The star anti-automorphism of the type-A Hecke algebra

Put $X:=\{1,\dots ,n-1\}$ and $F:=R^{(X^{\ast })}$, the free $R$-module with the finite words in $X$ as basis [F5]; write the basis word corresponding to $i_1,\dots ,i_k$ as $T_{i_1}\cdots T_{i_k}$ and the empty word as $1$. Extending concatenation of words $R$-bilinearly makes $F$ a unital associative $R$-algebra: concatenation is associative with two-sided identity the empty word by [F4], and distributivity holds by the bilinear extension. Let $I\subseteq F$ be the two-sided ideal generated by the relators $r_i:=T_i^2-(q-1)T_i-q\cdot 1$, $b_i:=T_i{T}_{i+1}{T}_i-T_{i+1}{T}_i{T}_{i+1}$ and $d_{ij}:=T_i{T}_j-T_j{T}_i$, and put $H:=F\mathrm{/}I$, which by [F1] is the presented algebra $H_{R,q}(S_n)$, the image of $T_i$ being again written $T_i$; every element of $H$ is the image of an $R$-linear combination of words in the $T_i$, so $H$ is generated as an $R$-algebra by $T_1,\dots ,T_{n-1}$.

Define $\rho :F\to F$ on the word basis by $\rho (T_{i_1}\cdots T_{i_k}):=T_{i_k}\cdots T_{i_1}$ and $\rho (1):=1$, extended $R$-linearly. Then $\rho$ is an $R$-linear involution and reverses products, $\rho (xy)=\rho (y)\rho (x)$, because reversal of a concatenation is the concatenation of the reversals in the opposite order and both sides are $R$-bilinear; in particular ${\rho }^2={\mathrm{i}\mathrm{d}}_F$ and $\rho (T_i)=T_i$ for each generator.

$\rho$ preserves $I$, i.e. $\rho (I)\subseteq I$: the relators are fixed up to sign, $\rho (r_i)=r_i$, $\rho (b_i)=b_i$ and $\rho (d_{ij})=-d_{ij}$ by step 1.2, so each lies in $I$; the set $J:=\{x\in F:\rho (x)\in I\}$ contains $0$, is closed under addition and additive inverses because $\rho$ is $R$-linear and $I$ is an additive subgroup, and for $x\in J$ and $a\in F$ one has $\rho (ax)=\rho (x)\rho (a)\in I$ and $\rho (xa)=\rho (a)\rho (x)\in I$ because $I$ is a two-sided ideal [F6]; hence $J$ is a two-sided ideal containing all relators, so $(S)\subseteq J$ and $I=(S)\subseteq J$, that is, $\rho (I)\subseteq I$.

Consequently $\bar{\rho }:H\to H$, $\bar{\rho }(x+I):=\rho (x)+I$, is well defined: if $x-y\in I$ then $\rho (x)-\rho (y)=\rho (x-y)\in I$ by step 2.1. It is an $R$-linear anti-automorphism of $H$ with ${\bar{\rho }}^2=\mathrm{i}\mathrm{d}$ and $\bar{\rho }(T_i)=T_i$: the quotient operations of [F7] and step 1.2 give $\bar{\rho }((x+I)(y+I))=\rho (xy)+I=\rho (y)\rho (x)+I=\bar{\rho }(y+I)\bar{\rho }(x+I)$, and similarly for addition, scalars and the unit.

$\bar{\rho }$ is the unique $R$-algebra anti-automorphism of $H$ with $\bar{\rho }(T_i)=T_i$ for all $i$: if $\alpha$ is any such anti-automorphism, then $\alpha$ and $\bar{\rho }$ agree on $1$, and for a word $T_{i_1}\cdots T_{i_k}$ one has $\alpha (T_{i_1}\cdots T_{i_k})=\alpha (T_{i_k})\cdots \alpha (T_{i_1})=T_{i_k}\cdots T_{i_1}=\bar{\rho }(T_{i_1}\cdots T_{i_k})$; the words span $H$ by step 1.1, so $\alpha =\bar{\rho }$ by $R$-linearity.

For $w\in S_n$ and a reduced expression $w=s_{i_1}\cdots s_{i_{\mathrm{\ell }}}$ one has $T_w=T_{i_1}\cdots T_{i_{\mathrm{\ell }}}$ in the generic algebra and hence $T_w=T_{i_1}\cdots T_{i_{\mathrm{\ell }}}\in H$; thus $T_w^{\ast }=\bar{\rho }(T_{i_1}\cdots T_{i_{\mathrm{\ell }}})=T_{i_{\mathrm{\ell }}}\cdots T_{i_1}$ by step 3.1. The reversed word $s_{i_{\mathrm{\ell }}}\cdots s_{i_1}$ represents $w^{-1}$ and has $\mathrm{\ell }$ letters, and $\mathrm{\ell }=\mathrm{\ell }(w)=\mathrm{\ell }(w^{-1})$ by [F3], so it is a reduced expression of $w^{-1}$ and therefore $T_{w^{-1}}=T_{i_{\mathrm{\ell }}}\cdots T_{i_1}\in H$; hence $T_w^{\ast }=T_{w^{-1}}$ for every $w$, and $T_w$ is well defined in $H$ independently of the chosen reduced expression by [F1] and [F2].

The assertions of the statement follow: $\ast :=\bar{\rho }$ is the unique extension of the assignment $T_i\mapsto T_i$, an involution by step 3.1; $T_w^{\ast }=T_{w^{-1}}$ is step 4.2; therefore $(T_w{T}_{w^{-1}}{)}^{\ast }=T_{w^{-1}}^{\ast }{T}_w^{\ast }=T_{(w^{-1}{)}^{-1}}{T}_{w^{-1}}=T_w{T}_{w^{-1}}$; and for the parabolic subgroup $S_{\lambda }$ the map $w\mapsto w^{-1}$ is a bijection of $S_{\lambda }$, so $m_{\lambda }^{\ast }={\sum }_{w\in S_{\lambda }}{T}_{w^{-1}}={\sum }_{v\in S_{\lambda }}{T}_v=m_{\lambda }$. No choice principle is used: the constructions are explicit on the word basis.