Anti-de Sitter space

Example

Let $n \ge 3$ and $\alpha > 0$. Equip $\R^{n+1}$ with coordinates $(u, v, x_1, \ldots, x_{n-1})$ and the form $\eta = -du^2 - dv^2 + dx_1^2 + \cdots + dx_{n-1}^2$ of signature $(2, n-1)$. The anti-de Sitter space of radius $\alpha$ is the quadric $$\set{-u^2 - v^2 + x_1^2 + \cdots + x_{n-1}^2 = -\alpha^2}$$ with the metric $g$ induced by $\eta$.

Check: the coordinates $u = \alpha\cosh\rho\cos\tau$, $v = \alpha\cosh\rho\sin\tau$, $x = \alpha\sinh\rho\,\omega$ with $\rho \ge 0$, $\tau$ an angle and $\omega \in S^{n-2}$ cover the quadric. A direct substitution gives $$g = \alpha^2\left(-\cosh^2\rho\, d\tau^2 + d\rho^2 + \sinh^2\rho\, d\Omega_{n-2}^2\right),$$ which has one negative and $n - 1$ positive directions, so $g$ is Lorentzian.

The sectional curvature is constant and equal to $-1/\alpha^2$. So $R_{ab} = -\frac{n-1}{\alpha^2} g_{ab}$, and the space solves $G_{ab} + \Lambda g_{ab} = 0$ with negative cosmological constant $\Lambda = -\frac{(n-1)(n-2)}{2\alpha^2}$. The group $O(2, n-1)$ acts by isometries, so the space is maximally symmetric.

The coordinate $\tau$ is periodic, so the curves $\rho = \text{const}$, $\omega = \text{const}$ are closed timelike curves. One usually passes to the universal cover, where $\tau \in \R$; this is also called anti-de Sitter space and has no closed timelike curves. Its slices $\tau = \text{const}$ are copies of hyperbolic space, as in the hyperboloid model.