Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

Question

The spatial part of the Minkowski metric, written in the Cartesian coordinates, $$d\vec{ x}^2=dx^2+dy^2+dz^2,$$ is invariant under spatial translations: $\vec{x}\to \vec{x}+\vec{a}$, where $\vec{a}$ is a constant vector, and also under spatial rotations: $\vec{x}\to R\vec{x}$ where $R$ is the orthogonal rotation matrix. The metric looks the same after these transformations.

Let us now come to the spatial part of the FRW metric, in spherical coordinates, $$d\vec{x}^2= a^2(t)\left[\frac{dr^2}{1-Kr^2}+r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)\right],$$ is based on spatial homogeneity and isotropy. Therefore, I expect that the metric maintains rotational and translational invariance. However, in the spherical coordinates, the metric depends on the coordinates themselves (in particular, $r$ and $\theta$) which perhaps hides the rotational or translational invariances.

Is there a simple way to understand that like the Minkowski case, the FRW metric also maintains its form under $\vec{x}\to \vec{x}+\vec{a}$ and $\vec{x}\to R\vec{x}$?

Answer 1

Roughly speaking, in addition to Einstein equations, the (spatial) FLRW metric is constructed by assuming that, at fixed time, in the Riemannian manifold defining space:

(a) metric properties are invariant under 3-rotations,

(b) metric properties are the same at every spatial point.

Hypothesis (b) is the spatial homogeneity notion that is assumed. There is no supposition about how to move from a point in the space (at given time) to another point in the space (at same time).

It turns out that one moves through the space using a maximal group of continuous isometries of a $3D$ Riemannian manifold with constant curvature.

Up to geometric identitifcations under discrete subgroups of isometries, only three well known possibilities exist. One is, an Euclidean (affine) space, where movements are described in terms of vectors. The remaining two cases are more subtle, but permitted by the assumed hypotheses.

Answer 2

If what you're confused about is why the metric seems to depend on the coordinates and thus might not be translationally or rotationally invariant, look at the Minkowski metric (flat spacetime) in both spherical and cartesian coordinates:

$$\mathrm{d}s^2=-c^2\mathrm{d}t^2+\mathrm{d}r^2+r^2\left(\mathrm{d}\theta^2+\sin^2\theta\mathrm{d}\phi^2\right)=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2.$$

It definitely still depends on coordinates, but is still flat. In the FLRW metric's case, there is an additional dependency on $K$ and $a(t)$ (because it's the FLRW metric) and another dependence on $r$ in the spatial term, but you will always still find that in any constant-time slice of the manifold the metric is roughly equivalent to the Minkowski metric in the sense that distances are equivalent (barring the scale factor being nontrivial).