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This gauge-invariant higher-derivative density has equationThe extra commutator comes from varying the connection inside the adjoint covariant derivative: $\delta (D_{\mu }{F}_{\nu \rho })=D_{\mu }\delta F_{\nu \rho }+[\delta A_{\mu },F_{\nu \rho }]$. Two uses of gauge-covariant integration by parts give the first term, and Killing-form invariance gives the second. In an abelian limit it reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$. Keeping the covariant derivative order is essential because their commutator is curvature.

Covariance of the fundamental gauge covariant derivative requires $D_{\mu }^{f\prime }(g\chi )=g{D}_{\mu }^f\chi$ for every column $\chi$. Expanding the left side and cancelling $g{\partial }_{\mu }\chi$ gives $({\partial }_{\mu }g)+A_{\mu }^{\prime }g=g{A}_{\mu }$. Therefore the necessary and sufficient Yang-Mills gauge transformation isIn operator notation $D_{\mu }^{f\prime }=g{D}_{\mu }^f{g}^{-1}$. Expanding its commutator on an arbitrary column shows $[D_{\mu }^f,D_{\nu }^f]\chi =F_{\mu \nu }\chi$: the second derivatives cancel, as do the terms containing a single derivative of $\chi$. Conjugating this identity proves the gauge field strength transformationThus the gauge field strength transforms homogeneously even though the potential has an inhomogeneous derivative term.

For the adjoint scalar field, write $B_{\mu }=({\partial }_{\mu }g)g^{-1}$ and ${\Phi }^{\prime }=g\Phi g^{-1}$. Differentiating the inverse matrix givesThe extra terms cancel, proving $D_{\mu }^{\prime }{\Phi }^{\prime }=g(D_{\mu }\Phi )g^{-1}$ for the adjoint covariant derivative. Its curvature is obtained by another direct expansion:where the Jacobi identity combines the last two terms.

For the equations of motion take compactly supported variations, and assume $e\ne 0$, as required by the Lagrangian coefficient. Invariance of the Killing form impliesIndeed, the two connection terms sum to $\kappa ([A_{\mu },X],Y)+\kappa (X,[A_{\mu },Y])=0$. This is the gauge-covariant integration by parts identity. The needed variations areAntisymmetry of the gauge field strength combines its two variation terms. Applying gauge-covariant integration by parts to the action $S=\int \mathcal{L}{d}^{4}x$ givesHere $\kappa ([\delta A_{\nu },\Phi ],D^{\nu }\Phi )=\kappa (\delta A_{\nu },[\Phi ,D^{\nu }\Phi ])$ fixes the sign of the scalar current. Since the Killing form is nondegenerate on a semisimple Lie algebra, independent variations give the coupled Yang-Mills equations and scalar equationEquivalently, $D_{\mu }{F}^{\mu \nu }=e^2[D^{\nu }\Phi ,\Phi ]$. These signs use the stated Minkowski metric, with upper spatial derivatives $D^i=-D_i$.

It remains to verify every equation for the static Bogomolny equations. The hypotheses imply $F_{0i}=0$ and $D_0\Phi =0$, so the temporal gauge equation is automatically zero. The gauge-theory Bianchi identityfollows, for example, from the Jacobi identity for the fundamental gauge covariant derivatives, using their commutator $F_{ij}$. Define $B_i=\frac{1}{2}{\epsilon }_{ijk}{F}_{jk}$. Contracting the gauge-theory Bianchi identity with ${\epsilon }_{ijk}$ gives $D_i{B}_i=0$. The Bogomolny equations say $B_i=e{D}_i\Phi$, so $e\ne 0$ implies $D_i{D}_i\Phi =0$. In the static configuration $D_{\mu }{D}^{\mu }\Phi =-D_i{D}_i\Phi$, verifying the scalar equation.

For the spatial gauge equation, use $F^{ij}=F_{ij}$ and $D^j\Phi =-D_j\Phi$. The Bogomolny equations, the curvature commutator proved above, and the contraction of two Levi-Civita symbols givesince ${\epsilon }_{ijk}{\epsilon }_{ik\ell }=-2{\delta }_{j\ell }$. This is exactly $D_i{F}^{ij}+e^2[\Phi ,D^j\Phi ]=0$. Thus the static first-order Bogomolny system satisfies all four gauge equations and the scalar equation, with no extra ansatz or boundary assumption needed for this local implication. This is the mechanism by which Bogomolny monopole equations imply Yang-Mills-Higgs equations.

Choose the convention ${\psi }^{\prime }=g\psi$ for a field in the defining representation and $D_{\mu }={\partial }_{\mu }+A_{\mu }$. Requiring $D_{\mu }^{\prime }{\psi }^{\prime }=g{D}_{\mu }\psi$ gives the Yang-Mills gauge transformationThe gauge field strength is the curvature $[D_{\mu },D_{\nu }]=F_{\mu \nu }$, soFor an adjoint field $Y$, the adjoint covariant derivative is $D_{\mu }Y={\partial }_{\mu }Y+[A_{\mu },Y]$, and direct substitution gives $D_{\mu }^{\prime }{Y}^{\prime }=g(D_{\mu }Y)g^{-1}$. Consequently $D_{\mu }{F}_{\nu \rho }$ transforms by conjugation too.

The Killing-form invariance under a Lie-group adjoint action gives $\kappa (gY{g}^{-1},gZ{g}^{-1})=\kappa (Y,Z)$. Its proof is that the adjoint matrices are conjugated by the same invertible map, leaving their trace product unchanged. With $s=\kappa (F_{\mu \nu },F^{\mu \nu })$, both $s^p$ and $\kappa (D_{\mu }{F}_{\nu \rho },D^{\mu }{F}^{\nu \rho })$ are therefore gauge invariant.

For their field equations, take a compactly supported variation $a_{\mu }=\delta A_{\mu }$. Curvature varies byThe invariant form permits gauge-covariant integration by parts, because ${\partial }_{\mu }\kappa (Y,Z)=\kappa (D_{\mu }Y,Z)+\kappa (Y,D_{\mu }Z)$. Antisymmetry of $F$ givesIntegrating, the power of the Yang-Mills invariant gives the equationFor $p=1$ this recovers the Yang-Mills equations. For larger $p$, keep the displayed product form, including at $s=0$.

For the derivative-squared action, write $B_{\mu \nu \rho }=D_{\mu }{F}_{\nu \rho }$. Varying the derivative as well as the curvature givesIts first variation isIntegrate the first term once, substitute $\delta F_{\nu \rho }=D_{\nu }{a}_{\rho }-D_{\rho }{a}_{\nu }$, and integrate again. It becomes $4\int \kappa (a_{\nu },D_{\mu }{D}^2{F}^{\mu \nu })d^{4}x$, where $D^2=D_{\lambda }{D}^{\lambda }$. Invariance converts the second term into $2\int \kappa (a_{\nu },[F_{\rho \sigma },D^{\nu }{F}^{\rho \sigma }])d^{4}x$. Nondegeneracy of the Killing form and arbitrary variations yield the derivative-squared Yang-Mills action equationThe derivative order matters because covariant derivatives generally do not commute. In an abelian limit the commutator term vanishes and the equation reduces to $\square {\partial }_{\mu }{F}^{\mu \nu }=0$, a useful check on the fourth-order equation for the potential.

For a semisimple gauge algebra and positive integer $p$, this gauge-invariant density yields $D_{\mu }(s^{p-1}{F}^{\mu \nu })=0$. The curvature variation $\delta F_{\mu \nu }=D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$ gives $\delta s=4\kappa (D_{\mu }\delta A_{\nu },F^{\mu \nu })$, and gauge-covariant integration by parts proves the equation. More generally a density $f(s)$ gives $D_{\mu }(f^{\prime }(s)F^{\mu \nu })=0$. The product form retains the equation at zeros of $s$; dividing by $s$ can discard solutions.