Three-dimensional gravity and BTZ black hole

Example

In a spacetime of dimension $3$, the Einstein field equations in vacuum with cosmological constant $\Lambda$ are $R_{ab} - \tfrac{1}{2} R g_{ab} + \Lambda g_{ab} = 0$. Taking the trace gives $R = 6\Lambda$ and then $R_{ab} = 2\Lambda g_{ab}$. In dimension $3$ the Weyl tensor vanishes identically, so the Riemann tensor is determined by the Ricci tensor. Hence every solution has constant sectional curvature $\Lambda$. So three-dimensional gravity has no local degrees of freedom: every vacuum solution is locally Minkowski, de Sitter or anti-de Sitter.

Let $\Lambda = -1/\ell^2$ with $\ell > 0$, and use units with $c = 1$ and $8G = 1$. For constants $M > 0$ and $J$ with $\abs{J} \le M\ell$, the BTZ black hole is the metric $$ds^2 = -f(r)\, dt^2 + \frac{dr^2}{f(r)} + r^2 \left( d\phi - \frac{J}{2r^2}\, dt \right)^2, \qquad f(r) = -M + \frac{r^2}{\ell^2} + \frac{J^2}{4r^2},$$ with $\phi$ an angle of period $2\pi$. It has constant curvature $-1/\ell^2$, so it solves the equations above, and it is locally isometric to anti-de Sitter space. The zeros of $f$ are at $r_\pm^2 = \tfrac{\ell^2}{2}\left( M \pm \sqrt{M^2 - J^2/\ell^2} \right)$, and the outer one, $r = r_+$, is an event horizon.