A crossing double transposition whose interval is Boolean, and the two incomparable maximal Coxeter elements of S3

(The interval below the crossing element.) By [F1], ${\mathrm{\ell }}_T(x)=4-2=2$. If $t$ is a transposition, then $t^{-1}x=tx$. For $t=(13)$ or $(24)$, $tx$ is the other transposition, so ${\mathrm{\ell }}_T(t^{-1}x)=1$ and $t{\le }_{T}x$. The remaining four transpositions join the two cycles of $x$; explicitly, $(12)x=(1324)$, $(14)x=(1342)$, $(23)x=(1243)$, and $(34)x=(1423)$, each a 4-cycle of reflection length $3$. None is below $x$. If $u{\le }_{T}x$, [F2] gives ${\mathrm{\ell }}_T(u)\le 2$; length $0$ forces $u=1$, and length $2$ forces ${\mathrm{\ell }}_T(u^{-1}x)=0$, hence $u=x$. Therefore the displayed four elements are the entire interval. Its two distinct atoms have meet $1$ and join $x$, so it is the Boolean lattice on two generators.

(The two maximal elements.) By [F1], the identity, the three transpositions, and the two 3-cycles are exactly the elements of $S_3$, with reflection lengths $0,1,2$, respectively. The products are $s_1{s}_2=(123)$ and $s_2{s}_1=(132)$. They are distinct and have equal length, so [F2] makes them incomparable. No element has length greater than $2$, so each is maximal. A common upper bound would have to be strictly above one of these distinct maximal elements; hence none exists and $\mathrm{Abs}(S_3)$ has no join for this pair.

(The crossing obstruction.) The diagonals joining $1$ to $3$ and $2$ to $4$ cross in the square, so the support partition of $x$ is crossing. Also direct right-to-left multiplication gives $x^{-1}c=(1432)$, which has length $3$ by [F1], while ${\mathrm{\ell }}_T(c)=3$ and ${\mathrm{\ell }}_T(x)=2$. Thus ${\mathrm{\ell }}_T(x)+{\mathrm{\ell }}_T(x^{-1}c)=5\mathrm{\ne }3={\mathrm{\ell }}_T(c)$, so $x\mathrm{̸}{\le }_{T}c$ by [F2]. This verifies directly the exclusion predicted by the type-A criterion.

(The Coxeter intervals are five-element lattices.) Fix either 3-cycle $d$. For $d=(123)$, the products $td$ for $t=(12),(23),(13)$ are $(23),(13),(12)$, respectively; for $d=(132)$ they are $(13),(12),(23)$. Hence every $t\in T$ satisfies ${\mathrm{\ell }}_T(d)=1+{\mathrm{\ell }}_T(t^{-1}d)=2$ and lies below $d$. If $w{\le }_{T}d$ has length $2$, [F2] forces ${\mathrm{\ell }}_T(w^{-1}d)=0$, so $w=d$. Therefore $[1,d{]}_{{\le }_T}=\{1\}\cup T\cup \{d\}$. Its three transpositions are incomparable atoms, any two have meet $1$ and join $d$, so this interval is a five-element lattice. Finally, $s_1(s_1{s}_2)s_1=s_2{s}_1$, so the two Coxeter elements are conjugate.