Substituting $\delta A_\mu=g^{-1}\partial_\mu\alpha-i[A_\mu,\alpha]$ into the curvature and collecting terms gives the covariant transformation
$$
\boxed{\delta F_{\mu\nu}=-i[F_{\mu\nu},\alpha]}.
$$
Equivalently, $\delta F_{\mu\nu}^a=f^{abc}F_{\mu\nu}^b\alpha^c$ in the stated conventions. Since $F_{\mu\nu}^aF^{\mu\nu,a}$ is proportional to $\operatorname{tr}(F_{\mu\nu}F^{\mu\nu})$, cyclicity of the <matrix trace> gives
$$
\delta\operatorname{tr}(F_{\mu\nu}F^{\mu\nu})
=-i\operatorname{tr}([F_{\mu\nu}F^{\mu\nu},\alpha])=0.
$$
Thus the <Yang-Mills action> is <gauge-invariant>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-304/4/a/solution