8.821 F2008 Lecture 12:Boundary of AdS; Poincar´e patch; wave equation in AdSLecturer: McGreevy Scribe: Francesco D’EramoOctober 16, 2008Today:1. the boundary of AdS2. Poincar´e patch3. motivate boundary value problem4. wave equation in AdS.1 The boundary of AdSWe defined the Lorentzian AdSp+2as the locus {ηabXaXb= −L2} ⊂ IRp+1,2, whereηabXaXb= −X20+p+1Xi=1X2i− X2p+2= −L2(1)The metric isds2AdS= ηabXaXb|(1)= L2−cosh2ρ dτ2+ dρ2+ sinh2ρ dΩ2p (2)1.1 Projective boundaryTake a solution V =X0,~X, Xp+2of equation (1). Reach the boundary by rescaling X, preserving(1). Let X = λ˜X, then equation (1) becomesηab˜Xa˜Xb= −L2λ2(3)We now take λ → ∞, the boun dary is{ηab˜Xa˜Xb= 0} / {˜X ∼ λ˜X} ≃ IRp,1(4)1Figure 1: Lorentzian AdS: The left-right axis is the ρ direction. At ρ = 0, the Spin the lower figureshrinks to zero size (like sinh ρ), while the radius of the τ direction, depicted in the top figure,approaches a constant (like cosh ρ).This relation can also be read as follows: the boundary of AdS is the set of lightrays in IRp+1,2,modulo the rescaling. Recall that this is exactly parametrized by points in IRp,1as:ρa= κXµ,12(1 − X2),12(1 + X2). (5)We used this fact earlier to make write the SO(p + 1, 2) action of the conformal group on IRp,1ina linear way. The fact that the conformal group of IRp,1has a n ice action on the bou ndary of AdSis very encouraging.2Alternative decomposition IFix λ by imposing 1 =~X2=Pp+1i=1X2i. Then we haveX20+ X2p+2=~X2= 1 ⇒ ∂AdS = S1× Sp(6)Alternative decomposition IILet u±= X0± iXp+1. Then (1) ⇒ −u+u−+~X2= 0.If u+6= 0 set u+= 1 ⇒ u−=~X2If u−6= 0 set u−= 1 ⇒ u+=~X2Then˜~X =~X~X2. u−is ’the point at ∞’. Th e boundary is compact.1.2 Penrose diagram (one more description of the boundary)Let dΘ =dρcosh ρ(this variable was called ‘squiggle’ in lecture). The metric in these new coord inatesresults inds2= cosh2ρ−dτ2+ dΘ2+ tan2Θ2dΩ2p(7)and thereforetanΘ2= tanhρ2Θ ∈ [0, π/2] (8)The boundary is {Θ = π/2} ∼ IR ×Sp. Note that the metric on the boundary is only specified upFigure 2: T he squiggle variable Θ runs from 0 to π/2 as ρ goes from 0 to ∞to rescaling, i.e. a Weyl transformation.But why do we care about this boundary more than say the conformal boundary of Minkowskispace? The answer is in th e next two subsections.31.3 Massless geodesicsThe massless geodesics are given by the condition ds2= 0, which implies0 = ds2= L2−cosh2ρ dτ2+ dρ2⇒ cosh ρ =dρdτ⇒ dτ =dρcosh ρ= dΘ (9)Θ is the time elapsed for a static observer. Whether the lightray reflects off the boundary dependson th e BC’s. Hence: Cauchy p roblem problem.Figure 3: Massless geodesics1.4 Massive geodesicsThe action for a massive relativistic point particle isS = mZds = mZqgµν˙Xµ˙Xν˙Xµ= ∂τXµ(10)The equation of motion isδSδXµ= 0 ⇒¨Xµ+ Γµνλ˙Xν˙Xλ= 0 (11)4where the second equation follows if˙X = ∂sX where s is proper time. If we assu me˙Ω = 0 theaction isS = mLZdτqcosh2ρ − (∂τρ)2.You will sh ow on problem set 3 that this has an oscillatory solution around ρ = 0, it never reaches∞.2 Poincar´e patchPick out Xp+1from among th e Xi. This will break the SO(p + 1) symmetry of the p-sphere. LetXµ=LzxµXp+2+ Xp+1=Lz−Xp+2+ Xp+1= v(12)Equation (1) and the metric becomeLτv −L2z2xµxµ= −L2ds2= L2dz2+ dxµdxµz2(13)(same cancellation as UHP). This is the metric which we showed hasRµν−12gµνR = ΛgµνΛ = −(p + 1)(p + 2)2L2(14)NOTE: it covers part of AdS. As z → ∞, ∂/∂t becomes NULL (Poincar´e horizon).CLAIM: relation between Poincar´e patch and global time is state-operator correspondence.EVIDENCE: symmetries → SO(p, 1) ×IRp+1and SO(p + 1) × SO(2).2.1 Towards CFT correlators from fields in AdSOur goal is to evaluate he−Rφ0OiCF T≡ e−WCF T[φ0].Conjecture: he−Rφ0OiCF T= Zstrings in AdS[φ0], but we cannot compute it. The pratical version isthe followingWCF T[φ0] = −lnheRφ0OiCF T≃ extremumφ|z=ǫ=φ0N2ISU GRA[φ]+ O1N2+ O1√λ(15)A few comments:• The supergravity description is valid for large N and large λ. In (15) we’ve made the N-dependence explicit: in units of the AdS radius, the Newton constant is1GN= N2. ISU GRAis some d imen sionless action.5Figure 4: Poincar´e patch• anticipating divergences at z → 0, we introduce a cutoff (which will be a UV cu toff in theCFT) and set boundary cond itions at z = ǫ.• Eqn (15) is written as if there is just one field in the bulk. Really there is a φ for everyoperator O in the dual field theory.We’ll say ‘φ couples to O’ at the boundary. How to match? We give four examples1. Dilaton field.Before near horizon limit, we have D3-branes in IR10; the asymptotic value of the dilatondetermines the string coupling constant gs= heφ(x→∞)i. The YM coupling on D3’s is g2Y M=gs.Changing φ → φ + δφ we getδS =Zδφg2sT rF2+ . . . (16)where the dots stand for all the CP-even term in the lagrangian. In conclusion we haveZstrings[φ → φ + δφ] ≃ he1g2sRδφT r[F2]iCF T(17)The dilaton couples to all the terms in th e lagrangian which are CP invariant.2. RR axion.We have that τstr=igs+χ2πtranforms u nder SL(2,C) nicely, like τ =igs+θ2π. Thereforeχ ↔ T r[F ∧ F ] (18)This time CP-odd terms63. Stress energy tensor.The tensor Tµνis the response of a local QFT to local change in the metric. SQF T⊃RγµνTµν.Here we are writing γµνfor the metric on the boundary. In this casegµν↔ Tµν(19)4. IIB in AdS5× S5.Isometry on S5→ SO(6) Kaluza-Klein (KK) gauge fields ↔ SO(6)R= SU(4)R. In this casethe correspondence is between these gauge fields and the R-current operatorsAKK aµ↔ Jµ aR(20)i.e. Sbdy∋RAaµJµa2.2 Useful visualizationFigure 5: Feynman graphs in AdS. We d o the one with two ext. legs firstClassical field theory in bulk (boundary value problem).Extr. of classical action (expand about quadratic solution in powers of φ0) = tree level SUGRAFeynman graphs.BUT: usually (QFT in IRD ,1), ext. legs of graphs = wavefunction of asymptotic states (example:plane waves).In AdS: ext. legs of graphs determined by boundary behavior of φ (‘bulk-to-boundary propagators’).3 Wave equation in AdSWe work in Poincar´e coordinates. The metric isds2= L2dz2+ dxµdxµz2≡ gABdzAdzBA = 0, . . . , p + 1 (21)The action for a scalar field isS =