and to the two–parameter case the results of [@BMMS; @sachs; @stewart]. Further, it then follows that $T$ is totally gauge invariant iff it is a combination of Kronecker deltas with coefficients depending only on $\lambda,\Omega$.

Gauge transformations
---------------------

If a tensor $T$ is not gauge invariant, it is important to know how its representation on $\cM_0$ changes under a gauge transformation. To this purpose it is natural to introduce, for each value of $(\lambda,\Omega)\in\RR^2$, the diffeomorphism $\Phi_{\lambda,\Omega}
:\cM_0\rightarrow\cM_0$ defined by \_[,]{}:=\_[-,-]{}\_[, ]{}. Given that from the geometrical point of view adopted here $\varphi$ and $\psi$ are two gauges, $\Phi$ represents the gauge transformation.

The action of $\Phi_{\lambda,\Omega}$ is illustrated in Fig. \[fig:GT\]. We must stress that $\Phi:\cM_0\times\RR^2\rightarrow\cM_0$ thus defined, [*is not*]{} a two–parameter group of diffeomorphisms in $\cM_0$. In fact, $\Phi_{\lambda_1,\Omega_1}\circ\Phi_{\lambda_2,\Omega_2}\neq
\Phi_{\lambda_1+\lambda_2,
\Omega_1+\Omega_2}$, essentially because the fields $(^{\varphi}\eta,\,^{\varphi}\zeta)$ and $(^{\psi}\eta,\,^{\psi}\zeta)$ have, in general, a non–vanishing commutator. However, it can be Taylor expanded, using the results of section (\[familydiffeo\]).

The tensor fields $T^\varphi_{\lambda,\Omega}$ and $T^\psi_{\lambda,\Omega}$, defined on $\cM_0$ by the gauges $\varphi$ and $\psi$, are connected by the linear map $\Phi_{\lambda,\Omega}^*$: T\^\_[,]{} & = & . \^\*\_[,]{}T|\_[\_0]{}= . (\^\*\_[,]{}\^\*\_[-,-]{} \^\*\_[,]{}T)|\_[\_0]{}\
& = & . \^\*\_[,]{}(\^\*\_[,]{}T )|\_[\_0]{} = \^\*\_[,]{}T\^\_[,]{}. Thus, the gauge 