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An Abelian gauge theory has a commutative gauge group. Its gauge field strength has no commutator term, so the pure gauge Lagrangian has no cubic or quartic gauge-boson self-interactions.

For $\mathcal{L}=-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$ with $\mathcal{N}>0$, rescale ${\mathcal{A}}^a=\sqrt{\mathcal{N}}{A}^a$ and $g_c=g/\sqrt{\mathcal{N}}$. The gauge field strength then has canonical component normalization. Masses inferred from matter couplings must use $g_c$, not an unnormalized coefficient. A matrix-trace convention $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ fixes this factor.

Write a matrix-valued gauge field ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ in the fundamental representation. Charge conjugation sends it to $-{\mathcal{A}}_{\mu }^T$ because reordering the fermions transposes both the spinor and internal matrices. The identity $[A^T,B^T]=-[A,B{]}^T$ then sends the gauge field strength to $-{\mathcal{F}}_{\mu \nu }^T$. This distinction prevents treating all color components as individually C-odd.

For a matrix-valued gauge field $A$, let $F=dA+A\wedge A$ be its gauge field strength; products below combine matrix multiplication with the wedge product of differential forms. The five-form can be written $Q_5=\mathrm{Tr}(A(dA{)}^2+\frac{3}{2}{A}^{3}dA+\frac{3}{5}{A}^5)$. Applying the exterior derivative, using its graded product rule and the graded cyclic property of the matrix trace, gives $d{Q}_5=\mathrm{Tr}(F^3)$. The six-form is an invariant polynomial in the gauge curvature, while its local potential depends on the connection.

If $D_{\mu }\Phi =0$ in the SO(3) vector Higgs model, antisymmetry of the gauge potential implies ${\partial }_{\mu }|\Phi {|}^2=0$. On a local patch with nonzero constant norm, unitary gauge aligns the field with a fixed axis. The connection and gauge field strength then take values in the axis stabilizer; curvature can remain nonzero. Global alignment can require several bundle patches.

Gauge boson self-interactions arise from the quadratic-field term in a non-Abelian gauge field strength. Squaring that field strength in the Lagrangian yields cubic and quartic pure-gauge interaction vertices. Higher-point scattering can occur by connecting these vertices, even without a corresponding fundamental higher-valence vertex.

A gauge field is a connection associated with a local symmetry in a gauge theory. Its gauge potential defines the internal-symmetry contribution to a covariant derivative. For an Abelian gauge field, the gauge field strength is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$.

A classical field theory is a gauge theory when its description has a local redundancy represented by gauge transformations. A gauge field defines the covariant derivative that compares matter fields at different points, and its gauge field strength measures the curvature of that comparison. Electromagnetism is the familiar Abelian example; Yang-Mills theory generalizes it to a non-Abelian internal group. Both classical equations and their quantum field theory can have this structure. Physically equivalent gauge descriptions must give the same observables, even though the gauge potential changes.

In $\mathcal{L}=-g_{ab}{F}_{\mu \nu }^a{F}^{b\mu \nu }/4$, only the symmetric part of $g$ contributes. A constant symmetric bilinear form gives local gauge invariance precisely when it is an invariant bilinear form on a Lie algebra, because the gauge field strength transforms in the Adjoint representation. A nondegenerate kinetic term requires a nondegenerate bilinear form; positivity of the Hamiltonian is an additional condition. For a compact simple real Lie algebra, $-\kappa$ is positive definite, and a positive multiple supplies the usual positive-energy convention with Minkowski signature $(+---)$.

An adjoint scalar has adjoint covariant derivative $D_{\mu }\varphi ={\partial }_{\mu }\varphi +[A_{\mu },\varphi ]$. A symmetric invariant bilinear form on a Lie algebra contracts the gauge field strength and scalar kinetic terms to give gauge invariance. On a compact real form the positive internal metric is $-\kappa$, because the Killing form itself is negative definite; complex scalar components require the associated Hermitian pairing. The Killing form is degenerate on Abelian factors, where another invariant metric is needed for a nondegenerate kinetic term.

For the convention $D_{\mu }={\partial }_{\mu }+g{A}_{\mu }$, a gauge group transformation $\Phi \mapsto R\Phi$ acts on the gauge potential by $A_{\mu }\mapsto R{A}_{\mu }{R}^{-1}-g^{-1}({\partial }_{\mu }R)R^{-1}$. The gauge field strength transforms by conjugation. This makes $D_{\mu }\Phi$ transform in the same gauge group representation as $\Phi$.

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

Choose a definite matrix convention: $A_{\mu }=A_{\mu }^a{T}_a$, $\omega ={\omega }^a{T}_a$, and $D_{\mu }={\partial }_{\mu }+e{A}_{\mu }$. Matter transforms as ${\psi }^{\prime }=U\psi$, so gauge covariance requires $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$. Applying both sides to a test field yields the Yang-Mills gauge transformationDefine the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=e{F}_{\mu \nu }$:Conjugating the gauge covariant derivative commutator proves $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$: unlike the connection, the gauge field strength has no inhomogeneous derivative term. The trace normalization supplies an invariant color metric; its Lie algebra structure constants with all indices lowered are antisymmetric.

For $U=1+e\omega +O({\omega }^2)$ and $U^{-1}=1-e\omega +O({\omega }^2)$, expand the finite transformations:In components this isA convention using the opposite matter transformation or gauge covariant derivative sign changes these signs consistently. The formulas above all follow from the explicitly chosen $D_{\mu }$.

Gauge reduction and the determinant. A direct integral $\int \mathcal{D}A{e}^{iS[A]}$ integrates repeatedly over gauge-equivalent configurations. The gauge group volume is infinite, and the ungauge-fixed quadratic operator has longitudinal zero directions, so it has no ordinary inverse quantum field theory propagator. Gauge-invariant observables require dividing out this redundancy rather than treating it as an ordinary physical fluctuation.

The finite-dimensional change-of-variables identity underlying gauge fixing is precise. Suppose a condition $\chi (x^{\alpha })=f$ has one solution in the group-coordinate patch and its derivative matrix at the solution is nonsingular. ThenFor several isolated solutions it instead counts those solutions. On the slice, the Jacobian determinant can be factored into the reciprocal of the orbit integral of the delta function. An invariant measure and invariant integrand then allow the group volume to factor out. In the functional version, group coordinates carry Haar measure and the corresponding operator determinant replaces the matrix determinant.

For example, choose the Lorenz gauge ${\chi }^a(A)={\partial }^{\mu }{A}_{\mu }^a=f^a$. Its variation is $\delta {\chi }^a=-{\partial }^{\mu }(D_{\mu }^{\mathrm{ad}}\omega {)}^a$. Hence on the gauge slice the Faddeev-Popov operator and Faddeev-Popov determinant areIts extension along each local gauge orbit yields the Faddeev-Popov gauge-orbit identityFor an unoriented finite-dimensional real integral the absolute determinant is required. In a perturbative gauge patch, its sign or phase is fixed and the constant convention is absorbed in normalization. Residual zero modes must be removed by boundary conditions or separate treatment. Multiple intersections, the Gribov ambiguity, prevent interpreting this as a globally unique gauge slice without additional work.

Insert the identity, change variables along the orbit using invariance of the action and measure, and divide out the common gauge group volume. Averaging $f$ with Gaussian weight $\exp [-i\int f^a{f}^a/(2\xi )]$ produces the covariant gauge-fixed functionalThe added term makes the quadratic gauge-field operator invertible after the stated residual-mode treatment.

Ghost representation. For a finite matrix $M$, the Grassmann Gaussian integral over independent anticommuting variables is proportional to its determinant:The constant is independent of $M$ and depends on ordering and factors of $i$. The functional counterpart introduces Faddeev-Popov ghost fields $c^a$ and antighost fields ${\overline{c}}^a$, withIntegration by parts writes its density as $({\partial }^{\mu }{\overline{c}}^a)(D_{\mu }c{)}^a$, displaying a kinetic term and a gauge-field/ghost interaction. These are Grassmann-valued fields with scalar Lorentz transformation law. They represent the orbit Jacobian determinant; their closed loops carry the fermionic minus sign, and they are not physical asymptotic particles. In the abelian case $D^{\mathrm{ad}}=\partial$, the determinant is gauge-field independent and the ghosts decouple. Nonabelian ghosts interact and are necessary for the consistent cancellation of unphysical gauge contributions.

Physical-state reduction. Let $Q$ be the Hermitian BRST charge. Its key algebraic property is BRST nilpotence, $Q^2=0$, and the gauge-fixed dynamics preserves it. Physical states are its BRST cohomology at ghost number zero:Here $|\chi ⟩$ has ghost number $-1$. BRST nilpotence ensures that exact vectors lie in the kernel, so the quotient is defined. Because $Q$ is Hermitian, an exact vector is orthogonal to every closed vector, $⟨\psi |Q\chi ⟩=⟨Q\psi |\chi ⟩=0$, and exact vectors are null, $⟨Q\chi |Q\chi ⟩=⟨\chi |Q^2\chi ⟩=0$. They therefore represent no independent physical state. Commutation of $Q$ with the Hamiltonian preserves these classes under time evolution.

The covariant gauge/ghost space before reduction has an indefinite inner product. This is essential: a Hermitian nilpotent BRST charge requires an indefinite auxiliary space if it is nonzero. On an ordinary positive-definite Hilbert space the same norm identity would force $Q=0$. Positivity of the resulting physical space uses the gauge-theory structure in addition to the abstract BRST nilpotence identities.

For a nonzero on-shell photon momentum, the one-particle conclusion isThe second relation preserves the first because $k^2=0$. In four dimensions the transverse kernel has dimension three, and quotienting by its null longitudinal direction leaves two physical photon polarizations. This is the positive one-particle BRST cohomology of the photon sector.

Hold the metric fixed when varying the gauge potential, put $F=dA$, and take variations with compact support or with appropriate fixed boundary data. The Hodge star operator then givesFor the Abelian Chern-Simons 5-form, the two occurrences of $F$ commute as even-degree differential forms. Since $dF=0$,The factor three comes from all three appearances of $A$, including those hidden in $F=dA$. The bulk variation is thereforeand the Abelian Chern-Simons five-form field equation is

Under an Abelian gauge transformation $A\mapsto A+d\lambda$, the gauge field strength stays fixed and the action changes byThus the local integrand is not gauge invariant, but its change is a boundary term by the Generalized Stokes theorem. With the stated boundary conditions the variational bulk equations are unchanged; they depend only on $F$ and the metric, so they are manifestly gauge invariant. On a spacetime with boundary, gauge transformations nonzero at the boundary require boundary conditions or boundary degrees of freedom; the bulk claim does not assert invariance of every possible boundary action.

For the metric variation, write the kinetic action in components,At fixed covariant components $A_{\mu }$ and $F_{\mu \nu }$,Substituting into the given definition of the stress-energy tensor yieldsThe Chern-Simons 5-form uses only wedge products of differential forms and orientation, with no metric or Hodge star operator. Its bulk metric derivative is zero, so the Chern-Simons term contributes no bulk stress-energy tensor. The displayed tensor is generally not traceless in five dimensions: $T^{\mu }{}_{\mu }=-F_{\rho \sigma }{F}^{\rho \sigma }/4$.

Geometric quantization. A classical symplectic manifold supplies observables as real functions, their Poisson bracket, and Hamiltonian flows. Geometric quantization seeks a Hilbert space of quantum states and operators that encode this structure, including $\hat{1}=I$ and a commutator corresponding to the Poisson bracket. A direct prescription cannot make every classical product become an operator product without qualification; operator ordering and the choice of physically admissible observables matter. The aim is to build the state space and operators from the geometry rather than from a preferred coordinate chart.

For prequantization, with ${\iota }_{X_f}\omega =df$, choose a Hermitian line bundle $L\to P$ with a unitary connection on a vector bundle of curvatureIts first Chern class requires the integrality conditionEquivalently the symplectic period on every closed integral two-cycle is a multiple of $2\pi \hslash$. Conversely an integral class supplies a line bundle, and a unitary connection can be adjusted by a one-form to have the required curvature. Locally choose $\theta$ with $\omega =-d\theta$; then $\nabla =d-i\theta /\hslash$. Potentials differing by an exact one-form change the local phase of the section. Integrality is precisely what makes these local descriptions glue consistently.

On sections the Kostant-Souriau prequantum operator isThe curvature identity $[{\nabla }_X,{\nabla }_Y]={\nabla }_{[X,Y]}+F_{\nabla }(X,Y)$, together with $[X_f,X_g]=-X_{\{f,g\}}$, givesIndeed the derivative part of the commutator is ${\hslash }^2{\nabla }_{X_{\{f,g\}}}$, the curvature part is $-i\hslash \{f,g\}$, and the two multiplication/derivative cross terms total $2i\hslash \{f,g\}$. These combine to the displayed operator identity. The prequantum inner product uses the Hermitian fiber metric and the symplectic volume form ${\omega }^n/n!$, with appropriate square-integrability and operator-domain conditions.

This is not yet the desired quantum state space: a prequantum section depends on all $2n$ phase-space variables, whereas a wavefunction normally depends on only $n$ configuration variables. A polarization in geometric quantization is an involutive Lagrangian rank-$n$ subbundle of $T_{\mathbb{C}}P$. Polarized sections obey ${\nabla }_{X}s=0$ along that distribution. The symplectic form vanishes on it, so its restricted curvature vanishes and these differential conditions are locally compatible.

For $T^{\ast }Q$, the vertical polarization is spanned locally by ${\partial }_{p_i}$. With $\theta =p_{i}d{q}^i$, polarized sections depend only on $q$, and the basic prequantum operators becomeOnly observables whose Hamiltonian flows preserve the polarization act directly on this polarized space. A generic quadratic kinetic energy does not preserve the vertical polarization, so obtaining its quantum operator needs an additional construction rather than simply restricting every prequantum operator. A compatible complex structure can instead provide a Kähler polarization, yielding holomorphic-type states, as for oscillator variables. The Blattner-Kostant-Sternberg pairing compares suitable state spaces of different polarizations and can implement some transported observables.

Finding a useful global polarization is a genuine problem: it may not exist as a nonsingular real foliation; leaf holonomy can obstruct globally nonzero flat sections; singular leaves and compact leaves impose further conditions. A metaplectic correction uses half-forms to improve the inner product and coordinate transformation law and account for familiar zero-point shifts, when its additional square-root data exist. Thus prequantum integrality, polarization, Hilbert-space completion and observable domains are separate steps, each with geometric constraints.

Applications of Stokes's theorem. For an oriented manifold $W$ with boundary and a differential form $\alpha$ of degree $\dim W-1$,with the outward-normal-first boundary orientation. Locally the identity is integration by parts in a half-space; a partition of unity reduces the general manifold to that local calculation, and contributions along internal patch boundaries cancel. The Generalized Stokes theorem unifies the fundamental theorem of calculus, circulation/flux formulas and the relation between local conservation laws and conserved integrated charges.

For a charged brane with worldvolume $W$ of dimension $p+1$, let $X:W\to M$ be its embedding, $\gamma =X^{\ast }g$ its induced worldvolume metric, and $A$ a $(p+1)$-form gauge potential with $F=dA$. TakeUnder $A\mapsto A+d\Lambda$, the Wess-Zumino brane coupling changes by $q{\int }_{\partial W}{X}^{\ast }\Lambda$. It is invariant for a closed worldvolume, or for suitable boundary conditions; open branes can require compensating boundary terms.

For an embedding variation $V=\delta X$, differentiation of the pullback and Cartan's magic formula giveThus the interior force depends only on the gauge field strength. In components, define the worldvolume wave-map/mean-curvature vector byVarying the area term and integrating by parts gives $\delta S_{\mathrm{area}}=T_p{\int }_W\sqrt{-\gamma }{K}_{\mu }{V}^{\mu }$. Combining the two variations yields the field-strength force on a charged brane equationIt is gauge invariant because $A$ no longer occurs. Its tangential projection vanishes identically by antisymmetry, consistently with worldvolume reparametrization invariance. At $p=0$, the chosen Lorentzian worldline convention has $K=-\ddot{X}$ in proper time and recovers the usual Lorentz-force sign.

For topological conservation, let the spatial manifold compactify to a closed oriented $n$-manifold $\Sigma$, and let a field $U:\Sigma \to N$ take values in a closed oriented $n$-dimensional target with normalized volume form $\Omega$, ${\int }_N\Omega =1$. ThenThis is the degree of a map between oriented manifolds: for a regular value it is the signed sum of inverse images. To relate the integral to that count, use a normalized target form supported near a regular value and change variables on its inverse-image neighborhoods. Any two normalized top forms on a connected closed oriented target differ by an exact form, whose pulled-back integral vanishes by the Generalized Stokes theorem. Hence the value is independent of this localization and equals the integer signed count.

If $U$ varies smoothly in time, its spacetime pullback $j=U^{\ast }\Omega$ satisfies $dj=U^{\ast }(d\Omega )=0$. Apply Stokes theorem to the spacetime slab between two slices: the difference of their charges is minus the flux through the spatial boundary. With no flux at infinity, $Q$ is conserved independently of the field equations. Equivalently a smooth homotopy cannot change the degree. Singularities, altered boundary conditions or nonzero boundary flux are the ways the argument can fail.

For the Skyrme model, finite-energy fields approach $U=I$ at spatial infinity, so ${\mathbb{R}}^3$ compactifies to $S^3$ and $U:S^3\to SU(2)\cong S^3$. With the conventional target orientation chooseThis is a normalized closed target volume form. The resulting topological baryon number in the Skyrme model isThe corresponding identically conserved spacetime current is $j^{\mu }=-(24{\pi }^2{)}^{-1}{ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}(L_{\nu }{L}_{\rho }{L}_{\sigma })$ in oriented flat coordinates. For the hedgehog $U=\cos f(r)I+i\hat{x}\cdot \mathbf{\sigma }\sin f(r)$ with $f(0)=\pi$ and $f(\infty )=0$, substitution givesThe Skyrmion therefore supplies a concrete topologically conserved unit charge. Topological conservation alone does not guarantee a stable finite-size static solution: the positive Skyrme term, with four derivatives, opposes collapse of the two-derivative energy. Under size rescaling, these contributions behave as $E_2\propto \lambda$ and $E_4\propto {\lambda }^{-1}$, allowing a nonzero equilibrium size while preserving the degree.

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.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.