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Use the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$. On the Adjoint representation, the adjoint covariant derivative is $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Set $a_{\mu }=\delta A_{\mu }$ and $\chi =\delta \Phi$. ThenThe trace obeys $\mathrm{Tr}(X[Y,Z])=\mathrm{Tr}([Z,X]Y)$, and covariant integration by parts has the ordinary boundary term because the trace of a commutator is zero. Thus compactly supported variations giveThe trace pairing on $\mathfrak{su}(2)$ is nondegenerate, so the Euler-Lagrange field equations areIn particular the sign of the scalar mass term follows from the plus sign in the supplied trace Lagrangian density; it should not be guessed from a different metric convention. This is the adjoint Yang-Mills-Higgs trace variation.

For the static Bogomolny-Prasad-Sommerfield monopole, take the usual purely magnetic extension ${\partial }_0{A}_i={\partial }_0\Phi =0$ and temporal gauge $A_0=0$. At $m=0$, the equations reduce towith the $\nu =0$ equation identically zero. We now verify both spatial equations from $B_i=D_i\Phi$ without assuming them. The gauge-theory Bianchi identity gives $D_i{B}_i=0$, hence immediately $D_i{D}_i\Phi =0$. Also $F_{ji}={\epsilon }_{jik}{B}_k$, soHere ${\epsilon }_{jik}{\epsilon }_{jk\ell }=-2{\delta }_{i\ell }$. This proves that the Bogomolny equations imply all static zero-mass Yang-Mills equations and the scalar equation. The specification of $A_0$ matters: the spatial Bogomolny equations by themselves do not constrain an arbitrary extra electric potential, so the implication is for their standard purely magnetic static extension.