# Why is GR renormalizable to one loop?

I have read in a few places that GR is renormalizable at one loop. (hep-th/9809169 for example, second sentence, although they don't seem to develop this point at all). Is this do to some hidden symmetry in the theory? Naively we need new counter terms at all orders, even one loop, in perturbation theory, right?

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Well, the analysis performed examines counterterms ... and the EH action requires an infinite number of counter-terms for two-loop corrections. See arXiv:0910.4110 for some details and references... –  Alex Nelson Jun 29 '12 at 5:43
Isn't GR divergent at the tree level if coupled to matter? –  Jerry Schirmer Aug 19 '12 at 18:59
DJBunk, could you un-accept Ron's answer so that he can delete it himself? –  dmckee Aug 19 '12 at 22:41

The counterterms at one loop would be $R^2$ operators, because loops are counted by powers of $G_N = 1/M_P^2$. The tree-level Lagrangian is the Einstein Hilbert action $M_P^2 R$, so the one-loop counterterms for logarithmic divergences should be terms that carry no powers of $M_P$ in front. Simply from dimensional analysis, then, these are $R^2$ terms, of which there are three linearly independent choices: $R^2$, $R_{\mu\nu}R^{\mu\nu}$, and $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$.

Two combinations can be eliminated by field redefinitions (of the form $g_{\mu\nu} \to g_{\mu\nu} + c_1 R g_{\mu\nu} + c_2 R_{\mu\nu}$), and the third is a total derivative and so has no local physical effect. (Namely, it's the Euler density $R_{\mu\nu\rho\sigma}^2-4 R_{\mu\nu}^2+R^2$, also known as the Gauss-Bonnet term, whose integral is the Euler characteristic, a topological invariant). The upshot is that you have to go to $R^3$ terms before you get nontrivial counterterms, and these come from two-loop diagrams.

As far as I can find, the original source for the argument is this paper of 't Hooft and Veltman.

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Although this is the standard argument, I'm feeling a little unsatisfied with it at the moment, because it still allows a divergent renormalization of $M_P^2 R$. In dim reg such power divergences would be absent, but I'm not very happy with a regulator-dependent statement. I'm not sure what to think about this. –  Matt Reece Aug 18 '12 at 19:39
I'd like to know this argument--- where does it come from? Is it Veltman's? I was guessing. –  Ron Maimon Aug 19 '12 at 6:44
+1 This is the conventional argument, I already knew it. What is more (and I don't if this is well-known), one can also renormalize gravity at 2 loops trough non-local metric redefinitions. –  drake Aug 19 '12 at 18:06
@RonMaimon: I learned the argument from hearing it repeatedly in talks about the possible finiteness of ${\cal N}=8$ supergravity, but it looks like the original source is the 't Hooft/Veltman paper I've now linked in the answer. (Maybe since 't Hooft seems to be hanging out here these days, he could clarify things...) –  Matt Reece Aug 19 '12 at 18:48
@drake: Please, could you explain how to do this? –  Ron Maimon Aug 20 '12 at 2:52