Anti – De Sitter Space is the maximally symmetric solution to field equations with negative cosmological constant.
The negative cosmological constant also shows that the spacetime has negative curvature. I don't understand how is the cosmological constant related to the curvature of spacetime? Also how do we show this mathematically that Anti-De Sitter spacetime has negative curvature?
Best Answer
The local radius of curvature $r(x)$ at a point $x$ in a spacetime is effectively given by the formula $$ R(x) = \frac{1}{r(x)^2} $$ where $R$ Ricci scalar.
AdS space is a maximally symmetric space satisfying Einstein's equations with a negative cosmological constant, $$ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R + \Lambda g_{\mu\nu} = 0 \, , \qquad \Lambda < 0 \,. $$ Then, taking a trace of the equation above, we find $$ R - \frac{1}{2} d R + \Lambda d = 0 ~~\implies ~~ R = \frac{2\Lambda d}{d-2} $$ Thus, $$ r^2 = \frac{d-2}{2\Lambda d} \, . $$ Here, $d$ is the dimension of the AdS space. For $d>2$, the RHS of the above equation is negative (since $\Lambda < 0$). Thus, the spacetime has constant (because $\Lambda$ is a constant) negative curvature.