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Lie's theorem
Statement
Let be a finite-dimensional solvable Lie algebra over an algebraically closed field of characteristic zero. Every nonzero finite-dimensional -module contains a common eigenvector: there are and such that for every .
Facts & Assumptions
Given: The algebraically closed characteristic-zero field , a finite-dimensional solvable -Lie algebra , and a nonzero finite-dimensional representation on .
A nonzero finite-dimensional solvable Lie algebra has a codimension-one ideal (Codimension-one ideal in a nonzero solvable Lie algebra).
A representation is linear and satisfies (Representations of Lie algebras).
Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue (Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue).
For finite-dimensional endomorphisms, (For and , ).
Proof
If , any nonzero is a common eigenvector with the zero functional.
Assume and the theorem for solvable Lie algebras of smaller dimension.
By [L1], choose a codimension-one ideal and write . Its derived series is contained termwise in that of , so is solvable. Step 1.2 gives and with for every .
Put and ; finite dimensionality makes finite-dimensional and -stable. Induction on , using and , shows and . Thus is -stable and every acts upper triangularly on a basis extracted from the cyclic list, with constant diagonal .
Let . Because is stable, [L2] and [L4] give for every , where the last equality uses the constant diagonal from step 3.1. Characteristic zero makes , so . This is the exact use of the characteristic hypothesis.
The nonzero common -weight space contains . It is -stable: for , [L2] and step 4.1 give .
By algebraic closure and [L3], has a nonzero eigenvector , say . Then for , and for we have . This defines the required linear functional . Algebraic closure is used only in [L3], characteristic zero only in step 4.1, and all choices are a finite sequence of existential choices rather than AC.
Depends on
- Codimension-one ideal in a nonzero solvable Lie algebra
- Representations of Lie algebras
- Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue
- For $A\in M_{m\times n}(F)$ and $B\in M_{n\times m}(F)$, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$
Used by
Dependency tree · two levels
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Sources
- Knapp, Lie Groups Beyond an Introduction, Theorem 1.25 (standard reference, not scraped)