The arbitrary-rank weighted Hecke basis and Bruhat-triangular bar

Finite-support spanning. Fix a finite $J\subseteq S$. Its odd-edge components have a universal Laurent coefficient ring with one unit parameter for each component. By [F2], sending that parameter to $v^{L(s)}\in A^{\times }$ is independent of $s$ in the component, so it defines a coefficient-ring homomorphism. By [F4] and [F5], the finite Hecke algebra on $J$, after this base change, has basis $\{T_u:u\in W_J\}$. Every word representing an element of the global presented algebra uses some finite $J$, so applying this finite spanning statement to that word shows that the global elements $T_w$ span $H_A(W,L)$.

The bar automorphism. The quadratic relation in [F3] gives $T_s^{-1}=T_s-(q_s-q_s^{-1})$, where $q_s=v^{L(s)}$. Define a semilinear multiplicative map on the free algebra by $v^{\gamma }\mapsto v^{-\gamma }$ and $T_s\mapsto T_s^{-1}$. The image of a quadratic relator is $(T_s^{-1}-q_s^{-1})(T_s^{-1}+q_s)$, which is zero because multiplying $T_s^2-(q_s-q_s^{-1})T_s-1=0$ by $T_s^{-2}$ gives that same identity. The image of a braid relation is the inverse of the reversed alternating relation; reversal of either alternating word is one of the two defining alternating words, so the relation is preserved. The map descends to $H_A(W,L)$. Its square fixes every coefficient and every $T_s$, hence is the identity.

Independence by finite support. Suppose a finite sum ${\sum }_{w\in F}{a}_w{T}_w$ is zero in $H_A(W,L)$. In the free associative presentation, a representative belongs to the two-sided ideal generated by the quadratic and braid relations. Such membership is a finite sum of products of generators and relations, so only finitely many simple generators occur in that expression and in the reduced words for $w\in F$. Put them in a finite set $J$. The same relation then holds in the finite Hecke presentation on $J$, where [F5] gives independence of its standard basis. Thus every $a_w=0$. Together with step 1.1 this proves that $\{T_w\}$ is an $A$-basis and that the reduced product $T_w$ is independent of the chosen expression.

Multiplication by a simple generator. By [F4], exactly one of $\mathrm{\ell }(sx)=\mathrm{\ell }(x)+1$ and $\mathrm{\ell }(sx)=\mathrm{\ell }(x)-1$ holds. In the first case, a reduced word for $x$ prefixed by $s$ is reduced, so $T_s{T}_x=T_{sx}$. In the second, $x=s(sx)$ is reduced and the quadratic relation gives $T_s{T}_x=T_{sx}+(q_s-q_s^{-1})T_x$. Thus, when a simple letter is prepended to a standard-basis term, each output label is either the value with that letter prepended or the value with that letter omitted.

Bruhat support. Write a reduced expression $w=s_1\cdots s_k$, and let $J\subseteq S$ be its finite set of letters. Then $\overline{T_w}=T_{s_1}^{-1}\cdots T_{s_k}^{-1}$. Expanding each inverse as $T_{s_i}-(q_{s_i}-q_{s_i}^{-1})$ gives scalar multiples of products along subwords. Build each selected product from its rightmost letter toward its left; at each multiplication step 3.1 shows that its standard-basis labels are still values of subsequences of the original reduced expression, though a selected subsequence need not itself be reduced. Each such label and its subsequence lie in the finite parabolic on $J$. By [F4], that parabolic is embedded with lengths preserved; applying the full arbitrary-subword clause of [F6] there gives $y\le w$ in the finite parabolic and hence in the global reflection-chain order. If the subsequence is proper, then $\mathrm{\ell }(y)$ is at most its number of letters, hence at most $k-1$, so $y\mathrm{\ne }w$ and $y<w$. The full reduced subsequence contributes only $T_w$ with coefficient one, since each of its suffixes is reduced and its standard-basis product is $T_w$. Hence $\overline{T_w}-T_w$ is supported strictly below $w$.

Steps 1.1 and 2.1 prove the standard-basis clause, step 1.2 constructs the involution, and step 4.1 proves its strict Bruhat support. All reductions take place in finite supports, so no finite-rank hypothesis or choice axiom is used.