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Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

Use the real Lie-algebra convention of the question, with a gauge covariant derivative $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The gauge field strength is its curvature:Write $a_{\mu }={\delta }_X{A}_{\mu }/ϵ=-{\partial }_{\mu }X+[X,A_{\mu }]$. To first order in $ϵ$,Substitute $a_{\mu }$. The mixed second derivatives of $X$ cancel. The terms containing first derivatives of $X$ cancel in pairs. The remaining terms areThe Jacobi identity combines the last two into $[X,[A_{\mu },A_{\nu }]]$. ConsequentlyThe transformation is homogeneous even though the connection transformation contains an inhomogeneous derivative term.

For a finite-dimensional Lie algebra, define its Adjoint representation by ${\mathrm{ad}}_X(Y)=[X,Y]$ and its Killing form byIt is a symmetric bilinear form by cyclicity of the trace. The Jacobi identity gives ${\mathrm{ad}}_{[Z,X]}=[{\mathrm{ad}}_Z,{\mathrm{ad}}_X]$. Set $A={\mathrm{ad}}_Z$, $B={\mathrm{ad}}_X$, $C={\mathrm{ad}}_Y$. ThenThis proves the invariant bilinear form on a Lie algebra property, without assuming simplicity or nondegeneracy.

For definiteness use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and take a real compact semisimple Lie algebra as the gauge algebra. Its positive internal metric is $B=-\kappa$. A Killing-form Yang-Mills Lagrangian with the coupling absorbed into the connection isThe spacetime metric is unchanged by internal gauge transformations. HenceThus gauge invariance follows directly from invariance of the Killing form. Other overall conventions are possible, but the energy sign must be checked rather than inferred from a prefactor in isolation.

For physical kinetic terms the internal form must be real, nondegenerate and positive definite after choosing the overall sign. If ${\mathcal{E}}_i=F_{0i}$ and ${\mathcal{B}}_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, the above convention has Lagrangian $[B({\mathcal{E}}_i,{\mathcal{E}}_i)-B({\mathcal{B}}_i,{\mathcal{B}}_i)]/(2g_{\mathrm{YM}}^2)$ and physical energy densityThe canonical Hamiltonian density also contains the nondynamical multiplier and a spatial divergence. With ${\Pi }_i={\mathcal{E}}_i/g_{\mathrm{YM}}^2$ and $D_i{\Pi }_i={\partial }_i{\Pi }_i+[A_i,{\Pi }_i]$, integration by parts using the invariant bilinear form on a Lie algebra givesThe Gauss law constraint in gauge theory sets $D_i{\Pi }_i=0$. If the boundary flux vanishes, or the appropriate boundary contribution is included, the integrated physical Hamiltonian is the positive energy displayed above.

An indefinite internal form would give gauge-field polarizations with opposite kinetic signs. A degenerate form would fail to supply a kinetic term for some directions. The compactness criterion from the Killing form says that negative-definiteness of the Killing form of a real finite-dimensional algebra is equivalent to compact semisimplicity; thus the pure Killing-form construction selects compact semisimple real forms. One cannot use a complex-bilinear Killing form on arbitrary complex field components as if it were a positive Hermitian metric.

This does not prohibit Abelian gauge theories. A compact Abelian factor has zero Killing form, so it needs a separately chosen positive invariant bilinear form on a Lie algebra, rather than the Killing form. More generally an algebra with a positive invariant metric on a Lie algebra is compact reductive, namely a direct sum of a compact semisimple algebra and an Abelian center. A noncompact group can also share the same compact Lie algebra through global covering choices in an Abelian factor; positivity is a statement about the algebra and internal metric, not by itself a classification of global gauge-group topology. Quantum matter anomalies and global restrictions would require additional input; no matter content is specified here.

For the finite matrix transformation, let $g(x)\in SU(N)$ and regard $g$ as a multiplication operator. The identity ${\partial }_{\mu }{g}^{-1}=-g^{-1}({\partial }_{\mu }g)g^{-1}$ givesTaking commutators of these differential operators cancels the adjacent multiplication operators $g^{-1}g$, yieldingThis argument keeps the derivatives acting on test fields and avoids treating $D_{\mu }$ as just a matrix.

There is a normalization issue in the printed last paragraph. The standard Killing form defined through the adjoint trace on $\mathfrak{su}(N)$ isnot simply $\mathrm{tr}(XY)$. For example, with $N=2$ and $X=Y=-i{\sigma }_3/2$, the defining trace is $-1/2$, whereas the adjoint trace is $-2$. The printed trace formula can be used as a rescaled invariant form, with the constant absorbed into the gauge coupling; it has exactly the invariance needed here. For either normalization,by cyclicity. This proves finite Yang-Mills gauge transformation invariance, with the healthy sign chosen for anti-Hermitian gauge fields. The normalization discrepancy is not a failure of gauge invariance.

Finally let $g(x)=e^{ϵX(x)}=I+ϵX(x)+O({ϵ}^2)$. Since $X\in \mathfrak{su}(N)$ is traceless and skew-Hermitian, this exponential lies in $SU(N)$. Expanding the finite formula givesThus the stated infinitesimal transformation is the derivative of the finite transformation at the identity. It describes transformations in the identity component; arbitrary global or large gauge transformations need not be generated by one globally defined infinitesimal parameter.

Write $F[A]={\partial }_{\mu }{A}_{\mu }+i{A}_{\mu }{A}_{\mu }$ and choose the U(1) gauge symmetry convention $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$. Its infinitesimal variation isA formal BRST symmetry construction uses $s{A}_{\mu }={\partial }_{\mu }c$, $sc=0$, $s\bar{c}=ih$, $sh=0$ and the gauge-fixing fermion $\Psi =\int \bar{c}(F-i\xi h/2)d^{4}x$. Acting with the odd differential givesFor a real admissible gauge functional, integration over $h$ gives $F^2/(2\xi )$ in the Euclidean action, and $\xi \to 0$ imposes the gauge condition. Overall constant phases in the Faddeev-Popov determinant can be absorbed in the measure convention. Here nonlinear Abelian gauge fixing makes ${\mathcal{O}}_A$ depend on $A$. The action for the Faddeev-Popov ghost fields contains the interaction $-2i\bar{c}{A}_{\mu }{\partial }_{\mu }c$, so the Faddeev-Popov ghost fields cannot be discarded as in a linear Abelian gauge, even though the $U(1)$ Adjoint representation is trivial.

There is a genuine reality obstruction for a complex Euclidean gauge condition in the source. For a Hermitian $U(1)$ gauge field, each Euclidean component $A_{\mu }$ is real. The real and imaginary parts of $F=0$ separately require $\partial \cdot A=0$ and ${\sum }_{\mu }{A}_{\mu }^2=0$, hence $A=0$. For example $A_2=B{x}_1$, with other components zero and $B\ne 0$, has curvature $F_{12}=B$; no gauge transformation can put it on this slice because curvature is gauge invariant. Thus the stated condition is not a gauge fixing of general real Euclidean configurations. The boxed action is the intended formal complex-gauge construction. A legitimate complexified contour prescription would be additional data, not an ordinary real delta-functional enforcing this printed condition.

Use $t_{\mu }=(1,-1,-1,-1{)}_{\mu }$ without summing in individual component transformations. The Dirac current transforms asSince the coupling is real, invariance of its contraction fixes the time reversal of an electromagnetic gauge field:These equations use a compatible gauge; an additional pure gauge transformation would not change the gauge field strength.

Coordinate differentiation at $x_T$ introduces $r_{\mu }=-t_{\mu }$, because the time coordinate is reversed and the spatial coordinates are unchanged. HenceThus the electric field is even and the magnetic field is odd. The chirality matrix is even by part (a). The two tensor signs cancel in the dipole contraction, but antiunitarity conjugates its explicit $i$. Therefore the time-reversal parity of a fermion electric dipole operator isIt is a time-reversal symmetry violating interaction, and, under the usual local relativistic quantum field theory hypotheses of the CPT theorem, it violates CP symmetry. A nonzero coefficient is absent from the renormalizable tree-level Standard Model Lagrangian: the broken-phase fermion electric dipole moment operator has mass dimension five, and its electroweak-invariant completion requires a Higgs field and has dimension six. However, it can arise as a radiatively induced effective interaction in the Standard Model, whose CKM matrix contains a CP-violating phase. It is not forbidden to all orders. An explicit primary calculation of quark dipoles is Czarnecki and Krause's Standard Model calculation.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

Use metric $(+,-,-)$ and the paper's normalization. A static multi-vortex solution is a smooth finite-energy field configuration of Abelian Higgs vortices whose complex Higgs field has winding at spatial infinity and zeros in the plane. In a positive $N$-vortex sector, the zeros have total multiplicity $N$ and the magnetic flux is $2\pi N$; they may be separated or coincident. Finite energy requires $|\varphi |\to 1$, $D_i\varphi \to 0$ and $B=f_{12}\to 0$ at infinity. Then $a_i$ asymptotically compensates the phase gradient, so Stokes' theorem relates flux to the phase winding. The positive Bogomolny vortex equations describe this same-sign sector; negative flux uses the opposite signs and is not covered by simply mixing positive and negative zero multiplicities.

For the critically coupled Abelian Higgs field equations, vary $\varphi$ and $\bar{\varphi }$ independently. After covariant integration by parts,Thus the scalar equation and its complex conjugate areVarying the gauge field gives ${\partial }_{\mu }{f}^{\mu \nu }\delta a_{\nu }$ from the Maxwell term. The matter variation isConsequentlyThese factors use the scalar kinetic coefficient $1/2$; importing the current from a convention with coefficient one would give an incorrect factor two. The current is invariant under gauge transformations, and its conservation follows from the field equations and antisymmetry of $f^{\mu \nu }$.

For the static purely magnetic solutions take temporal gauge $a_0=0$. The scalar equation reduces toSince $[D_1,D_2]=-iB$, applying $D_1-i{D}_2$ to the first Bogomolny vortex equation givesThe second equation sets $B=(1-|\varphi {|}^2)/2$, proving the scalar equation, including at smooth vortex cores without dividing by $\varphi$.

For the spatial gauge equations, put $k_i=\mathrm{Im}(\bar{\varphi }{D}_i\varphi )$. The first equation implies $D_1\varphi =-i{D}_2\varphi$, and ${\partial }_i|\varphi {|}^2=2\mathrm{Re}(\bar{\varphi }{D}_i\varphi )$. ThereforeWith the stated metric, $j^i=-k_i$ and the two static Maxwell equations are ${\partial }_{2}B=k_1$ and ${\partial }_{1}B=-k_2$. Both follow by differentiating $B=(1-|\varphi {|}^2)/2$. The time component is the Gauss law constraint in gauge theory; it is identically satisfied because both the electric field and $j^0$ vanish in this static temporal gauge. HenceAs an energy check, the Bogomolny square completion for an Abelian Higgs vortex gives $E\ge \pi N$ in this normalization, saturated by these equations. On the plane, the Abelian Higgs vortex moduli space admits arbitrary vortex positions, including coincidence; its static energy is $\pi N$, so there is no position-dependent static potential. Their slow dynamical interactions are a different question.

The logarithmic form of a Bogomolny vortex is only local away from zeros. Write $\varphi =e^{u+i\chi }$ there, with $u=\log |\varphi |$. ThenSeparating the real and imaginary parts of $(D_1+i{D}_2)\varphi =0$ yields $u_1-{\chi }_2+a_2=0$ and ${\chi }_1-a_1+u_2=0$. Hencewhere $J(v_1,v_2)=(-v_2,v_1)$ is the positive quarter-turn. Under a gauge transformation, $\chi \mapsto \chi +\omega$ and $\mathbf{a}\mapsto \mathbf{a}+\nabla \omega$, as this relation requires.

At regular contour points, the stipulated orthogonality means $\nabla \chi \cdot \nabla u=0$. Since $J\nabla u$ is also perpendicular to $\nabla u$,Thus, in that gauge, the gauge potential is tangent to contours of $u$, equivalently contours of $|\varphi |$, and is parallel to the phase gradient wherever that gradient is nonzero. This is a gauge-dependent statement about the vector potential, not an assertion that the magnetic field is tangent to these curves, nor that the potential is a pure gradient or divergence-free. The two gradient magnitudes are not fixed by contour orthogonality.

On a phase patch $B={\partial }_1{a}_2-{\partial }_2{a}_1=-\Delta u$. Combining with the second equation gives $\Delta u+(1-e^{2u})/2=0$ away from the cores. At a zero of multiplicity $n_j$, $u\sim n_j\log |x-X_j|$, and the phase has winding $2\pi n_j$; $u$ is not a finite real smooth field there. Distributionally, the full equality isThis is the Taubes equation in the paper's amplitude logarithm, rather than the often-used $\log |\varphi {|}^2=2u$ convention. No single smooth globally defined phase or regular contour description at every zero has been assumed.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

Under a gauge transformation, the electromagnetic field tensor changes byThe partial derivatives commute for a smooth gauge parameter, or in the distributional formulation. Thus the Maxwell Lagrangian depends only on a gauge-invariant tensor:Strictly, the pointwise calculation requires the second derivatives to exist; smooth gauge parameters are the usual convention. Mere first differentiability in the wording can instead be understood through commuting distributional derivatives.

For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_{\rho }=\beta {ϵ}^{\nu }{\partial }_{\nu }{A}_{\rho }$, and $\delta \mathcal{L}={\partial }_{\mu }(\beta {ϵ}^{\mu }\mathcal{L})$. Differentiating the Maxwell Lagrangian with respect to the field gradient givesThus Noether's theorem yields $j^{\mu }={ϵ}_{\nu }{T}^{\mu \nu }$, where the canonical stress-energy tensor of the Maxwell field isIt is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,need not vanish. For example, take $A_0=a{x}^1$, $A_2=b{x}^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.

Nor is it gauge-invariant: while $F$ and $\mathcal{L}$ are unchanged, a gauge transformation giveswhich is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the canonical stress-energy tensor.

Use $\delta F=ϵ\{F,\phi \}$ with the canonical Poisson bracket. The first-class constraint generatesIndeed the variation of the relativistic particle phase-space action isThe transformation is therefore a gauge invariance when its parameter vanishes at the time endpoints, or suitable boundary conditions remove the total derivative.

On a fixed interval of parameter length $\Delta t$, the proper-time modulusis unchanged by these gauge transformations. Choosing $ϵ(t)={\int }_{t_i}^t(s-e(u))du$ sets $e+\delta e=s$, with $ϵ(t_i)=ϵ(t_f)=0$. Thus the nonconstant part of the worldline einbein can be fixed, but its constant modulus must still be integrated over. Fixing $e=1$ as well would remove inequivalent values of the proper-time modulus; it is legitimate only if the parameter interval is allowed to vary instead. The name proper-time modulus refers to the Schwinger proper-time parameter; after eliminating $p$, the geometric proper length for a massive on-shell trajectory is $M\int edt$ in this normalization.

For the gauge fixing functional $F=e-s$, its variation is $\delta F=\dot{ϵ}$. The Faddeev-Popov determinant is consequentlyThe Grassmann Gaussian integral represents this determinant using anticommuting Faddeev-Popov ghost fields $b,c$:The phase and normalization of the determinant depend on the measure convention. Its domain carries the chosen boundary conditions: on an interval the constant modulus is excluded from the gauge-fixed directions, and on a periodic worldline the constant ghost zero mode in field theory is removed with the residual gauge volume treated separately. A bare determinant with such zero modes left in would vanish.

A collective coordinate describes a position, orientation, or another parameter of a family of static solitons. For a family $\Phi (\mathbf{x};q)$, promote $q$ to a slowly varying $q(t)$ and substitute into the field action. In a scalar theory with unit kinetic coefficient, this gives the collective-coordinate effective LagrangianFor a gauge-theory soliton, one also solves the Gauss law constraint in gauge theory constraint and projects out pure gauge transformations; arbitrary variations of gauge representatives do not define the physical metric. Tangent vectors to an exactly equal-energy family are zero modes in field theory. If $V=E_0$ is constant, the Euler-Lagrange equations of this moduli-space approximation arethe geodesic equations of its Riemannian metric. The approximation neglects radiation and deformation modes; it describes motion sufficiently slow that these omitted degrees of freedom remain unexcited to the required accuracy.

In collective-coordinate quantization, take the wavefunction measure $\sqrt{\det g}{d}^{n}q$ and the minimal scalar Hamiltonian operatorThe Laplace-Beltrami operator supplies coordinate-invariant kinetic energy. Global identifications and statistics must be imposed on the wavefunctions; the classical metric alone does not choose them. Curvature-ordering terms and loop corrections are additional quantum input.

For the phi-four kink, use the normalization and profile of Question 1. Substituting $\varphi (x,t)={\varphi }_K(x-X(t))$ givesThe metric is constant because of translation invariance. This proves the translational dynamics of a phi-four kink: classically the centre moves at constant velocity, and quantum mechanicallyPlane waves label the continuous translational momentum, with no position-dependent potential. Uniform-motion Lorentz invariance upgrades the dispersion to $\sqrt{M^2+p^2}$; the displayed collective Lagrangian is its small-velocity expansion. Small perturbations also include an internal shape mode and continuum radiation, which this single collective coordinate omits. The fluctuation operator of a phi-four kink in this normalization isthe translational eigenfunction has ${\omega }^2=0$, the shape mode has ${\omega }^2=3c^2$, and continuum modes in the spectrum start at ${\omega }^2=4c^2$. Thus the free-coordinate states describe the kink's low translational energies, not its full excitation spectrum or quantum mass correction.

For two Abelian Higgs vortices at critical coupling, the static energy is $2M_v$ and the Abelian Higgs vortex moduli space has four real dimensions. Let $z_1,z_2$ be their positions, $Z=(z_1+z_2)/2$, and $q=z_1-z_2$. The centre of mass decouples; the relative metric is rotationally symmetric and can be writtenAt large separation, $F(\rho )\to M_v/2$, recovering two free particles. Although there is no static separation potential, the nonconstant metric produces velocity-dependent interaction. Coincidence is smooth in the relative coordinate for two identical vortices $w=q^2$, not in the double-valued $q$. Smoothness gives $F(\rho )\sim C{\rho }^2$ for $C>0$ near zero. A head-on geodesic continues through $w=0$ to the opposite real ray, so $q$ changes its line by $\pi /2$: the vortices scatter through a right angle. This geometric argument does not require an explicit formula for $F$.

With ordinary bosonic exchange statistics, relative wavefunctions are single-valued in $w$ and smooth at coincidence. In the separated polar coordinate they obey $\Psi (\rho ,\theta +\pi )=\Psi (\rho ,\theta )$, with even integer angular labels. Their kinetic operator isThe apparent singularity at $\rho =0$ must be resolved with the smooth $w$ coordinate and regularity, rather than arbitrary boundary conditions on a punctured cone. The free centre-of-mass motion and the asymptotically free relative geometry give quantum scattering states; a flat static energy does not imply that the metric is flat or that scattering is absent. This is not a prediction of a discrete family of static two-vortex bound separations. The smooth collision geometry is developed in David Tong's arxiv.org/abs/hep-th/0509216.

For a Skyrmion of baryon number one, the Skyrmion hedgehog ansatz iswhere $\mathbf{\sigma }$ are the Pauli matrices. Include a centre $\mathbf{X}$ and an orientation $A\in SU(2)$ through $U=A{U}_0(\mathbf{x}-\mathbf{X})A^{-1}$. Hedgehog symmetry identifies spatial rotations with opposite internal rotations, so there are three independent orientation coordinates, not six. Since $A$ and $-A$ give the same field, the physical orientation space is $SO(3)$, with SU(2) group as its double cover. Write $A^{-1}\dot{A}=i\mathbf{\omega }\cdot \mathbf{\sigma }/2$. The leading collective Lagrangian has the formwhere $\Lambda >0$ is the rotational moment of inertia obtained by integrating the profile's field kinetic energy.

For the fermionic quantization appropriate to baryons, the Finkelstein-Rubinstein constraints on the double cover impose $\Psi (-A)=-\Psi (A)$. In SU(2) representations, the central element $-1$ acts by $(-1{)}^{2j}$, so $j$ must be half-integer. Left and right group actions supply isospin and spin angular momentum; hedgehog symmetry makes their magnitudes equal. The rotational quantization of a unit Skyrmion therefore givesThe $j=1/2$ level has four spin-isospin states and models the nucleon doublet, proton and neutron, each with two spin states. The $j=3/2$ level has sixteen states and models the Delta baryon quartet, each with four spin states. The rotor predicts a splitting $3{\hslash }^2/(2\Lambda )$. Without the fermionic sign, single-valued functions on $SO(3)$ would instead allow integer $j$, which is a different quantization. High rotor levels can couple to deformation and pion radiation; this semiclassical approximation does not establish that its entire formal tower consists of stable particles.

For a short Wilson line, a gauge transformation givesUse $V(x+an)=1+i\alpha (x)+ia{n}^{\mu }{\partial }_{\mu }\alpha (x)+O(a^2,{\alpha }^2)$ and $U=1+iga{n}^{\mu }{A}_{\mu }+O(a^2)$. Keeping terms of order $a$, including $a\alpha$, givesSince the Lie bracket is encoded by the Lie algebra structure constants, $i[\alpha ,A_{\mu }{]}^a=f^{abc}{A}_{\mu }^b{\alpha }^c$, and thereforeThis is the infinitesimal form of the Wilson-line gauge transformation.

FactorIntroduce $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$. Since $dt=dv-dr/f$,These are the charged analogue of Ingoing Eddington-Finkelstein coordinates. The $(v,r)$ block has determinant $-1$, so the metric and its inverse are nonsingular at $r=r_{+}$. For $M\ge Q$, $r_{+}$ is real and the metric extends analytically through it. The original potential becomes $A=-Qdv/r^2+Qdr/(r^{2}f)$; the radial term is locally exact, and a gauge transformation gives the horizon-regular potential $A=-Qdv/r^2$.

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.