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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 and .
The one-dimensional conclusion is proved over , under the complex form of Lie's theorem (Irreducible representations of solvable complex Lie algebras are one-dimensional).
Solvability is termination of the derived series (Derived series and solvable Lie algebras).
A Lie-algebra representation is a bracket-preserving linear map into the endomorphism algebra (Representations of Lie algebras).
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
Define . Since is abelian and , this is a representation by [L3]. Also , so is solvable by [L2].
Any nonzero proper subspace of is a real line. If such a line were -stable, a nonzero vector on it would be a real eigenvector of . But the characteristic polynomial of is , which has no real root. Thus no nonzero proper stable subspace exists, so is irreducible by [L4].
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 ; after complexification, has the two eigenlines with eigenvalues and . The witness and all calculations are explicit and choice-free.
Depends on
Used by
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Dependency tree · two levels
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Sources
- Knapp, Lie Groups Beyond an Introduction, field hypothesis in Lie's theorem (standard reference, not scraped)