Ricci scalar of AdS in $D$ spacetime dimensions from structure equations Starting from the AdS metric in $D$ spacetime dimensions in Poincare coordinates $ds^2 = \frac{R^2}{(x^3)^2}\eta_{\mu\nu}dx^\mu dx^\nu$ (R here is the AdS radius), I would like to compute the components of the Ricci tensor down to the Ricci scalar and get to the result $R_{Ricci} = \frac{-D(D-1)}{R^2}$. I want to do this using the structure equations: $d\theta^A + \omega^A_B \wedge \theta^B = 0$ and $\Omega^{A}_{B} = d\omega^A_B + \omega^A_C \wedge \omega^C_B$. We can easily see from the metric that we can set $\theta^A = \theta^A_\mu dx^\mu = \frac{R}{x^3}\delta^A_\mu dx^\mu$. The following link in 1)d) discusses this problem https://mcgreevy.physics.ucsd.edu/f13/225A-pset08-sol.pdf but I do not understand their notation and they get to a slightly wrong result missing a minus sign.
 A: Claim: For $AdS_D$, $R_{Ricci} = \frac{-D(D-1)}{R^2}$, $\Lambda = -\frac{(D-1)(D-2)}{2R^2}$.
Proof: For the proof we will use the structure equations in $D = 3$, generalise the curvature 2-form in $D$ dimensions and then use it to derive the Riemann tensor up to the Ricci scalar.
The metric for $AdS_3$ in Poincare coordinates is $ds^2 = \frac{R^2}{z^2}(-dt^2+dx^2+dz^2)$. The coframe is then read off to be $\theta^0 = \frac{R}{z}dt$, $\theta^1 = \frac{R}{z}dx$, $\theta^2 = \frac{R}{z}dz$, resulting in $ds^2 = \eta_{AB}\theta^A\theta^B$ and $\eta_{AB} = \text{diag}(-1,1,1)$. We have $d\theta_0 = \frac{R}{z^2}dz\wedge dt$, $d\theta_1 = -\frac{R}{z^2}dz\wedge dx$, $d\theta_2 = 0$. The first structure equation $d\theta_A + \omega_{AB} \wedge \theta^B = 0$ gives the following 3 equations, $\frac{R}{z^2}dz \wedge dt + \omega_{01} \wedge \theta^1 + \omega_{02} \wedge \theta^2 = 0$, $-\frac{R}{z^2}dz \wedge dx - \omega_{01} \wedge \theta^0 + \omega_{12} \wedge \theta^2 = 0$, $\omega_{02} \wedge \theta^0 = \omega_{12} \wedge \theta^1$, where we used $\omega_{AB} = -\omega_{BA}$. These are solved by $\omega_{01} = 0$, $\omega_{02} = \frac{1}{z}dt$, $\omega_{12} = -\frac{1}{z}dx$. The second structure equation $\Omega_{AB} = d\omega_{AB} + \omega_{AC} \wedge \omega^C_B$ gives the following 3 equations, $\Omega_{01} = \frac{1}{z^2}dt \wedge dx = \frac{R^2}{z^4}\theta^0 \wedge \theta^1$, $\Omega_{02} = \frac{1}{z^2}dt \wedge dz = \frac{R^2}{z^4}\theta^0 \wedge \theta^2$, $\Omega_{12} = \frac{1}{z^2}dz \wedge dx = \frac{1}{z^2}\theta^2 \wedge \theta^1$. By inspection and using $\Omega_{AB} = -\Omega_{BA}$, we have $\Omega_{AB} = -\frac{1}{2}\frac{R^2}{z^4}(\eta_{AC}\eta_{BD}-\eta_{AD}\eta_{BC})\theta^C \wedge \theta^D$. Given the symmetric form of the metric $ds^2 = \frac{R^2}{z^2}\eta_{\mu\nu}dx^\mu dx^\nu$, the metric indices can be extended to any number of dimensions and hence the expression for $\Omega_{AB}$ is true for any number of dimensions, we simply extend the range of the indices to run from $0$ to $D-1$ rather than from $0$ to $2$.
Now we derive the components of the Riemann tensor with the following identity, $\frac{1}{4}R_{\mu\nu\rho\sigma}(dx^\mu \wedge dx^\nu)(dx^\rho \wedge dx^\sigma) = -\frac{1}{2}\Omega_{AB}(\theta^A \wedge \theta^B) = \frac{R^2}{4z^4}(\eta_{AC}\eta_{BD}-\eta_{AD}\eta_{BC})(\theta^C \wedge \theta^D)(\theta^A \wedge \theta^B) = \frac{R^2}{4z^4}(\eta_{AC}\eta_{BD}-\eta_{AD}\eta_{BC})\theta^C_\mu \theta^D_\nu \theta^A_\rho \theta^B_\sigma(dx^\mu \wedge dx^\nu)(dx^\rho \wedge dx^\sigma) = \frac{1}{4R^2}(g_{\mu\rho} g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})(dx^\mu \wedge dx^\nu)(dx^\rho \wedge dx^\sigma) \implies R_{\mu\nu\rho\sigma} = \frac{1}{R^2}(g_{\mu\rho} g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})$. Then we proceed with the Ricci tensor and the Ricci scalar, $R_{\nu\rho} = g^{\mu\sigma}R_{\mu\nu\rho\sigma} = \frac{z^2}{R^2}\eta^{\mu\sigma}\frac{1}{R^2}(g_{\mu\rho} g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}) = \frac{1}{z^2}(\eta^{\mu\sigma}\eta_{\mu\rho}\eta_{\nu\sigma}-D\eta_{\nu\rho})$, $R_{Ricci} = g^{\nu\rho}R_{\nu\rho} = \frac{z^2}{R^2}\eta^{\nu\rho}\frac{1}{z^2}(\eta^{\mu\sigma}\eta_{\mu\rho}\eta_{\nu\sigma}-D\eta_{\nu\rho}) = \frac{-D(D-1)}{R^2}$.
Now we will use Einstein's field equations in vacuum $R_{\mu\nu} - \frac{1}{2}R_{Ricci}g_{\mu\nu} + \Lambda g_{\mu\nu} = 0$ to find $\Lambda$ in terms of $D, R$. We take the trace of the former to obtain $g^{\mu\nu}R_{\mu\nu} - \frac{1}{2}R_{Ricci}D + \Lambda D = R_{Ricci} - \frac{1}{2}R_{Ricci}D + \Lambda D = 0$. Using the above result for $R_{Ricci}$ gives $\Lambda = \frac{-(D-1)(D-2)}{2R^2}$, QED.
It would be more satisfying if somebody can solve this with the structure equations again but without resorting to $D = 3$ first.
