now have three tensor fields, i.e. $T_0$ and T\^\_[,]{}:= .\^\*\_[,]{} T|\_[\_0]{},         T\^\_[,]{}:= .\^\*\_[,]{} T|\_[\_0]{}. Since $\varphi$ and $\psi$ represent gauge choices for mapping a perturbed manifold $\cM_{\lambda,\Omega}$ into the unperturbed one $\cM_0$, $T^\varphi_{\lambda,\Omega}$ and $T^\psi_{\lambda,\Omega}$ are the representations, in $\cM_0$, of the perturbed tensor according to the two gauges. We can write, using (\[defperturbation\])–(\[defperturbationk\]) and the expansion (\[Taylorgroup\]), T\^\_[,]{} & = & \_[k=0]{}\^ \^[(k,k’)]{}\_T=\_[k,k’=0]{}\^ \^k\_[\^]{}\^[k’]{}\_[\^]{}T=T\_0+ \_0\^ T\_[,]{},\[defexpX\]\
T\^\_[,]{} & = & \_[k=0]{}\^ \^[(k,k’)]{}\_T = \_[k,k’=0]{}\^ \^k\_[\^]{}\^[k’]{}\_[\^]{}T=T\_0+\_0\^ T\_[,]{}\[defexpY\] where $\delta^{(k,k')}_\varphi T$, $\delta^{(k,k')}_\psi T$ are the perturbations (\[defperturbationk\]) in the gauges $\varphi$ and $\psi$ respectively, i.e. \^[(k,k’)]{}\_T & = & . \^k\_[\^]{} \^[k’]{}\_[\^]{}T|\_[\_0]{} ,\
\^[(k,k’)]{}\_T & = & . \^k\_[\^]{} \^[k’]{}\_[\^]{}T|\_[\_0]{}.

Gauge invariance {#gaugeinvariance}
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If $T^\varphi_{\lambda,\Omega}=T^\psi_{\lambda,\Omega}$, for any pair of gauges $\varphi$ and $\psi$, we say that $T$ is [*totally gauge invariant*]{}. This is a very strong condition, because then (\[defexpX\]) and (\[defexpY\]) imply that $\delta^{(k,k')}_\varphi T=
\delta^{(k,k')}_\psi T$, for all gauges $\varphi$ and $\psi$ and for any $(k,k')$. In any practical case, however, one is interested in perturbations up to a fixed order. It is thus convenient to weaken the definition above, saying that $T$ is [*gauge invariant up to order $(n,n')$*]{} iff for any two gauges $\varphi$ and $\psi$ 