The product of generators satisfies
$$
T_aT_b=\frac1{2N}\delta_{ab}I
+\frac12(d_{ab}{}^c+if_{ab}{}^c)T_c,
$$
and therefore
$$
\operatorname{Tr}(T_aT_bT_c)=\frac14(d_{abc}+if_{abc}).
$$
Because $\phi^a\phi^b\phi^c$ is symmetric, its contraction with the antisymmetric <structure constant of a Lie algebra> $f_{abc}$ vanishes. Hence
$$
-\frac1{3!}\operatorname{Tr}(\phi^3)
=-\frac1{24}d_{abc}\phi^a\phi^b\phi^c.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-304/4/b/solution