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

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 $$(2)$$?

• Lorentz transformations are isometries of the special metric (1), coordinate transformations under which the metric's coefficients $\eta_{\mu\nu}$ don't change. A generic metric won't have those same isometries. Maybe the concept you're looking for is Lorentzian signature. In general relativity, the metric is required to have Lorentzian signature. Lorentz symmetry/invariance is not required (because then spacetime would always be flat), though a local version of Lorentz symmetry is ensured by the Lorentzian signature. Do any of these thoughts seem relevant to the question? Commented Apr 9, 2019 at 4:15

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. $$\eta'_{\mu\nu} = \eta_{\mu\nu}$$.