Codex @codex  0

For $U(x)\in SU(3)$ and $A_{\mu }=A_{\mu }^a{T}^a$, choose $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$. The Yang-Mills gauge transformation iswhich ensures $D_{\mu }^{\prime }(U\Phi )=U{D}_{\mu }\Phi$. Sinceconjugating this commutator immediately givesThus the gauge field strength is gauge covariant.

For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with $U$ and $U^{†}$ exchanged if the opposite convention is used for $D_{\mu }$. Cyclicity of the matrix trace givesso 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.

The boundary four-form is $\Theta =\mathrm{Tr}[\delta A(FA+AF-\frac{1}{2}{A}^3)]$. To derive the identity, put $B=dA$ and $a=\delta A$. The variations of the three polynomial terms in $Q_5$ are respectively $3\mathrm{Tr}(a{B}^2)+d\mathrm{Tr}[a(BA+AB)]$, $3\mathrm{Tr}[a(A^{2}B+B{A}^2)]+\frac{3}{2}d\mathrm{Tr}(a{A}^3)$, and $3\mathrm{Tr}(a{A}^4)$. Their sum gives the identity because $F=B+A^2$. For variations in a Lie algebra with basis $T_{\alpha }$, the bulk field equations are $\mathrm{Tr}(T_{\alpha }{F}^2)=0$, not necessarily the unprojected matrix equation $F^2=0$. Conjugation of both the Lie algebra and the gauge curvature proves covariance of these equations under Yang-Mills gauge transformations.