Lie algebra of nilpotent endomorphisms

Let $𝔤 \leq 𝔤 𝔩 𝑛 𝕂$ be a linear Lie algebra for which every $𝑥 \in 𝔤$ is a nilpotent endomorphism of $𝕂 𝑛$ . Then there exists some nonzero $𝑣 \in 𝕂 𝑛$ for which $𝔤 𝑣 = 0$ .¹ #m/thm/lie

Proof

Let $𝔤 \leq 𝔤 𝔩 𝑛 𝕂$ be a linear Lie algebra. Assume that for $d i m 𝔤 < 𝑚$ , every $𝑥 \in 𝔤$ being nilpotent implies the existence of some nonzero $𝑣 \in 𝕂 𝑛$ such that $𝔤 𝑣 = 0$ . Note that this clearly holds for $𝑚 = 2$ .

Now take $d i m 𝔤 = 𝑚$ , and let $𝔥 < 𝔤$ be a strict subalgebra, so that $d i m 𝔥 < 𝑚$ . Then by Nilpotent transformations have nilpotent adjoint representations, $𝔤$ acts on $𝔤$ under the adjoint representation nilpotently, as does $𝔥$ on $𝔤 / 𝔥$ : Thus we have a Lie algebra homomorphism

𝜋 : 𝔥 \to 𝔤 𝔩 (𝔤 / 𝔥)

such that $𝜋 (𝔥)$ contains only nilpotent endomorphisms. Since $d i m 𝜋 (𝔥) < 𝑚$ satisfies the induction hypothesis, there exists a nonzero $𝑥 + 𝔥 \in 𝔤 / 𝔥$ such that $𝜋 (𝔥) (𝑥 + 𝔥) = 𝔥$ , or equivalently, the normalizer $𝔫 𝔤 (𝔥)$ is a strict superset of $𝔥$ .

Taking $𝔥$ to be a maximal strict subalgebra, it follows that $𝔫 𝔤 (𝔥) = 𝔥$ , thus $𝔥$ is an ideal of codimension one: Hence $𝔤 = 𝔥 + 𝕂 𝑧$ for any $𝑧 \in 𝔤 ∖ 𝔥$ . Let $𝑊 = {𝑣 \in 𝑉 : 𝔥 𝑣 = 0}$ be the subspace of vectors annihilated by $𝔥$ . Since $𝔥$ is an ideal, this is invariant under $𝔤 = 𝔥 + 𝕂 𝑧$ , and since $𝑧$ is nilpotent, it has an eigenvector $𝑣 \in 𝑊$ such that $𝑧 𝑣 = 0$ , and therefore $𝔤 𝑣 = 0$ as required.

#state/tidy | #lang/en | #SemBr

1972. Introduction to Lie Algebras and Representation Theory ↩

Lie algebra of nilpotent endomorphisms

Footnotes