# Sign convention with the $AdS$ metric

One would say that $AdS_n$ satisfies the equations for the scalar curvature (R) and Ricci tensor ($R_{\mu \nu}$), $R = - \frac{n(n-1)}{L^2}$ and $R_{ab} = - \frac{n-1}{L^2}g_{ab}$.

• But do the signs in the above depend on the sign of the metric convention?

I am confused when I look at these two metrics,

• $ds_1^2 = \frac{r^2}{Q^2}dt^2 - \frac{Q^2}{r^2}dr^2$

• $ds_2^2 = \left ( 1 - \frac{\Lambda r^2}{3} \right ) dt^2 - \frac{ dr^2 } { \left ( 1 - \frac{\Lambda r^2}{3} \right ) }$

Are both the above valid metrics of $AdS_2$? (with what sign of $\Lambda$?)

For the first one has, $R = 2/Q^2$ and $R_{ab} = \frac{1}{Q^2}g_{ab}$

For the second one has $R = - \frac{2\Lambda}{3}$ and $R_{ab} = -\frac{\Lambda}{3}g_{ab}$

Now I am not sure how to consistently define $L>0$ for these two metrics using the same formula for both!

• For anti-de Sitter space, it depends how you choose to define $\Lambda$, i.e. if you absorb the $\pm$ in the definition. Regardless, the energy density of $AdS$ is always negative. – JamalS Jun 27 '14 at 14:52
• @JamalS Can you kindly elaborate a bit? (1) Are you saying that the "n" formulas quoted initially are metric sign convention independent? (2) Can you give a prescription to define L for the last 2 AdS_ metrics with any choice of sign of $\Lambda$? (I can't see any consistent choice!) – user6818 Jun 27 '14 at 14:55
• See the different parametrizations of $AdS$ space in global coordinates and Poincaré coordinates – Trimok Jun 28 '14 at 11:16
• @Trimok I have seen these links - well - my "Q" metric is a Poincare coordinates AdS_2 in the time-positive convention. But to evaluate its "L" don't I need a sign reversed version of the Ricci tensor and scalar formula as given in en.wikipedia.org/wiki/Anti_de_Sitter_space#Geometric_properties – user6818 Jun 28 '14 at 16:47
• @Trimok Now with my $\Lambda$ metric what is going on? It seems to me that since it is in the time-positive convention I should use the sign reversed Ricci tensor and scalar formula as given in the link above - BUT for that to give any sensible value of "L" I need $\Lambda >0$. Right? (so what is this metric for $\Lambda < 0$?) [BUT if $\Lambda <0$ then it seems to fit the time-positive version of the global coordinates that you linked to! - and thats getting confusing!] – user6818 Jun 28 '14 at 16:51