Reading the book "Supergravity" from Freedman & van Proeyen I stumbled over the assertion that in 3D GR the vacuum solution $R_{\mu\nu} =0$ leads to a vanishing 4-rank curvature tensor $R_{\mu\nu\rho\sigma}=0$, therefore there are no gauge invariant degrees of freedom in the 3D GR vacuum case (I indeed found in Landau & Lifshitz's volume II chapter 93 a relation between the 3D Ricci-tensor and the 3D full 4-rank curvature tensor that confirms that).

But would that implicate that an axial-symmetric solution of the vacuum EFEs $R_{\mu\nu} =0$ in 3D=(1 time + 2space) would be trivial, i.e. not Schwarzschild ( replace if appropiate $r^2 d\Omega^2 \rightarrow r^2 d\phi^2$ instead of $r^2 d\Omega^2= r^2 (d\theta^2 +\sin^2 \theta d\phi^2$)) ?


2 Answers 2


No it is not. The Weyl tensor vanishes by definition in three dimensions, Einstein's equations (in the absence of matter) impose:

$$R_{\mu\nu} = 0 \rightarrow R=0$$

and since $Riemann = Weyl + Ricci$ no geometry can be formed.

The solution in three dimensional spacetime is the BTZ black hole (https://arxiv.org/abs/hep-th/9204099v3) which includes a cosmological constant thus the Ricci tensor is no longer equal to zero.

A derivation: Consider $2+1$ Gravity and a cosmological constant term: \begin{equation} S = \int d^3 x \sqrt{-g} \big(R -2Λ\big) \end{equation} Einstein's equation read: \begin{equation} G_{\mu\nu} + \Lambda g_{\mu\nu} = 0 \end{equation} and in the form of differential equations, imposing a two degree of freedom metric: \begin{equation} ds^2 = -b(r)dt^2 + f(r)dr^2 + r^2 d\theta^2 \end{equation} we get: \begin{equation} \frac{2 \Lambda -\frac{f'(r)}{r f(r)^2}}{2 b(r)} =0 \end{equation} \begin{equation} -\frac{\frac{b'(r)}{r b(r)}+2 \Lambda f(r)}{2 f(r)^2} =0 \end{equation} \begin{equation} \frac{b(r) \left(b'(r) f'(r)-2 f(r) b''(r)\right)+f(r) b'(r)^2-4 \Lambda b(r)^2 f(r)^2}{4 r^2 b(r)^2 f(r)^2} =0 \end{equation} The first one is a differential equation for $f(r)$: $$2 \Lambda -\frac{f'(r)}{r f(r)^2} =0 \Rightarrow \Big(\Lambda r^2 + \cfrac{1}{f(r)}\Big)' =0 \Rightarrow $$ \begin{equation} f(r) = \cfrac{1}{C - Λr^2} \end{equation} where $C$ is a constant of integration. Now we can obtain $b(r)$ from the second equation: $$\frac{b'(r)}{r b(r)}+2 \Lambda f(r)=0 \Rightarrow (\ln(C-\Lambda r^2))' - (\ln b(r))'=0 \Rightarrow $$ \begin{equation} b(r) = C - \Lambda r^2 \end{equation} Now, if we set $C=-M$ and $\Lambda = -1/l^2$, where $l$ the AdS radius we obtain the BTZ Black hole: \begin{equation} b(r) = \cfrac{r^2}{l^2} -M = \cfrac{1}{f(r)} \end{equation} We can see that this solution satisfies the gauge $g_{tt}g_{rr} = -1$. The obtained configurations satisfy the last Einstein equation.

  • $\begingroup$ Thank you for the quick answer. What does BTZ mean ? $\endgroup$ Oct 12, 2020 at 14:35
  • $\begingroup$ Bañados, Teitelboim & Zanelli $\endgroup$
    – Noone
    Oct 12, 2020 at 14:40
  • $\begingroup$ If i find time i'll sketch a derivation. $\endgroup$
    – Noone
    Oct 12, 2020 at 14:41

As mentioned in ApolloRa's answer, in 2+1 dimensions there exist no asympototically flat black hole solutions. However, you can still solve the Einstein Field Equations to find the metric of a non-spinning point mass $M$. The answer is given by

$$ ds^2 = -dt^2 +\frac{1}{(1-4GM)^2}dr^2 + r^2 d\phi^2$$

As you can easily check this metric is flat for all $r>0$. However, it has a singular curvature at $r=0$. This can be confirmed by calculation the holonomy along a curve around the origin. If you parallel transport a vector around the origin, you will find that it has been rotated by $8\pi GM$ radians when it returns to its original position.

The spatial part of this metric, is that of a cone, which gives this type of singularity it name, a conical singularity.

  • $\begingroup$ Thank you: Just a stupid question: in the denominator $1-4GM$ there is no $r$, is that meant like that ? And if there is a singular curvature in this metric, in a polar coordinates the curvature would be also singular ? $\endgroup$ Oct 12, 2020 at 15:19
  • $\begingroup$ @FredericThomas Yes, the coefficient of $dr^2$ is simply a constant. $\endgroup$
    – TimRias
    Oct 12, 2020 at 15:22
  • $\begingroup$ The curvature in this solution can be thought of (sorta) as a delta function at the origin. $\endgroup$
    – TimRias
    Oct 12, 2020 at 15:24
  • 1
    $\begingroup$ @FredericThomas With one fewer space dimension than usual, rhe Newtonian acceleration is $-GM/r$ instead of $-GM/r^2$, so $GM/c^2$ is dimensionless. $\endgroup$
    – J.G.
    Oct 13, 2020 at 11:32

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