Unequal-Parameter Hecke Cells and Representations

Put $B=A_{\mathfrak{p}}$. It is a nonzero Noetherian local domain by [F1]. Its residue is $\mathrm{Frac}(A\mathrm{/}\mathfrak{p})$: modulo $\mathfrak{p}$, the inverted elements are exactly the nonzero elements of $A\mathrm{/}\mathfrak{p}$. This identifies that field with $k_0\subseteq k$. Localizing a domain is injective. The elementary prime correspondence used below follows directly from fractions: a prime disjoint from the denominators extends to a prime, and contraction recovers it because a fraction belongs to the extended ideal exactly when its numerator does.

If $\mathfrak{p}\cap \mathbb{Z}=0$, every nonzero integer is a unit in $B$, so fraction arithmetic identifies $B$ with a prime localization of $\mathbb{Q}[v,v^{-1}]$. The latter is a PID: ideals in its localization are generated by the image of a generator of their contracted ideal in $\mathbb{Q}[v]$. Thus $B$ is a field or has a nonzero principal maximal ideal $(a)$. In the second case [F2] and the chain $(0)⊊(a)$ give dimension one. The class of $a$ modulo $(a{)}^2$ is nonzero, since $a=a^{2}b$ in the domain would imply $1=ab$ with $a$ a nonunit. Hence its embedding dimension is one. In the field case both dimensions are zero. Thus $B$ is regular in both cases.

Otherwise $\mathfrak{p}\cap \mathbb{Z}=(\mathrm{\ell })$ for a prime integer $\mathrm{\ell }$. The prime $\mathfrak{p}\mathrm{/}(\mathrm{\ell })$ in the PID ${\mathbb{F}}_{\mathrm{\ell }}[v,v^{-1}]$ is zero or $(f)$ for an irreducible Laurent polynomial $f$. If it is zero, $\mathfrak{p}=(\mathrm{\ell })$ and the principal-maximal-ideal argument of step 2.1 proves regularity. Otherwise choose a Laurent lift $G$ of $f$; then the maximal ideal of $B$ is $\mathfrak{m}=(\mathrm{\ell },G)$. The chain $(0)⊊(\mathrm{\ell })B⊊\mathfrak{m}$ and [F2] give $\dim B=2$. The quotient $B\mathrm{/}(\mathrm{\ell })$ is the domain ${\mathbb{F}}_{\mathrm{\ell }}[v,v^{-1}{]}_{(f)}$. If $\mathrm{\ell }=r{\mathrm{\ell }}^2+s\mathrm{\ell }G+t{G}^2$ in $B$, reduction modulo $\mathrm{\ell }$ gives $\bar{t}{f}^2=0$, hence $t=\mathrm{\ell }{t}^{\mathrm{\prime }}$. Cancellation of $\mathrm{\ell }$ then gives $1=r\mathrm{\ell }+sG+t^{\mathrm{\prime }}{G}^2\in \mathfrak{m}$, impossible. Thus $\mathrm{\ell }\mathrm{\notin }{\mathfrak{m}}^2$. If $a\mathrm{\ell }+bG\in {\mathfrak{m}}^2$, reducing modulo $\mathrm{\ell }$ gives $\bar{b}f\in (f^2)$, so $b\in \mathfrak{m}$. It follows that $a\mathrm{\ell }\in {\mathfrak{m}}^2$, and $\mathrm{\ell }\mathrm{\notin }{\mathfrak{m}}^2$ forces $a\in \mathfrak{m}$, since otherwise $a$ is a unit. The two displayed generators therefore give a basis of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$; its dimension is two. Thus $B$ is regular in this final case.

By [F3], $R=\widehat{B}$ is complete separated Noetherian local with residue $k_0$, is regular, and is a domain. Faithful flatness makes $B\to R$ injective: if $b$ maps to zero, flat tensoring the inclusion $bB\hookrightarrow B$ gives an injection $bB{\otimes }_{B}R\hookrightarrow R$ whose image is zero; faithfulness forces $bB=0$, so $b=0$. Therefore $A\hookrightarrow R$ induces $\mathbb{Q}(v)\hookrightarrow F$, with the required residue map. If $\mathfrak{p}=0$, this also covers the field $B=R=\mathbb{Q}(v)$ with zero maximal ideal.

For a finite projective module $P$ over any local ring, lift a basis of $P\mathrm{/}\mathfrak{m}P$ to $r$ elements of $P$. Nakayama on the finite cokernel makes $R^r\to P$ surjective. Projectivity splits it as $R^r=P\oplus Q$, with $Q$ finite. Reduction of this direct sum shows $Q\mathrm{/}\mathfrak{m}Q=0$, so Nakayama gives $Q=0$. Hence $P$ is free of rank $r$. Over our domain this rank equals both ${\dim }_F(F{\otimes }_{R}P)$ and ${\dim }_{k_0}(k_0{\otimes }_{R}P)$.

Let $x_0\in C$ lift an idempotent modulo ${\mathfrak{m}}_{R}C$. Define ${\delta }_i=x_i^2-x_i$ and $x_{i+1}=x_i+(1-2x_i){\delta }_i$. All factors here are polynomials in $x_i$, so integer-polynomial expansion gives ${\delta }_{i+1}={\delta }_i^2(4{\delta }_i-3)$. Thus ${\delta }_i\in {\mathfrak{m}}_R^{2^i}C$ and $x_{i+1}-x_i\in {\mathfrak{m}}_R^{2^i}C$. The finite-module completeness in [F4] gives a limit $q\equiv x_0$ modulo ${\mathfrak{m}}_{R}C$. Multiplication is continuous, since these ideal powers are two-sided ideals, so separatedness gives $q^2=q$. This construction involves no division and works in every characteristic.

For any right $C$-module $X$, the $R$-linear decomposition $X=Xq\oplus X(1-q)$ shows that projection $X\mapsto Xq$ commutes with every scalar extension. It is exact: under a surjection $X\to Y$, lift $y\in Y$ to $x$, and then $xq$ lifts $yq$; kernels and injections follow by restriction of the same projection. If $0\to I_{i+1}\to I_i\to B_i\to 0$ is a filtration sequence, projection gives $0\to I_{i+1}q\to I_{i}q\to B_{i}q\to 0$. When $B_i$ is finite free, $B_{i}q$ is finite projective and hence free by step 5.1; the sequence splits as $R$-modules. It therefore retains its exact layer after scalar extension, including reduction to $k_0$.

Finally let $J_R={⨁}_E{M}_{d_E}(R)$ act on an $R$-finite-free module $X$. With matrix units $e_{ij}^E$, put $Q_E=e_{11}^{E}X$. It is a direct summand of $X$, hence finite free by step 5.1. The map $R^{d_E}{\otimes }_R{Q}_E\to e_{E}X$ sending $u_i\otimes x$ to $e_{i1}^{E}x$ has inverse $y\mapsto {\sum }_i{u}_i\otimes e_{1i}^{E}y$, by the matrix-unit identities. Therefore the generic and residue multiplicities of the $E$-standard simple are both ${\mathrm{rank}}_R{Q}_E$. This proves the final assertion and completes the lemma.

Finite-generator parabolic embedding. The image of ${\varphi }_J$ is $⟨J⟩$, since its image is generated by the named generators. To prove injectivity, let $g\in W_J^{\ast }$ map to $1$, and represent $g$ by a word $u$ in $J$. In the free group $F(S)$, the word $u$ lies in the normal closure $N$ of the Coxeter relators by [F1]. By [F7], $N$ is the subgroup generated by all conjugates of those relators and their inverses: that subgroup is normal and contains the relators, and every normal subgroup containing the relators contains it. Therefore $u$ is a finite product of conjugates of relators and their inverses. Let $K$ contain $J$ and every letter appearing in these finitely many conjugating words and relators. The inclusion $F(K)\to F(S)$ has a retraction sending generators outside $K$ to $1$, so it is injective by the free-group universal property [F6]. Thus the same finite product identity holds in $F(K)$, and $u$ is the identity in the Coxeter group presented on $K$. Applying [F2] to the finite system on $K$ and its subset $J$ shows that the map from $W_J^{\ast }$ into this group is injective. Hence $g=1$.

Intrinsic and ambient lengths. Every $J$-word is an $S$-word, so ambient length is at most intrinsic $J$-length. Conversely, take $w\in W_J^{\ast }$ and a reduced ambient expression for ${\varphi }_J(w)$, of length $r$. Let $K$ contain $J$ and the letters in that expression. By step 1.1 applied to $K$, the presentation on $K$ embeds in $W$; thus the equality between a $J$-word for $w$ and the chosen ambient word already holds in the group on $K$. By [F2], intrinsic $J$-length equals $K$-length on $W_J^{\ast }$. The chosen expression gives $K$-length at most $r$, while every $K$-word is also an $S$-word and so has length at least $r$. Hence both lengths equal $r$, proving intrinsic and ambient length equality.

Simple-generator length changes. Fix $s\in S$ and $w\in W$. Choose a finite $K$ containing $s$ and the letters of a word for $w$. Steps 1.1 and 2.1 embed the group on $K$ and preserve lengths. Applying [F3] in that finite-generator system gives both asserted $\pm 1$ formulas in $W$.

Bruhat comparison implies a subword. Suppose $y\le w$, and fix a reduced expression of $w$. Choose a finite Bruhat chain from $y$ to $w$. Each element in this chain has finite word support, and each reflection used in an edge is a conjugate of a simple generator by a finite word. The union of these supports and the letters of the fixed expression of $w$ is contained in a finite $K\subseteq S$. By steps 1.1 and 2.1, the chain lies in the embedded finite-generator group on $K$ with all lengths preserved; its edges are still right multiplications by reflections. The finite-generator subword theorem [F5] now gives a reduced expression of $y$ as a subword of the fixed expression of $w$.

A reduced subword implies Bruhat comparison. Suppose a reduced expression of $y$ occurs as a subword of a fixed reduced expression of $w$. Let $J$ be the finite support of that expression of $w$. The subword theorem [F5] gives $y\le w$ in the finite-generator Coxeter group on $J$. By steps 1.1 and 2.1 its presentation embeds in $W$ and preserves lengths; each reflection in the finite Bruhat chain maps to a reflection in $W$. The chain therefore gives $y\le w$ in the global reflection-chain order.

Every element below $w$ is represented by a subword of a fixed reduced expression of $w$, by step 3.2. There are at most $2^{\mathrm{\ell }(w)}$ such subwords, so the lower interval is finite with the stated bound. A strict comparison has a nonempty length-increasing chain by [F1], hence $\mathrm{\ell }(y)<\mathrm{\ell }(w)$ when $y<w$. Steps 3.1–3.3 prove all remaining clauses. All supports, products in the normal closure, and chains used above are finite.

For $w\in W$, take a reduced expression $w=s_1\cdots s_k$ and set $L(w)={\sum }_{i=1}^k\lambda (s_i)$. This does not depend on the reduced expression: two reduced expressions have finite combined support $J\subseteq S$, and by [F2] are reduced expressions in the finite Coxeter system on $J$. By [F3] they are connected by braid moves. An even braid length uses each of its two letters equally often; an odd braid length uses one letter once more on one side and the other letter once more on the other, and those labels are equal by hypothesis. Thus each move preserves the sum, so $L(w)$ is well defined.

If $\mathrm{\ell }(xy)=\mathrm{\ell }(x)+\mathrm{\ell }(y)$, concatenating reduced expressions for $x$ and $y$ gives a reduced expression for $xy$. The sum defining $L$ therefore gives $L(xy)=L(x)+L(y)$. Any weight function with the prescribed simple labels must assign this same sum to every reduced expression (each prefix is reduced), proving both the weight-function property and uniqueness of the extension.

Let $s\mathrm{\ne }t$ have finite odd label $m(s,t)=2k+1$, and put $J=\{s,t\}$. By [F2], the restricted two-generator Coxeter presentation embeds as $W_J$ with intrinsic length equal to the ambient length. The two alternating words $(st{)}^{k}s$ and $t(st{)}^k$ have length $m(s,t)$ and are reduced in the finite-rank Coxeter system on $J$ by [F5], hence are reduced in $W$; they represent the same element by the braid relation. For any weight function $L$, additivity along the prefixes of each reduced word gives $(k+1)L(s)+kL(t)=kL(s)+(k+1)L(t)$. Cancelling the common terms yields $L(s)=L(t)$. This proves the necessary odd-edge condition.

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.

Fix $w\in W$. By [F3], its strict lower interval $\{y:y<w\}$ is finite and consists only of shorter elements, so construct $C_y$ for those $y$ by induction on length. Their triangular expansions show that the $C_y$ form an $A$-basis of $V_{<w}:={\mathrm{span}}_A\{T_y:y<w\}$: the transition matrix is finite, unitriangular, and ordered by length.

Set $D_w=\overline{T_w}-T_w$. By [F2], there are unique coefficients $a_y\in A$ with $D_w={\sum }_{y<w}{a}_y{C}_y$. Since bar is an involution and each $C_y$ is bar invariant by induction, $\overline{D_w}=-D_w$ implies ${\sum }_{y<w}\overline{a_y}{C}_y=-{\sum }_{y<w}{a}_y{C}_y$. The basis property in step 1.1 gives $\overline{a_y}=-a_y$ for each $y<w$.

Write $a_y={\sum }_{\gamma }{n}_{\gamma }{v}^{\gamma }$ with finite support. The equality $\overline{a_y}=-a_y$ forces $n_0=0$ and $n_{-\gamma }=-n_{\gamma }$. Let $p_y={\sum }_{\gamma <0}{n}_{\gamma }{v}^{\gamma }\in A_{<0}$; then $p_y-\overline{p_y}=a_y$. Define $C_w=T_w+{\sum }_{y<w}{p}_y{C}_y$. Applying bar and using $D_w=\sum a_y{C}_y$ gives $\overline{C_w}=T_w+{\sum }_{y<w}(a_y+\overline{p_y})C_y=C_w$. Expanding the $C_y$ in the standard basis shows that all coefficients below $T_w$ remain in $A_{<0}$, because a sum of two negative exponents is negative. This constructs a normalized bar-invariant $C_w$.

Suppose $C_w^{\mathrm{\prime }}$ is another element with the stated properties. Then $h=C_w^{\mathrm{\prime }}-C_w={\sum }_{y<w}{b}_y{T}_y$ is bar invariant with each $b_y\in A_{<0}$. Choose a nonzero $b_y$ with $\mathrm{\ell }(y)$ maximal. In $\bar{h}$, the coefficient of $T_y$ is $\overline{b_y}$: every other term of maximal length has a distinct leading basis vector, while all lower bar terms have strictly smaller length by [F2, F3]. Bar invariance gives $b_y=\overline{b_y}$, but $A_{<0}\cap \overline{A_{<0}}=\{0\}$, a contradiction. Hence $C_w^{\mathrm{\prime }}=C_w$.

The family $C_w$ is linearly independent: in any finite relation, choose a term of maximal length; its coefficient at the corresponding $T_w$ is unchanged by the unitriangular expansions and must vanish. It spans by induction on $\mathrm{\ell }(w)$, because $T_w=C_w-{\sum }_{y<w}{p}_{y,w}{T}_y$ and all terms on the right other than $C_w$ have smaller length. Thus it is an $A$-basis.