Mass Spectrum of D=11 Supergravity on AdS2 x S2 x T7

a rXiv:h ep-th/99615v321D ec1999hep-th/9906105KIAS-P99038Mass Spectrum of D =11Supergravity on AdS 2×S 2×T 7Julian Lee and Sangmin Lee 1School of Physics Korea Institute for Advanced Study Seoul,130-012,Korea ABSTRACT We compute the Kaluza-Klein mass spectrum of the D =11supergravity compactified on AdS 2×S 2×T 7and arrange it into representations of the SU (1,1|2)superconformal algebra.This ge-ometry arises in M theory as the near horizon limit of a D =4extremal black-hole constructedby wrapping four groups of M-branes along the T 7.Via AdS/CFT correspondence,our result gives a prediction for the spectrum of the chiral primary operators in the dual conformal quantum mechanics yet to be formulated.1IntroductionAmong all known examples of the AdS/CFT correspondence[1,2,3,4],the least understood is the AdS2/CFT1case.The D=1conformalfield theory(CFT),or conformal quantum mechanics (CQM),has not been formulated and therefore no quantitative comparison between the two sides of the duality has been made.See[5,6]for proposals on the CQM and[7,8,9,10]for progress made in the bulk theory.One of the most elementary check of the correspondence is to compare the spectrum of the two theories.In particular,the Kaluza-Klein(KK)mass spectrum of the supergravity(SUGRA) on AdS is identified with the spectrum of chiral primary operators in the dual CFT.One may hope that the KK spectrum of a SUGRA on AdS2may give a clue to formulate the dual CQM.The goal of this paper is to compute the KK spectrum in the cases where the AdS2is part of a string/M theory vacuum.We specialize in the example of D=11SUGRA compactified on AdS2×S2×T7.1We consider only the zero modes in T7.From the string theory point of view,this theory is a valid approximation when R>>r,˜r,where r,˜r are the radius and the dual radius of T7respectively,and R is the radius of the sphere.In what follows,we will put R to1for simplicity.To obtain this geometry from M theory,one begin with compactifying M theory on T7 with the following brane configuration[11].2Brane012345678910M2x x xM2x x xM5x x x x x xM5x x x x x xWith suitable choice of the orientation of the branes,this configuration breaks N=8super-symmetry(SUSY)of the D=4theory to N=1.When the number of branes in each group is all equal,the background metric describes a direct product of an extremal D=4Reissner-Nordstr¨o m black-hole and a T7.See section3.of[11]for more details.The AdS2×S2spacetime arising as the near horizon geometry of this black-hole is known as the Bertotti-Robinson metric[12].Note that the brane configuration at hand approaches the Bertotti-Robinson metric in the near horizon limit even when the four charges are not equal.The number of SUSY is doubled in the near-horizon limit as usual,so we have D=4,N=2 SUSY.The super-isometry group of the theory is SU(1,1|2).The KK spectrum form representa-tions of the SU(1,1|2)superalgebra.The methods of the computation used in this paper are well known from higher dimensional ex-amples.There are two approaches to the problem;one is direct SUGRA calculation[14]-[18],andthe other uses representation theory of superconformal algebra together with duality symmetry ofSUGRA[19]-[24].We will adopt thefirst approach and explicitly calculate the spectrum,startingfrom the D=11SUGRA lagrangian.Although we will be mainly interested in the modes whichhave bulk degrees of freedom.However,as was noted in ref.[29],we cannot ignore the boundarymodes completely because one of them forms a multiplet with bulk modes.We will make furthercomments on this point later.This paper is organized as follows.In section2,we review the SU(1,1|2)superalgebra andits representation theory following[19,20].In section3,as a warm-up exercise we compute thespectrum of a toy model,namely the minimal D=4,N=2SUGRA.This model illustratesmany important aspects of the compactification on AdS2×S2in a simple setting.In section4,we present a summary of our main result.In section5and6,we sketch the computation of bosonicand fermionic mass spectrum of the“realistic”model obtained from the D=11supergravity.We focus on the reduction from D=11to D=4and how the N=8supermultiplet break intoN=2multiplets.As this work was being completed,we received[29]which has overlap with section3of thispaper.While this paper was being submitted to hep-th e-print archive,we received[30]whichconsidered the same model in a manifestly U-duality covariant way.2Review of the SU(1,1|2)Superconformal AlgebraThe SU(1,1|2)superconformal algebra is defined by the following commutation relations[L m,L n]=(m−n)L m+n, J a,J b =iǫabc J c,[L m,J a]=0,(2.1a)(σa)αβGβ¯αr,(2.1b) [L m,Gα¯αr]=(12{Gα¯αr,Gβ¯βs}=ǫ¯α¯β{ǫαβL r+s−(r−s)(σaǫ)αβJ a}.(2.1c) and the Hermiticity conditionsL†m=L−m,(J a)†=J a,(Gα¯αr)†=ǫαβǫ¯α¯βGβ¯β−r.(2.2) The bosonic generators L+1,0,−1and J0,1,2generate the SL(2,I R)conformal group and the SU(2) R-symmetry group,respectively.We have eight supercharges all together;Gα¯αcarry L0charge±1/2∓12,1j L0Degeneracyn/2n/2n+1G++−1/2|0 ,G+−−1/2|0(n−2)/2(n+2)/2n−1( a1· a†1+ a†2· a2),L+1= a2· a1(2.3a)2J+= ψ†1· ψ†2,J0=1( a†1· a− a†2· a2)−123We thank Jan de Boer for a correspondence on this point.21Figure1:The complete KK spectrum of the toy model.Each circle in thefigure represents a state which has a definite value of h and j.The crossed circles correspond to the boundary states.The degeneracy(2j+1)of each state is included in the circle.The states belonging to the same SU(1,1|2)multiplet is connected by a dotted line.The two KK towers on the top row satisfy h=j and correspond to chiral primary states..3Toy ModelAs a warm-up exercise,we compute the mass spectrum of the minimal D=4,N=2SUGRA. It is the simplest SUGRA that admits the AdS2×S2solution with the SU(1,1|2)superalgebra. The theory contains a single N=2gravity multiplet whose componentfields are a graviton,a massless vector and a complex gravitino.3.1ResultThe computation in the following subsections show that the KK spectrum of the toy model contains the short multiplets in Table1for all even n.We have two copies of each multiplet for n≥4and one copy for n=2.The result is depicted in Figure1.From the point of view of the SUGRA computation,each physical degree of freedom of the fields in D=4give a KK tower.That explains why we have four bosonic and four fermionic series of states in the spectrum.Depending on the spin and the polarization of a givenfield,the low lying modes(j=0,13.2Bosonic mass spectrum3.2.1SetupWe normalize thefields such that the action reads2κ2S= d4x√4F2−¯ψmΓmnp∇nψp−i2F rsΓrsmn)ψn ,(3.1) up to terms quartic in fermions that are irrelevant to our computation.The bosonic equations of motion consist of the Einstein and Maxwell equations in vacuum.G mn F2,∇m F mn=0.(3.2)R mn=18These equations admit dyonic Reissner-Nordstr¨o m black-hole solutions.The near horizon geom-etry of an extremal black-hole gives the AdS2×S2solution.The radius of the S2is equal to the Schwarzchild radius of the black-hole.For simplicity,we consider an extremal electric black-hole with unit radius only.Then the AdS2×S2solution readsRµνλσ=−(gµλgνσ−gµσgνλ),¯Fµν=2ǫµν,Rαβγδ=gαγgβδ−gαδgβγ,¯Fαβ=0.(3.3) where the Greek lettersα,β···andµν···label two dimensional indices for AdS2and S2respec-tively.The fermions are set to zero.We are interested in the mass spectrum of thefluctuations of thefields around this background.We use the following parametrizations of thefluctuations.Gαβ=gαβ+h(αβ)+1h1gµν,Gµα=hµα,(3.4a)2Fαβ=∇αaβ−∇βaα,Fµν=2ǫµν+∇µaν−∇νaµ,(3.4b) where the paranthesis denotes the traceless part of a symmetric tensor.3.2.2Spherical harmonics decomposition and gauge choiceEachfield can be decomposed into spherical harmonics.Unlike higher dimensional spheres,S2 does not have genuine vector or tensor spherical harmonics.Vector and tensorfields are spanned by derivatives of the scalar spherical harmonics.h(αβ)=φI1∇(α∇β)Y I+φI2ǫγ(α∇β)∇γY I(j≥2),h2=h I2Y I,Y I,h1=h I1Y I,h(µν)=h I(µν)(3.5)hµα=w Iµ∇αY I+v Iµǫαβ∇βY I(j≥1),aα=q I∇αY I+b Iǫαβ∇βY I(j≥1),aµ=a IµY I.The composite index I specifies both the total angular momentum j and the J0eigenvalue m.The restrictions on the value of j for somefields are due to the fact that∇αY(j=0)=∇(α∇β)Y(j=1)=ǫγ(α∇β)∇γY(j=1)=0.(3.6) Not all the modes in the above expansion are physical since the(D=4)graviton and gaugefields are subject to the gauge transformations,δh mn=∇mΛn+∇nΛm,(3.7a)δa m=−¯F mnΛn+∇m(Σ+Λn¯A n).(3.7b)The functionsΛm andΣare also expanded in spherical harmonics.We need to make a choice of gauge.First,consider the case j≥2.We can gauge away h(αβ) completely by a suitable choice ofΛα.We then useΛµto eliminate the w Iµstly,we use Σto eliminate the q I terms.With this choice of gauge,we note that∇αhµα=0,∇αaα=0.(3.8) For j=1,h(αβ)modes are absent,soΛαcan be used to reduce other degrees of freedom.Wefind it convenient to parametrizeΛm andΣasΛ(1)µ=(Kµ+∇µX)·Y,(3.9a)Λ(1)α=P·ǫαβ∇βY−X·∇αY,(3.9b)Σ(1)=Q·Y,(3.9c)where the dot product means the sum over the three components of j=1spherical harmonics.We can use X,Kµand Q to gauge away h1,wµand q,respectively.Under the gauge transformation by P,vµis shifted by∇µP.This indicates that vµis a massless gaugefield in AdS2.Indeed,the mass term for vµis absent as we will see below.Being a gaugefield in D=2,vµhas no propagating degree of freedom in the bulk.Also h2can be locally gauged away by residual gauge symmetry.For j=0,h(µν),h1,h2and aµare the only modes that remain.The gauge parameterΛαis absent.We can useΛµto gauge away h(µν).The vector aµbecomes a gaugefield in AdS2withΣbeing the gauge transformation parameter,and again has no bulk degree of freedom.3.2.3Linearizedfield equationsThe linearized Einstein and Maxwell equations readR(1)mn(h)=¯F l n(∇m a l−∇l a m)+¯F l m(∇n a l−∇l a n)−g mn¯F kl∇k a l−¯F mk¯F nl h kl+14h mn¯F2,(3.10)∇m(∇m a n−∇n a m)−1Among the other six equations,(3.13c)and(3.16a)are constraints,and the other four are dynamical equations for each physicalfield.Wefirst use the constraints to set on shell.4vµ=2ǫµν∇νv,aµ=ǫµν∇νa(3.18) To simplify notations,we are suppressing the superscripts I in the equations from here to the end of this subsection.Inserting these in(3.13a)and(3.16b)immediately yields∇2x h2−(j2+j−2)h2−4∇2x a=0,(3.19a)∇2x b−j(j+1)b−4∇2x v=0.(3.19b) After some manipulations,the other two equations(3.15b)and(3.17)give∇2x v−(j2+j−2)v−b=0,(3.20a)∇2x a−j(j+1)a−h2=0.(3.20b) They are diagonalized by the following linear combinations of thefields.s1=b−2(j+2)v,s2=h2−2(j+1)a,(3.21a)t1=b+2(j−1)v,t2=h2+2ja.(3.21b)They satisfy∇2s i−j(j−1)s i=0,(3.22a)∇2t i−(j+1)(j+2)t i=0.(3.22b) In AdS2,the scaling dimension of the operator corresponding to a scalarfield is given by[2,3]h=11+4m2).(3.23)This implies that thefields s1,2have h=j and are chiral primaries,while t1,2have h=j+2.It remains to analyze(3.14b)and(3.15a).Inserting(3.15a)into(3.14b)and using(3.18),we find∇2x h I(µν)−j(j+1)h I(µν)+2h I(µν)=4∇(µ∇ν)a.(3.24) It is also possible to show that in two dimensions,(3.15a)implies(∇2+2)h(µν)=∇(µ∇ν)(h2+4a)(3.25) It can be derived most easily in a light-cone coordinate and a conformal bining these two equations,wefind that h(µν)is algebraically determined by h2and hence has no degree of freedom.This argument is valid for j=1also,but not for j=0.3.2.5j=1The computation for j=1differs from that for j≥2in two ways.First,h1is removed by a gauge choice rather than the constraint(3.13b)which is absent because∇(α∇β)Y(j=1)=0.Second,vµand h2has no bulk degree of freedom and can be eliminated.The equations can be diagonalized as before,and the three eigenstates are identified with the j=1points of the t1,t2and s2series. The absence of the corresponding point on the s1series is a consequence of the fact that vµis a gaugefield.Note also that s2can be gauged away on shell by a residual gauge degree of freedom. By X in(3.9b)with∇2X=0,s2is shifted to s2+X.Therefore it is also a boundary degrees of freedom.3.2.6j=0We have thefield equations for h1,h2and aµ∇2x(h1+h2)=−4ǫµν∇µaν−2h1,(3.26a)∇2x h2=4ǫµν∇µaν−2h2+4h1,(3.26b)∇ν(∇νaµ−∇µaν)=ǫµν∇ν(h2−h1).(3.26c) Recall that we gauged away h(µν).Its equation of motion then gives a“Gauss law”constraint,∇(µ∇ν)h2=0.(3.27) One can easily show that(in light cone coordinate,for example)the only normalizable solution to the constraint is h2=constant.It is consistent to set h2to zero.We can eliminate the gaugefield aµfrom the h1equation andfind that m2=2.This is identified with the j=0point of the t2 series.This completes the derivation of the bosonic spectrum in Figure1.3.3Fermionic mass spectrumThe linearizedfield equation for the fermion readsΓmnp∇nψp=−i2¯F rsΓmnrs ψn.(3.28) The linearized SUSY transformation law plays the role of a gauge symmetry,that is,the following variation leaves thefield equation invariant:δψm=∇mǫ−i¯F nl 18Γm nl ǫ.(3.29) It is convenient to separate the“trace”and the“traceless”part ofψµandψα.ψµ=ψ(µ)+1Γαη(Γµψ(µ)=Γαψ(α)=0).(3.30)2We decompose the D=4gamma matrices in terms of the D=2gamma matrices as follows:Γµ=γµ⊗1,Γα=¯γ⊗τα(¯γ=γ0γ1).(3.31) Now we can split(3.28)into four components(∇/x+¯γ∇/y)η+¯γ∇/yλ−2∇αψ(α)=−i¯γλ,(3.32a)(∇/x+¯γ∇/y)λ+∇/xη−2∇µψ(µ)=−i¯γη,(3.32b)∇(µ)η+¯γ∇/yψ(µ)=−i¯γψ(µ),(3.32c)∇(α)λ+∇/xψ(α)=+i¯γψ(α),(3.32d) where thefirst two equations are the traceless parts of theµandαcomponents of(3.28),respec-tively.The other two are the trace parts.Here,∇/x,∇/y are the two dimensional dirac operators.The gauge transformation law also divides into four pieces.δψ(µ)=∇(µ)ǫ,δλ=∇/xǫ−i¯γǫ,(3.33a)δψ(α)=∇(α)ǫ,δη=¯γ(∇/yǫ+iǫ).(3.33b)Consider the spherical harmonics decomposition.λ=λI+ΣI++λI−ΣI−,ψ(µ)=ψI(µ)+ΣI++ψI(µ)−ΣI−,η=ηI+ΣI++ηI−ΣI−,ψ(α)=ψI+∇(α)ΣI++ψI−∇(α)ΣI−,ǫ=ǫI+ΣI++ǫI−ΣI−.(3.34)See Appendix B for the definition and properties of spinor spherical harmonics.For j≥3/2,it is clear that one can gauge awayηcompletely.Then(3.32c)setsψI(µ)=0.In turn,wefind in (3.32b)thatλsatisfies the eom for a free massless spinor in d=4.Finally,(3.32a)determinesψ(α) algebraically in terms ofλ.As a consistency check,we substitute it into(3.32d)andfind the same eom forλ.Thus all that remains is tofind the mass spectrum ofλ.After the spherical harmonics decomposition,the equation reduces to∇/λ++i(j+12)¯γλ−=0.(3.35) The mass eigenstates areξ1=(1+i¯γ)E andξ2=(1−i¯γ)F both of which have m=j+12ταΣ±=⇒∇/yΣ±=±iΣ±.(3.36)It has three consequences.First,modes forψ(α)are absent.Second,(3.32d)is trivially satisfied. Finally,the gauge variation ofη−vanishes for arbitraryǫ−.We choose to gauge awayη+andψ(µ)−usingǫ+andǫ−,respectively.With these in mind,we analyze the threefield equations.From the coefficients ofΣ+in(3.32a)and(3.32c),wefind thatλ+=0,ψ(µ)+=0.(3.37)The coefficients ofΣ−of the same equations yield(∇/x−i¯γ)η−=0,∇(µ)η−=0.(3.38) These two equations together imply thatη−has no propagating degree of freedom and can be set to zero consistently.Finally(3.32b)gives(∇/x−i¯γ)λ−=0,(3.39) which we recognize as the j=1/2point of theξ2series.The scaling dimension of the operator corresponding to a spinorfield in AdS2is given byh=|m|+1210Hyper multiplet210Vector multiplet21Gravitino multipletGravity multiplet210Figure 2:The KK spectrum of the D =11SUGRA on AdS 2×S 2×T 7.12Boundary degrees of freedom can arise in the gravitino multiplet as well,but are not determined by the computation here.In the following two sections,we explain the dimensional reduction of the field equations from D =11to D =4,how different fields fall into N =2multiplets and how each multiplet produces the KK spectrum given in Figure 2.5Bosonic Mass Spectrum5.1SetupWe normalize the fields such that the action reads2κ2S = d 11x √2·4!F 2−¯ΨI ΓIJK ∇J ΨK +1−G12·3!F MIJK F N IJK −12·4!·4!ǫIJKL 1L 2L 3L 4M 1M 2M 3M 4F L 1L 2L 3L 4F M 1M 2M 3M 4.(5.2b)The AdS 2×S 2×T 7solution is given byds 211=g µνdx µdx ν+g αβdx αdx β+δab dz a dz b +δst dw s dw t,R µνλσ=−(g µλg νσ−g µσg νλ),R αβγδ=g αγg βδ−g αδg βγ,¯Fµν45=¯F µν67=ǫµν,¯F αβ46=¯F αβ75=ǫαβ.(5.3)All other fields are set to zero.This is the near horizon geometry of the brane configuration shownin the introduction.The notational conventions for the indices are summarized in Appendix A.Note that the background fields are self-dual in the SO (4)for the four coordinates along which the branes lie.Equivalently,it transforms in the (0,1)representation of SO (4)≃SU (2)+×SU (2)−.This follows from the requirement of partially unbroken supersymmetry.If any one of the relative sign for the gauge field is flipped,the supersymmetry is completely broken,even though it is still a solution to the equations of motion.6This background breaks the SO (7)isometry of the internal T 7down to SU (2)+×SU (2)3,where SU (2)3is the rotation of w 8,9,10.These internal symmetries will play a crucial role in grouping the fields,as will be shown in the next section.5.2Linearizedfield equations and reduction to D=4We linearize the equations in D=11influctuations around the background,G MN=g MN+h MN,F IJKL=¯F IJKL+4∇[I a JKL],(5.4)and then dimensionally reduce it to D=4by keeping only the zero modes of thefluctuations in internal T7=T4×T3.We then redefine some of thefluctuationfields,=h(4)mn−1h(11)mnǫklmn¯F ab kl W b mn,(5.8)4where W a mn is thefield strength of V a mn.The equation remains valid when a is replaced by s,except that in this case the right-hand side vanishes.The quantum numbers of the variousfluctuationfields with respect to the internal symmetries are summarized below,along with that of the background gaugefield¯F ab ing this table,one can divide thefields into small groups,where thefields belonging to the same group can couple to each other.Thefields within a group must have the same quantum numbers except the broken SU(2)−charge,which can be shifted by1by the backgroundfield.We label these groups by capital roman letters.Note that we separated the self-dual and the anti-self-dual parts of thefields which are rank two tensors in SO(4)byA ab±≡1SU(2)+SU(2)−Grouph mn0V a m000B11E 3B a a+2B s s01/21/2C B(st)200FA ab+m10EA as m100B1/21/2D C sab+110AC ast100F1/21/2DD s101The linearized equations of motion in D=4are given byGroup A∇2C sab+=14ǫklmn¯F ab kl F s mn,(5.11a)∇n F s nm=−12ǫstu F tu mn),(5.11b)∇n W s nm=1∇2C as=12ǫstu C atu),(5.12a)∇2B as=12ǫm nkl¯F ab nk∇l C bs.(5.12c) Group D2W[a mn¯F bc]mn,(5.13a)∇2D a=12ǫm nkl¯F ab nk∇l D b+1∇2B(ab)=12F bc−mn¯F ca mn+1∇2C=12¯F abmnF ab+mn−12¯F abmn(2∇k h kn−∇n h k k−∇n B s s)+12¯F abmkF ab+nk+16g mn¯F ab kl F ab+kl−16g mn¯F ab kj¯F ab kl h jl+16∇2B s s.(5.15d)165.3Computation of mass spectrum in each multipletWe have separatedfields which decouple from one another using their internal quantum numbers. We should now disentangle thefield equations further andfind out whichfield belong to which N=2multiplet.Obviously,the bosonicfields in the same multiplet satisfy the samefield equa-tions,and the same for the fermions.In this section,we jump to solve the equations of motion offields in each multiplet,except for the gravity multiplet which has been analyzed in detail in section3.The reduction of the equations obtained in the previous subsection to thefinal form require somewhat lengthy algebra, and we put it off until the next subsection.The complication partly arises from the fact that we chose a specific D=11configuration from the beginning.Although the M theoretic origin of the geometry is manifest in this framework,the U-duality invariance of the D=4theory and its symmetry breaking pattern is obscured.A manifestly duality invariant approach sketched in[21] could simplify the process to a large extent.5.3.1Hyper multipletMinimally coupled scalars in D=4belong to this multiplet.Clearly,the KK modes have m2= j(j+1).It follows that h=j+1.There is no gauge symmetry associated with the scalars.5.3.2Vector multipletA vector multiplet contains a vector A m and two real scalarsφ1,φ2.In the simplest case,φ1couples to Aαonly andφ2to Aµ.Thefield equations for thefirst group are(∇2x+∇2y−2)φ1=1For j=0,b is absent andφ1has m2=2,h=1.Thefield equations forφ2and Aµare(∇2x+∇2y+2)φ2=1All thefields in this group are minimally coupled scalars in D=4and belong to the hyper-multiplet.Group Bǫmn kl F s kl−¯F ab mn C sab+.(5.23)2Then the equation of motion of F s becomes the Bianchi identity for˜F s,and the Bianchi identity for F s become∇n˜F s nm=¯F ab mn∇n C sab+.(5.24)18Note that the right-hand side is the same as that of(5.11c).In terms of˜F s,thefield equation for C becomes∇2C sab+=1Writing down all the components of thefield equations and collecting those which couple to one another,onefinds twelve identical copies of the following set of coupled equations:∇2B=−∇2C=1If we defineC a=1First,B45+B67,B46−B75and B44−B55−B66+B77decouple and contribute to the hyper multiplet.The other equations fall into six groups each of which contains one scalar and one vector. Each group gives the spectrum for half a vector multiplet.Explicitly,the six groups are(B44+B55−B66−B77,A45−µ),(B44−B55+B66−B77,A46−α),(5.28)),(B45−B67,A47−α),(B46+B75,A47−µ(B47−B56,A46−µ),(B47+B56,A45−α).Group Fgrav vec Total 0018B442448D6301215F 20Sum2440Table 3:This table summarizes the number of degrees of freedom a group of bosonic equationscontribute to each of the four multiplets..First,h 2≡h αα,B ss ,A 45+µand A 46+αbelong to the gravity multiplet.Second,h µα,A 46+µand A 45+α,C belong to the vector multiplet.Finally,A 47+µand A 47+αdecouple and contribute to the gravitino multiplet.6Fermionic mass spectrum6.1Linearized field equations and reduction to D =4The linearized field equation for the gravitino in D =11readsΓIJK ∇J ΨK =18ΓJK ΨL F I JKL .(6.1)Throughout this section we suppress the bar on the background field strength.The linearized local SUSY transformation law plays the role of gauge symmetry for Fermions:δΨM =∇M ǫ+14Γa λ,χ=Γs Ψs ,Ψ(s )=Ψs −12Γm (λ+χ).(6.4)We then decompose the fermion into chiral and anti-chiral components with respect to SO (4)of T 4,Ψ±≡1SU (2)+SU (2)−Group 1/2IΨ−m Ψ+(a )1/21/21I 1/2G(3λ+2χ)−Ψ+(s )3/201/2H 1/2Iχ−Table 4:Internal quantum numbers of the fermionic fields..where ¯Γ≡1Γn ∇n Ψ+(s )=Γn ∇n (3λ+2χ)+=0,(6.6a)δΨ+(s )=δ(3λ+2χ)+=0.(6.6b)Group H4F ad ij Γij Ψ+(d ),(6.7a)Γn ∇n (3λ+2χ)−=116F cd ij Γij Γcd Ψ−(s ),(6.7c)δΨ+(a )=δ(3λ+2χ)−=δΨ−(s )=0.(6.7d)Group I8F ab ij Γb {Γij χ++(Γijk −2Γi δjk)Ψ+k },(6.8a)Γmnk ∇n Ψ+k=−14F ab ij Γij Γa Ψ−(b ),(6.8c)δΨ−(a )=1Group JF cd ijΓcd(Γijk−2Γiδjk)Ψ−k,(6.9a)16Γmnk∇nΨ−k=132F ab ij(Γmij−2δmiΓj)Γabχ−,(6.9b)δΨ−m=∇mǫ−−1F ab ijΓabΓijǫ−.(6.9d)166.2Computation of the mass spectrum in each multiplet6.2.1Hyper multipletSpinors with nonzero mass generated by the background gaugefield belong to this multiplet.After diagonalizing the mass matrix,they satisfy the equation of motion of the formΓm∇mψ+iΓ01ψ=0.(6.10) Onefinds that|m|=j+1=j+1for all j≥12transformation laws are given byδψ(µ)=∇(µ)ǫ,δλ=∇/xǫ,(6.12)δψ(α)=∇(α)ǫ,δη=¯γ∇/yǫ,δχ=−i¯γǫ.One can always gauge awayη.Then(6.11d)setsψ(µ)to zero.For j≥3/2,(6.11b)determineψ(α)algebraically.The only independent equations that remain are(6.11a)and(6.11c)withηandψ(µ)removed.The mass eigenvalues are the same as those of hyper multiplet:|m|=j+1All the spinors in this group are massless and minimally coupled in D=4and belong to the vector multiplet.Group H(σ1⊗Γ01+σ2⊗Γ23)Ψ.(6.15)2We can further decompose the equation by settingΨ= Ψ1Ψ2 .(6.16) After splitting each ofΨ1,2into two pieces according to their chirality in the non-compact D=4 spacetime,and recombining those pieces which couple to each other,onefinds that one linear combination belongs to the vector multiplet and the other one to the hyper multiplet.except that it has twice as many degrees of freedom The spectrum is exactly the same forΨ+(s)as3λ+2χ.Wefind the same result even forΨ+again except for the degeneracy.In counting(a)the degeneracy,one should remember the constraintΓaΨ(a)=0.In the basis we chose above,it reduces toΨ(4)−iσ1Ψ(5)−iσ2Ψ(6)−iσ3Ψ(7)=0.(6.17) Group I1.A half ofχ−decouple and belong to the vector multiplet.2.A half ofΨ−m decouple and satisfy the same equation as the gravitino in the toy model.Therefore they belong to the gravity multiplet.3.The other components ofχ−andΨ−m couple to each other.They contribute to the spin3/2multiplet.The following table summarizes the result of this section.ACKNOWLEDGEMENTWe are grateful to Seungjoon Hyun for many helpful discussions,Youngjai Kiem for carefully reading the manuscript and making useful comments,and Jan de Boer for a correspondence.SL would like to thank Jeremy Michelson for discussions at Carg`e se’99ASI.grav vec Total02424H0240840J80Sum2440Table5:Summary of the fermionic spectrum.A Notations and ConventionsWe consider D=11SUGRA on AdS2×S2×T4×T3.Each manifold in the product is parametrized by xµ(µ=0,1),yα(α=2,3),z a(a=4,5,6,7)and w s(s=8,9,10),re-spectively.We use the indices(M,N,···)to label all eleven coordinates together and(m,n,···) to label the coordinates of AdS2×S2.The signature of the metric is(−+···+).Thefield strength of a p-form potential in any dimension is defined byF M0···M p =p∇[MA M1···M p].(A.1)B Spherical HarmonicsThe spherical harmonics form a basis for thefields living on a sphere.In this appendix we consideronly the case of S2.We can construct them by considering the eigenstates of maximal commutingsubalgebra of SU(2)group,which are the total angular momentum J2=j(j+1),its z componentJ z=m,the orbital angular momentum L2=l(l+1),and the spin S2=s(s+1).The case forthe scalar is easiest since s=0and J2= L2,which we identify with the Laplacian on the sphere,∇2y,where y indicates two dimensional coordinates parametrizing the two-sphere.This expression for L2can be obtained by embedding S2into three dimensional space,writing down L2in termsof the Cartesian coordinates,which is quite well known,and reexpressing these in terms of polarcoordinates.Therefore,by construction,we have∇2y Y(j,m)(y)=−j(j+1)Y(j,m)(y),(B.1) where Y(j,m)(y)denotes the eigenstates with eigenvalues(j,m),which were defined above.Next consider the spinor spherical harmonics where s=1/2.The easiest way to consider itis to embed the sphere in the three dimensional space and use cartesian coordinates.We constructthem by taking the tensor product of scalar spherical harmonics with2-component spinor,and。