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Is gravity non-emergent?

  1. Sep 6, 2010 #1


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    Asymptotic safety is a scenario in which gravity is not emergent, ie. degrees of freedom at low energy are sufficient for a complete description at high energy.

    Calculations from Bern and colleagues have been showing that N=8 supergravity is less divergent than previously suspected. A new paper from Kallosh http://arxiv.org/abs/1009.1135 states "30 years ago N=8 supergravity in four dimensions was suspected to be UV divergent at higher loop orders. The suspicion was based on a construction of an infinite set of superinvariants [6, 7] in a Lorentz covariant on shell superspace geometry with 32 Grassmann coordinates. These counterterms were viewed as candidates for UV divergences. 2 years ago one of the accusers proposed to restore the presumption of innocence ... Thus, in this paper N=8 d = 4 supergravity is acquitted from the previous accusation in [6, 7] and is predicted to be UV finite if there are no anomalies violating the equivalence theorem for physical observables."

    What might the implications of this be for Asymptotic Safety, and for string theory?
  2. jcsd
  3. Sep 9, 2010 #2


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    E7(7) constraints on counterterms in N=8 supergravity
    Niklas Beisert, Henriette Elvang, Daniel Z. Freedman, Michael Kiermaier, Alejandro Morales, Stephan Stieberger

    " ... the question of whether the loop computations based on generalized unitarity [4] could yield a UV finite result to all orders ... This question is well-de fined whether or not N = 8 supergravity is sensible as a full quantum theory"

    So even if 4D N=8 supergravity is UV finite to all orders, it may not be sensible as a full quantum theory? Why?
  4. Sep 9, 2010 #3


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    http://golem.ph.utexas.edu/~distler/blog/archives/001235.html via http://motls.blogspot.com/2008/04/n8-supergravity-lance-dixons-puzzle.html " In other words, like it or not, the full quantum-mechanical theory of N=8 supergravity is M-theory, compactified on T 7, whether we put that in from the outset or not."

    Ultraviolet Behavior of N=8 Supergravity
    Lance J. Dixon

    "Suppose that N = 8 supergravity is finite to all loop orders. This still would not prove that it is a nonperturbatively consistent theory of quantum gravity. There are at least two reasons to think that it might need a nonperturbative ultraviolet completion:

    1. The (likely) L! or worse growth of the coefficients of the order L terms in the perturbative expansion, which for fixed-angle scattering, means a non-convergent behavior ~ L! (s/M2)L.

    2. The fact that the perturbative series seems to be E7(7) invariant, while the mass spectrum of black holes is non-invariant (see e.g. ref. [88] for recent discussions)."
    Last edited by a moderator: Apr 25, 2017
  5. Sep 10, 2010 #4


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    I am surprised by the hype regarding "perturbative finiteness" of D=4, N=8 SUGRA
    - as said perturbative finiteness order by order does not mean that the perturbation series itself is finite
    - the theory is phenomenologically not viable

    So what are the reasons to study this theory?
    Last edited: Sep 10, 2010
  6. Sep 10, 2010 #5


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    There are plenty of theoretical reasons, besides those related to string theory. The primary and obvious one being that this is the only known example of a nontrivial quantum field theory that is both pertubatively finite and contains gravity in 4 dimensions. It is not a realistic model in terms of particle and matter content, but it is a consistent theory and a very close cousin to the real thing in most respects.

    And its a very nice cousin at that, b/c you are allowed to ask a lot of questions, and here you can actually calculate a lot of answers (even nonperturbatively)

    Aside from that, it is an aesthetically beautiful theory and there is a lot of deep mathematics hiding in its structure.

    The modern interest lies in the application of the KLT relations and the strange but nontrivial relation with N=4 SYM
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