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Let $R_i(x)={\partial x_{g(\theta )}/\partial {\theta }_i|}_{\theta =0}$ be the infinitesimal orbit tangent vectors. Since $F_j(x_0)=0$, Taylor expansion givesThis explains the stated linearization. For a linear group representation, $R_i(x)=t_{i}x$ in the corresponding action convention.

Two assumptions are implicit in the advertised integration formula. The slice must be regular, and the determinant sign must be fixed if the absolute value is omitted. Uniqueness alone does not imply regularity: translations on the line with $F(x)=x^3$ have a unique gauge-fixed representative but $M(0)=0$. For ordinary real integration the Dirac delta distribution change-of-variables formula uses $|\det M|$. The choice $F(x)=-x$ instead has a unique regular zero and negative determinant, illustrating why its sign cannot simply be dropped.

Assume a regular gauge slice with one representative on each orbit, and choose the action convention $(x_{g_0}{)}_{g(\theta )}=x_{g(\theta )g_0}$, matching the specified right-invariance of Haar measure. For the unique $x_0=x_{g_0}$ with $F(x_0)=0$, put $g=g(\theta )g_0$. The Haar density is one at the identity, so changing from $\theta$ to $F$ near the unique zero givesThe integral is orbit invariant by the right-invariance of Haar measure. Thus $\Delta (x)=I(x{)}^{-1}$ is also orbit invariant, and on the gauge slice it equals $|\det M(x)|$. Insert $1=\Delta (x)I(x)$ into the original integral and use its invariant volume measure and invariant $f$:In the second line set $y=x_g$. This proves the finite-dimensional Faddeev-Popov gauge reductionChoose a positive orientation of the regular slice to obtain the question's $\det M$ form. If the group volume is infinite, its cancellation in normalized quantities is formal or regulated. Multiple intersections and residual zero modes invalidate the simple identity; in a gauge theory these are the Gribov ambiguity and residual gauge transformations.

The Yang-Mills action counts every gauge orbit redundantly in its path integral, and its quadratic gauge-field operator has gauge zero modes. A gauge fixing condition removes this redundancy, while the Faddeev-Popov determinant supplies the orbit-to-slice Jacobian. This is a local perturbative construction with the same regularity qualifications as the finite-dimensional derivation.

Take the linear condition $G[A]=F^{\mu }{A}_{\mu }$. Under an infinitesimal gauge transformation $\delta A_{\mu }=D_{\mu }\omega$, its change is $\delta G=F^{\mu }{D}_{\mu }\omega$. Hence the Faddeev-Popov operator is $\mathcal{M}[A]=F^{\mu }{D}_{\mu }$. For Lorenz gauge, $F^{\mu }={\partial }^{\mu }$. The gauge covariant derivative acts in the Adjoint representation.

Average the sharp constraint $G=h$ over $h$ with Gaussian weight $\exp [-i\int h^2/(2\xi )]$. This replaces it by $\exp [-i\int G[A{]}^2/(2\xi )]$. The Gaussian gauge fixing with an auxiliary field identity rewrites that factor asby completing the square. The bosonic Nakanishi-Lautrup field $b$ is nondynamical: its algebraic field equation is $b=-G[A]/\xi$. At $\xi =0$ it is a Lagrange multiplier enforcing $G[A]=0$.

The Grassmann Gaussian integral exponentiates the determinant, up to a field-independent phase, asThe independent anticommuting Faddeev-Popov ghost field $c$ and Faddeev-Popov antighost field $\bar{c}$ are spacetime scalars in the Adjoint representation. They represent the Jacobian and do not describe physical external fermions. Combining these factors givesThe parameter $\xi$ selects the Gaussian averaging of the gauge condition. In an anomaly-free regulated treatment, physical gauge-invariant observables are independent of this choice. One way to express this is BRST symmetry: with $sA=Dc$, $s\bar{c}=-b$ and $sb=0$, the gauge and ghost terms are $s[-\bar{c}\cdot (G+\xi b/2)]$. The ghost transformation supplies the off-shell nilpotence of the Yang-Mills BRST quartet, and gauge-fixing parameter independence from BRST symmetry explains why these auxiliary fields do not change the physical content.

For an oriented Euclidean string worldsheet with metric $h_{ab}$, use the string nonlinear sigma model actionHere ${ϵ}^{ab}$ is an antisymmetric density with ${ϵ}^{12}=1$. The factor $i$ in the Euclidean Kalb–Ramond field term follows from continuing the real Lorentzian two-form coupling; reversing orientation or the convention for $B$ reverses its sign. The conventional positive Euclidean dilaton coupling is being used. With worldsheet boundaries its curvature coupling also needs ${\int }_{\partial \Sigma }\Phi kds/(2\pi )$, where $k$ is the boundary geodesic curvature, to have the proper topological normalization for constant $\Phi$.

All three terms are invariant under worldsheet diffeomorphisms. Target-coordinate changes act on $X,G,B,\Phi$ by their ordinary scalar and tensor transformation rules, making the target description covariant as well. For the two-form there is also the target gauge symmetry $B\mapsto B+d\Lambda$. Its change in the closed-worldsheet action is an integral of an exact two-form and hence zero. With boundaries it is instead $i{\int }_{\partial \Sigma }{X}^{\ast }\Lambda /(2\pi {\alpha }^{\prime })$; an endpoint gauge field with $\delta A=-\Lambda /(2\pi {\alpha }^{\prime })$ cancels it.

Under a local Weyl transformation $h_{ab}\mapsto e^{2\omega }{h}_{ab}$, the metric term is invariant because the two-dimensional factors $\sqrt{h}$ and $h^{ab}$ cancel. The $B$ term is independent of $h$ and is separately invariant. The curvature transforms asOn a closed surface the dilaton variation is consequentlyThus a general nonconstant dilaton term is not separately invariant under Weyl transformations. Its variation is the Weyl variation of a nonconstant dilaton coupling; it vanishes for constant $\Phi$ after the boundary completion where necessary.

For $\Phi ={\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$. In the Euclidean path integral the resulting weight isFor a connected orientable genus-$g$ surface with $b$ boundaries, $\chi =2-2g-b$, so the weight is $g_s^{2g+b-2}$. A closed-string handle therefore adds two powers of string coupling. External vertex normalizations add their own coupling factors; they are distinct from this dilaton Euler-characteristic weighting. The constant dilaton controls the string genus expansion, rather than adding a force to the classical embedding equation.

Quantization treats the target fields as position-dependent couplings of a two-dimensional interacting quantum field theory. Expanding $X$ around a background embedding gives fluctuation vertices involving target curvature and derivatives of $B$ and $\Phi$. Short-distance contractions require local counterterms, so the target fields become renormalized couplings. In particular a curvature vertex contracted once yields a Ricci tensor counterterm proportional to ${\alpha }^{\prime }{R}_{\mu \nu }$. Each additional sigma-model loop brings another power of ${\alpha }^{\prime }$ and more target derivatives. The expansion is useful when the target variation scale is large compared with $\sqrt{{\alpha }^{\prime }}$; it is not the separate expansion in powers of $g_s$.

After gauge fixing, a worldsheet Weyl anomaly would make amplitudes depend on the discarded local conformal factor. Quantum consistency therefore requires the trace anomaly to vanish, including the reparameterization worldsheet ghost field contribution. In a conventional renormalization scheme the resulting bosonic sigma-model Weyl anomaly coefficients start asThe barred quantities include the dilaton stress-energy tensor improvement; ordinary coupling sigma-model beta functions can differ by field redefinitions or target diffeomorphisms. Setting these coefficients to zero gives differential equations for the background fields order by order, including the metric gravitational equation and the antisymmetric-field equation. These conventions for the leading coefficients are described in the background-field lecture notes. Higher orders supply higher-curvature and higher-derivative corrections. This explains how spacetime dynamics follows from quantum gauge consistency of the worldsheet, not from asking only for a classical stationary worldsheet in an arbitrarily chosen background.

The apparently noninvariant dilaton term is essential outside the critical dimension because its stress-energy tensor improvement can cancel the quantum central charge mismatch. A flat constant-dilaton background has matter central charge $d$ and worldsheet ghost field central charge $-26$. If $d\ne 26$, changing the constant value of $\Phi$ only changes $g_s$ and cannot cure that mismatch.

A concrete solution is the linear dilaton conformal field theory with $G=\eta$, $B=0$, and $\Phi ={\Phi }_0+V_{\mu }{X}^{\mu }$. Its metric and antisymmetric-field anomaly coefficients vanish, while the scalar condition isOne can also see this cancellation directly from the improved holomorphic stress-energy tensor. With the positive Euclidean curvature convention above,The free-field contraction is $X^{\mu }(z)X^{\nu }(w)\sim -{\alpha }^{\prime }{\eta }^{\mu \nu }\log (z-w)/2$. Differentiating it twice at each point gives a fourth-order pole $3{\alpha }^{\prime }{V}^2/(z-w{)}^4$ from the improvement. Since the fourth-order stress-tensor pole is $c/2$, the matter central charge becomes $c=d+6{\alpha }^{\prime }{V}^2=26$. The worldsheet ghost field contribution then cancels it. This is why a classical Weyl variation need not mean inconsistency: it can cancel the quantum variation of the other terms. For $d<26$ the required gradient is spacelike; for $d>26$ it is timelike in the Lorentzian target theory. Such backgrounds are not flat constant-dilaton Minkowski vacua with full target Lorentz symmetry, nor does a central charge cancellation by itself prove stability or unitarity of every noncritical background. If the conformal factor is instead retained dynamically, its Liouville field theory supplies the corresponding noncritical description.

Put $D={\partial }_{\rho }{A}^{\rho }$. The gauge-fixing variation isAfter integration by parts its contribution to the Euler-Lagrange field equations is $+{\partial }^{\nu }D$. It cancels the $-{\partial }^{\nu }D$ from the Maxwell Lagrangian, giving $\square A^{\nu }=0$ for all four components. This is the Feynman gauge equation; the physical Lorenz gauge condition must still be imposed on states rather than inferred as an additional classical equation from this gauge-fixed density.

Treating the covariant components $A_{\mu }$ as coordinates, the derivative of the actual displayed density with respect to ${\partial }_0{A}_{\mu }$ gives the canonical momentaIn terms of lower-index spatial coordinates,The gauge fixing has supplied the nonzero momentum of $A_0$, absent from the ungauge-fixed Maxwell Lagrangian.

There is a boundary-term convention needed for the next part's printed momentum expansion. Direct expansion gives the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction:To verify it, the difference of the densities is $\frac{1}{2}[({\partial }_{\rho }{A}_{\sigma })({\partial }^{\sigma }{A}^{\rho })-D^2]$; differentiating $K^{\rho }$ gives these two terms, while the remaining second-derivative terms cancel after renaming indices. The subtracted density has ${\tilde{\pi }}^{\mu }=-{\dot{A}}^{\mu }$, which is exactly the momentum used by the mode expansion in part (c). The literal momenta asked for here are the boxed expression, not $-{\dot{A}}^{\mu }$. Both densities have the same field equations, and their canonical commutation relations agree through the boundary-induced canonical transformation in Feynman gauge.

Without gauge fixing, gauge-equivalent configurations are integrated repeatedly, and the quadratic gauge-field operator has zero modes along infinitesimal gauge transformations. It therefore has no inverse with which to define perturbative propagators. A gauge condition removes these directions locally in the perturbative expansion. The associated Faddeev-Popov determinant can be represented by anticommuting Faddeev-Popov ghost fields and Faddeev-Popov antighost fields.

First make the normalization explicit. For the fields literally occurring in the action with overall $1/g^2$, putThe resulting kinetic terms are canonical, and the interaction terms carry powers of $g$. We use signature $(-,+,+,+)$ and incoming Fourier modes $e^{ipx}$, consistent with the displayed $p^2-i0$ denominators. The quadratic action in the canonical fields isColor indices are diagonal. Introduce the longitudinal and transverse projectors $P_{L,\mu \nu }=p_{\mu }{p}_{\nu }/p^2$ and $P_T=\eta -P_L$. The gauge kinetic matrix is $-p^2(P_T+{\xi }^{-1}{P}_L)$, with inverse $-(P_T+\xi P_L)/p^2$. The ghost kinetic operator is $-p^2$. The Gaussian functional integral and Grassmann integral therefore giveThe longitudinal formula is understood as the Feynman-prescribed inverse of the kinetic operator; away from its poles the projector inversion is ordinary algebra. The propagators for the original unrescaled fields are $g^2$ times these expressions. Using $P_{T}p=0$ and $P_{L}p=p$ giveswith either normalization. This is the longitudinal gauge propagator contraction.

The canonical ghost interaction comes directly from the covariant derivative:For incoming antighost momentum $p$, the derivative supplies $i{p}^{\mu }$; the expansion of $e^{iS}$ supplies another $i$. Thus the ghost-gluon vertex is $g{p}^{\mu }{f}_{abc}$, as requested. For unrescaled fields its coefficient is instead $p^{\mu }{f}_{abc}/g^2$. The printed order-$g$ rule presupposes canonical normalization; keeping the overall $1/g^2$ and an order-$g$ vertex for the same unrescaled variables would mix conventions. Choosing Fourier modes $e^{-ipx}$ instead reverses the vertex sign and must be done consistently for all momentum assignments.

In Feynman gauge, $\xi =1$, the one-loop ghost two-point graph has one internal ghost line and one internal gauge line joined by two such vertices. The external ghost line is open, so it does not carry a closed ghost-loop minus sign. With an orthonormal adjoint basis, the structure constants are totally antisymmetric, and the color product along this line is $f_{adc}{f}_{cdb}=-C{\delta }_{ab}$. The two internal propagators supply $(-i{)}^2=-1$, leaving a positive color-and-vertex factor for the amputated insertion. DefineThe amputated insertion is $i{\delta }_{ab}{g}^2{I}_d(p)C$. Restoring the two external canonical ghost propagators gives the one-loop contributionThe literal unrescaled correlator in the question is $g^2{G}^{(1)}$, hence of order $g^4$, whereas its tree term is $-i{\delta }_{ab}{g}^2/(p^2-i0)$. In $d\ne 4$ the dimensionless renormalized coupling convention adds ${\mu }^{4-d}$ multiplying $g^2$ in the canonical loop term.

To evaluate the Feynman integral, combine the denominators with a Feynman parameter:Shift $\ell =k-(1-\alpha )p$. The denominator becomes $[{\ell }^2+\alpha (1-\alpha )p^2-i0{]}^2$, and the numerator is $p\cdot \ell +(1-\alpha )p^2$. The odd term vanishes in translation-invariant dimensional regularization. For spacelike $p$, a Wick rotation converts the remaining integral into a Euclidean one and cancels the prefactor $1/i$. The needed radial integral isIndeed, spherical coordinates and $u={\ell }_E^2/M^2$ leave a beta integral ${\int }_0^{\infty }{u}^{d/2-1}(1+u{)}^{-2}du=\Gamma (d/2)\Gamma (2-d/2)$; the angular factor cancels $\Gamma (d/2)$. Establish this first in a convergent range and then analytically continue in $d$. Substituting $M^2=\alpha (1-\alpha )p^2$ yieldsThis is the printed result, including the factor $1/i$ in its Minkowski-space definition. Equivalently the parameter integral is $B(d/2-1,d/2)$. Symmetry under $k\leftrightarrow p-k$ also gives $I_d(p)$ as $p^2/2$ times the massless scalar bubble, independently checking the numerator factor.

Write $ϵ=4-d$. Since $\Gamma (ϵ/2)=2/ϵ+O(1)$ and the parameter integral tends to ${\int }_0^1(1-\alpha )d\alpha =1/2$, the pole isHere $p$ is off shell and nonzero, separating the ultraviolet pole from massless infrared issues. The unrescaled amplitude has an extra factor $g^2$. The divergence is proportional to the existing ghost kinetic operator, so the one-loop ghost kinetic counterterm in Feynman gauge is local:It inserts $-i\delta Z_c{p}^2$ into the inverse propagator, cancelling the divergent insertion $+i{g}^{2}C{I}_d(p)$. With the two external propagators attached, its contribution is $+i{\delta }_{ab}\delta Z_c/p^2$, cancelling the displayed connected amplitude pole. In the original variables the same action term is $-\delta Z_c\partial \bar{c}\cdot \partial c/g^2$. If instead one defines dimensional regularization by $d=4-2ϵ$, the identical counterterm is written $\delta Z_c=g^{2}C/(32{\pi }^2ϵ)$; the factor of two is purely the regulator convention.

Use the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and unit-weight antisymmetrization. The duality identity between ${\gamma }_5{\gamma }_{\nu }$ and three antisymmetrized gamma matrices turns the equation intoThe overall nonzero factor in this identity depends on the orientation and ${\gamma }_5$ convention and does not change its solutions. This Rarita-Schwinger field equation has the gauge symmetrysince ${\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\partial }_{\rho }ϵ=0$. It is the free-field limit of local supersymmetry, so it is suitable for the gravitino.

To check which spin it propagates, put $A={\gamma }^{\mu }{\Psi }_{\mu }$ and $B={\partial }^{\mu }{\Psi }_{\mu }$. Expanding the three gamma matrices using the Clifford algebra givesin four dimensions. Thus the Euler-Lagrange field equation requires $B=/\partial A$. Under the gauge transformation, $A\mapsto A+/\partial ϵ$; locally one may choose gamma-trace gauge fixing, $A=0$, by solving $/\partial ϵ=-A$. The Euler-Lagrange field equation then yieldsThese are the massless Dirac equation, transversality and gamma-trace constraints that remove the lower-spin part of a vector-spinor.

For a more explicit massless Rarita-Schwinger polarization count, take a positive-frequency plane wave with null momentum along the third axis, in signature $(-,+,+,+)$. Transversality relates its temporal and longitudinal components. Residual gauge transformations with $/pϵ=0$ remove both. Write the surviving components as $u_1,u_2$; their gamma-trace condition reads ${\gamma }^1{u}_1+{\gamma }^2{u}_2=0$, henceThe massless Dirac equation leaves two complex positive-frequency amplitudes in $u_1$. If $i{\gamma }^1{\gamma }^2{u}_1=s{u}_1$, $s=\pm 1$, then $(u_1,u_2)$ is proportional to $(1,-is)$ in the transverse vector index. With spatial rotation generators $J_3^{\mathrm{vec}}{}_{12}=-i$ and $J_3^{\mathrm{spin}}=-i{\gamma }^1{\gamma }^2/2$, the vector has helicity $-s$ and the Majorana spinor has helicity $-s/2$, giving total helicity $-3s/2$. The Majorana spinor condition fixes the negative-frequency coefficients from these amplitudes. Therefore a massless gravitino has spin $3/2$, with the two physical helicities $\pm 3/2$. A complex vector-spinor would also have independent antiparticle modes.

The requested count is instead the off-shell component count of minimal supergravity, before imposing any Euler-Lagrange field equations. A four-dimensional Majorana spinor has four real components. Its vector index supplies four copies, and local supersymmetry supplies four real gauge functions:One must not additionally subtract transversality and the massless Dirac equation in this count: those are on shell conditions. The gamma-trace gauge fixing already uses the same four gauge functions subtracted above. For the graviton, a symmetric metric tensor has ten real components, with four diffeomorphism gauge functions. Equivalently, a vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism functions:The gravitino and graviton each have two propagating states on shell, but their off-shell counts disagree by six.

An auxiliary field has an algebraic Euler-Lagrange field equation and no propagating kinetic term. In old-minimal supergravity, a complex scalar $M$ contributes two real bosonic components and a real vector $b_{\mu }$ contributes four, so the bosonic count becomes $6+2+4=12$, matching the gravitino. Their supersymmetry transformations also absorb the terms that would otherwise require the propagating Euler-Lagrange field equations to close the local algebra. Thus auxiliary fields allow off-shell supersymmetry closure and matching component counts without adding physical particles. Eliminating them recovers an on shell formulation.

Use target signature $(-,+,\dots ,+)$ and worldsheet signature $(-,+)$. The Polyakov action has local worldsheet diffeomorphism and Weyl invariance, together with global target-space Poincaré symmetry. Under a worldsheet coordinate change $\sigma \mapsto {\sigma }^{\prime }$, the embedding coordinates are scalars and the worldsheet metric is a tensor:A Weyl transformation acts as $g_{ab}\mapsto e^{2\rho (\sigma )}{g}_{ab}$ and leaves $X$ unchanged. In two dimensions $\sqrt{-g}$ scales by $e^{2\rho }$ and $g^{ab}$ by $e^{-2\rho }$, so the action is invariant. Target translations act by $X^{\mu }\mapsto X^{\mu }+a^{\mu }$, and Lorentz transformations by $X^{\mu }\mapsto {\Lambda }^{\mu }{}_{\nu }{X}^{\nu }$ with ${\Lambda }^{\mathsf{T}}\eta \Lambda =\eta$; they leave $g$ unchanged. Infinitesimally the latter have $\delta X^{\mu }={\omega }^{\mu }{}_{\nu }{X}^{\nu }$, where ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$.

Locally, the two diffeomorphism functions put the metric into the form $g_{ab}=e^{2\rho }{\eta }_{ab}$, and Weyl invariance removes the conformal factor. In this conformal gauge, the embedding coordinates are free scalar fields and satisfyThe metric equation must still be imposed after gauge fixing. Varying the ungauged metric gives the worldsheet stress-energy tensor constraintWith a dot and prime denoting the two coordinate derivatives, the independent Virasoro constraints areequivalently $(\dot{X}\pm X^{\prime }{)}^2=0$. Local conformal gauge leaves residual conformal transformations; on general worldsheet topologies it also leaves global moduli, rather than removing all metric information.

For target translations, differentiation of the Lagrangian with respect to ${\partial }_a{X}_{\mu }$ gives the Noether currentThe embedding equation is ${\partial }_a{j}^{a\mu }=0$. In conformal gauge, $j^{\tau \mu }={\dot{X}}^{\mu }/(2\pi {\alpha }^{\prime })$, so integration over the closed string givesPeriodic boundary conditions make its time derivative vanish. For an infinitesimal Lorentz transformation, the antisymmetric coefficient of ${\omega }_{\mu \nu }$ gives the current $j^{a,\mu \nu }=X^{\mu }{j}^{a\nu }-X^{\nu }{j}^{a\mu }$. Its divergence vanishes because the remaining product of embedding derivatives is symmetric in $\mu ,\nu$. Thus the Noether charge isThese are the target-space Noether charges of a string.

For the circular solution take $R>0$, and let ${\mathbf{e}}_r=(\cos \sigma ,\sin \sigma )$ and ${\mathbf{e}}_{\sigma }=(-\sin \sigma ,\cos \sigma )$. Its derivatives areEach spatial coordinate has equal second derivatives in $\tau$ and $\sigma$, while the time coordinate has both second derivatives zero. Furthermoreso both the field equations and the Virasoro constraints hold, including the collapse instants.

At target time $t=X^0=R\tau$, the string is a circle in the $X^1{X}^2$ plane of radius $R|\cos (t/R)|$. It contracts from radius $R$ to zero and re-expands; the negative cosine reverses material labels by a half-turn rather than creating a negative radius. The spatial configuration repeats after target time $\pi R$. This is a pulsating circular string. Its induced metric degenerates at collapse, but the conformal-gauge embedding and constraints give its continuation.

The integrals of $\sin \sigma$ and $\cos \sigma$ vanish, so the momentum isThe spatial velocity is radial, giving zero angular momentum density in the plane; all boost-charge terms integrate to zero as well. Hence $J^{\mu \nu }=0$ for every pair of indices.

At maximum expansion, all string elements are instantaneously at rest, and the length is $L_0=2\pi R$. Thus the rest mass per unit length, equal to the string tension, isAt other times the instantaneous circumference is $L=2\pi R|\cos \tau |$, and the elements have speed $|\sin \tau |$. Their Lorentz factor is $1/|\cos \tau |$, so the energy remains $TL/\sqrt{1-v^2}=2\pi TR$. The rest-energy density is $T$; the total rest-frame energy divided by the changing circumference is not constant. This distinguishes the requested intrinsic mass density from the larger energy density of moving string elements.

A Non-Abelian gauge theory describes fields with a local redundancy, not an independent physical degree of freedom for each component of its gauge field. Take a compact gauge group with Hermitian generators $T^a$ and Lie algebra structure constants defined by $[T^a,T^b]=i{f}^{abc}{T}^c$. In a mostly-plus metric signature, the pure Yang-Mills theory hasWith the matter gauge covariant derivative $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, a finite gauge transformation acts as $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$. Expanding $U=e^{ig{\omega }^a{T}^a}$ givesThe Yang-Mills action is constant along each gauge orbit. Consequently a naive functional integral $\int \mathcal{D}A{e}^{iS[A]}$ repeatedly integrates the same physical configuration and contains the infinite formal volume of the local gauge group. At quadratic order the same problem appears as noninvertibility: the free kinetic operator annihilates $A_{\mu }={\partial }_{\mu }\omega$, so no unique gauge-boson propagator exists. In canonical language the Gauss law constraint in gauge theory and the absence of an independent momentum for the time component show that the field components cannot all be independent oscillators. In covariant quantization, time-like and longitudinal polarizations must not be counted as additional physical states.

The Faddeev-Popov determinant implements a local quotient by the gauge orbit. Choose $G^a[A]={\partial }^{\mu }{A}_{\mu }^a$ and first impose $G^a[A]=f^a(x)$ for a specified function $f$. This is the Lorenz gauge condition when $f=0$. Its infinitesimal change isThe direct Faddeev-Popov operator is therefore $M_0^{ab}={\partial }^{\mu }(D_{\mu }{)}^{ab}$. Locally, if the condition cuts each orbit once and residual zero modes have been removed by boundary conditions, changing variables from $\omega$ to $G[A^{\omega }]-f$ proves the Faddeev-Popov gauge-orbit identityThe determinant is the functional Jacobian of this change of variables. A finite-dimensional version is $\int d\omega \delta (G(\omega )-f)=1/|\det (\partial G/\partial \omega )|$ near a unique root; the perturbative determinant has a fixed sign or phase, absorbed into normalization. This is the same change-of-variables mechanism used in Bastianelli's derivation of the Faddeev–Popov construction.

Insert the identity into the normalized functional integral for a gauge-invariant observable $\mathcal{O}[A]$:For each fixed $\omega$, change the integration variable from $A$ to $A^{\omega }$. The action, observable and measure are invariant, and the determinant is the orbit Jacobian at the resulting gauge slice. The integrand is now independent of the separately integrated orbit parameter. Its integral is $\mathrm{Vol}\mathcal{G}$, canceling the denominator. Thus, up to normalization,This assumes an anomaly-free measure and a local perturbative gauge slice; it is not an assertion of a globally unique representative on every gauge orbit.

Average over $f$ with weight $\exp [-i\int d^{4}x{f}^a{f}^a/(2\xi )]$. The delta functional sets $f=G[A]$, so the average produces the covariant gauge termNext represent the Faddeev-Popov determinant by a Grassmann Gaussian integral. Choose $M=-M_0=-{\partial }^{\mu }{D}_{\mu }$; changing $M_0$ to $-M_0$ changes the regulated determinant by a field-independent factor that is absorbed in normalization. For a Faddeev-Popov ghost field $c$ and independent antighost field $\bar{c}$,To see why it is a determinant rather than its reciprocal, diagonalize a finite-dimensional quadratic form formally: for each Grassmann pair only the term proportional to ${\bar{c}}_j{c}_j$ survives integration, supplying one eigenvalue. Multiplying the surviving eigenvalues gives $\det M$, with an overall factor of $i$ per pair. The functional version is defined with the same regulator as the rest of the theory.

Combining these steps gives the gauge-fixed Yang–Mills LagrangianThis is the gauge-fixed Yang-Mills Lagrangian in a covariant gauge. Integrating its ghost term by parts givesThe second term is the ghost-gluon vertex in the antighost convention chosen here. The Faddeev-Popov ghosts are anticommuting Lorentz scalars introduced to represent a determinant; they are not extra asymptotic particles. A closed ghost loop has a minus sign from their Grassmann parity. In a Non-Abelian gauge theory the determinant depends on $A$, so these loops are essential. In an Abelian gauge theory $D_{\mu }$ in the adjoint sector reduces to ${\partial }_{\mu }$, the determinant is field independent, and its ghost sector decouples.

The quadratic gauge field kernel at non-null momentum isDefine the mixed-index longitudinal projector of a vector field $P_L{}^{\mu }{}_{\nu }=k^{\mu }{k}_{\nu }/k^2$ and transverse projector of a vector field $P_T=1-P_L$. The kernel is $-k^2(P_T+{\xi }^{-1}{P}_L)$ and hence has inverse $-k^{-2}(P_T+\xi P_L)$. Multiplying by $i$ gives the gauge-boson propagator, off its poles,The vacuum Feynman i-epsilon prescription completes this off-pole formula; with the mostly-plus convention its physical transverse denominator is $k^2-i0$. The choice $\xi =1$ is Feynman gauge; $\xi \to 0$ is covariant Landau gauge. The invertible quadratic kernel supplies perturbation theory, while the ghost determinant ensures the gauge-orbit measure has been accounted for.

The consistency of the unphysical sector is organized by BRST symmetry. Introduce a bosonic Nakanishi-Lautrup field $B^a$ and a left-acting BRST differentialThe graded Leibniz rule and the Jacobi identity give $s^2=0$: for example $s^{2}A=D(sc)+g[Dc,c]=0$ since $D[c,c]=2[Dc,c]$, and $s^{2}c$ vanishes by the graded Jacobi identity. Nilpotence on $\bar{c}$ and $B$ is immediate. Using the odd antighost field, the same graded Leibniz rule givesEliminating $B^a$ by its algebraic field equation $B^a=-G^a/\xi$ recovers the displayed gauge-fixing and ghost terms. Thus they are BRST-exact together, and the total action preserves BRST symmetry. At the level of states, the BRST charge $Q$ implements physical states as the ghost-number-zero BRST cohomology $\ker Q/\mathrm{im}Q$. This prescription removes longitudinal, time-like and ghost excitations from physical amplitudes while allowing them on internal lines. For a BRST-closed observable, changing $\xi$ inserts $s({\bar{c}}^a{B}^a/2)$; its expectation vanishes by the BRST change-of-variables identity when the measure preserves the symmetry. This explains the independence of physical quantities from the arbitrary gauge parameter.

Finally, a local condition such as Lorenz gauge can have several solutions on the same gauge orbit, the Gribov ambiguity. The preceding Faddeev-Popov gauge-orbit identity and perturbative gauge fixing construction apply in a neighborhood where the linearized operator is invertible after residual modes are excluded. They resolve perturbative gauge overcounting and supply consistent covariant Feynman rules; they do not by themselves prove a globally unique nonperturbative gauge fixing.

Take ${\partial }_b$ of the Proca equation. Commuting partial derivatives and antisymmetry make ${\partial }_b{\partial }_a{F}^{ab}=0$, while current conservation gives ${\partial }_b{J}^b=0$. Thus $m^2{\partial }_b{A}^b=0$. For $m\ne 0$, $\boxed{{\partial }_b{A}^b=0}$. This Lorenz constraint in Proca theory is a consequence of the massive equations, rather than a gauge choice.