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In linearized gravity, where we take a general background metric $g$ with perturbation $h$, the linearized Einstein equations become

$$-\square h_{\alpha\beta}+\nabla^{\delta}\nabla_{\alpha}h_{\beta\delta}+\nabla^{\delta}\nabla_{\beta}h_{\alpha\delta}-\nabla_{\alpha}\nabla_{\beta}h-g_{\alpha\beta}\bigg(\nabla^{\gamma}\nabla^{\delta}h_{\gamma\delta}-\square h\bigg)=0$$

where the box operator is $\square=g^{\alpha\beta}\nabla_{\alpha}\nabla_{\beta}$. I was wondering, how to interpret the term $\square h_{\alpha\beta}$. Do I have to interpret is component-wise, i.e. in the sense of $\square $ is acting on each component function, which would yield

$$\square h_{\alpha\beta}=\square (\underbrace{h_{\alpha\beta}}_{\text{function}})=g^{\mu\nu}\nabla_{\mu}(\partial_{\nu}h_{\alpha\beta})=g^{\mu\nu}(\partial_{\mu}\partial_{\nu}h_{\alpha\beta}-\Gamma^{\rho}_{\mu\nu}\partial_{\rho}h_{\alpha\beta})$$

or really as the douple connection acting on a tensor (the so-called "connection Laplacian" in math terminology), which would least a much longer expression like

$$\square h_{\alpha\beta}=g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}h_{\alpha\beta}=g^{\mu\nu}(\partial_{\mu}\nabla_{\nu}h_{\alpha\beta}-\Gamma^{\rho}_{\mu\nu}\nabla_{\rho}h_{\alpha\beta}-\Gamma^{\rho}_{\mu\alpha}\nabla_{\nu}h_{\rho\beta}-\Gamma^{\rho}_{\mu\beta}\nabla_{\nu}h_{\alpha\rho})$$

with

$$\nabla_{\nu}h_{\alpha\beta}=\partial_{\nu}h_{\alpha\beta}-\Gamma^{\rho}_{\nu\alpha}h_{\rho\beta}-\Gamma^{\rho}_{\nu\beta}h_{\alpha\rho}$$ and etc for the others? I am pretty sure its the second, more complicated, option, but I did not found any reference confirming my guess. (Most references are discussing Minkowski background, where there is no difference, since all Christoffel symbols are anyway zero).

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2 Answers 2

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It's the second option, otherwise your equations of motion wouldn't be diffeomorphism-invariant. On the flat space-time background these options coincide.

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The second option is correct.

Notice that, in your first development of the D'Alembertian operator, you implicitly assume that $$\nabla_\mu h_{\alpha\beta} = \partial_\mu h_{\alpha\beta}. $$ You have rightly identified this only holds in a Minkowski background.

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