CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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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 $q = 1$ .

Depends on

The symmetric Macdonald difference operator of type A1

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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