Length parity, exchange, two-letter deletion, and faithfulness of the signed reflection action

The sign character and the parity laws. The map $f:S\to \{\pm 1\}$, $f(s)=-1$, satisfies $f(s{)}^2=1$ and $(f(s)f(t){)}^{m(s,t)}=(-1{)}^{2m(s,t)}=1$ for every finite edge, so the universal property in [F1] gives a unique homomorphism $\mathrm{sgn}:W\to \{\pm 1\}$ with $\mathrm{sgn}(s)=-1$. For any word $w=s_1\cdots s_k$ this gives $\mathrm{sgn}(w)=(-1{)}^k$; taking a word of length $\mathrm{\ell }(w)$ shows $\mathrm{sgn}(w)=(-1{)}^{\mathrm{\ell }(w)}$, so every word for $w$ has length congruent to $\mathrm{\ell }(w)$ modulo $2$. Next, $\mathrm{\ell }(sw)\le \mathrm{\ell }(w)+1$: a word of length $\mathrm{\ell }(w)$ for $w$ prefixed by $s$ is a word of length $\mathrm{\ell }(w)+1$ for $sw$, and $\mathrm{\ell }$ is a minimum; symmetrically $\mathrm{\ell }(w)=\mathrm{\ell }(s\cdot sw)\le \mathrm{\ell }(sw)+1$. Since $\mathrm{sgn}(sw)=\mathrm{sgn}(s)\mathrm{sgn}(w)=-\mathrm{sgn}(w)$ by [F3] and $(-1{)}^{\mathrm{\ell }(sw)}=\mathrm{sgn}(sw)$, the parities of $\mathrm{\ell }(sw)$ and $\mathrm{\ell }(w)$ are opposite, so $\mathrm{\ell }(sw)\mathrm{\ne }\mathrm{\ell }(w)$; with the two inequalities this forces $\mathrm{\ell }(sw)=\mathrm{\ell }(w)\pm 1$, and the congruence $\mathrm{\ell }(sw)\equiv \mathrm{\ell }(w)+1$ records the parity. The same argument with $ws$ in place of $sw$, using $\mathrm{\ell }(ws)\le \mathrm{\ell }(w)+1$ and $\mathrm{\ell }(w)=\mathrm{\ell }(ws\cdot s)\le \mathrm{\ell }(ws)+1$, gives $\mathrm{\ell }(ws)=\mathrm{\ell }(w)\pm 1$ and $\mathrm{\ell }(ws)\equiv \mathrm{\ell }(w)+1$ modulo $2$.

Faithfulness. Let $w\mathrm{\ne }1$. Then $\mathrm{\ell }(w)\mathrm{\ne }0$ because the only word of length $0$ is the empty word with value $1$ by [F1], so $\mathrm{\ell }(w)\ge 1$; choose a reduced expression $w=s_1\cdots s_k$ with $k=\mathrm{\ell }(w)\ge 1$. By [F2] the set $\mathrm{\Phi }(w)=\{r_1,\dots ,r_k\}$ has $\mathrm{\#}\mathrm{\Phi }(w)=\mathrm{\ell }(w)\ge 1$, so pick $r\in \mathrm{\Phi }(w)$, that is $\eta (r,w)=-1$. The action formula of [F2] then gives $(1,r)\cdot w=(\eta (r,w)\cdot 1,w^{-1}rw)=(-1,w^{-1}rw)\mathrm{\ne }(1,w^{-1}rw)$, so $w$ acts nontrivially on $\{\pm 1\}\times T$; hence the action is faithful and $W$ embeds in $\mathrm{Sym}(\{\pm 1\}\times T)$. In particular, for $s\mathrm{\ne }t$ in $S$ the elements $s,t$ have distinct images under this embedding because $U_s(1,s)=(-1,s)$ while $U_t(1,s)=(1,tst)$, and the first coordinates $-1$ and $1$ differ.

Exchange. Let $w=s_1\cdots s_k$ be reduced and let $\mathrm{\ell }(sw)=k-1$. Choose a reduced expression $sw=t_1\cdots t_{k-1}$; then $w=s{t}_1\cdots t_{k-1}$ is a word of length $k=\mathrm{\ell }(w)$, hence a reduced expression of $w$ whose first prefix reflection is $r_1=s$, so $\eta (s,w)=-1$ and $s\in \mathrm{\Phi }(w)$ by [F2]. By the expression-independence of $\mathrm{\Phi }(w)$ in [F2] applied to the reduced expression $w=s_1\cdots s_k$, there is $i\in \{1,\dots ,k\}$ with $s=r_i=w_{i-1}{s}_i{w}_{i-1}^{-1}$, where $w_{i-1}=s_1\cdots s_{i-1}$; multiplying this identity on the right by $w=w_{i-1}{s}_i{s}_{i+1}\cdots s_k$ gives $sw=w_{i-1}{s}_{i+1}\cdots s_k=s_1\cdots \widehat{s_i}\cdots s_k$, which is the asserted deletion. For the right-handed form, note first that $\mathrm{\ell }(u^{-1})=\mathrm{\ell }(u)$ for every $u$: reversing a reduced word for $u$ gives a word of the same length for $u^{-1}$, so $\mathrm{\ell }(u^{-1})\le \mathrm{\ell }(u)$, and applying this to $u^{-1}$ gives equality. If now $\mathrm{\ell }(ws)=k-1$, then $w^{-1}=s_k\cdots s_1$ is a reduced expression of $w^{-1}$ and $\mathrm{\ell }(s{w}^{-1})=\mathrm{\ell }((ws{)}^{-1})=\mathrm{\ell }(ws)=k-1$, so the left-handed form applied to $w^{-1}$ writes $s{w}^{-1}=s_k\cdots \widehat{s_j}\cdots s_1$ for some $j$; inverting both sides gives $ws=(s{w}^{-1}{)}^{-1}=s_1\cdots \widehat{s_j}\cdots s_k$.

Deletion. Let $(s_1,\dots ,s_k)$ be a word that is not reduced, and let $j$ be the least index such that the prefix $(s_1,\dots ,s_j)$ is not reduced; such $j$ exists and $j\ge 2$, and $(s_1,\dots ,s_{j-1})$ is reduced with value $u:=s_1\cdots s_{j-1}$. By the minimality of $j$ the element $u{s}_j$ has $\mathrm{\ell }(u{s}_j)<j=\mathrm{\ell }(u)+1$, so $\mathrm{\ell }(u{s}_j)=\mathrm{\ell }(u)-1=j-2$ by step 1.1, and the right-handed exchange of step 1.3 applied to the reduced expression $(s_1,\dots ,s_{j-1})$ and the letter $s_j$ gives $u{s}_j=s_1\cdots \widehat{s_i}\cdots s_{j-1}$ for some $i\le j-1$. Therefore $s_1\cdots \widehat{s_i}\cdots \widehat{s_j}\cdots s_k=(s_1\cdots \widehat{s_i}\cdots s_{j-1})s_{j+1}\cdots s_k=u{s}_j{s}_{j+1}\cdots s_k=s_1\cdots s_k$, so deleting the letters at positions $i$ and $j$ leaves the value unchanged. Iterating, the length strictly decreases by $2$ at each deletion and stops at length $\mathrm{\ell }(w)$, leaving a reduced expression of the same element. Conversely, a reduced word cannot be shortened by deleting two letters, since the deleted word is a strictly shorter word for the same element.