Fully commutative elements: the braid-factor criterion and the forbidden-chain heap criterion

Clause (1). Suppose first that some $s\in \mathcal{R}(w)$ contains the contiguous factor $F=⟨u,v{⟩}_m$ with $m:=m(u,v)\in [3,\mathrm{\infty })$, and let $s^{\ast }$ be obtained from $s$ by replacing $F$ with $⟨v,u{⟩}_m$. By [F2], $s^{\ast }$ represents $w$ and has the same length, so $s^{\ast }\in \mathcal{R}(w)$. Delete from a word all letters outside $\{u,v\}$; this projection is unchanged by every interchange of adjacent commuting letters, because such a pair consists of distinct letters with $m=2$, so either both letters lie outside $\{u,v\}$ and are deleted, or exactly one of them lies in $\{u,v\}$ and keeps its position among the surviving letters (both letters in $\{u,v\}$ is impossible since $m(u,v)\ge 3$). The projections have a common prefix and suffix outside the factor, while their middle blocks are the distinct alternating words $u,v,u,\dots$ and $v,u,v,\dots$; cancelling the common prefix and suffix shows that the full projections differ. Hence $s^{\ast }\mathrm{\notin }C(s)$, so $\mathcal{R}(w)$ is not a single commutativity class and $w$ is not fully commutative; this proves (a)$\Rightarrow$(b). Conversely, if $w$ is not fully commutative, choose $s,s^{\mathrm{\prime }}\in \mathcal{R}(w)$ with $s^{\mathrm{\prime }}\mathrm{\notin }C(s)$ and, by F3, a sequence of braid moves $s=q_0,q_1,\dots ,q_r=s^{\mathrm{\prime }}$; let $t$ be the first index with $q_t\mathrm{\notin }C(s)$. Then $q_{t-1}\in C(s)$ and the move $q_{t-1}\to q_t$ is not a commutation, so it replaces a contiguous factor $⟨x,y{⟩}_{m(x,y)}$ by $⟨y,x{⟩}_{m(x,y)}$ with $x\mathrm{\ne }y$ and $m(x,y)\ge 3$. The word $q_{t-1}$ is obtained from $s$ by braid moves, hence has length $k=\mathrm{\ell }(w)$ and represents $w$, so it is a reduced word of $w$ containing the forbidden factor; this proves (b)$\Rightarrow$(a).

Clause (2), (c)$\Rightarrow$(b). Assume $s$ is reduced and $w$ is fully commutative, so $C(s)=\mathcal{R}(w)$ by [F1]. If $P_s$ had a covering pair $i⋖j$ with $s_i=s_j$, then $\{i,j\}$ would be a two-element convex chain, so by [F4] some linear extension of $P_s$ has $i,j$ consecutive; its labeled word $s^{\mathrm{\prime }\mathrm{\prime }}$ lies in $L(P_s,s)=C(s)$ by [F5], hence in $\mathcal{R}(w)$, and contains two consecutive equal letters. Deleting those two letters gives an expression of $w$ with $k-2$ letters, because $s_i^2=1$ in $W$ by [F2], contradicting $\mathrm{\ell }(w)=k$. Hence (b) holds.

Clause (2), the braid class equals the commutation class under (a). Assume (a), and let $H(s)$ be the set of words obtained from $s$ by finitely many braid moves. Then $H(s)=C(s)$: otherwise choose a sequence of braid moves from $s$ to a word outside $C(s)$ with the fewest moves, so that its last move is applied to a word $q\in C(s)$ and leaves $C(s)$; that move is not a commutation, hence replaces a contiguous alternating factor $⟨x,y{⟩}_{m(x,y)}$, $x\mathrm{\ne }y$, $m(x,y)\ge 3$, of $q$. The positions of that factor form a chain in $P_q$, because consecutive positions of the factor carry the noncommuting pair $\{x,y\}$; they are convex, because the order of $P_q$ is contained in the position order, so an element lying between two positions of the factor is itself one of them. Thus $P_q$ contains a convex alternating chain of length $m(x,y)\ge 3$, and since $q\in C(s)$ gives $P_q\cong P_s$ by [F5] while containing such a chain is invariant under labeled isomorphism, condition (a) fails for $P_s$, a contradiction. Hence $H(s)=C(s)$.

Clause (2), (c)$\Rightarrow$(a). Assume $s$ is reduced and $w$ is fully commutative, so $C(s)=\mathcal{R}(w)$ by [F1]. If $P_s$ contained a convex chain with labels alternating between distinct $u,v$ of length $m=m(u,v)\in [3,\mathrm{\infty })$, then by [F4] some linear extension of $P_s$ has the chain's elements consecutive; its labeled word $s^{\mathrm{\prime }\mathrm{\prime }}$ lies in $L(P_s,s)=C(s)$ by [F5], hence in $\mathcal{R}(w)$, and its consecutive letters at those positions are the alternating factor $⟨u,v{⟩}_m$. This contradicts clause (1)(b), proved in step 1.1, so (a) holds.

Clause (2), (a) and (b) imply that $s$ is reduced. If $s$ were not reduced, then by F3 there is a sequence of braid moves from $s$ to a word $u$ containing a consecutive equal pair, so $u\in H(s)=C(s)$ by step 1.3. Hence $P_u\cong P_s$ by [F5]. The two consecutive equal positions of $u$ satisfy $i{\prec }_{u}i+1$; no element lies strictly between them, because the order of $P_u$ is contained in the position order and there is no integer strictly between $i$ and $i+1$; so they form a covering pair of $P_u$ with equal labels. Under the labeled isomorphism this gives a covering pair of $P_s$ with equal labels, contradicting (b). Hence $s$ is reduced.

Clause (2), conclusion of (a),(b)$\Rightarrow$(c). Assume (a) and (b). By step 2.2 the word $s$ is reduced, so $\mathrm{\ell }(w)=k$. Every word braid-equivalent to $s$ has length $k$, represents $w$ by [F2], and is therefore reduced; hence $H(s)$ consists of reduced words of $w$. By F3 every reduced word of $w$ is braid-equivalent to $s$, so $\mathcal{R}(w)\subseteq H(s)=C(s)$ by step 1.3, while conversely every member of $C(s)$ is obtained from $s$ by commutations and so has length $k$ and represents $w$, hence lies in $\mathcal{R}(w)$. Therefore $\mathcal{R}(w)=C(s)$ and $w$ is fully commutative; by [F5] this also gives $P_s\cong P_{s^{\mathrm{\prime }}}$ for every $s^{\mathrm{\prime }}\in \mathcal{R}(w)$.

Clause (3). If $P_s$ avoids (a) and (b), then (c) holds by step 3.1, so $w$ is fully commutative and $P_s$ is its heap. Conversely, if $P_s$ is the heap of a fully commutative element, then $P_s\cong P_{s^{\mathrm{\prime }}}$ for some $s^{\mathrm{\prime }}\in \mathcal{R}(w^{\mathrm{\prime }})$ with $w^{\mathrm{\prime }}$ fully commutative; by [F5], $s\sim s^{\mathrm{\prime }}$, so $s\in C(s^{\mathrm{\prime }})=\mathcal{R}(w^{\mathrm{\prime }})$; hence $s$ is reduced and represents the fully commutative element $w^{\mathrm{\prime }}$, and steps 1.2 and 2.1 give (a) and (b). Finally, clause (4) is the restriction already built into the definitions: condition (a) and clause (1)(b) quantify only over pairs with $3\le m(u,v)<\mathrm{\infty }$, and for $m(u,v)=\mathrm{\infty }$ no braid relator exists by [F2], so no finite alternating block is forbidden.