Is the metric $ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu$ Lorentz invariant?

Question

Postulate of Special Relativity leads to the conclusion that the metric in flat Minkowski space $$ds^2=c^2t^2-\textbf{r}^2=\eta_{\mu\nu}dx^\mu dx^\nu\tag{1}$$ is Lorentz invariant. This follows as a consequence of Lorentz transformations. Using this, we can derive that the metric tensor transforms as $$\eta_{\rho\sigma}=\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma}.\tag{2}$$ Now consider the metric of a curved space: $$ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu\tag{3}$$

$\bullet$ Do we say that $(3)$ is also Lorentz invariant? Do we assume it or does it follow?

$\bullet$ If so, does $g_{\mu\nu}$ transform in the same way under Lorentz transformation as $\eta_{\mu\nu}$ in $(2)$?

Answer 1

Every object without free indices is Lorentz invariant, provided that its components transform properly under Lorentz transformations. $T_{\mu\nu}\mathrm{d}x^\mu \mathrm{d}x^\nu$ as well as $V^\mu \partial_\mu$ or any other vector, tensor or form you have write down. Special relativity has nothing to do with it. In fact, these things are not only Lorentz invariant, but invariant under general coordinate transformations. All this means is that they describe "genuine" geometric objects (equivalently: they are "well-defined" vectors/tensors/forms in the sense of coordinate-free differential geometry), not things tied to a particular coordinate system. Note that we almost never use index notation such as $g_{\mu\nu}$ to denote objects that do not transform such that $g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ is invariant under coordinate transformations.

Therefore, if special relativity somehow said that the metric tensor is Lorentz invariant, it would say nothing at all - this behaviour is part of the definition of a metric and purely geometrical, not physical. Fortunately, it doesn't say that: It says that Lorentz transformations are an isometry of the metric. This is a much stronger claim: Not only does the metric transform as $\eta'_{\mu\nu} = \eta_{\sigma\rho}\Lambda^\sigma_\mu \Lambda^\rho_\nu$, but the components of $\eta'$ and $\eta$ are numerically equal, i.e. $\eta'_{\mu\nu} = \eta_{\mu\nu}$.

General relativity does not make such a claim, but instead promotes the metric to a dynamical field that may or may not have isometries. Vector fields that generate isometries are still significant as they are related to conserved quantities (Noether's theorem doesn't stop working just because we're in curved space!) and are called Killing vectors.