Sum of order polynomials of a set of posets

Question

Let $n\in \mathbb{Z}_{>0}$. For every subset $S\subseteq \left[ n-1\right]$ we define a poset $P_S=\left([n],\le_{P_S}\right)$ given by the covering relation $\lessdot$ which is defined as \begin{align*} &\forall i\in S & i+1&\lessdot i \\ &\forall i\in\left[n-1\right]\setminus S & i&\lessdot i+1\\ \end{align*}

For instance if $n=3$ we have the posets

Useful Definitions

• A function $f\colon P_S\to [m]$ is called order-preserving for $P_S$, if for any $x,y\in P_S$ $$x<_{P_S}y\implies f(x)\le f(y),$$ where $m\in\mathbb{Z_{>0}}$.
• We denote with $\Omega(P_S,m)$ the order polynomial of $P_S$ for $m$, i.e. $$\Omega(P_S,m)=\#\left\{ f\colon P_S\to [m]\,\mid f\;\; \text{is order-preserving}\right\}$$
• We call an order-preserving bijection $\omega\colon P_S\to[n]$ natural labeling of $P_S$.
• For any permutation $w\in\mathcal{S}_n$ we denote with $\mathsf{Des}(w)$ the descent set of $w$, i.e. $$\mathsf{Des}(w)=\left\{ i\in\left[n\right]\mid w\left(i\right)>w\left(i+1\right)\right\} $$
• We denote with $A_S$ the set of permutations that corresponds to the natural labelings of $P_S$, i.e. $$A_{S}=\left\{ w\in\mathcal{S}_{n}\mid w=\left(\omega\left(1\right),\omega\left(2\right),\dots,\omega\left(n\right)\right)\text{ for some natural labeling }\omega\text{ of }P_{S}\right\} $$

---

The question

We want to calculate, for any given $m\in\mathbb{Z}_{>0}$ the sum of the order polynomials on the subsets $S\subseteq[n-1]$, i.e. $$\sum_{S\subseteq[n-1]}\Omega\left(P_{S},m\right).$$

---

My progress

It seems that the requested sum is equal to

$$m(1+m)^{n-1},$$

but I did not manage to show it.

---

It is not hard to show that the set of all the natural labelings for a given subset $S$ has the same cardinality as the set of all permutations of $[n]$ with descent set equal to $S$, i.e. $$\#A_S=\#\left\{ \omega\colon P_{S}\to[n]\mid\omega\;\text{natural labeling}\right\} =\#\left\{ w\in\mathcal{S}_{n}\mid\mathsf{Des}\left(w\right)=S\right\} .$$

It is well known that $$\Omega\left(P_{S},m\right)=\sum_{w\in A_{S}}\binom{m+n-\mathsf{des}\left(w\right)-1}{n}.$$

I tried to use these facts in order to calculate the requested sum but I failed...

(P.S. I posted this question on Math Stack Exchange some days ago, without getting an answer. If this question does not fit here, I will agreeable delete it.)

Answer 1

The key result is equation (1) of the paper here, which expresses the chromatic polynomial $\chi(G,m)$ of a graph $G$ as a sum of strict order polynomials $\overline{\Omega}(\overline{\mathcal{O}},m)$ of the transitive (and reflexive) closures $\overline{\mathcal{O}}$ of the acyclic orientations $\mathcal{O}$ of $G$. If we take $G$ to be an $n$-vertex path, then we get $$ \chi(G,m) = \sum_{S\subseteq[n-1]}\overline{\Omega}(P_S,m). $$ Now $\chi(G,m)=m(m-1)^{n-1}$. Moreover, by the reciprocity theorem for order polynomials, $$ \overline{\Omega}(P_S,m) = (-1)^n\Omega(P_S,-m), $$ and the result follows.