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For a finite-dimensional Lie algebra representation $\rho$ on $V$, the Trace form of a Lie algebra representation isWrite again $A=\rho (a)$, $B=\rho (b)$, and $C=\rho (c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the cyclic property of the trace givesIn the last line, cyclicity and $AC=CA$ turn $\mathrm{tr}(ABC)$ into $\mathrm{tr}(BAC)$. Thus the nonzero vector $c\in \mathfrak{g}$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak{g}$. The bilinear form is therefore degenerate.

Solved by gpt-5.6-sol high.

The Trace form of a Lie algebra representation is invariant becauseChoose a basis orthonormal for the positive-definite form $-\kappa$ of the compact simple algebra. Any invariant bilinear form determines an endomorphism commuting with the irreducible adjoint action; Schur lemma makes it scalar. Thus $H(T_a,T_b)=c{\delta }_{ab}$. Since $d(T_a)$ is anti-Hermitian,The inequality is strict because the kernel of the nontrivial irreducible representation is an ideal and hence zero. Therefore $c=-\mu$ with $\mu >0$.

Solved by gpt-5.6-sol high.