A rank-three chain labeling translated into facets of the order complex

The cover labeling is well defined and ordinary: by [F1] the covers of $B_3$ are exactly the covers $S⋖S\cup \{i\}$ with $i\mathrm{\notin }S$, so the assignment is a labeling of all covers, and the label $i$ of a cover depends only on that cover, not on any descending chain above it; hence it is an ordinary edge labeling and so, in particular, a descending rooted-chain labeling of $[\mathrm{\varnothing },\{1,2,3\}]$ in the sense of [F2].

(N) and (L). Let $\mathrm{\varnothing }\subseteq S\subseteq T\subseteq \{1,2,3\}$ and let $m$ be a maximal chain of $[S,T]$; its steps add the elements of $T\setminus S$ one at a time, so the labels on $m$ are exactly the distinct elements of $T\setminus S$ and are pairwise distinct, which is (N) for the rooted interval $([S,T],c)$. Every maximal chain of $[S,T]$ corresponds to just such an order of adding the $r:=\mathrm{\mid }T\setminus S\mathrm{\mid }$ elements of $T\setminus S$, and its word read from the top is the reverse addition order; hence increasing words correspond exactly to adding the elements of $T\setminus S$ in decreasing order, and the increasing arrangement of $T\setminus S$ is the lexicographically first of the words; so there is exactly one increasing maximal chain, namely the one that adds the elements in decreasing order, and it is lexicographically first, which is (L). Since $S,T$ and $c$ were arbitrary, (N) and (L) hold on every rooted interval of $[\mathrm{\varnothing },\{1,2,3\}]$.

The six maximal chains. A maximal chain of $[\mathrm{\varnothing },\{1,2,3\}]$ is an order of adding $1,2,3$, and its word is the reverse of that order; hence there are exactly $6$ maximal chains and their words are the six permutations, obtained as follows: adding $2,3,1$ gives $\{1,2,3\}⋗\{2,3\}⋗\{2\}⋗\mathrm{\varnothing }$ with word $(1,3,2)$; adding $3,1,2$ gives $\{1,2,3\}⋗\{1,3\}⋗\{3\}⋗\mathrm{\varnothing }$ with word $(2,1,3)$; adding $1,3,2$ gives $\{1,2,3\}⋗\{1,3\}⋗\{1\}⋗\mathrm{\varnothing }$ with word $(2,3,1)$; adding $2,1,3$ gives $\{1,2,3\}⋗\{1,2\}⋗\{2\}⋗\mathrm{\varnothing }$ with word $(3,1,2)$; adding $1,2,3$ gives $\{1,2,3\}⋗\{1,2\}⋗\{1\}⋗\mathrm{\varnothing }$ with word $(3,2,1)$; and adding $3,2,1$ gives $\{1,2,3\}⋗\{2,3\}⋗\{3\}⋗\mathrm{\varnothing }$ with word $(1,2,3)$. Among the six permutations only $(1,2,3)$ is increasing and only $(3,2,1)$ is (strictly) falling, so the increasing chain is $\{1,2,3\}⋗\{2,3\}⋗\{3\}⋗\mathrm{\varnothing }$ and the strictly falling chain is $\{1,2,3\}⋗\{1,2\}⋗\{1\}⋗\mathrm{\varnothing }$, as stated.

Facets of the open interval. By [F6] the facets of $\mathrm{\Delta }((\mathrm{\varnothing },\{1,2,3\}))$ are the maximal chains of the open interval, that is, the sets obtained from the six maximal chains of step 2.1 by deleting the two endpoints: $(1,2,3)\mapsto \{2,3\}\supset \{3\}$, $(1,3,2)\mapsto \{2,3\}\supset \{2\}$, $(2,1,3)\mapsto \{1,3\}\supset \{3\}$, $(2,3,1)\mapsto \{1,3\}\supset \{1\}$, $(3,1,2)\mapsto \{1,2\}\supset \{2\}$ and $(3,2,1)\mapsto \{1,2\}\supset \{1\}$, ordered by the words of their parent chains: this is the list of six two-element chains in the stated order.

The Möbius value. Since $(3,2,1)$ is the only falling word among the six by step 2.1, the falling-chain formula of [F5] gives $\mu (\mathrm{\varnothing },\{1,2,3\})=(-1{)}^3\cdot 1=-1$ in $B_3$. The recurrence [F7] gives $\mu (\mathrm{\varnothing },\mathrm{\varnothing })=1$, $\mu (\mathrm{\varnothing },\{i\})=-\mu (\mathrm{\varnothing },\mathrm{\varnothing })=-1$ for each singleton, $\mu (\mathrm{\varnothing },\{i,j\})=-(\mu (\mathrm{\varnothing },\mathrm{\varnothing })+\mu (\mathrm{\varnothing },\{i\})+\mu (\mathrm{\varnothing },\{j\}))=-(1-1-1)=1$ for each two-element subset, and $\mu (\mathrm{\varnothing },\{1,2,3\})=-(1+3\cdot (-1)+3\cdot 1)=-1$, in agreement with the formula.

The replacement step. Take $m^{\mathrm{\prime }}$ the chain with word $(1,3,2)$, namely $\{1,2,3\}⋗\{2,3\}⋗\{2\}⋗\mathrm{\varnothing }$, and $m$ the chain with word $(2,1,3)$, namely $\{1,2,3\}⋗\{1,3\}⋗\{3\}⋗\mathrm{\varnothing }$; then $\lambda (m^{\mathrm{\prime }})\prec \lambda (m)$. The first divergence is at index $1$ and the first reunion at index $3$, so $d=0$, $g=3$, the window word of $m$ is $(2,1,3)$, and $m^{\mathrm{\prime }}$ and $m$ share exactly the vertices $\{1,2,3\}$ and $\mathrm{\varnothing }$, i.e. $m^{\mathrm{\prime }}\cap m=\{\{1,2,3\},\mathrm{\varnothing }\}$. The window word has its descent at position $e=1$: ${\lambda }_1(m)=2>1={\lambda }_2(m)$; the rooted rank-two interval $([\{3\},\{1,2,3\}],\{1,2,3\})$ has the two maximal chains $\{1,2,3\}⋗\{1,3\}⋗\{3\}$ and $\{1,2,3\}⋗\{2,3\}⋗\{3\}$ with words $(2,1)$ and $(1,2)$, so the increasing one is $\{1,2,3\}⋗\{2,3\}⋗\{3\}$ and replacing the segment $\{1,2,3\}⋗\{1,3\}⋗\{3\}$ of $m$ by it gives $k=\{1,2,3\}⋗\{2,3\}⋗\{3\}⋗\mathrm{\varnothing }$, the chain with word $(1,2,3)$. Then $\lambda (k)=(1,2,3)\prec (2,1,3)=\lambda (m)$, the intersection $k\cap m=\{\{1,2,3\},\{3\},\mathrm{\varnothing }\}$ has $\mathrm{\mid }k\cap m\mathrm{\mid }=3=\mathrm{\mid }m\mathrm{\mid }-1$, and $m^{\mathrm{\prime }}\cap m=\{\{1,2,3\},\mathrm{\varnothing }\}\subseteq k\cap m$; the replaced two-step segment is $\{1,2,3\}⋗\{1,3\}⋗\{3\}$, as in the shelling lemma [F4].

The computations verify every claim of the Example: the element-added cover labeling is an ordinary edge labeling and hence a descending rooted-chain labeling; it satisfies (N) and (L) on every rooted interval of $[\mathrm{\varnothing },\{1,2,3\}]$; the six maximal chains have the six permutation words, with $(\{1,2,3\}⋗\{2,3\}⋗\{3\}⋗\mathrm{\varnothing })$ the unique increasing chain and $(\{1,2,3\}⋗\{1,2\}⋗\{1\}⋗\mathrm{\varnothing })$ the unique strictly falling chain; the facets of $\mathrm{\Delta }((\mathrm{\varnothing },\{1,2,3\}))$ are the six listed two-element chains in the stated order; the exhibited chain $k$ realizes the shelling replacement for the pair $(1,3,2)\prec (2,1,3)$; and the falling-chain formula returns $\mu (\mathrm{\varnothing },\{1,2,3\})=-1$, which the recurrence confirms.