The heap of $s_1{s}_3{s}_2$ in type $A_3$: a V-shaped heap with exactly two linear extensions

The heap relations are computed pair by pair from [F1]: positions $1,2$ carry the distinct commuting letters $s_1,s_3$ with $m(s_1,s_3)=2$, so there is no relation between them; position $3$ is above position $1$ because $1<3$ and $m(s_1,s_2)=3\ge 3$; and position $3$ is above position $2$ because $2<3$ and $m(s_3,s_2)=m(s_2,s_3)=3$. These are therefore all generating relations, and the transitive closure adds nothing, so $P$ is the three-element poset with exactly $1\prec 3$ and $2\prec 3$. By [F5] its covering pairs are $1⋖3$ and $2⋖3$: the only elements are $1,2,3$, and no $z$ satisfies $1\prec z\prec 3$ or $2\prec z\prec 3$, since $1$ and $2$ are incomparable. The covering labels are $(s_1,s_2)$ and $(s_3,s_2)$.

The order ideals are the subsets $I\subseteq \{1,2,3\}$ that contain $1$ and $2$ whenever they contain $3$, that is, $\mathrm{\varnothing },\{1\},\{2\},\{1,2\},\{1,2,3\}$: downward closure of a subset of the three-element poset is a condition only on the predecessors of $3$. These five sets are exactly $J(P)$, and by [F6] they form a finite distributive lattice under inclusion, with meet intersection and join union.

The criterion of [F3] applies: every chain of $P$ has at most two elements, since the only relations are $1\prec 3$ and $2\prec 3$, so there is no convex chain of any length $m\ge 3$, and in particular none of length $m(s_1,s_2)=m(s_2,s_3)=3$; hence condition (a) of Fully commutative elements: the braid-factor criterion and the forbidden-chain heap criterion (2) holds. The covering pairs $1⋖3$ and $2⋖3$ have labels $(s_1,s_2)$ and $(s_3,s_2)$, and $s_1,s_2,s_3$ are pairwise distinct, so condition (b) holds. By [F3] the word $(s_1,s_3,s_2)$ is reduced, $w$ is fully commutative and $P=P_w$; since a reduced word of $w$ has length $\mathrm{\ell }(w)$, this gives $\mathrm{\ell }(w)=3$.

A listing of $\{1,2,3\}$ is a linear extension of $P$ exactly when $3$ is last, because both relations point to $3$ and $1,2$ are incomparable; the two linear extensions are therefore $(1,2,3)$ and $(2,1,3)$, with labeled words $s_1{s}_3{s}_2$ and $s_3{s}_1{s}_2$. By [F4], $L(P,s)=C(s)$, and by 2.1 the word is reduced with $w$ fully commutative, so $C(s)=\mathcal{R}(w)$; hence the reduced words of $w$ are exactly these two words, which differ by interchanging the adjacent commuting letters $s_1,s_3$.