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What is the best way to see that the number of propagating degrees of freedom or gravitons in 3 dimensions is $0$ ? By graviton I mean the metric and NOT some topologically massive graviton that one can include in the Lagrangian. Also is the number in 4-dimensions 2?

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This question (v3) is also addressed in e.g this and this Phys.SE answers. – Qmechanic Jan 31 '13 at 13:04

The Weyl tensor vanishes in 3 dimensions, so the only remaining gravitational degrees of freedom are the Ricci coefficients, but they're tied to matter sources by the Einstein equations so aren't "propagating" degrees of freedom.

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Is there a way to see it through the little group of massless particles? Like you can say that there are 2 helicity states of photon in 4d by looking at the little group which is Euclidean group E(2). – dbranes Jan 31 '13 at 12:56
I suspect that there will indeed be an argument along these lines for linearized gravity, but I'm not so familiar with that approach. Hopefully someone might contribute an answer based on that reasoning. – twistor59 Jan 31 '13 at 13:31

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