LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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 one-root gl2 Bethe equation cancels the unwanted transfer term

Statement

In $Y_{i} ({g l}_{2})$ write $A (u) = t_{11} (u)$ , $B (u) = t_{12} (u)$ , and $D (u) = t_{22} (u)$ . Let $H$ be a representation equipped with a common spectral domain $D \subseteq C$ on which the represented series $A (u), B (u), D (u)$ have defined, regular operator values and the component RTT identities from [F1] hold pointwise for $u, v \in D$ with $u \neq v$ . Let $Ω \in H$ and scalar-valued functions $a, d$ , regular on $D$ , satisfy $A (u) Ω = a (u) Ω$ and $D (u) Ω = d (u) Ω$ on $D$ . For $v \in D$ with $B (v) Ω \neq 0$ , if the one-root Bethe equation $a (v) = d (v)$ holds, then for every $u \in D ∖ {v}$ the vector $B (v) Ω$ is an eigenvector of $A (u) + D (u)$ with eigenvalue

$Λ_{v} (u) = \frac{u - v - i}{u - v} a (u) + \frac{u - v + i}{u - v} d (u) .$

This one-root cancellation gives no completeness conclusion.

Facts & Assumptions

Given: The RTT presentation for $N = 2$ , a representation with common spectral domain $D$ as specified in the statement, and a vector satisfying the displayed vacuum eigenvalue conditions there.

[F1]

The RTT relation has an explicit component commutator form: $x [t_{i j} (u), t_{k l} (v)] = i (t_{k j} (v) t_{i l} (u) - t_{k j} (u) t_{i l} (v))$ for $x = u - v$ .

Proof

1.1givenF1algebra

For $u, v \in D$ with $x = u - v \neq 0$ , apply [F1] to $(i, j, k, l) = (1, 2, 1, 1)$ and interchange $u, v$ ; rearranging gives $A (u) B (v) = \frac{x - i}{x} B (v) A (u) + \frac{i}{x} B (u) A (v)$ . Applying [F1] to $(i, j, k, l) = (2, 2, 1, 2)$ gives $D (u) B (v) = \frac{x + i}{x} B (v) D (u) - \frac{i}{x} B (u) D (v)$ . The common-domain assumption supplies these pointwise identities, and all divisions are by the nonzero scalar $x$ .

2.1givenstep 1.1algebra∎

Apply these identities to $Ω$ and use its eigenvalue conditions on $D$ . The coefficient of $B (u) Ω$ in $(A (u) + D (u)) B (v) Ω$ is $i (a (v) - d (v)) / x$ , which vanishes under $a (v) = d (v)$ . The remaining coefficient of $B (v) Ω$ is exactly $Λ_{v} (u)$ , so the claimed eigenvector identity follows; nonzero-ness of $B (v) Ω$ makes it an eigenvector rather than merely a zero-vector identity.

Depends on

Used by

A one-root Bethe equation need not produce a complete set of vectors Counterexample

Dependency tree · two levels

5 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

L. D. Faddeev, How Algebraic Bethe Ansatz Works for Integrable Model (standard reference, not scraped)
A. P. Isaev, Lectures on Quantum Groups and Yang-Baxter Equations (standard reference, not scraped)