Define
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu].
$$
The gauge-field transformation is precisely the one for which the <gauge covariant derivative> $D_\mu=\partial_\mu-igA_\mu$ transforms by $D_\mu' =UD_\mu U^\dagger$. Since $[D_\mu,D_\nu]=-igF_{\mu\nu}$, it follows immediately that
$$
\boxed{F_{\mu\nu}'=UF_{\mu\nu}U^\dagger}.
$$
For $U=1+i\alpha^cT_c+O(\alpha^2)$,
$$
\delta F_{\mu\nu}=i[\alpha^cT_c,F_{\mu\nu}^bT_b].
$$
Define the <structure constant of a Lie algebra> by $[T_b,T_c]=if_{bc}{}^aT_a$. Antisymmetry then gives
$$
\boxed{\delta F_{\mu\nu}^a=f_{bc}{}^aF_{\mu\nu}^b\alpha^c}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 304 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-304/4/a/solution