ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 $P_{1} = m_{1}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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.