Bernstein PBW basis for the affine Hecke algebra of type A

The images of the Laurent generators commute and are units by [F1], so their finite Laurent monomials define the coefficient map from $A$. Every element of the presented algebra is represented by a finite linear combination of words in $T_i$ and $X_j^{\pm 1}$ by [F6]. To show spanning, take such a word and count its pairs in which a $T_i$ occurs to the left of an $X_j^{\pm 1}$. If the count is positive, an adjacent out-of-order pair can be rewritten by the Bernstein relation of [F3] as a Laurent term with that pair reversed plus a finite Laurent sum with one fewer $T$; the second assertion uses that $D_i$ maps $A$ to $A$. Induction first on the number of $T$ factors and then on this pair count moves every Laurent factor to the left. The finite $T_i$ generators in [F1] satisfy the presentation of $H_{n,Q}^{\mathrm{f}\mathrm{i}\mathrm{n}}$, so its universal property maps $H_{n,Q}^{\mathrm{f}\mathrm{i}\mathrm{n}}$ to $H_n^{\mathrm{a}\mathrm{f}\mathrm{f}}$; its basis in [F2] then expresses the remaining finite word as an $R_0$-linear combination of $T_w$. Thus the displayed family spans.

For a reduced expression $w=s_{i_1}\cdots s_{i_m}$, expand the image of $T_w$ under [F3] as a product of $m$ factors $c_{i_k}[1]+d_{i_k}[s_{i_k}]$. Each term is supported at a permutation expressed using at most $m$ adjacent transpositions, so its Coxeter length is at most $m$. The coefficient of $[w]$ can only come from choosing every reflection term: omitting one leaves fewer than $m=\mathrm{\ell }(w)$ factors. Thus that coefficient is $b_w={\prod }_{k=1}^m(s_{i_1}\cdots s_{i_{k-1}})(d_{i_k})$, with the empty prefix acting as the identity. Every $d_{i_k}$ is nonzero by [F3] and each prefix acts by a field automorphism, so $b_w\mathrm{\ne }0$ in $K$.

Suppose ${\sum }_w{f}_w{T}_w=0$ with $f_w\in A$ and finite support. If any coefficient is nonzero, choose the maximum length $L$ of a $w$ with $f_w\mathrm{\ne }0$. Apply the homomorphism of [F3] and compare the coefficient of $[w]$ for each such $w$ with $\mathrm{\ell }(w)=L$. Terms indexed by $v$ with $\mathrm{\ell }(v)<L$ cannot contribute to $[w]$ by step 1.2; terms with $\mathrm{\ell }(v)=L$ contribute to $[w]$ only when $v=w$, since a length-$L$ permutation in the expansion requires choosing all $L$ reflection terms and then equals $v$. Thus the coefficient of $[w]$ is $f_w{b}_w=0$ in $K$. Since $b_w\mathrm{\ne }0$ and $A\hookrightarrow K$ by [F4], $f_w=0$, a contradiction. Hence the family $(T_w)$ is $A$-linearly independent.

Every $h\in H_n^{\mathrm{a}\mathrm{f}\mathrm{f}}$ has a finite expansion ${\sum }_w{f}_w{T}_w$ by step 1.1; step 2.1 makes that expansion unique. Expanding each $f_w$ in the Laurent monomial basis of [F4] proves that $(X^{\lambda }{T}_w{)}_{\lambda ,w}$ is an $R_0$-basis. In particular, the coefficient map $A\to H_n^{\mathrm{a}\mathrm{f}\mathrm{f}}$ is injective, since a finite relation among its Laurent monomials would be an $R_0$-linear relation among the basis elements with $w=1$.

For a commutative $R_0$-algebra $R^{\mathrm{\prime }}$, [F5] identifies $R^{\mathrm{\prime }}{\otimes }_{R_0}{H}_n^{\mathrm{a}\mathrm{f}\mathrm{f}}$ with the quotient of the free associative $R^{\mathrm{\prime }}$-algebra by the images of the defining relations, and carries the $R_0$-basis of step 3.1 to the stated $R^{\mathrm{\prime }}$-basis. No generic-faithfulness argument after specialization is used; no flatness assumption on $R^{\mathrm{\prime }}$ is needed.

The preceding steps prove spanning, independence over the universal domain, the faithful Laurent coefficient map, and transport of the explicit basis to every commutative base change. The proof is choice-free: it selects only a maximum from a finite support and uses the explicit monomial and operator bases.