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? 
 A: 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.
A: GR have nothing to do with renormalization. You mean: “one loop semiclasical quantum gravity”. The way this is worked out is well explained in the classical book by Birell & Davis. What is done is to consider only closed graviton loops. The metric tensor is decomposed in a classical part + small fluctuation part that is quantized only to first loop level. The Field equations of GR are solved with the Einstein tensor equated to the average value the quantum mass-energy tensor of some quantum fields that are present. This average mass-energy tensor has to be regularized some way because is formally infinite. The book B&D contains many worked examples of this one loop semiclassical quantum gravity. The paper is calling SQG Einstein's gravity, but GR is a well-tested theory while SQG is not. You are right in principle there must be counter terms to all orders. The problem is that GR can't be quantized to all orders because is fundamentally not renormalizable. This means that the theory actually truncates quantum GR to all orders after one loop, and renormalize it to only this one loop level.
