2018
DOI: 10.1007/jhep02(2018)133
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Unmixing supergravity

Abstract: Abstract:We examine the double-trace spectrum of N = 4 super Yang-Mills theory in the supergravity limit. At large N double-trace operators exhibit degeneracy. By considering free-field and tree-level supergravity contributions to four-point functions of half-BPS operators we resolve the degeneracy for a large family of double-trace operators. The mixing problem reveals a surprisingly simple structure which allows us to obtain their three-point functions at leading order in the large N expansion as well as the… Show more

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Cited by 110 publications
(207 citation statements)
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“…The relevant explicit formulas for the anomalous dimensions and three-point functions of doubletrace operators were obtained by considering multiple correlators which exhibit the same exchanged operators in their OPE decompositions [18]. Such data is available due to a remarkably compact formula [12] for all four-point tree-level scattering processes of graviton multiplets or their associated Kaluza-Klein modes which are present in the five-dimensional bulk due to the reduction from ten dimensions on S 5 .…”
Section: Jhep05(2018)056mentioning
confidence: 99%
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“…The relevant explicit formulas for the anomalous dimensions and three-point functions of doubletrace operators were obtained by considering multiple correlators which exhibit the same exchanged operators in their OPE decompositions [18]. Such data is available due to a remarkably compact formula [12] for all four-point tree-level scattering processes of graviton multiplets or their associated Kaluza-Klein modes which are present in the five-dimensional bulk due to the reduction from ten dimensions on S 5 .…”
Section: Jhep05(2018)056mentioning
confidence: 99%
“…Here we will summarise the results obtained in [17,18] which allowed for the bootstrapping of the one-loop four-graviton amplitude. Firstly we may consider the following set of double-trace operators K t,l,n,i , labelled by i which runs from 1 to (t − n − 1), t,l,n,i = − 2(t − n − 1)(t) 2 (t + n + 2)(t + l − n)(t + l + 1) 2 (t + l + n + 3) (l + 2i + n − 1) 6 , (…”
Section: Jhep05(2018)056mentioning
confidence: 99%
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