I would like to know (1) what exactly is a UV-complete theory and (2) what is a confirmatory test of that?

  • Is asymptotic freedom enough to conclude that a theory is UV-complete?

    Does it become conclusive a test if the beta-function is shown to be non-perturbatively negative?

    If the one-loop beta-function is negative then does supersymmetry (holomorphy) immediately imply that the beta-function is non-perturbatively negative and hence the theory is proven to be UV-complete?

  • Or is vanishing of the beta-function enough to conclude that theory is UV-complete?

    Again similarly does supersymmetry (holomorphy) guarantee that if the 1-loop beta function is vanishing then it is non-perturbatively so and hence the theory is UV-complete?

  • Does superconformal theory necessarily mean UV-complete? We know that there exists pairs of superconformal theories with different gauge groups - related by S-duality - which have the same partition function - what does this say about UV-completeness?

  • Does a theory being UV-complete necessarily mean that it has a Maldacena dual? (...and isn't that the meaning of AdS/CFT that every UV-complete theory has a string dual and vice-versa?..)


1 Answer 1


A UV complete theory is one whose correlation functions or amplitudes may be calculated and yield unambiguously finite results for arbitrarily high energies.

Yes, asymptotic freedom is enough for UV completeness because the UV limit of the theory is free and therefore well-defined. Whenever the coupling constant is small and the beta-function is negative at one-loop level, the higher-loop corrections should be small so that the exact beta-function should be negative, too. Such implications become even easier to make with SUSY.

Yes, exactly scale-invariant (and especially conformal) theories are UV complete if they're consistent at any scale because they predict the same thing at all scales due to the scale invariance. However, scale invariance at the leading order doesn't imply exact scale invariance. So no, one-loop vanishing of the beta-function may be coincidental and – even in a SUSY theory – the full beta-function may still have both signs.

Yes, superconformal theories are a subset of conformal theories so they're UV-complete. The S-duality exchanges descriptions with a different value of the coupling which is a different quantity, and therefore an independent operation/test, from the UV-completeness and scale invariance that relates different dimensionful scales.

It is believed that all conformal theories must have "some" Maldacena dual in the bulk although whether this dual obeys all the usual conditions of a "theory of quantum gravity" or even a "stringy vacuum" is unknown, especially because we can't even say what all these conditions are. The non-AdS/non-CFT correspondence would in principle work for all UV-complete theories but it's much less established and more phenomenological than the proper AdS/CFT correspondence that works with the exactly conformal theories only.

  • $\begingroup$ Motl Thanks for the help. (1) So how often is it true that thanks to holomorphy/SUSY the 1-loop beta-function is non-perturbatively exact? You seem to suggest that neither asymptotic freedom nor scale invariance is necessarily proven by the 1-loop result..(2) So if the Lagrangian is known to be invariant under superconformal variations of the fields and the 1-loop beta-function vanioushes then does it follow the theory is non-perturbatively scale-invariant? (conformal?) $\endgroup$
    – user6818
    Sep 24, 2013 at 13:47
  • $\begingroup$ (3) Can you give some examples of non-AdS/non-CFT correspondences which happen? - and these necessarily need the QFT side to be UV-complete? $\endgroup$
    – user6818
    Sep 24, 2013 at 13:48

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