The rank-two block computation, exact dihedral orders, the signed reflection action, and ambient reducedness

Involutions and nonidentity for every generator. Fix any $s\in S$. The primitive formulas of [F2] give ${\sigma }_s^2({\alpha }_s)={\sigma }_s(-{\alpha }_s)={\alpha }_s$ and, for $u\mathrm{\ne }s$, ${\sigma }_s^2({\alpha }_u)={\sigma }_s({\alpha }_u+c_{su}{\alpha }_s)={\alpha }_u+c_{su}{\alpha }_s-c_{su}{\alpha }_s={\alpha }_u$. By [F3] the square is the identity on $E$, so ${\sigma }_s$ is invertible with inverse itself. Also ${\sigma }_s({\alpha }_s)=-{\alpha }_s\mathrm{\ne }{\alpha }_s$, since the basis vector ${\alpha }_s$ is nonzero and $\mathrm{char}K=0$; hence ${\sigma }_s\mathrm{\ne }{\mathrm{i}\mathrm{d}}_E$. These computations require no second generator; when $S=\mathrm{\varnothing }$, the assertions indexed by $s$ are vacuous.

The rank-two block. Fix any distinct $s,t\in S$ and use the notation of (2). Since ${\sigma }_t({\alpha }_s)={\alpha }_s+c{\alpha }_t$ and ${\sigma }_t({\alpha }_t)=-{\alpha }_t$ with $c=c_{st}$, linearity of ${\sigma }_s$ gives $A({\alpha }_s)={\sigma }_s({\alpha }_s+c{\alpha }_t)=(c^2-1){\alpha }_s+c{\alpha }_t$ and $A({\alpha }_t)={\sigma }_s(-{\alpha }_t)=-c{\alpha }_s-{\alpha }_t$, so $A(P)\subseteq P$ and the matrix of $B=A{\mathrm{\mid }}_P$ in the basis $({\alpha }_s,{\alpha }_t)$ is $\left(\begin{array}{cc}{c}^2-1& -c\\ c& -1\end{array}\right)$, whose trace is $(c^2-1)+(-1)=c^2-2$ and whose determinant is $(c^2-1)(-1)-(-c)c=1$. For $u\mathrm{\notin }\{s,t\}$ one has $A({\alpha }_u)={\sigma }_s({\alpha }_u+c_{tu}{\alpha }_t)={\alpha }_u+(c_{su}+c_{tu}c){\alpha }_s+c_{tu}{\alpha }_t$, so $A({\alpha }_u)-{\alpha }_u\in P$; by linearity $A(v)-v\in P$ for every $v\in E$.

The two-block recursion. The subspace $Q=\mathrm{span}\{{\alpha }_u:u\mathrm{\notin }\{s,t\}\}$ satisfies $P\cap Q=\{0\}$ and $P+Q=E$ because the basis of $E$ splits into the basis $({\alpha }_s,{\alpha }_t)$ of $P$ and the basis of $Q$; hence $E=P\oplus Q$, and every $v\in E$ has unique coordinates $v=p+q$ with $p\in P$, $q\in Q$. I claim that for every $k\ge 1$ and all such $p,q$ one has $A^k(p+q)=B^{k}p+S_k(A(q)-q)+q$, where $S_k:={\mathrm{i}\mathrm{d}}_P+B+\cdots +B^{k-1}$. For $k=1$ this reads $A(p)+A(q)=Bp+(A(q)-q)+q$, which holds because $A{\mathrm{\mid }}_P=B$. If it holds for $k$, then $B^{k}p+S_k(A(q)-q)\in P$ and $A(q)-q\in P$ by step 1.2, so applying $A$ and using $A{\mathrm{\mid }}_P=B$ gives $A^{k+1}(p+q)=B(B^{k}p+S_k(A(q)-q))+A(q)=B^{k+1}p+(B{S}_k+{\mathrm{i}\mathrm{d}}_P)(A(q)-q)+q$, and $B{S}_k+{\mathrm{i}\mathrm{d}}_P=B+B^2+\cdots +B^k+{\mathrm{i}\mathrm{d}}_P=S_{k+1}$, which completes the induction.

The cases $m=2$ and $m=\mathrm{\infty }$. If $m=2$, then $\zeta$ has order $4$, so ${\zeta }^2=-1$ and ${\zeta }^{-1}=-\zeta$, whence $c=\zeta +{\zeta }^{-1}=0$; substituting into the matrix of step 1.2 leaves $B=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-{\mathrm{i}\mathrm{d}}_P$, so $B^2={\mathrm{i}\mathrm{d}}_P$ and ${\mathrm{i}\mathrm{d}}_P+B=0$. If $m=\mathrm{\infty }$, then $c=2$ by the convention in [F2], and the matrix of step 1.2 gives $B={\mathrm{i}\mathrm{d}}_P+M$ with $M=B-{\mathrm{i}\mathrm{d}}_P=\left(\begin{array}{cc}2& -2\\ 2& -2\end{array}\right)\mathrm{\ne }0$ and $M^2=0$. The binomial expansion in the commutative algebra of endomorphisms then gives $B^k=({\mathrm{i}\mathrm{d}}_P+M{)}^k={\mathrm{i}\mathrm{d}}_P+kM$ for every $k\ge 1$, and $kM\mathrm{\ne }0$ because $M\mathrm{\ne }0$ and $k\cdot 1_K\mathrm{\ne }0$ in the characteristic-zero field $K$; hence $B^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$ for all $k\ge 1$, and in particular $B\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$.

The finite case $2<m<\mathrm{\infty }$. By step 1.2 and [F5] the characteristic polynomial of $B$ is ${\chi }_B(x)=x^2-(c^2-2)x+1$; since $c=\zeta +{\zeta }^{-1}$ satisfies $c^2-2={\zeta }^2+{\zeta }^{-2}$ and ${\zeta }^2{\zeta }^{-2}=1$, one has ${\chi }_B(x)=x^2-({\zeta }^2+{\zeta }^{-2})x+1=(x-{\zeta }^2)(x-{\zeta }^{-2})$. The two roots are distinct: ${\zeta }^2={\zeta }^{-2}$ would give ${\zeta }^4=1$, impossible because $\zeta$ has order $2m>4$. Hence the characteristic polynomial splits into distinct linear factors, and [F6] makes $B$ diagonalisable, with eigenvalues ${\zeta }^2$ and ${\zeta }^{-2}$. Consequently $B^m={\mathrm{i}\mathrm{d}}_P$, because on an eigenvector for $\lambda \in \{{\zeta }^2,{\zeta }^{-2}\}$ the operator $B^m$ scales it by ${\lambda }^m={\zeta }^{\pm 2m}=1$ by [F7] and [F8]; the operator $S_m={\mathrm{i}\mathrm{d}}_P+B+\cdots +B^{m-1}$ is zero, because the polynomial $1+x+\cdots +x^{m-1}$ vanishes at both eigenvalues, where ${\zeta }^{\pm 2m}=1$ and ${\zeta }^{\pm 2}\mathrm{\ne }1$ since $0<2<2m$; and $B^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$ for $0<k<m$, because on the eigenvector for ${\zeta }^2$ it scales by ${\zeta }^{2k}\mathrm{\ne }1$, again since $0<2k<2m$ and $2m$ is the least positive exponent killing $\zeta$.

Order of $A$, the relators, and the exact dihedral orders. For any distinct pair $s,t$ with finite $m=m(s,t)$, step 2.1 gives $A^m(p+q)=B^{m}p+S_m(A(q)-q)+q=p+q$ for all $p,q$, using $B^m={\mathrm{i}\mathrm{d}}_P$ and $S_m=0$ from steps 2.2 and 2.3; hence $A^m={\mathrm{i}\mathrm{d}}_E$. If $0<k<m$, choose $p\in P$ with $B^{k}p\mathrm{\ne }p$, which exists because $B^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$; then $A^k(p)=B^{k}p\mathrm{\ne }p$, so $A^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_E$ and $A$ has order exactly $m$. For $m=\mathrm{\infty }$, $B^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$ for every $k\ge 1$ gives $A^k\mathrm{\ne }{\mathrm{i}\mathrm{d}}_E$ for every $k\ge 1$, so $A$ has infinite order. Since this verifies each arbitrary distinct finite pair and step 1.1 proves ${\sigma }_s^2={\mathrm{i}\mathrm{d}}_E$ for every generator, the map $f:S\to {\mathrm{G}\mathrm{L}}_K(E)$, $f(s)={\sigma }_s$, satisfies the hypotheses of the universal property in [F1], so there is a unique homomorphism $\sigma :W\to {\mathrm{G}\mathrm{L}}_K(E)$ with $\sigma (s)={\sigma }_s$ for all $s$. For $\mathrm{\#}S\le 1$ there are no distinct-pair relators, so the same universal property applies using only step 1.1; for $S=\mathrm{\varnothing }$ it gives the unique homomorphism from the trivial group. For every $s\in S$, step 1.1 now gives $\sigma (s)\mathrm{\ne }{\mathrm{i}\mathrm{d}}_E$, hence $s\mathrm{\ne }1$ and $\mathrm{\ell }(s)\ge 1$ by [F1]; the one-letter expression gives $\mathrm{\ell }(s)\le 1$, so $\mathrm{\ell }(s)=1$. If $s\mathrm{\ne }t$ then ${\sigma }_s\mathrm{\ne }{\sigma }_t$, because ${\sigma }_s({\alpha }_s)=-{\alpha }_s$ while ${\sigma }_t({\alpha }_s)={\alpha }_s+c_{st}{\alpha }_t$ and $-2{\alpha }_s-c_{st}{\alpha }_t\mathrm{\ne }0$ by linear independence of ${\alpha }_s,{\alpha }_t$ and $2\mathrm{\ne }0$ in $K$; hence $s\mathrm{\ne }t$ in $W$. Finally, $st$ has order exactly $m$ when $m<\mathrm{\infty }$: writing $n$ for the order of $st$ in $W$, one has $n\le m$ because $(st{)}^m=1$ in $W$ and $n$ is least, and $m\le n$ because $A^n=\sigma ((st{)}^n)={\mathrm{i}\mathrm{d}}_E$ and $m$ is the least positive exponent killing $A$; hence $n=m$. When $m=\mathrm{\infty }$, a positive exponent $k$ with $(st{)}^k=1$ would give $A^k={\mathrm{i}\mathrm{d}}_E$, so no such exponent exists and $st$ has infinite order.

The signed action. First, $U_s$ is a bijection of $\{\pm 1\}\times T$ with $U_s^2=\mathrm{i}\mathrm{d}$: for $r=s$ one has $U_s(\epsilon ,s)=(-\epsilon ,s)$ and applying $U_s$ again gives $(+\epsilon ,s)$, while for $r\mathrm{\ne }s$ one has $srs\mathrm{\ne }s$ — if $srs=s$ then multiplying on the left by $s$ gives $rs=1$ and hence $r=s^{-1}=s$ using $s^2=1$ in $W$ — so $\delta (s,srs)=0$ and the two applications of $U_s$ return $(\epsilon ,r)$ from $U_s(\epsilon ,srs)=(\epsilon ,r)$. Next, fix a word $(s_1,\dots ,s_k)$ in $S$ with $w_j:=s_1\cdots s_j$ and prefix reflections $r_i:=w_{i-1}{s}_i{w}_{i-1}^{-1}$; applying $U_{s_1},U_{s_2},\dots ,U_{s_k}$ successively to $(\epsilon ,r)$ returns $(\epsilon (-1{)}^{n(r)},w_k^{-1}r{w}_k)$ with $n(r):=\mathrm{\#}\{i:r_i=r\}$, by induction on $k$: the $i$-th step multiplies the sign by $(-1{)}^{\delta (s_i,w_{i-1}^{-1}r{w}_{i-1})}$, whose exponent is $1$ exactly when $r=w_{i-1}{s}_i{w}_{i-1}^{-1}=r_i$, and it conjugates the coordinate to $s_i{w}_{i-1}^{-1}r{w}_{i-1}{s}_i=(w_{i-1}{s}_i{)}^{-1}r(w_{i-1}{s}_i)$. Now fix any distinct $a,b\in S$ with $m=m(a,b)<\mathrm{\infty }$; consider the alternating word $(a,b,a,b,\dots )$ of length $2m$ and write $r_i=(ab{)}^{i-1}a$ for $1\le i\le 2m$ for its prefix reflections. The closed form is proved by induction on $i$: the conjugating prefix is $w_{i-1}=(ab{)}^{(i-1)\mathrm{/}2}$ for odd $i$ and $w_{i-1}=(ab{)}^{(i-2)\mathrm{/}2}a$ for even $i$, and the identities $(ab{)}^{-j}=(ba{)}^j$ and $a(ba{)}^j=(ab{)}^{j}a$ rewrite $w_{i-1}{s}_i{w}_{i-1}^{-1}$ as $(ab{)}^{i-1}a$ in both cases. Hence $r_{i+m}=(ab{)}^{i-1}(ab{)}^{m}a=(ab{)}^{i-1}a=r_i$ for $1\le i\le m$, while $r_1,\dots ,r_m$ are pairwise distinct: an equality $r_i=r_j$ with $i<j\le m$ gives $(ab{)}^{j-i}=1$ after right multiplication by $a$, contradicting $0<j-i<m$ and the fact from step 3.1 that $ab$ has order exactly $m$. Therefore the sequence of $2m$ prefix reflections is $r_1,\dots ,r_m,r_1,\dots ,r_m$, so $n(r)\in \{0,2\}$ for every $r\in T$ and every accumulated sign $(-1{)}^{n(r)}$ is $1$; the conjugating coordinate returns to $r$ because $w_{2m}=(ab{)}^m=1$. So the permutation effected by each alternating word of length $2m$ is the identity, which verifies $(U_a{U}_b{)}^m=\mathrm{i}\mathrm{d}$ and $(U_b{U}_a{)}^m=\mathrm{i}\mathrm{d}$ for this arbitrary finite pair. There are no distinct-pair relators when $\mathrm{\#}S\le 1$, so the same verification applies in those cases using only $U_s^2=\mathrm{i}\mathrm{d}$. By [F1]'s universal property applied to $f:S\to \mathrm{Sym}(\{\pm 1\}\times T)$, $f(s)=U_s$, there is then a homomorphism $\rho :W\to \mathrm{Sym}(\{\pm 1\}\times T)$ with $\rho (s)=U_s$ for all $s$. Define $(\epsilon ,r)\cdot w:=\rho (w^{-1})(\epsilon ,r)$; this is a right action, because $(wv{)}^{-1}=v^{-1}{w}^{-1}$ gives $(\epsilon ,r)\cdot (wv)=\rho (v^{-1})(\rho (w^{-1})(\epsilon ,r))=((\epsilon ,r)\cdot w)\cdot v$, and $(\epsilon ,r)\cdot s=\rho (s{)}^{-1}(\epsilon ,r)=U_s(\epsilon ,r)$.

Deletion. Suppose $r_i=r_j$ for some $i<j$, that is $w_{i-1}{s}_i{w}_{i-1}^{-1}=w_{j-1}{s}_j{w}_{j-1}^{-1}$. Multiplying on the left by $w_{i-1}^{-1}$ and on the right by $w_{j-1}$ and using $w_{i-1}^{-1}{w}_{j-1}=s_i{s}_{i+1}\cdots s_{j-1}$ gives $s_i{s}_i{s}_{i+1}\cdots s_{j-1}=s_i{s}_{i+1}\cdots s_{j-1}{s}_j$, hence $s_i{s}_{i+1}\cdots s_j=s_{i+1}\cdots s_{j-1}$. Substituting this into the word $(s_1,\dots ,s_k)$ shows that deleting the two letters $s_i$ and $s_j$ leaves the value unchanged: $s_1\cdots \widehat{s_i}\cdots \widehat{s_j}\cdots s_k=w$.

Expression independence and reduced words. For a word of $w$ the sign accumulated in step 4.1 is $(-1{)}^{n(r)}$, and the right action $(\epsilon ,r)\cdot w$ is independent of the word; comparing with the displayed formula for two words of $w$ shows that $(-1{)}^{n(r)}=\eta (r,w)$ depends only on $w$ and $r$, and that $(\epsilon ,r)\cdot w=(\epsilon \eta (r,w),w^{-1}rw)$ for every word of $w$. If the word $(s_1,\dots ,s_k)$ is reduced, then $n(r)\le 1$ for every $r$: if $n(r)\ge 2$, two indices would have $r_i=r_j$ with $i<j$, and step 5.1 would produce a word of length $k-2$ for $w$, contradicting $k=\mathrm{\ell }(w)$. Consequently the map $i\mapsto r_i$ is injective, so the set $\{r_1,\dots ,r_k\}=\{r\in T:\eta (r,w)=-1\}$ has exactly $k=\mathrm{\ell }(w)$ elements; by the independence of $\eta$ from the word, this set is independent of the reduced expression chosen.

Ambient reducedness of alternating words. Let $s\mathrm{\ne }t$, $m:=m(s,t)$, and let $w_q$ be the value of the alternating word $(s,t,s,\dots )$ of length $q$ beginning with $s$, where $q\le m$ if $m<\mathrm{\infty }$ and $q\ge 1$ is arbitrary if $m=\mathrm{\infty }$. Its prefix reflections are $r_i=(st{)}^{i-1}s$ by the closed form of step 4.1. They are pairwise distinct: an equality $r_i=r_j$ with $i<j\le q$ gives $(st{)}^{j-i}=1$ after right multiplication by $s$, which is impossible when $m<\mathrm{\infty }$ because then $0<j-i<q\le m$ and $m$ is the least positive exponent killing $st$ by step 3.1 and [F7], and equally impossible when $m=\mathrm{\infty }$ because by step 3.1 no positive power of $st$ is $1$. Hence exactly $q$ elements $r$ satisfy $\eta (r,w_q)=-1$, namely the $r_i$; on the other hand, for any word of length $k$ in $S$ representing $w_q$ [F1], step 6.1 gives $\mathrm{\#}\{r:\eta (r,w_q)=-1\}=\mathrm{\#}\{r:n^{\mathrm{\prime }}(r)\text{ odd}\}\le {\sum }_r{n}^{\mathrm{\prime }}(r)=k$, where $n^{\mathrm{\prime }}(r)$ counts the occurrences of $r$ among that word's prefix reflections. Therefore $k\ge q$ for every word for $w_q$, so $\mathrm{\ell }(w_q)=q$; in particular the values $w_1,\dots ,w_q$ are pairwise distinct, having distinct lengths.