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(I found this related Phys.SE post: Why is GR renormalizable to one loop?)

I want to know explicitly how it comes that Einstein-Hilbert action in 3+1 dimensions is not renormalizable at two loops or more from a QFT point of view, i.e., by counting the power of perturbation terms. I tried to find notes on this, but yet not anything constructive. Could anybody give an explanation with some details, or a link to some paper or notes on it?

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you're quite right that Einstein gravity is not renormalizable by powercounting. Be careful though, this is not a rigorous proof, it's a mere estimation. In fact there is not proof to this date which once and for all proves that gravity is really not renormalizable. If you think in terms of Feynman diagrams (which are a nightmare for Einstein gravity), there might be non-trivial cancellations hidden within the sum of graph which tame divergences. It might also be that the potential counterterms are related by some non-obvious symmetry, so that in the end only a finite number of field redefinitions is necessary to get rid of the divergences -- or in other words that a sensible implementation of renormalization is possible. In fact, the question about UV finiteness is currently being addressed by Zvi Bern and friends who could show using sophisticated techniques that maximally supersymmetric quantum gravity is much less divergent than one would naively think. The buzzwords here are color-kinematics duality and the double copy construction which basically says that a gravity scattering amplitude is in some sense the square of a gauge theory amplitude. Check the arxiv, there's a plethora about this.

Now, regarding powercounting the reasoning is roughly as follows: the EH action is basically $$\mathcal{L} = \frac{1}{\kappa} \int d^4x \sqrt{-g}R $$ with $g$ the determinant of the spacetime metric $g^{\mu\nu}$. The mass dimension of the Ricci scalar $R$ is $[m^2]$, that of the integral measure $[m^{-4}]$, order for the whole expression to be dimensionless $\kappa$ has to have mass dimension $[m^{-2}]$. If you now do a perturbative expansion around a flat background of the metric, you'll encounter at each step more and more powers of one over $\kappa$. Graphically, this expansion is an expansion in numbers of loops in Feynman diagrams. At each step, i.e. at each loop level the whole expression should be dimensionless, i.e. at each step you need more and more powers of loop momentum (at each loop level two more powers, to be precise), s.t. in the end your expressions become the more divergent the higher you go in the perturtabive expansion. In order to cancel these ever sickening divergences you'd have to introduce an infinite number of counterterms which -- in terms of renormalization -- makes no sense, hence this theory is by powercouting non-renormalizable.

There are nice lecture notes about this, cf (Notes by Lance Dixon about supergravity but the introductory bit is quite general).

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You mention " there is not proof to this date which once and for all proves that gravity is really not renormalizable". Was this not proved by Goroff and Sagnotti? or is it the case that some miracle might happen beyond two loops? – twistor59 Apr 16 '13 at 6:17
@twistor59: Thanks for mentioning the paper. They say in their conclusions that they don't expect any more cancellations for Einstein gravity beyond one loop based on a 2-loop computation. While I believe this is true and don't think that Einstein gravity is renormalizable, what I don't like about their argument is that it is not an all-loop proof but just a 2-loop computation. I might be too strict here in my sense of proof but I would really like to see an all-loop proof... – A friendly helper Apr 16 '13 at 7:12
...given the magic happening in amplitude computations, why not expect a miracle beyond two loops? It happened in supergravities as well. Again, I don't see how this could happen though :) – A friendly helper Apr 16 '13 at 7:13
I wouldn't disagree. Don't you just hate perturbation theory? +1 btw – twistor59 Apr 16 '13 at 7:19
@twistor59: I sure do :). – A friendly helper Apr 16 '13 at 21:34

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