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Every invariant subspace of the Adjoint representation of a Lie algebra is an ideal. Hence the adjoint representation of a Simple Lie algebra is irreducible.

The Killing form of a complex Simple Lie algebra is nondegenerate. Since $B$ is also nondegenerate, there is a unique endomorphism $T$ of $\mathfrak{g}$ satisfyingInvariance of both forms givesso $T$ intertwines the adjoint representation. That representation is irreducible because its invariant subspaces are ideals. The Schur lemma therefore gives $T=cI$. Since $\kappa$ is nondegenerate, $c\ne 0$, andThis is the uniqueness of an invariant bilinear form on a simple Lie algebra.

Solved by gpt-5.6-sol high.

Let $W\subseteq \mathfrak{g}$ be invariant under the Adjoint representation of a Lie algebra. Then $[\mathfrak{g},W]\subseteq W$, so $W$ is an ideal. If $\mathfrak{g}$ is a Simple Lie algebra, its only ideals are $0$ and $\mathfrak{g}$. Hence the adjoint representation is irreducible. Compact type is compatible with the anti-Hermitian realization used below, although simplicity alone proves this adjoint irreducibility of a simple Lie algebra.

Solved by gpt-5.6-sol high.