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The Bogomolny equations are first-order equations obtained by expressing a static energy as nonnegative squares plus a term fixed by the topological charge. Vanishing of the squares minimizes the energy in that topological sector. In particular, the solutions satisfy the full second-order field equations, since their first variation vanishes for compactly supported variations preserving the boundary data. Both monopole and vortex realizations illustrate this mechanism.

For a Yang-Mills theory with gauge group SU(2) and an adjoint Higgs field ${\Phi }^a$, take the static, purely magnetic energyHere $B_i^a=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}^a$, and $e$ is the gauge coupling. In the Bogomolny limit ${\lambda }_H=0$, retain the boundary condition $|\Phi |\to v_H>0$; it breaks SU(2) to U(1) at infinity even though the scalar potential has been removed. The normalized Higgs field on the sphere at infinity defines a map $S_{\infty }^2\to S^2$ of topological degree $n$. In a consistent magnetic orientation the asymptotic flux is $g_m=4\pi n/e$, andFor example, this equality follows by projecting the asymptotic gauge curvature on the Higgs direction: its flux is the integral of the target-sphere area form, divided by $e$.

The gauge-theory Bianchi identity $D_i{B}_i=0$ makes the cross term a divergence:For either sign $s=\pm 1$, complete the square:Choosing $s=\mathrm{sgn}n$ givesThese are the Bogomolny-Prasad-Sommerfield monopole equations. A nonzero scalar potential would add a positive term; a nontrivial smooth monopole with a core cannot generally saturate this same bound at ${\lambda }_H>0$. The Bogomolny monopole equations imply Yang-Mills-Higgs equations: the Bianchi identity already gives $D_i{D}_i\Phi =0$, and the square-completion argument gives stationarity with respect to the gauge field as well.

An explicit charge-one solution is furnished by the hedgehog ansatz for a monopole. To fix its signs, use component conventions $D_i\Phi ={\partial }_i\Phi -e{A}_i\times \Phi$ and $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i-e{A}_i\times A_j$. With $\rho =e{v}_{H}r$ and $\hat{x}=x/r$, putThe radial and transverse parts of $D_i\Phi$ are $e{v}_H^2{H}^{\prime }{\hat{x}}_i{\hat{x}}^a$ and $v_{H}HK({\delta }_{ia}-{\hat{x}}_i{\hat{x}}^a)/r$. The corresponding parts of $B_i^a$ are $(1-K^2){\hat{x}}_i{\hat{x}}^a/(e{r}^2)$ and $-v_H{K}^{\prime }({\delta }_{ia}-{\hat{x}}_i{\hat{x}}^a)/r$. Therefore $B=D\Phi$ reduces toThe Prasad-Sommerfield radial monopole solution isDirect differentiation verifies both equations. Near $\rho =0$, $K=1-{\rho }^2/6+O({\rho }^4)$ and $H=\rho /3+O({\rho }^3)$, so the apparent angular singularities disappear and the fields are smooth. At infinity $K$ decays exponentially while $H=1-1/\rho +O(e^{-2\rho })$, leaving an Abelian $1/(e{r}^2)$ magnetic field and mass $4\pi v_H/e$.

Translations of this magnetic monopole give three collective coordinates; a residual U(1) phase provides another in the framed description, where gauge transformations are fixed at infinity. Multimonopole solutions of a common sign also saturate the linear bound. Their widely separated constituents have position and phase parameters, with no static force: the magnetic repulsion and massless-Higgs attraction cancel in this limit. Relative motion is described at low speeds by the moduli-space approximation. An opposite-sign monopole pair does not solve one common sign of the Bogomolny equations and is not protected by that no-force argument.

For the Abelian Higgs model, a second realization occurs in the plane. Let $\psi$ be a complex Higgs field, $D_i={\partial }_i-ie{A}_i$, $B={\partial }_1{A}_2-{\partial }_2{A}_1$, and choose the critically coupled static energyDifferent conventions for the scalar kinetic coefficient change the numerical coefficients of both the bound and the equations. This normalization gives scalar and gauge masses $\sqrt{2}e{v}_H$, explaining the critical balance of their long-distance forces. In three spatial dimensions the planar solutions describe straight strings with this energy per unit length.

Finite energy requires $|\psi |\to v_H$ and $D_i\psi \to 0$. On a large circle the Higgs phase winds by $2\pi n$, so the magnetic flux quantization of an Abelian Higgs vortex givesWriting $J_i=\mathrm{Im}(\bar{\psi }{D}_i\psi )$, expansion and integration by parts giveThe divergence integrates to zero for the usual localized vortex boundary conditions. Thus the Bogomolny square completion for an Abelian Higgs vortex isFor $n\ge 0$ the Bogomolny vortex equations and bound areFor $n<0$, reverse the sign in both first-order equations and obtain $E=2\pi v_H^2|n|$.

For a rotationally symmetric positive vortex number, use $\psi =v_{H}f(r)e^{in\theta }$ and the physical angular component $A_{\theta }=na(r)/(er)$. The Bogomolny vortex equations becomewith $f(0)=a(0)=0$ and $f(\infty )=a(\infty )=1$. The regular core behaves as $f\sim c{r}^n$ and $a\sim e^2{v}_H^2{r}^2/(2n)$. Linearization at infinity gives exponentially decaying massive tails. Unlike the radial monopole, these planar radial profiles have no elementary closed form in general.

The general multi-vortex solution can be encoded by the Taubes equation. Away from zeros, write $\psi =v_H{e}^{h/2+i\chi }$. The first equation givesAt zeros $z_a$ of multiplicity $m_a$, the phase contributes circulation $2\pi m_a$. The second equation becomes, including these Dirac delta terms,For prescribed zeros the planar existence result gives a solution; the prescribed-zero construction of planar Abelian Higgs vortices reconstructs smooth fields, with $h=2m_a\log |x-z_a|+O(1)$ near each zero. Uniqueness is visible directly: for two solutions with the same zeros, their nonsingular difference $w$ satisfies $\Delta w=2e^2{v}_H^2(e^{h_1}-e^{h_2})$. Multiplying by $w$ and integrating giveshence $w=0$. The Abelian Higgs vortex moduli space has arbitrary unordered positions of the $n$ zeros, giving $2n$ real position parameters. The saturated energy is independent of their positions, so critically coupled vortices have no static interaction energy. Away from critical coupling this cancellation is lost.

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.

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.