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For zero scalar potential, the static field equations are $D_i{D}_i\Phi =0$ and $D_i{F}_{ij}=[\Phi ,D_j\Phi ]$. If $B_i=-D_i\Phi$, the gauge-theory Bianchi identity gives the first equation. For the second,using $F_{ik}={ϵ}_{ik\ell }{B}_{\ell }$ and ${ϵ}_{ijk}{ϵ}_{ik\ell }=-2{\delta }_{j\ell }$. Thus the first-order equations imply the full second-order equations locally, independently of the radial ansatz.

For $F=dA+A\wedge A$, the gauge-theory Bianchi identity is $DF=0$ and the variation is $\delta F=D\delta A$. The graded product rule for the covariant exterior derivative gives the displayed boundary integral. Thus a constant-coefficient Yang-Mills theta term leaves the bulk Yang-Mills equations unchanged, but can change the boundary variational condition. Its four-form density is exactly gauge-invariant by conjugation of $F$ and cyclicity of the matrix trace, even if the gauge parameter is nonzero on the boundary. Gauge changes of a local Chern-Simons 3-form must not be confused with gauge changes of this globally defined curvature density.

Apply $D^{\rho }$ to the gauge-theory Bianchi identity and commute it through the other derivatives using $[D_{\mu },D_{\nu }]X=[F_{\mu \nu },X]$. The Yang-Mills equations remove the differentiated divergences. The two remaining commutator terms both equal $[F_{\mu }{}^{\rho },F_{\nu \rho }]$, proving the displayed nonlinear covariant wave equation for the Yang-Mills field strength on flat spacetime.

// Target: special-relativity.bigb

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 an SU(2) connection one-form $A=A_{\mu }d{x}^{\mu }$ with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, so $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Absorb the gauge coupling into $A$, and write its gauge curvature as $F=dA+A\wedge A$. The Yang-Mills gauge transformation convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.

In four dimensions the Second Chern form and Second Chern number of the associated fundamental rank-two bundle areHere $M$ is a closed oriented four-manifold. This sign follows by expanding the total Chern class $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. The first Chern class vanishes for SU(2); the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual instanton number isBoth sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.

The form is gauge-invariant by conjugation and cyclicity of the matrix trace. The gauge-theory Bianchi identity $D_{A}F=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:On a closed $M$ the integral of this exact form vanishes. This is Chern-Weil connection transgression, showing that $k_C$ depends on the bundle's topology rather than on a particular gauge field.

Locally the same four-form is an exterior derivative. The Chern-Simons 3-formsatisfiesFor verification, differentiating gives $\mathrm{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\mathrm{tr}{A}^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\mathrm{tr}{A}^4=0$. This local exactness does not make the Second Chern number zero on every closed manifold: on a nontrivial bundle the gauge potentials and ${\omega }_3$ are only patchwise defined.

For the usual finite-action sector on ${\mathbb{R}}^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The Maurer-Cartan equation gives $d(h^{-1}dh)=-(h^{-1}dh{)}^2$, henceWith the outward boundary orientation, Stokes theorem therefore givesThese are opposite choices of the winding number, and either becomes the topological degree of $h$ after specifying the orientation of the target SU(2) sphere. Thus integrality also follows from ${\pi }_3(SU(2))=\mathbb{Z}$. Smooth deformations preserving the asymptotic sector cannot change it.

The positive Euclidean Yang-Mills action isDefine $\Vert X{\Vert }^2=-\int \mathrm{tr}(X\wedge \ast X)$. Since the Hodge star has square one on two-forms in Euclidean dimension four,Choosing $s=\mathrm{sgn}Q$ gives the Yang-Mills instanton Bogomolny boundThe self-dual Yang-Mills equations and their anti-self-dual version imply the Yang-Mills equations $D_A\ast F=0$ by the Bianchi identity. Their finite-action solutions are Yang-Mills instantons.

For a concrete self-dual example, the BPST instanton with center $a$ and size $\rho >0$ haswhere ${\eta }_{\mu \nu }^a$ are self-dual 't Hooft symbols. Their contraction is ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$. Thus $F_{\mu \nu }^a{F}_{\mu \nu }^a=192{\rho }^4/(|x-a{|}^2+{\rho }^2{)}^4$, andThis gives $S=8{\pi }^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the instanton moduli space.

On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the Chern-Simons number of a gauge field, in the instanton-charge conventionIt is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta =h^{-1}dh$ and $\bar{\theta }=dh{h}^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta =-{\theta }^2$ and graded trace cyclicity, gives the gauge change of the Chern-Simons three-formThe mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,Thus $N_{\mathrm{CS}}\mathrm{mod}\mathbb{Z}$, or $\exp (2\pi i{N}_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.

For a Yang-Mills vacuum on ${\mathbb{R}}^3$, zero magnetic energy requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the theta vacuum. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact no static finite-energy lump in three-dimensional pure Yang-Mills theory exists with the usual decay conditions: $A_i^{(s)}(x)=s{A}_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.

Finally, on an oriented spacetime slab $[t_{-},t_{+}]\times \Sigma$, with no side-boundary contribution, the transgression identity givesA Yang-Mills instanton therefore interpolates between vacuum representatives whose Chern-Simons numbers differ by its integer charge. In a four-dimensional quantum theory a Yang-Mills theta term weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi \ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.

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.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

Take a local gauge transformation $g(x)$ acting on the adjoint field by ${\Phi }^{\prime }=g\Phi g^{-1}$. The connection convention $D_{\mu }={\partial }_{\mu }+[A_{\mu },\cdot ]$ is compatible withIndeed, differentiating ${\Phi }^{\prime }$ gives the extra commutator $[({\partial }_{\mu }g)g^{-1},{\Phi }^{\prime }]$, exactly cancelled by the inhomogeneous term in the transformed gauge potential.

For a direct proof of the Yang-Mills gauge transformation of curvature, put ${\Omega }_{\mu }=({\partial }_{\mu }g)g^{-1}$ and $B_{\mu }=g{A}_{\mu }{g}^{-1}$. Differentiation givesSubstituting $A_{\mu }^{\prime }=B_{\mu }-{\Omega }_{\mu }$ into the Yang-Mills field strength cancels all terms involving $\Omega$ and leavesThis direct calculation also applies when the adjoint action has a kernel.

Expand $D_{\mu }{F}_{\nu \rho }+D_{\nu }{F}_{\rho \mu }+D_{\rho }{F}_{\mu \nu }$. The mixed second derivatives of $A$ cancel in pairs. The first-derivative terms from differentiating $[A_{\nu },A_{\rho }]$ cancel the terms from $[A_{\mu },{\partial }_{\nu }{A}_{\rho }-{\partial }_{\rho }{A}_{\nu }]$. The remaining sum isby the Jacobi identity. Thus the gauge-theory Bianchi identity is

For the variation of the Yang-Mills action, write $a_{\mu }=\delta A_{\mu }$. Differentiating the curvature gives $\delta F_{\mu \nu }=D_{\mu }{a}_{\nu }-D_{\nu }{a}_{\mu }$. Invariance of the Killing form impliesThis is invariant integration by parts for a gauge covariant derivative. Using symmetry of the Killing form, antisymmetry of $F$, and compactly supported variations,Since the Killing form of a semisimple Lie algebra is nondegenerate and the variations are arbitrary, the Yang-Mills equations are

For the constant Yang-Mills theta term, symmetry under exchanging the two antisymmetric index pairs givesup to the boundary variation of the Yang-Mills theta term. Contracting the gauge-theory Bianchi identity with ${ϵ}^{\mu \nu \rho \sigma }$ makes the last integrand vanish. Therefore a constant theta term does not alter the bulk Yang-Mills equations: $D_{\mu }{F}^{\mu \nu }=0$. Its variation is a boundary term, so this conclusion uses compact support or boundary conditions that remove that term.

Finally apply $D^{\rho }$ to the gauge-theory Bianchi identity, obtainingThe adjoint covariant derivative obeys $[D_{\alpha },D_{\beta }]X=[F_{\alpha \beta },X]$. Commute $D^{\rho }$ past the other derivative. The differentiated divergences vanish by the Yang-Mills equations, leavingIn the last line the second commutator needs its sign tracked carefully: directly,Thus the simplified sum gives the covariant wave equation for the Yang-Mills field strengthAll spacetime derivatives and index manipulations here use the fixed flat metric.

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 $D^i=-D_i$.