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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

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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.)

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    $\begingroup$ It should probably follow from the fact that every permutation of $[n]$ is a linear extension of a unique one of your $P_S$'s. Ah, well, this is basically what you wrote with $A_S$ being the set of permutations with descent set $S$. $\endgroup$ – Sam Hopkins Jan 26 at 23:40
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    $\begingroup$ I deleted an answer which maybe had some ideas that could be made to work to explain your formula but wasn't right- Richard's answer is much better and very nice! $\endgroup$ – Sam Hopkins Jan 27 at 3:07
  • $\begingroup$ @SamHopkins Thank you very much for the effort. Your ideas were also very useful! Of course Richard Stanley’s solution is very elegant! $\endgroup$ – asknohope Jan 27 at 12:41
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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.

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