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For the left graded Leibniz rule and $s\bar{c}=-b$, $sb=0$, the displayed gauge-fixing fermion gives $b\partial \cdot A+\xi b^2/2+\bar{c}\partial \cdot Dc$. The ghost signs follow from odd parity. BRST nilpotence implies invariance of this density without using field equations. Eliminating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$ for nonzero $\xi$.

For the convention $s\bar{c}=-b$, the action $S_{\mathrm{YM}}-s\Psi$ contains $-b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$ after integration by parts. Integrating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$. The Grassmann Gaussian integral represents the Faddeev-Popov determinant.

The Grassmann-odd partner of the Faddeev-Popov ghost field in the gauge-fixing determinant representation. In an auxiliary-field BRST symmetry description it obeys $sb=h$, where $h$ is the Nakanishi-Lautrup field. Its ordinary spacetime derivative remains odd, which matters when moving an anticommuting BRST parameter past it.

For gauge fixing with $F^a=\partial \cdot A^a+(n\cdot \partial {\varphi }^a)/\sqrt{2}$, integrating out the Nakanishi-Lautrup field produces gauge-scalar mixing. The Schur complement of the free gauge-field kernel cancels the extra $(n\cdot p{)}^2/2$ in the scalar kernel, leaving $⟨{\varphi }^a(p){\varphi }^b(-p)⟩={\delta }^{ab}/p^2$. This is independent of the gauge vector $n$, but still depends on the full momentum $p$.

Use $F_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a+g{f}^{abc}{A}_{\mu }^b{A}_{\nu }^c$ and $(D_{\mu }c{)}^a={\partial }_{\mu }{c}^a+g{f}^{abc}{A}_{\mu }^b{c}^c$. The Faddeev-Popov gauge-orbit identity with Lorenz gauge function $G^a={\partial }^{\mu }{A}_{\mu }^a$, followed by Gaussian averaging of $G^a=f^a$, supplies the quadratic gauge-fixing term. Represent the determinant of $-{\partial }^{\mu }{D}_{\mu }$ using a Grassmann Gaussian integral over a Faddeev-Popov ghost field and antighost field to obtain the ghost term. The minus sign in this operator differs by a field-independent determinant factor from the direct gauge-condition Jacobian; it is a consistent antighost convention.

The ghost term equals $({\partial }^{\mu }{\bar{c}}^a)(D_{\mu }c{)}^a$ up to a boundary term. Thus it contains a ghost-gluon vertex with the sign matching this kinetic convention. An equivalent auxiliary-field form replaces the gauge-fixing term by $B^a{G}^a+\xi B^a{B}^a/2$. The Nakanishi-Lautrup field equation gives $B^a=-G^a/\xi$ and recovers the displayed Lagrangian. With left-acting BRST transformations $sA=Dc$, $sc=-g[c,c]/2$, $s\bar{c}=B$, and $sB=0$, the gauge-fixing and ghost terms together are $s[{\bar{c}}^a(G^a+\xi B^a/2)]$.

Complete the square as $bF+\xi b^2/2=\xi (b+F/\xi {)}^2/2-F^2/(2\xi )$. Translation of the Gaussian functional integral gives the displayed identity for nonzero $\xi$, with the appropriate oscillatory contour prescription. At $\xi =0$, the Nakanishi-Lautrup field is a Lagrange multiplier imposing the gauge-condition delta functional. This field has no kinetic term and no independent propagating degree of freedom.

The Nakanishi-Lautrup field is an auxiliary bosonic field used to write gauge fixing as a BRST-exact term. Its algebraic field equation produces the usual quadratic gauge-fixing term.

Using the graded Leibniz rule, $s^{2}A=D(sc)+g(Dc)\times c=0$ because $D(c\times c)=2(Dc)\times c$. The graded Jacobi identity gives $s^{2}c=g^2(c\times c)\times c/2=0$. BRST nilpotence on the antighost and Nakanishi-Lautrup field is immediate. No field equations are required.

Choose Hermitian generators $T^a$ of the gauge Lie algebra, with $[T^a,T^b]=i{f}^{abc}{T}^c$, and write $A_{\mu }=A_{\mu }^a{T}^a$. For the usual compact gauge algebra choose $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$. More generally use a nondegenerate invariant symmetric bilinear form on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.

A matter vector is useful to fix conventions: with ${\psi }^{\prime }=U(x)\psi$ and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Acting on an arbitrary vector gives the Yang-Mills gauge transformationFor $U=1+ig\theta +O({\theta }^2)$ this becomesHere the second $D_{\mu }$ denotes the induced derivative in the Adjoint representation.

Define the curvature by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$. Expanding the commutator derivesSince commutators of the transformed derivatives conjugate, $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$. Cyclicity of the trace proves the invariant bilinear-form construction of a Yang-Mills action:For a general invariant form $h$, the corresponding proof is $h([\theta ,X],Y)+h(X,[\theta ,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.

Introduce the adjoint Faddeev-Popov ghost $c^a$ and antighost ${\bar{c}}^a$, independent Grassmann-odd scalar fields of ghost numbers $+1$ and $-1$, together with the Grassmann-even Nakanishi-Lautrup field $b^a$ of ghost number zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left BRST symmetry differential:It is an odd derivation: $s(XY)=(sX)Y+(-1{)}^{|X|}X(sY)$ for homogeneous Grassmann parity $|X|$. The two relevant properties are $s{\mathcal{L}}_{\mathrm{YM}}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the Lie algebra's graded Jacobi identities. ThusAlgebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd gauge-fixing fermion of ghost number $-1$ so that $s\Psi$ is even and has ghost number zero.

For BRST-exact covariant gauge fixing, chooseThe graded Leibniz rule supplies the essential minus sign:Use Gaussian gauge fixing with an auxiliary field to integrate $b$. Pointwise,Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is thereforeThis is covariant Feynman gauge, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell BRST symmetry formulation.

Finally, the ungauge-fixed quadratic momentum kernel isGauge directions are null vectors, so there is no inverse defining a vector propagator, and the functional integral also counts an infinite gauge orbit volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu \nu }^{ab}=-{\delta }^{ab}{p}^2{\eta }_{\mu \nu }$. With the vacuum prescription, the propagators areThe ghost action represents the Faddeev-Popov determinant and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative Feynman rules. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.

The gauge-invariant Yang-Mills action integrates over physically equivalent configurations along each gauge orbit. At the quadratic level its gauge-field kernel has longitudinal zero directions: in momentum space $K_{\mu \nu }(k)=-k^2{\eta }_{\mu \nu }+k_{\mu }{k}_{\nu }$ satisfies $K_{\mu \nu }{k}^{\nu }=0$. It therefore has no inverse from which to construct a gauge-boson propagator. Gauge fixing both removes the redundant orbit integration and supplies an invertible quadratic operator away from the ordinary physical propagator poles.

For the covariant condition ${\partial }^{\mu }{A}_{\mu }=0$, the Faddeev-Popov determinant is that of ${\partial }^{\mu }{D}_{\mu }$. Its representation by odd Faddeev-Popov ghost fields and antighost fields produces the displayed ghost term. In a nonabelian theory $D_{\mu }$ depends on the gauge field, so the determinant is not an irrelevant constant: ghost interactions must be retained. The even Nakanishi-Lautrup field $b$ writes gauge fixing in a form with off-shell BRS symmetry. For $\xi \ne 0$, its algebraic equation is $b=-\partial \cdot A/\xi$, giving $-(\partial \cdot A{)}^2/(2\xi )$ after elimination. At $\xi =0$ it imposes the gauge condition as a Lagrange multiplier.

For the nilpotence proof, write the adjoint bracket as $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$, with coefficients in a Grassmann algebra. If $|X|,|Y|$ are their parities, this is a graded Lie bracket:Thus two odd fields have a symmetric bracket, in particular $[c,c]$ need not vanish. The transformations become $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $s\bar{c}=-b$, $sb=0$, and $s$ obeys the left graded Leibniz rule.

Because $D_{\mu }$ is even, $D_{\mu }[c,c]=[D_{\mu }c,c]+[c,D_{\mu }c]=2[D_{\mu }c,c]$. Varying both the field and the connection in $D_{\mu }c$ now givesFor the ghost,The last equality is the graded Jacobi identity applied to three copies of the odd field $c$. In components the remaining expression is $\frac{1}{2}{f}^{abc}{f}^{bde}{c}^d{c}^e{c}^c$; its coefficient is antisymmetrized over $c,d,e$, and the ordinary Jacobi identity for the structure constants makes that antisymmetrization zero. Finally $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. HenceThis is off-shell nilpotence of the Yang-Mills BRST quartet.

To verify invariance of the complete Lagrangian, $s{F}_{\mu \nu }=[F_{\mu \nu },c]$, obtained from $[D_{\mu },D_{\nu }]c=[F_{\mu \nu },c]$. Thus $s(F_{\mu \nu }^a{F}^{\mu \nu a})=0$ because the product of the two curvature components is symmetric in their color labels while $f^{abc}$ is antisymmetric. The remaining terms vary asThe last term vanishes by the already proved $s^{2}A=0$, and the first and third terms cancel pointwise. ThereforeEquivalently, the covariant gauge-fixing density as a BRST variation is $s\Psi$ with $\Psi =-{\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded sign from moving $s$ past $\bar{c}$ produces the positive ghost term, and $s^2\Psi =0$ proves its invariance. No integration-by-parts qualification is needed for the density as written.

Let $Q=Q_{\mathrm{BRS}}$ be the BRS charge. The physical state space is its BRST cohomology, usually at ghost number zero:Nilpotence ensures $\mathrm{im}Q\subseteq \ker Q$. A state $Q|\chi ⟩$ is orthogonal to every closed state because $⟨\psi |Q|\chi ⟩=⟨Q\psi |\chi ⟩=0$, and its norm is $⟨\chi |Q^2|\chi ⟩=0$. Thus adding an exact state changes neither physical inner products nor amplitudes. Self-adjointness here is relative to the stated indefinite Hermitian form; a nonzero self-adjoint nilpotent operator could not exist on a positive-definite state space. Positivity must be proved on the quotient, not assumed on the gauge/ghost space.

Fix a color label and a nonzero momentum $k$, and take the usual nontrivial asymptotic charge, $\alpha \ne 0$. Nilpotence on a vector state gives $Q|c⟩=0$, because at least one component of $k$ is nonzero. Nilpotence on the antighost state then givesThe two coefficients are complex conjugates as required by self-adjointness: using the given ghost pairing,Thus the antighost-to-vector coefficient is ${\alpha }^{\ast }$, as the printed bar notation indicates. These conditions are the massless shell condition and the consistency of the charge with the inner products.

A general one-particle vector representative is $|\epsilon ⟩={\epsilon }^{\mu }|A_{\mu }⟩$. It is closed precisely when $k_{\mu }{\epsilon }^{\mu }=0$. The vector state $k^{\mu }|A_{\mu }⟩=Q|\bar{c}⟩/\bar{\alpha }$ is exact, so representatives obeyThe ghost state is also exact: choose $v^{\mu }$ with $\alpha v\cdot k=1$, then $Q(v^{\mu }|A_{\mu }⟩)=|c⟩$. The antighost state is not closed because its image is a nonzero longitudinal vector. More explicitly, closure of a general sum of vector, ghost and antighost states forces the antighost coefficient to vanish and the vector polarization to be orthogonal to $k$; its ghost component disappears in the quotient. Thus the full one-particle cohomology already reduces toWith the mostly-plus Minkowski metric, choose a Lorentz frame with $k^{\mu }=(\omega ,0,0,\omega )$, $\omega >0$. Closure gives ${\epsilon }^3={\epsilon }^0$, and the norm isSubtracting a multiple of $k$ sets ${\epsilon }^0={\epsilon }^3=0$. Consequently positive one-particle BRST cohomology consists of two transverse positive-norm polarizations per color, represented by $|A_1⟩,|A_2⟩$. A nonzero class has strictly positive norm; the zero-norm longitudinal direction was precisely the exact direction removed.

The qualifications $k\ne 0$ and $\alpha \ne 0$ are essential to this one-particle argument. The zero-momentum sector requires separate treatment, and the trivial-charge case cannot yield the asserted removal of negative-norm vector states from the stated relations alone. Abstract nilpotence and self-adjointness by themselves do not guarantee positive cohomology.

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.

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.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.

Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.