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After integration by parts, the quadratic Proca action with an external source isSince $M>0$, this kernel is invertible with the Feynman i-epsilon prescription; the massless gauge redundancy is absent in this unitary gauge quadratic action. Complete the square by $A\mapsto A+K_F^{-1}J$ in the Gaussian functional integral. Its source-independent determinant cancels in the normalized generating functional:In Fourier space, the kernel isFor $k^2\ne 0$, use the transverse projector of a vector field $P_T^{\mu \nu }=g^{\mu \nu }-k^{\mu }{k}^{\nu }/k^2$ and the longitudinal projector of a vector field $P_L^{\mu \nu }=k^{\mu }{k}^{\nu }/k^2$. Their mixed-index products obey $P_T{P}_L=0$, soThe last expression also extends to $k^2=0$; that point only obstructs the intermediate projector decomposition. Two source functional derivatives yield the gauge-boson propagator $D_F=i{K}_F^{-1}$:The pole residue is the physical massive-vector massive vector polarization sum. The longitudinal term is essential for inverting the quadratic kernel, and makes the naive $M\to 0$ limit of this unitary gauge Proca propagator singular.

The same invariant interval can be written as $g_{\mu \nu }d{x}^{\mu }d{x}^{\nu }={\tilde{g}}_{\alpha \beta }d{\tilde{x}}^{\alpha }d{\tilde{x}}^{\beta }$. Applying the chain rule to the differentials proves the metric transformation lawFor first-order scalar cosmological perturbations and ${\xi }^{\mu }=(T,{\partial }^{i}L)$, compare perturbations at the same background coordinate label. Expanding the Jacobians and the shifted background gives the linear metric gauge-transformation lawFor the background ${\bar{g}}_{00}=-a^2$, ${\bar{g}}_{ij}=a^2{\delta }_{ij}$, the Lie derivative components areHere primes denote conformal time and $\mathcal{H}$ is the conformal Hubble parameter. These expressions establish all four alternatives, including any chosen pair.

The $00$ component is $\delta g_{00}=-2a^{2}A$, immediately giving the lapse function perturbation transformation. The $0i$ component is $a^2{\partial }_{i}B$, giving the shift vector scalar-potential transformation. The trace of the spatial perturbation is $6a^{2}C$, so tracing the spatial Lie derivative gives the transformation of $C$. Its trace-free scalar part is $2a^2({\partial }_i{\partial }_j-{\delta }_{ij}{\nabla }^2/3)E$, giving the transformation of $E$. ThusAs usual, identifying scalar potentials from their derivatives uses the standard boundary conditions, or nonzero Fourier modes, to remove homogeneous ambiguities.

At first order, tensor cosmological perturbations are spatial transverse-traceless tensors. The coordinate-generated spatial perturbation consists of a trace term and symmetrized derivatives of the displacement. In Fourier space, the latter terms carry a factor $k_i$ or $k_j$. The transverse-traceless projector removes these longitudinal terms and the trace, including the derivative of a transverse vector displacement. Therefore the tensor perturbation is gauge invariant at linear order around the homogeneous background. This is a first-order statement, not a claim of automatic invariance at arbitrary perturbative order.

A survey window function describes the finite spatial coverage and weights of a galaxy survey. Multiplication by this window in position space mixes Fourier modes by convolution in Fourier space. A dimension of extent $L$ cannot robustly measure wavelengths much larger than $L$; the fundamental wavenumber of a periodic box of that size is $2\pi /L$.