Yang-Mills gauge transformation

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

For $D_{\mu }={\partial }_{\mu }+A_{\mu }$ and $\delta \varphi =\Lambda \varphi$, covariance requires $\delta A_{\mu }=-{\partial }_{\mu }\Lambda +[\Lambda ,A_{\mu }]$. Expanding $[D_{\mu },D_{\nu }]$ gives $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$, and operator Jacobi gives $\delta F_{\mu \nu }=[\Lambda ,F_{\mu \nu }]$. An invariant internal quadratic form makes its Yang-Mills kinetic contraction gauge invariant. Couplings and factors of i are absorbed into the connection in this convention.

With $D_{\mu }={\partial }_{\mu }+\rho ({\mathcal{A}}_{\mu })$ and $\Phi \mapsto R(U)\Phi$, a local $U=1+ϵ$ gives $\delta {\mathcal{A}}_{\mu }=-{\partial }_{\mu }ϵ+[ϵ,{\mathcal{A}}_{\mu }]$, $\delta \Phi =\rho (ϵ)\Phi$ and $\delta {\mathcal{F}}_{\mu \nu }=[ϵ,{\mathcal{F}}_{\mu \nu }]$. Derivative terms cancel in $\delta (D_{\mu }\Phi )=\rho (ϵ)D_{\mu }\Phi$. Here the gauge coupling is absorbed into the connection.