The heap of $s_1{s}_2{s}_1$ in type $A_2$: a convex alternating chain and two commutation classes

The heap relations are computed pair by pair from [F1]. Positions $1,2$ carry the noncommuting letters $s_1,s_2$ with $m(s_1,s_2)=3$, so $1\prec 2$; positions $2,3$ carry the noncommuting letters $s_2,s_1$, so $2\prec 3$; and positions $1,3$ carry the equal label $s_1$, so $1\prec 3$ as well. Hence the generating relations are contained in the chain $1\prec 2\prec 3$, and the transitive closure of all three relations is exactly that chain; its labels are $s_1,s_2,s_1$. By [F5] the covering pairs are $1⋖2$ and $2⋖3$, since no element lies strictly between consecutive positions of the chain, and their labels $(s_1,s_2)$ and $(s_2,s_1)$ are distinct.

Reducedness and the commutation classes. By F3 the alternating words of length $3=m(s_1,s_2)$ in $⟨s_1,s_2⟩$ are reduced in $W$, so $s_1{s}_2{s}_1$ and $s_2{s}_1{s}_2$ are reduced of length $3=\mathrm{\ell }(w_0)$, and by [F2] they represent the same element $w_0$. The only braid moves between words in $\{s_1,s_2\}$ replace an alternating subword of length $m(s_1,s_2)=3$ by the other alternating word, that is, they exchange the two displayed words; by F3 every reduced word of $w_0$ is braid-equivalent to $s_1{s}_2{s}_1$, hence $\mathcal{R}(w_0)=\{s_1{s}_2{s}_1,s_2{s}_1{s}_2\}$. Neither word contains two adjacent commuting letters, since every adjacent pair is $\{s_1,s_2\}$ with $m(s_1,s_2)=3\mathrm{\ne }2$; hence no commutation applies and $C(s_1{s}_2{s}_1)=\{s_1{s}_2{s}_1\}$, $C(s_2{s}_1{s}_2)=\{s_2{s}_1{s}_2\}$ are two distinct classes whose union is $\mathcal{R}(w_0)$.

The order ideals are the prefixes of the chain $1\prec 2\prec 3$: a subset $I$ is downward closed exactly when $2\in I\Rightarrow 1\in I$ and $3\in I\Rightarrow 2\in I$, which gives $\mathrm{\varnothing }$, $\{1\}$, $\{1,2\}$, $\{1,2,3\}$ and no further subset.

The chain $(1,2,3)$ is convex in $P$: it exhausts the three elements of $P$, so there is no element outside it lying between two of its members. It has length $3=m(s_1,s_2)$ and its labels $s_1,s_2,s_1$ alternate between the distinct letters $s_1,s_2$, so it is a convex alternating chain of the forbidden length and condition (a) of [F4] clause (2) fails. Its covering pairs are $1⋖2$ and $2⋖3$ by 1.1, with labels $(s_1,s_2)$ and $(s_2,s_1)$ distinct, so condition (b) holds.

Since $s_1{s}_2{s}_1$ is a reduced word of $w_0$ by 1.2 and contains the contiguous factor $⟨s_1,s_2{⟩}_3$ (its three letters), [F4] clause (1) shows that $w_0$ is not fully commutative. Moreover $P$ is not the heap of any fully commutative element: if $P\cong P_{s^{\mathrm{\prime }}}$ for some $s^{\mathrm{\prime }}\in \mathcal{R}(w^{\mathrm{\prime }})$ with $w^{\mathrm{\prime }}$ fully commutative, then by [F4] clause (2) applied to the reduced word $s^{\mathrm{\prime }}$, the heap $P_{s^{\mathrm{\prime }}}$ contains no convex alternating chain of length $m(u,v)$ with $3\le m(u,v)<\mathrm{\infty }$, while $P$ contains the convex alternating chain exhibited in 2.1 and convexity, length and labels are preserved by the labeled isomorphism, a contradiction.