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The title says it all. I think this should be a pretty simple question but I just couldn't find the answer.

Ok -- I'll give a bit more context to my question. I'm encountering this in the context of general relativity, when we're trying to solve for the Schwarzschild solution. We write down a metric that respects spherical symmetry. According to the textbook (Carroll), "we argue that a spherically symmetric space-time can be foliated by two-spheres - in other words, that (almost) every point lies on a unique sphere that is left invariant by the generators of spherical symmetry." Some time later he mentions the Killing vectors of $S^2$, which are

$$R=\partial_\phi$$

$$S=\cos \phi \partial_\theta - \cot \theta \sin \phi \partial_\phi$$

$$T=-\sin\phi\partial_\theta - \cot\theta \cos\phi\partial_\phi$$

and he observes that these vectors obey the following algebra:

$$[R,S]=T\quad [S,T]=R\quad [T,R]=S.$$

I'm not sure how the Killing vectors are exactly linked to spherical symmetry, and why such an algebra between the Killing vectors is important (specifically with regard to spherical symmetry). And I also don't know what he means when he says "generators of spherical symmetry", hence the original question.

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The first thing it's important to understand is the notion of symmetry in general relativity. It is subtly different from the concept in Hamiltonian mechanics.

We say that a one parameter family of diffeomorphisms $\phi_t$ with velocity vector field $X$ preserves a tensor field $T$ iff

$$\mathcal{L}_X T = 0$$

If you aren't sure about the terminology above then read this handy intro. The vector field $X$ is often referred to as the generator of the transformations $\phi_t$.

We say that $X$ generates a symmetry of a metric spacetime if the associated $\phi_t$ preserve the metric $g$, or equivalently

$$\mathcal{L}_Xg=0$$

This is exactly the condition that $X$ is a Killing vector field for the metric $g$. So the Killing vector fields are exactly those vector fields which generate spacetime symmetries.

Back to your example. In a spacetime with spherical symmetry you should be able to identify the Killing vectors above, using spherical polar coordinates.

Conversely, if (no subset of) the Killing vectors of your manifold obeys the Killing algebra of $S^2$ then you may conclude that your manifold doesn't have spherical symmetry. This is because the Killing vectors determine the Lie algebra of the maximal symmetry group preserving the metric, by definition.

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  • $\begingroup$ @EdwardHuges what if you do not have polar coordinates. Imagine a two dimensional manifold with constant Gaussian curvature - a two-sphere that is not embedded in any other space. Lets call it a static two-dimensional universe with constant Riemann (one component only) curvature. Is the above definition of spherically symmetric spacetime valid, too? $\endgroup$
    – JanG
    Jul 25, 2022 at 18:33
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If we represent the sphere as the surface $x^2+y^2+z^2=1$ in $R^3$, then its simple to check that the vector fields $S=(-z,0,x)$, $T=(0,z,-y)$ and $R=(y,0,-x)$ preserve the surface.

When written in spherical polar coordinates these are the ones written above.

One way to check that the vector fields leave the surface unchanged (i.e., map points on the surface to other points on the same surface) is to observe that these vector fields generate rotations - physicists might like to think of $R=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$ for instance.

To check they are Killing vector fields requires the metric compatibility check as detailed above.

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