The Trace form of a Lie algebra representation is invariant because
Choose a basis orthonormal for the positive-definite form of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; Schur lemma makes it scalar. Thus . Since is anti-Hermitian,
The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore with .
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