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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.

Absorb the gauge coupling into the connection $A_{\mu }=A_{\mu }^a{T}_a$ and use $D_{\mu }={\partial }_{\mu }+A_{\mu }$. The anti-Hermitian generators make $U(x)=\exp [{\omega }^a(x)T_a]$ unitary for real parameters. With a matter field transforming as ${\psi }^{\prime }=U\psi$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Applying both sides to an arbitrary field givesExpanding $U=1+\omega +O({\omega }^2)$ and $U^{-1}=1-\omega +O({\omega }^2)$ yields the infinitesimal gauge transformation with an anti-Hermitian connectionThe adjoint covariant derivative is $D_{\mu }\omega ={\partial }_{\mu }\omega +[A_{\mu },\omega ]$. In components,The trace normalization makes the structure constants totally antisymmetric: $f_{abc}=-\mathrm{Tr}([T_a,T_b]T_c)$ and cyclicity of the trace supply antisymmetry in the remaining indices. Replacing $U=e^{\omega }$ by $e^{-\omega }$ reverses the infinitesimal parameter convention and gives $\delta A=D\omega$; mixing the two conventions would change the ghost operator incorrectly. The trace normalization also gives the invariant quadratic form $-\mathrm{Tr}(A_{\mu }{A}_{\nu })=A_{\mu }^a{A}_{\nu }^a$.

A gauge-invariant functional integral counts all representatives along each gauge orbit. The gauge condition $\mathcal{F}[A]=0$ selects a representative, and its delta functional enforces that selection. Assume a local patch where each orbit intersects the gauge slice once and the gauge-condition derivative has no residual zero modes. DefineChanging the orbit integration variable from $\omega$ to $\mathcal{F}[A^{\omega }]$ givesInsert this Faddeev-Popov gauge-orbit identity into the unfixed path integral. The action and measure are invariant under a change $A\mapsto A^{\omega }$, so integration over $\omega$ factors out as the gauge-group volume. Dividing by that redundant volume leaves the fixed integral and its Faddeev-Popov determinant. In a perturbative patch with fixed determinant sign or phase, the absolute value is replaced by the oriented determinant, up to a field-independent normalization:Here $D_{\mu }^{cb}={\delta }^{cb}{\partial }_{\mu }+f_{dbc}{A}_{\mu }^d$. This is a functional determinant of the gauge-condition Jacobian, not an arbitrary extra weight. The local invertibility assumption matters: residual gauge transformations or multiple orbit intersections, described by the Gribov ambiguity, prevent treating this derivation as a global unique gauge slice without further qualifications.

Both extra factors can be incorporated into an enlarged action. An auxiliary real field $b^a$ Fourier-represents the delta functional, while a Grassmann Gaussian integral over independent Faddeev-Popov ghost fields $c,\bar{c}$ represents the determinant:Thus the original fixed integral can be written with action $S+\int b\mathcal{F}+\int \bar{c}Mc$ and ordinary integration over the extra fields, with no explicit delta functional or determinant. The anticommuting fields give a determinant rather than the inverse determinant obtained from commuting Gaussian fields.

For the usual Gaussian family of gauges, add $\xi b^a{b}^a/2$ and integrate out $b$. Completing the square producesThis is Gaussian gauge fixing with an auxiliary field; equivalently one averages the condition $\mathcal{F}=f$ with its Gaussian weight. Gauge-invariant observables are unchanged apart from normalization. For the Lorenz gauge ${\mathcal{F}}_a={\partial }^{\mu }{A}_{\mu }^a$, the chosen transformation convention gives $M=-{\partial }^{\mu }{D}_{\mu }$. The ghost action is $-\int \bar{c}{\partial }^{\mu }{D}_{\mu }c$, or $\int ({\partial }^{\mu }\bar{c})D_{\mu }c$ after integration by parts. The dependence of $D$ on $A$ accounts for the ghost interactions.

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.