Universal parameters, the generic Coxeter Hecke algebra and generator conjugacy

The relation is reflexive (the one-term chain with $k=0$), symmetric (a chain $s=s_0,\dots ,s_k=t$ with all $m(s_i,s_{i+1})$ odd reverses to a chain from $t$ to $s$, because $m$ is symmetric) and transitive (two chains are concatenated at their common endpoint). It is therefore the connected-component relation of the graph on $S$ whose edges are the unordered pairs $\{s,t\}$, $s\mathrm{\ne }t$, with $m(s,t)$ odd; its classes are the components, so $0\le c\le \mathrm{\mid }S\mathrm{\mid }<\mathrm{\infty }$, with $c=0$ exactly when $S=\mathrm{\varnothing }$ (then $W$ is trivial and $R=\mathbb{Z}$).

Since $\mathbb{Z}$ is a commutative ring, applying the Laurent construction of [F2] gives that $R$ is a commutative ring in which each $v_i$ is a unit; $F$ is a unital associative $R$-algebra with central $R$-image and $I$ is the two-sided ideal generated by the displayed finitely many relations ([F1], [F3]). The quotient $H=F\mathrm{/}I$ is a unital associative $R$-algebra ([F1], [F3]); an $R$-algebra homomorphism $F\to A$ is exactly an assignment of the generators $T_s$, and by the quotient universal property ([F1]) it factors uniquely through $H$ precisely when it kills the relations, which is the stated universal property.

Suppose $s\mathrm{\ne }t$ and $m:=m(s,t)$ is finite odd, say $m=2k+1$ with $k\ge 1$. In the free group on $S$ one has $(st{)}^{2k+1}=(st{)}^{k}s\cdot t(st{)}^k$, and the two words $t(st{)}^k$ and $(ts{)}^{k}t$ coincide. Since $s^2=t^2=1$ and $(st{)}^m=1$ are relators of $W$ ([F4]), the product $(st{)}^{k}s\cdot t(st{)}^k=(st{)}^{2k+1}$ is trivial in $W$, so $(st{)}^{k}s=(t(st{)}^k{)}^{-1}=(st{)}^{-k}t=(ts{)}^{k}t=t(st{)}^k$ there; thus the elements $x:=(st{)}^k$ and $t$ satisfy $xs=tx$ in $W$, so $xs{x}^{-1}=t$: the generators joined by an odd edge are conjugate ([F5]).

Fix a class $C$ of $\sim$ and define ${\phi }_C:S\to \{\pm 1\}$ by ${\phi }_C(u)=-1$ for $u\in C$ and ${\phi }_C(u)=1$ for $u\in S\setminus C$. Then ${\phi }_C(u{)}^2=1$, and for $u\mathrm{\ne }v$ with $m(u,v)<\mathrm{\infty }$ one has ${\phi }_C((uv{)}^{m(u,v)})=(-1{)}^{m(u,v)({\delta }_u+{\delta }_v)}$, where ${\delta }_w=1$ for $w\in C$ and $0$ otherwise: if exactly one of $u,v$ lies in $C$ then $m(u,v)$ is even, because odd $m(u,v)$ would make $u$ and $v$ odd-connected and hence place them in the same class $C$, a contradiction. So every relator of the presentation ([F4]) is killed and ${\phi }_C$ extends to a group homomorphism $W\to \{\pm 1\}$ ([F4]). If $s=xt{x}^{-1}$ in $W$, then ${\phi }_C(s)={\phi }_C(x){\phi }_C(t){\phi }_C(x{)}^{-1}={\phi }_C(t)$ because $\{\pm 1\}$ is abelian ([F6]); hence conjugate generators lie in one class: for $s\nsim t$, the class $C=[s]$ gives ${\phi }_C(s)=-1\mathrm{\ne }1={\phi }_C(t)$, so $s$ and $t$ are not conjugate.

By the universal property of the Laurent ring ([F2]) the assignment $v_i\mapsto u_i$ extends uniquely to a ring homomorphism $R\to A$ for every commutative ring $A$ and units $u_1,\dots ,u_c\in A^{\times }$; combining it with the universal property of $H$ established in 1.2 ([F3], [F1]) exhibits $R$ as the universal coefficient ring for a unit-valued parameter that is constant on each class.

Expanding in the free algebra, using that the coefficients $v_s$ and $v_s^{-1}$ are central ([F1]), gives $(T_s-v_s)(T_s+v_s^{-1})=T_s^2-(v_s-v_s^{-1})T_s-1$, and the reverse product $(T_s+v_s^{-1})(T_s-v_s)$ is the same element; replacing $v_s$ by the unit $-v_s^{-1}$ leaves $v_s-v_s^{-1}$ unchanged, so the quadratic relation depends on $v_s$ only through that difference.

By 1.3 an odd edge joins conjugate generators, so transitivity of conjugacy ([F5]) gives that $s\sim t$ implies conjugacy of $s$ and $t$ in $W$; by 1.4 a pair with $s\nsim t$ is separated by the homomorphism ${\phi }_{[s]}$, so it is not conjugate. This proves the generator-conjugacy criterion, and with 1.1, 1.2, 1.5, 1.6 all the assertions of the definition are established; every construction and every separating homomorphism above is explicit, so no choice is used.