Generic Coxeter Hecke Algebras and the Standard Basis

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.

For $x\in W$ and $s\in S$, [F1] gives $\mathrm{\ell }(sx)=\mathrm{\ell }(x)\pm 1$, so exactly one clause of the displayed definition of $P_s$ applies to $e_x$, and likewise exactly one clause of the definition of $Q_s$ applies to $e_x$; each basis vector therefore has exactly one prescribed image, and extending $R$-linearly defines $P_s,Q_s\in {\mathrm{End}}_R(E)$ ([F6], [F7]). Both clauses have the form $e_x\mapsto e_{\sigma x}+c{e}_x$ with $\sigma$ the multiplication by $s$ from the appropriate side and $c$ the appropriate unit difference, so no selection is involved.

Write $u_s:=v_s-v_s^{-1}$. If $\mathrm{\ell }(sw)=\mathrm{\ell }(w)+1$, then $P_s^2(e_w)=P_s(e_{sw})=e_w+u_s{e}_{sw}=e_w+u_s{P}_s(e_w)$, because $\mathrm{\ell }(s\cdot sw)=\mathrm{\ell }(w)=\mathrm{\ell }(sw)-1$. If $\mathrm{\ell }(sw)=\mathrm{\ell }(w)-1$, then $P_s(e_w)=e_{sw}+u_s{e}_w$ and $P_s(e_{sw})=e_w$, so $P_s^2(e_w)=e_w+u_s{P}_s(e_w)$; in both cases $P_s^2=u_s{P}_s+{\mathrm{id}}_E$ on each basis vector, hence on $E$ ([F6], [F7]). Therefore $P_s(P_s-u_s{\mathrm{id}}_E)={\mathrm{id}}_E=(P_s-u_s{\mathrm{id}}_E)P_s$, so $P_s$ is invertible with inverse $P_s-u_s{\mathrm{id}}_E$. Multiplying all length data on the right gives the same computation for $Q_s$.

(two-length lemma) Let $x\in W$ and $s,t\in S$ satisfy $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$ and $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)$; then $sx=xt$. Indeed, let $x=s_1\cdots s_q$ be a reduced expression. If $\mathrm{\ell }(xt)=q+1$, then $(s_1,\dots ,s_q,t)$ is a reduced expression of $xt$ of length $q+1=\mathrm{\ell }(xt)$, and $\mathrm{\ell }(s\cdot xt)=\mathrm{\ell }(x)=q$, so exchange [F1] applied to this expression and the letter $s$ gives an index $i\in \{1,\dots ,q+1\}$ with $s{s}_1\cdots s_{i-1}=s_1\cdots s_i$, where $s_{q+1}:=t$. If $i=q+1$, this reads $sx=xt$ and we are done; if $i\le q$, then $sx=(s{s}_1\cdots s_{i-1})s_i{s}_{i+1}\cdots s_q=(s_1\cdots s_i)s_i{s}_{i+1}\cdots s_q=s_1\cdots s_{i-1}{s}_{i+1}\cdots s_q$ is represented by a word of $q-1$ letters, so $\mathrm{\ell }(sx)\le q-1<q+1=\mathrm{\ell }(xt)=\mathrm{\ell }(sx)$, a contradiction. If $\mathrm{\ell }(xt)=q-1$, put $x^{\mathrm{\prime }}:=xt$; then $x^{\mathrm{\prime }}t=x$, so $\mathrm{\ell }(x^{\mathrm{\prime }}t)=\mathrm{\ell }(x)=q=\mathrm{\ell }(xt)+1=\mathrm{\ell }(x^{\mathrm{\prime }})+1$, $\mathrm{\ell }(s{x}^{\mathrm{\prime }}t)=\mathrm{\ell }(sx)=q-1=\mathrm{\ell }(x^{\mathrm{\prime }})$ and $\mathrm{\ell }(s{x}^{\mathrm{\prime }})=\mathrm{\ell }(sxt)=\mathrm{\ell }(x)=q=\mathrm{\ell }(x^{\mathrm{\prime }}t)$, so $x^{\mathrm{\prime }}$ satisfies the hypotheses of the case just treated with the roles of the lengths interchanged, and that case yields $s{x}^{\mathrm{\prime }}=x^{\mathrm{\prime }}t$; multiplying by $t$ on the right gives $sx=xt$.

Fix $s,t\in S$ and $x\in W$ and expand both $P_s{Q}_t(e_x)$ and $Q_t{P}_s(e_x)$ from the definitions. In the four length configurations (i) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)+2$, where both sides equal $e_{sxt}$; (ii) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)-2$, where both sides equal $e_{sxt}+u_t{e}_{sx}+u_s{e}_{xt}+u_s{u}_t{e}_x$; (iii) $\mathrm{\ell }(sx)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, where both sides equal $e_{sxt}+u_s{e}_{xt}$; (iv) $\mathrm{\ell }(sx)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, where both sides equal $e_{sxt}+u_t{e}_{sx}$; the two expansions agree, where $u_s=v_s-v_s^{-1}$ and $u_t=v_t-v_t^{-1}$.

(reduced products at $e_1$) Let $w=s_1\cdots s_k$ be a reduced expression. Every contiguous subword is reduced: replacing a subword by a shorter representative would shorten the entire expression of $w$, contradicting $\mathrm{\ell }(w)=k$ ([F5]). Put $z_i:=s_i\cdots s_k$ for $1\le i\le k$ and $z_{k+1}:=1$; then $\mathrm{\ell }(z_i)=k-i+1$ and $s_i{z}_{i+1}=z_i$, so $P_{s_i}(e_{z_{i+1}})=e_{z_i}$. Applying the operators from right to left gives $P_{s_1}\cdots P_{s_k}(e_1)=e_w$. For right multiplication put $w_i:=s_1\cdots s_i$, $w_0:=1$; the same reduced-subword argument gives $\mathrm{\ell }(w_i)=i$ and $w_{i-1}{s}_i=w_i$, so $Q_{s_i}(e_{w_{i-1}})=e_{w_i}$. Thus $Q_{s_k}\cdots Q_{s_1}(e_1)=e_w$. Both identities also hold for the empty expression.

In the two remaining length configurations, (v) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)-1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, the two expansions are $P_s{Q}_t(e_x)=e_{sxt}+u_t{e}_{sx}+u_s{u}_t{e}_x$ and $Q_t{P}_s(e_x)=e_{sxt}+u_s{e}_{xt}+u_s{u}_t{e}_x$; (vi) $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)=\mathrm{\ell }(x)+1$, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$, the two expansions are $P_s{Q}_t(e_x)=e_{sxt}+u_s{e}_{xt}$ and $Q_t{P}_s(e_x)=e_{sxt}+u_t{e}_{sx}$. In both, $\mathrm{\ell }(sxt)=\mathrm{\ell }(x)$ and $\mathrm{\ell }(sx)=\mathrm{\ell }(xt)$, so $sx=xt$ by 1.3. Hence $s$ and $t$ are conjugate in $W$ (with $x$ as conjugating element) and therefore $v_s=v_t$ by the parameter rule of the definition ([F2]), so $u_s=u_t$ and the two expansions coincide: also $P_s{Q}_t(e_x)=Q_t{P}_s(e_x)$. Together with the four configurations of 1.4 this proves $P_s{Q}_t(e_x)=Q_t{P}_s(e_x)$ for every basis vector $e_x$, so $P_s{Q}_t=Q_t{P}_s$ because $(e_x)$ spans $E$ ([F6]).

Let $s\mathrm{\ne }t$ with $m:=m(s,t)<\mathrm{\infty }$, and put $a:=P_s{P}_t{P}_s\cdots$ and $b:=P_t{P}_s{P}_t\cdots$, each product having $m$ alternating factors. Let $w_1$ be the alternating product of $m$ factors beginning with $s$ and $w_2$ that beginning with $t$. Each successive factor is length-increasing along the alternating word by [F3] (the partial alternating words of length $\le m$ are reduced), so the computation of 1.5 gives $a(e_1)=e_{w_1}$ and, with the roles of $s,t$ exchanged, $b(e_1)=e_{w_2}$. In $W$ one has $w_1=w_2$: since $(st)(ts)=stts=s^2=1$ in $W$, we have $(ts)=(st{)}^{-1}$ ([F5]), and the relator $(st{)}^m=1$ gives $(st{)}^{-k}=(st{)}^{m-k}$ for $0\le k\le m$; hence if $m=2k+1$ then $w_2=(ts{)}^{k}t=(st{)}^{-k}t=(st{)}^{k+1}t=(st{)}^{k}s=w_1$, while if $m=2k$ then $w_2=(ts{)}^k=(st{)}^{-k}=(st{)}^k=w_1$. Now let $w=u_1\cdots u_k$ be any reduced expression. By 1.5, $Q_{u_k}\cdots Q_{u_1}(e_1)=e_w$, and by 2.1 every $P_s$ commutes with every $Q_t$, so $a(e_w)=a{Q}_{u_k}\cdots Q_{u_1}(e_1)=Q_{u_k}\cdots Q_{u_1}a(e_1)=Q_{u_k}\cdots Q_{u_1}b(e_1)=b(e_w)$; since $(e_w{)}_{w\in W}$ is a basis, $a=b$ ([F6]). The same argument with $P$ replaced by $Q$ throughout gives the braid relations for the $Q$'s.

By [F4] any two reduced expressions of $w$ are braid-equivalent, and a braid move replaces a block $P_s{P}_t{P}_s\cdots$ of $m(s,t)$ factors by $P_t{P}_s{P}_t\cdots$ with the same product by 3.1, so the product $P_{s_1}\cdots P_{s_k}$ depends only on $w$; writing $P_w$ for this common endomorphism, $P_w(e_1)=e_w$ by 1.5. Hence part 4 holds, and with part 1 (2.1), part 2 (1.2) and part 3 (3.1) all four assertions are proved. No finiteness of $W$ and no regularity of $R$ was used, and no choice: the operators are defined by explicit length conditions and every product above runs along a reduced expression that exists by the definition of $\mathrm{\ell }$.

If $w=s_1\cdots s_k$ and $w=s_1^{\mathrm{\prime }}\cdots s_k^{\mathrm{\prime }}$ are reduced expressions, then by [F1] they are connected by finitely many braid moves. A braid move replaces a consecutive block $s,t,s,\dots$ of $m(s,t)$ alternating letters by $t,s,t,\dots$, and the corresponding block $T_s{T}_t{T}_s\cdots$ of the product is replaced by $T_t{T}_s{T}_t\cdots$, which equals it in $H$ by the braid relation ([F2]); letters outside the block are untouched. Reading the moves one at a time, the two products are equal, so $T_w:=T_{s_1}\cdots T_{s_k}$ is a well-defined element of $H$; for $w=1$ the empty product equals $1$.