[Physics] Are all maximally symmetric spacetimes constant curvature spacetimes

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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?

Best Answer

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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