Gauge invariance of the quadratic graviton Lagrangian I've been reading up on how the linearised Einstein field equations can be derived from a field theoretic perspective, by constructing the most general Lagrangian for a massless spin-2 field, $h_{\mu\nu}$, compatible with locality and Lorentz invariance. This is given by $$\mathcal{L}=a\partial_{\lambda}h^{\mu\nu}\partial^{\lambda}h_{\mu\nu}+b \partial_{\mu}h^{\mu\lambda}\partial^{\nu}h_{\lambda\nu}+c\partial_{\mu}h\partial_{\nu}h^{\mu\nu}+d\partial_{\mu}h\partial^{\mu}h$$ Now, the requirement of Lorentz invariance requires that $\mathcal{L}$ is invariant under the transformation $$h_{\mu\nu}\rightarrow h_{\mu\nu}+2\partial_{(\mu}\xi_{\nu)}$$ where $\xi^{\mu}$ is a generic 4-vector. This transformation corresponds exactly to a gauge transformation of the spin-2 field.
In principle, one should be able to determine the forms of the coefficients $a$, $b$ $c$ and $d$. However, it seems impossible to do so without relaxing the condition of strict Lorentz invariance of $\mathcal{L}$ to requiring that it is invariant up to a boundary term. What does this mean for the gauge invariance of the theory? If it is to correspond to GR then the theory should be invariant under these linearised diffeomorphisms, right?
What am I missing? If someone could enlighten me on this it would be much appreciated.
 A: First of all, as soon as one neglects curvature by resorting to a linear perturbation around the Minkowski metric, one is then forced to use the tenets of classical field theory. One of these is that the fields (in this case the 10 independent $h$ functions) are Schwartz test functions, i.e functions whose modulus and spacetime differentials of all orders go fast to 0, as the „radius of Minkowski spacetime" is pushed to $+\infty$. In this simple perspective, there is no discussion of "Lorentz invariance of the Lagrangian (density)". The true invariance is imposed upon the Lagrangian action, i.e. to the integral on the flat spacetime continuum of the Lagrangian density. So there's no "relaxing" needed, it is all there in the standard assumptions of classical field theory. 
Feynman in his GR lectures explains how one derives the a,b,c,d factors. That is an elegant method reminding me of the Pauli and Fierz one. On the other hand, one can linearize by brute force the Hilbert-Einstein action and get the 4 coefficients. This works, too.   
