I have a doubt linked with the definition of spherically symmetric manifold: when we talk about the isometry group on $\mathbb{R}^3$, considering that its orbit are 2-dimensional spheres (from the definition of spherically symmetric spacetime), what means this in the view of defining the metric in $S^2$ as the metric induced by $\mathbb{R^3}$.

I mean: the orbits are the equivalence classes induced by an equivalence relation for which $p\sim q\iff Rp=q$, with $R\in SO(3)$. If we now consider the quotient space of the manifold $M$ (the spacetime) with respect to this relation then we have that if $N=M/\sim\implies M\cong N \times S^2$.

Now what I have not understand is: if I can consider $S^2$ embedded in $\mathbb{R}^3$, what is the link of the fact that the orbits of the isometries of $\mathbb{R^3}$ are isomorphic to $S^2$, with the fact that we can consider in $S^2$ a metric proportional to the euclidean (so a metric that must be a multiple of the metric of a 2-sphere, as suggested in https://en.wikipedia.org/wiki/Spherically_symmetric_spacetime)?

Also in my book in fact seems that these two facts are in a relation of implication! Thanks in advance!

  • $\begingroup$ What do you mean by "proportional to the Euclidean?" $\endgroup$ Mar 12, 2021 at 16:33
  • $\begingroup$ Something like $g{\big|}_{S^2}= R^2(d\theta^2+(\sin{\theta})^2d\phi^2)$, so there is a coefficient that multiply the euclidean metric in polar coordinates (where g is the euclidean metric in $\mathbb{R}^3$ $\endgroup$
    – pawel
    Mar 12, 2021 at 17:02
  • $\begingroup$ Hmm. At this point someone more knowledgeable than me should comment, but I'm not sure how I would generalize a statement like that. The Euclidean metric is $dr^2 + r^2 d\theta^2 + r^2(\sin\theta)^2 d\phi^2$, so I wouldn't say that $g|_{S^2}$ as you've written it is proportional to the euclidean metric. We had to fix a coordinate to be constant and remove its corresponding differential to go from the Euclidean metric to the spherical metric. $\endgroup$ Mar 12, 2021 at 17:58
  • $\begingroup$ A deeper problem is that the orbits of the isometry group of $\mathbb{R}^3$ are not spheres. That is only true if we restrict to isometries that fix a given point. $\endgroup$ Mar 12, 2021 at 18:00
  • $\begingroup$ @CharlesHudgins the orbits are spheres not in general but in particular case of definition of a spherically symmetric spacetime, not? $\endgroup$
    – pawel
    Mar 12, 2021 at 18:06

1 Answer 1


I'm not entirely sure I understand your precise question, but I'll attempt to clear up a couple of things.

As people pointed out in the comments, the isometry group of $\mathbb{R}^3$ acts transitively on $\mathbb{R}^3$ (meaning there is exactly one orbit, namely all of $\mathbb{R}^3$). Indeed, translations are isometries, and any point can be mapped to any other point using a translation.

On the Wikipedia page you link to, it says

A spherically symmetric spacetime is a spacetime whose isometry group contains a subgroup which is isomorphic to the rotation group SO(3) and the orbits of this group are 2-spheres (ordinary 2-dimensional spheres in 3-dimensional Euclidean space). The isometries are then interpreted as rotations and a spherically symmetric spacetime is often described as one whose metric is "invariant under rotations". The spacetime metric induces a metric on each orbit 2-sphere (and this induced metric must be a multiple of the metric of a 2-sphere).

This does not mention the full group of isometries of $\mathbb{R}^3$, but only the subgroup $\textrm{SO}(3)$, which consists of all rotations fixing the origin. We have a spacetime on which $\textrm{SO}(3)$ acts, and the orbits are spheres. The claim is that the induced metric on each orbit is a multiple of the usual metric on the sphere. This is not entirely obvious, and I believe this claim is what you're asking about. Here it is, rephrased:

Claim. If $\textrm{SO}(3)$ acts transitively and isometrically on $S^2$, the metric on $S^2$ is a multiple of the usual one.

To understand why this claim is true, you should learn about homogeneous spaces (https://en.wikipedia.org/wiki/Homogeneous_space). An example of a document in which this claim is proved is "Homogeneous Metrics on Spheres" by Kerin and Wraith (which I admittedly found by simply googling). Quoting from page 64:

By the above it follows that [...], and hence there is a unique homogeneous metric on $\frac{\textrm{SO}(n+1)}{\textrm{SO}(n)} \cong S^n$ up to scaling by a positive constant.


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