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If I'm not mistaken, one of the properties of maximally symmetric spacetimes is that the Riemann tensor can be written as $R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd} - g_{ad}g_{bc})$, which would imply that the Weyl tensor vanishes.

And Wikipedia states that a d-dimensional spacetime with $d \geq 4$ is conformally flat if and only if the Weyl tensor vanishes.

This is telling me that the answer to my first question is "Yes", but what about the second one? I guess that one could have a spacetime with vanishing Weyl tensor but with a Riemann tensor that's not only "trace" but also has "symmetric" components coming from $R_{ij} \neq \frac{1}{d} R g_{ab}$. Are there any easy examples of such spacetimes?

The usual examples given for both concepts (MS and CF) are Minkowski, AdS, and dS; and so I am a bit confused.

Thank you!

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All FLRW spacetimes are conformally flat, and they are not maximally symmetric unless $ρ=-p=\text{constant}$ (in which case they are Minkowski or (anti) de Sitter).

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  • $\begingroup$ That’s not a counterexample. Op is asking whether maximally symmetric spacetimes are necessarily conformally flat, not the converse. $\endgroup$ Jun 4, 2021 at 12:41
  • $\begingroup$ @Prof.Legolasov There are two questions in the title, and the question body correctly answers the first one (third paragraph, first sentence). $\endgroup$
    – benrg
    Jun 4, 2021 at 14:02

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