Counterexample to spherical symmetry definition in general relativity

Question

In practical terms we say a spacetime is spherically symmetric in GR when we have coordinates in which the spacetime metric takes the form: $$ds^2 = -f(r,t)dt^2 +g(r,t)dr^2+h(r,t)d\Omega^2 \tag{*}$$ If $h(r,t)$ is not a constant, further transformations can be carried out to obtain the metric in the following form: $$ds^2 = -f(r,t)dt^2 +g(r,t)dr^2+r^2d\Omega^2 \tag{1}$$ where $d\Omega^2$ is the metric of a 2-sphere.

However one can also provide an "algebraic" definition based on the symmetry group of the spacetime: a spacetime is spherically symmetric if its isometry group contains a subgroup isomorphic to $SO(3)$ whose orbits are 2-spheres.

Some authors (i.e. Carroll) do not mention this second requirement on orbits, and one might wonder whether it is necessary. Can someone provide a counterexample of a spacetime whose isometry group contains a copy of $SO(3)$ yet cannot be brought into the form $(*)$, with proof?

Note: this question has been edited to reflect the fact that the most general form of the metric is $(*)$ and not $(1)$, for which there is a simple counterexample provided in the answer by user A.V.S.

Answer 1

A simple counterexample: any spacetime that has structure $\mathcal{M}_2 \times S_2 $ for some 2-dimensional space $\mathcal{M}_2 $ with constant radius of $S_2$ fibers. A proof that such spacetime cannot be brought into form (1), is noting that orbits of metric (1) at different values of $r$ are necessarily spheres of different radii.

A physically interesting spacetime with such a structure is a Bertotti–Robinson spacetime which is simply $AdS_2 \times S_2$: $$ ds^2 =− (1 + x^2) \,dτ^2 + (1 + x^2)^{-1} dx^2 + dΩ_2^2.$$

Answer 2

...a spacetime is spherically symmetric if its isometry group contains a subgroup isomorphic to SO(3) whose orbits are 2-spheres. Some authors (i.e. Carroll) do not mention this second requirement on orbits, and one might wonder whether it is necessary.

Here are two examples showing that the condition on orbits is not redundant:

First example: start with Minkowski spacetime $ds^2=-dt^2+dx^2+dy^2+dz^2$, delete the timelike worldline defined by $(x,y,z)=(0,0,0)$, and impose the equivalence relation $(-x,-y,-z)\cong(x,y,z)$. The resulting spacetime is geodesically incomplete because of the deleted worldline, but the Schwarzschild metric is also geodesically incomplete, so I assume this is acceptable. Everywhere else, the quotient leaves the metric well-defined and locally flat, and $SO(3)$ is still a subgroup of the isometry group (still has spacelike rotational symmetry about the origin), but the orbits are no longer 2-spheres; they are copies of $\mathbb{R} P^2$ instead.

Second example: This one is smooth everywhere; no points are excluded. Consider the case $\mathbb{R}\times S^3$ with the standard metric on $S^3$, and identify antipodal points of $S^3$ to get $\mathbb{R} P^3$. The metric is locally unchanged, and the isometry group includes $SO(3)$ as a subgroup, but the orbits are again copies of $\mathbb{R} P^2$, not $S^2$.

In both of these examples, the metric can still be written as in the OP's equation (1), so these are not counterexamples in that sense. However, they do show that the condition that the orbits be 2-spheres is an independent condition, not implied by having $SO(3)$ as a subgroup of the isometry group.

Maybe the $AdS_2\times S_2$ example described by A.V.S. can be modified in the same way (replacing $S_2$ by $\mathbb{R}P^2$) to obtain an example in which the orbits of $SO(3)$ are not 2-spheres and the metric cannot be written locally as in (1).

I assumed here that the isometry group of $\mathbb{R}P^2$ is isomorphic to $SO(3)$. According to https://en.wikipedia.org/wiki/Projective_orthogonal_group, the isometry group of $\mathbb{R}P^{2k}$ is $SO(2k+1)$ even though the isometry group of $\mathbb{R}P^{2k+1}$ is not $SO(2k+2)$.