Double-Affine Hecke Algebras and Macdonald Theory

For $m>0$, multiplying ${\sum }_{j=0}^{m-1}{z}^{m-2j}$ by $1-z^{-2}$ cancels all intermediate terms and leaves $z^m-z^{-m}$. The negative-index formula is its negative, so it gives $(z^{-m}-z^m)\mathrm{/}(1-z^{-2})$; the case $m=0$ has zero numerator. Hence the first quotient equals $B_{1,m}$ for every integer $m$.

For $m>0$, the geometric identity $z^{2m}-q^{-2m}=(z^2-q^{-2}){\sum }_{j=0}^{m-1}{z}^{2(m-1-j)}{q}^{-2j}$ and $1-q^2{z}^2=-q^2(z^2-q^{-2})$ give $\frac{z^m-q^{-2m}{z}^{-m}}{1-q^2{z}^2}=-{\sum }_{j=0}^{m-1}{q}^{-2j-2}{z}^{m-2-2j}=B_{0,m}$. For $m=-r<0$, $z^{-r}-q^{2r}{z}^r=z^{-r}(1-(q^2{z}^2{)}^r)$, so division by $1-q^2{z}^2$ gives ${\sum }_{j=0}^{r-1}{q}^{2j}{z}^{-r+2j}=B_{0,-r}$. The zero case is again zero. Thus the second quotient is Laurent for every integer $m$.

Write ${\gamma }_j=g_j(z)\in K(z)$. Distinct elements of $G$ give distinct ${\gamma }_j$: if the signs differ, one is a nonzero scalar multiple of $z$ and the other of $z^{-1}$; if the signs agree, equality would give $q^r=q^s$, hence $r=s$ because $q$ is indeterminate. Thus all differences ${\gamma }_j-{\gamma }_i$ for $i\mathrm{\ne }j$ are nonzero in $K(z)$.

Apply the assumed operator relation to $h=1,z,\dots ,z^{N-1}$. The resulting linear system for $(f_j)$ has matrix $({\gamma }_j^k{)}_{0\le k<N,1\le j\le N}$, whose determinant is the Vandermonde product ${\prod }_{i<j}({\gamma }_j-{\gamma }_i)\mathrm{\ne }0$ by step 1.1. The matrix is invertible over $K(z)$, so every $f_j=0$.

From ${\pi }^2=1$ and $\pi T_0\pi =T_1$ one obtains $\pi T_0=T_1\pi$ and $\pi T_1=T_0\pi$. Repeatedly use these identities to move every occurrence of $\pi$ in an affine word to the right, and then use ${\pi }^2=1$ to leave at most one $\pi$. The quadratic relation gives $T_i^2=(t-t^{-1})T_i+1$, so induction on affine-word length reduces any word in $T_0,T_1$ to a finite linear combination of alternating strings. This proves the asserted form for words containing no $Z$ letters.

From ${\pi }^2=1$ and $\pi Z\pi =q^{-1}{Z}^{-1}$, conjugation gives $\pi Z^m\pi =q^{-m}{Z}^{-m}$ for every integer $m$, hence $\pi Z^m=q^{-m}{Z}^{-m}\pi$. In a generator word, move each lattice letter left across $\pi$ by this identity and across $T_i$ by $T_i{Z}^m=s_i(Z^m)T_i+(t-t^{-1})B_{i,m}(Z)$. Use lexicographic induction on (the number of $T_0,T_1$ letters, the number of out-of-order lattice/reflection pairs): a main term decreases the second coordinate, while a correction term is a finite Laurent sum with one fewer Hecke letter and decreases the first. After all lattice letters are on the left, move every $\pi$ right using $\pi T_0=T_1\pi$, $\pi T_1=T_0\pi$ and ${\pi }^2=1$; each move decreases the number of $\pi$-before-Hecke pairs. Finally apply step 1.1 to reduce the remaining Hecke words to alternating strings. The process terminates and gives the asserted spanning form.

Multiplication by $z$ and $z^{-1}$ are inverse operators, so the Laurent-inverse relator holds. Direct substitution gives ${\mathrm{\Pi }}^2=1$ and $\mathrm{\Pi }{M}_z\mathrm{\Pi }=M_{q^{-1}{z}^{-1}}$, which is the $\pi Z\pi$ relation. It also gives $\mathrm{\Pi }{\sigma }_0\mathrm{\Pi }={\sigma }_1$ and $\mathrm{\Pi }{M}_{u_0}\mathrm{\Pi }=M_{u_1}$; since both Hecke formulas use the same $t$, conjugating the formula for ${\mathsf{T}}_0$ yields $\mathrm{\Pi }{\mathsf{T}}_0\mathrm{\Pi }={\mathsf{T}}_1$.

Put $c=t-t^{-1}$, $a_i=(t{u}_i-t^{-1})\mathrm{/}(u_i-1)$, and $b_i=-c\mathrm{/}(u_i-1)$. Since ${\sigma }_i(u_i)=u_i^{-1}$, one has ${\mathsf{T}}_i=a_i{\sigma }_i+b_i$, ${\sigma }_i(a_i)=(t^{-1}{u}_i-t)\mathrm{/}(u_i-1)$, and ${\sigma }_i(b_i)=c{u}_i\mathrm{/}(u_i-1)$. Multiplication in the crossed-product operator algebra gives $a_i{\sigma }_i(b_i)+b_i{a}_i=c{a}_i$ and $a_i{\sigma }_i(a_i)+b_i^2=1+c{b}_i$, so ${\mathsf{T}}_i^2=c{\mathsf{T}}_i+1$, equivalent to $({\mathsf{T}}_i-t)({\mathsf{T}}_i+t^{-1})=0$.

For multiplication $M_f$ by a Laurent polynomial, direct expansion of ${\mathsf{T}}_i{M}_f-M_{s_i(f)}{\mathsf{T}}_i$ gives multiplication by $c(f-s_i(f))\mathrm{/}(1-u_i)$. With $f=z^m$, [F3] identifies this Laurent polynomial with $c{B}_{i,m}$, which is exactly each Bernstein relation in [F2]. The finite-sum formulas in [F3] also show that ${\mathsf{T}}_i$ preserves $\mathcal{P}$.

The assignments in [F5], checked in steps 1.1, 1.2, and 1.3, satisfy all relators of the free-algebra quotient. By [F4] they descend to a unital $K$-algebra homomorphism ${\mathcal{H}}_{q,t}(A_1)\to {\mathrm{End}}_K(\mathcal{P})$.

Write each ${\mathsf{T}}_i=a_i{\sigma }_i+b_i$ as in [F1], with $b_i(z)\in K(z)$. The rational function $a_i(z)$ is nonzero: its numerator $t{u}_i(z)-t^{-1}$ is nonzero because $u_i$ is a nonconstant monomial and $t\mathrm{\ne }0$. In the expansion of an alternating word $T_{i_1}\cdots T_{i_r}{\pi }^{ϵ}$, choosing the substitution term $a_{i_j}{\sigma }_{i_j}$ from every ${\mathsf{T}}_{i_j}$ gives the affine substitution $g=s_{i_1}\cdots s_{i_r}{\pi }^{ϵ}$ with a nonzero rational coefficient ${\gamma }_g(z)$, a product of transforms of the nonzero $a_{i_j}$. Every other term omits at least one reflection and therefore has affine Coxeter length strictly less than $r$. The semidirect-product normal form in [F4] makes the top substitutions for distinct alternating words and ${\pi }^{ϵ}$ distinct.

Suppose a finite linear combination of the spanning words in [F2] acts as zero, and choose the largest affine length $r$ among words with a nonzero coefficient. For any fixed top substitution $g$ of length $r$, step 1.1 shows that its coefficient in the crossed-product expansion is exactly ${\gamma }_g(z){\sum }_m{c}_m{z}^m$: shorter words and all lower terms have smaller length. By [F3], the coefficient of each substitution must vanish. Since ${\gamma }_g\mathrm{\ne }0$, the Laurent polynomial ${\sum }_m{c}_m{z}^m$ is zero, so all its coefficients vanish. Repeating for the finitely many top substitutions of length $r$ and then descending in length shows that every coefficient in the original relation is zero.

The spanning family in [F2] has linearly independent images by step 2.1. Every element of the representation kernel is a finite linear combination of that spanning family, so it is zero. Thus the representation is injective and hence faithful.

The coefficient identity $A+B=t+t^{-1}$ rewrites the formula as $(\mathcal{D}f)(z)=(t+t^{-1})f(q^{-1}z)+A(z)(f(qz)-f(q^{-1}z))$. For $f(z)=f(z^{-1})$, the Laurent polynomial $h(z)=f(qz)-f(q^{-1}z)$ vanishes at $z=1$ and $z=-1$, since $f(q)=f(q^{-1})$ and $f(-q)=f(-q^{-1})$. Multiplying $h$ by a power of $z$ gives an ordinary polynomial still vanishing at both distinct roots $1,-1$, so $z^2-1$ divides $h$; hence $A(z)h(z)$ is Laurent. Thus $\mathcal{D}f$ is a Laurent polynomial.

One has $A(z^{-1})=B(z)$ and $B(z^{-1})=A(z)$. For symmetric $f$, $f(q{z}^{-1})=f(q^{-1}z)$ and $f(q^{-1}{z}^{-1})=f(qz)$. Substituting $z^{-1}$ in [F1] therefore gives $(\mathcal{D}f)(z^{-1})=(\mathcal{D}f)(z)$, so the result is Weyl-invariant.

The coefficients $A,B$ have finite limits $t,t^{-1}$ at $z=\mathrm{\infty }$, so $\mathcal{D}{m}_n$ has no exponent above $n$. Its coefficient of $z^n$ is $t{q}^n+t^{-1}{q}^{-n}$, obtained from the leading terms of $m_n(qz)$ and $m_n(q^{-1}z)$. Weyl invariance from step 2.1 gives the same coefficient at $z^{-n}$. Subtracting ${\lambda }_n{m}_n$ leaves a symmetric Laurent polynomial $g$ with no exponent of absolute value $n$ or greater. Write $g={\sum }_{k=-r}^r{b}_k{z}^k$; symmetry gives $b_k=b_{-k}$, so $g=b_0{m}_0+{\sum }_{k=1}^r{b}_k{m}_k$. Thus $g$ is a linear combination of $m_0,\dots ,m_{n-1}$. When $n=0$, $A+B=t+t^{-1}$ gives the formula directly.

Let $f\in V$. Since ${\sigma }_{1}f=f$, the divided-difference term in [F1] vanishes and $T_{1}f=tf$. Since $f$ is symmetric, $(\pi f)(z)=f(q^{-1}{z}^{-1})=f(qz)$. Hence $(Yf)(z)=tf(qz)$.

Write $c=t-t^{-1}$ and $T_1=a(z){\sigma }_1+b(z)$, where $a(z)=(t^{-1}{z}^2-t)\mathrm{/}(z^2-1)$ and $b(z)=c{z}^2\mathrm{/}(z^2-1)$. By [F4], $T_1^{-1}=T_1-c$, so $Y^{-1}=T_1^{-1}\pi$. For $f\in V$, ${\sigma }_1\pi f(z)=f(q^{-1}z)$ and $\pi f(z)=f(qz)$, giving $(Y^{-1}f)(z)=\frac{t^{-1}{z}^2-t}{z^2-1}f(q^{-1}z)+\frac{c}{z^2-1}f(qz)$. Adding the formula from step 1.1 yields exactly [F2], so $(Y+Y^{-1}){\mathrm{\mid }}_V=\mathcal{D}$.

The $Y$-Laurent algebra is commutative because it is generated by the single invertible element $Y$. By [F5], the finite reflection negates the rank-one translation exponent, so the involution in the Statement replaces $Y$ by $Y^{-1}$; this is an action on translation exponents, not conjugation by $T_1$. If $p(Y)={\sum }_{k=-N}^N{a}_k{Y}^k$ is invariant, then, since $K$ has characteristic zero, $p(Y)=\frac{1}{2}(p(Y)+p(Y^{-1}))=a_0+{\sum }_{k=1}^N\frac{a_k+a_{-k}}{2}(Y^k+Y^{-k})$. This identity needs no linear independence of the powers of $Y$. The elements $S_n=Y^n+Y^{-n}$ satisfy $S_0=2$, $S_1=Y+Y^{-1}$, and $S_{n+1}=(Y+Y^{-1})S_n-S_{n-1}$ by multiplication; induction therefore expresses each $S_n$ as a polynomial in $Y+Y^{-1}$. By [F3] and step 2.1, $Y+Y^{-1}$ preserves $V$ and restricts to $\mathcal{D}$. Thus every invariant $p(Y)$ restricts to a polynomial in $\mathcal{D}$, and these restrictions commute.

For distinct $n,m\ge 0$, ${\lambda }_n-{\lambda }_m=(q^n-q^m)(t-t^{-1}{q}^{-(n+m)})$. If $n\mathrm{\ne }m$, the first factor is nonzero because $q$ is not a root of unity. Also $n+m\ge 1$, and the second factor is nonzero because $t^2\mathrm{\ne }{q}^{-(n+m)}$. Thus all eigenvalues ${\lambda }_n$ are pairwise distinct.

Write $f={\sum }_{k=-r}^r{a}_k{z}^k\in V$. Comparing coefficients in $f(z)=f(z^{-1})$ gives $a_k=a_{-k}$, so $f=a_0{m}_0+{\sum }_{k=1}^r{a}_k{m}_k$; uniqueness follows from the linear independence of Laurent monomials. Thus $m_0,\dots ,m_r$ form a basis of the symmetric Laurent polynomials of degree at most $r$. Construct the $P_n$ by induction: set $P_0=m_0$, an eigenvector by [F2], and suppose $P_0,\dots ,P_{n-1}$ form a basis of the degree-at-most-$n-1$ subspace. By [F2], write $(\mathcal{D}-{\lambda }_n)m_n={\sum }_{j<n}{r}_{n,j}{P}_j$. Set $P_n=m_n+{\sum }_{j<n}\frac{r_{n,j}}{{\lambda }_n-{\lambda }_j}{P}_j$. The coefficient of each $P_j$ in $(\mathcal{D}-{\lambda }_n)P_n$ is $r_{n,j}+({\lambda }_j-{\lambda }_n)r_{n,j}\mathrm{/}({\lambda }_n-{\lambda }_j)=0$, so $P_n$ is an eigenvector with leading term $m_n$.

If another monic eigenvector $Q$ with leading term $m_n$ existed, write $Q-P_n={\sum }_{j<n}{a}_j{P}_j$. Then $0=(\mathcal{D}-{\lambda }_n)(Q-P_n)={\sum }_{j<n}{a}_j({\lambda }_j-{\lambda }_n)P_j$. The $P_j$ are linearly independent by their triangular leading terms, and every ${\lambda }_j-{\lambda }_n$ is nonzero by step 1.1, so each $a_j=0$ and $Q=P_n$. Since $(m_n)$ is a basis and the change to $(P_n)$ is triangular with diagonal entries $1$, the $P_n$ form a basis. Finally [F3] makes each symmetric DAHA translation polynomial act as a polynomial in $\mathcal{D}$; for every polynomial $r$, $r(\mathcal{D})P_n=r({\lambda }_n)P_n$. Thus each $P_n$ is a simultaneous eigenfunction of the full symmetric translation family.