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.

  • 2
    $\begingroup$ IMHO there is a better way. Define AdS as a hypersurface in a $D+1$ dimensional flat space of appropriate signature. The isometries of AdS are the isometries of the ambient space which map AdS back to itself. You then find out the dimension of this group is the maximum number of Killing vectors for a $D$-dimensional space. Therefore AdS is maximally symmetric. For maximally symmetric spaces there is a simple expression for the Riemann tensor and from there the Ricci scalar follows easily. $\endgroup$
    – Gold
    Aug 22, 2021 at 22:41

1 Answer 1


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.


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