The greedy scan computes the c-sorting word; commutation, conjugation and rank-two alignment

Left descents have the root test $s\in D_L(x)⟺\rho (x^{-1})e_s\in {\mathrm{\Phi }}_{-}$: inversion invariance gives $\mathrm{\ell }(sx)=\mathrm{\ell }(x^{-1}s)$ and $\mathrm{\ell }(x)=\mathrm{\ell }(x^{-1})$, while [F6] applied to $x^{-1}$ gives the stated equivalence.

If $w=1$, the empty set is the unique admissible set, the scan makes no selection, and its remainder is already $1$.

Induction hypothesis for clause (1): for a fixed $w$ of positive length, assume for every $r$ with $\mathrm{\ell }(r)<\mathrm{\ell }(w)$ and every suffix of $c^{\mathrm{\infty }}$ that the greedy scan selects $\mathrm{\ell }(r)$ positions, ends at remainder $1$, and produces the lexicographically least admissible position set for $r$ in that suffix. Every letter of $S$ occurs infinitely often in every such suffix, even when its first block is partial.

Let $s$ be initial in the chosen word $c=s{s}_2\cdots s_n$ and put $c^{\ast }=s_2\cdots s_{n}s$. Then $\rho (s)e_s=-e_s$ and $\rho (s)e_p=e_p-K(e_p,e_s)e_s$ for $p\mathrm{\ne }s$, where $K=2B$. Since $s$ is first in $c$ and last in $c^{\ast }$, $E_{c^{\ast }}(e_p,e_s)=0$, $E_{c^{\ast }}(e_s,e_q)=K(e_s,e_q)$ for $p,q\mathrm{\ne }s$, and $E_{c^{\ast }}(e_p,e_q)=E_c(e_p,e_q)$ when $p,q\mathrm{\ne }s$. Expanding by bilinearity gives $E_{c^{\ast }}(\rho (s)e_p,\rho (s)e_q)=E_c(e_p,e_q)$: for $p=q=s$ both sides are $1$; for $p=s\mathrm{\ne }q$ the left side is $-K(e_s,e_q)+K(e_s,e_q)=0=E_c(e_s,e_q)$; for $q=s\mathrm{\ne }p$ it is $K(e_p,e_s)=E_c(e_p,e_s)$; and for $p,q\mathrm{\ne }s$ the two added terms $-K(e_p,e_s)K(e_s,e_q)$ and $+K(e_p,e_s)K(e_s,e_q)$ cancel. Bilinearity extends the identity to all vectors, and subtracting the transposed identity gives the one for $\omega$. By F2, the identities are independent of the reduced Coxeter words chosen.

For $J\subseteq S$, deleting the letters outside $J$ gives a word $c_J$ containing each letter of $J$ once. By [F9], $(W_J,J)$ is a Coxeter system with the restricted length; by F2, $c_J$ is a reduced Coxeter word for its value. For $p,q\in J$ the relative order and the entries $K(e_p,e_q)$ are unchanged, so the triangular definitions give $E_{c_J}(e_p,e_q)=E_c(e_p,e_q)$; bilinearity on the basis of $V_J$ gives both restriction identities in (3).

For clause (4), assume finite type and write the given generalized rank-two parabolic as $W^{\mathrm{\prime }}=v{W}_J{v}^{-1}$ with $\mathrm{\mid }J\mathrm{\mid }=2$, so $P=\rho (v)V_J$. The vectors $\rho (v)e_j$ are roots for $j\in J$, so $P$ is root-spanned. Let $f_J\in C_J$ be the face point with $f_J(e_s)=0$ for $s\in J$ and $f_J(e_s)=1$ otherwise, and put $y=b^{-1}(f_J)$ and $x=\rho (v)y$, where $b(z)=B(z,\cdot )$. By [F14], $b$ is an isomorphism; by [F10], it is equivariant for the reflection and dual actions, so $b(x)=v\cdot f_J$ and ${\mathrm{Stab}}_W(x)={\mathrm{Stab}}_W(v\cdot f_J)$. The point-stabilizer formula [F16] gives ${\mathrm{Stab}}_W(f_J)=W_J$; since the dual action is a group action [F15], ${\mathrm{Stab}}_W(v\cdot f_J)=v{W}_J{v}^{-1}=W^{\mathrm{\prime }}$. For every $j\in J$, $B(x,\rho (v)e_j)=B(y,e_j)=f_J(e_j)=0$, hence $x\in P^{\perp }$. If a root $\alpha \mathrm{\notin }P$ satisfied $B(x,\alpha )=0$, [F10] and its unit norm would make $t_{\alpha }$ fix $x$, so $t_{\alpha }\in W^{\mathrm{\prime }}$. Then $v^{-1}{t}_{\alpha }v=t_{\rho (v^{-1})\alpha }\in W_J$ by [F5], and [F11] together with $t_{-\gamma }=t_{\gamma }$ forces $\rho (v^{-1})\alpha \in V_J$, hence $\alpha \in P$, a contradiction. Thus $x$ meets the hypotheses of [F3] for the plane $P$ and the subgroup $W^{\mathrm{\prime }}$.

If a remainder $r\mathrm{\ne }1$, choose a reduced expression $r=a_1\cdots a_k$; since $a_1^2=1$, $a_{1}r=a_2\cdots a_k$ has length at most $k-1$, and [F8] makes it exactly $k-1$, so $a_1\in D_L(r)$. Each letter occurs infinitely often in any suffix, so the scan eventually reaches a letter in the nonempty set $D_L(r)$. Every selected letter lowers the remainder length by one by [F8]; once the remainder is $1$, no later letter is selected because [F8] gives $\mathrm{\ell }(u)=1$ for every $u\in S$. Thus from any starting remainder $r$ the scan makes exactly $\mathrm{\ell }(r)$ selections and ends at $1$.

If distinct $s,t$ commute, then $m(s,t)=2$ by [F13, F22], so [F12] gives $B(e_s,e_t)=0$ and [F20, F21] give $\rho (t)e_s=r_{e_t}{e}_s=e_s$; symmetrically $\rho (s)e_t=e_t$. By step 1.1, $s\in D_L(tx)$ iff $\rho ((tx{)}^{-1})e_s=\rho (x^{-1})\rho (t)e_s=\rho (x^{-1})e_s$ is negative, iff $s\in D_L(x)$; likewise $t\in D_L(sx)$ iff $t\in D_L(x)$.

Put ${\alpha }_1:={\beta }_{u_1}={\beta }_{r_1}$ and ${\alpha }_m:={\beta }_{u_m}={\beta }_{r_2}$, where $r_1,r_2$ are the canonical reflections in the statement. By F3,(4), their order can be chosen so that ${\omega }_c({\alpha }_1,{\alpha }_m)\ge 0$. Use the orientation on $P$ determined by the ordered basis $({\alpha }_1,{\alpha }_m)$. Write ${\beta }_{u_i}=a_i{\alpha }_1+b_i{\alpha }_m$ with $a_i,b_i\ge 0$, as all roots lie in the pointed sector. The strict angular order then gives $a_i{b}_j-b_i{a}_j>0$ for $i<j$. Bilinearity and skew-symmetry yield ${\omega }_c({\beta }_{u_i},{\beta }_{u_j})=(a_i{b}_j-b_i{a}_j){\omega }_c({\alpha }_1,{\alpha }_m)$, which is zero for all pairs if the endpoint value is zero and positive for every $i<j$ if it is positive. Since ${\alpha }_1,{\alpha }_m$ span $P$, endpoint value zero is equivalent to ${\omega }_c$ vanishing on all of $P$, hence on $\mathrm{\Phi }\cap P$. This proves (4)(i) without assuming equally spaced roots.

For $w\mathrm{\ne }1$ on any suffix, let $h$ be the first position whose letter $u_h$ lies in $D_L(w)$. The first letter of any admissible word for $w$ is a left descent, since if that word is $a_1\cdots a_k=w$ then $a_{1}w=a_2\cdots a_k$ has length at most $k-1$ and [F8] makes it exactly $k-1$; hence no admissible set starts before $h$. Also $\mathrm{\ell }(u_{h}w)=\mathrm{\ell }(w)-1$. A reduced expression of $u_{h}w$ can be embedded after $h$ in the suffix because each letter occurs infinitely often, so an admissible set starting at $h$ exists (if $u_{h}w=1$, use the empty tail). The admissible sets starting at $h$ are exactly $\{h\}\cup Q$ with $Q$ admissible for $u_{h}w$ in the later suffix. By the induction hypothesis, the continued greedy scan gives the lexicographically least such $Q$; therefore the full scan is the lexicographically least admissible set for $w$. Along with the base case and termination this proves (1), including existence and uniqueness.

Compare the scans for Coxeter words differing by an adjacent swap of commuting letters $s,t$. At the pair, both scans have the same remainder $x$. If neither letter is a descent both skip; if only one is a descent both select that letter, since the other remains a non-descent by step 2.2; if both are descents both select both, since each remains a descent after left multiplication by the other. In every case the selected subset of the pair is the same and the remainders after the pair agree; when both are selected, the equality is $stx=tsx$.

By F2, any two reduced Coxeter words for $c$ are connected by adjacent swaps of commuting letters. Repeat the comparison of step 3.2 in every block of the two periodic words, carrying the common remainder through the identical positions between swapped pairs; induction over positions shows that the selected letter subsets in corresponding blocks agree. Thus their block sequences are equal, and the sorting words can differ only by adjacent commuting swaps inside a block, never across a divider. This proves (2).

Fix a reduced expression $w=s_1\cdots s_k$ and let ${\beta }_i=\rho (s_1\cdots s_{i-1})e_{s_i}$ and $t_i=s_1\cdots s_{i-1}{s}_i{s}_{i-1}\cdots s_1$. By F5, the ${\beta }_i$ are exactly the positive roots of $N(w^{-1})$, and F5 gives $t_{{\beta }_i}=t_i$. Direct cancellation gives $t_{i}w=s_1\cdots \widehat{s_i}\cdots s_k$, so $\mathrm{\ell }(t_{i}w)\le k-1<\mathrm{\ell }(w)$. Conversely every reflection $t$ with $\mathrm{\ell }(tw)<\mathrm{\ell }(w)$ has its positive root in $N(w^{-1})$ by [F18]. Thus these positive roots correspond exactly to the left inversions in Reading--Speyer's c-alignment definition following Proposition 4.1. When an initial order has negative endpoint value, reverse the canonical pair and its list as in F3; this gives the same convention used in the statement. No Axiom of Choice is used: the only witnesses are single instantiations (a reduced word for a fixed element and the explicit face point $f_J$), and the scan, word comparisons and finite-dimensional calculations are deterministic. Clauses (1)-(4) are proved.