\^[(k,k’)]{}\_T =\^[(k,k’)]{}\_T        (k,k’)    kn,k’n’.\[charatGI\]

We have that a tensor field $T$ is gauge invariant to order $(n,n')$ iff in a given gauge $\varphi$ we have that $\cL_{\xi}\delta^{(k,k')}_\varphi T=0$, for any vector field $\xi$ defined on $\cM$ and for any $(k,k')<(n,n')$.

To prove this statement, let us first show that it is true for $(n,n')=(1,0)$. In fact, if $\delta^{(1,0)}_\varphi T = \delta^{(1,0)}_\psi T$ for two arbitrary gauges $\varphi,\psi$, we have $\cL_{\,^{\varphi}\!\eta-^{\psi}\!\eta}T|_{\cM_0}
=0$. But since $\varphi$ and $\psi$ are arbitrary gauges, it follows that $\,^{\varphi}\!\eta-\,^{\psi}\!\eta$ is an arbitrary field $\xi$, and $\xi^m=\xi^{m+1}=0$ because $\,^{\varphi}\!\eta^m=\,{}^{\psi}\!\eta^m=1,\,
\,^{\varphi}\!\eta^{m+1}=\,^{\psi}\!\eta^{m+1}=0$, so $\xi$ is tangent to $\cM$. In the same way one proves the statement for $(n,n')=(0,1)$. Now let us suppose that the statement is true for some $(n,n')$. Then, if one also has $\delta^{(n+1,n')}_\varphi T|_{\cM_0}=\delta^{(n+1,n')}_\psi T|_{\cM_0}$, it follows that $\cL_{\,^{\varphi}\!\eta-^{\psi}\!\eta}\delta^{(n,n')}_\varphi T=0$, while if $\delta^{(n,n'+1)}_\varphi T|_{\cM_0}=\delta^{(n,n'+1)}_\psi T|_{\cM_0}$, it follows that $\cL_{\,^{\varphi}\!\zeta-^{\psi}\!\zeta}\delta^{(n,n')}_\varphi T=0$, and we establish the result by induction over $(n,n')$.

As a consequence, $T$ is gauge invariant to order $(n,n')$ iff $T_0$ and all its perturbations of order lower than $(n,n')$ are, in any gauge, either vanishing or constant scalars, or a combination of Kronecker deltas with constant coefficients. Thus, this generalizes to an arbitrary order $(n,n')$ 