# How understand the metric of $S^2$ by definition of a spherically symmetric spacetime

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!

• What do you mean by "proportional to the Euclidean?" Mar 12, 2021 at 16:33
• 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 $\mathbb{R}^3$ Mar 12, 2021 at 17:02
• 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. Mar 12, 2021 at 17:58
• 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. Mar 12, 2021 at 18:00
• @CharlesHudgins the orbits are spheres not in general but in particular case of definition of a spherically symmetric spacetime, not? Mar 12, 2021 at 18:06

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.

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.
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.