The <Trace form of a Lie algebra representation> is invariant because
$$
\begin{aligned}
H([X,Y],Z)
&=\operatorname{Tr}([d(X),d(Y)]d(Z))\\
&=\operatorname{Tr}(d(X)[d(Y),d(Z)])
=H(X,[Y,Z]).
\end{aligned}
$$
Choose 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,
$$
H(T_a,T_a)=-\operatorname{Tr}(d(T_a)^\dagger d(T_a))<0.
$$
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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 3 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-302/3/e/solution