# Are all maximally symmetric spacetimes constant curvature spacetimes?

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?

• 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? – CR Drost Nov 28 '15 at 19:40
• 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 ;) – LeastSquare Nov 28 '15 at 20:14
• Wouldn't a constant curvature spacetime with a point removed still be of constant curvature but not maximally symmetric? – Slereah May 31 '16 at 12:46