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Double-Affine Hecke Algebras and Macdonald Theory: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These calculations use the A-page type- $A_{1}$ normalization and test its parameter boundary.

Worked computations

The first symmetric A1 Macdonald eigenfunctions verifies $P_{0} = 1$ and $P_{1} = z + z^{- 1}$ directly from the difference operator and its eigenvalue theorem.
At q=1 the rank-one Macdonald eigenvalues are not distinct shows that $q = 1$ collapses the operator to a scalar and destroys pairwise eigenvalue separation.

The B items are checks of the A-page formulas; they do not supply general DAHA or Macdonald results.

Companion: double-affine-hecke-algebras-and-macdonald-theory.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

The first symmetric A1 Macdonald eigenfunctions

Example

Let $m_{0} = 1$ and $m_{1} = z + z^{- 1}$ . The first two monic symmetric rank-one Macdonald eigenfunctions are

P_{0} (z) = 1, P_{1} (z) = z + z^{- 1}

with eigenvalues $t + t^{- 1}$ and $t q + t^{- 1} q^{- 1}$ , respectively.

Facts & Assumptions

Given: $q \in K^{\times}$ is not a root of unity, $t \in K^{\times}$ , $t^{2} \neq q^{- N}$ for every positive integer $N$ , and $D$ is the rank-one symmetric Macdonald operator.

[F1]

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

[F2]

The symmetric rank-one Macdonald polynomials are DAHA eigenfunctions: the monic symmetric eigenfunction with leading orbit sum $m_{n}$ is unique when $q$ is not a root of unity and $t^{2} \neq q^{- N}$ for every positive integer $N$ .

Verification

1.1F1F2givenalgebra

For $f = 1$ , [F1] gives $D 1 = (A + B) 1 = (t + t^{- 1}) 1$ . Since there is no lower orbit sum below $m_{0} = 1$ , uniqueness in [F2] yields $P_{0} = 1$ .

2.1F1F2algebra∎

Applying [F1] to $m_{1}$ and collecting over the denominator $z^{2} - 1$ gives $D m_{1} = \frac{(t z^{2} - t^{- 1}) (q z + q^{- 1} z^{- 1}) + (t^{- 1} z^{2} - t) (q^{- 1} z + q z^{- 1})}{z^{2} - 1} = (t q + t^{- 1} q^{- 1}) (z + z^{- 1})$ . The numerator equality follows by expansion: the $z$ terms cancel and the remaining numerator is $(t q + t^{- 1} q^{- 1}) (z^{3} - z^{- 1})$ . Thus $m_{1}$ is the monic eigenfunction with leading orbit sum $m_{1}$ , and [F2] gives $P_{1} = m_{1}$ .

CounterexampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

At q=1 the rank-one Macdonald eigenvalues are not distinct

Statement refuted

For every pair of nonzero complex parameters $q, t$ , the scalars $λ_{n} = t q^{n} + t^{- 1} q^{- n}$ are pairwise distinct as $n$ ranges over the nonnegative integers.

Facts & Assumptions

Given: The displayed scalar sequence and the type-A1 difference operator $D$ .

[F1]

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

Counterexample

Take $q = 1$ and any $t \in C^{\times}$ . Then both shifts in [F1] are the identity and the two coefficients sum to $t + t^{- 1}$ . Given: $q = 1$ , $t \neq 0$ , and $m_{0} = 1$ , $m_{1} = z + z^{- 1}$ .

1.1F1givenalgebra

The coefficient sum is $\frac{t z^{2} - t^{- 1} + t^{- 1} z^{2} - t}{z^{2} - 1} = t + t^{- 1}$ , so $D = (t + t^{- 1}) i d$ on symmetric Laurent polynomials.

2.1step 1.1algebra∎

The distinct polynomials $m_{0}, m_{1}$ are both eigenvectors with eigenvalue $t + t^{- 1}$ ; correspondingly $λ_{0} = λ_{1} = t + t^{- 1}$ . Thus the claimed pairwise distinction for all nonzero $q, t$ fails at $q = 1$ .