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Assume an invariant volume measure, an invariant integrand and one regular gauge slice representative per orbit. At that representative, let $M$ be the Jacobian matrix of $F$ along the group coordinates, whose Haar measure is normalized to ordinary coordinate measure at the identity. Changing coordinates from the group parameter to $F$ gives ${\int }_{G}d\mu (g)\delta (F(x_g))=1/|\det M(x_0)|$. Multiply by its reciprocal and insert this identity into the original integral. Invariance permits a change of variables $x\mapsto x_g$ for each $g$, factoring out the group volume. The absolute determinant is required for ordinary unoriented real integration; a consistently positive orientation permits writing $\det M$. Infinite group volume is divided out formally or treated with a regulator.

Let a gauge transformation act infinitesimally as $\delta A_{\mu }^a=(D_{\mu }\omega {)}^a$, and define the Faddeev-Popov operator by variation of the chosen gauge functional:Work in a perturbative regular gauge slice, with boundary conditions that remove residual zero modes and with one local representative of each gauge orbit. The Faddeev-Popov gauge-orbit identity is $1={\Delta }_{\mathrm{FP}}[A]\int \mathcal{D}\omega \delta [\mathcal{F}[A^{\omega }]]$. Inserting it and factoring out the formal gauge-group volume givesThe functional delta imposes the gauge condition at every point and for every color. The determinant compensates the gauge-orbit Jacobian. For ordinary unoriented real gauge-orbit integration the local Jacobian is $|\det \mathcal{M}|$; perturbatively its sign is fixed and absorbed into $\mathcal{N}$, leaving the displayed determinant. Multiple global intersections, or a Gribov ambiguity, require more care; the local perturbative formula is not a global uniqueness assertion.

The complex Grassmann Gaussian integral now generalizes to spacetime and color indices. Introduce independent anticommuting Faddeev-Popov ghost fields $c^a(x),{\bar{c}}^a(x)$ and writeThe factors $i$ in the functional determinant differ from a real exponential only by a regulated, field-independent constant. These Grassmann fields are Lorentz scalars in the adjoint color representation; the antighost is independent of the ghost rather than an ordinary complex-conjugate commuting variable. They are unphysical fields, not external observable particles. Representing the functional determinant by a local ghost action when $\mathcal{F}$ is local makes ordinary Feynman rules available. Anticommutation gives the minus sign of a closed ghost loop, enabling cancellation of unphysical gauge contributions and maintaining the gauge identities in perturbation theory.

For example, the linear covariant choice ${\mathcal{F}}^a={\partial }^{\mu }{A}_{\mu }^a$ gives ${\mathcal{M}}^{ab}={\partial }^{\mu }{D}_{\mu }^{ab}$, so non-abelian ghost-gauge interactions remain. In an axial gauge ${\mathcal{F}}^a=n^{\mu }{A}_{\mu }^a$ with constant $n$, however,on the exact gauge slice. Its determinant is field-independent, so axial-gauge ghosts decouple and contribute only a normalization factor. This statement uses the strict delta-functional constraint; a finite-width Gaussian gauge weight is not the same exact slice.

The disadvantage is that choosing $n$ obscures manifest Lorentz covariance and the axial-gauge propagator contains spurious $1/(n\cdot k)$ poles. An axial-gauge pole prescription and control of residual transformations satisfying $n\cdot \partial \omega =0$ are necessary; formal decoupling alone does not fix those problems.

Abelian ghosts do not always decouple. In a field-independent linear Lorenz gauge, $\delta A_{\mu }={\partial }_{\mu }\omega$ gives $\mathcal{M}=\square$ and they are free. But the quadratic Abelian gauge fixing $\mathcal{F}[A]=\partial \cdot A+\kappa A_{\mu }{A}^{\mu }$ givesIts Faddeev-Popov operator depends on $A$, and the ghost action contains $2\kappa \bar{c}{A}^{\mu }{\partial }_{\mu }c$. This is an explicit abelian ghost-gauge interaction. Abelianity removes the commutator term in the gauge transformation, not the field dependence of an arbitrary nonlinear gauge condition.

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.