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False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Every irreducible real representation of a solvable Lie algebra is one-dimensional

Statement

Every finite-dimensional irreducible real representation of a finite-dimensional solvable real Lie algebra is one-dimensional.

Facts & Assumptions

Given: The one-dimensional abelian real Lie algebra a=Rt and V=R2.

[L1]

The one-dimensional conclusion is proved over C, under the complex form of Lie's theorem (Irreducible representations of solvable complex Lie algebras are one-dimensional).

[L2]

Solvability is termination of the derived series (Derived series and solvable Lie algebras).

[L3]

A Lie-algebra representation is a bracket-preserving linear map into the endomorphism algebra (Representations of Lie algebras).

[L4]

A nonzero Lie-algebra representation is irreducible when it has no stable subspace other than zero and the whole module (Irreducible, completely reducible, and faithful representations).

Refutation

technique · direct
1.1

Define ρ(t)=R=(0110). Since a is abelian and [R,R]=0, this is a representation by [L3]. Also a(1)=0, so a is solvable by [L2].

givenL2L3algebra
2.1

Any nonzero proper subspace of V is a real line. If such a line were R-stable, a nonzero vector on it would be a real eigenvector of R. But the characteristic polynomial of R is T2+1, which has no real root. Thus no nonzero proper stable subspace exists, so V is irreducible by [L4].

L4step 1.1algebra
3.1

The representation in steps 1.1–2.1 is irreducible and two-dimensional, contradicting the proposed one-dimensional conclusion. It does not contradict [L1], whose scalar field is C; after complexification, R has the two eigenlines with eigenvalues i and i. The witness and all calculations are explicit and choice-free.

L1step 1.1step 2.1algebra

Depends on

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Sources