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A $d$ dimensional maximally symmetric spacetime is a spacetime with the maximum allowed number of Killing vectors. This number is $\frac{d(d+1)}{2}$.

Constant curvature spacetimes are spacetimes whose Weyl tensor is zero. Thus, the Riemann tensor $R_{abcd}$ is written as $$\frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc}).$$

If the metric is Euclidean, constant curvature spacetimes are either spherical, hyperbolic or flat and one can check explicitly for each of these examples that indeed spacetime is maximally symmetric. Similarly, if the metric is Lorentzian, constant curvature spacetimes are either deSitter, antideSitter or flat, and one can check that they are maximally symmetric. Thus, constant curvature spacetimes are maximally symmetric. Is the reverse statement also true? Also, is it possible to prove that constant curvature spacetimes are maximally symmetric without resorting to examples?

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  • $\begingroup$ Just as a comment: your title question and your question question are different; the title is "does maximal symmetry imply constant curvature?" while the question is "is there a way to prove that constant curvature implies maximal symmetry in general?"... Which question did you mean to ask? $\endgroup$ – CR Drost Nov 28 '15 at 19:40
  • $\begingroup$ Thanks. I edited the question to be more clear. I meant both questions, but I consider the one in the title to be more important ;) $\endgroup$ – LeastSquare Nov 28 '15 at 20:14
  • $\begingroup$ Wouldn't a constant curvature spacetime with a point removed still be of constant curvature but not maximally symmetric? $\endgroup$ – Slereah May 31 '16 at 12:46
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If the metric is Riemannian (positive) your conjecture (a maximally symmetric spacetime is a constant curvature spacetimes) is a known theorem: Theorem 3.1 in Transformation Groups in Differential Geometry by S. Kobayashi. From the proof, it seems to me that the result should hold in the Lorenzian case too, but without a closer scrutiny I am not completely sure.

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  • $\begingroup$ Thanks a lot! I took a look at the reference you provided, but, without further study, I couldn't really understand it. I might ask this question in the mathematics forum, just because I am curious if this is true (even though I might not understand the proofs). $\endgroup$ – LeastSquare Nov 30 '15 at 14:26
  • $\begingroup$ Just to clarify, there are multiple theorems 3.1 in the book, the referenced theorem is the one in chapter II. $\endgroup$ – Sjorszini Sep 29 '19 at 13:41

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