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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.