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The symmetric DAHA translation restricts to the A1 Macdonald operator

Statement

Let $Y = π T_{1}$ be the fundamental coweight translation in the polynomial representation of The rank-one DAHA polynomial representation. On the symmetric Laurent polynomials $V$ , the restriction of $Y + Y^{- 1}$ is the operator $D$ of The symmetric Macdonald difference operator of type A1. Here Weyl-invariant means fixed by the involution $Y \mapsto Y^{- 1}$ induced by the rank-one finite reflection. Every such Laurent polynomial in $Y$ restricts to a polynomial in $D$ . Thus the symmetric rank-one DAHA translations give a commuting family of difference operators on $V$ .

Facts & Assumptions

Given: The rank-one DAHA polynomial representation, the symmetric subspace $V$ , and the operator $D$ .

[F1]

The rank-one DAHA polynomial representation: on $P = K [z, z^{- 1}]$ , $Y = π T_{1}$ , $π f (z) = f (q^{- 1} z^{- 1})$ , and $T_{1} = t σ_{1} + (t - t^{- 1}) (σ_{1} - 1) / (z^{- 2} - 1)$ .

[F2]

The symmetric Macdonald difference operator of type A1: $D f = (t z^{2} - t^{- 1}) (z^{2} - 1)^{- 1} f (q z) + (t^{- 1} z^{2} - t) (z^{2} - 1)^{- 1} f (q^{- 1} z)$ on $V$ .

[F3]

The A1 Macdonald operator is triangular on Weyl orbit sums: $D$ maps $V$ to $V$ .

[F4]

The equal-parameter double-affine Hecke algebra of type A1: $T_{1}$ satisfies $(T_{1} - t) (T_{1} + t^{- 1}) = 0$ , and $Y = π T_{1}$ is the lift of the fundamental translation.

[F5]

The rank-one root datum and parameter conventions for DAHA: $z = X^{ρ}$ , and the finite Weyl reflection sends $z$ to $z^{- 1}$ , so it negates the rank-one exponent.

Proof

1.1F1givenalgebra

Let $f \in V$ . Since $σ_{1} f = f$ , the divided-difference term in [F1] vanishes and $T_{1} f = t f$ . Since $f$ is symmetric, $(π f) (z) = f (q^{- 1} z^{- 1}) = f (q z)$ . Hence $(Y f) (z) = t f (q z)$ .

2.1F1F2F4step 1.1algebra

Write $c = t - t^{- 1}$ and $T_{1} = a (z) σ_{1} + b (z)$ , where $a (z) = (t^{- 1} z^{2} - t) / (z^{2} - 1)$ and $b (z) = c z^{2} / (z^{2} - 1)$ . By [F4], $T_{1}^{- 1} = T_{1} - c$ , so $Y^{- 1} = T_{1}^{- 1} π$ . For $f \in V$ , $σ_{1} π f (z) = f (q^{- 1} z)$ and $π f (z) = f (q z)$ , giving $(Y^{- 1} f) (z) = \frac{t^{- 1} z^{2} - t}{z^{2} - 1} f (q^{- 1} z) + \frac{c}{z^{2} - 1} f (q z)$ . Adding the formula from step 1.1 yields exactly [F2], so $(Y + Y^{- 1}) ∣_{V} = D$ .

3.1F1F3F4F5step 2.1algebra∎

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 $D$ . Thus every invariant $p (Y)$ restricts to a polynomial in $D$ , and these restrictions commute.

Depends on

Used by

The symmetric rank-one Macdonald polynomials are DAHA eigenfunctions Theorem

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alexander A. Kirillov Jr., Lectures on the affine Hecke algebras and Macdonald conjectures (standard reference, not scraped)
I. G. Macdonald, Affine Hecke algebras and orthogonal polynomials (standard reference, not scraped)