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For the adjoint covariant derivative $D_i={\partial }_i+[A_i,\cdot ]$, the gauge curvature isThe matrix-valued wedge product gives one commutator in each $d{x}^i\wedge d{x}^j$ coefficient; $[D_i,D_j]\Phi =[F_{ij},\Phi ]$ follows directly. The given trace defines a real positive inner product on the anti-Hermitian algebra $\mathfrak{su}(2)$.

The additional terms in $⟨D_j\Phi ,\Psi ⟩+⟨\Phi ,D_j\Psi ⟩$, compared with ordinary differentiation, areThus trace cyclicity proves the compatibility identityWith compactly supported variations, this gives covariant integration by parts without boundary contributions.

Use $F_{ij}={ϵ}_{ijk}{B}_k$, so $|B{|}^2=\frac{1}{2}{\sum }_{i,j}|F_{ij}{|}^2$. Write the independent variations as $\delta A_i=a_i$ and $\delta \Phi =\chi$. The curvature and Higgs derivative variations areThe variation of the Yang-Mills theory magnetic energy isFor the Higgs field derivative energy, trace cyclicity also gives $⟨X,[a,\Phi ]⟩=⟨[\Phi ,X],a⟩$. HenceArbitrariness of the variations proves the Euler-Lagrange field equationsThe gauge equation can equivalently be written $(D\times B{)}_j+[\Phi ,D_j\Phi ]=0$, where $(D\times B{)}_j={ϵ}_{jik}{D}_i{B}_k$. This equivalent form also checks the sign of the commutator source.

Now suppose the positive-sign Bogomolny equations $B_i=D_i\Phi$ hold. The gauge-theory Bianchi identity $D_i{F}_{jk}+D_j{F}_{ki}+D_k{F}_{ij}=0$ implies ${\sum }_i{D}_i{B}_i=0$, and therefore ${\sum }_i{D}_i{D}_i\Phi =0$. For the gauge equation,Consequently $(D\times B{)}_i+[\Phi ,D_i\Phi ]=[B_i,\Phi ]+[\Phi ,B_i]=0$. Thus the first-order Bogomolny equations imply both second-order field equations. The negative-sign convention works as well, with both occurrences of that sign tracked consistently.

Finally differentiate the norm twice, using the trace compatibility identity at each step:The second term vanishes, and $|B{|}^2=|D\Phi {|}^2$ for the Bogomolny equations. Thus the Higgs norm identity for a Bogomolny monopole givesThere is no extra factor 2 multiplying $v$: the printed energy density already contains both equal terms without an overall $1/2$.