Breaking of diffeomorphism invariance after fixing a background metric The Lagrangian for the gravitational field in absence of matter is the following $$L=1/k\int dx^4 \sqrt g R,$$ where $k=\sqrt G$, $g$ is the determinant of the metric and $R$ the Ricci scalar. It's possible to fix a background metric like $\eta_{uv}$ and then study the perturbations $h_{uv}$ around it by $$g_{uv}=\eta_{uv}+kh_{uv}$$
The Lagrangian becomes $$L=L^{0}+kL^{1}+k^{2}L^{2}+.......$$
which can be interpreted as an effective field theory of self-interacting particles called gravitons.
Now, given the transformation law of $h_{uv}$, how is it possible to say the entire Lagrangian is invariant, order by order, under local diffeomorphisms? Of course the symmetry is still there, but I was wondering if there is some kind of Spontaneus Symmetry Breaking associated with the perturbation field $h_{uv}$ and the diffeomorphisms group.
The procedure resemble the SSB for the Higgs Boson, where the Lagrangian is $$L=\partial_{u}\phi\partial^{u}\phi - m^{2}\phi^{2}+\lambda\phi^{4}$$
This Lagrangian is invariant under parity in $\phi$, but after the redefinition around the vacuum $v$, the minimum of the potential, you deal with $\phi=v+\delta\phi$ and the Lagrangian in δϕ is no more parity invariant. Does this happen in the previous example after fixing a background? 
 A: No, there is no SSB after a field redefinition. Recall that SSB is a dynamical effect. You cannot trigger dynamical effects by changing your coordinates. In the scalar case the SSB is triggered by a quadratic term with "the wrong sign", not by the change of variables $\phi=v+\delta\phi$. The new physics are more transparent in the new coordinates; but the SSB is not caused by changing into  the new coordinates: the symmetry breaks whether you define $\phi=v+\delta\phi$ or not.
In other words, physics are independent of coordinates. Using $g\to\eta+h$ leaves the physics invariant. You can also change $g\to g_0+g_1$, and choose any background $g_0$. The flat background is convenient, but in the literature you'll also find people that consider more general backgrounds; for example, $g_0$ can be taken to be the metric of an asymptotically flat space-time. In any case, the dynamics are determined by the Lagrangian, not by the coordinates.
A: Schematically, the Einstein-Hilbert action is given by,
$$S \sim \int d^D x \, \sqrt{|\det g_{\mu\nu}|} \, \mathcal R$$
for a metric, $g_{\mu\nu}$. Now, as noted in previous questions and by the OP, one can expand the field as,
$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$
and since the inverse metric is an infinite series in $h_{\mu\nu}$, one obtains an infinite number of terms in the Einstein-Hilbert action expressed in this form. This procedure does not violate diffeomorphism invariance, as it is a mere field redefinition and we know all the terms sum to give $S$.
We can express any metric in the form $\eta_{\mu\nu} + h_{\mu\nu}$, it is as trivial as expressing a scalar $\phi$ in terms of two scalars, one of which we can choose completely freely, e.g. $\phi = \eta + h$.
