Approaches to Quantum Gravity

336 Questions and answersspectrum in the Riemannian theory and a continuous spectrum (for spacelikeintervals) in the Lorentzian case. In four space-time dimensions, thegauge group of LQG is truly the complexification of SU(2) and the realityconditions might actually select a non-compact section of the complexgroup from which we would then derive a contiuous spectrum. Finally,these results are only at the kinematical level. They do not use the physicalHilbert space and inner-product, so we can not be sure of their physicalrelevance. Actually, the area operator is itself only defined in the kinematicalHilbert space (not invariant under diffeomorphism and not in the kernelof the Hamiltonian constraint) and we have not been able to lift it to aphysical operator acting on physical state. Nevertheless, in three space-timedimensions, work by Noui & Perez (2004) suggests that we can constructa physical length operator by introducing particles in the theory and wethen recover the kinematical results i.e a continuous length spectrum forthe Lorentzina theory. The issue is, however, still open in four space-timedimensions.• Q - L. Crane - to D. Oriti:It seems an awful shame to get to the point where each Feynman diagram in aGFT model is finite, then to describe the final theory as an infinite sum of suchterms. Have you ever thought of the possibility that by specifying the structureof the observer including its background geometry we limit the numberof simplicial complexes we need to sum over, or at least make most of thecontributions small, thereby rendering the answer to any genuinely physicalquestion finite?– A-D.Oriti:I agree. I would be careful in distinguishing the “definition of the theory”,given by its partition function (or its transition amplitudes), and the quantitiesthat, in the theory itself, corresponds to physical observables and are thusanswers to physical questions. The partition function itself may be defined,in absence of a better way, through its perturbative expansion in Feynmandiagrams, and thus involve an infinite sum that is most likely beyond reachof practical computability, and most likely divergent. However, I do believethat, once we understand the theory better, the answer to physical questionswill require only finite calculations. This can happen in three ways, I think.As you suggest, the very mathematical formulation of the question, involvingmaybe the specification of an observer or of a reference frame, or referringto a finite spacetime volume only, or some other type of physical restriction,will allow or even force us to limit the sum over graphs to a finitenumber of them, thus making the calculation finite. Another possibility isthat, as in ordinary QFT, the answer to a physical question (e.g. the result