# Verifying whether the vanishing term in the Quasi-minkowskian metric is indeed a tensor

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 $$\eta$$?

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*}

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.