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Jan 15, 2023 at 22:32 comment added Matheus Andrade Correcting my previous comments: the scalar curvature is indeed constant and equal to $n(n+1)a$. But the Ricci tensor of $g$ is given by: $$ \mathrm{Ric}(g) = \bigl(n\,a - \tfrac{1}{2}\,b\,(n-1) \, u^{-n-1}\bigr)\,g + \frac{(n^2{-}1)b\,\mathrm{d}u^2}{2\bigl(b\,u^2+k\,u^{n+1}-a\,u^{n+3}\bigr)}, $$ So there was only a minor typo in the original answer, and indeed $g$ is Einstein if, and only if, $b = 0$.
Jan 13, 2023 at 17:52 comment added Matheus Andrade Alright, I'm looking forward to it, thanks for the reply! I think maybe $\text{Ric}(g)$ or the definition of $g$ is not right.
Jan 13, 2023 at 7:29 comment added Robert Bryant @MatheusAndrade: Hmmm. You are right that something is wrong with my answer, but I'm not sure what. I'm traveling now and don't have those calculations with me so I can't check them. I'll have a look at them when I get back home next week, check them over, and see what I can do about correcting the formulae.
Jan 12, 2023 at 13:35 comment added Matheus Andrade Could you perhaps have made a typo when defining $g$? I calculated the scalar curvature of a metric $$g = \frac{\mathrm{d}u^2}{(f(u))^2} + u^2 h$$ and found it to be equal to $$\text{Scal}_{g} = \frac{n f'}{fu} - \frac{n f f'}{u} + \frac{n(n-1)}{u^2}\left(k - \frac{1}{f^2} \right) $$ Setting $f = \sqrt{k - au^2 + bu^{1-n}}$ as in your example, I found that $\text{Scal}_g$ is not constant.
Dec 2, 2022 at 13:55 history edited Robert Bryant CC BY-SA 4.0
Simplified my answer and rearranged it to improve readability.
Dec 1, 2022 at 19:36 comment added Matheus Andrade I just saw your addendum. What a wonderful answer! Thanks a lot. This completely explicit construction goes above and beyond what I expected.
Dec 1, 2022 at 19:14 history edited Robert Bryant CC BY-SA 4.0
Added a little more information to the addendum that details explicit constructions.
Nov 30, 2022 at 11:34 history edited Robert Bryant CC BY-SA 4.0
Added a simplified construction of explicit examples wiithout needing to solve ODE
Nov 28, 2022 at 15:50 comment added Matheus Andrade Thanks very much! This confirmed a few suspicions of mine. When you have the time I'd love to see all the other examples you mentioned, but this is already pretty good :)
Nov 28, 2022 at 15:45 vote accept Matheus Andrade
Nov 28, 2022 at 14:24 history edited Robert Bryant CC BY-SA 4.0
Fixed a few typos and put in a little more detail.
Nov 28, 2022 at 11:56 history answered Robert Bryant CC BY-SA 4.0