the number of times $D$ acts
by comparing the parameter of the monomial with lowest degree and the monomial
of highest degree. Geometrically, $D$ modifies the Hasse diagram by increasing
the number of holes.

  5. (5)

If the word $D$ occurs $d$ times in $w$, the only way to obtain the term
$\mathfrak{Z}\left(i\right)$ is if the word $\mu$ occurs $i-2d-1$ times in
$w$.

  6. (6)

Consider the representation of $w$ as a binary tree with some marked edges. If
the tree has $m$ binary branches, then it has $m+1$ leaves.

  7. (7)

Follows from Equation (18) and the fact that $h_{0}^{*}=1$

  8. (8)

Follows from Equation (16).

∎

###  3.1. The inverse problem

When is a power series $f(x)$ represented as $\mathfrak{Z}\left(X\right)$ for
some poset $X$? It is worth remembering that several posets can be associated
to the same power series. They are called Doppelgänger posets.

Note that if we order the chains in the decomposition
$\mathfrak{Z}\left(X\right)=a\mathfrak{Z}\left(i\right)+\cdots+b\mathfrak{Z}\left(j\right),\mathfrak{Z}\left(Y\right)=a_{2}\mathfrak{Z}\left(i_{2}\right)+\cdots+b_{2}\mathfrak{Z}\left(j_{2}\right)$,
the terms $a,i,b,j$ change under the rules:

$\mu(\zeta[X],\zeta[Y])=aa_{2}\mathfrak{Z}\left(i+i_{2}\right)+\cdots+bb_{2}\mathfrak{Z}\left(j+j_{2}\right)$

and

$D(\mathfrak{Z}\left(X\right))=ia\mathfrak{Z}\left(i+2\right)+\cdots+(j+1)b\mathfrak{Z}\left(j+3\right).$

If we have a candidate poset $X$, to reduce the computational time, we first
compute the lowest and highest order term of the factorization
$\mathfrak{Z}\left(X\right)=a\mathfrak{Z}\left(i\right)+\cdots+b\mathfrak{Z}\left(j\right),$
and compare it with the lowest and highest order term of $f(x)$. To select
candidate posets, we do not need to explore all possible words $w$. Since
$\mu$ is associative, we can pick $\mu(\mu(\cdots\mu(\leavevmode\nobreak\
,\leavevmode\nobreak\ ),\leavevmode\nobreak\ ),\cdots\leavevmode\nobreak\ )$
as a representative of every composition of $n$ operations $\mu$. We call this
representative an _$n$ -corolla_. In other words, the $2$–corolla is
$\mu(\leavevmode\nobreak\ ,\leavevmode\nobreak\ )$, the $3$–corolla is
$\mu(\mu(\leavevmode\nobreak\ ,\leavevmode\nobreak\ ),\leavevmode\nobreak\ )$,
and so on. If a word $w$ is a solution to our problem, then any permutation of
the inputs of the corollas will also be a solution to our problem. In our
setting, some of those Doppengängers are explained because the poset
concatenation is not commutative, but the power series concatenation is
commutative.

The language of operads allows us to choose representatives from equivalence
classes of trees. Consider the two-colored symmetric operad 2–$\mathcal{P}$.
This operad has a binary operation $D$, colored in red, and for every $n>1$ an
$n$–ary operation, given by an $n$-corolla $\mu$, colored in green. Red
colored operations have no restrictions, but green ones can only be composed
with red ones. In other words, a green corolla cannot be grafted into a green
corolla.

######  Proposition 3.2.

Given a finite sum $f(x)=\sum a_{j}\mathfrak{Z}\left(j\right)$ where $a_{j}$
are non-negative integers, there is an algorithm to determine if
$f(x)=\mathfrak{Z}\left(X\right)$ for some Wixarika poset $X$ and, in the
positive case, the algorithm returns all possible posets representing $f(x)$.

###### Proof.

Let $f(x)=\sum_{j=i}^{k}a_{j}\mathfrak{Z}\left(j\right)$.

We first restrict the space of posets that can represent $f$. From Proposition
3.1, we know the number of times the characters $\mu,D$ occurs in the
candidate words as well as the possible value of $|X|$.

Now, we verify that $f$ satisfy the restrictions from Ehrhart theory listed on
[14] as well as Lemma 2.23.

The next step is to choose representatives of Doppelgängers. We restrict the
number of candidate words by using the 2–$\mathcal{P}$ structure under the
rule: Any $n$-corolla counts as $n-1$ uses of $\mu$.

After that, in a loop:

  * •

we evaluate a candidate words $\\{w\\}$ only on the terms $(i,a_{i},k,a_{k})$
using Algorithm 1.

  * •

for $w$ that passed the algorithm we compute $w(\mathfrak{Z}\left(1\right))$.
This can be done evaluating from top to bottom as in [10].

  * •

If any poset passes this step, we use [7] to find the remaining Doppelgängers.
Then we stop the algorithm, otherwise we continue the loop.

Input: A candidate 2-$\mathcal{P}$ tree with $i$ leaves and $d$ red marks. A
value $(a_{i},i,a_{k},k)$ ;

// The four numbers stand for
$a\mathfrak{Z}\left(i\right)+\cdots+b\mathfrak{Z}\left(j\right),(1,1,1,1)=\mathfrak{Z}\left(1\right)$

// The $n$-corollas represent the composition of $n$ copies of $\mu$, and the
red marks represent $D$

Put $(1,1,1,1)$ on the leaves;

Let $B$ be the set of red marks ordered by inverse Deep First Search on the
tree;

for  _$b\in B$_ do

compute the input of $b$, lets say $(r_{b},t_{b},s_{b},u_{b})$;

// only compute the lowest and highest term

if  _$t_{b}$ does not divide $a_{i}$_ then

return False;

else

store $t_{b}$ in Cache; // Use Cache to find the root’s value

end if

end for

if  _the root has a value different to $a_{i}$ _ then

return False

else

True.

end if

Algorithm 1 Is $w(1,1,1,1)==(a_{i},i,a_{k},k)$?

∎

The algorithm has as input a description of $f(x)$ as a finite combination of
elements $\mathfrak{Z}\left(i\right)$ for some natural numbers $i$. Note that
every coefficient of
$\sum_{i=1}^{\infty}\mathfrak{Z}\left(i\right)=x+3x^{2}+\cdots$ is finite.
Because of this, we require the input to be written already as a finite sum of
terms $\\{x^{i}/(1-x)^{i+1}\\}$. We don’t have an algorithm, that is
guaranteed to finish, to discover the factorization of an arbitrary power
series $f$.

For arbitrary series-parallel posets, we need to adapt Proposition 3.1 and
increase the search space as now we are required to work with different
connected components and handles can consist of several elements.

######  Proposition 3.3.

Let $Y$ be a series-parallel poset. If
$\mathfrak{Z}\left(Y\right)=\sum_{j=i}^{k}a_{j}\mathfrak{Z}\left(j\right)$ is
a finite sum, then

  1. (1)

the Betti number of $Y$ is smaller or equal to $k-i$,

  2. (2)

the number $i$ is the length of any maximal chain of the largest connected
component of the Hasse diagram of $Y$,

  3. (3)

the number $k$ is the total number of points in the poset,

  4. (4)

$a_{i}$ and $a_{k}$ can be factor in a product of terms obtained by the
application of disjoint union on the subposets of $Y$.

  5. (5)

$\sum_{u=1}^{k}(-1)^{k-u}a_{u}=1$

###### Proof.

  1. (1)

Geometrically, when we apply $D$ to a disjoint union of posets, we increase
the Betti number of the Hasse diagram. But according to Proposition (2.10),
the index $k$ reflects not only the number of handles in the Hasse diagram but
also the number of points on those handles and the number of connected
components on the Hasse diagram. If

$\sum_{j_{1}=i_{1}}^{k_{1}}a_{j_{1}}\mathfrak{Z}\left(j_{1}\right)\odot\sum_{j_{2}=i_{2}}^{k_{2}}a_{j_{2}}\mathfrak{Z}\left(j_{2}\right)=\sum_{j_{3}=i_{3}}^{k_{3}}a_{j_{3}}\mathfrak{Z}\left(j_{3}\right)$

then we can verify that $k_{3}-i_{3}\geq(k_{1}-i_{1})+(k_{2}-i_{2})$. Here
$(k_{1}-i_{1})$ and $(k_{2}-i_{2})$ are bounds for the Betti numbers of each
connected component, and $(k_{3}-i_{3})$ bounds the Betti number of the union.
Hence the inequality.

  2. (2)

We use the following observation: If $m\geq k$, then, by the definition of
$\odot$ (see also Proposition 2.10), we find that

$\mathfrak{Z}\left(k\right)\odot\mathfrak{Z}\left(m\right)={m\choose
k}{k\choose 0}\mathfrak{Z}\left(m\right)+\cdots.$

The proof is similar to the proof of item 2 of Proposition 3.1.

  3. (3)

If $m>n$, then by Proposition 2.10 the last non-zero coefficient of $\langle
m\rangle\bigsqcup\langle n\rangle$ is ${m+n\choose n}{n\choose
n}\mathfrak{Z}\left(m+n\right)$. Since concatenation satisfies Equation (12),
it also preserves the number of elements. These results imply that the term
$\mathfrak{Z}\left(k\right)$ is obtained by adding all elements involved in
all disjoint unions as well as those elements involved in concatenation.

  4. (4)

It follows from Proposition 2.10.

  5. (5)

Follows from Equation (18) and the fact that $h_{0}^{*}=1$

∎

The coefficients of the power series associated to a series-parallel poset are
not topological invariants. For example the posets $\langle 1\rangle*(\langle
1\rangle\bigsqcup\langle 1\rangle)*\langle 1\rangle$ and $\langle
2\rangle*(\langle 1\rangle\bigsqcup\langle 1\rangle)$ have the same number of
labeling maps, but different Betti numbers.

To adapt Algorithm 1 for series-parallel posets we need to increase the search
space to allow disjoint union of posets. The algorithm depends on [7] which
already works at the level of series-parallel posets.

######  Remark 3.4.

Given the $h^{*}$ series of an order polytope $P(X)$, where $X$ is series-
parallel, we can use Proposition 19 to compute $i=\min\\{s|a_{s}\neq 0\\}$,
$k=\max\\{s|a_{s}\neq 0\\}$. Then from Proposition 3.1 and Proposition 3.3 we
obtain topological properties of $X$.

##  4\. Combinatorics

In the first subsection, we apply our results to study binomial identities. We
could not find identities (22), the identities of Proposition 4.1 and
Proposition 4.2 in the book [12]. In Section 4.2 we introduce a combinatorial
approach to study the multivariate negative hypergeometric distribution.

###  4.1. Identities

The existence of the operad structure on order series implies Proposition 4.1.
The associativity of the concatenation on order series, implies Proposition
4.2. We call this two identities, structural binomial identities.

######  Proposition 4.1.

Given $p,q,s\leq p+q$. Then for $t\leq s$:

$\displaystyle{p+q+t\choose s}{s\choose t}=\sum_{\begin{subarray}{c}a+c=s\\\
n\leq a\\\ r\leq c\\\ n+r=t\end{subarray}}{p+n\choose a}{a\choose
n}{q+r\choose c}{c\choose r}-\sum_{\begin{subarray}{c}a+c=s-1\\\ n\leq a\\\
r\leq c\\\ n+r=t-1\end{subarray}}{p+n\choose a}{a\choose n}{q+r\choose
c}{c\choose r}.$

###### Proof.

Consider Equation (17) with
$f=\mathfrak{Z}\left(q\right),g=\mathfrak{Z}\left(p\right)$.

$\displaystyle\mathfrak{Z}\left(s\right)\odot(\mathfrak{Z}\left(q\right)*\mathfrak{Z}\left(p\right))$
$\displaystyle=$
$\displaystyle\mathfrak{Z}\left(s\right)\odot(\mathfrak{Z}\left(q+p\right))$
$\displaystyle=$ $\displaystyle\sum_{t=0}^{s}{p+q+t\choose s}{s\choose
t}\mathfrak{Z}\left(p+q+t\right)$
$\displaystyle\mathfrak{Z}\left(s\right)\odot(\mathfrak{Z}\left(q\right)*\mathfrak{Z}\left(p\right))$
$\displaystyle=$
$\displaystyle\sum_{a+c=s}(\mathfrak{Z}\left(a\right)\odot\mathfrak{Z}\left(p\right))*(\mathfrak{Z}\left(c\right)\odot\mathfrak{Z}\left(q\right))$
$\displaystyle-$
$\displaystyle\left(\sum_{a+c=s-1}(\mathfrak{Z}\left(a\right)\odot\mathfrak{Z}\left(p\right))*(\mathfrak{Z}\left(c\right)\odot\mathfrak{Z}\left(q\right))\right)*\mathfrak{1}$
$\displaystyle=$ $\displaystyle\sum_{a+c=s}\sum_{n=0}^{a}{p+n\choose
a}{a\choose n}\mathfrak{Z}\left(p+n\right)*\sum_{r=0}^{c}{q+r\choose
c}{c\choose r}\mathfrak{Z}\left(q+r\right)$ $\displaystyle-$
$\displaystyle\sum_{a+c=s-1}\sum_{n=0}^{a}{p+n\choose a}{a\choose
n}\mathfrak{Z}\left(p+n\right)*\sum_{r=0}^{c}{q+r\choose c}{c\choose
r}\mathfrak{Z}\left(q+r\right)*1$ $\displaystyle=$
$\displaystyle\sum_{a+c=s}\sum_{n=0}^{a}\sum_{r=0}^{c}{p+n\choose a}{a\choose
n}{q+r\choose c}{c\choose r}\mathfrak{Z}\left(p+n+q+r\right)$ $\displaystyle-$
$\displaystyle\sum_{a+c=s-1}\sum_{n=0}^{a}\sum_{r=0}^{c}{p+n\choose
a}{a\choose n}{q+r\choose c}{c\choose r}\mathfrak{Z}\left(p+n+q+r+1\right)$

The result follows by comparing the coefficients. ∎

######  Proposition 4.2.

Let $n$ be a non-negative integer and $k\in\mathbb{N}$. For any $k$-partition
$n=n_{1}+\cdots+n_{k}$, with $n_{i}\geq 0$, and for $m\geq n+k-1$ we have

${m\choose n}={k-1\choose 0}\sum_{\begin{subarray}{c}\sum m_{i}=m\\\ m_{i}\geq
n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose n_{i}}-{k-1\choose
1}\sum_{\begin{subarray}{c}\sum m_{i}=m-1\\\ m_{i}\geq
n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose n_{i}}\pm\cdots\\\
+(-1)^{k-2}{k-1\choose k-2}\sum_{\begin{subarray}{c}\sum m_{i}=m-(k-2)\\\
m_{i}\geq n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose
n_{i}}+(-1)^{k-1}{k-1\choose k-1}\sum_{\begin{subarray}{c}\sum
m_{i}=m-(k-1)\\\ m_{i}\geq n_{i}\end{subarray}}\prod_{i=1}^{k}{m_{i}\choose
n_{i}}.$

###### Proof.

Decompose $\langle n\rangle$ as $\langle n_{1}\rangle*\langle
n_{2}\rangle*\cdots*\langle n_{k}\rangle$. Then the result is obtained by
applying Proposition 2.7 and computing the degree $m$ coefficient of the last
row in the sequence of identities below:

$\displaystyle\sum_{m=n}^{\infty}{m\choose n}x^{m}$
$\displaystyle=\mathfrak{Z}\left(n\right)$
$\displaystyle=\mathfrak{Z}\left(n_{1}\right)(1-x)\mathfrak{Z}\left(n_{2}\right)(1-x)\cdots(1-x)\mathfrak{Z}\left(n_{k}\right)$
$\displaystyle=\mathfrak{Z}\left(n_{1}\right)\mathfrak{Z}\left(n_{2}\right)\cdots\mathfrak{Z}\left(n_{k}\right)(1-x)^{k-1}$
$\displaystyle=\sum_{m_{1}=n_{1}}^{\infty}{{m_{1}}\choose
n_{1}}x^{m_{1}}\cdots\sum_{m_{k}=n_{k}}^{\infty}{{m_{k}}\choose
n_{k}}x^{m_{k}}\sum_{j=0}^{k-1}(-1)^{j}{k-1\choose j}x^{j}.$

∎

Using $\mathfrak{Z}^{+}\left(n\right)$ and $*^{+}$ from Equation (13), we
obtain the equivalent identity:

######  Proposition 4.3.

Let $n=n_{1}+\cdots+n_{k}$. Then for $v\geq n+k-1$

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k+1}{n}\right)\kern-3.00003pt\right)={k-1\choose
0}\sum_{\begin{subarray}{c}\sum v_{i}=v\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{n_{i}}\right)\kern-3.00003pt\right)-{k-1\choose
1}\sum_{\begin{subarray}{c}\sum v_{i}=v-1\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{n_{i}}\right)\kern-3.00003pt\right)+\cdots$
$+(-1)^{k-2}{k-1\choose k-2}\sum_{\begin{subarray}{c}\sum v_{i}=v-(k-2)\\\
v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{n_{i}}\right)\kern-3.00003pt\right)+{k-1\choose
k-1}\sum_{\begin{subarray}{c}\sum v_{i}=v-(k-1)\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{n_{i}}\right)\kern-3.00003pt\right)$

Following the ideas of the previous section, we define
$f(x)*^{-}g(x)=f(x)\frac{1}{x}g(x)$ and generate the monoid over
$\frac{x}{(1-x)^{2}}$. We associate power series
$\mathfrak{Z}^{-}\left(1\right)=\frac{x}{(1-x)^{2}}$ to the chain $\langle
1\rangle$ and extend by
$\mathfrak{Z}^{-}\left(n\right)=\mathfrak{Z}^{-}\left(n-1\right)*^{-}\mathfrak{Z}^{-}\left(1\right)$
to get $\mathfrak{Z}^{-}\left(n\right)=\frac{x}{(1-x)^{2n}}$. Applications of
this structure to probability will be explained in Section 4.2.

Using the expansion of the Maclaurin series, we obtain the Vandermonde
identity for negative integers:

######  Proposition 4.4.

Let $n=n_{1}+\cdots+n_{k}$. Then for $v\geq k$ :

(20)
$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k+1}{2n-1}\right)\kern-3.00003pt\right)=\sum_{\begin{subarray}{c}\sum
v_{i}=v\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{2n_{i}-1}\right)\kern-3.00003pt\right).$

###### Proof.

It follows from
$\mathfrak{Z}^{-}\left(n\right)=\mathfrak{Z}^{-}\left(n_{1}\right)*^{-}\mathfrak{Z}^{-}\left(n_{2}\right)\cdots\mathfrak{Z}^{-}\left(n_{k}\right)$.
∎

From (20) and $n=1+\cdots+1$ we obtain for $v\geq n$:

(21) ${v+n-1\choose 2n-1}=\sum_{\begin{subarray}{c}\sum v_{i}=v\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}{v_{i}},$

which gives us a formula to compute certain binomial coefficients without
using any division.

######  Proposition 4.5.

Given $m\in\mathbb{N}$, then for $n\leq\frac{m+1}{2}$ we have

(22) ${m\choose 2n-1}=\sum_{\begin{subarray}{c}\sum v_{i}=m-(n-1)\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}{v_{i}}.$

###### Proof.

It follows from Equation (21). ∎

The latter identity requires that the lower entry in the binomial coefficient
must be odd. Consider instead $n=1+\cdots+1$. Then using Proposition 4.2, for
$m\geq 2n-1$,

${m\choose n}={n-1\choose 0}\sum_{\begin{subarray}{c}\sum m_{i}=m\\\ m_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}m_{i}-{n-1\choose
1}\sum_{\begin{subarray}{c}\sum m_{i}=m-1\\\ m_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}m_{i}\pm\cdots$ (23) $+(-1)^{n-2}{n-1\choose
n-2}\sum_{\begin{subarray}{c}\sum m_{i}=m-(n-2)\\\ m_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}{m_{i}}+(-1)^{n-1}{n-1\choose
n-1}\sum_{\begin{subarray}{c}\sum m_{i}=m-(n-1)\\\ m_{i}\geq
1\end{subarray}}\prod_{i=1}^{n}{m_{i}}$

Define $\tilde{n}(m,k)=\sum_{\sum_{i=1}^{k}m_{i}=m}1$. In words,
$\tilde{n}(m,k)$ is the number of partitions of $m$ points into $k$ sets that
can be empty. Let $n=0$, $k\in\mathbb{N}$, $m\geq k-1$. From Proposition 4.2
we obtain the identity:

(24) $1={k-1\choose 0}\tilde{n}(m,k)-{k-1\choose 1}\tilde{n}(m-1,k)+\cdots\\\
+(-1)^{k-2}{k-1\choose k-2}\tilde{n}(m-(k-2),k)+(-1)^{k-1}{k-1\choose
k-1}\tilde{n}(m-(k-1),k).$

The Stirling number of the second kind ${m\brace k}$ is defined as the number
of partitions of $m$ points into $k$ non-empty subsets. It follows that

$\tilde{n}(m,k)={k\choose 0}{m\brace k}+{k\choose 1}{m\brace
k-1}+\cdots+{k\choose k-1}{m\brace 1}$

If we use $n^{2}=n+\cdots+n$, then using Proposition 4.2, for $m\geq
n^{2}+n-1$,

${m\choose n^{2}}={n-1\choose 0}\sum_{\begin{subarray}{c}\sum m_{i}=m\\\
m_{i}\geq n\end{subarray}}\prod_{i=1}^{n}{m_{i}\choose n}-{n-1\choose
1}\sum_{\begin{subarray}{c}\sum m_{i}=m-1\\\ m_{i}\geq
n\end{subarray}}\prod_{i=1}^{n}{m_{i}\choose n}\pm\cdots$ (25)
$+(-1)^{n-2}{n-1\choose n-2}\sum_{\begin{subarray}{c}\sum m_{i}=m-(n-2)\\\
m_{i}\geq n\end{subarray}}\prod_{i=1}^{n}{m_{i}\choose
n}+(-1)^{n-1}{n-1\choose n-1}\sum_{\begin{subarray}{c}\sum m_{i}=m-(n-1)\\\
m_{i}\geq n\end{subarray}}\prod_{i=1}^{n}{m_{i}\choose n}.$

Similar identities allow us to compute ${m\choose n^{k}}$ in terms of
${m\choose n^{r}}$ for $r<k$.

On Equation (21), for every partition $\sum v_{i}=v$, if $v_{i}$ appears
$s_{i}$ times then the term $\prod_{i}v_{i}$ will appear as many times as the
_multinomial coefficient_ ${n\choose s_{1},\cdots,s_{j}}$. We compute

(26) ${v+n-1\choose 2n-1}=\sum_{\begin{subarray}{c}\sum v_{i}=v\\\
\\#v_{i}=s_{i}\end{subarray}}{n\choose
s_{1},\cdots,s_{j}}\prod_{i=1}^{n}{v_{i}}$

where the sum is over different $n$-partitions of $v$, and each term $v_{i}$
appears $s_{i}$ times.

Finally, we apply Proposition 4.3 to the case
$2n-k=\sum^{k}_{i=1}2n_{i}-1,v\geq 2n-1$.

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k+1}{2n-k}\right)\kern-3.00003pt\right)={k-1\choose
0}\sum_{\begin{subarray}{c}\sum v_{i}=v\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{2n_{i}-1}\right)\kern-3.00003pt\right)-{k-1\choose
1}\sum_{\begin{subarray}{c}\sum v_{i}=v-1\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{2n_{i}-1}\right)\kern-3.00003pt\right)\pm$
$\cdots+(-1)^{k-2}{k-1\choose k-2}\sum_{\begin{subarray}{c}\sum
v_{i}=v-(k-2)\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{2n_{i}-1}\right)\kern-3.00003pt\right)+{k-1\choose
k-1}\sum_{\begin{subarray}{c}\sum v_{i}=v-(k-1)\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{2n_{i}-1}\right)\kern-3.00003pt\right).$

We then substitute Equation (20) to obtain, for $v\geq 2n-1+k-1$,

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k+1}{2n-k}\right)\kern-3.00003pt\right)={k-1\choose
0}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k+1}{2n-1}\right)\kern-3.00003pt\right)-{k-1\choose
1}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-k}{2n-1}\right)\kern-3.00003pt\right)\pm\cdots$
$+(-1)^{k-2}{k-1\choose
k-2}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-2k+3}{2n-1}\right)\kern-3.00003pt\right)+(-1)^{k-1}{k-1\choose
k-1}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v-2k+2}{2n-1}\right)\kern-3.00003pt\right)$

or

$\displaystyle{v-2k+2n\choose 2n-k}={k-1\choose 0}{v-k-1+2n\choose
2n-1}-{k-1\choose 1}{v-k-2+2n\choose 2n-1}\pm\cdots$ (27)
$\displaystyle+(-1)^{k-2}{k-1\choose k-2}{v-2k+1+2n\choose
2n-1}+(-1)^{k-1}{k-1\choose k-1}{v-2k+2n\choose 2n-1}.$

###  4.2. Probability

Consider $N$ classes of objects. Take $V$ objects with repetition. Divide the
$N$ classes into groups, the first with $n_{1}$ classes, the next with $n_{2}$
classes, and so on. We assume $N=n_{1}+\cdots+n_{k}$. Let
$V=v_{1}+\cdots+v_{k}$. What is the probability that $v_{1}$ objects belong to
the group of $n_{1}$ fixed classes, and $v_{2}$ objects belong to the group of
$n_{2}$ fixed classes, and so on? The answer comes from the multivariate
negative hypergeometric distribution

$P(X=(v_{1},\cdots,v_{k}))=\frac{\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n_{1}}{v_{1}}\right)\kern-3.00003pt\right)\dots\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n_{k}}{v_{k}}\right)\kern-3.00003pt\right)}{\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{V}\right)\kern-3.00003pt\right)}.$

As an application of our methods, we provide a direct proof that $P$ is a
distribution.

######  Proposition 4.6.

The function $P$ is a distribution.

###### Proof.

From
$\frac{x}{(1-x)^{N}}=\frac{x}{(1-x)^{n_{1}}}\frac{1}{x}\cdots\frac{1}{x}\frac{x}{(1-x)^{n_{k}}}$
and Proposition 4.4 we obtain

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{V-(k-1)}{N-1}\right)\kern-3.00003pt\right)=\sum_{\begin{subarray}{c}\sum
v_{i}=V\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{v_{i}}{n_{i}-1}\right)\kern-3.00003pt\right).$

Equivalently,

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{V-k}\right)\kern-3.00003pt\right)=\sum_{\begin{subarray}{c}\sum
v_{i}=V\\\ v_{i}\geq
1\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n_{i}}{v_{i}-1}\right)\kern-3.00003pt\right),$

and by changing variables we obtain

$\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{W}\right)\kern-3.00003pt\right)=\sum_{\begin{subarray}{c}\sum
w_{i}=W\\\ w_{i}\geq
0\end{subarray}}\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n_{i}}{w_{i}}\right)\kern-3.00003pt\right),\quad
1=\sum_{\begin{subarray}{c}\sum w_{i}=W\\\ w_{i}\geq
0\end{subarray}}\frac{\prod_{i=1}^{k}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n_{i}}{w_{i}}\right)\kern-3.00003pt\right)}{\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{W}\right)\kern-3.00003pt\right)}.$

∎

We also provide a method to compute the expectation value of the negative
hypergeometric distribution.

$\displaystyle\sum_{W=1}x^{W}\sum_{w=1}^{W}w\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{w}\right)\kern-3.00003pt\right)\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N-n}{W-w}\right)\kern-3.00003pt\right)$
$\displaystyle=$
$\displaystyle\left(\sum_{i=1}i\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{i}\right)\kern-3.00003pt\right)x^{i}\right)\left(\sum_{j=1}\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N-n}{j}\right)\kern-3.00003pt\right)x^{j}\right)$
$\displaystyle=$
$\displaystyle\left(x\frac{d}{dx}\left(\frac{1}{(1-x)^{n}}\right)\right)\frac{1}{(1-x)^{N-n}}$
$\displaystyle=$ $\displaystyle\frac{nx}{(1-x)^{N+1}}$ $\displaystyle=$
$\displaystyle\sum_{W=1}n\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N+1}{W-1}\right)\kern-3.00003pt\right)x^{W}.$

It follows that

$\mathbb{E}(P)=\sum_{w=0}^{W}\frac{w\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{n}{2}\right)\kern-3.00003pt\right)\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N-n}{W-w}\right)\kern-3.00003pt\right)}{\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{W}\right)\kern-3.00003pt\right)}=\frac{n\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N+1}{W-1}\right)\kern-3.00003pt\right)}{\left(\kern-3.00003pt\left(\genfrac{}{}{0.0pt}{}{N}{W}\right)\kern-3.00003pt\right)}.$

##  5\. Open problems

We conclude with a list of open problems.

  * •

Do the polynomials in Remark 2.18 have a combinatorial interpretation?

  * •

Is there a combinatorial proof of Proposition 4.1?

  * •

Is there a proof of Proposition 4.3 by computing the Euler characteristic or
using the inclusion-exclusion principle?

  * •

Given an order series $\sum c_{i}\mathfrak{Z}\left(i\right)$ of a series-
parallel poset, can we find further topological information from the chains
that appear, that is, from the coefficients $a_{i}\neq 0$? are the
coefficients $c_{i}$ non-negative for all finite posets?

  * •

What is the topological story behind the $*^{-}$ structure linked to the
negative hypergeometric distribution?

  * •

What are the generators of the operad of finite posets?

## Acknowledgement

We thank Elton P. Hsu for carefully reading the section on probability. We
thank Max Hopkings and Isaias Marin Gaviria for several discussions. Further
we thank Raman Sanyal for suggesting the reference [4] and the anonymous
referee(s) for pointing us to the papers [10], [18] and [22]. MB was funded
through the Royal Society grant URF$\setminus$R1$\setminus$20147\. ERDC was
funded by the National Research Foundation of Korea (NRF) grant funded by the
Korea government (MSIT) (No. 2020R1C1C1A0100826). This project started while
ERDC worked at NewSci Labs and he thanks them for their hospitality.

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