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In Weinberg's Gravitation and Cosmology, on page 165 he notes that $h_{\mu \nu}$ is lowered and raised with the $\eta$'s since unlike $R_{\mu\kappa}$ it is not a true tensor (or at least implies it). I may be misreading this but isn't this not true as $h$ is indeed a tensor since it transforms as one? Is this simply for convenience of notation? Since if $h$ is indeed a true tensor then shouldn't we be raising/lowering the indices with $g$ not $\eta$?

Page 165 of Weinberg's Gravitation and Cosmology

For a quick proof as to why $h_{\mu \nu}$ is indeed a tensor we take $h_{\mu \nu} = g _{\mu \nu} - \eta_{\mu \nu}$ which then follows: $$\begin{align*} h_{\mu \nu} \frac{\partial x^\mu}{\partial \tilde{x}^\alpha} \frac{\partial x^\nu}{\partial \tilde{x}^\beta} &= (g _{\mu \nu} - \eta_{\mu \nu})\frac{\partial x^\mu}{\partial \tilde{x}^\alpha} \frac{\partial x^\nu}{\partial \tilde{x}^\beta} \\ &= \tilde{g} _{\alpha \beta} - \tilde{\eta}_{\alpha \beta}\\ &= \tilde{h}_{\alpha \beta} \end{align*}$$

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The metric $g_{\mu\nu}$ is a tensor, but the Minkowski quantity $\eta_{\mu\nu}$ is not.* Therefore, their difference $h_{\mu\nu}\equiv g_{\mu\nu}-\eta_{\mu\nu}$ is not a tensor either. That is the formal justification, but there is also a physical reasoning behind the fact that we use $\eta$ to raise and lower indices on $h$. The situations in which $h$ is a useful quantity are those in which the gravitational field is weak and so it makes sense to linearize the theory. For the equations of motion under those circumstances, it is a good approximation to neglect anything of $\mathcal{O}(h^{2})$. Any corrections to $h^{\mu}{}_{\nu}=\eta^{\mu\alpha}h_{\alpha\nu}$ would be of $\mathcal{O}(h^{2})$, so it is no problem to drop them in this regime.

*If this puzzles you, just notice that $\eta^{\mu\nu}=\eta_{\mu\nu}\neq g^{\mu\alpha}g^{\nu\beta}\eta_{\alpha\beta}$ if spacetime is not flat.

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