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Substituting $\delta A_{\mu }=g^{-1}{\partial }_{\mu }\alpha -i[A_{\mu },\alpha ]$ into the curvature and collecting terms gives the covariant transformationEquivalently, $\delta F_{\mu \nu }^a=f^{abc}{F}_{\mu \nu }^b{\alpha }^c$ in the stated conventions. Since $F_{\mu \nu }^a{F}^{\mu \nu ,a}$ is proportional to $\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$, cyclicity of the matrix trace givesThus the Yang-Mills action is gauge-invariant.

The Yang-Mills action is BRST invariant because its field strength transforms covariantly, and the gauge-fixing contribution is BRST exact. Nilpotence therefore givesThe gauge-fixing functional itself is generally not closed: $sL[A]=L[Dc]$. Applying the graded Leibniz rule givesThe Nakanishi-Lautrup field $B^a$ is auxiliary. Its algebraic equation $B^a=L[A^a]/\xi$ turns the middle terms into $L[A{]}^2/(2\xi )$, while the final term is the Faddeev-Popov ghost field action.

Because ${}^{\star }{}^{\star }=1$, the exterior algebra of two-forms has the orthogonal decompositionwhere ${}^{\star }H=H$ for a self-dual differential form and ${}^{\star }G=-G$ for an anti-self-dual differential form. The Hodge star is self-adjoint, soIt follows that

For an $SU(n)$ connection, use the positive normand define the Second Chern number byOrthogonality givesTherefore the Euclidean Yang-Mills actionobeys the Yang-Mills instanton Bogomolny boundEquality holds precisely when $F_{-}=0$ or $F_{+}=0$, according to the sign of $k$; these are the self-dual and anti-self-dual Yang-Mills instantons. Other trace and orientation conventions may reverse $k$ but leave the absolute-value bound unchanged.