LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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 standard SU(3) representation has no nonzero fixed vector

Statement

Let $S U (3) = {U \in M_{3} (C) : U^{*} U = I, \det U = 1}$ act on $C^{3}$ by matrix multiplication. If $v \in C^{3}$ satisfies $U v = v$ for every $U \in S U (3)$ , then $v = 0$ .

Facts & Assumptions

Given: The standard matrix action of the displayed group $S U (3)$ on $C^{3}$ .

Proof

technique · direct coordinate check

Given: A vector $v = (v_{1}, v_{2}, v_{3})$ fixed by every matrix in $S U (3)$ .

1.1givenalgebra

For every real $t$ , $D_{t} = diag (e^{i t}, e^{- i t}, 1)$ is unitary with determinant one. From $D_{t} v = v$ for all $t$ , taking any $t$ with $e^{i t} \neq 1$ gives $v_{1} = v_{2} = 0$ .

2.1givenstep 1.1algebra∎

For every real $s$ , $E_{s} = diag (1, e^{i s}, e^{- i s})$ also lies in $S U (3)$ . Since $v_{1} = v_{2} = 0$ and $E_{s} v = v$ for all $s$ , choosing $e^{- i s} \neq 1$ gives $v_{3} = 0$ . Hence $v = 0$ .

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Sophie Morel, Representation Theory, notes for Topics in Representation Theory (Princeton University, 2016 course; March 26, 2019)