Probing holographic dualities through Penrose limits and dual spin chains. Horatiu Nastase IFT-UNESP

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1 Probing holographic dualities through Penrose limits and dual spin chains Horatiu Nastase IFT-UNESP Gauge/gravity duality, Wurzburg August 018 based on: T. Araujo, G. Itsios, HN and E. O Colgain (JHEP 18) and G. Itsios, HN, C. Nunez, K. Sfetsos and S. Zacarias (JHEP 18) 1

2 Summary: 0. Introduction 1. GJV duality and +1d CS-SYM in IR. Penrose limits of GJV and string quantization 3.Spin chains in CS-SYM field theory IR limit 4. Penrose limits of Abelian and Nonabelian T-duals of AdS 5 S 5 5. Operators in field theory, RG flow and deconstruction 6. Conclusions.

3 0. Introduction Penrose limit of gravity dual background: pp wave a large J charge limit. AdS 5 S 5 max. susy pp wave BMN operators Pp wave calculable for strings lightcone Hamiltonian H J calculable for BMN operators Want to use the same logic for less symmetric, and more complicated holographic dual pairs. We use it for GJV duality, and for T-duals of AdS 5 S 5. 3

4 GJV duality: warped, squashed AdS 4 S 6 vs. +1d N = SYM-CS theory in the IR. T-duals of AdS 5 S 5 less susy for the background. Abelian: known, but dual field theory only conjectured: quiver field theory. Nonabelian: recent, less known about it also a quiver. Find spin chains in the Penrose limit, but mismatch in conformal dimensions. Fixable for GJV, not so much for T-duals: field theory issue here? 4

5 1. GJV duality and +1d CS-SYM in the IR GJV find geometry (in string frame) (. ds = e ϕ/+a ds AdS dα + 6 sin α (3 + cos α) ds CP + 9 ) sin α (5 + cos α) η, (ds AdS4 + 3 ) dα + Ξds CP + Ωη L string e ϕ = e ϕ (5 + cos α)3/4 0 (3 + cos α), B = e ϕ0/ sin α cos α 6L J 3L e ϕ 0/ sin αdα η, (3 + cos α) 1 F 0 =, 3L e 5ϕ 0 /4 ( 6L 4 sin F ) α cos α 3(3 cos α) = J + sin α dα η, e 3ϕ 0/4 (3 + cos α)(5 + cos α) (5 + cos α) F 4 = L3 e ϕ 0/4 ( 6vol(AdS 4 ) 1 3 (7 + 3 cos α) (3 + cos α) sin4 α vol(cp ) (9 + cos α) sin3 α cos α (3 + cos α)(5 + cos α) J dα η where η = dψ + ω, dω = J, ω = 1 sin λ(dσ + cos θdϕ), and ds CP = dλ sin λ { dθ + sin θdϕ + cos λ(dσ + cos θdϕ) } ) 5

6 Quantizing changes, n p = ˆF p, we find that the parameters of the solutions are n 6 = N and n 0 = k = πl s m, and that π 3/8 l s L = 7/48 3 7/4(kN 5 ) 1/4 ; e ϕ 0 = 11/1 π 1/ 1 3 1/6 (k 5 N) 1/6 L string = 1/6 ( ) π N 1/3 3 /3 l s 5 + cos α k Field theory: N = SYM-CS in the IR: -chiral superfield Φ = ϕ + θψ + θθf -vector superfield (in WZ gauge) V(x) = iθ θσ + θγ µ θa µ + i θ θ χ i θ θχ + θ θ D Is dimensional reduction of N = 4 SYM in 3+1d. superpotential W = gtr (Φ 1 [Φ, Φ 3 ]) = g 1 ϵ ijkf abc Φ a i Φb j Φc k Then, 6

7 Action (in the IR), CS + matter + superpotential, S CS = k ( d 3 xtr [ϵ µνρ A µ ν A ρ + i ) ] 4π 3 A µa ν A ρ + i χχ Dσ L m = Tr [ (D µ ϕ) i D µ ϕ i + i ψ i γ µ D µ ψ i F i F i + ϕ i Dϕ i + ϕ i σ ϕ i i ψ i σψ i +iϕ i χψ i + i ψ i χϕ i] L sp = d θw(φ) d θw(φ) ( ) W(ϕ) = Tr F i + W(ϕ) F ϕ i ϕ i 1 W(ϕ) i ϕ i ϕ j ψi ψ j 1 W(ϕ) ψ ϕ i ϕ i ψ j j Solve for σ a, D and find interaction potential terms V conf = 4π k Tr ( [[ϕ i, ϕ i ], ϕ k ][[ϕ j, ϕ j ], ϕ k ] ) V sp = g Tr ( [ϕ i, ϕ j ][ϕ i, ϕ j ] ) Obs.: in the IR, SYM subleading to CS drop it. 7

8 Moreover, in the IR, conformal term in the potential remains, while nonconformal one is in the vacuum. Indeed, classical conformal dimensions: [g] = 1/, [ϕ] = 1/ = [Φ], ([V ] = 3, [W ] = ), [ψ] = 1, [θ] = 1/. Then, V sp must be put to 0 (vacuum) in the IR [ϕ i, ϕ j ] = 0, i j. Obs: one still has [ϕ i, ϕ j ] 0. 8

9 Alternative explanation: Gaiotto+Yin: D/D6 in massive IIA. D:Φ 1 D6, Φ, Φ 3 D6, Q, Q : D-D6 coords. Then in the IR, W = Tr [Φ 1 [Φ, Φ 3 ]] + QΦ 1 Q [Φ 1 ] = 1, [Q] = [ Q] = [Φ,3 ] = 1/. Why? Φ 1 is auxiliary (no d 4 θ Φ 1 Φ 1 ). Can add ϵtr Φ 1 / W = 1 ϵ Tr [([Φ 3, Φ 3 ] + QQ) ] In our symmetric case (GJV): [Φ 1 ] = [Φ ] = [Φ 3 ] = /3; then only W/ ϕ i F i in action ([F ] = 5/3), so δs/δf i again W/ ϕ i = 0 [ϕ i, ϕ j ] = 0. 9

10 . Penrose limits of GJV and their quantization. Penrose limit: close to null geodesic for geodesic parametrized by λ, in direction x λ, need no acceleration, Γ i λλ = 0. For isometry direction, Also ds = 0 (null). g ij j g λλ = 0 GJV: σ, ψ, ϕ isometries: σ g µν = ϕ g µν = ψ g µν = 0. Motion in ψ: α = π/, ρ = 0, λ = 0. Motion in σ: λ = π/4, α = π/, θ = π/, ρ = 0. Motion in ϕ: λ = π/, α = π/, θ = π/, ρ = 0. Motion in ϕ + σ: θ = 0, λ = π/4, α = π/, ρ = 0, ψ = 0. 10

11 Motion in ψ: pp wave:. ds pp = 4dx + dx + du + e ϕ = e ϕ 0, H 3 = 0, 3 (dy i ) + i=1 dw j d w j u + j=1 3 yi F 0 = 0, F = e ϕ 0 du dx +, F4 = 3e ϕ 0 dx + dy 1 dy dy 3 i=1 w j (dx + ) j=1 Symmetries U(1) ± SO(3) r U(1) u SO(3) U(1) R SU() r U(1) u SU() L Closed string in background find Hamiltonian H = 4 n= A=1 N (A) n n (α p + ) + and eigenenergies at n/(α p + ) 1 E A n B=5 N (B) n (α p + ), EB 1 + n (α p + ) n (α p + ) 11

12 Motion in σ: pp wave ds pp = 4dx + dx + 3 dx i + 8 dy k ( i=1 k=4 ) 1 3 x i + y y 5 + y 6 + y 7 + y 8 (dx + ) 8 6 i=1 1 [ ( ) ( )] x + x + y 5 cos + y 6 sin (dx + ) [ ( ) ( )] x + x + B = y 5 cos + y 6 sin dy 4 dx +, F 4 = 3e ϕ 0 dx+ dx 1 dx dx 3 [ ( ) ( ) ] + e ϕ 0 x dx + + x + cos dy 5 + sin dy 6 dy 7 dy

13 Motion in ϕ: pp wave ds = dx + dx + dρ + ρ ds (S ) + dv + dx + x dσ + dw [ 1 + dw 1 3 v ρ x + 1 ( ) ] cos x+ 1 4 w 1 sin x+ 4 w (w 1 + w ) (dx + ), e ϕ = e ϕ 0, 3 F 4 = dx + ρ dρ vol(s ) + 1 dx + xdx dz dσ, e ϕ 0 6e ϕ 0 H 3 = 1 ( ) dx + cos x+ 3 4 dw 1 sin x+ 4 dw dv Motion in ϕ + σ: pp wave is (modulo rescalings and signs) the same as for motion in σ. 13

14 . Open string quantization Open strings attached to D4-brane [ wrapping R t CP : (t, λ, θ, ϕ, σ) (x ±, x, y, z) x ± ], Y 7, Y 8, Y σ Y 6 sin σ 0 4. String action is S pp = 1 4πα πα dσ 0 p ( + X dσ [η 1 ab ( a X i b X i + a Y k b Y k ) + µ i 0 + Y Y 5 + Y 6 + Y 7 + Y ] σ 0 σ 0 ) [Y 5 + Y 6 sin ] ] σ 0 σ 0 + [Y 3 µ 5 + Y 6 sin 1 Y X 4, Y 5, Y 6 hard. X i, Y 7, Y 8 simple. Eigenmodes ω (i) n = µ + n (α p + ), ω(i ) n = and same for ϕ, σ, or ϕ + σ pp waves. µ 6 + n (α p + ), I = 7, 8. 14

15 3. Spin chains in CS-SYM field theory IR limits Interactions in IR involving scalars ϕ i : [ϕ i, ϕ j ] = 0, and. H int,1 = 4π k Tr ( [[ϕ i, ϕ i ], ϕ k ][[ϕ j, ϕ j ], ϕ k ] ) Closed string spin chain: pick out Z among ϕ i rest ϕ m, m = 1,. Z e iα Z corresponds to U(1) of pp wave in ψ. Define J = 0 for Z J Z = 1/. Then, Z Z ϕ m ϕ m A µ D µ 1/ 1/ 1/ 1/ 1 1 J 1/ 1/ J 0 1 1/ 1/ 1 1 Unique object with J = 0, Z, defines the vacuum 0, p + 1 JN J/ Tr [ZJ ] 15

16 Oscillators: trickier. Since [ϕ i, ϕ j ] = 0, one possibility is [ϕ m, ϕ m ]. Also (ϕ i ϕ j ) (like for ABJM) doesn t work. Only possibility for oscillators such that a M n 0, p + Φ M = {ϕ m, ϕ m, Z, D a } J 1 l=0 e πinl J Tr [Z l Φ M Z J l ] Then classical dimensions match n = 0 pp wave results: 4 with J = 1 and 4 with J = 1/. 16

17 Insertion of Φ M = Z Feynman diagrams give factor f Z (p) = 8 sin ( ) p 1 sin p string Hamiltonian (ϕ ) (ϕ ).. Insertion of Φ M = ϕ m f ϕ (p) = 8 sin p string Hamiltonian 10(ϕ ) 3(ϕ ). Insertion of Φ M = ϕ m f ϕ (p) = [ 16 sin p string Hamiltonian 4(ϕ ) 3 (ϕ ). ( ) 5 6 sin p ( 1 3 sin )] p Feynman diagram factor is F(x) F tree (x) = 1+f i(p) N k ln x Λ+finite = 1+f i(p) λ to be compared with ln x Λ+finite F(x) F tree (x) = (1 + tree finite) x 1 δ (λ) ln x Λ + finite x (λ) 17

18 Thus J ( J)(tree) + δ (λ) = ( J)(tree) f i (p) λ In N = 4 SYM, we had no f(λ) max. susy. In N = 6 ABJM (3/4 maximal susy) one f(λ). Now, f i (λ) for each insertion, and at small coupling, f i (λ)g(p) = f i (p)λ. Moreover, valid only at small p. At strong coupling, n (α p + ) = L string α n J λ/3n J 18

19 Z insertion:. J 1 4λ sin p ( 1 sin p ) vs. 1 f Z(λ) sin p/ 1 1 f Z(λ) sin p gives f Z(λ) 8λ to f Z (λ) λ/3. ϕ m insertion: J 1 4λ sin p ( 1 sin p ) vs. 1 f Z(λ) sin p/ 1 1 f Z(λ) sin p gives f ϕ (λ) 40λ vs. f ϕ (λ) λ/3. ϕ m insertion: ( ) 1 J 1 8λ sin p 1 3 sin p, vs. 4 1 f ϕ(λ) sin p 1 1 f ϕ(λ) sin p gives f ϕ (λ) 16λ vs. f ϕ (λ) λ /3. Open strings on D-branes: not even J at n = 0 works out. 19

20 4. Penrose limits of Abelian and Nonabelian T-duals of AdS 5 S 5 Apply the same technique to understand field theories AdS/CFT. dual to T-duals of AdS 5 S 5. Abelian T-dual on ψ of AdS 5 S 5 /Z k : coordinate acted on by Z k : ψ. ds = 4 L ds (AdS 5 ) + 4 L dω (α, β) + L dψ cos α + L cos α dω (χ, ξ), B = L ψ sin χ dχ dξ, F 4 = 8 L4 e Φ = L cos α g s α g s α cos 3 α sin α sin χ dα dβ dχ dξ Here we rescaled ψ = L α ψ, meaning ψ = α ψ L [ 0, πkα /L ]. Also define g s = L g s. α For Penrose limits, 3 isometries: ξ, β, ψ. For geodesic in ξ: at χ = π/, α = 0, r = 0, ψ = 0. ds = 4 dx + dx + d r + r dω 3 + dx + x dβ + dz + dy ( r + x + 4 z ) (dx + ) B = y dz dx +, e Φ = 1 g s F 4 = 4 x g s dx dβ dz dx +. 0

21 Geodesic in β: at α = π/, r = 0, ψ = ψ 0, χ = χ 0, ξ = ξ 0,. ds = 4 dx + dx + d r + r dω dy + y dω (χ, ξ) + α d ψ ( r + 4 y )(dx + ) y B = α ψ sin χ dχ dξ, e ϕ = y α g s F 4 = 8 y3 g s α sin χ dx+ dy dχ dξ doesn t test T-duality. Is not in Brinkmann form unclear. In ψ: at α = 0, χ = π/. But motion just in ψ is pathologic. Geodesic must move in ψ AND ξ, with ξ = dξ du = J (at dt/du = 1/4). From null condition, find ψ = 1 4 which means J 1/. ( 1 4 J ) = ψ = 1 4 J u 1

22 Pp wave is [ r ds = du dv + d r + r dω 3 + dz + dx + x dβ + dw. e Φ = g s J 1 16 α L g s, B = u dz dw, F 4 = J x du dz dx dβ g s Non-tachyonic mode 8J J 1. x + J z ] du Closed string in pp wave gives ( κ 1 + κ = 1) S = 1 [ ( ) X dτ dσ X i X i 1 ( ) + X ( ) + X 3 ( ) + X π α 16 ( ) X 5 ( ) + X 6 + (8J 1) + J ( X 7) ( ) ] κ 1 X 7 σ X 8 κ X 8 σ X 7 16 and frequencies ωn,i = n , i = 1,..., 4 ωn,i = n + 8 J 1 16 ωn,± = n + J ± 1, i = 5, 6 n + J 4

23 Nonabelian T-dual of AdS 5 S 5 /Z k on direction becoming ρ [0, πk] is. ds = 4 L ds (AdS 5 ) + 4 L dω (α, β) + α d ρ L cos α + α 3 ρ 3 B = α ρ sin χ dχ dξ, + L 4 cos 4 α e Φ F = 8 L4 g s α 3/ sin α cos3 α dα dβ, F 4 = 8 α 3/ L 4 g s ρ 3 cos 3 α α ρ + L 4 cos 4 α α L ρ cos α α ρ + L 4 cos 4 α dω (χ, ξ) = L cos α g s α 3 ( α ρ + L 4 cos 4 α ) sin α sin χ dα dβ dχ dξ For Penrose limits, isometries: on β and ξ. For geodesic on β, at α = π/, ρ = ρ 0, χ = χ 0, ξ = ξ 0, is ds = 4 dx + dx + d r + r dω 3 + dy ( x 4 + y ) dx + B = g s F = +4 α y d ρ + 4 α ρ y 16 α ρ + y 4dΩ (χ, ξ) 16 α 3 ρ 3 16 α ρ + y sin χ dχ dξ, 4 e Φ = gs y ( 16 α ρ + y 4) 64α 3 y 3 dy, g α 3/dx+ s F 4 = 8 α 3/ y 3 ρ 3 16 α ρ + y sin χ 4 dx+ dy dχ dξ 3

24 For geodesic in ξ, find no solution ned also motion in ρ.. Define ρ = L α ρ and redefine also coupling as g s = g α 3/ s L 3. Then pp wave near geodesic at χ = π/, α = 0, r = 0 is L ds = 4 ds (AdS 5 ) + 4 dω dρ (α, β) + cos α + ρ cos α ρ + cos 4 α dω (χ, ξ) L ρ 3 B = sin χ dχ dξ, e Φ = cos α( ρ + cos 4 α ) ρ + cos 4 α F = 8 L g s sin α cos 3 α dα dβ, F 4 = 8 L3 g s g s ρ 3 cos 3 α ρ + cos 4 α sin α sin χ dα dβ dχ dξ Particle on the null geodesic ṫ = 1/4 and p ξ Then, find which means again J 1/. ρ = 1 4 ρ + 1 ρ J L = ρ ρ +1 ξ = J. 4

25 Moreover, ρ L α L /α.. Pp wave becomes J 1 4J needs to fit in (0, πk], meaning k ds = du [ dv + d r + r dω 3 + dx + x dβ + dz + dw ] r 16 + x 16 (8J 1) + (ρ + 1) J z F ρ 4 z z F w w du H = db = 1 e Φ = ρ + 1 g s ρ + 3 du dz dw ρ + 1 F 4 = J x ρ + 1 g s du dx dz dβ, F = 0 which means that non-tachyonic 8J J 1. 5

26 Closed string on pp wave S = 1 [ (( ) X dτ dσ X i X i 1 ( ) + X ( ) + X 3 ( ) + X π α 16 ( ) X 5 ( ) + X 6 + (8J 1) + (ρ + 1) J ( X 7) ( ) Fz X 7 ( ) ] Fw X 8 16 ρ 4 ρ + 3 ( ) ] κ ρ 1 X 7 σ X 8 κ X 8 σ X Frequencies are ωn,i = n , i = 1,..., 4, ω n,i = n + 8 J 1 16 n 1: Abelian T-dual n 1 gives ω n,i = n , i = 1,..., 4 ωn,i = n + 1 a, i = 5, 6 16 ωn,± = n + 1 ( a + ) ± 1 ( ) a n + ( a 1 ) 16 8, i = 5, 6 Abelian and Nonabelian pp waves both preserve just the minimum (for pp waves) 1/ susy. 6

27 5. Operators in field theory, RG flow and deconstruction AdS/CFT map Conjecture for field theories: quivers: Abelian: SU(N) k N = susy. Each node: N = vector multiplet of SU(N), each adjoining nodes: N = bifundamental hypers. Nonabelian: infinitely long quiver, SU(N) SU(N)... SU(kN)... Terminates only for a completion of the background. Quiver dual AdS 5 S 5 /Z k and its Penrose limit was considered before [Alishahiha+Sheikh-Jabbari 00 & Mukhi+Rangamani+Verlinde 00]. 7

28 Field theory limit Limit is different than theirs. ψ or ρ [0, πk] gives k L /α = 4πg B s N, meaning g Y M N k = 4πgB s N k k unlike the case in previous papers. T-duality acts as gs A = gb s and since gs A = L g s, we find g s k/n 1. α α L, Nonabelian case: g s = L 3 /α 3/ g s, so we find g s 1/N 1. Then strings on pp waves are classical need only compute eigenvalues. 8

29 Field theory has superpotential W = and kinetic terms L kin = k i=1 with symmetries: k i=1 d θ Tr i+1 [V i X i W i ] d θ d θ Tr i [ V i e V V i + W i e +V W i + X i ev X i ] -SU() R acting on V i and W i -U(1) R acting on X i and d θ as X i e iα X i, d θ e iα d θ. -non-r U(1) acting on V i, W i as V i e iα V i, W i e iα W i. Same in dual: SU() χ,ξ U(1) β U(1) ψ. 9

30 Charge assignment. X V W X V W J kj H Large charge (BMN) operators for U(1) ψ = U(1) extra, but also U(1) ξ SU() R, with ( ) J J them = ( J1 J ) us = ξ ψ = J 1 4J We can t vary J 1 /J, and J 1 /J = 1, for J = 1/, is only possibility. Vacuum, T-dual to one of Mukhi et al. quiver us winds around the p = 1, m = 0 them = m = 0, p = 1 us = O k = 1 N Tr [V 1 V...V k ] 30

31 Oscillators of energy H = 1: D a, a = 0, 1,, 3, W i, W i, X i, X i, O Dp = a D,0 m = 0, p = 1 us = 1 Nk 1 N O Xp = a X,0 m = 0, p = 1 us = 1 Nk 1 N O X p = a X,0 m = 0, p = 1 us = 1 Nk 1 N O W,0 = a W,0 m = 0, p = 1 us = 1 1 N k N O W,0 = a W,0 m = 0, p = 1 us = 1 k 1 N a X,n m, p = 1 us = 1 Nk 1 N k Tr [V 1...V i 1 (D a V i )...V k ] i=1 k Tr [V 1...V i 1 X i V i...v k ] i=1 k Tr [V 1...V i 1 X i V i...v k ] i=1 k Tr [V 1...V i 1 V i W i V i...v k ] i=1 k Tr [V 1...V i 1 W i V i+1...v k ] i=1 k l=1 Tr [V 1...V l 1 X l V l...v k ]e πiln k 31

32 Note that now, momentum m = i n i. Eigenenergies: for pp waves, should be ω 0,a = 1 4, a = 1,, 3, 4, ω 0,i = 8J 1 4, i = 5, 6, ω 0,+ = J, ω 0, = 0 D a matches: H = 1 ω 0,a = 1/4 (rescale). But X, X, W, W give H = 1 also. Origin: extra interactions in g YM N k k limit. From interaction V g Y M Tr i W i V i = Tr i ( W i W i V i V i ) mixing O W,0 and O W,0 and V g Y M Tr k[ X i V i V i X i ] mixing O X,0 and O X,0. 3

33 Nonabelian case: RG flow in ρ. Match it to Abelian case if ρ 0 g s = 1 g s. Eigenenergies ω = ω k (u) flow in lightcone time u, between u = 0 and u =. flow generic in Gaiotto-Maldacena backgrounds. Conformal symmetry broken in dual by winding modes of strings? Or dual to conconformal theory in higher dimensions via deconstruction: -Wilson loops show deviation from conformal behaviour. -pp wave limit of Janus solution (dual to defect CFT) has similarity with Nonabelian case. 33

34 . Conclusions We can use Penrose limits to probe difficult holographic dualities For GJV, Penrose limit with closed strings matches BMN sectors in field theory, but only to leading order. Open strings don t match. We find several functions f i (λ) between weak and strong coupling. For abelian and nonabelian T-duals of AdS 5 S 5 /Z k, BMN-type operators are clear, but naive calculations don t match large contributions from Feynman diagrams that can t be neglected, unlike N = 4 SYM or ABJM cases. Take pp wave calculations as predictions that probe unknown field theories. 34