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I heard a couple of times that there is no dynamics in 3D (2+1) GR, that it's something like a topological theory. I got the argument in the 2D case (the metric is conformally flat, Einstein equations trivially satisfied and the action is just a topological number) but I don't get how it is still true or partially similar with one more space dimension.

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The absence of physical excitations in 3 dimensions has a simple reason: the Riemann tensor may be fully expressed via the Ricci tensor. Because the Ricci tensor vanishes in the vacuum due to Einstein's equations, the Riemann tensor vanishes (whenever the equations of motion are imposed), too: the vacuum has to be flat (no nontrivial Schwarzschild-like curved vacuum solutions). So there can't be any gravitational waves, there are no gravitons (quanta of gravitational waves). In other words, Ricci flatness implies flatness.

Counting components of tensors

The reason why the Riemann tensor is fully determined by the Ricci tensor is not hard to see. The Riemann tensor is $R_{abcd}$ but it is antisymmetric in $ab$ and in $cd$ and symmetric under exchange of the index pairs $ab$, $cd$. In 3 dimensions, one may dualize the antisymmetric index pairs $ab$ and $cd$ to simple indices $e,f$ using the antisymmetric $\epsilon_{abe}$ tensor and the Riemann tensor is symmetric in these new consolidated $e,f$ indices so it has 6 components, just like the Ricci tensor $R_{gh}$.

Because the Riemann tensor may always be written in terms of the Ricci tensor and because they have the same number of components at each point in $D=3$, it must be true that the opposite relationship has to exist, too. It is $$ R_{abcd} = \alpha(R_{ac}g_{bd} - R_{bc}g_{ad} - R_{ad}g_{bc} + R_{bd}g_{ac} )+\beta R(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ I leave it as a homework to calculate the right values of $\alpha,\beta$ from the condition that the $ac$-contraction of the object above produces $R_{bd}$, as expected from the Ricci tensor's definition.

Counting polarizations of gravitons (or linearized gravitational waves)

An alternative way to prove that there are no physical polarizations in $D=3$ is to count them using the usual formula. The physical polarizations in $D$ dimensions are the traceless symmetric tensor in $(D-2)$ dimensions. For $D=3$, you have $D-2=1$ so only the symmetric tensor only has a unique component, e.g. $h_{22}$, and the traceless condition eliminates this last condition, too. So just like you have 2 physical graviton polarizations in $D=4$ and 44 polarizations in $D=11$, to mention two examples, there are 0 of them in $D=3$. The general number is $(D-2)(D-1)/2-1$.

In 2 dimensions, the whole Riemann tensor may be expressed in terms of the Ricci scalar curvature $R$ (best the Ricci tensor itself is $R_{ab}=Rg_{ab}/2$) which is directly imprinted to the component $R_{1212}$ etc.: Einstein's equations become vacuous in 2D. The number of components of the gravitational field is formally $(-1)$ in $D=2$; the local dynamics of the gravitational sector is not only vacuous but it imposes constraints on the remaining matter, too.

Other effects of gravity in 3D

While there are no gravitational waves in 3 dimensions, it doesn't mean that there are absolutely no gravitational effects. One may create point masses. Their gravitational field creates a vacuum that is Riemann-flat almost everywhere but creates a deficit angle.

Approximately equivalent theories

Due to the absence of local excitations, this is formally a topological theory and there are maps to other topological theories in 3D, especially the Chern-Simons theory with a gauge group. However, this equivalence only holds in some perturbative approximations and under extra assumptions, and for most purposes, it is a vacuous relationship, anyway.

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Thanks for the very detailed explanation Lubos, should have counted the components of the Riemann and the Ricci tensors. –  toot May 24 '12 at 11:21
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I like this nice explanation too :-) –  Dilaton May 24 '12 at 11:25
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In the expression for the Riemann tensor in terms of Ricci tensor, $\alpha=1,\beta=-1$, right? –  c.p. Jun 14 '12 at 23:06

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