de Sitter space and Holography
de Sitter space and Holography de Sitter space and Holography
Let’s consider X 0 − X d which turns out to beX 0 − X d = −le −t/l ≤ 0Therefore this coordinates does not cover the entire spacetime.In fact it just cover only half of it which in our convention it isgiven by the following figure.X 0X 0 = −X d X 0 = X dlX d14
The corresponding metric isdS 2 = −dt 2 + e −2t/l dx i dx iIn this coordinates system•∂∂x ifor i = 1,.., d − 1 are the Killing vectors•∂∂tis NOT a Killing vector• This breaks conservation of the energy so the Hamiltonianis not well-defined15
- Page 1 and 2: de Sitter space and HolographyThird
- Page 3: 1. Historical review and motivation
- Page 7 and 8: 5. Technologies:There are a number
- Page 9 and 10: Let us know study different coordin
- Page 11 and 12: Conformal coordinates: T, θ i , i
- Page 13: Planer (Inflationary) coordinates:
- Page 17 and 18: It is easy to see thatX 0 + X d =
- Page 20 and 21: 1.2 Causal structure (Penrose diagr
- Page 22 and 23: IINorth PoleOSouth PoleNorth PoleOS
- Page 24 and 25: The entire region of the dS space i
- Page 26 and 27: For example• In static coordinate
- Page 28 and 29: An observer moving along a timelike
- Page 30 and 31: 3. dS/CFT correspondenceFrom what w
- Page 32 and 33: In d dimensions the Einstein equati
- Page 34 and 35: One can generalize it for a general
- Page 36 and 37: • One can always decompose the me
- Page 38 and 39: Consider the Brown-York stress tens
- Page 40 and 41: Using the definition of Brown-York
- Page 42 and 43: Using this diffeomorphism one may s
- Page 44 and 45: lim dtdtr→∞∫I ′ dφdφ ′
- Page 46 and 47: dS static patch (spatial)dg =0 g =1
- Page 48 and 49: • Lower dimensional Plank mass is
- Page 50 and 51: One can use AdS/CFT correspondence
- Page 52 and 53: Under this conformal map one has:
- Page 54 and 55: Conformal AnomalyAdS/CFT instructs
- Page 56 and 57: • Naively one would think that gr
- Page 58: This talk is based on the following
Let’s consi<strong>de</strong>r X 0 − X d which turns out to beX 0 − X d = −le −t/l ≤ 0Therefore this coordinates does not cover the entire <strong>space</strong>time.In fact it just cover only half of it which in our convention it isgiven by the following figure.X 0X 0 = −X d X 0 = X dlX d14
Change language