 Solution

Choose an infinitesimal gauge transformation $\delta A_{\mu }^a=D_{\mu }^{ab}{\omega }^b$, where $D_{\mu }^{ab}={\delta }^{ab}{\partial }_{\mu }+g{f}^{acb}{A}_{\mu }^c$. Varying the axial gauge function at fixed $\sigma$ givesThis is the Faddeev-Popov operator. The Faddeev-Popov determinant has a Grassmann representation $\det M\propto \int \mathcal{D}\bar{c}\mathcal{D}c{e}^{i\int \bar{c}Mc}$. Choosing this normalization for the antighost givesA simultaneous antighost rephasing changes the overall ghost-action sign and vertex conventions, but not the determinant's field-dependent content. The essential facts are the kinetic operator $n\cdot \partial$ and a ghost-gluon vertex proportional to $n^{\mu }$.

Now average the fixed-$\sigma$ constraint with the stated Gaussian:up to a field-independent normalization. ConsequentlyThe Faddeev-Popov ghost determinant is not generally field independent before imposing the strict axial condition: its interaction term contains $n\cdot A$. In the $\xi \to 0$ gauge this quantity vanishes on the integration slice, giving the decoupling established next. Defining the inverse of $n\cdot \partial$ still requires boundary conditions for residual gauge transformations.