The reversal anti-involution, generator invertibility, the bar operator and the multiplicative normalization

Define ${\mathrm{\#}}_F:F\to F$ by $R$-linear extension of word reversal, $(T_{s_1}\cdots T_{s_k}{)}^{\mathrm{\#}}=T_{s_k}\cdots T_{s_1}$ and the empty word fixed. Reversal is an involutive anti-automorphism of the word monoid, so ${\mathrm{\#}}_F$ is an $R$-linear involutive anti-automorphism of $F$ ([F1]). The ideal $I$ is preserved: the quadratic generator $T_s^2-(v_s-v_s^{-1})T_s-1$ is a polynomial in $T_s$ with central coefficients and is fixed by ${\mathrm{\#}}_F$, and for $s\mathrm{\ne }t$ with $m:=m(s,t)<\mathrm{\infty }$ the alternating products $a_1=T_s{T}_t{T}_s\cdots$ and $a_2=T_t{T}_s{T}_t\cdots$ of $m$ factors are each reversed into one of the two: the alternating word $sts\cdots$ of length $m$ is a palindrome exactly when $m$ is odd, so $a_1^{\mathrm{\#}}=a_1$, $a_2^{\mathrm{\#}}=a_2$ for odd $m$ and $a_1^{\mathrm{\#}}=a_2$, $a_2^{\mathrm{\#}}=a_1$ for even $m$, and in both cases $(a_1-a_2{)}^{\mathrm{\#}}=\pm (a_1-a_2)$; since the two-sided ideal generated by these elements consists of finite sums ${\sum }_i{c}_i{x}_i{g}_i{y}_i$ with $g_i$ among the generators ([F1]) and ${\mathrm{\#}}_F$ reverses products, ${\mathrm{\#}}_F$ maps each such sum to a sum of the same shape (a generator $g_i$ being replaced by $\pm$ itself), so ${\mathrm{\#}}_F(I)\subseteq I$. Therefore $\mathrm{\#}(x+I):={\mathrm{\#}}_F(x)+I$ is a well-defined $R$-linear anti-automorphism of $H$ with ${\mathrm{\#}}^2=\mathrm{i}\mathrm{d}$: if $x-y\in I$ then ${\mathrm{\#}}_{F}x-{\mathrm{\#}}_{F}y={\mathrm{\#}}_F(x-y)\in I$. It is determined by $T_s^{\mathrm{\#}}=T_s$ because $H$ is generated as a ring by the $T_s$ and the coefficients ([F2]), and for a reduced expression $w=s_1\cdots s_k$ one has $\mathrm{\#}(T_w)=T_{s_k}\cdots T_{s_1}=T_{s_k\cdots s_1}=T_{w^{-1}}$, since reversing a word of length $k$ representing $w$ represents $w^{-1}$, whence $\mathrm{\ell }(w^{-1})=\mathrm{\ell }(w)=k$ by [F4] and the reversed word is a reduced expression of $w^{-1}$ ([F3]).

By the quadratic relation of [F2], $T_s^2-(v_s-v_s^{-1})T_s-1=0$ in $H$, so $T_s(T_s-(v_s-v_s^{-1}))=1=(T_s-(v_s-v_s^{-1}))T_s$ and $T_s^{-1}=T_s-(v_s-v_s^{-1})$. Inverses multiply in reverse order, so for $T_w=T_{s_1}\cdots T_{s_k}$ one has $T_w^{-1}=T_{s_k}^{-1}\cdots T_{s_1}^{-1}$, and $T_w$ is a unit.

The assignment $v_i\mapsto v_i^{-1}$ gives units of $R$, so by the universal property of the Laurent ring it extends to a unique ring endomorphism $\sigma :R\to R$ with $\sigma (v_i)=v_i^{-1}$ ([F5]); note that $\sigma$ is not $R$-linear (it does not fix the coefficients), only a ring endomorphism, which is all that is used below. Since ${\sigma }^2$ is again a ring endomorphism fixing every $v_i$, uniqueness of the extension identifies ${\sigma }^2=\mathrm{i}\mathrm{d}$, so $\sigma$ is an involution.

Extend $\sigma$ to $\stackrel{ˉ}{}$ on the free algebra: define $\stackrel{ˉ}{}:F\to F$ on words by $\overline{T_{s_1}\cdots T_{s_k}}:={\bar{T}}_{s_1}\cdots {\bar{T}}_{s_k}$ with ${\bar{T}}_s:=T_s-(v_s-v_s^{-1})$, and on ${\sum }_w{r}_{w}w$ by ${\sum }_w\sigma (r_w)\bar{w}$; this is the unique $\sigma$-semilinear ring endomorphism of $F$ with those generator images, because $F$ is free on the words ([F1]). It maps $I$ into itself: the quadratic generator satisfies $\overline{(T_s-v_s)(T_s+v_s^{-1})}=({\bar{T}}_s-v_s^{-1})({\bar{T}}_s+v_s)=(T_s-v_s)(T_s+v_s^{-1})\in I$, using $\sigma (v_s)=v_s^{-1}$ and $\sigma (v_s^{-1})=v_s$; and for the braid generator, ${\bar{a}}_1=T_s^{-1}{T}_t^{-1}\cdots$ equals, by step 1.2, the inverse of the product along the reversed alternating word, and the reversed alternating word has the same length $m$ and alternates between $s$ and $t$, so its product is the same element $A:=a_1=a_2$ of $H$ by relation (B) ([F2]); hence ${\bar{a}}_1=A^{-1}={\bar{a}}_2$ and $\overline{(a_1-a_2)}\in I$; the general element of $I$ is a finite sum as above and $\stackrel{ˉ}{}$ is additive, so $\stackrel{ˉ}{}(I)\subseteq I$. Hence $\stackrel{ˉ}{}(x+I):=\bar{x}+I$ is well defined on $H$, is a $\sigma$-semilinear ring endomorphism, and satisfies $\overline{T_s}=T_s-(v_s-v_s^{-1})=T_s^{-1}$ by step 1.2. It is involutive: ${\stackrel{ˉ}{}}^2$ is ${\sigma }^2$-semilinear, hence $R$-linear, and ring-endomorphic, it fixes $R$ pointwise because ${\sigma }^2=\mathrm{i}\mathrm{d}$ (step 1.3), and it fixes each generator because ${\stackrel{ˉ}{}}^2(T_s)=\stackrel{ˉ}{}(T_s^{-1})=\stackrel{ˉ}{}(T_s{)}^{-1}=(T_s^{-1}{)}^{-1}=T_s$; since $H$ is generated by these ([F2]), ${\stackrel{ˉ}{}}^2=\mathrm{i}\mathrm{d}$. Finally, multiplicativity of $\stackrel{ˉ}{}$ gives $\overline{T_w}=\overline{T_{s_1}}\cdots \overline{T_{s_k}}=T_{s_1}^{-1}\cdots T_{s_k}^{-1}=(T_{s_k}\cdots T_{s_1}{)}^{-1}=T_{w^{-1}}^{-1}$ for a reduced expression, using step 1.2 and $T_{s_k\cdots s_1}=T_{w^{-1}}$ ([F3]).

Put $S_s:=v_s{T}_s$ and $Q_s:=v_s^2$. Substituting $T_s=v_s^{-1}{S}_s$ in $(T_s-v_s)(T_s+v_s^{-1})=0$ ([F2]) gives $v_s^{-1}(S_s-Q_s)\cdot v_s^{-1}(S_s+1)=0$, and multiplying by the unit $v_s^2$ gives $(S_s-Q_s)(S_s+1)=0$, equivalently $S_s^2=(Q_s-1)S_s+Q_s$. Since $v_s$ is a unit of $R$ ([F2]) and $T_s$ is a unit with inverse $T_s-(v_s-v_s^{-1})$ (step 1.2), also $S_s$ is a unit with $S_s^{-1}=v_s^{-1}(T_s-(v_s-v_s^{-1}))$, and $T_s=v_s^{-1}{S}_s$ by construction. If $v_s=v$ for every $s$ then $Q_s=Q=v^2$ for every $s$, so the displayed relation is uniform in $s$. No other identification between the parameters is made.

Assembly: the reversal anti-involution of part 1 is step 1.1, the invertibility statement of part 2 is step 1.2, the bar operator of part 3 is steps 1.3 and 2.1, and the normalization of part 4 is step 2.2. No choice is used: word reversal and the monomial substitution $v_i\mapsto v_i^{-1}$ are explicit, and the standard basis is used only to name the elements $T_w$, never to select them.