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  Is there a simple way to understand why SUGRA is two-loop renormalisable?

+ 7 like - 0 dislike
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Naïve quantum gravity is one-loop renormalisable. There is a very simple way to argue that this is true: one lists all possible counter-terms that may appear at one loop, and show that they are, up to boundary terms, identical to terms already appearing in the original Lagrangian. This requires a non-trivial cancellation that results from the fact that a certain combination of terms happens to be topological (the Euler-Poincaré characteristic, cf. this PSE post).

Question: Can a similar analysis show that ($\mathcal N=4,8$) SUGRA is two-loop renormalisable?

As far as I know, the two-loops (and three- and four-loops) renormalisability of supersymmetric quantum gravity has been established by computing certain tree-level graphs and using the optical theorem or similar techniques. I guess that enumerating all possible counter-terms to three and four-loops is very cumbersome, but to two-loops it seems feasible. I don't know whether this has been attempted and the analysis was inconclusive (not enough symmetries to rule out all possibilities), or whether the computation is just so cumbersome it is now worth it. It seems to be a very direct approach, so it would be nice if it could be done.

This post imported from StackExchange Physics at 2018-03-25 20:23 (UTC), posted by SE-user AccidentalFourierTransform
asked Mar 23 in Theoretical Physics by AccidentalFourierTransform (480 points) [ no revision ]
retagged Mar 25

2 Answers

+ 4 like - 0 dislike

I think you're either misusing the word "renormalizable" or using it in a very old-fashioned way. Perturbative quantum gravity is renormalizable at all loop orders in the sense that divergences can be removed by the addition of local counterterms. Like all EFTs for which the leading interaction is irrelevant (in this case the Einstein-Hilbert term) such counterterms are necessarily higher order in the derivative expansion. The claim about pure gravity is that it is finite at one-loop, meaning that there are no UV divergent one-loop amplitudes. 

One can show that in \(d=4\) for \(\mathcal{N}\geq 1\)supersymmetry, pure supergravity is also finite at two-loops. Let's be clear what this means, it says that any two loop amplitude calculated with vertex factors from the supersymmetrization of the Einstein-Hilbert action is non-divergent if calculated with a supersymmetry preserving regulator. One way to show this is by a combination of counterterm analysis and on-shell methods. You begin by showing that the only possible counterterm would have to be a supersymmetrization of \((R_{\mu\nu\rho\sigma})^3\)and then show that no such supersymmetrization exists. A very efficient way to do the second step is to first see that such a counterterm would produce tree-level on-shell 3-point amplitudes for three gravitons with helicity assignments \((+2,+2,+2)\). However, such an amplitude can never satisfy the on-shell supersymmetry Ward identities and therefore cannot be present in a supersymmetric model. 

Similar arguments can be extended to higher loop orders. The most interesting results have been obtained in \(\mathcal{N}\geq 5\) supergravity. At leading order in the derivative expansion all supergravities have an electromagnetic duality symmetry which acts linearly on the graviphotons and chiral fermions and non-linearly on the scalars (if present). At tree-level the non-linear symmetry gives rise to low-energy theorems which say that any amplitude vanishes in the limit that the momentum of a scalar is taken to zero. For \(\mathcal{N}<5\) there is a non-vanishing \((\text{Diff})^2(\text{EM-duality})\) anomaly which leads to a breakdown of the low-energy theorem at loop order. For \(\mathcal{N}\geq 5\) this anomaly vanishes and so we can regularize loop integrals in such a way that the duality symmetry is unbroken. Consequently the on-shell amplitudes arising from any potential counterterms must satisfy the vanishing scalar low-energy theorem. 

For example, in https://arxiv.org/abs/1009.1643 it was shown explicitly in \(\mathcal{N}=8\) supergravity that the first on-shell amplitudes which satisfy \(\mathcal{N}=8\) supersymmetry, \(\text{SU}(8)_R\) symmetry and \(E_{7(7)}\) duality symmetry (importantly the low-energy theorems for the scalars) correspond to an interaction term of the form \(D^8R^4\) which in turn corresponds to a 7-loop counterterm. So we expect \(\mathcal{N}=8\) supergravity in \(d=4\) to be UV finite up to six-loops. 

answered Mar 26 by OldManAndTheSeaQuark (40 points) [ revision history ]
+ 3 like - 0 dislike

This does seem to be the way that 2-loop finiteness of $\mathcal{N}=1$ SUGRA was discovered. A discussion of the construction of possible counterterms is given in e.g. this reference: [arXiv1506.03757]. They show there simply isn't a supersymmetric counterterm at 2-loop order, and hence find that there will be no divergence in $\mathcal{N}=1$ SUGRA (and therefore, this means no counterterm will exist for $\mathcal{N}=4$ or $\mathcal{N}=8$ SUGRA as well, since these theories of course have $\mathcal{N}=1$ supersymmetry).

For higher loop order in the more supersymmetric theories, they appear to have better UV behavior beyond what you would expect even from counterterm arguments. There is some explanation in this reference: [arXiv:1703.08927]. There is also a conjecture that $\mathcal{N}=8$ SUGRA may be perturbatively finite at all loop orders, which, if true, does not seem to be able to follow from some symmetry principle. So in particular, the absence of an appropriately symmetric counterterm will fail at some loop order to explain the good UV behavior of SUGRA scattering amplitudes.

This post imported from StackExchange Physics at 2018-04-01 10:35 (UTC), posted by SE-user asperanz
answered Mar 26 by asperanz (175 points) [ no revision ]
Informative and to the point, great! Thank you.

This post imported from StackExchange Physics at 2018-04-01 10:35 (UTC), posted by SE-user AccidentalFourierTransform

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