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In three space dimensions, rescale $A_i(x)$ to $\lambda A_i(\lambda x)$ and $\Phi (x)$ to $\Phi (\lambda x)$. The gauge curvature then scales with two derivatives, and the adjoint covariant derivative of the Higgs field with one. Magnetic, Higgs-gradient and potential energies give the displayed formula. Stationarity requires $V_B=V_D+3V_U$, which permits nonzero energies. In the Bogomolny-Prasad-Sommerfield monopole limit $V_U=0$, the Bogomolny equations give $V_B=V_D$.

If a four-dimensional Euclidean connection is independent of one coordinate and that connection component is renamed $\Phi$, the self-dual Yang-Mills equation reduces to the three-dimensional Bogomolny-Prasad-Sommerfield monopole equation $B_i=\pm D_i\Phi$.

For the trace inner product and adjoint covariant derivative, compatibility gives $\Delta |\Phi {|}^2=2|D\Phi {|}^2+2⟨\Phi ,D_i{D}_i\Phi ⟩$. The gauge-theory Bianchi identity and $B_i=\pm D_i\Phi$ imply $D_i{D}_i\Phi =0$. Therefore $\Delta |\Phi {|}^2=2|D\Phi {|}^2$, equal to the energy density $|B{|}^2+|D\Phi {|}^2$ on a Bogomolny-Prasad-Sommerfield monopole. If the energy density is instead defined with an overall factor $1/2$, this Laplacian is twice that density.

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.

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.

Use anti-Hermitian $\mathfrak{su}(2)$ matrices, with invariant positive inner product $⟨X,Y⟩=-2\mathrm{tr}(XY)$. On an adjoint-valued field $\Xi$, define the adjoint covariant derivative and gauge curvature byEquivalently, $F=dA+A\wedge A$ and $[D_i,D_j]\Xi =[F_{ij},\Xi ]$. Put $B_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, with ${ϵ}_{123}=1$. The monopole Bogomolny equations are $B_i=D_i\Phi$.

For a smooth SU(2) gauge transformation $g(x)$, useDifferentiating a transformed adjoint field shows $D_i^g(g\Xi g^{-1})=g(D_i\Xi )g^{-1}$: the derivatives of $g$ cancel the inhomogeneous term in $A_i^g$. Applying the commutator of these derivatives gives $F_{ij}^g=g{F}_{ij}{g}^{-1}$, and therefore $B_i^g=g{B}_i{g}^{-1}$. Both sides of $B_i=D_i\Phi$ transform identically, establishing gauge invariance of the solution condition. Individual matrix components are covariant, rather than invariant numbers.

The monopole boundary condition is a nonzero asymptotic Higgs norm,together with finite three-dimensional Yang-Mills-Higgs energy. Standard monopole asymptotics have $D\Phi =O(r^{-2})$ and $F=O(r^{-2})$. The normalized Higgs field at infinity defines a map $S_{\infty }^2\to SU(2)/U(1)\simeq S^2$, whose topological degree is the magnetic charge. A nonzero charge gives a genuine non-Abelian magnetic monopole; the nonzero Higgs boundary value alone also allows the zero-charge vacuum. Thus the relevant nontrivial solutions are Bogomolny-Prasad-Sommerfield monopoles in the zero-Higgs-potential limit.

Now form the four-dimensional connection with $\tau$-independent components. Its gauge curvature hasThe self-duality sign requires an orientation. Choose the Euclidean metric and orientation $d\tau \wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$. For cyclic $(i,j,k)$, the Hodge star operator satisfiesUsing $F_{jk}={ϵ}_{jki}{B}_i$ and $d{x}^i\wedge d\tau =-d\tau \wedge d{x}^i$, the curvature isEach term has negative star eigenvalue, soThis proves the Anti-self-dual Yang-Mills equations in the stated convention. The orientation of a monopole lift matters: with $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d\tau$ instead, the same connection is self-dual. With that opposite orientation, the anti-self-dual lift would use $-\Phi d\tau$.

To fix the action normalization, take the Euclidean Yang-Mills actionBoth sums are over ordered index pairs. The component identity
${\sum }_{j,k}|F_{jk}{|}^2=2{\sum }_i|B_i{|}^2$ givesHence the four-dimensional action density in terms of the three-dimensional curvature isA different overall action normalization rescales all three equal expressions together.

A Yang-Mills instanton is a smooth finite-Euclidean-action self-dual or anti-self-dual connection in pure Yang-Mills theory, with the usual asymptotic pure-gauge behavior. Physically the term generally refers to a nontrivial field; its asymptotic gauge winding labels an instanton number. The Bianchi identity $D\mathcal{F}=0$, together with $\ast \mathcal{F}=-\mathcal{F}$, implies $D\ast \mathcal{F}=0$, so the first-order equations already imply the second-order Yang-Mills equations. Finite action is a separate requirement.

Here ${\mathcal{L}}_4$ is independent of $\tau$. If the monopole has nonzero curvature, the density is positive on an open set in ${\mathbb{R}}^3$, and integrating its infinite $\tau$-extent givesFor a finite-energy monopole this can also be written as its positive action per unit length times ${\int }_{\mathbb{R}}d\tau$. Thus a nontrivial monopole lift is not a Yang-Mills instanton on ${\mathbb{R}}^4$. The flat solution $A_i=0$, $\Phi ={\Phi }_0$ constant is the zero-action exception: it is a vacuum, or a trivial charge-zero instanton if that terminology includes flat connections. It is not a counterexample to the nontrivial-monopole assertion.

More generally, finite Yang-Mills action forbids nontrivial translation symmetry, even when the translation preserves a connection only up to gauge. Such a symmetry would preserve its positive gauge-invariant action density. A nonflat field has a ball on which this density is bounded below by some positive constant. Repeated translations by a nonzero symmetry vector produce infinitely many disjoint copies of a sufficiently small such ball, all with the same positive action contribution, contradicting finiteness. Continuous translation invariance is a special case. Therefore a nonflat ASDYM instanton is localized in all four Euclidean directions and has no nonzero translational symmetry. Translating an instanton still produces another instanton with a shifted center; these position parameters belong to the instanton moduli space, rather than being invariances of one localized solution.

Let $T=\frac{1}{2}\int |\nabla \varphi {|}^2{d}^{n}x$ and $W=\int U(\varphi )d^{n}x$. For a sufficiently regular finite-energy field configuration with vacuum asymptotics, the scaled field ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$ stays in the same boundary sector. Changing variables gives the Derrick scaling formulaA static solution must be stationary under this variation, so the Derrick virial identity isFor $n>2$, both terms are nonnegative, so both vanish and the field is constant at a vacuum. Hence there are no nonconstant smooth finite-energy static solitons of this scalar type for $n>2$. For $n=1$ the identity is $T=W$, allowing kinks. For $n=2$ it requires $W=0$: a nonzero potential contribution excludes a static soliton, but the scaling identity alone does not exclude a purely derivative energy.

There is a useful refinement specific to the unconstrained flat target ${\mathbb{R}}^l$. If $U$ is differentiable and nonnegative, $W=0$ implies $U(\varphi (x))=0$ and ${\nabla }_{\varphi }U=0$ on the image. The static Euler-Lagrange field equations then give $\Delta \varphi =0$. Each first derivative is an entire harmonic function in L2 space. Its mean-value bound over a disk of radius $R$ is at most its total squared norm divided by $\pi R^2$; letting $R\to \infty$ shows that derivative is zero. Thus even the exceptional two-dimensional case has only constant smooth finite-energy solutions into this flat target. This two-dimensional flat-target Derrick obstruction uses the field equations in addition to the scale identity.

The Yang-Mills theory with a Higgs field escapes the original scaling because the gauge curvature is quadratic in spatial scaling. In three dimensions use $A_i^{\lambda }(x)=\lambda A_i(\lambda x)$ and ${\Phi }^{\lambda }(x)=\Phi (\lambda x)$. Then $F_{ij}^{\lambda }={\lambda }^2{F}_{ij}(\lambda x)$ and $D_i^{\lambda }{\Phi }^{\lambda }=\lambda D_i\Phi (\lambda x)$. Writing the magnetic, Higgs-gradient and Higgs-potential energies as $V_B,V_D,V_U$, the Derrick scaling of Yang-Mills-Higgs energy givesat stationarity. These terms can balance. In particular, the zero-potential Bogomolny-Prasad-Sommerfield monopole has $V_B=V_D\ne 0$.

For a nonlinear sigma model in two dimensions, the harmonic map energy is purely quadratic in first derivatives and is scale invariant: $V_{\lambda }=V$. Its target is a curved manifold, not the unconstrained Euclidean target of the refinement above. Its equation is the nonlinear harmonic-map equation, not componentwise $\Delta \varphi =0$. The Derrick theorem therefore gives no obstruction. Nonconstant finite-energy maps into $S^2$, as in the O3 nonlinear sigma model, provide examples with nonzero degree.

For the complex theory, regard $\varphi$ and $\bar{\varphi }$ as independent variables in the variation, and write $s=|\varphi {|}^2$, $U=U(s)$. Variation in $\bar{\varphi }$ givesThusThe factor 2 follows from the printed $1/2$ normalization of the complex kinetic term; equivalently vary its two real components.

A non-topological soliton is a localized finite-energy persistent field solution with no protecting winding or other nontrivial topological sector. In this complex theory a Q-ball has a stationary spatial profile and a rotating phase,with real $f$. It is not static as a field even though its energy density is time-independent. The global phase symmetry gives the Noether chargeThe profile equation is $-\Delta f+2U^{\prime }(f^2)f-{\omega }^{2}f=0$. Its energy is $T+W+{\omega }^{2}I/2$. To vary the size at fixed charge one must also vary the frequency, because scaling $f$ at fixed $\omega$ would change $Q$. Eliminating $\omega =Q/I$ gives charge-constrained scalar-field scaling:The new positive-power term can balance the other terms for $n>1$, including $n>2$, so the static Derrick theorem does not rule out these solutions. Equivalently, at fixed frequency the profile extremizes $T+\int [U(f^2)-{\omega }^2{f}^2/2]$, whose effective potential need not be nonnegative. This explains both the time dependence and the charge constraint behind the evasion; a complex field by itself is not enough.

For an explicit example take the sextic unique-vacuum Q-ball potentialIt is nonnegative, since $U(s)=s[(s-1/2{)}^2+3/4]$, and its only vacuum is zero. Its small-field mass squared is $2U^{\prime }(0)=2$, while ${\min }_{s>0}2U(s)/s=3/2$. It supports localized Q-balls, for example radial ones in three dimensions with $3/2<{\omega }^2<2$. In that range the profile potential is positive quadratic near zero but negative somewhere away from zero, permitting a localized nontrivial profile. This example uses a conserved phase charge and no vacuum winding.

The curvature of $D=d+A$ isor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }+[A_{\mu },A_{\nu }]$. The covariant exterior derivative and the Jacobi identity give the Bianchi identity

Since the Hodge star operator satisfies ${\ast }^2=1$ on Euclidean two-forms,Using the opposite choice of sign as well givesEquality holds precisely for a self-dual or anti-self-dual Yang-Mills instanton, with the sign selected by that of $k$.

Now assume ${\partial }_0{A}_{\mu }=0$ and write $\Phi =A_0$. ThenThe spatial Bianchi identity and the self-duality equations becomewhere the upper four-dimensional sign gives the first displayed reduced sign under the orientation used here. Consequently

Gauge invariance of the inner product givesFor $w=1-|\Phi {|}^2$,Since $w\to 0$ at infinity, the weak maximum principle for elliptic operators excludes a negative interior minimum. Hence

With $d^{4}x=d{x}^0{d}^{3}x$, direct decomposition givesBecause $D_i{B}_i=0$,so the Pontryagin density reduces to the surface charge $4\pi N$ per unit $x^0$. The conventional three-dimensional energyobeys the Bogomolny bound $E_3\ge 4\pi |N|$, saturated by the Bogomolny-Prasad-Sommerfield monopole equation. The four-dimensional action density per unit $x^0$ is $2E_3$; signs relating $k$ and $N$ depend on the self-duality and orientation convention.

For the hedgehog ansatz for a monopole in the question, direct differentiation with $[T_a,T_b]=-{ϵ}_{abc}{T}_c$ givesOn a sphere of radius $r$,ThereforeThe boundary conditions $f\to 1$ and $r^2\alpha \to 1$ make the integrand equal to $r^{-2}+o(r^{-2})$. Thus

For a Bogomolny-Prasad-Sommerfield monopole, the dimensionless radial Bogomolny equations are $dK/d\rho =-KH$ and $dH/d\rho =(1-K^2)/{\rho }^2$. The displayed profiles solve both equations: differentiating $\rho /\sinh \rho$ gives $K(1/\rho -\coth \rho )=-KH$, while differentiating $H$ gives $1/{\rho }^2-{\mathrm{csch}}^2\rho =(1-K^2)/{\rho }^2$. At the origin $K=1-{\rho }^2/6+O({\rho }^4)$ and $H=\rho /3+O({\rho }^3)$, giving a regular core. At infinity $K$ decays exponentially and $H=1-1/\rho +O(e^{-2\rho })$, giving the vacuum Higgs magnitude and a long-range Abelian magnetic field.

A smooth finite-energy monopole in an $SU(2)$ Yang-Mills theory with an adjoint Higgs field uses the normalized Higgs direction at infinity to define a map $S_{\infty }^2\to S^2$. Its nonzero degree supplies the magnetic topological sector. Unlike a singular Abelian Dirac monopole, the non-Abelian gauge and Higgs fields can have a regular core. In the zero-potential limit the Bogomolny-Prasad-Sommerfield monopole equations are $B_i=\pm D_i\Phi$.