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The A1 Macdonald operator is triangular on Weyl orbit sums

Statement

Let $m_{0} = 1$ and $m_{n} = z^{n} + z^{- n}$ for $n \geq 1$ be the A1 Weyl orbit sums. The degree of a nonzero Laurent polynomial $\sum_{k} a_{k} z^{k}$ is $\max {∣ k ∣ : a_{k} \neq 0}$ . For every $n \geq 0$ , the operator $D$ of The symmetric Macdonald difference operator of type A1 sends $m_{n}$ to

D m_{n} = λ_{n} m_{n} + \sum_{0 \leq j < n} d_{n, j} m_{j}, λ_{n} = t q^{n} + t^{- 1} q^{- n},

with coefficients $d_{n, j} \in K$ . In particular, $D$ preserves the symmetric Laurent-polynomial space.

Facts & Assumptions

Given: The operator $D$ on $V$ and the rank-one Laurent variable $z$ .

[F1]

The symmetric Macdonald difference operator of type A1: $D f = A (z) f (q z) + B (z) f (q^{- 1} z)$ , with $A (z) = (t z^{2} - t^{- 1}) / (z^{2} - 1)$ and $B (z) = (t^{- 1} z^{2} - t) / (z^{2} - 1)$ .

Proof

1.1F1givenalgebra

The coefficient identity $A + B = t + t^{- 1}$ rewrites the formula as $(D f) (z) = (t + t^{- 1}) f (q^{- 1} z) + A (z) (f (q z) - f (q^{- 1} z))$ . For $f (z) = f (z^{- 1})$ , the Laurent polynomial $h (z) = f (q z) - f (q^{- 1} z)$ 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 $D f$ is a Laurent polynomial.

2.1F1step 1.1algebra

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 $f (q^{- 1} z^{- 1}) = f (q z)$ . Substituting $z^{- 1}$ in [F1] therefore gives $(D f) (z^{- 1}) = (D f) (z)$ , so the result is Weyl-invariant.

3.1F1step 2.1algebra∎

The coefficients $A, B$ have finite limits $t, t^{- 1}$ at $z = \infty$ , so $D m_{n}$ has no exponent above $n$ . Its coefficient of $z^{n}$ is $t q^{n} + t^{- 1} q^{- n}$ , obtained from the leading terms of $m_{n} (q z)$ and $m_{n} (q^{- 1} z)$ . Weyl invariance from step 2.1 gives the same coefficient at $z^{- n}$ . Subtracting $λ_{n} m_{n}$ 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.

Depends on

Used by

Cited to discharge well-definedness by The symmetric Macdonald difference operator of type A1.

Dependency tree · two levels

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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)