Strings in AdS 4 CP 3, and three paths to h(λ)

Transcription

1 Strings in AdS 4 CP 3, and three paths to h(λ) Michael Abbott, TIFR Work (and work in progress) with Ines Aniceto & Diego Bombardelli (Lisbon & Porto) and with Per Sundin (Cape Town) Puri, 7 January 2011

2 size. page, should I get there... J 2 = Q 2 Exact dispersion relation E(p) for: h(λ)2 sin 2 p 2 Magnons of the spin chain for ABJM, h(λ) = λ + bλ (λ 1) String excitations in AdS 4 CP 3, (λ 1) λ h(λ) = 2 + c + d +... c = log 2 λ 2π or possibly c = 0? Will discuss three AdS/CFT tests which tell us about c.

3 t. be about the same size. a fancier contents page, should I get there... Programme 1. Integrability and the AdS/CFT spectral problem 2. The new example of ABJM 3. One-loop energy corrections for spinning strings and for giant magnons, using algebraic curves [June 2010] 5. Extension to the case J < [MA/IA/DB, i.p.] 6. The near-flat-space limit and its uses [MA/PS, i.p.] 7. And two loops?

4 c 2010 Niklas Beisert c 2010 Niklas Beisert c 2010 Niklas Beisert 1 The Spectral Problem for N = SYM hard interacting quantum strings loop quantum gauge theory expansion genus expansion gauge loops Figure from review [Beisert et. al. 2010] free classical strings worldsheet loops free quantum strings planar limit 0 interacting classical strings gstr handles perturbative gauge theory 1/Nc handles 0 classical gauge theory Figure 1: Planar and non-planar Feynman graph (top), free and interacting string worldsheet (bottom), Feynman graph corresponding to a patch of worldsheet (right). λ c 2010 Niklas Beisert Figure 2: Map of the parameter space of N = 4 SYM or strings on AdS5 S 5. In the strictly planar limit of AdS/CFT, now know the spectrum of tofor all λ. growth, such that the ces the we complexity of graphs from factorial exponential us of convergence of the perturbative series grows to a finite size. Moreover, the ace on which the Feynman graphs are drawn introduces a two-dimensional structure ello this is a test. his text should be about the same size. age one will be a fancier contents page, should I get there... c 2010 Niklas Beisert c 2010 Niklas Beisert

5 Hello this is a test. Easiest case is J =, where we have: Very long operators O = Tr(ZZZZZZZZ... J ) + impurities with J a spin-chain Hamiltonian: J = λh i,i+1 + λ 2 H i,i+1,i+2 + λ 3 H i,i+1,i+2,i i Strings with infinite SO(6) angular momentum, thus decompactified worldsheet X µ (σ R, τ = t) and semiclassical corrections, O(1/ λ) Asymptotic Bethe equations give the spectrum as solution of connecting large and small λ. B(, λ) = 0

6 c 2010 Niklas Beisert J = is easy because excitations can be widely separated: Dispersion relation E(p) = 1 + π λ sin2 p 2 for isolated particle, Energies are additive: E {i} = i E(p i ) Two-particle S-matrix S(p i, p i ) Factorised scattering: S {i}{j} = i j S(p i, p j ) Ansätze and the R-Matrix Formalism rcise Bethe s and toansatz gather forexperience, N-particle state Chapter is a superposition III.1 solves {p i } uantum constrained mechanical by ψ(0, systems x 2, x 3... x N ) the = ψ(j, Heisenberg x 2, x 3... xspin N ) ne along on a circle the of lines J of size. Bethe s original work, using onscatteringpicture,butalsoinseveralvariants z. This introduces us with ubiquitous concepts of suchsimilar as R-matrices, equations fortransfer J < : Thermodynamic matrices and Bethe famous Ansatz / Yang Baxter Y-system Hello this is [Gromov, a test. Kazakov, Vieira] [Arutyunov, Frolov, Suzuki] 2009 R 12 R 13 R 23 = R 23 R 13 R 12.

7 Giant magnons are classical string solutions dual to spin chain magnons: [Hofman & Maldacena, 2006] X 1 + ix 2 = e it [cos p 2 + i sin p 2 tanh(u)] X 3 = sin p 2 sech(u) where u = (x t cos p 2 )/ sin p 2. Charges E(p) = J = λ π sin p 2 Turning on 2nd charge Q λ in the X 3 -X 4 plane gives: E(p, Q) = J = Q 2 + λ π 2 sin2 p 2 Dyonic giant magnon in R S 3 [Dorey] [Chen, Dorey, Okamura] 2006 dual to a bound state of Q spin chain magnons. Hello this is a test.

8 2 ABJM and AdS 4 CP 3 ABJM [2008] is 3+1-dim. N = 6 superconformal Chern-Simons theory. Dual to M2-branes in AdS 4 S 7 /Z k, (also [BL & G, ] etc.) KK reduction as k leads to IIA strings on CP 3. Planar limit has t Hooft coupling N k = λ = R 4 32π 2 α 2 New example of integrable AdS/CFT. Almost everything can be copied across, with slight modifications. Hello this is a test.

9 er of integrability will show up only in Part III. c 2010 Niklas Beisert Scalar fields are in (N, N) of U(N) (rather than adjoint), and the spin chain vacuum is = 4 SYM O = Tr (Y 1 Y 4 )J ge theory, its local operators, and it outtrum of their planar one-loop anomalous Can excite even or odd chain, decoupled at leading order: ow to map one-to-one local operators to J pin chain. The operator 2 = (H i,i+2 + H which i+1,i+3 ) + four loops i measures ous dimensions corresponds to the spin cture. Symmetries Importantly, fix the exact this dispersion Hamiltonian relation: is of the integrable can be viewed as a generalisation of the Heisenberg spin J Q spectrum is solved efficiently 2 = by h(λ)2 sin the 2 p corresponding 2 Bethe lanar butanomalous leave the function dimensions h(λ) unknown. δd is encoded in the solutions quations for the variables u k C (k =1,...,M) (True in AdS 5 S 5 too, but there h(λ) = λ thanks to experiments and an argument from S-duality [Berenstein & Trancanelli, 2009]) Hello this is a test. j + i, 1= M u j + i2, δd = λ M 1.

10 At λ 1, leading term is 2 loops. [Minahan & Zarembo, 2008] Next term comes from 4 loops: h(λ) 2 = λ 2 4 ζ(2) λ [Leoni, Mauri, Minahan, Ohlsson Sax, Santambrogio, Sieg, Tartaglino-Mazzucchelli 2010] (and earlier papers by bold names) Their all-λ guess is: h(λ) 2 = 1 2π 2 ± ±(1 ± 2πiλ) log(1 ± 2πiλ) S r3 = V r31a = V r31b = V r32a = V r32b = 2(4π)4 k 4 (4π)4 2k 4 M 3 (4π)4 k 4 M 3 (4π)4 k 4 M 3 (4π)4 2k 4 M M consistent with c = 0 when λ 1: λ h(λ) = 2 + c +... (leading term from PP-wave ). Hello this is a test. V r33a = V r33b = V r34 = (4π)4 (M k 4 = (λˆλ) 2 ( 16 (4π)4 (MN k 4 (4π)4 (MN k 4

11 3 String Solitons at One Loop First example: folded spinning strings in AdS 3 subspace. Classical solution ϕ = τ, ρ = σ, with charges S = 2λ log S when S. One-loop disagreement : δ = 3 log 2 2π log S = 5 log 2 2π log S from sl(2) Bethe equations explicit string calculation [Gromov & Vieira] vs. [McLoughlin & Roiban] + [Alday, Arutyunov, Bykov] + [Krishnan], 2008 Hello this is a test.

12 Hello this is a test. Two resolutions: Modify the summation prescription, keeping c = 0 like S 5 case. [Gromov & Mikhaylov, 2008] Turn on c = log 2 2π, and keep naïve mode sum. [McLoughlin, Roiban, Tseytlin, 2008] Summary: S log S 2 = 2h(λ) 3log 2π + o( 1 h ) = 5 log 2 2π 2λ + using the 3 log 2 2π from sl(2) Bethe equations old sum new sum with c old = log 2 2π c new = 0

13 Hello this is a test. String calculations are ħ δe = n 2 ω n Prototype is sine-gordon: compare one-soliton to no-solitons. This tends to be very infinite... but ( 1) F will save us. Modes are of course perturbations like this X µ classical + t e iωn δx n µ becoming plane waves e ikx iωt±iδ/2 as x ±. In AdS 5 S 5, all of these modes have the same mass: ω 2 = k But in AdS 4 CP 3, instead ω 2 = k heavy ω 2 = k 2 + 1/2 light ( subspaces radius R and R/2)

14 Hello this is a test. The two choices of cutoff are: Heavy modes... δe old = lim δe new = lim N N n= N (ω light n c = log 2 2π + ω heavy n ) N ( ω light 2N n + N n= N n= 2N c = 0 ω heavy n ) are simply 4 of the 8 directions in space! (and fermions) do not appear in the Bethe ansatz (they are superpositions, or stacks ) are perhaps composite at one loop? [Zarembo, 2009]

15 Hello this is a test. 4 Corrections for Giant Magnons Compare one-loop corrections to exact dispersion relation: J Q 2 = h(λ)2 sin 2 p 2 Q 2 = 4 + 2λ p sin2 2 + c 8λ sin 2 p 2 Q λ sin2 p 2 + O ( 1 λ ) An early result [Shenderovich, 2008] gave c = 0, confusingly before [G&V] s new sum.

16 Hello this is a test. Giant magnon in CP 1 is identical to S 2 case. [Gaiotto, Giombi, Yin, 2008] But the dyonic version is new, [MCA, Aniceto, Ohlsson Sax, 2009] explores CP 2 by turning on ξ π/2, φ 1 = ωt +...: z = sin ξ cos(ϑ 2 /2) e iφ 2/2 cos ξ e iφ 1/2 0 sin ξ sin(ϑ 2 /2) e iφ 2/2 Other giant magnons in RP 2 (and dyonic RP 3 ) are superpositions of two elementary magnons. [Hollowod & Miramontes, 2009] In principle we could compute δx µ (x, t) from worldsheet solutions by hand (like S 5 case [Papathanasiou & Spradlin, 2007]) but it is less work to use power tools...

17 Hello this is a test. Use some integrable systems technology called the algebraic curve: Classical string solutions Riemann surfaces one-to-one Construction from Lax connection is like this: M(x) = P exp dσ J σ (x) eigm = {e ip 1(x), e ip 2(x), e ip 3(x),...} x = ±1 AdS } q 1 = q 10 q 2 = q 9 } CP q 3 = q 8 q 4 = q 7 x < 1 q 5 = q 6

18 Hello this is a test. Well-developed scheme for semiclassical perturbations: Add cut connecting sheets (i, j) at point y solving q i (y) q j (y) = 2πn with filling fraction S i j = g iπ Ci j dx (1 1 x 2 ) q i (x) = 1 [Beisert, Kazakov, Sakai, Zarembo, 2005] After constructing mode δq i (x), [Gromov, Vieira, 2007] you can read off its perturbation of the energy: The you add all of these up... δ = Ω(y) = ω Light polarisations (i, j) connect to sheet 5(=6) Heavy ones do not.

19 Hello this is a test. For giant magnons, this gives simple off-shell frequencies: Ω(y) = n plane: 1 y 2 1 (1 y X+ + X 1 + X + X ) 1 (i, j) light 2 heavy Not easy to find positions x i j n, hence on-shell frequencies ω i j n = Ω(x i j n ). Can still add them up, with some complex analysis: [Schäfer-Nameki 2006] δe = 1 2 n = 1 4i R Ω i j (x i j n ) R(N) dn ( 1) F i j cot(πn)ω i j (xn i j ) i j n = N

20 Hello this is a test. x plane: x =1 X + U() x =1+ X Using q i (x i j n ) q j (x i j n ) = 2πn, write in x: δe = 1 4i U q dx ( 1) F i (x) q j (x) i j i j 2π cot ( q i(x) q j (x) ) Ω i j (x) 2 x =1+ For now (J = ) can ignore other contour components. But we can t ignore details of the cutoff n < N, which is x > 1 + є...

21 New sum is simplest: x heavy 2N δe new = lim = 0 є 0 i j Old sum is more work, x heavy N x light, thus cut off at same x = 1+є for both: N U(є) dx ( 1)F i j 4i 2x light N δe old = lim є 0 dx + i j light U(є) = log 2 2π 2 sin p 2 Dyonic case: δe old = log 2 2π 8λ sin 2 p 2 so i j heavy Q λ sin2 p 2 All consistent with previous AdS results... Hello this is a test. q i q j 2π cot( q i q j 2 )Ω i j (x) dx U(2є) ( 1) F i j 4i q i q j 2π cot( q i q j 2 )Ω i j (x. [MCA, Aniceto, Bombardelli, 2010]

22 c 2010 Niklas Beisert (Review [Klose, 2010] latest [APGHO,2011] ) Figure 1: Planar and non-p Arguments about cutoff prescriptions: string worldsheet (bottom), F sheet (right). New: Old: N N N (light + heavy) perturbative gauge theory c= 1/Nc log π Easiest in worldsheet calculations. c= classical gauge theory heavy + light Because heavy mode is composite? heavy ω N ω N + ω N. handles inte Equivalent to hard energy cutoff: ω N N Λ same for both types. Easier to match all=λ guess? loop q expansion (Freq. w.r.t. AdS time.) (light + heavy) є In the spectral plane,... hence easiest in algebraic curve є heavy + є light calculations. plan 0 gauge loops 0 Hello this is a test.

23 Hello this is a test. 5 Giant magnons at J < [in progress] Corrections are organised like this: E = m,n=0,1,2... a m,n (e / 2λ ) m (e 2 /E ) n a 0,0 = E class. + δe is the case J = from before. Corrections a 0,1 are F-terms, zero classically. Corrections a 1,0 are µ-terms, classical + one-loop, so we can make a comparison: a 0,1 e 2 /E = h(λ) a class. (p, Q) e 2 /E0(h,p,Q) + a subl. e 2 /E 0 + O( 1 h ) λ = 2 a class.(p, Q) e 2 /E 0( λ/2,p,q) + δe µ + O( 1 ) λ

24 Hello this is a test. Gauge side: Lüsher terms, which are wrapped Feynman diagrams:. F-term: µ-term: F-terms: computed by [Bombardelli & Fioravanti, 2008], classically zero. µ-terms: Unsolved (order-of-limits?) issues for single elementary magnon [Lukowski & Ohlsson Sax, 2008] [Bombardelli & Fioravanti, 2008] For a bound state (dyonic magnon): classical µ-term OK, (S 5 case: [Hatsuda & Suzuki, 2008]) one-loop term not certain...

25 string tension α and the AdS5 /S 5 radius R) and the string coupling gstr. The AdS/CFT classical planar limit strings 8 0 gauge loops λ worldsheet loops Classical string solutions: Map to kink train in sine-gordon [Okamura & Suzuki, 2006] or construct X µ (σ, τ) directly. Algebraic curve: this is the more natural case! Giant magnon is a two-cut solution: x plane Y +X + re-organise cuts to: pole log Y X length δ e /E Semiclassical calculation is an expansion in δ of previous integral U(є) dx..., plus some discrete terms. (S case: [Gromov, Scha fer-nameki, Vieira, 2008]) Hello this is a test.

26 Hello this is a test. Old vs. New: Earlier, I claimed old physical : all modes to same energy. Here, old leads to divergences, but physical like this works: δe n plane: phys = lim ( 1) F 1 i j Λ i j 2 ωi n j R(N) ω i j n <Λ = lim 1 Λ 4i ( 1) F i j dx i j U(є i j,є + i j ) Needs cutoffs for every polarisation: Ω i j ( 1 є i j ) = Ω i j(1+є n = + N i j ) = Λ x plane: x =1 X + U(, + ) x = X + ± ε x = 1 x =1+ + X

27 Cut structure is different: In S 5 case, cuts always connect sheets p 2(x) and p 3(x) = p 2(x), allowing ansatzae of the form p (x) = 1 α f (1) (K + (x X+ )(x Y + )(x X )(x Y ) x 1 + But for CP 3 they connect q 4 (x) and q 6 (x)... The RP 3 magnon is like S 5 in this regard, and for this we can compute both new c = 0 and physical c = log 2 2π. Both match match Lüscher corrections from [Bombardelli & Fioravanti, 2008].

28 Hello this is a test. 6 The Near-Flat-Space Limit [in progress] Intermediate limit: [Maldacena & Swanson, 2006] BMN: p ws 1/ λ Near-flat-space: p ws 1/λ 1/4 Magnons: p ws 1 Write the dispersion relation as follows: E 2 = h(λ)2 sin 2 p ws 2 = p2 1 + [ c p2 p4 2λ 96λ ] + O ( 1 ) λ

29 Hello this is a test. p 2 = E 2 p 2 1 = [δm2 ] Thus mass corrections to the propagator will teach us about c: G 2 (p) = i p i p A i p = i p δm2 Near-BMN Lagrangian computed by [Sundin, 2009], from coset model. Taking a large boost of this p λ 1/4, p + λ 1/4 0 and shifting fields to get canonical L 2 leads us to the following theory:

30 Hello this is a test. However, for the off diagonal <s L 2 = ± s > we get opposite sign compared to the two (2.10 spinor propagator. This is 1 2 +y y + 1 rather odd. What 2 +z i z i + 1 is more, when 4 +ω α ω α + 1 summing the diagrams 4 ω α + ω α 1 ( ) y 2 + zi 2 1 together one should keep this minus sign. I.e., <s 2 8 ω α ω α + s >= i 2 M, otherwise the divergent However, parts for + i does the off not diagonal cancel. <s ± s > we get ( ψ+a ψ +a + 2 ψ a + ψ a) i opposite sign compared to the two spinor ( The cubic propagator. Lagrangian This is is given rather by (s ) a 2 α + (s ) α a +(s + ) a α (s + ) a) 7 odd. What is more, when summing the diagrams α together one + 1 should keep this minus sign. I.e., <s + s >= i 2 M, otherwise the divergent L ψ a ψ+ 2( a + ψ +a ψ ) a + i(s+ ) a α(s ) α 3 = parts i does not cancel. 4 (s ( ) aα + ψa ω α + ψ a ω α) + 3i a. The cubic Lagrangian is given by 7 16 (s ( ) aα ψa ω α ψ ω a α) (2.14) The cubic As Lagrangian is given by 7 ( +(s it turns ) aα ψa out, both the cubic and quartic Lagrangian exhibits left moving fermion For Lthe 3 = quartic i ω α ψ a ω α) + i 4 (s ( 2 (s ( +) aα ψa ω α ) aα piece, + ψa these ω α can + readily ψ ω α) be+ shifted 3i ψ ω a α) L3 = i ( 4 (s )aα + ψa ω α +ψ ω a α) + 3i ( 16 (s ( ) away aα ψa while ω α ψthe ω a α) cubic (2.14) 1 16 (s )aα ψa ω α ψ ω a α) (2.14) ones are mo complicated. 4 ( i ( (s ) aα ψa In principle these can also be shifted away, but the shifts will, howeve i + ω α ψ+ ω a α) + 1 ψ a ψ+ induce 4 ( 2( b + ψ +a ψ ) b Z a +(s additional ) aα ψa ω α ψ a quartic terms ω α) + i b i which 2 (s ( 8 yω 4 +(s )aα ψa α ω α ω α ψ ω a α) + i ( 2 (s+)aα ψa ω α ψ ω a α) 1 ( 4 (s )aα +) have aα ψa + i ψa to be ω α addressed ψ ω a α) + ω 2( α + ψ+ ω a α) + 1 ψ aψ ψ a + ψ b + ψ a ψ b + 2( ) b + ψ+a ψ ) Z a b Z a b i 8 yωα ω α + i ψ a +ψ b with a second shift of rather 1 2( b + ψ aψ ) b Z involved form. Thus, it is probably easiest to calculate loops using the origin cubic 4 ( (s ) aα ψa + ω α + ψ+ ω a α) + 1 ψ a ψ+ Lagrangian. For this reason 2( b + ψ +a ψ ) b Z a b i it is convenient to introduce 8 yω where Z α ω α b a = b a. Compared i z i(σ i ) a b. tocompared the cubic BMNto Lagrangian the cubic in ([4]) BMN the above Lagrangian is in ([4]) the above is very simple. a two spinor notatio for + 2( i The quartic the ψ a fermions. + ψ b Lagrangian + Introducing ψ a ψ ) b have Zb a pure boson, pure fermi and mixed terms, L 4 = L BB +L FF +L BF. The pure boson term is very simple, and in direct analogue to the AdS ( ) ( ) ( ) ( ψ γ 0 = σ 1 =, γ 1 = iσ 2 =, Ψ a a where Z a = ψ a, χ a (s+ ) b = 5 case, LBB = 1 z 2 8( i z i(σ i ) a i y 2 1 b. Compared to the cubic BMN Lagrangian in ([4]) the above is α = very simple. L The quartic Lagrangian have pure boson, pure fermi and mixed terms, i(s BB = 1 4 ωα ωα)( ( y) 2 + ωα ω α +( zi) 2). (2.15) z 2 ) L 4 = L BB +L FF +L 8( i y 2 1 BF. The4pure ω α ω α)( ( y) 2 + ω α ω α terms in the +( z i ) 2) Lagrangian, but for the case above the additional t. (2.15) against the mass term (or vice versa) so the entire contribution boson where the spinors conjugates as Ψ term is = Ψ very γ 0 simple, and χ α Forand the pure in a =(χ a fermi direct α) piece γ 0 ofanalogue the Lagrangian we have The. Using this we ca to pure the AdS fermi 5 case, and mixed piece is, however, more involved. For the mixing piece we ) have write the fermionic ( ω piece of the quadratic Lagrangian as L BB = 1 z 2 L8( α ω γ ω α ω γ 2 F i y 2 = i 1 i LFF = ( ) 1 ) 2 i ( )( ψ ψ 8( 8 ψ ψ ψ ψ+ ψ+ ψ ) 2 χα aγ 4 ω 32 ψ ψ ψ ψ ω ω α ω α)( i ) i ( ψ ψ ( χ a α i Ψ y) 2 ) 4( ψ ψ+ + a ω α ω γ Ψ a + 1 α +( 2 χα aχ a z α + 1 i ) 2 Ψ 2) ψ ψ ψ+ ψ 2 ψ ψ ψ 1 ( (. (2.15) 8 L a BF = + ψ ψ ) ψ ) 1 ( ψ Ψ 4 a. ψ ψ + ψ ψ + (2.16) (2.11 ψ The i ( pure fermi and mixed piece is, however, more involved. For the mixing piece we 8 y2 (s ) aα (s ) aα i ) The reason we introduce 8 (s ) aα (s ) a γ ( ω α ω γ ω α ω γ i ( ψ ψ the two spinor formalism 32 is because the ψ ) 16 y(s )aα ψ a ω α + ψ ω a α) i 8 ψ aψ Z b c a Zb c + i 8 (s )aα (s )aα zi ) 1 { + ψ ψ 2( ψ ψ + ψ +ψ ψ ψ + (s ) α c 16 (s ) α c + ) ψ ψ a (s )aα c (s ) ψ α ψ c +ψ a ω (s )aα c + +(s ) ω α ψ c cubic Lagrangia have + i ( i + ψ contains ψ ω ω both + ψ left ψand ω ω right ψ 8 (s )aα(s )α b Zd a Z db + (s ) α ψ moving ψω ω fields, ψ soψω ω we need 1 c ψ a +(s )aα c +(s ) α c ) ψ +ψ a (s )aα c +(s ) α c + ψ ) 2(s ) i (sthe ) α ψ c ψ a +(s )aα aα propagators c (s ) aα 2 +(s ) ω ω that α c ψ ψ a (s )aα c + +(s ) mix} them Quadratic: However, for the off diagonal <s± s > we get opposite sign compared to the two spinor propagator. This is rather odd. What is more, when summing the diagrams together one should keep this minus sign. I.e., <s+ s >= i 2 M, otherwise the divergent parts does not cancel. Cubic: where Zb a = i zi(σi)a b very simple. The quartic Lagrangian have pure boson, pure fermi and mixed terms, L4 = LBB+LFF+LBF. The pure boson term is very simple, and in direct analogue to the AdS5 case, The pure fermi and mixed piece is, however, more involved. For the mixing piece we have Quartic: LBF = (2.16) i 8 y2 (s )aα (s ) aα i 8 (s )aα(s )a γ + i 8( ψ ψ ω ω + ψ ψ ω ω ψ ψω ω ψ ψω ω 1 2 (s )aα (s )aα ω ω y(s )aα (s )α c Z ca + i 4 (s )aα ( ψ cω α Z ca ψ b ω α Z a b + i 8 (s )aα ( ψ cω α Z ca ψ b ω α Z a b y2 ψ ψ. 8 + i { (s ) α ψ a ψ c (s ) +(s ) α ψ a ψ c (s ) + 1 (s )

31 x ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu ugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui o cia deserunt mollit anim id est laborum. β Diagrams for correction to light boson w α w β = δ α For talk? AB = AT = ω α (p) ω α (k) ω β (k) ω α (p) + y(q) + ω α (p) y(k) or z i (k) ω α (p) p i are: / ψ a (k) s αa (q) + ψ b (k) ω α (p) Bubble diagrams always contain both heavy and light, so there is no way for the cutoff to discriminate? It is easiest to use dimensional regularisation... Hello this is a test.

32 AB = ω α (k) ψ a (k) + ω (p) ω (p) For the tadpoles things are perfect: y(q) s (q) α ω β (k) Nonsense α a α y(k) or z i (k) ψ b (k) A T =sit amet, consectetur +adipisicing elit, sed do + eiusmod tempor incididunt ut labore et Lorem ipsum dolor dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ω α (p) ω α (p) ω α (p) ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu λ fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui o cia deserunt mollit anim id est laborum. 5 δm = log π For talk? But for the bubbles... AB = AT = ω α (p) ω α (k) ω β (k) + y(q) + ω α (p) y(k) or z i (k) + ψ a (k) s αa (q) ψ (k) b δm = λ π z (p)i z j, ψ a ψωb(p) Also check other,... ω (p) modes ω and the S-matrix... [Klose & Zarembo, 2007] S case: [Klose, Minahan, Zarembo, McLoughlin, 2007] α α α Hello this is a test.

33 7 Two-Loops? We ve discussed essentially two kinds of one-loop calculation. Both kinds done in AdS 5 S 5 to two loop accuracy: Soliton energy corrections: Three papers and three years? [Roiban & Tseytlin, 2007] [Giombi, Ricci, Roiban, Tseytlin, Vergu, 2010] Near-flat-space: One sunset diagram, half a page! [Klose, Minahan, Zarembo, McLoughlin, 2007] There is also an all-loop argument that h(λ) = λ, using S-duality, which fails for AdS 4 CP 3. [Berenstein & Trancanelli, 2009] One further complication: relation N/k = λ = R 4 /32π 2 α 2 gets modified, starting at two loops λ 1. [Bergman Hirano, 2009] Hello this is a test.

34 Hello this is a test. The End. With thanks to Olof, Nikolay, Shiraz, Valentina, and collaborators Diego, Ines, Per.