All maximal chains of a rank-three interval in S4, their deleted-position labels, and the lexicographically first chain

The eight elements. The fixed word $s_1{s}_2{s}_3$ is reduced with $\mathrm{\ell }(c)=3$, because $2341$ has exactly the three inversions $(3,4)$, $(2,4)$, $(1,4)$ [F3]; here the displayed letters $\{1,2,3,4\}$ are those of the published $S_4$ on $\{0,1,2,3\}$ shifted by $j\mapsto j-1$ [F4]. By [F2] an element $x$ of $W$ satisfies $x\le c$ if and only if $x$ is the product of a subword of $s_1{s}_2{s}_3$, so it remains to note that each of the eight subwords, with positions $\mathrm{\varnothing },\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}$, is reduced: its product $1234,2134,1324,1243,2314,2143,1342,2341$ has inversion number equal to its number of letters, as displayed [F3]. The eight products are distinct one-line forms, so $[e,c]$ has exactly these eight elements, of ranks $0,1,1,1,2,2,2,3$.

The covers and the six maximal chains. By [F1] the elements covered by $c$ are the single-letter deletions of the reduced word $s_1{s}_2{s}_3$ whose remaining word is reduced: deleting positions $1,2,3$ leaves $s_2{s}_3=1342$, $s_1{s}_3=2143$, $s_1{s}_2=2314$, each of length $2=\mathrm{\ell }(c)-1$, so these three and no others are covered by $c$, because every element covered by $c$ has length $2$ and the length-two elements of $[e,c]$ are exactly these three. Each atom covers $e$ by the cover criterion, and each atom lies below $c$ by [F2], so the three pairs $s⋗e$ are covers. For a length-two element $x=t_i{t}_j$ with $i<j$, the products of subwords of the reduced word $t_i{t}_j$ are exactly $e,t_i,t_j,x$, so by [F2] the elements of $[e,x]$ are exactly these four and the atoms covered by $x$ are $t_i$ and $t_j$; applying this to $1342=s_2{s}_3$, $2143=s_1{s}_3$, $2314=s_1{s}_2$ gives the six covers $1342⋗1243$, $1342⋗1324$, $2143⋗1243$, $2143⋗2134$, $2314⋗1324$, $2314⋗2134$, and shows that the remaining three pairs of adjacent ranks are incomparable (for instance $2134\le ̸1342$, since $2134=s_1$ is not one of $e,s_2,s_3,s_2{s}_3$). Since every maximal chain of $[e,c]$ has $\mathrm{\ell }(c)-\mathrm{\ell }(e)=3$ steps [F11], the maximal chains are the paths of covers from $2341$ to $1234$, namely the six chains displayed in (ii).

The label words. At the first step the retained expression is $c=s_1{s}_2{s}_3$ and the deleted position is read off from the cover by [F5]: $2341⋗1342$ deletes position $1$, $2341⋗2143$ deletes position $2$, $2341⋗2314$ deletes position $3$. In the rooted intervals the retained expressions are $1342=s_2{s}_3$, $2143=s_1{s}_3$, $2314=s_1{s}_2$, with their original positions; deleting the letter $s_1$, $s_2$ or $s_3$ from such a retained word gives the corresponding atom, so the second and third labels are the original positions of the deleted letters. Reading the six chains of step 1.2 gives exactly the six words $(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)$. These are pairwise distinct and, as the six permutations of $\{1,2,3\}$, exhaust all label words; the only strictly falling one is $(3,2,1)$, and $(1,2,3)$ is increasing.

The lexicographically first chain. By step 2.1 the six label words are distinct permutations of $\{1,2,3\}$, so the lexicographically first maximal chain is the one with word $(1,2,3)$, namely $2341⋗1342⋗1243⋗1234$, and this word is increasing; by [F7] the increasing maximal chain of $[e,c]$ is unique and lexicographically first, in agreement.

The local descent replacement. Consider the chain $m\text{\hspace{0.17em}⁣}:2341⋗1342⋗1324⋗1234$, whose word $(1,3,2)$ has its descent at position $2$. Its part above $m_1=1342$ is the single cover $2341⋗1342$, and the rooted rank-two interval $([m_3,m_1],c_m)=([1234,1342],2341⋗1342)$ has retained expression $s_2{s}_3$, with the two maximal chains $1342⋗1324⋗1234$ and $1342⋗1243⋗1234$ and label words $(3,2)$ and $(2,3)$, by [F8] and step 2.1 (the words are computed from the retained expression with its original positions $2,3$). The second chain is the unique increasing one, so replacing the falling segment by it gives the maximal chain $2341⋗1342⋗1243⋗1234$ with word $(1,2,3)\prec (1,3,2)$, which is the lexicographically first chain of step 3.1; this is the instance for $e=2$ of the local descent replacement [F9], and it agrees with the earlier/later comparison [F10] with $m^{\mathrm{\prime }}$ the lexicographically first chain, for which $m^{\mathrm{\prime }}\cap m=\{2341,1342,1234\}$, $\mathrm{\mid }{m}^{\mathrm{\prime }}\cap m\mathrm{\mid }=3=\mathrm{\mid }m\mathrm{\mid }-1$ and $\lambda (m^{\mathrm{\prime }})\prec \lambda (m)$.