General linear Lie algebra

Lie algebra of nilpotent endomorphisms

Let be a ^subalgebra for which every is a nilpotent endomorphism of . Then there exists some nonzero for which .¹ lie

Proof

Let be a linear Lie algebra. Assume that for , every being nilpotent implies the existence of some nonzero such that . Note that this clearly holds for .

Now take , and let be a strict subalgebra, so that . Then by ^P1, acts on under the adjoint representation nilpotently, as does on : Thus we have a Lie algebra homomorphism

such that contains only nilpotent endomorphisms. Since satisfies the induction hypothesis, there exists a nonzero such that , or equivalently, the normalizer is a strict superset of .

Taking to be a maximal strict subalgebra, it follows that , thus is an ideal of codimension one: Hence for any . Let be the subspace of vectors annihilated by . Since is an ideal, this is invariant under , and since is nilpotent, it has an eigenvector such that , and therefore as required.

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Footnotes

1. 1972. Introduction to Lie Algebras and Representation Theory ↩