Write $U(x)=e^{i\alpha^a(x)T^a}$ and $A_\mu=A_\mu^aT^a$. With the convention
$$
D_\mu\psi=(\partial_\mu-igA_\mu)\psi,
$$
the <gauge field> must transform as
$$
\boxed{A'_\mu=UA_\mu U^{-1}-\frac{i}{g}(\partial_\mu U)U^{-1}}.
$$
Then $D'_\mu\psi'=U D_\mu\psi$. The conjugate field has $D_\mu\bar\psi=\partial_\mu\bar\psi+ig\bar\psi A_\mu$. For the adjoint scalar,
$$
\boxed{D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi]},
$$
or $(D_\mu\phi)^a=\partial_\mu\phi^a+g\epsilon^{abc}A_\mu^b\phi^c$. It transforms as $D'_\mu\phi'=U(D_\mu\phi)U^{-1}$. The <gauge field strength>
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu]
$$
similarly transforms by conjugation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 304 / 4 / a / Solution

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.

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

Direct expansion gives
$$
[D_\mu,D_\nu]\phi
=\{\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu]\}\phi
=F_{\mu\nu}\phi.
$$
Since $D'_\mu(g\phi)=gD_\mu\phi$,
$$
F'_{\mu\nu}(g\phi)
=[D'_\mu,D'_\nu](g\phi)
=g[D_\mu,D_\nu]\phi
=gF_{\mu\nu}\phi.
$$
Therefore the non-Abelian <gauge field strength> transforms covariantly:
$$
F'_{\mu\nu}=gF_{\mu\nu}g^{-1}.
$$

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 4 / c / Solution

{title2=$F_{\mu\nu}$}

The curvature of a non-Abelian gauge field is $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu]$. It satisfies $[D_\mu,D_\nu]=F_{\mu\nu}$ and transforms as $F_{\mu\nu}\mapsto gF_{\mu\nu}g^{-1}$.


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