Affine Hecke Algebras Beyond Type A

For every integer $n\ge 2$, $1+q+\cdots +q^{n-1}$ is a sum of positive real numbers and is therefore positive. No finite $n$ satisfies the defining vanishing condition in [F1], so $e(\mathbb{C},q)=\mathrm{\infty }$.

Since $\xi \mathrm{\ne }1$, the identity $(1-\xi )(1+\xi +\cdots +{\xi }^{n-1})=1-{\xi }^n$ shows that the geometric sum vanishes exactly when ${\xi }^n=1$. Its least such positive exponent is the multiplicative order, which is at least $2$; hence [F1] gives $e(\mathbb{C},\xi )=\mathrm{ord}(\xi )<\mathrm{\infty }$. Also $\mathrm{\mid }\xi \mathrm{\mid }=1$ while $\mathrm{\mid }q\mathrm{\mid }=q>1$, so $q\mathrm{\ne }\xi$.

First derive the residue and valuation consequences needed below from [F2]. If $0\mathrm{\ne }x\in \mathcal{O}$, its leading exponent $r$ is nonnegative because $\mathrm{\mid }x{\mathrm{\mid }}_3\le 1$; the expansion gives $x=a_0+\pi y$ with $a_0\in \{0,1,2\}$ and $y\in \mathcal{O}$, since the shifted tail is a limit of partial sums in the closed unit ball. Distinct such digits represent distinct residue classes, since a nonzero difference in $\{\pm 1,\pm 2\}$ has absolute value $1$, whereas every element of $\pi \mathcal{O}$ has absolute value at most $1\mathrm{/}3$. Thus $\mathcal{O}\mathrm{/}\pi \mathcal{O}$ has exactly the three classes $0,1,2$ with the operations of ${\mathbb{F}}_3$. For a nonzero $x\in \mathcal{O}$ with leading exponent $r$, the shifted expansion gives $x={\pi }^{r}u$ with $u=a_r+\pi z$, $a_r\in \{1,2\}$ and $z\in \mathcal{O}$. The leading digit gives $\mathrm{\mid }u{\mathrm{\mid }}_3=\mathrm{\mid }{u}^{-1}{\mathrm{\mid }}_3=1$, so [F1] gives $u,u^{-1}\in \mathcal{O}$ and $u$ is a unit. Every nonzero ideal of $\mathcal{O}$ therefore contains ${\pi }^r$ for the minimum valuation $r$ of its elements, and every element of that ideal is divisible by ${\pi }^r$ in $\mathcal{O}$. Thus every ideal is principal and $\mathcal{O}$ is a PID. After scaling any lattice $M$ into ${\mathcal{O}}^2$, [F3] gives an $\mathcal{O}$-basis $f_1,f_2$ of ${\mathcal{O}}^2$ with $M=\mathcal{O}{\pi }^a{f}_1+\mathcal{O}{\pi }^b{f}_2$, where $0\le a\le b$ after absorbing unit factors into the basis. Scaling to $a=0\le b$ gives a primitive representative and an exponent $n=b-a$. This exponent is well defined on the homothety class: a primitive representative has minimum coordinate valuation zero, so any scalar multiple that is also primitive differs by an $\mathcal{O}$-unit. For a basis matrix of the primitive representative, the determinant valuation is $n$; changing either $\mathcal{O}$-basis multiplies the determinant by a unit. Neighbors of $[L]$ correspond to the four lines in $L\mathrm{/}\pi L\cong {\mathbb{F}}_3^2$, because the inverse image of a line is an intermediate lattice. Distinct lines give distinct lattices, and they cannot become homothetic: all four inverse images have the same index $3$ in $L$, so a scalar relating two of them has valuation zero and is a unit, which preserves each lattice.

Relative to the base vertex $[{\mathcal{O}}^2]$, a vertex with normalized elementary divisors $(0,n)$ has a chain of length $n$ through $\mathcal{O}{f}_1+{\pi }^j\mathcal{O}{f}_2$. If $n>0$, exactly one of its four neighbors has distance parameter $n-1$: in the basis $f_1,{\pi }^n{f}_2$ of its representative, the line spanned by ${\pi }^n{f}_2$ has inverse image $\pi \mathcal{O}{f}_1+{\pi }^n\mathcal{O}{f}_2$, whose homothety class normalizes to parameter $n-1$. Every other line has inverse image with a primitive vector $f_1+t{\pi }^n{f}_2$ and second elementary divisor $n+1$, hence parameter $n+1$. At $n=0$ all neighbors have parameter $1$. Therefore the parameter changes by one at each step, and the displayed chain is a shortest path. A cycle would have a vertex of maximal parameter and two distinct adjacent vertices of parameter one less, contradicting uniqueness of the decreasing neighbor. Thus the graph is connected and has no cycle.

Replacing a lattice basis multiplies its determinant by an $\mathcal{O}$-unit; homothety adds twice an integer to its determinant valuation. Hence type is well defined, and determinant-one matrices preserve it. For an edge with the endpoint of type $0$ first, choose a basis $f_1,f_2$ of its first lattice with second lattice $\mathcal{O}{f}_1+\pi \mathcal{O}{f}_2$. Its determinant has even valuation. Rescale both vectors by a power of $\pi$ to make that determinant a unit and then multiply $f_1$ by its inverse; neither operation changes the edge, and the resulting change-of-basis matrix has determinant one. This proves transitivity. A determinant-one matrix fixes $[L_0]$ exactly when it preserves $L_0$ (homothety would change determinant valuation unless its scalar is a unit). Within ${\mathrm{SL}}_2(\mathcal{O})$, preservation of $L_1$ is exactly $c\in \pi \mathcal{O}$, proving the stated stabilizer.

The unique shortest edge gallery from $E$ to any other edge changes endpoint types alternately: changing twice at the same endpoint would shorten it. For a gallery proceeding along $L_0,L_1,\dots ,L_n$, the matrices $\left(\begin{array}{cc}1& 0\\ {\pi }^{n}b& 1\end{array}\right)$ for $b\in \mathcal{O}$ fix every earlier vertex and act transitively on the three possible next edges. In the basis $e_1,{\pi }^n{e}_2$, the backward edge corresponds to the line spanned by the second basis vector; the other lines are spanned by $e_1+t{\pi }^n{e}_2$ for $t\in {\mathbb{F}}_3$, and the displayed matrices send $t$ to $t+b$. For the other initial direction the apartment chain is $L_1,L_0,L_{-1},\dots ,L_{-n}$; use instead $\left(\begin{array}{cc}1& {\pi }^{n}b\\ 0& 1\end{array}\right)$ at $L_{-n}$, which fixes that earlier chain and is transitive on the three forward lines there. Inductively move each gallery onto the corresponding apartment segment while fixing its earlier segment. Hence $I$ is transitive on the edges with each alternating type word. There are three choices at each extension, so the orbit size is $3^n$. Different words have different type sequences or distances and cannot be in the same orbit. By step 2.2 edges identify with $G\mathrm{/}I$, so these orbits are exactly the double cosets.

The affine simple set is finite by [F1]. Specialize [F2] to $v_s=\sqrt{q_s}$; it gives the basis $\{N_w:w\in W_{\mathrm{a}\mathrm{f}\mathrm{f}}\}$ of the normalized presentation, where $N_w$ is a reduced product of the $N_s$. For a reduced expression $w=s_1\cdots s_k$, the unnormalized product is $T_w=({\prod }_{j=1}^k{v}_{s_j})N_w$. By [F4], changing reduced expressions uses braid moves. An even braid preserves the product of the $v_s$ factors because both sides have equal counts of each generator; an odd braid preserves it because its two generators are conjugate in $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$ and [F1] makes their parameters equal. Thus the scalar is well defined and nonzero, so the unnormalized products $T_w$ are nonzero scalar multiples of the basis elements $N_w$. Hence $\{T_w:w\in W_{\mathrm{a}\mathrm{f}\mathrm{f}}\}$ is a $\mathbb{C}$-basis of ${\mathcal{H}}_{\mathrm{a}\mathrm{f}\mathrm{f}}$.

By [F3], every element of the crossed product has a unique finite sum ${\sum }_{\omega \in \mathrm{\Omega }}{h}_{\omega }{U}_{\omega }$ with $h_{\omega }\in {\mathcal{H}}_{\mathrm{a}\mathrm{f}\mathrm{f}}$. Expanding each coefficient uniquely in the basis from step 1.1 gives a unique finite $\mathbb{C}$-linear combination of $T_w{U}_{\omega }$. Hence the displayed family is a basis. No choice principle is used.

Average any inner product on $V=Y\otimes \mathbb{R}$ over the finite group $W_0$. Each reflection is then orthogonal, has normal line $\mathbb{R}{\alpha }^{\vee }$, and fixes $\ker \alpha$, so $⟨\alpha ,v⟩=2(v,{\alpha }^{\vee })\mathrm{/}({\alpha }^{\vee },{\alpha }^{\vee })$. Put $a_{\alpha }=2{\alpha }^{\vee }\mathrm{/}({\alpha }^{\vee },{\alpha }^{\vee })$. Its metric coroot is exactly ${\alpha }^{\vee }$, and $(v,a_{\alpha })=⟨\alpha ,v⟩$. Thus [F2] identifies the $a_{\alpha }$ as a reduced crystallographic root system on $E=\mathbb{R}{\mathrm{\Phi }}^{\vee }$, with the chosen positive system. The orthogonal complement $C=E^{\perp }$ is exactly the common root kernel and is fixed by $W_0$. Its presence adds a Cartesian factor $C$ to every alcove and changes neither the reflecting walls nor their combinatorics on $E$. Apply [F3]–[F4] on $E$ to obtain the asserted Coxeter system, simple transitivity and length for $Q^{\vee }⋊W_0$ on $V=E\oplus C$. The base alcove is equivalently $0<⟨\alpha ,v⟩<1$ for all positive $\alpha$: no integer wall cuts this region; every alcove in the positive chamber touching $0$ must satisfy these inequalities, and the supplied fundamental-alcove theorem identifies it with this region.

Every translation by $y\in Y$ sends $H_{\alpha ,j}$ to $H_{\alpha ,j+⟨\alpha ,y⟩}$ and every Weyl element permutes root directions, so $W_{\mathrm{e}}$ preserves the arrangement. Its conjugation preserves the group generated by all wall reflections, namely $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$, hence that subgroup is normal. For $h\in W_{\mathrm{e}}$, simple transitivity gives a unique $w\in W_{\mathrm{a}\mathrm{f}\mathrm{f}}$ with $wA=hA$; then $w^{-1}h\in \mathrm{\Omega }$. The intersection is trivial by simple transitivity, proving the semidirect decomposition. Separating-wall length agrees with Coxeter length on the affine subgroup and vanishes exactly on $\mathrm{\Omega }$. If the roots are empty, $E=0$, $A=V$, $W_0=W_{\mathrm{a}\mathrm{f}\mathrm{f}}=1$, and $\mathrm{\Omega }=Y$, so every assertion still holds.

Every wall is a facet wall of some alcove by [F4]. Write that alcove as $wA$ with $w\in W_{\mathrm{a}\mathrm{f}\mathrm{f}}$; its reflection is $ws{w}^{-1}$ for a unique wall type $s$ of $A$. If two such descriptions give $ws{w}^{-1}=ut{u}^{-1}$, then $s$ and $t$ are conjugate in $W_{\mathrm{e}}$, so [F5] gives the same parameter. This defines $v(H)$ independently of the alcove used and makes it invariant under $W_{\mathrm{e}}$. Translation by ${\alpha }^{\vee }$ changes the integer level by $2$, so only parity can matter for fixed $\alpha$. If $\alpha \mathrm{\notin }2X$, perfect duality supplies $y\in Y$ with $⟨\alpha ,y⟩$ odd: otherwise evaluating on a lattice basis would put $\alpha \mathrm{/}2$ in $X$. Translation by that $y$ identifies odd and even walls, forcing equal parameters. If $\alpha \in 2X$, all pairings are even, and the formulas asserted are exactly the two potentially distinct parity values. The argument uses finitely many basis vectors and a finite Weyl average, without a choice axiom.

A backtrack crosses the same wall twice in opposite directions, so its signed product is $N_s{N}_s^{-1}=1$ or $N_s^{-1}{N}_s=1$. To check a braid move of length $m<\mathrm{\infty }$, examine the associated $2m$-sector dihedral residue. Project a point sufficiently far in $D$ to a plane transverse to its common codimension-two face. The $m$ wall lines through the origin separate that plane into the $2m$ sectors of [F1]. On traversing their cycle once, the signs are $m$ consecutive plus signs followed by $m$ consecutive minus signs, up to cyclic rotation: the roots defining the crossed sides run consecutively through a half-circle, and the sign of their pairing with the projected point changes exactly at the two bounding rays of a half-plane. Starting at the beginning of the plus block, the positive product is one alternating word of length $m$ and the negative product is the inverse of the other alternating word. The Hecke braid relation makes their product $1$. Every cyclic rotation also has product $1$, since every factor is invertible. Cutting the cycle into the two gallery arcs of the braid move proves equality of their signed products. This also proves the assertion for a residue translated away from the base alcove.

For two expressions $g=s_1\cdots s_r\omega =t_1\cdots t_k\eta$ in the stated form, the semidirect decomposition in [F1] gives $\omega =\eta$ and $s_1\cdots s_r=t_1\cdots t_k$ in $W_{\mathrm{a}\mathrm{f}\mathrm{f}}$. By [F2], reduce the affine words by braid moves and backtrack deletions and compare the resulting reduced expressions. Step 1.1 preserves their signed products for the fixed $D$; multiplying by the common terminal $U_{\omega }$ proves the asserted independence. The same backtrack and dihedral-cycle argument works with any initial alcove. Common root-kernel directions do not meet a wall and are recorded only in the terminal frame, so introduce no additional gallery relation.

Choose a reduced expression for $g$. Replacing each inverse factor by $N_s-c_s$ gives the term $N_g$ from retaining every generator. Every other term uses fewer than $\mathrm{\ell }(g)$ simple factors, so repeated simple multiplication from [F3] expresses it as a finite sum of $N_h$ with $\mathrm{\ell }(h)<\mathrm{\ell }(g)$. Thus $E_D(g)=N_g+{\sum }_{\mathrm{\ell }(h)<\mathrm{\ell }(g)}{a}_{h,g}{N}_h$, with finitely many nonzero coefficients. Induction on $\mathrm{\ell }(g)$ shows that the $E_D(g)$ span: length-zero elements are exactly $U_{\omega }$, and the formula expresses $N_g$ through $E_D(g)$ and shorter elements.

In a finite relation among the $E_D(g)$, choose the largest length with a nonzero coefficient. At that length the expansion in step 3.1 has distinct leading terms $N_g$, and all its other terms have smaller length. Their coefficients cannot cancel because the $N_g$ are a basis. Thus no nonzero relation exists. This proves the asserted basis, even when $\mathrm{\Omega }$ is infinite, because every expansion and relation considered has finite support. All choices and inductions here concern finite words and finite-dimensional residue planes.

The determinant equation makes $G$ a closed subspace of ${\mathbb{Q}}_3^4$, hence it is LCH by [F4]; matrix operations and inversion are continuous. The determinant and congruence conditions make $I$ closed in the compact space ${\mathbb{Z}}_3^4$, hence compact. It is open in $G$ because its entry conditions $a,b,d\in {\mathbb{Z}}_3$ and $c\in \pi {\mathbb{Z}}_3$ are open in the matrix topology. By [F5] choose left Haar measure. Its value on $I$ is finite and positive: if it were zero, every compact support of a continuous test function would have a finite cover by translates of the open subgroup $I$, making every Haar integral zero. Rescale to $\mu (I)=1$. This is the only use of arbitrary choice beyond the countable compactness input in [F4]. For $m\ge 1$, let $K_m=\{g\in {\mathrm{SL}}_2({\mathbb{Z}}_3):g\equiv I_2(\mathrm{m}\mathrm{o}\mathrm{d}{\pi }^m)\}$. These are kernels of reduction homomorphisms, hence subgroups. Since ${\mathbb{Z}}_3$ is compact open, ${\mathrm{SL}}_2({\mathbb{Z}}_3)$ is compact open in $G$; the entrywise congruence conditions make each $K_m$ compact open, and the matrix topology shows that $(K_m)$ is a neighborhood basis at the identity. For a locally constant compactly supported integrand, each point of its compact support has a right $K_m$-coset on which it is constant; a finite subcover and the nested subgroup basis give a finite coset partition on which it is constant. Thus convolution integrals and their associativity calculation reduce to finite sums; the change of variables $y=zu$ uses left invariance of $\mu$. Compact support and local constancy are preserved.

Averaging on the left and right by $e$ makes a function invariant under both $I$ actions. Indeed $e\ast f(g)={\int }_{I}f(i^{-1}g)d\mu (i)$ and $f\ast e(g)={\int }_{I}f(gi)d\mu (i)$. The restriction of $\mu$ to $I$ is a left Haar probability on the compact Hausdorff group $I$, so [F7] makes it right invariant. Thus $e\ast e=e$, and every bi-invariant $f$ has $e\ast f=f=f\ast e$. This proves the corner equality. The double cosets are open, and a compact support meets only finitely many of them, so their characteristic functions form a vector-space basis.

The displayed $s_1$ fixes $[L_0]$ and sends the base edge through that type-$0$ vertex; $s_0=D{s}_1{D}^{-1}$ with $D=\mathrm{diag}(1,\pi )$ fixes $[L_1]$ and changes the base edge through type $1$. Their squares are $-I\in I$. The three right cosets in each simple double coset are explicitly $I{s}_{1}I={\coprod }_{b\in {\mathbb{F}}_3}\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)s_{1}I,I{s}_{0}I={\coprod }_{b\in {\mathbb{F}}_3}\left(\begin{array}{cc}1& 0\\ \pi b& 1\end{array}\right)s_{0}I\mathrm{.}$ For the first, reduce the action on the four lines at $[L_0]$: $I$ fixes the line $e_1$ and upper unipotents translate the other three. For the second, use the basis $e_1,\pi e_2$ at $[L_1]$, where lower unipotents translate the three lines other than the backward one. These matrices have entries in $I$, their three images are distinct, and [F1] shows they exhaust the orbit.

Under $G\mathrm{/}I$ identified with type-ordered edges, the operator given by $1_{I{s}_{i}I}$ sums over the three other edges at the endpoint of the indicated type. Two such moves at the same vertex return to the original edge in three ways and reach any specified other edge in two ways. Equivalently the finite right-coset formula for convolution gives $1_{I{s}_{i}I}\ast 1_{I{s}_{i}I}=3e+21_{I{s}_{i}I}$. A product with alternating endpoint types follows the unique shortest gallery in the tree. There is exactly one intermediate gallery to any specified final edge of that type word, so the product of the simple characteristic functions is the characteristic function of its entire double coset, with coefficient one. By [F1] these exhaust all double cosets. The infinite dihedral group has no finite braid relation between the two generators, so the quadratic identities give an algebra map from [F2] onto the corner. The basis of [F3] maps bijectively to the distinct double-coset characteristic functions, proving injectivity as well as surjectivity. Rescaling its quadratic identities yields the displayed symmetric normalization.

If $V$ is smooth, each vector $v$ is fixed by an open subgroup, which contains some $K_m$ by step 1.1. For compactly supported locally constant $f$, shrink $m$ further so that $f$ is constant on every right $K_m$-coset meeting its support; compactness gives a finite such partition. Since $\pi (gk)v=\pi (g)v$ for $k\in K_m$, the integral ${\int }_{G}f(g)\pi (g)vd\mu (g)$ is a finite sum. Refinement preserves the sum by additivity of Haar measure, so it is well defined. Finite-sum convolution and the representation law prove multiplicativity. Averaging by $e$ is the projection onto $V^I$ because left translation by $I$ preserves its integral and averaging fixes every $I$-fixed vector. Hence the corner acts on $V^I$ with identity $e$, including $V^I=0$. Finally [F6] applies to $3>1$, giving the precise parameter boundary. No equivalence assertion is used in this construction.

By [F1], $s_{\alpha }(y)=y-n{\alpha }^{\vee }$, so by [F2] the numerator is ${\theta }_y(1-u^n)$. If $n>0$, then $(1-u^n)\mathrm{/}(1-u)=1+u+\cdots +u^{n-1}$; if $n=0$, the quotient is zero; and if $n=-k<0$, then $(1-u^n)\mathrm{/}(1-u)=-u^{-k}(1+u+\cdots +u^{k-1})$. Each expression is a finite Laurent polynomial, proving the first assertion.

If $\alpha \in 2X$, write $\alpha =2\beta$ with $\beta \in X$. Then $n=⟨2\beta ,y⟩=2⟨\beta ,y⟩=2m$ for an integer $m$. With $u={\theta }_{-{\alpha }^{\vee }}$, the numerator is ${\theta }_y(1-u^{2m})$. For $m>0$, $(1-u^{2m})\mathrm{/}(1-u^2)=1+u^2+\cdots +u^{2m-2}$; for $m=0$ it is zero; and for $m<0$ the same finite negative-power identity from step 1.1, applied to $u^2$, is a Laurent polynomial. Since $1-u^2=1-{\theta }_{-2{\alpha }^{\vee }}$, the second assertion follows.