The right weak order interval below a fully commutative element is the lattice of order ideals of its heap

Clause (1), construction of the ideal. Let $x{\le }_{R}w$, in the right weak order of [F2], and let $s^{\mathrm{\prime }}\in \mathcal{R}(x)$. By the length identity F3, $\mathrm{\ell }(w)=\mathrm{\ell }(x)+\mathrm{\ell }(x^{-1}w)$; fix a reduced word $u^{\mathrm{\prime }}$ of $x^{-1}w$. Then $s^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$ represents $x\cdot x^{-1}w=w$ and has length $\mathrm{\ell }(x)+\mathrm{\ell }(x^{-1}w)=\mathrm{\ell }(w)$, so $s^{\mathrm{\prime }}{u}^{\mathrm{\prime }}\in \mathcal{R}(w)$. Since $w$ is fully commutative, $\mathcal{R}(w)=C(s)=L(P,s)$ by [F5] and [F6], so $s^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$ is the labeled word of a linear extension $\pi$ of $P$. In any linear extension the elements of the chain $C_u$ occur in increasing order $u(1)\prec \cdots \prec u(n_u)$, so the first $\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\ell }(x)$ letters of $s^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$, namely the letters of $s^{\mathrm{\prime }}$, are exactly $u(1),\dots ,u(\nu (u,s^{\mathrm{\prime }}))$ for each $u$. Hence $\nu (u,s^{\mathrm{\prime }})\le n_u$, and $I(s^{\mathrm{\prime }})$ is the set of the first $\mathrm{\ell }(x)$ entries of $\pi$, an initial segment of a linear extension; a prefix of a linear extension is downward closed, so $I(s^{\mathrm{\prime }})$ is an order ideal of $P$.

Clause (2), the heap of an ideal. Let $I\in J(P)$. By [F7], choose a linear extension $\pi$ of $P$ whose initial block ${\pi }_I=(a_1,\dots ,a_m)$ lists exactly $I$, where $m=\mathrm{\mid }I\mathrm{\mid }$. Let $s_I$ be the word of labels on this block, let $x_I$ be its product in $W$, and define $\phi :[m]\to I$ by $\phi (j)=a_j$. This bijection preserves labels. For any strict relation $a{\prec }_{P}b$ with $a,b\in I$, choose a chain $a=z_0\prec z_1\prec \cdots \prec z_r=b$ of maximal length between $a$ and $b$ (such a chain exists because $P$ is finite). Every $z_i$ lies in $I$, since $z_i⪯b$ and $b\in I$. Each consecutive pair is a cover in $P$; it must be a generating pair of the heap, because a generating path with an intermediate element would contradict the cover property. Since ${\pi }_I$ is a linear extension, each such pair occurs in the same order in $s_I$, and its labels are equal or noncommuting. Thus the corresponding positions are related in $P_{s_I}$, and transitivity shows that $a{\prec }_{P}b$ in $I$ implies ${\phi }^{-1}(a){\prec }_{P_{s_I}}{\phi }^{-1}(b)$.

Clause (1), well-definedness and basic properties. If $s^{\mathrm{\prime }\mathrm{\prime }}\in \mathcal{R}(x)$ is a second reduced word, then with the same suffix $u^{\mathrm{\prime }}$ the word $s^{\mathrm{\prime }\mathrm{\prime }}{u}^{\mathrm{\prime }}$ also has length $\mathrm{\ell }(w)$ and represents $w$, so $s^{\mathrm{\prime }\mathrm{\prime }}{u}^{\mathrm{\prime }}\in \mathcal{R}(w)=L(P,s)$; a word in $L(P,s)$ is the labeled word of a linear extension of $P$, hence contains each $u\in S$ exactly $n_u$ times, and this holds for $s^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$ as well. Subtracting the common multiplicity $\nu (u,u^{\mathrm{\prime }})$ of the suffix gives $\nu (u,s^{\mathrm{\prime }})=\nu (u,s^{\mathrm{\prime }\mathrm{\prime }})$ for every $u$, so $I(x):=I(s^{\mathrm{\prime }})$ is well defined. Moreover $\mathrm{\ell }(x)=\mathrm{\mid }{s}^{\mathrm{\prime }}\mathrm{\mid }={\sum }_u\nu (u,s^{\mathrm{\prime }})=\mathrm{\mid }I(x)\mathrm{\mid }$ because the sets $\{u(1),\dots ,u(\nu (u,s^{\mathrm{\prime }}))\}$ are disjoint; $I(1)=\mathrm{\varnothing }$ because the empty word has $\nu (u,())=0$; and $I(w)=P$ because for $s^{\mathrm{\prime }}=s$ one has $\nu (u,s)=n_u$.

Conversely, each generating relation of $P_{s_I}$ joins positions with equal or noncommuting labels. Their corresponding elements of $I$ are comparable in $P$ by [F5], and the order is the one in ${\pi }_I$, so every generating relation of $P_{s_I}$ respects the induced order on $I$. Together with 1.2, this proves that $\phi$ is a labeled poset isomorphism. If a different linear extension of $I$ is used, transport its listing through ${\phi }^{-1}$ to a linear extension of $P_{s_I}$; the labels are unchanged, so its labeled word lies in $L(P_{s_I},s_I)$ and [F6] makes it commutation-equivalent to $s_I$, and [F1] says these interchanges preserve the product. Therefore $x_I$ depends only on $I$, and $\psi (I):=x_I$ is well defined.

Clause (2), $\psi$ is order-preserving. Let $I\subseteq J$ be order ideals of $P$. Apply [F7] twice: extend a linear extension of $I$ (with first $\mathrm{\mid }I\mathrm{\mid }$ entries $I$) to a linear extension of the induced poset on $J$, which therefore has first $\mathrm{\mid }I\mathrm{\mid }$ entries $I$ and first $\mathrm{\mid }J\mathrm{\mid }$ entries $J$, and extend that in turn to a linear extension $\pi$ of $P$; then the first $\mathrm{\mid }I\mathrm{\mid }$ entries of $\pi$ are $I$ and its first $\mathrm{\mid }J\mathrm{\mid }$ entries are $J$. The full labeled word of $\pi$ lies in $L(P,s)=C(s)=\mathcal{R}(w)$ by [F5] and [F6]. The word $s_I$ constructed in 1.2 is a prefix of the word $s_J$, and both are reduced: replacing either prefix by a shorter word for its product would shorten the full reduced word of $w$; by the prefix property F3 applied to $s_I$ and $s_J$ we get $\psi (I){\le }_R\psi (J)$.

Clause (2), monotonicity of $x\mapsto I(x)$. If $x{⋖}_{R}y$, then by [F4] $y=xs$ with $s\in S$ and $\mathrm{\ell }(y)=\mathrm{\ell }(x)+1$; for $s^{\mathrm{\prime }}\in \mathcal{R}(x)$ the word $s^{\mathrm{\prime }}s$ represents $y$ and has length $\mathrm{\ell }(x)+1=\mathrm{\ell }(y)$, hence lies in $\mathcal{R}(y)$, and $\nu (u,s^{\mathrm{\prime }}s)=\nu (u,s^{\mathrm{\prime }})+[u=s]\ge \nu (u,s^{\mathrm{\prime }})$ for all $u$. By the well-definedness 2.1 the ideals may be computed from these words, so $I(x)\subseteq I(y)$. For arbitrary $x{\le }_{R}y{\le }_{R}w$, [F4] joins $x$ to $y$ by a chain of covers and inclusion is transitive along that chain; hence $x{\le }_{R}y$ implies $I(x)\subseteq I(y)$.

The full labeled word $q$ of $\pi$ belongs to $L(P,s)$ by definition. By [F6] and full commutativity [F5], $L(P,s)=C(s)=\mathcal{R}(w)$, so $q$ represents $w$ and is reduced of length $k=\mathrm{\ell }(w)$. Its prefix $s_I$ is also reduced, since a shorter expression for that prefix would shorten $q$ as an expression of $w$. The prefix property F3 gives $\psi (I)=x_I{\le }_{R}w$. Conversely, for any $x{\le }_{R}w$, step 1.1 gives a reduced word of $x$ as the initial segment of a linear extension of $P$ with initial ideal $I(x)$; using that extension in the definition of $\psi$ gives $\psi (I(x))=x$. Finally, because $I$ is an order ideal and each $C_u$ is a chain, $I\cap C_u$ is an initial segment of $C_u$ with exactly $\nu (u,s_I)$ elements; hence $I(x_I)=I$. Thus $\psi :J(P)\to [1,w{]}_R$ is a two-sided inverse of $x\mapsto I(x)$.

Clauses (3) and (4). By steps 3.1 and 3.2, $x\mapsto I(x)$ is an order-preserving bijection with inverse $\psi$, and by step 2.3 the inverse is order-preserving; hence this is an order isomorphism $[1,w{]}_R\to J(P)$. For $x,y{\le }_{R}w$, put $m_0:=\psi (I(x)\cap I(y))$ and $j_0:=\psi (I(x)\cup I(y))$. Since $\psi$ preserves order and is inverse to $I$, $m_0{\le }_{R}x,y$ and $x,y{\le }_R{j}_0$. If $z\in [1,w{]}_R$ satisfies $z{\le }_{R}x,y$, then monotonicity of $I$ gives $I(z)\subseteq I(x)\cap I(y)$, so $z=\psi (I(z)){\le }_R{m}_0$; therefore $m_0=x\wedge y$. Dually, if $z\in [1,w{]}_R$ is a common upper bound, then $I(x)\cup I(y)\subseteq I(z)$, so $j_0{\le }_{R}z$ and $j_0=x\vee y$. Applying $I$ gives the displayed intersection and union formulas. By [F9], $J(P)$ is a finite distributive lattice with least element $\mathrm{\varnothing }$ and greatest element $P$, so the order isomorphism transports this structure to $[1,w{]}_R$, whose least and greatest elements are $1$ and $w$. Clause (4) holds because the argument uses only the right weak order and its prefix property, and it assumes that $w$ is fully commutative; no Bruhat-interval or non-fully-commutative claim is made.