Specialization of the generic Hecke algebra to the group ring

By [F4] the specialization $H_A=A{\otimes }_{R}H$ is the quotient of the free associative $A$-algebra on $(T_s{)}_{s\in S}$ by the images under $\phi$ of the relations (Q) and (B) of [F1]. When $u_s=1$ for all $s$, those images are $T_s^2-1$ and the braid relations. The assignment $T_s\mapsto s$ extends uniquely to a unital $A$-algebra homomorphism from the free algebra to $A[W]$ ([F8], first assertion); it kills $T_s^2-1$ because $s^2=1$ in $W$, and it kills each braid relation because the defining relator $(st{)}^{m(s,t)}=1$ holds in $W$ ([F7]). By the quotient universal property ([F8], second assertion) it therefore factors uniquely through $H_A$, giving a unital $A$-algebra homomorphism $\mathrm{\Phi }:H_A\to A[W]$ with $\mathrm{\Phi }(T_s)=s$.

Conversely, in $H_A$ the images of the relations give $T_s^2=1$ (the image of (Q) at $u_s=1$) and the braid relations, and each $T_s$ is a unit ([F3]). For $s\mathrm{\ne }t$ with $m:=m(s,t)<\mathrm{\infty }$ put $a:=T_s$, $b:=T_t$ and $x:=ab$; then $a^2=b^2=1$, so $x^{-1}=ba$ and $x^{-k}=(ba{)}^k$ for every $k\ge 0$. The braid relation says that the two alternating products of $m$ factors coincide: if $m=2k$ it reads $x^k=x^{-k}$, whence $x^{2k}=1$, while if $m=2k+1$ it reads $x^{k}a=x^{-k}b$, and multiplying on the right by $b=b^{-1}$ gives $x^{k+1}=x^{-k}$, whence $x^{2k+1}=1$. Thus $(T_s{T}_t{)}^{m(s,t)}=x^m=1$; with $T_s^2=1$ this shows that the assignment $s\mapsto T_s$ kills every defining relator of $W$ ([F7]), so it extends to a group homomorphism $\phi :W\to H_A^{\times }$ with $\phi (w)=T_w$ for every $w\in W$ (the product along a reduced expression). Since $([w]{)}_{w\in W}$ is an $A$-basis of $A[W]$ and $[w][w^{\mathrm{\prime }}]=[w{w}^{\mathrm{\prime }}]$ ([F6]), the formula $\mathrm{\Psi }({\sum }_w{a}_w[w]):={\sum }_w{a}_w\phi (w)$ defines a unital $A$-algebra homomorphism $\mathrm{\Psi }:A[W]\to H_A$ with $\mathrm{\Psi }(w)=T_w$: it is $A$-linear by construction, multiplicativity reduces on basis elements to $\phi (w{w}^{\mathrm{\prime }})=\phi (w)\phi (w^{\mathrm{\prime }})$, and $\mathrm{\Psi }([1])=\phi (1)=1$.

The two maps are mutually inverse: $\mathrm{\Phi }(\mathrm{\Psi }(s))=\mathrm{\Phi }(T_s)=s$ for every $s\in S$ and $\mathrm{\Psi }(\mathrm{\Phi }(T_s))=\mathrm{\Psi }(s)=T_s$, and both composites fix $A$ because the maps are $A$-linear; they fix every $T_w$ since each is a product of the $T_s$, and every group basis element $w$ since each is a product of the generators $s$. The $A$-bases ([F2], [F6]) therefore make the composites the respective identities. Hence $\mathrm{\Phi }$ is an isomorphism of $A$-algebras. On bases, $\mathrm{\Phi }(T_w)=s_1\cdots s_k=w$ for a reduced expression $w=s_1\cdots s_k$ by multiplicativity, so $\mathrm{\Phi }$ carries the $A$-basis $(1\otimes T_w)$ of $H_A$ ([F2]) to the $A$-basis $(w)$ of $A[W]$ ([F6]) and is an isomorphism of free $A$-modules.

For arbitrary units $u_s$, the image of (Q) under $\phi$ reads $T_s^2-(u_s-u_s^{-1})T_s-1=0$ in $H_A$ ([F4]), and $H_A$ is free with basis $(1\otimes T_w)$ ([F2]). If an $A$-algebra homomorphism $H_A\to A[W]$ with $T_s\mapsto s$ existed, applying it to that relation would give $0=s^2-(u_s-u_s^{-1})s-1=-(u_s-u_s^{-1})s$ in $A[W]$; since the elements $w$ form an $A$-basis of $A[W]$ ([F6]), this forces $u_s-u_s^{-1}=0$, i.e. $u_s^2=1$, and over a field $u_s=\pm 1$. Conversely, if every $u_s^2=1$, all specialized quadratics are $T_s^2-1$, so the constructions of 1.1–2.1 apply and give the same basis-preserving isomorphism $H_A\cong A[W]$. Consequently, when $R=\mathbb{Z}[v^{\pm 1}]$ is the one-component coefficient ring, the specializations $v\mapsto 1$ and $v\mapsto -1$ both satisfy $u-u^{-1}=0$; for $u=\pm 1$ the image of (Q) is $T_s^2-1$ in either case, so the construction of 1.1-2.1 applies verbatim and both give the group ring, whereas any specialization to a unit with $u_s^2\mathrm{\ne }1$ admits no such map $T_s\mapsto s$: the standard basis $(1\otimes T_w)$ still exists by [F2] but the quadratic relation is not the group relation.

Assembly: part 1 is steps 1.1, 1.2 and 2.1, the specialization mechanism of part 2 is steps 1.1 and 2.2 (the case $Q_s=1$ of the normalization [F3]), and part 3 is step 2.2 together with the base-change freeness of [F2]. No choice is used: the coefficient homomorphism is unique by [F5], both maps are defined on explicit generators, and no selection occurs. The two application pages named in the statement are reading pointers only; no item of this pair depends on them.