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For a gauge-field configuration, spatial symmetry means symmetry up to gauge transformation: the pullback of the fields by a spatial isometry lies in their original gauge orbit. In the anti-Hermitian convention,The potential and charged fields need not be strictly fixed in a particular gauge; their gauge-invariant observables must be fixed.

For example, the Abelian potential $A=Bxdy$ describes a uniform magnetic field. Translating $x$ by $a$ adds $Bady=d(Bay)$, a gauge change, although the potential itself changes. For an axially symmetric vortex $\Phi =f(r)e^{iN\theta }$, $A=a(r)d\theta$, rotation through $\beta$ multiplies $\Phi$ by $e^{iN\beta }$; a constant $U(1)$ transformation compensates. These illustrate why literal invariance of the representative is too strong.

For the 't Hooft-Polyakov monopole, a spatial rotation $R\in SO(3)$ has an $SU(2)$ lift $g_R$ with adjoint action $R$. The two lifts differ by a central sign and act identically on adjoint fields. The hedgehog ansatz for a monopole satisfiesThe second identity follows from invariance of ${ϵ}_{iaj}$ under simultaneous rotations of all three indices. Thus the pulled-back fields are obtained by the constant gauge transformation $g_R$; applying $g_R^{-1}$ restores the representative. In particular $|\Phi {|}^2$ and the energy density are radial even though ${\Phi }^a=f(r)x^a/r$ is not literally invariant under rotating space alone.

For the field equations, use the invariant inner product with $[e_a,e_b]={ϵ}_{abc}{e}_c$. Since $B_i^a{B}_i^a=\frac{1}{2}{F}_{ij}^a{F}_{ij}^a$, a variation $a_j=\delta A_j$, $\psi =\delta \Phi$ givesHere the scalar kinetic contribution to the gauge variation uses
$⟨D_j\Phi ,[a_j,\Phi ]⟩=⟨[\Phi ,D_j\Phi ],a_j⟩$. The Euler-Lagrange field equations for the energy actually printed areThe PDF has $\lambda (1-|\Phi {|}^2)$, without a square. This is not the usual nonnegative monopole potential with vacuum $|\Phi |=1$. If the usual $\lambda (1-|\Phi {|}^2{)}^2$ were intended, only the scalar equation would change toThe final requested proof at $\lambda =0$ is identical for either interpretation; we do not silently alter the printed energy.

To derive the covariant derivative of the radial fields, first differentiate the Higgs profile:The commutator term isAdding the two terms gives exactly

The radial eigenvalue of the supplied magnetic tensor is $2\alpha +r^2{\alpha }^2$, while that of this covariant derivative is $f^{\prime }$. Its transverse coefficients are respectively $r{\alpha }^{\prime }+2\alpha$ and $f/r+r\alpha f$. Consequently $B=-D\Phi$ reduces toEquivalently, the hedgehog monopole equations in radial profile variables with $K=1+r^2\alpha$ areThese are two equations rather than an overdetermined system: radial and transverse projections span the two invariant tensor structures. For a regular monopole in this hedgehog sector, smoothness at the origin gives $f(0)=0$, $K(0)=1$, while the normalized vacuum conditions give $f\to 1$, $K\to 0$ at infinity; solving the ODEs is not needed here.

Finally, prove that the Bogomolny monopole equations imply Yang-Mills-Higgs equations, without restricting to the radial ansatz. The three-dimensional gauge-theory Bianchi identity is $D_i{B}_i=0$. If $B_i=-D_i\Phi$, thenwhich is the scalar equation at $\lambda =0$. For the gauge equation,In the second line antisymmetry eliminates the symmetric derivative part; the curvature commutator is $[D_i,D_k]\Phi =[F_{ik},\Phi ]$. The next equality uses
${ϵ}_{ijk}{ϵ}_{ik\ell }=-2{\delta }_{j\ell }$ and $F_{ik}={ϵ}_{ik\ell }{B}_{\ell }$. These are exactly both field equations at $\lambda =0$, so every smooth Bogomolny solution satisfies them, not merely those of the hedgehog form.

For finite-energy monopoles, the associated energy argument is also transparent:The gauge-theory Bianchi identity converts $⟨B_i,D_i\Phi ⟩$ into this boundary term. At fixed asymptotic magnetic charge, $B=-D\Phi$ saturates the Bogomolny bound. This complements the local differential proof above and explains the energy-minimizing role of the first-order equations.