= Solution
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>.
Back to article page
