The standard basis of the type-A Hecke algebra and its multiplication rule

Since $x{s}_i{s}_i=x$ and $\mathrm{\ell }(x{s}_i)\le \mathrm{\ell }(x)+1$, $\mathrm{\ell }(x)\le \mathrm{\ell }(x{s}_i)+1$ by the same inequality for $x{s}_i$ and the involution $s_i^2=1$, we have $\mathrm{\ell }(x{s}_i)=\mathrm{\ell }(x)\pm 1$ for every $x\in S_n$; define the free $A$-module $E:={⨁}_{w\in S_n}A{e}_w$ and, for each $i$, the $A$-linear endomorphism ${\rho }_i$ by $e_w{\rho }_i:=e_{w{s}_i}$ when $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$ and $e_w{\rho }_i:=(q-1)e_w+q{e}_{w{s}_i}$ when $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)-1$. In the first case we say $i$ is an ascent at $w$ and in the second a descent.

${\rho }_i^2=(q-1){\rho }_i+q$: if $i$ is an ascent at $w$ then $w=(w{s}_i)s_i$ and the last step is a descent, so $e_w{\rho }_i^2=e_{w{s}_i}{\rho }_i=(q-1)e_{w{s}_i}+q{e}_w=(q-1)e_w{\rho }_i+q{e}_w$; if $i$ is a descent at $w$ then $i$ is an ascent at $w{s}_i$, so $e_w{\rho }_i^2=(q-1)e_w{\rho }_i+q{e}_{w(s_i{s}_i)}=(q-1)((q-1)e_w+q{e}_{w{s}_i})+q{e}_w$ and $(q-1)e_w{\rho }_i+q{e}_w$ takes the same value.

Distant braid: if $\mathrm{\mid }i-j\mathrm{\mid }>1$ then $s_i,s_j$ commute and by [F3] they fix each other's simple roots, so if ${\gamma }_i:=x{\beta }_i,{\gamma }_j:=x{\beta }_j$ then $(x{s}_i){\beta }_j={\gamma }_j$, $(x{s}_j){\beta }_i={\gamma }_i$ and $(x{s}_i{s}_j){\beta }_i={\gamma }_i,(x{s}_i{s}_j){\beta }_j={\gamma }_j$; hence by [F4] the descent behaviour at $x{s}_i$ in direction $j$ agrees with that at $x$ in direction $j$, and likewise with $i,j$ exchanged. Consequently in the four cases according to whether $i,j$ are ascents or descents at $x$, the two compositions ${\rho }_i{\rho }_j$ and ${\rho }_j{\rho }_i$ expand to the same combination of $e_x,e_{x{s}_i},e_{x{s}_j},e_{x{s}_i{s}_j}$: both give $e_{x{s}_i{s}_j}$ if both are ascents, $(q-1)e_{x{s}_i}+q{e}_{x{s}_i{s}_j}$ if $i$ climbs and $j$ descends, the mirror expression $(q-1)e_{x{s}_j}+q{e}_{x{s}_i{s}_j}$ if $i$ descends and $j$ climbs, and $(q-1{)}^2{e}_x+q(q-1)e_{x{s}_i}+q(q-1)e_{x{s}_j}+q^2{e}_{x{s}_i{s}_j}$ if both descend.

Adjacent braid: for adjacent $i,j$ one has $s_i{\beta }_j={\beta }_i+{\beta }_j$ and $s_j{\beta }_i={\beta }_i+{\beta }_j$ by [F3], so writing ${\gamma }_i:=x{\beta }_i,{\gamma }_j:=x{\beta }_j$ for an element $x$, appending $s_i$ sends the pair to $(-{\gamma }_i,{\gamma }_i+{\gamma }_j)$ and appending $s_j$ sends it to $({\gamma }_i+{\gamma }_j,-{\gamma }_j)$; by [F4] the ascent/descent behaviour along the six elements $x,x{s}_i,x{s}_j,x{s}_i{s}_j,x{s}_j{s}_i,x{s}_i{s}_j{s}_i=x{s}_j{s}_i{s}_j$ of the right coset of $⟨s_i,s_j⟩$ is therefore determined by the signs of ${\gamma }_i$, of ${\gamma }_j$ and, when these are mixed, of the root $x({\beta }_i+{\beta }_j)={\gamma }_i+{\gamma }_j$ (for ${\gamma }_i,{\gamma }_j$ both positive the sum is positive and for both negative it is negative, since it is a root equal to the image of the positive root ${\beta }_i+{\beta }_j$, and it lies in the closed positive or negative cone according to the signs of its two summands). Writing $v_e=e_x$, $v_i=e_{x{s}_i}$, $v_j=e_{x{s}_j}$, $v_{ij}=e_{x{s}_i{s}_j}$, $v_{ji}=e_{x{s}_j{s}_i}$ and $v_{iji}=e_{x{s}_i{s}_j{s}_i}$, both ${\rho }_i{\rho }_j{\rho }_i$ and ${\rho }_j{\rho }_i{\rho }_j$ applied to $v_e$ expand within the span of these six vectors, and evaluating the two expansions in the four cases gives the same result: $v_{iji}$ when ${\gamma }_i>0<{\gamma }_j$; $[(q-1{)}^3+q(q-1)]v_e+q(q-1{)}^2(v_i+v_j)+q^2(q-1)(v_{ij}+v_{ji})+q^3{v}_{iji}$ when ${\gamma }_i<0>{\gamma }_j$; $(q-1)v_{ij}+q{v}_{iji}$ when ${\gamma }_i>0>{\gamma }_j$ and ${\gamma }_i+{\gamma }_j>0$; and $(q-1{)}^2{v}_i+q(q-1)v_e+q(q-1)v_{ij}+q^2{v}_{iji}$ when ${\gamma }_i>0>{\gamma }_j$ and ${\gamma }_i+{\gamma }_j<0$. The remaining two cases have ${\gamma }_j>0>{\gamma }_i$, and exchanging the names $i$ and $j$ carries each of the four computed cases to one of these while swapping the two compositions.

$E$ is a right $H_n$-module: steps 1.1, 2.1, 2.2 and 3.1 verify the defining relations of [F1] for the operators ${\rho }_i$, so $e_w(T_{i_1}\cdots T_{i_r}):=e_w{\rho }_{i_1}\cdots {\rho }_{i_r}$ is well defined. If $i_1\dots i_k$ is a reduced word for $w$, then each step of its prefix chain is an ascent, so $e_e{T}_{i_1}\cdots T_{i_k}=e_w$ for every $w$, and by [F2] the element $T_w:=T_{i_1}\cdots T_{i_k}$ is independent of the chosen reduced word (the braid and commutation relations used to pass between reduced words are relations of $H_n$ by [F1]).

Independence of $\{T_w\}$: if ${\sum }_w{a}_w{T}_w=0$ in $H_n$ with $a_w\in A$, then applying the module action of step 4.1 to $e_e$ gives ${\sum }_w{a}_w{e}_w=0$ in the free module $E$, so $a_w=0$ for every $w$.

Multiplication rule and spanning: let $u:=w{s}_i$. If $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)+1$ then $w{s}_i$ has the reduced word (reduced word for $w$) followed by $i$, so $T_w{T}_i=T_{w{s}_i}$; if $\mathrm{\ell }(w{s}_i)=\mathrm{\ell }(w)-1$ then $w=(w{s}_i)s_i$ is a reduced factorisation, so $T_w=T_{w{s}_i}{T}_i$ and $T_w{T}_i=T_{w{s}_i}{T}_i^2=(q-1)T_{w{s}_i}{T}_i+q{T}_{w{s}_i}=(q-1)T_w+q{T}_{w{s}_i}$. Since every element of $H_n$ is a finite $A$-combination of monomials in the $T_i$, the rule just proved rewrites any such monomial, by induction on the number of letters, as an $A$-combination of the $T_w$; hence the $T_w$ span $H_n$.

Triangular normalized products and the standard normalized basis: put ${\tilde{T}}_w:=v^{\mathrm{\ell }(w)}{T}_w$; since multiplication by the unit $v^{\mathrm{\ell }(w)}$ is an $A$-linear automorphism and $\{T_w\}$ is an $A$-basis by step 5.1, the family $\{{\tilde{T}}_w\}$ is an $A$-basis of $H_n$ as well, and ${\tilde{T}}_w$ does not depend on the reduced word because $T_w$ does not. For a reduced word $i_1\dots i_k$ for $w$ one has $H_{i_1}\cdots H_{i_k}={\prod }_r(v{T}_{i_r}+v)$, and by step 5.2 each successive multiplication by $v{T}_{i_r}+v$ replaces a combination ${\sum }_u{a}_u{T}_u$ with $\mathrm{\ell }(u)\le r-1$ by a combination of $T_u$ and $T_{u{s}_{i_r}}$ with $\mathrm{\ell }(u{s}_{i_r})\le \mathrm{\ell }(u)+1\le r$, the coefficient of $T_w$ being $v^k\mathrm{\ne }0$ and no term of length $k$ other than $T_w$ appearing; so $H_{i_1}\cdots H_{i_k}=v^k{T}_w+{\sum }_{\mathrm{\ell }(u)<k}{a}_u{T}_u={\tilde{T}}_w+{\sum }_{\mathrm{\ell }(u)<k}{a}_u{\tilde{T}}_u$ with $a_u\in A$, which is triangularity with unit diagonal against $\{{\tilde{T}}_u\}$. Consequently a family with one such product for each element $w$, chosen reduced word by chosen reduced word, is an $A$-basis of $H_n$: its transition matrix to $\{{\tilde{T}}_w\}$ is triangular with unit diagonal, hence invertible over $A$. The product does depend on the chosen reduced word in general: for adjacent colours $s,t$ with $m_{st}=3$ the relation $H_s{H}_t{H}_s-H_s=H_t{H}_s{H}_t-H_t$ of Elias–Williamson's example for the failure of well-definedness rewrites as $H_s{H}_t{H}_s-H_t{H}_s{H}_t=H_s-H_t\mathrm{\ne }0$, while both products have leading term ${\tilde{T}}_{sts}$; this is why the well-defined normalized basis is $\{{\tilde{T}}_w\}$ and not the family of reduced-word products itself. Adding clause 1, $\{T_w\}$ is an $A$-basis of $H_n$ and the displayed product rule of clause 2 holds. ∎