For $U(x)\in SU(3)$ and $A_\mu=A_\mu^aT^a$, choose $D_\mu=\partial_\mu-igA_\mu$. The <Yang-Mills gauge transformation> is
$$
A'_\mu=UA_\mu U^{-1}-\frac{i}{g}(\partial_\mu U)U^{-1},
$$
which ensures $D'_\mu(U\Phi)=U D_\mu\Phi$. Since
$$
[D_\mu,D_\nu]=-igF_{\mu\nu},
$$
conjugating this commutator immediately gives
$$
\boxed{F'_{\mu\nu}=UF_{\mu\nu}U^{-1}}.
$$
Thus the <gauge field strength> is gauge covariant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 305 / 3 / b / i / Solution

For the convention in the question, a finite <Yang-Mills gauge transformation> acts covariantly on the field strength:
$$
F_{\mu\nu}\mapsto F'_{\mu\nu}=UF_{\mu\nu}U^\dagger
$$
or with $U$ and $U^\dagger$ exchanged if the opposite convention is used for $D_\mu$. Cyclicity of the <matrix trace> gives
$$
\operatorname{Tr}(F'_{\mu\nu}F'^{\mu\nu})
=\operatorname{Tr}(UF_{\mu\nu}F^{\mu\nu}U^\dagger)
=\operatorname{Tr}(F_{\mu\nu}F^{\mu\nu}),
$$
so the <Yang-Mills theory> Lagrangian is gauge invariant.

The quadratic gauge-field operator has zero directions $A_\mu\sim A_\mu+D_\mu\alpha$ along each <gauge orbit>. It therefore has no inverse on the full field space. <Gauge fixing> removes this degeneracy and produces a propagator, while the <Faddeev-Popov determinant> accounts for the corresponding Jacobian.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 304 / 3 / a / Solution

{c}

For $D_\mu=\partial_\mu-igA_\mu$ and matter transforming as $\psi\mapsto U\psi$, a Yang-Mills connection transforms inhomogeneously while its curvature transforms by conjugation, $F_{\mu\nu}\mapsto UF_{\mu\nu}U^{-1}$.


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