The <Killing form> is invariant under the adjoint action:
$$
\kappa(\operatorname{Ad}_gX,\operatorname{Ad}_gY)=\kappa(X,Y).
$$
Together with $F'_{\mu\nu}=\operatorname{Ad}_gF_{\mu\nu}$, this immediately gives
$$
\kappa(F'_{\mu\nu},F'^{\mu\nu})
=\kappa(F_{\mu\nu},F^{\mu\nu}),
$$
so every positive integral power $\mathcal L_1$ is gauge invariant.

The <adjoint covariant derivative> obeys
$$
(D'_\mu F'_{\nu\rho})
=g(D_\mu F_{\nu\rho})g^{-1}.
$$
This follows either by substituting the transformation laws or by applying the covariance of $D'_\mu$ to an adjoint-valued field. A second use of invariance of the Killing form gives
$$
\kappa(D'_\mu F'_{\nu\rho},D'^{\mu}F'^{\nu\rho})
=\kappa(D_\mu F_{\nu\rho},D^{\mu}F^{\nu\rho}),
$$
which proves gauge invariance of $\mathcal L_2$.

Solution

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

On an adjoint-valued field $X$, the covariant derivative is $D_\mu X=\partial_\mu X+[A_\mu,X]$. If $X\mapsto gXg^{-1}$, then $D_\mu X\mapsto g(D_\mu X)g^{-1}$.


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