aa r X i v : . [ h e p - t h ] N ov Wolf Space Coset Spectrum in the Large N = 4 Holography
Changhyun Ahn C.N. Yang Institute for Theoretical Physics, Stony Brook University, Stony Brook, NY 11794-3840, USADepartment of Physics, Kyungpook National University, Taegu 41566, Korea [email protected]
Abstract
After reviewing the four eigenvalues (the conformal dimension, two SU (2) quantum num-ber, and U (1) charge) in the minimal (and higher) representations in the Wolf space cosetwhere the N = 4 superconformal algebra is realized by 11 currents in nonlinear way, these foureigenvalues in the higher representations up to three boxes (of Young tableaux) are examinedin detail. The eigenvalues associated with the higher spin-1 , , N, k ) where the Wolf space coset contains the group SU ( N + 2) and theaffine Kac-Moody spin 1 current has the level k . Under the large ( N, k ) ’t Hooft-like limit,they are simply linear combinations of the eigenvalues in the minimal representations. Forthe linear case where the N = 4 superconformal algebra is realized by 16 currents in linearway, the eigenvalues, corresponding to the spin 2 current and the higher spin 3 current, whichare the only different quantities (compared to the nonlinear case), are also obtained at finite( N, k ). They coincide with the results for the nonlinear case above after the large (
N, k )’t Hooft-like limit is taken. As a by product, the three-point functions of the higher spincurrents with two scalar operators can be determined at finite (
N, k ). On leave from the Department of Physics, Kyungpook National University, Taegu 41566, Korea andaddress until Aug. 31, 2018: C.N. Yang Institute for Theoretical Physics, Stony Brook University, StonyBrook, NY 11794-3840, USA ontents SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf space coset . . . . . 92.1.1 The (0; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 102.1.2 The ( f ; 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 The higher representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.1 The ( f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 122.2.2 The ( f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.3 The (symm; 0) representation . . . . . . . . . . . . . . . . . . . . . . 132.2.4 The (antisymm; 0) representation . . . . . . . . . . . . . . . . . . . . . 142.2.5 The (0; symm) representation . . . . . . . . . . . . . . . . . . . . . . . 152.2.6 The (0; antisymm) representation . . . . . . . . . . . . . . . . . . . . . 172.2.7 The (0; antisymm) representation with three boxes . . . . . . . . . . . 182.2.8 New result: the (0; mixed) representation . . . . . . . . . . . . . . . . 192.2.9 The (0; symm) representation with three boxes . . . . . . . . . . . . . . 212.2.10 Summary of this section . . . . . . . . . . . . . . . . . . . . . . . . . . 21 SU ( N +2) SU ( N ) × SU (2) × U (1) Wolfspace coset 21 + with two boxes . . . . . . . . . . . . . . . 223.1.1 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . 233.1.2 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . 243.1.3 The (symm; 0) representation . . . . . . . . . . . . . . . . . . . . . . . 253.1.4 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . 253.1.5 The (symm; antisymm) representation . . . . . . . . . . . . . . . . . . 263.1.6 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . 273.2 The antisymmetric representations Λ + with two boxes . . . . . . . . . . . . . 283.2.1 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . 293.2.2 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . 303.2.3 The (antisymm; 0) representation . . . . . . . . . . . . . . . . . . . . . 313.2.4 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . 313.2.5 The (antisymm; symm) representation . . . . . . . . . . . . . . . . . . 323.2.6 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . 331.3 The symmetric representations Λ + with three boxes . . . . . . . . . . . . . . 343.3.1 The (symm; symm) representation with three boxes . . . . . . . . . . . 363.3.2 The (symm; symm) representation with three and two boxes . . . . . . 373.3.3 The (symm; f ) representation with three boxes . . . . . . . . . . . . . . 383.3.4 The (symm; 0) representation with three boxes . . . . . . . . . . . . . 393.3.5 The (symm; f ) representation with three boxes . . . . . . . . . . . . . . 403.3.6 The (symm; antisymm) representation with three and two boxes . . . . 413.3.7 The (symm; symm) representation with three and two boxes . . . . . . 423.3.8 The (symm; antisymm) representation with three and two boxes . . . . 433.3.9 The (symm; mixed) representation with three boxes . . . . . . . . . . . 443.3.10 The (symm; antisymm) representation with three boxes . . . . . . . . . 453.4 The antisymmetric representations Λ + with three boxes . . . . . . . . . . . . 463.4.1 The (antisymm; antisymm) representation with three boxes . . . . . . . 483.4.2 The (antisymm; antisymm) representation with three and two boxes . . 493.4.3 The (antisymm; f ) representation with three boxes . . . . . . . . . . . 493.5 The mixed representations Λ + with three boxes . . . . . . . . . . . . . . . . . 503.5.1 The (mixed; mixed) representation with three boxes . . . . . . . . . . . 533.5.2 The (mixed; antisymm) representation with three and two boxes . . . . 543.5.3 The (mixed; symm) representation with three and two boxes . . . . . . 553.5.4 The (mixed; f ) representation with three boxes . . . . . . . . . . . . . 563.5.5 The (mixed; 0) representation with three boxes . . . . . . . . . . . . . 573.5.6 The (mixed; f ) representation with three boxes . . . . . . . . . . . . . 583.5.7 The (mixed; symm) representation with three and two boxes . . . . . . 593.5.8 The (mixed; antisymm) representation with three and two boxes . . . . 603.5.9 The (mixed; antisymm) representation with three boxes . . . . . . . . . 613.5.10 The (mixed; mixed) representation with three boxes . . . . . . . . . . . 623.5.11 Summary of this section . . . . . . . . . . . . . . . . . . . . . . . . . . 63 , , currents in the SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf space coset 64 f ) and (0; f ) representations . . . . . . . . . . . . 674.2 The eigenvalues in the ( f ; 0) and ( f ; 0) representations . . . . . . . . . . . . . 69 , , cur-rents in the SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf space coset 70 f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702.2 The ( f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 715.3 The ( f ; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.4 The ( f ; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 745.5 The ( f ; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 755.6 The ( f ; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 765.7 The (symm; 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.8 The (antisymm; 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . 785.9 The (0; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.10 The (0; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 805.11 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 815.12 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.13 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.14 The (symm; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . 835.15 The (symm; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . 845.16 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 855.17 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . . . . . 875.18 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 875.19 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 885.20 The (antisymm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . 895.21 The (antisymm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . 905.22 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . . . . . 91 SU ( N +2) SU ( N ) coset 937 Review of eigenvalues for the minimal representations with the higherspin- , , currents in the SU ( N +2) SU ( N ) coset 95 f ) and (0; f ) representations . . . . . . . . . . . . . 987.2 The eigenvalues in the ( f ; 0) and ( f ; 0) representations . . . . . . . . . . . . . 100 , , cur-rents in the SU ( N +2) SU ( N ) coset 101 f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018.2 The ( f ; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1028.3 The ( f ; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1058.4 The ( f ; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1068.5 The ( f ; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1073.6 The ( f ; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1088.7 The (symm; 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1108.8 The (antisymm; 0) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1118.9 The (0; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1128.10 The (0; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1138.11 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1148.12 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1158.13 The (symm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1178.14 The (symm; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . 1188.15 The (symm; antisymm) representation . . . . . . . . . . . . . . . . . . . . . . . 1198.16 The (symm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1208.17 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . . . . . 1218.18 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1228.19 The (antisymm; f ) representation . . . . . . . . . . . . . . . . . . . . . . . . . 1238.20 The (antisymm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . 1258.21 The (antisymm; symm) representation . . . . . . . . . . . . . . . . . . . . . . . 1268.22 The (antisymm; antisymm) representation . . . . . . . . . . . . . . . . . . . . 127 SU (5) generators in the symmetric 15 representation of SU (5) withtwo boxes 134B The SU (5) generators in the antisymmetric 10 representation of SU (5) with two boxes 140C The SU (5) generators in the symmetric 35 representation of SU (5) withthree boxes 142D The SU (5) generators in the mixed 40 representation of SU (5) with threeboxes 144E The SU (5) generators in the antisymmetric 10 representation of SU (5) with three boxes 149 Introduction
It is known in [1] that the conformal dimension of a coset GH primary can be described as h (Λ + ; Λ − ) = C ( N +2) (Λ + )( k + N + 2) − C ( N ) (Λ − )( k + N + 2) − ˆ u N ( N + 2)( k + N + 2) + n. (1.1)The quantity C ( N +2) (Λ + ) is the quadratic Casimir of SU ( N +2) of G on the representation Λ + .Similarly, the quantity C ( N ) (Λ − ) is the quadratic Casimir of SU ( N ) of H on the representationΛ − . These Casimirs depend on N and have explicit forms for any representations. The ˆ u charge is related to the U (1) (of H ) current of large N = 4 “linear” superconformal algebra[2, 3, 4, 5, 6]. Finally, the last quantity n is known as the excitation number which canbe positive integer or half-integer. That is, n = , , , · · · . When the representation Λ − appears in the branching of Λ + , then this excitation number vanishes. Otherwise (when therepresentation occurs in the product of (Λ + ; 0) and (0; Λ − )), in general, the excitation numberis nonzero. Because the ˆ u charge behaves as N under the large ( N, k ) ’t Hooft-like limit anddue to the fact that the denominator behaves as cubic term, the third term in (1.1) behavesdifferently when one compares to the first two terms of (1.1). Note that the quadratic Casimirbehaves as N under the large ( N, k ) ’t Hooft-like limit. The quantity k appearing in (1.1) isthe level of spin 1 current having the adjoint index of SU ( N + 2).In this paper, we would like to examine the above conformal dimensions closely, obtainthem at finite ( N, k ), identify the BPS representations (as well as non BPS representations),and observe some relation between the above formula (1.1) and the BPS bound [7, 8, 9],up to three boxes of Young tableaux. The BPS bound is described in terms of two SU (2)quantum numbers l ± corresponding to six spin 1 currents of the large N = 4 “nonlinear”superconformal algebra [10, 11, 7, 12]. Therefore, the four quantum numbers, h correspondingto the stress energy tensor of the large N = 4 nonlinear superconformal algebra, l ± , and ˆ u willbe presented explicitly. One of the reasons why one should reexamine these quantum numbersis that there is no known formula like as (1.1) for the higher spin currents. In particular, whenthe excitation number is nonzero (the representation appears in the product of (Λ + ; 0) and(0; Λ − )), then the conformal dimensions and other eigenvalues corresponding to the higherspin currents are not sum of the ones in (Λ + ; 0) and the ones in (0; Λ − ). Instead, there is acontribution, at finite ( N, k ), from the commutator between the zeromode of spin 2 current(or the zeromode of higher spin current) and other mode from the multiple product of spin curents having the adjoint index of SU ( N +2). After one makes sure that the two approaches,from the conformal formula (1.1) directly and from the collection of three contributions above,give the same answer, one can go to the eigenvalues for the higher spin currents.5ecause there is a higher spin symmetry in the Wolf space coset [13, 14, 15, 16], it isnatural to ask what are the eigenvalues for the higher spin currents on the (minimal and)higher representations. How one can obtain the eigenvalues of the higher spin zero modesacting on the higher representations? Recall that in the above eigenvalues, h , l ± and ˆ u , thefirst three quantum numbers for the any representation remains the same by going to itscomplex conjugated representation. However, the last one, ˆ u charge can have an extra minussign under this process. In particular, when one considers the vanishing excitation number(when the representation Λ − appears in the branching of Λ + ), the SU ( N + 2) generatorsplay the crucial role in the calculations of these eigenvalues. One can associate the quadratic SU ( N + 2) generators with the zeromodes of the three quantities (corresponding to h and l ± ) while linear SU ( N + 2) generators with the zeromode of spin 1 current corresponding toˆ u . Then it is obvious to see that the former does not change the sign and the latter doeschange the sign when one changes the sign of SU ( N + 2) generators because the complexconjugated representation is the minus of the original representation. We will see that somestates contain the same quantum number h but they will have different quantum numbersfrom the higher spin currents.Recall that the lowest 16 currents of the N = 4 multiplet have the spin contents, (1,( ) , 2 , ( ) , 3). For the higher spin 1 and 3 currents which transform as SO (4) singlet,one expects that there should be minus sign between the representation and its complexconjugated representation. On the other hand, there is no sign change for the higher spin2 currents, which transform as two SU (2) adjoints, between these two representations. Onecan make two SU (2) singlets by squaring of each higher spin 2 current and summing overthe adjoint indices, in analogy of two quantities corresponding to the quantum numbers l ± .Of course, their conformal dimensions are 4. For the minimal representations, ( f ; 0) (or( f ; 0)) and (0; f ) (or (0; f )), the corresponding eigenvalues for the higher spin currents wereobtained in [17]. In this paper, the eigenvalues for the higher spin currents in the higherrepresentations, which are obtained from the various products of the minimal representations(which will correspond to single or multi particle states in the AdS bulk theory in the large( N, k ) ’t Hooft like limit), are determined explicitly by considering the eigenvalues for eachminimal representation (and some additional eigenvalue contributions).One of the reasons why one should obtain the eigenvalues for the higher spin currentsis that one can determine the three-point functions (in the higher representations) for thehigher spin (which is fixed) current with two scalar operators at finite (
N, k ). It will turn outthat although the various three-point functions for the higher spin currents with two scalaroperators depend on (
N, k ) explicitly (so far, it is not known how to write them down in6erms of the data of the representations, contrary to (1.1)), the large (
N, k ) ’t Hooft like limitof them leads to very simple form. That is, the ’t Hooft coupling dependent piece of thethree-point functions in the higher representations (the coefficient of the two point functionof two scalar operators) is a multiple of the ones in the minimal representations. For example,for the higher spin 3 current, the fundamental quantities are given by the eigenvalues of therepresentations, (0; f ) (or (0; f )), ( f ; 0) (or ( f ; 0)) and ( f ; f ) (or ( f ; f )). The first two of thesebehave as the functions of ’t Hooft coupling while the last one behaves as N . Although weconsider the two boxes (of Young tableaux) in this paper, we expect that the above behaviorholds for any boxes of the Young Tableaux .In section 2, the work of Gaberdiel and Gopakumar [1] is reviewed. The four eigenvaluesmentioned in the abstract are obtained. There are also new observations for the eigenvalueswith the mixed higher representation.In section 3, further eigenvalues are determined for the other higher representations upto three boxes of Young tableaux. In particular, we summarize the conformal dimensions wehave found under the large ( N, k ) ’t Hooft like limit with Tables.In section 4, the eigenvalues for the higher spin currents in the minimal representations[17] are reviewed.In section 5, as in sections 2 and 3, the eigenvalues for the higher spin currents in thehigher representations up to two boxes of Young tableaux are obtained. This section is oneof the main results of this paper. Some Tables under the large (
N, k ) ’t Hooft like limitsummarize this section.In section 6, some results [22] corresponding to sections 2 and 3 in the linear version arereviewed.In section 7, some results [17] corresponding to section 4 in the linear version are reviewed.In section 8, some results corresponding to section 5 in the linear version are obtained.This section is also one of the main results of this paper. Section 5 is necessary to understandthe results of this section.In section 9, some open problems related to this paper are presented. Some expectationsfor the particular eigenvalues for the higher spin currents are given.In Appendices, the SU ( N + 2 = 5) generators in various higher representations are given Recently [18], by analyzing the BPS spectrum of string theory and supergravity theory on
AdS × S × S × S , it has been found that the BPS spectra of both descriptions agree (where the world sheet approach isused). See also [19, 20, 21] for further studies along this direction. It would be interesting to see how the large N = 4 superconformal higher spin and CFT duality arises in the context of these world sheet approaches. The higher representations can be obtained from the minimal representations and have more than twoboxes of Young tablueaux. The currents of the N = 4 nonlinear superconformal algebra can act on theminimal representations as well as the higher representations. SU ( N + 2 = 7 , , ,
13) generators in higher representations(with two boxes) can be presented, but they are omitted here.The mathematica [23] package by Thielemans [24] is used all the time.Although some presentations are repetitive, most of the subsections (for example, sections3 or 5) can be read independently and are self-contained.
The unitary Wolf space coset [13, 14, 15, 16] is given byWolf = GH = SU ( N + 2) SU ( N ) × SU (2) × U (1) . The adjoint group indices in the complex basis are divided into G indices : a, b, c, · · · = 1 , , · · · ,
12 [( N + 2) − , ∗ , ∗ , · · · ,
12 [( N + 2) − ∗ ,GH indices : ¯ a, ¯ b, ¯ c, · · · = 1 , , · · · , N, ∗ , ∗ , · · · , N ∗ . (2.1)For given ( N + 2) × ( N + 2) unitary matrix, one can associate the above 4 N Wolf space cosetindices as follows [1]: ∗ ∗∗ ∗ ... ... ∗ ∗∗ ∗∗ ∗ · · · ∗ ∗∗ ∗ · · · ∗ ∗ ( N +2) × ( N +2) . (2.2)Note that the adjoint subgroup H indices run over 1 , , · · · , [ N + 3] , ∗ , ∗ , · · · , [ N + 3] ∗ .The operator product expansions (OPEs) between the spin-1 current V a ( z ) and the spin- current Q a ( z ) are described as [25] V a ( z ) V b ( w ) = 1( z − w ) k g ab − z − w ) f abc V c ( w ) + · · · ,Q a ( z ) Q b ( w ) = − z − w ) ( k + N + 2) g ab + · · · ,V a ( z ) Q b ( w ) = + · · · . (2.3)The positive integer k is the level of the spin 1 current. The metric g ab in (2.3) is givenby g ab = Tr( T a T b ) and the structure constant f abc is given by f abc = Tr( T c [ T a , T b ]) where T a
8s the SU ( N + 2) generator. Note that the nonvanishing metric components are given by g AA ∗ = g A ∗ A = 1 where A = 1 , , · · · , [( N + 2) − SU ( N + 2) adjointlower index A , one has the SU ( N + 2) adjoint upper index A ∗ and vice versa. Note that theabove adjoint indices a, b, · · · for the group G are further divided into index A and index A ∗ .The explicit 11 currents of large N = 4 nonlinear superconformal algebra with (2.1) aregiven by G ( z ) = i ( k + N + 2) Q ¯ a V ¯ a ( z ) , G i ( z ) = i ( k + N + 2) h i ¯ a ¯ b Q ¯ a V ¯ b ( z ) ,A + i ( z ) = − N f ¯ a ¯ bc h i ¯ a ¯ b V c ( z ) , A − i ( z ) = − k + N + 2) h i ¯ a ¯ b Q ¯ a Q ¯ b ( z ) ,T ( z ) = 12( k + N + 2) h ( k + N + 2) V ¯ a V ¯ a + k Q ¯ a ∂ Q ¯ a + f ¯ a ¯ bc Q ¯ a Q ¯ b V c i ( z ) − k + N + 2) X i =1 ( A + i + A − i ) ( z ) , (2.4)where i = 1 , ,
3. The G µ ( z ) with µ = (0 , i ) currents are four supersymmetry currents, A ± i ( z )are six spin-1 generators of SU (2) k × SU (2) N and T ( z ) is the spin-2 stress energy tensor. Thethree almost complex structures h i ¯ a ¯ b are given by 4 N × N matrices as in [17]. Note that thespin-1 current A + i ( z ) with level k depends on the spin-1 current V a ( z ) only while the spin-1current A − i ( z ) with level N depends on the spin- current Q ¯ a ( z ) only. Then it is obviousthat the OPEs between them are regular. SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf spacecoset
Let us describe the eigenvalues of1) the zero mode of the stress energy tensor spin-2 current : T ,
2) the zero mode of sum of the square of spin-1 current : − " X i =1 ( A + i ) ,
3) the zero mode of sum of the square of spin-1 current : − " X i =1 ( A − i ) ,
4) the zero mode of other spin-1 current : 12 U , acting on the minimal representations [1]. The corresponding eigenvalues are denoted by h , l + ( l + + 1), l − ( l − + 1) and ˆ u respectively. Note that the U (1) current U ( z ) is equiva-lent to the one U ( z ) of the large N = 4 linear superconformal algebra. That is, U ( z ) =2 i q N ( N + 2) U ( z ) [17] appearing in section 6.9 .1.1 The (0; f ) representation One of the minimal representations is given by the fundamental representation f (or ) of SU ( N ) living in the denominator of the Wolf space coset. The corresponding state is givenby [1] | (0; f ) > = 1 √ k + N + 2 Q ¯ A ∗ − | >, ¯ A ∗ = 1 ∗ , ∗ , · · · , N ∗ . (2.5)They correspond to the rectangular N × N + 2) × ( N + 2) unitarymatrix in (2.2). Under the decomposition of SU ( N + 2) into the SU ( N ) × SU (2), the adjointrepresentation of SU ( N + 2) contains ( N , ) which corresponds to this (0; f ) representation.Furthermore, the representation (0; ¯ f ) corresponds to ( N , ) associated with √ k + N +2 Q ¯ A − | > with ¯ A = 1 , , · · · , N in the other rectangular 2 × N unitary matrix in (2.2).The four eigenvalues are summarized by [1] h (0; ) = (2 k + 3)4( N + k + 2) ,l + ( l + + 1)(0; ) = 0 ,l − ( l − + 1)(0; ) = 34 , ˆ u (0; ) = ( N + 2)2 . (2.6)The eigenvalue h can be obtained by using the relation (1.1). The excitation number n isgiven by . Note that the excitation number for the fermion is given by and for the multiplefermions, one can have the excitation number which is given by the number of fermions dividedby two. Or one can calculate the OPE between the “reduced” T ( z ), where the spin-1 current V a ( z ) dependence is ignored completely, and Q ¯ A ∗ ( w ) and read off the second-order pole.For the l − quantum number, the eigenvalue is obtained from the OPE between the corre-sponding operator − P i =1 ( A − i ) ( z ) and Q ¯ A ∗ ( w ) and read off the second-order pole. There-fore, one obtains l − = which is consistent with the fact that the above state transforms asa doublet under the SU (2) N .For the l + quantum number, the relevant spin-1 current is given by A + i ( z ) which containsonly the spin-1 current V a ( z ). Therefore, the corresponding eigenvalue vanishes and l + = 0.Similarly, the U (1) charge ˆ u can be determined by calculating the first order pole in theOPE between the U (1) current U ( z ) and Q ¯ A ∗ ( w ).It is known that the BPS bound for the conformal dimension [7, 8, 9] is1( N + k + 2) " ( k + 1) l − + ( N + 1) l + + ( l + − l − ) . (2.7)10ne can easily check that by substituting the quantum numbers l + = 0 and l − = , the aboveconformal dimension becomes the BPS bound . The large ( N, k ) ’t Hooft like limit for theBPS bound can be obtained. ( f ; 0) representation Other minimal representation is given by ( f ; 0) representation. In other words, the singletswith respect to the SU ( N ) can be obtained from the fundamental representation in the SU ( N + 2). It is known that the branching rule for the fundamental representation in the SU ( N + 2) k with respect to the SU ( N ) k × SU (2) k × U (1) is characterized by → ( , ) + ( , ) − N . (2.8)The indices 1 and − N are the U (1) charges ˆ u [26, 27]. Then the state | ( f ; 0) > correspondsto ( , ) − N .The four eigenvalues are described as [1] h ( ; 0) = (2 N + 3)4( N + k + 2) ,l + ( l + + 1)( ; 0) = 34 ,l − ( l − + 1)( ; 0) = 0 , ˆ u ( ; 0) = − N . (2.9)One can obtain the conformal dimension using the formula (1.1). Or after substituting the SU ( N +2) generator T a ∗ into the zero mode of spin-1 current V a in the quadratic of the reducedspin-2 current T ( z ) where all the Q a dependent terms are ignored, one obtains ( N +2) × ( N +2)unitary matrix acting on the state | ( f ; 0) > . Then the conformal dimension for this state canbe determined by the diagonal elements of the last 2 × l + quantum number, as described in the conformal dimension, one can find ( N +2) × ( N + 2) unitary matrix for the zero mode of − P i =1 ( A + i ) which contains only the spin-1current V a ( z ). It turns out that the diagonal element of the last 2 × which implies l + = . This is consistent with the fact that this state is a doubletunder the SU (2) k ( , ) − N .For the l − quantum number, because the spin- current Q a ( z ) does not contribute to theeigenvalue equation associated with this state, one has l − = 0. For the state | (0; f ) > , one can obtain similar eigenvalues where the only difference appears in theeigenvalue ˆ u . That is, ˆ u (0; ) = − ( N +2)2 . Because the conformal dimension depends on ˆ u in (1.1), the signchange for the ˆ u does not change the conformal dimension. U (1) charge, one can construct ( N + 2) × ( N + 2) unitary matrix for the zeromode of the reduced U ( z ) where the spin- current Q a ( z ) dependence is removed. Then thediagonal element of the last 2 × − N .One can check the above conformal dimension satisfies the BPS bound by substituting l + = and l − = 0.For the state | ( f ; 0) > , one can obtain similar eigenvalues where the only difference appearsin the eigenvalue ˆ u . That is, ˆ u ( ; 0) = N . Let us describe the four eigenvalues for the higher representations which arise in the variousproducts of the above minimal representations (0; f ), ( f ; 0), (0; f ) and ( f ; 0). ( f ; f ) representation According to the previous branching rule in (2.8), this higher representation corresponds to( , ) transforming as a fundamental representation f (or ) under the SU ( N ) and a singletunder the SU (2) k . The nonzero ˆ u charge is given by 1 from the subscript.One can summarize the following eigenvalues [1] h ( ; ) = 1( N + k + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 1 . (2.10)In this case, the conformal dimension can be determined by reading off the diagonalelements in the N × N subdiagonal unitary matrix inside of ( N + 2) × ( N + 2) unitary matrixobtained in previous subsection. Or the formula in (1.1) can be used also. From the diagonalelements in the N × N subdiagonal unitary matrix inside of ( N + 2) × ( N + 2) unitary matrixfor the zero mode of − P i =1 ( A + i ) , one can determine the l + quantum number which is equalto 0. This is consistent with the singlet under the SU (2) k in ( , ) . For the l − quantumnumber, it is the same as before and l − = 0. From the diagonal elements in the N × N subdiagonal unitary matrix inside of ( N + 2) × ( N + 2) unitary matrix for the zero modeof the reduced U ( z ), the above ˆ u charge can be obtained. It is easy to see that the aboverepresentation does not satisfy the BPS bound because the conformal dimension for the BPSbound is equal to 0 by substituting l + = l − = 0.12 .2.2 The ( f ; f ) representation This representation can be obtained from the product between the minimal representations( f ; 0) and (0; f ). Then one can write down the state as √ k + N +2 Q ¯ A − | ( f ; 0) > where ¯ A =1 , , · · · , N (2.5). The four eigenvalues are given by [1] h ( ; ) = 12 ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . (2.11)For the conformal dimension, there exists other contribution in addition to the sum of h ( ; 0) and h (0; ) (which is equal to h (0; )). The extra contribution coming from the lowerorder pole in the commutator [ T , Q ¯ A − ] takes the form N + k +2) N × N − N + k +2) × ! . By realizing the diagonal elements in the lower 2 × N + 3)4( N + k + 2) + (2 k + 3)4( N + k + 2) − N + k + 2) = 12 . Of course, this counting can be seen from the formula (1.1). For the l ± quantum numbers,one can add each contribution from (2.9) and (2.6). For the ˆ u charge, one can add eachcontribution and it turns out that − N − N +22 = − N −
1. By substituting l ± = , the aboveconformal dimension satisfies the BPS bound. (symm; 0) representation This representation can be obtained from the product of minimal representations ( f ; 0) and( f ; 0). By taking the product between the two identical branching rules in (2.8), we obtain[1, 26, 27] ⊗ = + → " ( , ) + ( , ) − N ⊗ " ( , ) + ( , ) − N = " ( , ) + ( , ) − N + ( , ) − N + " ( , ) + ( , ) − N + ( , ) − N . SU ( N ) k × SU (2) k × U (1) → ( , ) + ( , ) − N + ( , ) − N , → ( , ) + ( , ) − N + ( , ) − N . (2.12)The (symm; 0) representation, which is the singlet in SU ( N ) k , corresponds to ( , ) − N .Then one can describe the four eigenvalues as follows: h ( ; 0) = ( N + 2)( N + k + 2) ,l + ( l + + 1)( ; 0) = 2 ,l − ( l − + 1)( ; 0) = 0 , ˆ u ( ; 0) = − N. (2.13)One can calculate the conformal dimension by using the formula or by substituting the SU ( N + 2) generator T a ∗ into the zero mode of spin-1 current V a in the reduced spin-2current T ( z ), one obtains ( N + 2)( N + 3) × ( N + 2)( N + 3) unitary matrix acting on thestate | (symm; 0) > . See Appendix A for the generators for N = 3. Then the conformaldimension for this state can be determined by the diagonal elements of the last 3 × N ( N + 1) × N ( N + 1) unitarymatrix or 2 N × N unitary matrix will be explained in next subsection.From the whole unitary matrix for the zero mode of − P i =1 ( A + i ) , the diagonal elementof the last 3 × SU (2) k ) implies l + = 1. This is consistentwith the fact that this state is a triplet under the SU (2) k ( , ) − N . For the l − quantumnumber, one has trivial l − = 0. For the U (1) charge, one can construct the whole unitarymatrix for the zero mode of the reduced U ( z ). Then the diagonal element of the last 3 × − N which is simply the sum of two U (1) charge − N for .One can check the above conformal dimension satisfies the BPS bound by substituting l + = 1and l − = 0. For the state | (symm; 0) > , one can obtain similar eigenvalues where the onlydifference appears in the eigenvalue ˆ u . That is, ˆ u ( ; 0) = N . (antisymm; 0) representation The other representation can arise from the product of minimal representations ( f ; 0) and( f ; 0). The (antisymm; 0) representation, which is the singlet in SU ( N ) k , corresponds to( , ) − N in the branching rule (2.12).Then the four eigenvalues are given as follows: h ( ; 0) = N ( N + k + 2) , + ( l + + 1)( ; 0) = 0 ,l − ( l − + 1)( ; 0) = 0 , ˆ u ( ; 0) = − N. (2.14)The conformal dimension formula can be used here or by substituting the SU ( N +2) generator T a ∗ into the zero mode of spin-1 current V a in the reduced spin-2 current T ( z ), one obtains ( N + 2)( N + 1) × ( N + 2)( N + 1) unitary matrix acting on the state | (antisymm; 0) > .See also Appendix B for the generators with N = 3. The conformal dimension for this statecan be obtained by the last diagonal element (singlet of SU (2) k ). The detailed descriptionsof remaining diagonal elements in the N ( N − × N ( N −
1) unitary matrix or 2 N × N unitary matrix will be given in next subsection.By reading off the last diagonal element in the whole matrix for the zero mode of theoperator corresponding to l + quantum number, one obtains l + = 0 which is a singlet underthe SU (2) k ( , ) − N . As before, one has a trivial l − = 0 quantum number. The similaranalysis for the ˆ u charge can be done. One can check the above conformal dimension doesnot satisfy the BPS bound by substituting l ± = 0. For the state | (antisymm; 0) > , one hasˆ u ( ; 0) = N . (0; symm) representation One can also construct the product of two minimal representations (0; f ) and (0; f ). The(0; symm) representation can arise. Let us focus on N = 3 case for simplicity. One canvisualize the spin- current Q ¯ a ( z ) in the following 5 × Q ≡ Q ∗ Q ≡ Q ∗ Q ≡ Q ∗ Q ≡ Q ∗ Q ≡ Q ∗ Q ≡ Q ∗ Q Q Q Q Q Q . (2.15)The Q ∗ , · · · , Q ∗ appear in (2.5).Let us construct the symmetric combinations between the spin- currents Q ¯ A ∗ ( z ). Thereare six states and the corresponding operators are as follows: Q Q ( z ) , ( Q Q + Q Q )( z ) , ( Q Q + Q Q )( z ) ,Q Q ( z ) , ( Q Q + Q Q )( z ) , Q Q ( z ) , (2.16)up to some overall normalizations. First of all, the two SU (2) indices should be different fromeach other because of the fermionic property of Q ¯ a ( z ). One should have one factor from the15lements, Q ( z ), Q ( z ) or Q ( z ) and the other factor from the elements, Q ( z ), Q ( z ) or Q ( z ). When the SU ( N = 3) indices are equal (i.e., if one takes two operators in the samerow of the matrix (2.15)), the interchange of these indices gives the original term. Then oneobtains the single terms in (2.16). When the SU ( N = 3) indices are not equal to each other,then there are extra terms. Then we have the remaining terms in (2.16).The four eigenvalues are given by h (0; ) = k ( N + k + 2) ,l + ( l + + 1)(0; ) = 0 ,l − ( l − + 1)(0; ) = 0 , ˆ u (0; ) = N + 2 . (2.17)One can interpret the conformal dimension by calculating the following second order poleof the OPE T ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = k ( N + k + 2) Q Q ( w ) . Of course, the stress energy tensor spin-2 current does not contain the spin-1 current V a ( z ).The N -dependence appearing in the denominator can be easily generalized from N = 3 result.It is easy to check that the similar calculations for other symmetric combinations in (2.16)lead to the same results.For the l + quantum number, we have trivial l + = 0. For the l − quantum number, one cancompute the following OPE and read off the coefficient of second order pole − X i =1 ( A − i ) ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 0 . Similarly, the ˆ u charge can be added and leads to i q N ( N + 2) U ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w ) = ( N + 2) Q Q ( w ) . One can check the above conformal dimension does not satisfy the BPS bound by substituting l ± = 0 . One has ˆ u (0; ) = − ( N + 2). We have also checked that other symmetric combinations in (2.16) leadto the corresponding same quantum numbers. .2.6 The (0; antisymm) representation Let us construct the antisymmetric combinations between the spin- currents Q ¯ a ( z ). Thereare six states and the corresponding operators are as follows: Q Q ( z ) , Q Q ( z ) , Q Q ( z ) ,Q Q ( z ) Q Q ( z ) , Q Q ( z ) . (2.18)The two SU (2) indices should be equal to each other because of the fermionic property of Q ¯ a ( z ). One should take the two operators in (2.15) from the same column.The four eigenvalues are summarized by h (0; ) = ( k + 2)( N + k + 2) ,l + ( l + + 1)(0; ) = 0 ,l − ( l − + 1)(0; ) = 2 , ˆ u (0; ) = N + 2 . (2.19)For the conformal dimension, one can calculate the following OPE T ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = ( k + 2)( N + k + 2) Q Q ( w ) . Similarly, one can check that the conformal dimension for other operators in (2.18) leads tothe same quantum number. The denominator for general N can be expected.For l + quantum number, we have trivial l + = 0. For the l − quantum number, one cancompute the following OPE and read off the second order pole − X i =1 ( A − i ) ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 2 Q Q ( w ) . This implies that the l − quantum number is l − = 1. Similarly, the ˆ u charge can be obtainedfrom the following OPE i q N ( N + 2) U ( z ) Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w ) = ( N + 2) Q Q ( w ) . By substituting l + = 0 and l − = 1 into the BPS bound for the conformal dimension, one seesthat this state satisfies the BPS bound . We see that ˆ u (0; ) = − ( N + 2). .2.7 The (0; antisymm) representation with three boxes For the antisymmetric representation with three boxes, one has the following operators Q Q Q ( z ) , Q Q Q ( z ) . The four eigenvalues are h (0; ) = 3(2 k + 5)4( N + k + 2) ,l + ( l + + 1)(0; ) = 0 ,l − ( l − + 1)(0; ) = 154 , ˆ u (0; ) = 32 ( N + 2) . (2.20)The conformal dimension can be obtained from the following OPE and the coefficient ofsecond order pole leads to T ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 3(2 k + 5)4( N + k + 2) Q Q Q ( w ) . Or the conformal dimension formula (1.1) is given by − N − N + 1)2 N ( k + N + 2) − ( ( N + 2)) N ( N + 2)( k + N + 2) + 32 = 3(2 k + 5)4( k + N + 2) . As before, the l + quantum number is trivial and l + = 0. For the l − quantum number, onecan calculate the following OPE − X i =1 ( A − i ) ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 154 Q Q Q ( w ) . This implies that the l − quantum number is given by l − = . For the ˆ u charge, the followingresult holds i q N ( N + 2) U ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w ) = 32 ( N + 2) Q Q Q ( w ) . With l + = 0 and l − = , the above conformal dimension satisfies the BPS bound . The following quantum number ˆ u (0; ) = − ( N + 2) can be checked also. .2.8 New result: the (0; mixed) representation In the triple product of three minimal representations (0; f ), (0; f ) and (0; f ), one has alsothe (0; mixed) representation. There are two fundamental representations, ( Q , Q , Q ) forfixed SU (2) index and ( Q , Q , Q ) for fixed other SU (2) index, under the SU ( N = 3).Let us select one of the three quantities in ( Q , Q , Q ). Then the next quantity is chosenfrom the one of the three quantities ( Q , Q , Q ). Then the final quantity is selected fromthe first SU (3) fundamental ( Q , Q , Q ). The mixed representation is either the one thatis antisymmetric in the first two indices or the other one that is symmetric in the first twoindices. For the first choice, one has the following states (with the conventions in Appendices)ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ E − E ) , (2.21)up to some overall normalizations. The three numbers stand for each fundamentals. Forexample, the operator correpsonding to (211 − Q Q Q − Q Q Q )( z ).Of course the second term is identically zero by moving Q to the left. See also Appendix D for SU (5) generators in mixed representation.Then after rewriting (2.21) with the help of adjoint spin- current, one has the followingoperators corresponding to mixed representations Q Q Q ( z ) , Q Q Q ( z ) , ( Q Q Q − Q Q Q + Q Q Q − Q Q Q )( z ) , Q Q Q ( z ) ,Q Q Q ( z ) , Q Q Q ( z ) , (2.22)( Q Q Q − Q Q Q + Q Q Q − Q Q Q )( z ) , Q Q Q ( z ) . The four eigenvalues are described as h (0; ) = 3(1 + 2 k )4( N + k + 2) ,l + ( l + + 1)(0; ) = 0 ,l − ( l − + 1)(0; ) = 34 , ˆ u (0; ) = 32 ( N + 2) . (2.23)19y using the quadratic Casimir for the mixed representation [28] C ( N ) ( ) = C ( N ) ([1 , , , · · · , N − N , and inserting the ˆ u charge ˆ u (0; ) = ( N + 2), one can calculate the conformal dimensionformula and it is given by − N − N ( k + N + 2) − ( ( N + 2)) N ( N + 2)( k + N + 2) + 32 = 3(2 k + 1)4( k + N + 2) . Here the numerical value comes from the three products of spin- current Q ¯ a ( z ). Or onecan calculate the following OPE and focus on the second order pole T ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 3(2 k + 1)4( N + k + 2) Q Q Q ( w ) , where we also put the N dependence in the denominator. For the l + quantum number, wehave l + = 0 and for the l − quantum number, one can calculate the following OPE and readoff the second order pole − X i =1 ( A − i ) ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w )2 = 34 Q Q Q ( w ) . This implies that we have l − = . For the ˆ u charge, the similar OPE calculation leads to i q N ( N + 2) U ( z ) Q Q Q ( w ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z − w ) = 32 ( N + 2) Q Q Q ( w ) , where also the general N dependence is included. With l + = 0 and l − = , the aboveconformal dimension does not satisfy the BPS bound . It is easy to check that the abovefour eigenvalues can be obtained for other mixed representations in (2.22) as follows:ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √
12 (2 E + 2 E − E − E − E − E ) , ˆ u = 1 √ E − E + E ) , ˆ u = 1 √ E − E + E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ − E + 2 E − E ) , ˆ u = 1 √ − E + E + E ) . These are symmetric in the first two indices. For the higher representation (0; ), the similar analysis can be done. .2.9 The (0; symm) representation with three boxes Are there any states? There are no such states because of the property of the fermions Q ¯ a .There is a minus sign when two of them are interchanged each other. Let us summarize what has been done in this section. The conformal dimension for anyrepresentation can be encoded in the formula (1.1). For the representation (Λ + ; 0), the ˆ u charge is additive and is given by − N (which is the ˆ u charge in ( ; 0)) times the number ofboxes. Its l − quantum number is trivial l − = 0. For the l + quantum number, the maximumnumber, which is the times the number of boxes, can arise in the symmetric representation.Other l + quantum numbers arise in the mixed representation which will appear in next section.For the representation (0; Λ − ), the ˆ u charge is additive and is given by ( N + 2) (which isthe ˆ u charge in (0; )) times the number of boxes. Its l + quantum number is trivial l + = 0.For the l − quantum number, the maximum number, which is the times the number of boxes,can arise in the antisymmetric representation. Other l − quantum numbers arise in the mixedrepresentation which will appear in next section.For the representation (Λ + ; Λ − ), there are two cases.1) The case where the representation Λ − appears in the branching of Λ + under the SU ( N ) k × SU (2) k × U (1). There is a trivial l − = 0 quantum number. For the l + quan-tum number and ˆ u charge can be read off from the multiple product of ( , ) + ( , ) − N in(2.8).2) The case where the representation arises in the product of (Λ + ; 0) and (0; Λ − ). The l ± and ˆ u quantum numbers are additive. In other words, the l + quantum number of (Λ + ; Λ − )comes from the one of (Λ + ; 0) while the l − quantum number of (Λ + ; Λ − ) comes from theone of (0; Λ − ). The ˆ u charge of (Λ + ; Λ − ) is the sum of the one in (Λ + ; 0) and the one in(0; Λ − ). Note that this is not true for the conformal dimension because there exists the extracontribution. SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf space coset
In this section, we would like to construct the four eigenvalues for the zero modes of previousstress energy tensor spin-2 current, the sum of square of spin-1 current, the sum of squareof other spin-1 current and other spin-1 current, acting on other higher representations by21onsidering the multiple products of (0; f ), ( f ; 0), (0; f ) or ( f ; 0). Λ + with two boxes One of the simplest higher representation is given by the symmetric representation symm or. It is known that the product of the minimal representation ( f ; 0) and itself ( f ; 0) impliesthat there are symmetric and antisymmetric representations. From the branching rule for the SU ( N + 2) under the SU ( N ) k × SU (2) k × U (1), one can identify the following branching rules ⊗ = + → " ( , ) + ( , ) − N ⊗ " ( , ) + ( , ) − N = " ( , ) + ( , ) − N + ( , ) − N + " ( , ) + ( , ) − N + ( , ) − N . The subscript stands for the U (1) charge ˆ u . The final ˆ u charge is obtained by adding each ˆ u charge. Then one obtains the following branching rules for the symmetric and antisymmetricrepresentations under the SU ( N ) k × SU (2) k × U (1) (again from (2.12)) → ( , ) + ( , ) − N + ( , ) − N , → ( , ) + ( , ) − N + ( , ) − N . (3.1)Note that the last representations in (3.1) correspond to the symmetric and antisymmetricrepresentations | ( ; 0) > and | ( ; 0) > repspectively.The two-index symmetric parts of the SU ( N + 2) representation can be obtained from thegenerators of the fundamental representation of SU ( N + 2) by using the projection operator ( δ ik δ jl + δ jk δ il ) where i ≤ j and k ≤ l and i, j, k, l = 1 , , · · · , ( N + 2) [29, 30]. Then by actingon the space T a ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ T a , one has the generators for the symmetricrepresentation for the SU ( N + 2)( T a ) ik δ jl + ( T a ) jk δ il + δ ik ( T a ) jl + δ jk ( T a ) il . For N = 3, one has ( N + 2)( N + 3) × ( N + 2)( N + 3) = 15 ×
15 unitary matrix, andthe row and columns are characterized by the following double index notations11 , , , , ,
33; 14 , , , , ,
35; 44 , , , corresponding to ˆ u , · · · , ˆ u in Appendix A . The first six elements correspond to the sym-metric representation for SU (3). The next six elements correspond to the fundamental rep-resentation of SU (3) with SU (2) k doublet. The last three elements correspond to the singlet22f SU (3) with SU (2) k triplet according to (3.1). The explicit form for the 24 generators of SU (5) is presented in Appendix A .Let us calculate the zero mode for the reduced stress energy tensor spin-2 current actingon the state | ( ; 0) > . It turns out that the 15 ×
15 matrix is given by k ) × k ) ×
00 0 k ) × . (3.2)There are three block diagonal elements. The last block diagonal elements correspond to theeigenvalue on the state | ( ; 0) > . We will describe the detailed quantum numbers for theother eigenvalues soon.One can also calculate the zero mode for the sum of the square for the spin-1 current withminus sign acting on the above state | ( ; 0) > and the explicit result is given by × ×
00 0 2 × . (3.3)The diagonal elements correspond to the eigenvalues and in particular, the last one is theeigenvalue for the state | ( ; 0) > which behaves as a singlet under the SU (3).Similarly, one can also compute the zero mode for the spin-1 current acting on the state | ( ; 0) > and one obtains × − ×
00 0 − × . (3.4)In this case, also the last elements are the eigenvalues for the state | ( ; 0) > . (symm; symm) representation Let us consider the higher representation where the symmetric representation in SU ( N )( , ) survives in the branching of (3.1). The four eigenvalues are given by h ( ; ) = 2( N + k + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 . (3.5)One can calculate the conformal dimension for this representation using the previous for-mula or one performs the explicit form for the matrix in this particular representation as in233.2). Therefore, one obtains k +5) in the first diagonal elements. One can apply for other N values where N = 5 , , , , , · · · . It turns out that the numerator of the above quan-tity does not depend on N and takes the common value and the denominator is generalizedto ( k + N + 2). By realizing that the quadratic Casimir of SU ( N + 2) for the symmet-ric representation C ( N +2) ( ) = ( N +1)( N +4)( N +2) (and the one for the symmetric representation C ( N ) ( ) = ( N − N +2) N ) and the correct ˆ u charge is given by 2, the following relation can beobtained2( N + 1)( N + 2 + 2)2( N + 2)( k + N + 2) − N − N + 2)2 N ( k + N + 2) − N ( N + 2)( k + N + 2) = 2( N + k + 2) . Similarly, the quantum number for the l + can be determined by the above matrix calcula-tion in (3.3). From the zero eigenvalues appearing in the first block diagonal matrix in (3.3),one can see the above l + quantum number, a singlet under the SU (2) k . This also canbe seenfrom the previous expression ( , ) . The trivial l − quantum number l − = 0 arises. For thelast eigenvalue corresponding to ˆ u charge, one uses the previous matrix calculation given in(3.4). The eigenvalues appearing in the first block diagonal matrix in (3.4) imply that theˆ u charge is given by 2. This is also consistent with the representation ( , ) where thesubscript denotes the ˆ u charge. One observes that the above conformal dimension does notsatisfy the vanishing BPS bound with l ± = 0. (symm; f ) representation Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N survives in the branching of (3.1). The four eigenvalues can be summarized by h ( ; ) = (2 N + 11)4( N + k + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N . (3.6)The explicit form for the matrix in this particular representation of conformal dimension isgiven by (3.2). Then, one obtains k +5) in the second block diagonal elements. One can applyfor other N values and it turns out that the numerator of the above quantity does depend on N linearly as well as the constant term while the denominator is generalized to 4( k + N + 2).One can also use the formula with the correct ˆ u charge2( N + 1)( N + 2 + 2)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − (1 − N ) N ( N + 2)( k + N + 2) = (2 N + 11)4( N + k + 2) , C ( N ) ( ) = ( N − N ) is used. The quantum number for the l + can be obtained by the above matrix calculation in (3.3). From the eigenvalues appearing inthe second block diagonal matrix in (3.3), one can see the above l + quantum number l + = ,a doublet under the SU (2) k . This also canbe seen from the previous expression ( , ) − N .The trivial l − = 0 quantum number holds in this representation. For the last eigenvaluecorresponding to ˆ u charge, the previous matrix calculation given in (3.4) can be used. Theeigenvalues appearing in the second block diagonal matrix in (3.4) imply that the ˆ u charge isgiven by − . Varying the N values, one finds that the ˆ u charge is linear in N as well as theconstant term. This is also consistent with the representation ( , ) − N where the subscriptdenotes the ˆ u charge. One observes that the above conformal dimension does not satisfy theBPS bound. (symm; 0) representation The eigenvalues in this higher representation are given in the subsection 2.2.3. One seesthat there are eigenvalues in (2.13). From the three matrices in (3.2), (3.3) and (3.4), therelevant eigenvalues are given by k +5) , 2 and − ( N +2)( k + N +2) , 2 and − N respectively. The conformal dimension can be checkedfrom N +1)( N +2+2)2( N +2)( k + N +2) − N N ( N +2)( k + N +2) also. (symm; f ) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.3 and the latter occurs in the subsection 2.1.1 togetherwith the complex conjugation in the footnote 4.In this case, the corresponding four eigenvalues are described by h ( ; ) = (4 N + 2 k + 7)4( N + k + 2) ,l + ( l + + 1)( ; ) = 2 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . (3.7)First of all, one can obtain the following 15 ×
15 matrix by calculating the commutator[ T , Q ¯ A − ] as in the subsection 2.2.2 k ) × − k ) ×
00 0 − k ) × . (3.8)25he last three eigenvalues (the N generalization is straightforward to obtain) appearing inthe last block diagonal matrix in (3.8) provide the extra contribution as well as the sum ofconformal dimensions of ( ; 0) and (0; ). They are given in (2.13) and (2.6) respectively.Then one obtains the final conformal dimension by adding the above contribution appearingin (3.8) as follows( N + 2)( N + k + 2) + (2 k + 3)4( N + k + 2) − N + k + 2) = (4 N + 2 k + 7)4( N + k + 2) , as in (3.7). It is also useful to interpret the above result from the conformal dimension formula.One determines the following result2( N + 1)( N + 2 + 2)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 12 = (4 N + 2 k + 7)4( N + k + 2) . Here we used the quadratic Casimirs for C ( N +2) ( ) and C ( N ) ( ). The correct ˆ u charge isinserted. The excitation number is given by .For the l + quantum number, due to the vanishing of l + in (0; ), it turns out that the l + is the same as the one( l + = 1) in ( ; 0). For the l − quantum number, due to the vanishingof l − in ( ; 0), it turns out that the l − is the same as the one( l − = ) in (0; ). It is easyto see that the above conformal dimension satisfies the BPS bound by substituting l + = 1and l − = . One can add each ˆ u charge and it is obvious that the total ˆ u charge is given by − N − ( N +2)2 which leads to the above result. Note that the ˆ u charge for (0; ) is opposite tothe one for (0; ). (symm; antisymm) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.3 and the latter occurs in the subsection 2.2.6.The four eigenvalues can be summarized by h ( ; ) = ( N + k + 6)( N + k + 2) ,l + ( l + + 1)( ; ) = 2 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = 2 . (3.9)The following 15 ×
15 matrix by calculating the commutator [ T , Q − Q − ] as in the sub-section 3.1.4 can be obtained − k ) × k ) ×
00 0 k ) × . (3.10)26he last three eigenvalues (the N generalization is simply N + k +2) ) appearing in the lastblock diagonal matrix in (3.10) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (2.13) and (2.19) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.10)as follows ( N + 2)( N + k + 2) + ( k + 2)( N + k + 2) + 2( N + k + 2) = ( N + k + 6)( N + k + 2) . Furthermore, this analysis can be seen from the conformal dimension formula2( N + 1)( N + 2 + 2)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − (( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = ( N + k + 6)( N + k + 2) . Here we also use the quadratic Casimir C ( N ) ( ) = ( N − N +1) N and the excitation number isequal to 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishingof l + in (0; ), the l + is the same as the one( l + = 1) in ( ; 0). For the l − quantum number,due to the vanishing of l − in ( ; 0), the l − is the same as the one( l − = 1) in (0; ). Thetotal ˆ u charge is given by − N + ( N + 2) which leads to the above result. One can easily seethat the above conformal dimension does not lead to the BPS bound with l ± = 1 . (symm; symm) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.3 while the latter occurs in the subsection 2.2.5 withcomplex conjugation in the footnote 5.The four eigenvalues are given by h ( ; ) = ( k + N )( k + N + 2) ,l + ( l + + 1)( ; ) = 2 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N − . (3.11) For the higher representation ( ; ), the similar analysis can be done. We have h ( ; ) = 1 , l + ( l + + 1)( ; ) = 2 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = − N − . This conformal dimension does lead to the BPS bound with l ± = 1. Note that different ˆ u charge leads to thedifferent h value. T , Q − Q − ] and it turns out that k ) × − k ) ×
00 0 − k ) × . (3.12)The last three eigenvalues (the N generalization is simply − N + k +2) ) appearing in the lastblock diagonal matrix in (3.12) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (2.13) and (2.17) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.12)as follows ( N + 2)( k + N + 2) + k ( k + N + 2) − k + N + 2) = ( k + N )( k + N + 2) . The conformal dimension formula implies that2( N + 1)( N + 2 + 2)2( N + 2)( k + N + 2) − N − N + 2)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = ( k + N )( k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equalto 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + in (0; ), the l + is the same as the one( l + = 1) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0) and (0; ), the total l − is given by l − = 0, a singlet. Thetotal ˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u chargefor (0; ) is opposite to the one for (0; ). One can easily see that the above conformaldimension does not lead to the BPS bound with l + = 1 and l − = 0. Λ + with two boxes The two-index antisymmetric parts of the SU ( N + 2) representation can be obtained from thegenerators of the fundamental representation of SU ( N + 2) by using the projection operator ( δ ik δ jl − δ jk δ il ) where i < j and k < l and i, j, k, l = 1 , , · · · , ( N + 2) [29, 30]. Then byacting on the space T a ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ T a , one has the generators for theantisymmetric representation for the SU ( N + 2)( T a ) ik δ jl − ( T a ) jk δ il + δ ik ( T a ) jl − δ jk ( T a ) il . For N = 3, one has ( N + 2)( N + 1) × ( N + 2)( N + 1) = 10 ×
10 unitary matrix, andthe row and columns are characterized by the following double index notations12 , ,
23; 14 , , , , ,
35; 45 , u , ˆ u , · · · , ˆ u in Appendix B . The first three elements correspond to theantisymmetric representation for SU (3). The next six elements correspond to the fundamentalrepresentation of SU (3) with SU (2) k doublet. The last element corresponds to the singlet of SU (3) with SU (2) k singlet. The explicit form for the 24 generators of SU (5) is presented inAppendix B . See also (3.1).Let us calculate the zero mode for the reduced stress energy tensor spin-2 current actingon the state | ( ; 0) > . It turns out that the 10 ×
10 matrix is given by k ) × k ) ×
00 0 k ) . (3.13)The last block diagonal element gives us the eigenvalue on the state | ( ; 0) > and its N generalization is straightforward.One can also calculate the zero mode for the spin-1 current acting on the state | ( ; 0) > and one obtains × ×
00 0 0 . (3.14)Again, the last block diagonal element gives us the eigenvalue associated with the aboveoperator on the state | ( ; 0) > and its N generalization is straightforward.Similarly, one can also compute the zero mode for the spin-1 current acting on the state | ( ; 0) > and it turns out that × − ×
00 0 − . (3.15)The last block diagonal element gives us the eigenvalue associated with the above spin-1operator on the state | ( ; 0) > and its N generalization is straightforward. (antisymm; antisymm) representation Let us consider the higher representation where the antisymmetric representation in SU ( N )( , ) survives in the branching of (3.1). The four eigenvalues are given by h ( ; ) = 2( N + k + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 . (3.16)29ne can calculate the conformal dimension for this representation using the previousformula or one performs the explicit form for the matrix in this particular representation asin (3.13). Therefore, one obtains k +5) in the first block diagonal elements. It turns out thatthe numerator of the above quantity does not depend on N and takes the common value andthe denominator is generalized to ( k + N + 2). By realizing that the quadratic Casimir of SU ( N + 2) for the antisymmetric representation C ( N +2) ( ) = N ( N +3)( N +2) (and the one for theantisymmetric representation C ( N ) ( ) = ( N − N +1) N ) and the correct ˆ u charge is given by 2,the following relation can be obtained2( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − N ( N + 2)( k + N + 2) = 2( N + k + 2) . Similarly, the quantum number for the l + can be determined by the above matrix calcu-lation in (3.14). From the zero eigenvalues appearing in the first block diagonal matrix in(3.14), one can see the above l + quantum number which is a singlet under the SU (2) k . Thisalso can be seen from the previous expression ( , ) . The trivial l − quantum number l − = 0occurs. For the last eigenvalue ˆ u charge, one uses the previous matrix calculation given in(3.15). The eigenvalue appearing in the first block diagonal matrix in (3.15) implies that theˆ u charge is given by 2. This is also consistent with the representation ( , ) . The aboveconformal dimension does not satisfy the vanishing BPS bound with l ± = 0. (antisymm; f ) representation Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N survives in the branching of (3.1). The four eigenvalues can be summarized by h ( ; ) = (2 N + 3)4( N + k + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N . (3.17)The explicit form for the matrix in this particular representation is given by (3.13). Then,one obtains k +5) in the second block diagonal elements. One can apply for other N valuesand it turns out that the numerator of the above quantity does depend on N linearly as wellas the constant term while the denominator is generalized to 4( k + N + 2). One can also usethe formula with the correct ˆ u charge as follows:2( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − (1 − N ) N ( N + 2)( k + N + 2) = (2 N + 3)4( N + k + 2) . C ( N ) ( ) is used. The quantum number for the l + can be obtainedby the above matrix calculation in (3.14). From the eigenvalues appearing in the secondblock diagonal matrix in (3.14), one can see the above l + quantum number l + = , a doubletunder the SU (2) k . This also can be seen from the previous expression ( , ) − N . The trivial l − = 0 quantum number holds in this representation. For the last eigenvalue ˆ u charge, theprevious matrix calculation given in (3.15) can be used. The eigenvalues appearing in thesecond block diagonal matrix in (3.15) imply that the ˆ u charge is given by − . One findsthat the ˆ u charge is linear in N as well as the constant term. This is also consistent with therepresentation ( , ) − N . One observes that the above conformal dimension does satisfy theBPS bound. (antisymm; 0) representation The eigenvalues in this higher representation are given in the subsection 2.2.4. One sees thatthere are eigenvalues in (2.14). From the three matrices in (3.13), (3.14) and (3.15), therelevant eigenvalues are given by k +5) , 0 and − N ( k + N +2) , 0 and − N respectively. The conformal dimension can be checkedfrom N ( N +2+1)2( N +2)( k + N +2) − N N ( N +2)( k + N +2) also. (antisymm; f ) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.4 and the latter occurs in the subsection 2.1.1 togetherwith the complex conjugation in the footnote 4.In this case, the corresponding four eigenvalues are described by h ( ; ) = (4 N + 2 k − N + k + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . (3.18)One can obtain the following 10 ×
10 matrix by calculating the commutator [ T , Q ¯ A − ] asin the subsection 2.2.2 k ) × − k ) ×
00 0 − k ) . (3.19)31he last eigenvalue (the N generalization is straightforward to obtain) appearing in the lastdiagonal element in (3.19) provides the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (2.14) and (2.6) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.19)as follows N ( N + k + 2) + (2 k + 3)4( N + k + 2) − N + k + 2) = (4 N + 2 k − N + k + 2) , as described in (3.7). It is also useful to interpret the above result from the conformaldimension formula. One determines the following result2( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 12 = (4 N + 2 k − N + k + 2) . Here we used the quadratic Casimirs for C ( N +2) ( ) and C ( N ) ( ). The correct ˆ u charge isinserted. The excitation number is given by .For the l + quantum number, due to the vanishing of l + in (0; ) and ( ; 0), it turns outthat the total l + is trivial. For the l − quantum number, due to the vanishing of l − in ( ; 0), itturns out that the l − is the same as the one( l − = ) in (0; ). It is easy to see that the aboveconformal dimension does not satisfy the BPS bound by substituting l + = 0 and l − = . Onecan add each ˆ u charge and it is obvious that the total ˆ u charge is given by − N − ( N +2)2 whichleads to the above result. Note that the ˆ u charge for (0; ) is opposite to the one for (0; ). (antisymm; symm) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.4 and the latter occurs in the subsection 2.2.5.The four eigenvalues can be summarized by h ( ; ) = 1 ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 . (3.20)The following 15 ×
15 matrix by calculating the commutator [ T , Q − Q − ] as in the pre-vious subsections can be obtained − k ) × k ) ×
00 0 k ) (3.21)32he last eigenvalue (the N generalization is simply N + k +2) ) appearing in the last element in(3.21) gives the extra contribution as well as the sum of conformal dimensions of ( ; 0) and(0; ). They are given in (2.14) and (2.17) respectively. Then one obtains the final conformaldimension by adding the above contribution appearing in (3.21) as follows N ( N + k + 2) + k ( N + k + 2) + 2( N + k + 2) = 1 . Furthermore, this analysis can be seen from the conformal dimension formula2 N ( N + 2 + 1)2( N + 2)( k + N + 2) − N − N + 2)2 N ( k + N + 2) − (( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = 1 . Here we also use the quadratic Casimirs for C ( N +2) ( ) and C ( N ) ( ) and the excitationnumber is equal to 1. The correct ˆ u charge is inserted. For the l ± quantum number, due tothe vanishing of l ± in ( ; 0) and (0; ), the total l ± is trivial. The total ˆ u charge is given by − N + ( N + 2) which leads to the above result. One can easily see that the above conformaldimension does not lead to the BPS bound with l ± = 0 . (antisymm; antisymm) representation Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 2.2.4 while the latter occurs in the subsection 2.2.6 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = ( k + N )( k + N + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = − N − . (3.22)One should calculate the commutator [ T , Q − Q − ] and it turns out that k ) × − k ) ×
00 0 − k ) (3.23) For the higher representation ( ; ), the similar analysis can be done by changing the ˆ u charge correctly. N generalization is simply − N + k +2) ) appearing in the last elementin (3.23) gives the extra contribution as well as the sum of conformal dimensions of ( ; 0) and(0; ). They are given in (2.14) and (2.19) respectively. Then one obtains the final conformaldimension by adding the above contribution appearing in (3.23) as follows N ( k + N + 2) + k + 2( k + N + 2) − k + N + 2) = ( k + N )( k + N + 2) . The conformal dimension formula implies that N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − N − N + 1)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = ( k + N )( k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equalto 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishingof l + in (0; ) and ( ; 0), the total l + is trivial. For the l − quantum number, due to thevanishing of l − in ( ; 0), the total l − is given by l − = 1 in (0; ). The total ˆ u charge is givenby − N − ( N + 2) which leads to the above result. Again, the ˆ u charge for the representation(0; ) is opposite to the one for the representation (0; ). One can easily see that the aboveconformal dimension does not lead to the BPS bound with l + = 0 and l − = 1. Λ + with three boxes From the triple product of the fundamental representation of SU ( N + 2), one obtains thefollowing branching under the SU ( N ) × SU (2) × U (1) ⊗ ⊗ = " + ⊗ = + 2 + → " ( , ) + ( , ) − N ⊗ " ( , ) + ( , ) − N ⊗ " ( , ) + ( , ) − N = " ( , ) + ( , ) − N + ( , ) − N + ( , ) − N + 2 " ( , ) + ( , ) − N + ( , ) − N + ( , ) − N + ( , ) − N + ( , ) − N + " ( , ) + ( , ) − N + ( , ) − N . , the three representations, symmetric, mixed and antisymmetric ones havethe following decompositions under the SU ( N ) × SU (2) × U (1) [26, 27] → ( , ) + ( , ) − N + ( , ) − N + ( , ) − N , → ( , ) + ( , ) − N + ( , ) − N + ( , ) − N + ( , ) − N + ( , ) − N , → ( , ) + ( , ) − N + ( , ) − N . (3.24)Note that there is no SU ( N ) singlet in the antisymmetric representation contrary to thesymmetric and mixed ones.The three-index symmetric parts of the SU ( N +2) representation can be obtained from thegenerators of the fundamental representation of SU ( N + 2) by using the projection operator ( δ il δ jm δ kn + δ il δ km δ jn + δ kl δ im δ jn + δ jl δ im δ kn + δ jl δ km δ in + δ kl δ jm δ in ) where i ≤ j ≤ k and l ≤ m ≤ n and i, j, k, l, m, n = 1 , , · · · , ( N + 2) [31]. Then by acting on the space T a ⊗ ( N +2) × ( N +2) ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ T a ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ ( N +2) × ( N +2) ⊗ T a , one hasthe generators for the symmetric representation for the SU ( N + 2) as follows: ( T a ) il δ jm δ kn + ( T a ) il δ km δ jn + ( T a ) kl δ im δ jn + ( T a ) jl δ im δ kn + ( T a ) jl δ km δ in + ( T a ) kl δ jm δ in + δ il ( T a ) jm δ kn + δ il ( T a ) km δ jn + δ kl ( T a ) im δ jn + δ jl ( T a ) im δ kn + δ jl ( T a ) km δ in + δ kl ( T a ) jm δ in + δ il δ jm ( T a ) kn + δ il δ km ( T a ) jn + δ kl δ im ( T a ) jn + δ jl δ im ( T a ) kn + δ jl δ km ( T a ) in + δ kl δ jm ( T a ) in . For N = 3, one has ( N + 2)( N + 3)( N + 4) × ( N + 2)( N + 3)( N + 4) = 35 ×
35 unitarymatrix, the row and columns are characterized by the following triple index notations111 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . (3.25)See also Appendix C in ˆ u , · · · , ˆ u . The first ten elements correspond to the symmetric rep-resentation for SU (3). The next twelve elements correspond to the symmetric representationof SU (3) with SU (2) k doublet. The next nine elements correspond to the fundamental rep-resentation of SU (3) with SU (2) k triplet. The last four elements correspond to the singlet of SU (3) with SU (2) k quartet. The explicit form for the 24 generators of SU (5) is presented inAppendix C . Note that the orderings given in (3.25) are different from the ones in Appendix C . The higher representations on the remaining three subsections, 3 .
3, 3 . . SU ( N + 2) generators with these particular representations having three boxes for several N values. | ( ; 0) > . It turns out that the 35 ×
35 matrix is given by1(5 + k ) diag(3 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
334 ) . (3.26)If one takes the ordering from (3.25), then all the diagonal elements will look nicer. There arefour block diagonal elements. The last block diagonal elements correspond to the eigenvalueon the state | ( ; 0) > . We will describe the detailed quantum numbers for the othereigenvalues soon.One can also calculate the zero mode for the sum of the square for the spin-1 current withminus sign acting on the above state | ( ; 0) > and the explicit result is given bydiag(0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,
154 ) . (3.27)The diagonal elements correspond to the eigenvalues and in particular, the last one is theeigenvalue for the state | ( ; 0) > which behaves as a quartet under the SU (2).Similarly, one can also compute the zero mode for the spin-1 current acting on the state | ( ; 0) > and one obtains (twice of the ˆ u )diag(6 , , , , , , , , , , , , , , , , − , − , − , , , , , − , − , − , , , − , − , − , − , − , − , − . (3.28)In this case, also the last elements are the eigenvalues for the state | ( ; 0) > . (symm; symm) representation with three boxes Let us consider the higher representation where the symmetric representation in SU ( N )( , ) survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = 3( N + k + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 3 . One can calculate the conformal dimension for this representation using the previous formulaor one performs the explicit form for the matrix in this particular representation as in (3.26).36herefore, one obtains k +5) in the first block diagonal elements. One can apply for other N values where N = 5 , , , , , · · · . It will turn out that the numerator of the above quantitydoes not depend on N and takes the common value and the denominator is generalizedto ( k + N + 2). By realizing that the quadratic Casimir of SU ( N + 2) for the symmetricrepresentation C ( N +2) ( ) = N +1)( N +5)2( N +2) (and the one for the symmetric representation C ( N ) ( ) = N − N +3)2 N ) and the correct ˆ u charge is given by 3, the following relation canbe obtained3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 3)2 N ( k + N + 2) − N ( N + 2)( k + N + 2) = 3( N + k + 2) . Similarly, the quantum number for the l + can be determined by the above matrix calcu-lation in (3.27). From the zero eigenvalues appearing in the first block diagonal matrix in(3.27), one can see the above l + quantum number, a singlet under the SU (2) k . This also canbe seen from the previous expression ( , ) . The trivial l − quantum number l − = 0 arises.For the last eigenvalue corresponding to ˆ u charge, one uses the previous matrix calculationgiven in (3.28). The eigenvalues appearing in the first block diagonal matrix in (3.28) implythat the ˆ u charge is given by 3. This is also consistent with the representation ( , ) wherethe subscript denotes the ˆ u charge. One observes that the above conformal dimension doesnot satisfy the vanishing BPS bound with l ± = 0. (symm; symm) representation with three and two boxes Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues can be summarized by h ( ; ) = (2 N + 19)4( N + k + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N . The explicit form for the matrix in this particular representation is given by (3.26). Then,one obtains k +5) in the second block diagonal elements. One can apply for other N valuesand it will turn out that the numerator of the above quantity does depend on N linearly aswell as the constant term while the denominator is generalized to 4( k + N + 2). One can alsouse the formula with the correct ˆ u charge3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 2)2 N ( k + N + 2) − (2 − N ) N ( N + 2)( k + N + 2) = (2 N + 19)4( N + k + 2) , C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.27). From the eigenvalues appearing in the second block diagonal matrix in (3.27), one can see the above l + quantumnumber l + = , a doublet under the SU (2) k . This also can be seen from the previousexpression ( , ) − N . The trivial l − = 0 quantum number holds in this representation. Forthe last eigenvalue corresponding to ˆ u charge, the previous matrix calculation given in (3.28)can be used. The eigenvalues appearing in the second block diagonal matrix in (3.28) implythat the ˆ u charge is given by . Varying the N values, one finds that the ˆ u charge is linearin N as well as the constant term. This is also consistent with the representation ( , ) − N where the subscript denotes the ˆ u charge. One observes that the above conformal dimensiondoes not satisfy the BPS bound. (symm; f ) representation with three boxes Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = ( N + 5)( N + k + 2) ,l + ( l + + 1)( ; ) = 2 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N + 1 . The explicit form for the matrix in this particular representation is given by (3.26). Then,one obtains k +5) in the third block diagonal elements. One can apply for other N values andit will turn out that the numerator of the above quantity does depend on N linearly as wellas the constant term while the denominator is generalized to ( k + N + 2). One can also usethe formula with the correct ˆ u charge3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − (1 − N ) N ( N + 2)( k + N + 2) = ( N + 5)( N + k + 2) . where the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.27). From the eigenvalues 2appearing in the third block diagonal matrix in (3.27), one can see the above l + quantumnumber l + = 1, a triplet under the SU (2) k . This also can be seen from the previous expression( , ) − N . The trivial l − = 0 quantum number holds in this representation. For the lasteigenvalue corresponding to ˆ u charge, the previous matrix calculation given in (3.28) can be38sed. The eigenvalues appearing in the third block diagonal matrix in (3.28) imply that theˆ u charge is given by −
2. Varying the N values, one finds that the ˆ u charge is linear in N aswell as the constant term. This is also consistent with the representation ( , ) − N wherethe subscript denotes the ˆ u charge. One observes that the above conformal dimension doesnot satisfy the BPS bound. (symm; 0) representation with three boxes Let us consider the higher representation where the singlet representation in SU ( N ) ( , ) − N survives in the branching of (3.24). The four eigenvalues are given by h ( ; 0) = 3(2 N + 5)4( N + k + 2) ,l + ( l + + 1)( ; 0) = 154 ,l − ( l − + 1)( ; 0) = 0 , ˆ u ( ; 0) = − N . (3.29)The explicit form for the matrix in this particular representation is given by (3.26). Then,one obtains k +5) in the last block diagonal elements. One can apply for other N values andit will turn out that the numerator of the above quantity does depend on N linearly as wellas the constant term while the denominator is generalized to 4( k + N + 2). One can also usethe formula with the correct ˆ u charge3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − ( − N ) N ( N + 2)( k + N + 2) = 3(2 N + 5)4( N + k + 2) , where the quadratic Casimir C ( N +2) ( ) is used. The quantum number for the l + can beobtained by the above matrix calculation in (3.27). From the eigenvalues appearing inthe last block diagonal matrix in (3.27), one can see the above l + quantum number l + = ,a quartet under the SU (2) k . This also can be seen from the previous expression ( , ) − N .The trivial l − = 0 quantum number holds in this representation. For the last eigenvaluecorresponding to ˆ u charge, the previous matrix calculation given in (3.28) can be used. Theeigenvalues appearing in the last block diagonal matrix in (3.28) imply that the ˆ u charge isgiven by − . Varying the N values, one finds that the ˆ u charge is linear in N . This is alsoconsistent with the representation ( , ) − N where the subscript denotes the ˆ u charge. Oneobserves that the above conformal dimension does satisfy the BPS bound.39 .3.5 The (symm; f ) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 and the latter occurs in the subsection 2.1.1 togetherwith the complex conjugation.In this case, the corresponding four eigenvalues are described by h ( ; ) = ( k + 3 N + 6)2( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . First of all, one can obtain the following 35 ×
35 matrix by calculating the commutator[ T , Q ¯ A − ] as in the subsection 2.2.21(5 + k ) diag(1 , , , , , , , , , , , , , , , , − , − , − , , , , , − , − , − , , , − , − , − , − , − , − , −
32 ) . (3.30)The last four eigenvalues (the N generalization is straightforward to obtain) appearing inthe last block diagonal matrix in (3.30) provide the extra contribution as well as the sum ofconformal dimensions of ( ; 0) and (0; ). They are given in (3.29) and (2.6) respectively.Then one obtains the final conformal dimension by adding the above contribution appearingin (3.30) as follows3(2 N + 5)4( k + N + 2) + (2 k + 3)4( k + N + 2) − k + N + 2) = ( k + 3 N + 6)2( k + N + 2)as in (3.7).It is also useful to interpret the above result from the conformal dimension formula. Onedetermines the following result3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 12 = ( k + 3 N + 6)2( k + N + 2) . Here we used the quadratic Casimirs for C ( N +2) ( ) and C ( N ) ( ). The correct ˆ u charge isinserted. The excitation number is given by .For the l + quantum number, due to the vanishing of l + in (0; ), it turns out that the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, due to the vanishing40f l − in ( ; 0), it turns out that the l − is the same as the one( l − = ) in (0; ). It is easyto see that the above conformal dimension satisfies the BPS bound by substituting l + = and l − = . One can add each ˆ u charge and it is obvious that the total ˆ u charge is given by − N − ( N +2)2 which leads to the above result. Note that the ˆ u charge for (0; ) is opposite tothe one for (0; ). (symm; antisymm) representation with three and two boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 and the latter occurs in the subsection 2.2.6.The four eigenvalues can be summarized by h ( ; ) = (4 k + 6 N + 35)4( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = − N . The following 35 ×
35 matrix by calculating the commutator [ T , Q − Q − ] as in the pre-vious subsection can be obtained1(5 + k ) diag( − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , , , , − , − , − , − , , , , − , − , , , , , , , . (3.31)The last four eigenvalues (the N generalization is simply N + k +2) ) appearing in the last blockdiagonal matrix in (3.31) give the extra contribution as well as the sum of conformal dimen-sions of ( ; 0) and (0; ). They are given in (3.29) and (2.19) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.31)as follows 3(2 N + 5)4( k + N + 2) + ( k + 2)( k + N + 2) + 3( k + N + 2) = (4 k + 6 N + 35)4( k + N + 2) . Furthermore, this analysis can be seen from the conformal dimension formula N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − (( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = (4 k + 6 N + 35)4( k + N + 2) . Here we also use the quadratic Casimir C ( N ) ( ) = ( N − N +1) N and the excitation number isequal to 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing41f l + in (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number,due to the vanishing of l − in ( ; 0), the l − is the same as the one( l − = 1) in (0; ). Thetotal ˆ u charge is given by − N + ( N + 2) which leads to the above result. One can easily seethat the above conformal dimension does not lead to the BPS bound with l + = and l − = 1. (symm; symm) representation with three and two boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 while the latter occurs in the subsection 2.2.5 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = (4 k + 6 N + 3)4( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − ] and it turns out that1(5 + k ) diag(2 , , , , , , , , , , , , , , , , − , − , − , , , , , − , − , − , , , − , − , − , − , − , − , − . (3.32)The last four eigenvalues (the N generalization is simply − N + k +2) ) appearing in the lastblock diagonal matrix in (3.32) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (3.29) and (2.17) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.32)as follows 3(2 N + 5)4( k + N + 2) + k ( k + N + 2) − k + N + 2) = (4 k + 6 N + 3)4( k + N + 2) . The conformal dimension formula implies that N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 2)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = (4 k + 6 N + 3)4( k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equalto 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l +
42n (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0) and (0; ), the total l − is given by l − = 0, a singlet. Thetotal ˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u chargefor (0; ) is opposite to the one for (0; ). One can easily see that the above conformaldimension does not lead to the BPS bound with l + = and l − = 0. (symm; antisymm) representation with three and two boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 while the latter occurs in the subsection 2.2.6 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = (4 k + 6 N + 11)4( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − ] and it turns out that1(5 + k ) diag(2 , , , , , , , , , , , , , , , , − , − , − , , , , , − , − , − , , , − , − , − , − , − , − , − . (3.33)The last four eigenvalues (the N generalization is simply − N + k +2) ) appearing in the lastblock diagonal matrix in (3.33) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (3.29) and (2.19) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.33)as follows 3(2 N + 5)4( k + N + 2) + ( k + 2)( k + N + 2) − k + N + 2) = (4 k + 6 N + 11)4( k + N + 2) . The conformal dimension formula implies that N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 1)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = (4 k + 6 N + 11)4( k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equal to1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l +
43n (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0), the total l − is given by l − = 1 in ( ), a triplet. The totalˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u charge for(0; ) is opposite to the one for (0; ). One can easily see that the above conformal dimensiondoes lead to the BPS bound with l + = and l − = 1. (symm; mixed) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 while the latter occurs in the subsection 2.2.8.The four eigenvalues are given by h ( ; ) = 3( k + N + 6)2( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = 3 . One should calculate the commutator [ T , Q − Q − Q − ] and it turns out that1(5 + k ) diag( − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , , , , − , − , − , − , , , , − , − , , , , , , ,
92 ) . (3.34)The last four eigenvalues (the N generalization is simply N + k +2) ) appearing in the lastblock diagonal matrix in (3.34) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (3.29) and (2.23) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.34)as follows 3(2 N + 5)4( k + N + 2) + 3(2 k + 1)4( k + N + 2) + 92( k + N + 2) = 3( k + N + 6)2( k + N + 2) . The conformal dimension formula implies that3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N )( k + N + 2) − ( N +2)2 − N ) N ( N + 2)( k + N + 2) + 32 = 3( k + N + 6)2( k + N + 2) . The quadratic Casimir C ( N ) ( ) is used and the excitation number is equal to . The correctˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + in (0; ), the l +
44s the same as the one( l + = ) in ( ; 0). For the l − quantum number, due to the vanishingof l − in ( ; 0), the total l − is given by l − = in ( ), a doublet. The total ˆ u charge isgiven by − N + ( N + 2) which leads to the above result. One can easily see that the aboveconformal dimension does not lead to the BPS bound with l + = and l − =
12 12 . (symm; antisymm) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.3.4 while the latter occurs in the subsection 2.2.7.The four eigenvalues are given by h ( ; ) = 3( k + N + 8)2( k + N + 2) ,l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 154 , ˆ u ( ; ) = 3 . (3.35)One should calculate the commutator [ T , Q − Q − Q − ] and it turns out that1(5 + k ) diag( − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , − , , , , − , − , − , − , , , , − , − , , , , , , ,
92 ) . (3.36)The last four eigenvalues (the N generalization is simply N + k +2) ) appearing in the lastblock diagonal matrix in (3.36) give the extra contribution as well as the sum of conformaldimensions of ( ; 0) and (0; ). They are given in (3.29) and (2.20) respectively. Then oneobtains the final conformal dimension by adding the above contribution appearing in (3.36)as follows 3(2 N + 5)4( k + N + 2) + 6 k + 154( k + N + 2) + 92( k + N + 2) = 3( k + N + 8)2( k + N + 2) . The conformal dimension formula implies that3( N + 1)( N + 3 + 2)2( N + 2)( k + N + 2) − N − N + 1)(2 N )( k + N + 2) − ( N +2)2 − N ) N ( N + 2)( k + N + 2) + 32 = 3( k + N + 8)2( k + N + 2) . For the higher representation ( ; ), one can do similar analysis with an appropriate ˆ u charge. C ( N ) ( ) is used and the excitation number is equal to . The correctˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + in (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, due to the vanishingof l − in ( ; 0), the total l − is given by l − = in ( ), a quartet. The total ˆ u charge isgiven by − N + ( N + 2) which leads to the above result. One can easily see that the aboveconformal dimension does not lead to the BPS bound with l ± =
32 13 . Λ + with three boxes The three-index antisymmetric parts of the SU ( N +2) representation can be obtained from thegenerators of the fundamental representation of SU ( N + 2) by using the projection operator ( δ il δ jm δ kn − δ il δ km δ jn + δ kl δ im δ jn − δ jl δ im δ kn + δ jl δ km δ in − δ kl δ jm δ in ) where i < j < k and l < m 10 unitary matrix,and the row and columns are characterized by the following triple index notations123; 124 , , , , , , , . (3.37)See also Appendix E for ˆ u , · · · , ˆ u . See also (3.24). The first element corresponds to theantisymmetric (singlet) representation for SU (3). The next six elements correspond to theantisymmetric representation of SU (3) with SU (2) k doublet. The last three elements corre-spond to the fundamental representation of SU (3) with SU (2) k singlet. The explicit form forthe 24 generators of SU (5) is presented in Appendix E . Note that the ordering in (3.37) isdifferent from the ones in Appendix E . One can do the similar analysis for the higher representation ( ; ). We have h ( ; ) = 32 , l + ( l + + 1)( ; ) = 154 ,l − ( l − + 1)( ; ) = 154 , ˆ u ( ; ) = − N − . This conformal dimension does lead to the BPS bound with l ± = . Due to the different ˆ u charge, theconformal dimension is different from the one in (3.35) | ( ; 0) > . It turns out that the 10 × 10 matrix is given by k ) k ) × k ) k ) × 00 0 0 0 k ) × . (3.38)If one takes the ordering in (3.37), then the diagonal elements will look nicer. There arefive block diagonal elements. The last block diagonal elements (together with the third one)correspond to the eigenvalue on the state | ( ; ) > . We will describe the detailed quantumnumbers for the other eigenvalues soon.One can also calculate the zero mode for the sum of the square for the spin-1 current withminus sign acting on the above state | ( ; ) > and the explicit result is given by × × 00 0 0 0 0 × . (3.39)The diagonal elements correspond to the eigenvalues and in particular, the last one (togetherwith the third one) is the eigenvalue for the state | ( ; ) > which behaves as a triplet(fundamental) under the SU (3).Similarly, one can also compute the zero mode for the spin-1 current acting on the state | ( ; ) > and one obtains (twice of ˆ u ) × − × 00 0 0 0 − × . (3.40)In this case, also the last elements (together with the third one) are the eigenvalues for thestate | ( ; ) > .Note that there is no SU ( N ) singlet in the antisymmetric representation in (3.24).47 .4.1 The (antisymm; antisymm) representation with three boxes Let us consider the higher representation where the antisymmetric representation in SU ( N )( , ) survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = 3( k + N + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 3 . One can calculate the conformal dimension for this representation using the previous formulaor one performs the explicit form for the matrix in this particular representation as in (3.26).Therefore, one obtains k +5) in the first diagonal elements in (3.38). One can apply for other N values where N = 5 , , , , , · · · . It will turn out that the numerator of the above quantitydoes not depend on N and takes the common value and the denominator is generalized to( k + N + 2). By realizing that the quadratic Casimir of SU ( N + 2) for the antisymmetricrepresentation C ( N +2) ( ) = N − N +3)2( N +2) (and the one for the antisymmetric representation C ( N ) ( ) = N − N +1)2 N ) and the correct ˆ u charge is given by 3, the following relation can beobtained3( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − N ( N + 2)( k + N + 2) = 3( k + N + 2) . Similarly, the quantum number for the l + can be determined by the above matrix calcu-lation in (3.39). From the zero eigenvalues appearing in the first block diagonal matrix in(3.39), one can see the above l + quantum number, a singlet under the SU (2) k . This also canbe seen from the previous expression ( , ) . The trivial l − quantum number l − = 0 arises.For the last eigenvalue corresponding to ˆ u charge, one uses the previous matrix calculationgiven in (3.40). The eigenvalues appearing in the first block diagonal matrix in (3.40) implythat the ˆ u charge is given by 3. This is also consistent with the representation ( , ) wherethe subscript denotes the ˆ u charge. One observes that the above conformal dimension doesnot satisfy the vanishing BPS bound with l ± = 0.48 .4.2 The (antisymm; antisymm) representation with three and two boxes Let us consider the higher representation where the antisymmetric representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues can be summarized by h ( ; ) = (2 N + 3)4( k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 − N . The explicit form for the matrix in this particular representation is given by (3.38). Then,one obtains k +5) in the second (and fourth one) diagonal elements. One can apply for other N values and it will turn out that the numerator of the above quantity does depend on N linearly as well as the constant term while the denominator is generalized to 4( k + N + 2).One can also use the formula with the correct ˆ u charge3( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − (2 − N ) N ( N + 2)( k + N + 2) = (2 N + 3)4( k + N + 2)where the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number for the l + can be obtained by the above matrix calculation in (3.39). From the eigenvalues appearingin the second (and fourth one) block diagonal matrix in (3.39), one can see the above l + quantum number l + = , a doublet under the SU (2) k . This also can be seen from the previousexpression ( , ) − N . The trivial l − = 0 quantum number holds in this representation. Forthe last eigenvalue corresponding to ˆ u charge, the previous matrix calculation given in (3.40)can be used. The eigenvalues appearing in the second block diagonal matrix in (3.40) implythat the ˆ u charge is given by . Varying the N values, one finds that the ˆ u charge is linearin N as well as the constant term. This is also consistent with the representation ( , ) − N where the subscript denotes the ˆ u charge. One observes that the above conformal dimensiondoes satisfy the BPS bound. (antisymm; f ) representation with three boxes Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = ( N − k + N + 2) , + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 1 − N. The explicit form for the matrix in this particular representation is given by (3.38). Then,one obtains k +5) in the third (and the last) diagonal elements. One can apply for other N values and it will turn out that the numerator of the above quantity does depend on N linearly as well as the constant term while the denominator is generalized to ( k + N + 2). Onecan also use the formula with the correct ˆ u charge3( N + 2 − N + 2 + 1)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − (1 − N ) N ( N + 2)( k + N + 2) = ( N − k + N + 2)where the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.39). From the eigenvalue 0appearing in the third (and the last) block diagonal matrix in (3.39), one can see the above l + quantum number l + = 0, a singlet under the SU (2) k . This also can be seen from the previousexpression ( , ) − N . The trivial l − = 0 quantum number holds in this representation. For thelast eigenvalue corresponding to ˆ u charge, the previous matrix calculation given in (3.40) canbe used. The eigenvalues appearing in the third (and the last) block diagonal matrix in (3.40)imply that the ˆ u charge is given by − 2. Varying the N values, one finds that the ˆ u chargeis linear in N as well as the constant term. This is also consistent with the representation( , ) − N where the subscript denotes the ˆ u charge. One observes that the above conformaldimension does not satisfy the BPS bound. Λ + with three boxes Let us describe the appropriate projection operators which pick up the mixed parts of thethird-rank tensor representation of SU ( N + 2). The symmetric projection operator S actingon the first two indices of the general rank 3 tensor X abc is described as S X abc = 12 ( X abc + X bac ) . (3.41)Then let us introduce the antisymmetric projection operator A acting on the last two indicesof the above quantity (3.41) as follows: A S X abc = 14 ( X abc + X bac − X acb − X bca ) . (3.42)50urthermore, one projects (3.42) by using the symmetric projection operator S . Then oneobtains, with appropriate an overall normalization,43 S A S X abc = 16 (2 X abc + 2 X bac − X bca − X cba − X acb − X cab ) . (3.43)Note that the quantity (3.43) is symmetric under the interchange of the first two indices. Thisrepresentation comes from the product of the representation and the representation .Similarly, one can try to do other mixed projection operator. First of all, by acting theantisymmetric projection operator on the rank 3 tensor with the first two indices A X abc = 12 ( X abc − X bac ) . (3.44)Then one can act the symmetric projection operator on (3.44) with the last two indices S A X abc = 14 ( X abc − X bac + X acb − X bca ) . (3.45)Finally, one acts the antisymmetric projection operator on (3.45) with the first two indices asfollows: 43 A S A X abc = 16 (2 X abc − X bac − X bca + X cba + X acb − X cab ) . (3.46)Note that the quantity (3.46) is antisymmetric under the interchange of the first two indices.This representation comes from the product of the representation and the representation. One can easily see that the sum of the two projection operators, (3.43) and (3.46), leadsto 16 (4 X abc − X cab − X bca ) . (3.47)It is known that the totally symmetric and antisymmetric projection operators acting on the X abc are given by S X abc = 16 ( X abc + X acb + X cab + X bac + X bca + X cba ) ,A X abc = 16 ( X abc − X acb + X cab − X bac + X bca − X cba ) . (3.48)By adding these two operators in (3.48), one obtains16 (2 X abc + 2 X cab + 2 X bca ) . (3.49)As we expect, one obtains X abc by adding (3.49) to (3.47).51he three-index mixed parts of the SU ( N + 2) representation corresponding to (3.43) canbe obtained from the generators of the fundamental representation of SU ( N + 2) by usingthe projection operator (2 δ il δ jm δ kn + 2 δ jl δ im δ kn − δ jl δ km δ in − δ kl δ jm δ in − δ il δ km δ jn − δ kl δ im δ jn )where i ≤ j ≤ k and l ≤ m ≤ n and i, j, k, l, m, n = 1 , , · · · , ( N + 2) [31]. Then by acting onthe space T a ⊗ ( N +2) × ( N +2) ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ T a ⊗ ( N +2) × ( N +2) + ( N +2) × ( N +2) ⊗ ( N +2) × ( N +2) ⊗ T a , one has the generators for the mixed representation for the SU ( N + 2) asfollows: T a ) il δ jm δ kn + 2( T a ) jl δ im δ kn − ( T a ) jl δ km δ in − ( T a ) kl δ jm δ in − ( T a ) il δ km δ jn − ( T a ) kl δ im δ jn +2 δ il ( T a ) jm δ kn + 2 δ jl ( T a ) im δ kn − δ jl ( T a ) km δ in − δ kl ( T a ) jm δ in − δ il ( T a ) km δ jn − δ kl ( T a ) im δ jn +2 δ il δ jm ( T a ) kn + 2 δ jl δ im ( T a ) kn − δ jl δ km ( T a ) in − δ kl δ jm ( T a ) in − δ il δ km ( T a ) jn − δ kl δ im ( T a ) jn . See also Appendix D . Let us calculate the zero mode for the reduced stress energy ten-sor spin-2 current acting on the state | ( ; 0) > . It turns out that the ( N +2)( N +3)( N +1)3 × ( N +2)( N +3)( N +1)3 = 40 × 40 matrix (see also [32]) is given by k ) diag(3 , , , , , , , , , , , , , , , , , , , , (3.50)3 , , , , , , , , , , , , , , − √ − √ 32 154 − √ − √ 32 154 , , . There are four block diagonal elements after diagonalizing the 35 , , , 38 matrix elements(the eigenvalues are given by k +5) , k +5) , k +5) and k +5) and the eigenfunctions are(0 , , −√ , , ( −√ , , , , (0 , , √ , , ( 1 √ , , , . The 19, 20 elements of (3.50), k +5) and k +5) , correspond to the eigenvalue on the state | ( ; 0) > corresponding to ( , ) − N . See also the ˆ u and ˆ u in Appendix D . We willdescribe the detailed quantum numbers for the other eigenvalues soon.One can also calculate the zero mode for the sum of the square for the spin-1 current withminus sign acting on the above state | ( ; 0) > and the explicit result is given bydiag(0 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , √ √ 32 32 √ √ 32 32 , , , , , , . (3.51)The eigenvalues for the 31, 32, 33 and 34 elements are given by 2, 2, 0 and 0. The correspond-ing eigenfunctions are (0 , , √ , √ , , , 0) (0 , , −√ , −√ , , , 0) as before. The529, 20 elements of (3.51), and , correspond to the eigenvalue on the state | ( ; 0) > . whichbehaves as a doublet under the SU (2).Similarly, one can also compute the zero mode for the spin-1 current acting on the state | ( ; 0) > corresponding to ( , ) − N and one obtains (twice of ˆ u )diag(6 , , , , , , − , − , , , , , − , − , , , − , − , − , − , , , , , , , , , , , − , − , − , − , , , , , − , − . (3.52)In this case, also the 19, 20 elements are the eigenvalues for the state | ( ; 0) > . Then onecan generalize the above eigenvalues for general N as follows: − → − N , ) − N , − → − N ) : ( , ) − N + ( , ) − N , → − N , ) − N + ( , ) − N , → × , ) , by carefully analyzing (3.24) . (mixed; mixed) representation with three boxes Let us consider the higher representation where the mixed representation in SU ( N ) ( , ) survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = 3( k + N + 2) ,l + ( l + + 1)( ; ) = 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 3 . One can calculate the conformal dimension for this representation using the previous formulaor one performs the explicit form for the matrix in this particular representation as in (3.26). The row and columns are characterized by the following triple index notations112 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . See also Appendix D for ˆ u , · · · , ˆ u . For example, the eight ( , ) comes from112 , , , , , , 123 and 132 which do not contain the indices 4 or 5. Similarly, the twelve( , ) − N + ( , ) − N comes from 144 , , , , , , , , , , 345 and 354 which contain asingle SU (3) fundamentals (1 or 2 or 3). The two ( , ) − N comes from 445 and 455 which do not containthe indices 1 or 2 or 3. Finally, the eighteen ( , ) − N + ( , ) − N comes from the remaining ones whichcontain two SU (3) indices. k +5) in the first diagonal elements in (3.50). One can apply for other N values where N = 5 , , , , , · · · . It will turn out that the numerator of the above quantitydoes not depend on N and takes the common value and the denominator is generalized to( k + N +2). By realizing that the quadratic Casimir of SU ( N +2) for the mixed representation C ( N +2) ( ) = N +2) − N +2) (and the one for the mixed representation C ( N ) ( ) = N − N ) andthe correct ˆ u charge is given by 3, the following relation can be obtained3( N + 4 N + 1)2( N + 2)( k + N + 2) − N − N ( k + N + 2) − N ( N + 2)( k + N + 2) = 3( k + N + 2) . Similarly, the quantum number for the l + can be determined by the above matrix calcu-lation in (3.51). From the zero eigenvalues appearing in the first diagonal matrix elements in(3.51), one can see the above l + quantum number, a singlet under the SU (2) k . This also canbe seen from the previous expression ( , ) . The trivial l − quantum number l − = 0 arises.For the last eigenvalue corresponding to ˆ u charge, one uses the previous matrix calculationgiven in (3.52). The eigenvalues appearing in the first diagonal matrix elements in (3.52)imply that the ˆ u charge is given by 3. This is also consistent with the representation ( , ) where the subscript denotes the ˆ u charge. One observes that the above conformal dimensiondoes not satisfy the vanishing BPS bound with l ± = 0. (mixed; antisymm) representation with three and two boxes Let us consider the higher representation where the antisymmetric representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues can be summarized by h ( ; ) = (2 N + 15)4( k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 − N . The explicit form for the matrix in this particular representation is given by (3.50). Then,one obtains k +5) in the second line elements. One can apply for other N values and it willturn out that the numerator of the above quantity does depend on N linearly as well as theconstant term while the denominator is generalized to 4( k + N + 2). One can also use theformula with the correct ˆ u charge3( N + 4 N + 1)2( N + 2)( k + N + 2) − N − N + 1)2 N ( k + N + 2) − (2 − N ) N ( N + 2)( k + N + 2) = (2 N + 15)4( k + N + 2)54here the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.51). From the eigenvalues appearing in the 24, 26, 28, 30 matrix elements (and some linear combination between 35and 36 matrix elements and the linear combination between 37 and 38 matrix elements) in(3.51), one can see the above l + quantum number l + = , a doublet under the SU (2) k . Thisalso can be seen from the previous expression ( , ) − N . The trivial l − = 0 quantum numberholds in this representation. For the last eigenvalue corresponding to ˆ u charge, the previousmatrix calculation given in (3.52) can be used. The eigenvalues appearing in the 24, 26, 28,30 matrix elements (and other two from the linear combinations in the 35, 36, 37 and 38matrix elements) in (3.52) imply that the ˆ u charge is given by . Varying the N values, onefinds that the ˆ u charge is linear in N as well as the constant term. This is also consistentwith the representation ( , ) − N where the subscript denotes the ˆ u charge. One observesthat the above conformal dimension does not satisfy the BPS bound. (mixed; symm) representation with three and two boxes Let us consider the higher representation where the symmetric representation in SU ( N )( , ) − N survives in the branching of (3.24). The four eigenvalues can be summarizedby h ( ; ) = (2 N + 7)4( k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 2 − N . The explicit form for the matrix in this particular representation is given by (3.50). Then,one obtains k +5) in the 3, 4, 10, 11, 15, 16, 23, 25, 27, 29 (and two from 35, 36, 37 and 38)elements. One can apply for other N values and it will turn out that the numerator of theabove quantity does depend on N linearly as well as the constant term while the denominatoris generalized to 4( k + N + 2). One can also use the formula with the correct ˆ u charge3( N + 4 N + 1)2( N + 2)( k + N + 2) − N − N + 2)2 N ( k + N + 2) − (2 − N ) N ( N + 2)( k + N + 2) = (2 N + 7)4( k + N + 2) , where the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.51). From the eigenvalues l + quantum number l + = ,a doublet under the SU (2) k . This also can be seen from the previous expression ( , ) − N .The trivial l − = 0 quantum number holds in this representation. For the last eigenvaluecorresponding to ˆ u charge, the previous matrix calculation given in (3.52) can be used. Theeigenvalues appearing in the previous matrix elements in (3.52) imply that the ˆ u charge isgiven by . Varying the N values, one finds that the ˆ u charge is linear in N as well as theconstant term. This is also consistent with the representation ( , ) − N where the subscriptdenotes the ˆ u charge. One observes that the above conformal dimension does not satisfy theBPS bound. (mixed; f ) representation with three boxes Let us consider the higher representation where the fundamental representation in SU ( N )( , ) − N or ( , ) − N survives in the branching of (3.24). The four eigenvalues are given by h ( ; ) = ( N + 2)( k + N + 2) ,l + ( l + + 1)( ; ) = 2 , or 0 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = 1 − N. The explicit form for the matrix in this particular representation is given by (3.50). Then,one obtains k +5) in the 7, 8, 13, 14, 17, 18, 31, 32, 33, 34, 39 and 40 elements. One canapply for other N values and it will turn out that the numerator of the above quantity doesdepend on N linearly as well as the constant term while the denominator is generalized to( k + N + 2). One can also use the formula with the correct ˆ u charge3( N + 4 N + 1)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − (1 − N ) N ( N + 2)( k + N + 2) = ( N + 2)( k + N + 2) , where the quadratic Casimir C ( N +2) ( ) and C ( N ) ( ) are used. The quantum number forthe l + can be obtained by the above matrix calculation in (3.51). From the eigenvalue 2appearing in the 7, 8, 13, 14, 17, 18, two linear combinations (between 31, 32, 33, and 34),and 39 elements in (3.51), one can see the above l + quantum number l + = 1, a triplet underthe SU (2) k . This also can be seen from the previous expression ( , ) − N . Furthermore,from the eigenvalue 0 appearing in the two linear combinations (between 31, 32, 33, and34) and 40 elements in (3.51), one can see the above l + quantum number l + = 0, a singletunder the SU (2) k . This also can be seen from the previous expression ( , ) − N . The trivial56 − = 0 quantum number holds in this representation. For the last eigenvalue correspondingto ˆ u charge, the previous matrix calculation given in (3.52) can be used. The eigenvaluesappearing in the previous elements in (3.52) imply that the ˆ u charge is given by − 2. Varyingthe N values, one finds that the ˆ u charge is linear in N as well as the constant term. Thisis also consistent with the representation ( , ) − N or ( , ) − N where the subscript denotesthe ˆ u charge. One observes that the above conformal dimension does satisfy the BPS boundwith l + = 1 and l − = 0. There is no BPS bound for l ± = 0. (mixed; 0) representation with three boxes Let us consider the higher representation where the singlet representation in SU ( N ) ( , ) − N survives in the branching of (3.24). The four eigenvalues are given by h ( ; 0) = 3(2 N + 1)4( k + N + 2) ,l + ( l + + 1)( ; 0) = 34 ,l − ( l − + 1)( ; 0) = 0 , ˆ u ( ; 0) = − N . (3.53)The explicit form for the matrix in this particular representation is given by (3.50). Then,one obtains k +5) in the 19 and 20 elements. One can apply for other N values and it willturn out that the numerator of the above quantity does depend on N linearly as well as theconstant term while the denominator is generalized to 4( k + N + 2). One can also use theformula with the correct ˆ u charge3( N + 4 N + 1)2( N + 2)( k + N + 2) − ( − N ) N ( N + 2)( k + N + 2) = 3(2 N + 1)4( k + N + 2) , where the quadratic Casimir C ( N +2) ( ) is used. The quantum number for the l + can beobtained by the above matrix calculation in (3.51). From the eigenvalues appearing inthe 19, 20 elements in (3.51), one can see the above l + quantum number l + = , a doubletunder the SU (2) k . This also can be seen from the previous expression ( , ) − N . The trivial l − = 0 quantum number holds in this representation. For the last eigenvalue correspondingto ˆ u charge, the previous matrix calculation given in (3.52) can be used. The eigenvaluesappearing in the 19, 20 elements in (3.52) imply that the ˆ u charge is given by − . Varyingthe N values, one finds that the ˆ u charge is linear in N . This is also consistent with therepresentation ( , ) − N where the subscript denotes the ˆ u charge. One observes that theabove conformal dimension does not satisfy the BPS bound.57 .5.6 The (mixed; f ) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.5.5 and the latter occurs in the subsection 2.1.1 togetherwith the complex conjugation.In this case, the corresponding four eigenvalues are described by h ( ; ) = ( k + 3 N )2( k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . First of all, one can obtain the following 40 × 40 matrix by calculating the commutator[ T , Q ¯ A − ] as in the subsection 2.2.21(5 + k ) diag(1 , , , , , , − , − , , , , , − , − , , , − , − 23 ; − , − 32 ;1 , , , , , , , , , , − , − , − , − , , , , , − , − 23 ) . (3.54)The eigenvalues (the N generalization is straightforward to obtain) appearing in the 19, 20elements in (3.54) provide the extra contribution as well as the sum of conformal dimensionsof ( ; 0) and (0; ). They are given in (3.53) and (2.6) respectively. Then one obtains thefinal conformal dimension by adding the above contribution appearing in (3.54) as follows(2 k + 3)4( k + N + 2) + 3(2 N + 1)4( k + N + 2) − k + N + 2) = ( k + 3 N )2( k + N + 2) , as in (3.7).It is also useful to interpret the above result from the conformal dimension formula. Onedetermines the following result3( N + 4 N + 1)2( N + 2)( k + N + 2) − ( N − N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 12 = ( k + 3 N )2( k + N + 2) . Here we used the quadratic Casimirs for C ( N +2) ( ) and C ( N ) ( ). The correct ˆ u charge isinserted. The excitation number is given by .For the l + quantum number, due to the vanishing of l + in (0; ), it turns out that the l + isthe same as the one( l + = ) in ( ; 0). For the l − quantum number, due to the vanishing of58 − in ( ; 0), it turns out that the l − is the same as the one( l − = ) in (0; ). It is easy to seethat the above conformal dimension does not satisfy the BPS bound by substituting l ± = .One can add each ˆ u charge and it is obvious that the total ˆ u charge is given by − N − ( N +2)2 which leads to the above result. Note that the ˆ u charge for (0; ) is opposite to the one for(0; ). (mixed; symm) representation with three and two boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.5.5 while the latter occurs in the subsection 2.2.5 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = (4 k + 6 N − k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 0 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − ] and it turns out that1(5 + k ) diag(2 , , , , , , − , − , , , , , − , − , , , − , − 43 ; − , − , , , , , , , , , , − , − , − , − , , , , , − , − 43 ) . (3.55)The two eigenvalues (the N generalization is simply − N + k +2) ) appearing in the 19, 20 el-ements in (3.55) give the extra contribution as well as the sum of conformal dimensions of( ; 0) and (0; ). They are given in (3.53) and (2.17) respectively. Then one obtains thefinal conformal dimension by adding the above contribution appearing in (3.55) as follows3(2 N + 1)4( k + N + 2) + k ( k + N + 2) − k + N + 2) = (4 k + 6 N − k + N + 2) . The conformal dimension formula implies that N + 4 N + 1)2( N + 2)( k + N + 2) − N − N + 2)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = (4 k + 6 N − k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equalto 1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + 59n (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0) and (0; ), the total l − is given by l − = 0, a singlet. Thetotal ˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u chargefor (0; ) is opposite to the one for (0; ). One can easily see that the above conformaldimension does not lead to the BPS bound with l + = and l − = 0. (mixed; antisymm) representation with three and two boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.5.5 while the latter occurs in the subsection 2.2.6 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = (4 k + 6 N − k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 2 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − ] and it turns out that1(5 + k ) diag(2 , , , , , , − , − , , , , , − , − , , , − , − 43 ; − , − , , , , , , , , , , − , − , − , − , , , , , − , − 43 ) . (3.56)The two eigenvalues (the N generalization is simply − N + k +2) ) appearing in the 19, 20 el-ements in (3.56) give the extra contribution as well as the sum of conformal dimensions of( ; 0) and (0; ). They are given in (3.53) and (2.19) respectively. Then one obtains thefinal conformal dimension by adding the above contribution appearing in (3.56) as follows3(2 N + 1)4( k + N + 2) + ( k + 2)( k + N + 2) − k + N + 2) = (4 k + 6 N − k + N + 2) . The conformal dimension formula implies that N + 4 N + 1)2( N + 2)( k + N + 2) − N − N + 1)(2 N )( k + N + 2) − ( − ( N + 2) − N ) N ( N + 2)( k + N + 2) + 1 = (4 k + 6 N − k + N + 2) . The quadratic Casimir C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equal to1. The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + 60n (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0), the total l − is given by l − = 1 in ( ), a triplet. The totalˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u charge for(0; ) is opposite to the one for (0; ). One can easily see that the above conformal dimensiondoes not lead to the BPS bound with l + = and l − = 1. (mixed; antisymm) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.5.5 while the latter occurs in the subsection 2.2.7 withcomplex conjugation.The four eigenvalues are given by h ( ; ) = 3( k + N )2( k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 154 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − Q − ] and it turns out that1(5 + k ) diag(3 , , , , , , − , − , , , , , − , − , , , − , − − , − 92 ;3 , , , , , , , , , , − , − , − , − , , , , , − , − . (3.57)The two eigenvalues (the N generalization is simply − N + k +2) ) appearing in the 19, 20elements in (3.57) give the extra contribution as well as the sum of conformal dimensions of( ; 0) and (0; ). They are given in (3.53) and (2.20) respectively. Then one obtains thefinal conformal dimension by adding the above contribution appearing in (3.57) as follows3(2 N + 1)4( k + N + 2) + 3(2 k + 5)4( k + N + 2) − k + N + 2) = 3( k + N )2( k + N + 2) . The conformal dimension formula implies that3( N + 4 N + 1)2( N + 2)( k + N + 2) − N − N + 1)2( k + N + 2) − ( − N + 2) − N ) N ( N + 2)( k + N + 2) + 32 = 3( k + N )2( k + N + 2) . C ( N ) ( ) is the same as C ( N ) ( ) and the excitation number is equal to . The correct ˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + in (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, dueto the vanishing of l − in ( ; 0), the total l − is given by l − = in ( ), a quartet. The total ˆ u charge is given by − N − ( N + 2) which leads to the above result. Again, the ˆ u charge for(0; ) is opposite to the one for (0; ). One can easily see that the above conformal dimensiondoes not lead to the BPS bound with l + = and l − = 32 15 . (mixed; mixed) representation with three boxes Let us consider the higher representation which arises from the product of ( ; 0) and (0; ).The former occurs in the subsection 3.5.5 while the latter occurs in the subsection 2.2.8.The four eigenvalues are given by h ( ; ) = 3( k + N − k + N + 2) ,l + ( l + + 1)( ; ) = 34 ,l − ( l − + 1)( ; ) = 34 , ˆ u ( ; ) = − N − . One should calculate the commutator [ T , Q − Q − Q − ] and it turns out that1(5 + k ) diag(3 , , , , , , − , − , , , , , − , − , , , − , − − , − 92 ;3 , , , , , , , , , , − , − , − , − , , , , , − , − . (3.58)The two eigenvalues (the N generalization is simply − N + k +2) ) appearing in the 19, 20 matrixelements in (3.58) give the extra contribution as well as the sum of conformal dimensions of( ; 0) and (0; ). They are given in (3.53) and (2.23) respectively. Then one obtains thefinal conformal dimension by adding the above contribution appearing in (3.58) as follows3(2 N + 1)4( k + N + 2) + 3(2 k + 1)4( k + N + 2) − k + N + 2) = 3( k + N − k + N + 2) . For the higher representation ( ; ), the similar analysis can be done by calculating the ˆ u chargecorrectly. N + 4 N + 1)2( N + 2)( k + N + 2) − N − N − ( − N + 2) − N ) N ( N + 2)( k + N + 2) + 32 = 3( k + N − k + N + 2) . The quadratic Casimir C ( N ) ( ) is used and the excitation number is equal to . The correctˆ u charge is inserted. For the l + quantum number, due to the vanishing of l + in (0; ), the l + is the same as the one( l + = ) in ( ; 0). For the l − quantum number, due to the vanishingof l − in ( ; 0), the total l − is given by l − = in ( ), a doublet. The total ˆ u charge isgiven by − N − ( N + 2) which leads to the above result. One can easily see that the aboveconformal dimension does not lead to the BPS bound with l ± = . Let us describe the conformal dimension for the higher representation arising from the productof (symm; 0) and (0; antisymm) where the number of box for the symm representation isgiven by p and the number of box for the antisymm representation is given by q . Then bysubstituting the quadratic Casimirs into the formula and the excitation number is given by q , one obtains the following expression( N + 1) p ( N + p + 2)2( N + 2)( k + N + 2) − ( N + 1) q ( N − q )2 N ( k + N + 2) − ( ( N + 2)( − q ) − Np ) N ( N + 2)( k + N + 2) + q 2= 2 kq + 2 N p + p − pq + 2 p + q + 2 q k + N + 2) . (3.59)According to the conditions p = 2 l + , q = 2 l − , the above expression (3.59) reduces to the BPS bound in (2.7).Now one can classify the possible combinations as follows: p = 0 , q = 1 , , 3; (0; ) , (0; ) , (0; ) ,p = 1 , q = 0 , , , 3; ( ; 0) , ( ; ) , ( ; ) , ( ; ) ,p = 2 , q = 0 , , , 3; ( ; 0) , ( ; ) , ( ; ) , ( ; ) ,p = 3 , q = 0 , , , 3; ( ; 0) , ( ; ) , ( ; ) , ( ; ) . (3.60)Furthermore, there are also complex conjugated representations for (3.60). The quadraticCasimirs do not change and the ˆ u does not change.63et us describe the conformal dimension for the higher representation where the represen-tation Λ − appears in the branching rule of Λ + . The representation arises from (antisymm; antisymm)where the number of box for the first antisymm representation is given by p and the numberof box for the second antisymm representation is given by q . Then the formula implies( N + 2 + 1) p ( N − p + 2)2( N + 2)( k + N + 2) − ( N + 1) q ( N − q )2 N ( k + N + 2) − ( q − N ) N ( N + 2)( k + N + 2)= (2 N p − N q − N p + 10 N p + 2 N q − N q − N − p + 12 p + 6 q )4( N + 2)( k + N + 2) . (3.61)Under the further condition q = p − , the above result (3.61) reduces to (2 N + 3)4( k + N + 2) , which is equal to the BPS bound with l + = and l − = 0: p = 1 , , 3; ( , , ( ; ) , ( ; ) . In this case also, the complex conjugated representations are possible. The quadratic Casimirsdo not change and the ˆ u does not change.In summary of this section, the conformal dimensions for the higher representations upto three boxes are described explicitly. In next two Tables, its large ( N, k ) ’t Hooft like limitis written and the particular ones with “BPS” notation are specified. The large ( N, k ) ’tHooft-like limit is defined by N, k → ∞ , λ ≡ ( N + 1)( N + k + 2) fixed , (3.62)which will be used in Tables. , , currents in the SU ( N +2) SU ( N ) × SU (2) × U (1) Wolf space coset Let us describe the eigenvalues of1) the zero mode of the higher spin-1 current : (Φ (1)0 ) , + ; Λ − ) 00 0 ( − λ ) bps ( − λ ) bps − λ (1 − λ ) bps − λ ( − λ ) bps ( λ ) bps λN ( ) bps − λ − λ − λ ( − λ ) bps ( λ ) bps ( ) bps λ N 2 − λ ( − λ ) bps 2 − λ − λ ( λ ) bps λ ( λ +12 ) bps λN ) bps λ ( λ ) bps λ +12 λN λ ) bps ( λ + ) bps λ ( ) bps 2 λ N λ λ + ( λ ) bps λ N ( λ ) bps λ ( λ +12 ) bps λ λ +22 λ +22 ( λ +22 ) bps λ λ bps λ +12 λ λ λ +22 λ +22 λ ( λ ) bps ( λ ) bps ( λ + ) bps λ λ + ( λ + ) bps λ λ + λ λ + λ bps λ + λ + λ λ λ ( λ ) bps Table 1: The eigenvalue h under the large ( N, k ) ’t Hooft-like limit (3.62). All the eigenvaluesdescribed in the section 3 are presented in this Table and next one. Those in the footnotesin the section 3 and some eigenvalues appearing in next sections are denoted by the boldfacenotation. The subscript “bps” stands for the conformal dimension satisfying the BPS bound(1.1) at finite ( N, k ). According to the last one of the branching rule (3.24) (and its com-plex conjugated one), there is no singlet under the SU ( N ). Therefore, there are no higherrepresentations corresponding to the blanks that can be obtained from the product of zeroand (0; Λ − ) in this Table and next one. Of course, there are higher representations where therepresentation Λ − appears in the branching of Λ + = antisymm or antisymm with three boxesin these Tables. Note that there is a BPS bound for mixed representation described in thesubsection 3 . . V + ) , 3) the zero mode of sum of the square of other higher spin-2 current : ( V − ) , 4) the zero mode of the higher spin-3 current : (Φ (1)2 ) , where the higher spin-2 currents V ± i ( z ) in the SU (2) k × SU (2) N basis are related to the onesΦ (1) ,µν ( z ) in the SO (4) basis V ± ( z ) ≡ i (Φ (1) , ∓ Φ (1) , )( z ) ,V ± ( z ) ≡ − i (Φ (1) , ± Φ (1) , )( z ) ,V ± ( z ) ≡ i ( − Φ (1) , ± Φ (1) , )( z ) . + ; Λ − )0 − λ )2 ( − λ )2 ) bps ( − λ ) ( ( − λ ) ) bps3 − λ − λ − λ ( − λ ) bps3 − λ ( − λ ) bps 3 − λ − λ − λ − λ − λ ( − λ ) bps3 − λ − λ − λ − λ − λ ( − λ ) bps 3 − λ − λ − λ − λ − λ − λ λN 32 32 ( ) bps λN 32 323 λN ( ) bps 3 λ N 32 3232 32 3 λ N 323 λ N Table 2: The (continued) eigenvalue h under the large ( N, k ) ’t Hooft-like limit (3.62). Allthe eigenvalues described in the section 3 are presented in this Table and previous one. Thosein the footnotes in the section 3 and some eigenvalues appearing in next sections are denotedby the boldface notation. The subscript “bps” stands for the conformal dimension satisfyingthe BPS bound (1.1) at finite ( N, k ). Because there are no (0; symm) or (0; symm) repre-sentations with three boxes from the description of the subsection 2 . . 9, there are no higherrepresentations corresponding to the blanks that can be obtained from the product of (Λ + ; 0)and zero in this Table. Of course, there are higher representations where the representationΛ − appears in the branching of Λ + = symm or symm with three boxes. Note that thereis a BPS bound for mixed representation described in the subsection 3 . . SU (2) adjointindices V + ( z ) ≡ X i =1 ( V + i ) ( z ) , V − ( z ) ≡ X i =1 ( V − i ) ( z ) , which have the conformal dimension (or spin) of 4. The corresponding eigenvalues are de-scribed by φ (1)0 , v + , v − and φ (1)2 respectively. See also the relevant works in [33, 34].The higher spin 1 current is described as [35] (see also [36, 37, 38] for fixed N )Φ (1)0 ( z ) = − k + N + 2) d a ¯ b f ¯ a ¯ bc V c ( z ) + k k + N + 2) d a ¯ b Q ¯ a Q ¯ b ( z ) , (4.1)66here the antisymmetric d tensor of rank 2 is given by 4 N × N matrix as follows: d a ¯ b = − − 11 0 0 00 1 0 0 . Each element is N × N matrix. The locations of the nonzero elements of this matrix are thesame as the previous almost complex structure h a ¯ b but numerical values are different fromeach other. Note that the summation over c index in (4.1) runs over the whole range of SU ( N + 2) adjoint indices.This higher spin 1 current plays the role of the ‘generator’ of the next higher spin currentsbecause one can construct them using the OPEs between the spin currents of the large N = 4 nonlinear superconformal algebra and the higher spin 1 current. That is, from thefirst order pole of the OPE between G µ ( z ) and the higher spin 1 current Φ (1)0 ( w ), one obtainsΦ (1) ,µ ( w ) with minus sign [22]. After that, one can calculate the OPE between G µ ( z ) and thehigher spin current Φ (1) ,ν ( w ). Then the first order pole will provide the next higher spin 2current Φ (1) ,µν ( w ). One can go further. The OPE between G µ ( z ) and the higher spin 2 currentΦ (1) ,νρ ( w ) contains the first order pole where the next higher spin current δ µν Φ (1) ,ρ ( w ) occurs.Finally, one can calculate the OPE between G µ ( z ) and the higher spin current Φ (1) ,ν ( w )and then the first order pole gives us to the last higher spin 3 current term δ µν Φ (1)2 ( w ). Oncethe normalization of the higher spin 1 current is fixed, then the normalization for the higherspin 3 current can be fixed in this way.So far, although the complete closed form for the higher spin currents in terms of theadjoint spin 1 and currents is not known, but their expressions for several N values areknown explicitly. They (which are written explicitly for N = 3 , , , , 11 and maybe for N = 13 for some other cases) are enough to obtain all the results of this paper. (0; f ) and (0; f ) representations The relevant subsection is given by the subsection 2.1.1. The above four eigenvalues associatedwith one of the minimal representations can be summarized by φ (1)0 (0; ) = − k ( N + k + 2) ,v + (0; ) = 24 k ( k + N + 2) ,v − (0; ) = 12 k (5 k + 4 N + 2)( k + N + 2) , (1)2 (0; ) = 4 k (12 + 28 N + 5 N + 14 k + 39 kN + + 4 k + N )3(2 + k + N ) (4 + 5 k + 5 N + 6 kN ) . (4.2)For the first eigenvalue, one should calculate the OPE between the “reduced” Φ (1)0 ( z ) and Q ( w ) (in SU (5)) and read off the first order pole. See also (2.5) and (2.15). The coefficientof Q ( w ) in the right hand side of this OPE is the corresponding eigenvalue. In other words,the zero mode of d a ¯ b Q ¯ a Q ¯ b in (4.1) acting on this state gives − N + k + 2). For the secondand third eigenvalues, one calculates the OPEs between the “reduced” V ( ± ) ( z ) and Q ( w )(in SU (5)) and read off the fourth order pole respectively. The coefficients of Q ( w ) in theright hand side of these OPEs are the corresponding eigenvalues respectively. For the lasteigenvalue, one computes the OPE between the “reduced” Φ (1)2 ( z ) and Q ( w ) (in SU (5)) andread off the third order pole. Of course, all the higher spin currents do not contain the spin-1currents V a ( z ).By counting the highest powers of k or N (the sum of powers in k and N for the expressionscontaining both dependences) in the numerators and the denominators appearing in the aboveeigenvalues, one can observe the behaviors under the large ( N, k ) ’t Hooft like limit. Exceptthe v + eigenvalue having N dependence, the remaining three eigenvalues approach to thefinite λ dependent values .Similarly, the other four eigenvalues can be also obtained from φ (1)0 (0; ) = k ( N + k + 2) ,v + (0; ) = 24 k ( k + N + 2) ,v − (0; ) = 12 k (5 k + 4 N + 2)( k + N + 2) ,φ (1)2 (0; ) = − k (12 + 28 N + 5 N + 14 k + 39 kN + + 4 k + N )3(2 + k + N ) (4 + 5 k + 5 N + 6 kN ) . (4.3)Because the generators for the complex conjugated (antifundamental) representation havean extra minus sign compared to the fundamental representation , the eigenvalue for theodd higher spin currents (corresponding to the first and the last ones) have an extra minussign and the ones for the even higher spin currents (corresponding to the two middle ones)remain the same compared to the results of the previous section in (4.2).More explicitly, one can obtain the OPEs between the “reduced” higher spin currents andthe current Q ( w ). By reading off the corresponding coefficients in the appropriate poles, Note that we make some boldface notation for the highest power of ( N, k ) in the numerator of the higherspin 3 current in (4.2). We will observe that they will play the role of the fundamental quantity in the sensethat the eigenvalues of the higher spin 3 current for any representation (0; Λ − ) will be a multiple of thisquantity, under the large ( N, k ) ’t Hooft like limit. N, k ) ’t Hooft likelimit for these eigenvalues. ( f ; 0) and ( f ; 0) representations The relevant subsection is given by the subsection 2.1.2. The four eigenvalues can be describedas φ (1)0 ( ; 0) = − N ( N + k + 2) ,v + ( ; 0) = 12 N (4 k + 5 N + 2)( k + N + 2) ,v − ( ; 0) = 24 N ( k + N + 2) ,φ (1)2 ( ; 0) = − N (12 + 28 k + 5 k + 14 N + 39 kN + N + 4 N + )3(2 + k + N ) (4 + 5 k + 5 N + 6 kN ) . (4.4)One obtains these eigenvalues by substituting the SU ( N + 2) generators T a ∗ into the zeromode of the spin 1 current V a in the corresponding “reduced” higher spin currents whereall the Q a ( z ) dependent terms are ignored. Then one has the unitary matrix acting on thecorresponding state and the diagonal elements of the last 2 × d a ¯ b f ¯ a ¯ bc V c acting on this state is equalto 2 N . The large ( N, k ) ’t Hooft like limit can be analyzed similarly .As observed in [17], under the symmetry N ↔ k and 0 ↔ , the eigenvalues become φ (1)0 ( ; 0) → φ (1)0 (0; ), v + ( ; 0) → v − (0; ), v − ( ; 0) → v + (0; ) and φ (1)2 ( ; 0) → − φ (1)2 (0; ).When we consider the complex conjugated representation, the following results hold φ (1)0 ( ; 0) = N ( N + k + 2) ,v + ( ; 0) = 12 N (4 k + 5 N + 2)( k + N + 2) ,v − ( ; 0) = 24 N ( k + N + 2) ,φ (1)2 ( ; 0) = 4 N (12 + 28 k + 5 k + 14 N + 39 kN + N + 4 N + )3(2 + k + N ) (4 + 5 k + 5 N + 6 kN ) . (4.5) In this case, the highest power of ( N, k ) in the numerator of the higher spin 3 current, denoted by theboldface notation will play the role of the fundamental quantity in the sense that the eigenvalues of the higherspin 3 current for any representation (Λ + ; 0) will be a multiple of this quantity, under the large ( N, k ) ’t Hooftlike limit. , , currents in the SU ( N +2) SU ( N ) × SU (2) × U (1) Wolfspace coset In this section, there are 22 subsections where we consider the explicit 22 higher representa-tions and the same number of subsections with coincident higher representation will appearin section 8 for the linear case. ( f ; f ) representation The relevant subsection on this higher representation is given by 2 . . 1. In this case, whenone takes the N × N subdiagonal unitary matrix inside of ( N + 2) × ( N + 2) unitary matrix,the corresponding diagonal elements for the higher spin currents provide the following foureigenvalues φ (1)0 ( ; ) = 2( k + N + 2) ,v + ( ; ) = 96 k ( k + N + 2) ,v − ( ; ) = 96 N ( k + N + 2) ,φ (1)2 ( ; ) = 8( k − N )( + 5 k + 5 N + 16)3( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.1)Under the symmetry N ↔ k (with ↔ ), the first eigenvalue in (5.1) remains the same,the second eigenvalue becomes the third one, the third eigenvalue becomes the second oneand the last eigevalue remains the same with an extra sign change. By power counting of N and k , one sees that the above eigenvalues behave as N dependence under the large ( N, k )’t Hooft like limit. However, this will play the role of next leading order and moreover thisterm will be the fundamental quantity in the sense that the eigenvalues of the higher spin 3current for any representation (Λ + ; Λ + ) will be a multiple of this quantity. Here Λ + is thesymmetric or antisymmetric representation and the number of boxes is arbitrary. Note the70resence of factor ( k − N ) in the above . ( f ; f ) representation The relevant subsection on this higher representation is given by 2 . . 2. The four eigenvaluescorresponding to the zero modes of the higher spin currents of spins 1 , , φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = ( k − N )( k + N + 2) ,v + ( ; ) = 12(4 kN + 2 k + 5 N − N + 5)( k + N + 2) ,v − ( ; ) = 12(5 k + 4 kN − k + 2 N + 5)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N + 2 k + N + k N − k + + 7 kN − kN − k + 2 N − N − N − . (5.2) The previous relations in (4.3) and (4.4) are used. Note that the eigenvalue in (5.2) forthe higher spin 1 current does not have any contribution from the commutator [(Φ (1)0 ) , Q ¯ A − ]because the OPE between the corresponding higher spin 1 current and the spin currenthas only the first order pole. See also (2.5). This provides only the eigenvalue for therepresentation (0; ). Also the term Q ¯ A − (Φ (1)0 ) acting on the representation ( ; 0) gives theeigenvalue φ (1)0 ( ; 0) with Q ¯ A − acting on the state | ( ; 0) > . By inserting the overall factorinto this state, one has the final state associated with the representation ( ; ). Therefore,one arrives at the above eigenvalue for the higher spin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are the con-tributions from the lower order poles appearing in the commutators, [( V + ) , Q ¯ A − ], [( V − ) , Q ¯ A − ]and [(Φ (1)2 ) , Q ¯ A − ] . They can be summarized by δv + ( ; ) = − N − k + N + 2) , One obtains φ (1)0 ( ; ) = − φ (1)0 ( ; ) , v ± ( ; ) = v ± ( ; ) , φ (1)2 ( ; ) = − φ (1)2 ( ; ) . Note that there are sign changes in the higher spin currents with odd spin. In other words, for example, one has [(Φ (1)2 ) , Q ¯ A − ] = H c dw πi w − +( − H C w dz πi z − Φ (1)2 ( z ) Q ¯ A ( w ).One sees that the OPE between Φ (1)2 ( z ) and Q ¯ A ( w ) contains the first and second order poles as well as thethird order pole. They all contribute to the eigenvalue. v − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 8( N + 6 k + + 20 kN + 16 k + 6 N + 12 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.3)Once again, the large ( N, k ) ’t Hooft like limits for these extra contributions lead to the N behavior. Then the above eigenvalues are obtained from the relations, v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) (5.4)respectively . The previous relations (4.4) and (4.3) can be used. ( f ; symm ) representation This higher representation can be obtained from the product of ( ; 0) and (0; ) . It turnsout that the four eigenvalues are given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − ( N + 2 k )( N + k + 2) ,v + ( ; ) = 12(4 kN + 8 k + 5 N + 38 N + 20)( k + N + 2) ,v − ( ; ) = 24(4 k + 4 kN + 12 k + N )( k + N + 2) ,φ (1)2 ( ; ) = 43( k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N + 73 k N − k − − kN − kN − k − N − N − N − . (5.6) One obtains δv + ( ; ) = 12(18 N + 5)( k + N + 2) , δv − ( ; ) = 60(6 k + 1)( k + N + 2) ,δφ (1)2 ( ; ) = − N + 3 k + kN + 10 kN + 11 k + 3 N + 3 N + 4)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.5)Here the commutators, [( V + ) , Q ¯ A ∗ − ], [( V − ) , Q ¯ A ∗ − ] and [(Φ (1)2 ) , Q ¯ A ∗ − ] are used. Then one also has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with (4.2), (4.5), and (5.5). Note that φ (1)2 ( ; ) is not equal to φ (1)2 ( ; ) even in the large ( N, k )’t Hooft like limit. The subsections 5 . . (1)0 ) , Q − Q − ] (see also the subsection 2.2.5) because the OPE between the correspondinghigher spin 1 current and the product of spin currents has only the first order pole. Thisprovides only the eigenvalue for the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting on the representation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on thestate | ( ; 0) > . By inserting the overall factor into this state, one has the final state associatedwith the representation ( ; ). Therefore, one arrives at the above eigenvalue for the higherspin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],[( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = 48(9 N + 5)( k + N + 2) ,δv − ( ; ) = 384 k ( k + N + 2) ,δφ (1)2 ( ; ) = 8( N + − kN − k − N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.7)It is easy to see that the large ( N, k ) ’t Hooft like limit for these extra contributions lead tothe N behavior . Then the relation (5.6) can be obtained from the previous ones in (4.4) and(5.20) together with (5.7). The highest power terms of ( N, k ) in the higher spin 3 current canbe described by the corresponding terms in (4.4) and (5.20) respectively via simple additionof them. From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.8)together with (5.7), one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; )with (4.5), (5.20) and (5.8). .4 The ( f ; symm ) representation This higher representation can be obtained from the product of ( ; 0) and (0; ). It turnsout that the four eigenvalues are given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − ( N − k )( N + k + 2) ,v + ( ; ) = 12(4 kN + 8 k + 5 N − N − k + N + 2) ,v − ( ; ) = 24(4 k + 4 kN − k + N )( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N − k N − k + − kN − kN − k + 4 N − N + 48 N + 48) . (5.9) The relations (4.4) and the footnote 28 can be used. Note that the eigenvalue for the higherspin 1 current in (5.9) does not have any contribution from the commutator [(Φ (1)0 ) , Q − Q − ](see also the subsection 2.2.5) because the OPE between the corresponding higher spin 1current and the spin current has only the first order pole. This provides only the eigenvaluefor the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting on the representation( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By insertingthe overall factor into this state, one has the final state associated with the representation( ; ). Therefore, one arrives at the above eigenvalue for the higher spin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],[( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = − N − k + N + 2) ,δv − ( ; ) = − k ( k + N + 2) , From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.10)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with (4.5), the footnote 28 and (5.10). φ (1)2 ( ; ) = 8( N + 6 k + + 19 kN + 2 k + 3 N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.11)Then the relation (5.9) can be obtained from the previous ones in (4.4) and the footnote 28together with (5.11). The highest power terms of ( N, k ) in the last eigenvalue of (5.11) istwice of the ones in (5.3). It is easy to see that the highest power terms of ( N, k ) in the higherspin 3 current can be described by the corresponding terms in (4.4) and the footnote (28)respectively and it turns out that it is given by simple addition of two contributions. ( f ; antisymm ) representation It turns out that the four eigenvalues are given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − ( N + 2 k )( N + k + 2) ,v + ( ; ) = 12(4 kN + 8 k + 5 N + 38 N + 20)( k + N + 2) ,v − ( ; ) = 8(16 k + 12 kN + 108 k + 3 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = 43( k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N + 145 k N − k − − kN + 196 kN + 156 k − N + 40 N + 24 N + 48) . (5.12) See also the subsection 2.2.6. The first and the second eigenvalues in (5.12) are the same asthe ones in (5.6) respectively. This implies that there is no difference between or as longas these eigenvalues are concerned. Furthermore, the last eigenvalue has common behaviorwith the one in (5.6) because they contain (24 k N + 6 k N − kN ) in the numerators.Note that the eigenvalue for the higher spin 1 current does not have any contributionfrom the commutator [(Φ (1)0 ) , Q − Q − ] because the OPE between the corresponding higherspin 1 current and the product of spin currents has only the first order pole. This providesonly the eigenvalue for the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting onthe representation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By inserting the overall factor into this state, one has the final state associatedwith the representation ( ; ). Therefore, one arrives at the above eigenvalue for the higherspin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],75( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = 48(9 N + 5)( k + N + 2) ,δv − ( ; ) = 32(26 k + 5)( k + N + 2) ,δφ (1)2 ( ; ) = 8( N + + 23 kN + 22 k + 9 N + 6 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.13)The first eigenvalue in (5.13) is the same as the one in (5.7). For the last eigenvalue, thebehavior of highest power of ( N, k ) is the same as the one in (5.7) . Then the relations(5.12) can be obtained from (4.4) and (5.21) together with (5.13). ( f ; antisymm ) representation Similarly, the four eigenvalues are given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − ( N − k )( N + k + 2) ,v + ( ; ) = 12(4 kN + 8 k + 5 N − N − k + N + 2) ,v − ( ; ) = 8(16 k + 12 kN − k + 3 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N + 11 k N − k + − kN − kN − k + 4 N − N − N − . (5.15) The relations in (4.4) and the footnote 29 are needed. The first and the second eigenvalues in(5.15) are the same as the ones in (5.9). Furthermore, the last eigenvalue has common behaviorwith the one in (5.9) because they contain (24 k N + 18 k N + 12 kN ) in the numerators.See also the subsection 2.2.6. From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.14)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with (4.5), (5.21), and (5.14). (1)0 ) , Q − Q − ] because the OPE between the corresponding higher spin 1current and the spin current has only the first order pole. This provides only the eigenvaluefor the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting on the representation ( ; 0)gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By inserting theoverall factor into this state, one has the final state associated with the representation ( ; ).Therefore, one arrives at the above eigenvalue for the higher spin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],[( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = − N − k + N + 2) ,δv − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 8( N + 6 k + + 43 kN + 32 k + 15 N + 24 N + 16)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.16)The first eigenvalue in (5.16) is the same as the one in (5.11). For the last eigenvalue, thebehavior of highest power of ( N, k ) is the same as the one in (5.11) . Then one obtains therelations in (5.15) from the relations in (4.4) and the footnote 29 with (5.16). ( symm ; 0) representation The relevant subsection on this higher representation is given by 2 . . 3. This higher represen-tation can be obtained from the product of the minimal representation ( f ; 0) and itself. The From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.17)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with (4.5), (5.17) and the footnote 29. φ (1)0 ( ; 0) = − N ( k + N + 2) ,v + ( ; 0) = 32 N (3 k + 4 N + 1)( k + N + 2) ,v − ( ; 0) = 96 N ( k + N + 2) ,φ (1)2 ( ; 0) = − N ( N + 5 k + + 45 kN + 43 k − N − N + 12)3( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.18)It is easy to see that the first eigenvalue is the twice of the one in (4.4) at finite ( N, k ). Exceptthe third eigenvalue, the remaining ones in (5.18) behave as λ dependent constant values underthe large ( N, k ) ’t Hooft like limit. Furthermore, if one sees the last eigenvalue closely, oneobserves that the highest power terms in the numerator are given by − N (6 k N + 12 kN )which is the twice of the ones in (4.4). Note that the denominators in both expressions arethe same at finite ( N, k ). This implies that the eigenvalue for the higher spin 3 current in thishigher representation can be interpreted as the additive quantum number and it is given bythe sum of each eigenvalue for the higher representation in the minimal representation ( ; 0)in (4.4) under the large ( N, k ) ’t Hooft like limit . ( antisymm ; 0) representation The relevant subsection on this higher representation is given by 2 . . 4. The remaining higherrepresentation obtained from the product of the minimal representation ( f ; 0) and itself isgiven by this higher representation and the four eigenvalues can be summarized by φ (1)0 ( ; 0) = − N ( k + N + 2) ,v + ( ; 0) = 96 N ( k + N − k + N + 2) ,v − ( ; 0) = 96 N ( k + N + 2) ,φ (1)2 ( ; 0) = − N ( N + 5 k + + 9 kN − k − N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.19) For the similar higher representation (symm; 0), one has the following results φ (1)0 ( ; 0) = − φ (1)0 ( ; 0) , v ± ( ; 0) = v ± ( ; 0) , φ (1)2 ( ; 0) = − φ (1)2 ( ; 0)together with (5.18). + . The third eigenvalue behaves as N under the large ( N, k ) ’t Hooft likelimit. The last eigenvalue shares the common behavior with the one in (5.18). Although theexact expressions in both cases are different from each other at finite (N,k), the highest powerterms in the numerator given by − N (6 k N + 12 kN ) are the same as the ones in (5.18). Inother words, they are twice of the ones in (4.4). Therefore, this eigenvalue is the sum of eacheigenvalue in the minimal representation ( ; 0) in (4.4) under the large ( N, k ) ’t Hooft likelimit . (0; symm ) representation The relevant subsection on this higher representation is given by 2 . . 5. This higher represen-tation can be obtained from the product of the minimal representation (0; f ) and itself. Thefour eigenvalues with this representation can be described as φ (1)0 (0; ) = − k ( N + k + 2) ,v + (0; ) = 96 k ( k + N + 2) ,v − (0; ) = 96 k ( k + N − k + N + 2) ,φ (1)2 (0; ) = 8 k ( N − k + + 9 kN − k + 5 N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.20)It is easy to see that the first eigenvalue is the twice of the one in (4.2) at finite ( N, k ). Exceptthe second eigenvalue, the remaining ones in (5.20) behave as λ dependent constant valuesunder the large ( N, k ) ’t Hooft like limit. The first two eigenvalues are related to the first andthe third eigenvalues in (5.18) by considering the exchange of N and k . Furthermore, oneobserves that the highest power terms in the numerator are given by 8 k (12 k N + 6 kN ) whichis the twice of the ones in (4.2). Compared to the expression in (5.18), this quantity can beobtained from the corresponding one by using the symmetry under the N ↔ k with extraminus sign. Note that the denominators in both expressions are the same at finite ( N, k ).This implies that the eigenvalue for the higher spin 3 current in this higher representation can For the similar higher representation (antisymm; 0), the following results are satisfied φ (1)0 ( ; 0) = − φ (1)0 ( ; 0) , v ± ( ; 0) = v ± ( ; 0) , φ (1)2 ( ; 0) = − φ (1)2 ( ; 0)together with (5.19). 79e interpreted as the additive quantum number and it is given by the sum of each eigenvaluefor the higher representation in the minimal representation (0; ) in (4.2) under the large( N, k ) ’t Hooft like limit . (0; antisymm ) representation The relevant subsection on this higher representation is given by 2 . . 6. The remaining higherrepresentation obtained from the product of the minimal representation (0; f ) and itself isgiven by this higher representation and the four eigenvalues can be summarized by φ (1)0 (0; ) = − k ( N + k + 2) ,v + (0; ) = 96 k ( k + N + 2) ,v − (0; ) = 32 k (4 k + 3 N + 1)( k + N + 2) ,φ (1)2 (0; ) = 8 k ( N − k + + 45 kN − k + 5 N + 43 N + 12)3( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.21)One observes that the first eigenvalue and the second eigenvalue are the same as the onesin (5.20) respectively. The second eigenvalue behaves as N under the large ( N, k ) ’t Hooftlike limit. The last eigenvalue in (5.21) shares the common behavior with the one in (5.20).Although the exact expressions in both cases are different from each other at finite ( N, k ),the highest power terms in the numerator given by 8 k (12 k N + 6 kN ) are the same as theones in (5.20). They are twice of the ones in (4.2). Therefore, this eigenvalue is the sum ofeach eigenvalue in the minimal representation (0; ) in (4.2) under the large ( N, k ) ’t Hooftlike limit . For the similar complex conjugated higher representation, one obtains the following results φ (1)0 (0; ) = − φ (1)0 (0; ) , v ± (0; ) = v ± (0; ) , φ (1)2 (0; ) = − φ (1)2 (0; )together with (5.20). For the similar higher representation, the following results can be obtained φ (1)0 (0; ) = − φ (1)0 (0; ) , v ± (0; ) = v ± (0; ) , φ (1)2 (0; ) = − φ (1)2 (0; )together with (5.21). .11 The ( symm ; symm ) representation The relevant subsection on this higher representation is given by 3 . . 1. One can describe thefollowing eigenvalues φ (1)0 ( ; ) = 4( N + k + 2) ,v + ( ; ) = 192( k − k + N + 2) ,v − ( ; ) = 192( N + 1)( k + N + 2) ,φ (1)2 ( ; ) = 16( N + 5 k − − kN + 13 k − N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.22)One can construct the matrices (3.2), (3.3) and (3.4) for the corresponding (higher spin)currents. Under the symmetry N ↔ k (with ↔ ), the first eigenvalue remains thesame, and the second eigenvalue becomes the third one and the third eigenvalue becomesthe second one by ignoring the constant term in the numerator. The last eigenvalue in (5.22)remains the same with an extra sign change if we consider only the case where the total powerof N and k is given by 3: (6 k N − kN ) = 6( k − N ) kN . Then one can see that the eigenvalueof the higher spin 3 current is the twice of the one in (5.1) under the large ( N, k ) ’t Hooftlike limit. By power counting of N and k , one sees that the above eigenvalues behave as N dependence under the large ( N, k ) ’t Hooft like limit . ( symm ; f ) representation The relevant subsection on this higher representation is given by 3 . . 2. The following eigen-values in this higher representation can be determined by φ (1)0 ( ; ) = − ( N − N + k + 2) ,v + ( ; ) = 12(4 kN + 16 k + 5 N + 14 N + 12)( k + N + 2) ,v − ( ; ) = 24(5 N − k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) The following relations hold φ (1)0 ( ; ) = − φ (1)0 ( ; ) , v ± ( ; ) = v ± ( ; ) , φ (1)2 ( ; ) = − φ (1)2 ( ; )with (5.22). ( N − k N − k + + 87 kN + 106 kN − k + 4 N + 42 N + 140 N + 24) . (5.23)One observes that the first, the second and the last eigenvalues in (5.23) coincide with theones in (4.4) if one takes the higher order terms in the numerators respectively. The thirdeigenvalue behaves as N under the large ( N, k ) ’t Hooft like limit . ( symm ; f ) representation The relevant subsection on this higher representation is given by 3 . . 4. The four eigenvaluescorresponding to the zero modes of the higher spin currents of spins 1 , , φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = ( k − N )( k + N + 2) ,v + ( ; ) = 8(12 kN + 3 k + 16 N − N + 20)( k + N + 2) ,v − ( ; ) = 12(5 k + 4 kN − k + 8 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N + 4 k + N − k N − k + + 35 kN − kN − k − N − N − N − . (5.24) The relations (4.3) and (5.18) are used. Note that the eigenvalue for the higher spin 1current in (5.24) does not have any contribution from the commutator [(Φ (1)0 ) , Q ¯ A − ] becausethe OPE between the corresponding higher spin 1 current and the spin current has onlythe first order pole. This provides only the eigenvalue for the representation (0; ). Also theterm Q ¯ A − (Φ (1)0 ) acting on the representation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q ¯ A − acting on the state | ( ; 0) > . By inserting the overall factor into this state, one hasthe final state associated with the representation ( ; ). Therefore, one arrives at the aboveeigenvalue for the higher spin 1 current. All the eigenvalues survive even under the large( N, k ) ’t Hooft like limit. As expected, one can rewrite the highest power terms in terms of(24 kN + 12 k N ) and k (12 k N + 6 kN ). The former comes from (5.18) and the latter comesfrom (4.3). The following relations can be obtained φ (1)0 ( ; ) = − φ (1)0 ( ; ) , v ± ( ; ) = v ± ( ; ) , φ (1)2 ( ; ) = − φ (1)2 ( ; )with (5.23). V + ) , Q ¯ A − ], [( V − ) , Q ¯ A − ]and [(Φ (1)2 ) , Q ¯ A − ]. They can be summarized by δv + ( ; ) = − N − k + N + 2) ,δv − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 8( N + 13 k + + 41 kN + 32 k + 10 N + 24 N + 16)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.25)In particular, the second and last eigenvalues in (5.25) are twice of the ones in (5.3) as longas the highest power of ( N, k ) is concerned. Then as done in (5.4), the final eigenvalues aredetermined in (5.24). Of course, the eigenvalues in (5.25) vanish under the large ( N, k ) ’tHooft like limit . As before, the relations (5.24) can be determined by (4.3) and (5.18)together with (5.25). ( symm ; antisymm ) representation The relevant subsection on this higher representation is given by 3 . . 5. The four eigenvaluesare characterized by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − k + N )( k + N + 2) ,v + ( ; ) = 32(3 kN + 3 k + 4 N + 29 N + 20)( k + N + 2) ,v − ( ; ) = 32(4 k + 3 kN + 53 k + 3 N + 20)( k + N + 2) , One obtains δv + ( ; ) = 32(14 N + 5)( k + N + 2) , δv − ( ; ) = 240(3 k + 1)( k + N + 2) ,δφ (1)2 ( ; ) = − N + 5 k + + 19 kN + 22 k + 8 N + 6 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.26)Here the commutators, [( V ± ) , Q ¯ A ∗ − ] and [(Φ (1)2 ) , Q ¯ A ∗ − ] are used. Then one also has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 26, (4.2), and (5.26). (1)2 ( ; ) = 83( k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + 100 k N − k − − kN + 132 kN + 144 k + 2 N + 67 N + 24 N + 48) . (5.27)The relations in (5.18) and (5.21) are used. In the third eigenvalue of (5.27), the quantitiesin the highest power of ( N, k ) are the same as the ones in (5.12). For the highest power of( N, k ) in the last eigenvalue, the 12 k N is the same as the one in (5.12) while the − kN isthe twice of the one in (5.12).Here the relevant extra contributions from the various commutators are used in thesecalculations δv + ( ; ) = 128(7 N + 5)( k + N + 2) ,δv − ( ; ) = 128(13 k + 5)( k + N + 2) ,δφ (1)2 ( ; ) = 16( N − k + + 22 kN + 22 k + 11 N + 6 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.28)The last eigenvalue in (5.28) for the highest power of ( N, k ) is twice of the one in (5.13) .In (5.27), the relations (5.18), (5.21) and (5.28) are combined together. ( symm ; antisymm ) representation The foue eigenvalues are given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − − k + N )( k + N + 2) ,v + ( ; ) = 32(3 kN + 3 k + 4 N − N + 20)( k + N + 2) , From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.29)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 26, (5.29) and (5.21). One can analyze the eigenvalues on this particular higher representation in the context of sections 2 and3. − ( ; ) = 32(4 k + 3 kN − k + 3 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N − k N − k + − kN − kN − k − N − N − N − . (5.30)The third eigenvalue of (5.30) in the highest power of ( N, k ) is the same as the one in (5.15).The highest power terms in the last eigenvalue of (5.30) are twice of the ones in (5.2).We have δv + ( ; ) = − N − k + N + 2) ,δv − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 16( N + 7 k + + 44 kN + 32 k + 13 N + 24 N + 16)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.31)The highest powers of ( N, k ) in the second and third eigenvalues in (5.31) are twice of theones in (5.16) . In (5.30), the relations (5.18), the footnote 29 and (5.31) are combinedtogether. ( symm ; symm ) representation The relevant subsection on this higher representation is given by 3 . . 6. The four eigenvaluesare φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = 2( − N + k )( k + N + 2) ,v + ( ; ) = 32(3 kN + 3 k + 4 N − N + 20)( k + N + 2) , From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.32)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 26, (5.32) and the footnote 29. − ( ; ) = 96( k + kN − k + N )( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N − k N − k + − kN − kN − k − N − N + 48 N + 48) . (5.33)The last eigenvalue with boldface notation in (5.33) is the same as the one in (5.30). The thirdeigenvalue has the common behavior with the one in (5.9) under the large ( N, k ) ’t Hooft likelimit. Note that the eigenvalue for the higher spin 1 current does not have any contributionfrom the commutator [(Φ (1)0 ) , Q − Q − ] because the OPE between the corresponding higherspin 1 current and the spin current has only the first order pole. This provides onlythe eigenvalue for the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting on therepresentation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By inserting the overall factor into this state, one has the final state associatedwith the representation ( ; ). Therefore, one arrives at the above eigenvalue for the higherspin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],[( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = − N − k + N + 2) ,δv − ( ; ) = − k ( k + N + 2) ,δφ (1)2 ( ; ) = 16( N + 7 k + + 20 kN + 2 k + N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.34)In particular, the last eigenvalue in (5.34) for the highest power of ( N, k ) in the numeratoris exactly four times the ones in (5.3) or twice of the ones in (5.25) . Note that the second One obtains δv + ( ; ) = 128(7 N + 5)( k + N + 2) , δv − ( ; ) = 768 k ( k + N + 2) ,δφ (1)2 ( ; ) = − N − k + − kN − k − N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.35)The commutators [( V ± ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ] are used. Then one also has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) , ( antisymm ; antisymm ) representation The relevant subsection on this higher representation is given by 3 . . 1. The four eigenvaluesare given by φ (1)0 ( ; ) = 4( k + N + 2) ,v + ( ; ) = 192( k + 1)( k + N + 2) ,v − ( ; ) = 192( N − k + N + 2) ,φ (1)2 ( ; ) = 16( N + 5 k − + 18 kN + 43 k − N − N + 12)3( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.36)Under the symmetry N ↔ k (with ↔ ), the first eigenvalue in (5.36) remains the same,and the second eigenvalue becomes the third one, and the third eigenvalue becomes the secondone by ignoring the constant term in the numerator. The last eigenvalue in (5.36) remainsthe same with an extra sign change if we consider only the case where the total power of N and k is given by 3: (6 k N − kN ). Compared to the ones in (5.22), the eigenvalues lookssimilar to each other. The large ( N, k ) behavior is the same. By power counting of N and k ,one sees that the above eigenvalues behave as N dependence under the large ( N, k ) ’t Hooftlike limit . ( antisymm ; f ) representation The relevant subsection on this higher representation is given by 3 . . 2. The four eigenvaluesare φ (1)0 ( ; ) = − ( N − k + N + 2) , v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 26, (5.20), and (5.35). It turns out that the following relations hold φ (1)0 ( ; ) = − φ (1)0 ( ; ) , v ± ( ; ) = v ± ( ; ) , φ (1)2 ( ; ) = − φ (1)2 ( ; )with (5.36). + ( ; ) = 12(4 kN + 5 N − N + 12)( k + N + 2) ,v − ( ; ) = 24(5 N − k + N + 2) , (5.37) φ (1)2 ( ; ) = − N − N + 5 k + + 39 kN + 28 k + 4 N + 14 N + 12)3( k + N + 2) (6 kN + 5 k + 5 N + 4) . One observes that the first, the second and the last eigenvalues in (5.37) coincide with theones in (4.4) if one takes the higher order terms in the numerators respectively. The thirdeigenvalue behaves as N under the large ( N, k ) ’t Hooft like limit . The large ( N, k ) behaviorof the eigenvalue for the higher spin 3 current is the same as the one in (5.23). ( antisymm ; f ) representation The relevant subsection on this higher representation is given by 3 . . 4. The four eigenvaluesare given by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = ( k − N ) k + N + 2 ,v + ( ; ) = 24(4 kN + k + 4 N − N )( k + N + 2) ,v − ( ; ) = 12(5 k + 4 kN − k + 8 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N + 4 k + N − k N − k + − kN − kN − k − N − N − N − . (5.38) The relations (5.19) and (4.3) are used. The large ( N, k ) behavior of the eigenvalue in (5.38)for the higher spin 3 current is the same as the one in (5.24). Note that the eigenvalue forthe higher spin 1 current does not have any contribution from the commutator [(Φ (1)0 ) , Q ¯ A − ]because the OPE between the corresponding higher spin 1 current and the spin currenthas only the first order pole. This provides only the eigenvalue for the representation (0; ).Also the term Q ¯ A − (Φ (1)0 ) acting on the representation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0)with Q ¯ A − acting on the state | ( ; 0) > . By inserting the overall factor into this state, onehas the final state associated with the representation ( ; ). Therefore, one arrives at the It turns out that the following relations hold φ (1)0 ( ; ) = − φ (1)0 ( ; ) , v ± ( ; ) = v ± ( ; ) , φ (1)2 ( ; ) = − φ (1)2 ( ; )with (5.37). N, k ) ’t Hooft like limit.For the eigenvalues corresponding to the remaining higher spin currents, there are the con-tributions from the lower order poles appearing in the commutators, [( V + ) , Q ¯ A − ], [( V − ) , Q ¯ A − ]and [(Φ (1)2 ) , Q ¯ A − ]. They can be summarized by δv + ( ; ) = − N ( k + N + 2) ,δv − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 8( N + 9 k + + 33 kN + 22 k + 10 N + 14 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.39)In particular, the second and last eigenvalues in (5.39) are twice of the ones in (5.3), and theones in (5.25) as long as the highest power of ( N, k ) is concerned. Then as done in (5.4), thefinal eigenvalues are determined in (5.38). Of course, the eigenvalues in (5.39) vanish underthe large ( N, k ) ’t Hooft like limit . ( antisymm ; symm ) representation The relevant subsection on this higher representation is given by 3 . . 5. The four eigenvaluesare summarized by φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − k + N ) k + N + 2 ,v + ( ; ) = 96( kN + k + N + 7 N )( k + N + 2) , One obtains δv + ( ; ) = 384 N ( k + N + 2) , δv − ( ; ) = 240(3 k + 1)( k + N + 2) ,δφ (1)2 ( ; ) = − N + 9 k + + 27 kN + 32 k + 8 N + 16 N + 16)( k + N + 2) (6 kN + 5 k + 5 N + 4) , (5.40)where the commutators [( V ± ) , Q ¯ A ∗ − ] and [(Φ (1)2 ) , Q ¯ A ∗ − ] are used. Then one also has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 27, (4.2), and (5.40). − ( ; ) = 96( k + kN + 7 k + N )( k + N + 2) ,φ (1)2 ( ; ) = 83( k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + 64 k N + 11 k − + 8 kN + 36 kN + 2 N + N − N − . (5.41)Again the relations (5.19) and (5.20) are used. The third eigenvalue (5.41) in the highestpower of ( N, k ) is the twice of the one in (5.6). The large ( N, k ) behavior of the eigenvaluefor the higher spin 3 current is the same as the one in (5.27).The following results can be obtained similarly δv + ( ; ) = 768 N ( k + N + 2) ,δv − ( ; ) = 768 k ( k + N + 2) ,δφ (1)2 ( ; ) = 16( N + 3 k + + 6 kN + 2 k − N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.42)The second eigenvalue in (5.42) is the twice as the one in (5.7) while the last eigenvalue sharesthe common value in the highest power of ( N, k ) with the twice of the one in (5.7) . Thenthe above expressions (5.41) can be obtained explicitly by using the relations (5.19), (5.20)and (5.42). ( antisymm ; symm ) representation The four eigenvalues are φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = − − k + N ) k + N + 2 , From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.43)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 27, (5.43) and (5.20). One can also analyze the eigenvalues on this particular higher representation in the context of sections 2and 3. + ( ; ) = 96( kN + k + N − N )( k + N + 2) ,v − ( ; ) = 96( k + kN − k + N )( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N − k N − k + − kN − kN + 36 k − N − N + 84 N + 96) . (5.44)The last eigenvalue with boldface notation in (5.44) is the same as the one in (5.30) or (5.33).Here the previous relations (5.19) and the footnote 28 are used. In particular, the large ( N, k )behavior in the eigenvalue for the higher spin 3 current is the same as the one in (5.30).The following relations hold δv + ( ; ) = − N ( k + N + 2) ,δv − ( ; ) = − k ( k + N + 2) ,δφ (1)2 ( ; ) = 16( N + 3 k + + 12 kN − k + N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.45)The highest power of ( N, k ) in the last eigenvalue of (5.45) is the twice of the ones in (5.16) . ( antisymm ; antisymm ) representation The relevant subsection on this higher representation is given by 3 . . 6. The four eigenvaluesare φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) = 2( k − N )( k + N + 2) , From the relations δv ± ( ; ) = δv ± ( ; ) , δφ (1)2 ( ; ) = − δφ (1)2 ( ; ) , (5.46)one has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 27, (5.46) and the footnote 28. + ( ; ) = 96( kN + k + N − N )( k + N + 2) ,v − ( ; ) = 32(4 k + 3 kN − k + 3 N + 20)( k + N + 2) ,φ (1)2 ( ; ) = − k + N + 2) (6 kN + 5 k + 5 N + 4) × ( N − k + N − k N − k + − kN − kN − k − N − N − N − . (5.47)The last eigenvalue with boldface notation in (5.47) is the same as the one in (5.30), (5.33)or (5.44). See also (3.22). Note that the eigenvalue for the higher spin 1 current does nothave any contribution from the commutator [(Φ (1)0 ) , Q − Q − ] because the OPE between thecorresponding higher spin 1 current and the spin current has only the first order pole.This provides only the eigenvalue for the representation (0; ). Also the term Q − Q − (Φ (1)0 ) acting on the representation ( ; 0) gives the eigenvalue φ (1)0 ( ; 0) with Q − Q − acting on thestate | ( ; 0) > . By inserting the overall factor into this state, one has the final state associatedwith the representation ( ; ). Therefore, one arrives at the above eigenvalue for the higherspin 1 current.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutators, [( V + ) , Q − Q − ],[( V − ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ]. They can be summarized by δv + ( ; ) = − N ( k + N + 2) ,δv − ( ; ) = − k − k + N + 2) ,δφ (1)2 ( ; ) = 16( N + 3 k + + 36 kN + 22 k + 13 N + 14 N + 8)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.48)It is obvious that the last eigenvalue for the highest power of ( N, k ) in the numerator in (5.48)is exactly four times the ones in (5.3), twice of the ones in (5.25), or the ones in (5.34) .Then the above result (5.47) can be obtained from (5.19), the footnote 29 and (5.48). The One obtains δv + ( ; ) = 768 N ( k + N + 2) , δv − ( ; ) = 128(13 k + 5)( k + N + 2) ,δφ (1)2 ( ; ) = − N + 3 k + + 30 kN + 32 k + 11 N + 16 N + 16)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (5.49) N, k ) behavior in the eigenvalue for the higher spin 3 current is the same as the one in(5.30).In summary of this section, we present the relevant Tables where the large ( N, k ) ’t Hooftlimit for the eigenvalues is taken. We will observe that for the linear case, the correspondingeigenvalues behave as exactly same as these Tables under the large ( N, k ) ’t Hooft like limit.At finite ( N, k ), the eigenvalues, φ (1)0 and v ± in linear case coincide with the ones in thenonlinear case. Only the eigenvalues φ (1)2 are different from each other at finite ( N, k ).For the eigenvalue for the higher spin 1 current on the representation (Λ + ; Λ − ) which canbe obtained the product of (Λ + ; 0) and (0; Λ − ), one obtains φ (1)0 (Λ + ; Λ − ) = φ (1)0 (Λ + ; 0) + φ (1)0 (0; Λ − ) = " ∓| Λ + | N ∓ | Λ − | k ( N + k + 2) (Λ + ; Λ − ) . (5.50)There are four cases depending on whether the representation Λ ± is given by the multipleproduct of or . We have minus (plus) sign for the former (latter) in (5.50). We also denotethe number of boxes as | Λ ± | . SU ( N +2) SU ( N ) coset The explicit relation between the 11 currents of the large N = 4 “nonlinear” superconformalalgebra and the 16 currents (with boldface notation) of the large N = 4 “linear” supercon-formal algebra is described by [10] T µν ( z ) = T µν ( z ) − i (2 + k + N ) Γ µ Γ ν ( z ) ,G µ ( z ) = G µ ( z ) − i (2 + k + N ) UΓ µ ( z ) − ε µνρσ " i k + N ) Γ ν Γ ρ Γ σ − k + N ) T νρ Γ σ ( z ) ,T ( z ) = T ( z ) + 1(2 + k + N ) " UU − ∂ Γ µ Γ µ ( z ) . (6.1) The commutators [( V ± ) , Q − Q − ] and [(Φ (1)2 ) , Q − Q − ] are used. Then one also has φ (1)0 ( ; ) = φ (1)0 ( ; 0) + φ (1)0 (0; ) ,v ± ( ; ) = v ± ( ; 0) + v ± (0; ) + δv ± ( ; ) ,φ (1)2 ( ; ) = φ (1)2 ( ; 0) + φ (1)2 (0; ) + δφ (1)2 ( ; ) , together with the footnote 27, (5.21), and (5.49). + ; Λ − ) 00 0 − (1 − λ ) (1 − λ ) − − λ ) − − λ ) 2(1 − λ ) 2(1 − λ ) − λ λN (1 − λ ) − (2 − λ ) − (2 − λ ) (2 − λ ) (2 − λ ) λ − (1 − λ ) − λN − (2 − λ ) − (2 − λ ) (2 − λ ) (2 − λ ) − λ − λ (1 − λ ) λN − − λ ) 2(1 − λ ) − λ − λ (1 − λ ) − λN − λ ) 2(1 − λ )2 λ − (1 − λ ) λ − − λ ) − − λ ) − λN λ − (1 − λ ) λ − − λ ) − − λ ) 2 − λN Table 3: The eigenvalue φ (1)0 under the large ( N, k ) ’t Hooft-like limit (3.62). The eigenvaluewith Λ + = Λ − can be written in terms of the multiple of the eigenvalue of ( f ; f ) or ( f ; f ).The N behavior in this case is written explicitly in this Table and next ones. The generalstructure for the eigenvalue with Λ + = Λ − (in the product of (Λ + ; 0) and (0; Λ − )) is given bythe linear combinations of the one of (0; f ) (or (0; f )) and the one of ( f ; 0) (or ( f ; 0)). Theneach coefficient depends on the the number of boxes in Λ + and Λ − . When the representationΛ − appears in the branching of Λ + , the eigenvalue leads to the representation ( | Λ + | − | Λ − | ; 0)where | Λ ± | denotes the number of boxes. We also present the eigenvalues in terms of the ’tHooft coupling constant λ . See also the first Table in section 9.Here the four fermionic spin currents are given by [6, 17] Γ ( z ) = − i N + 1) h j ˜ a ˜ b f ˜ a ˜ b ˜ c h j ˜ c ˜ d Q ˜ d ( z ) , Γ j ( z ) = − i N + 1) h j ˜ a ˜ b f ˜ a ˜ b ˜ c Q ˜ c ( z ) , (6.2)where j = 1 , , j in the first equation of (6.2) . The bosonic spin1 current is given by U ( z ) = − N + 1) h j ˜ a ˜ b f ˜ a ˜ b ˜ c h j ˜ c ˜ d " V ˜ d − k + N + 2) f ˜ d ˜ e ˜ f Q ˜ e Q ˜ f ( z ) , where there is no sum over the index j . Of course, the 11 currents in (2.4) or (6.1) are regularin the OPEs between them and the spin currents Γ µ ( z ) and the spin 1 current U ( z ).The spin 2 current in the N = 4 linear superconformal algebra has extra two terms in(6.1), compared to the one in the N = 4 nonlinear superconformal algebra. Moreover, theOPE between the current Γ µ ( z ) and the adjoint spin current Q ¯ A ∗ ( w ) is regular. Therefore,the Γ µ term does not contribute to the eigenvalue for the representation (0; Λ − ). Furthermore,the current Γ µ ( z ) contains only the adjoint spin current Q ¯ A ∗ ( w ) from (6.1). This implies One should change the index structures as follows: Γ → − i Γ , Γ → − i Γ , Γ → i Γ and Γ → − i Γ in order to use (6.1) from (6.2). We introduce the coset GH = SU ( N +2) SU ( N ) notation ˜ a = (¯ a, ˆ a ) where the ¯ a indexruns over 4 N values as before and the index ˆ a associates with the 2 × SU (2) × U (1)and runs over 4 values. The 4 × h i ˆ a ˆ b appearing in (6.2) are given in [17]. + ; Λ − ) 00 0 N (1 − λ ) λ N (1 − λ ) λ λ (4 + λ ) N (1 − λ ) λ λ (4 + λ )12 λ (4 + λ ) 12 λ (4 + λ ) N (1 − λ ) λ λ ( λ + 3) 12 λ ( λ + 4) 32 λ ( λ + 3)96 λ λ ( λ + 4) 96 λ λ ( λ + 3) 32 λ ( λ + 3) 12 λ ( λ + 4)96 λ λ λ ( λ + 4)Table 4: The eigenvalue v + under the large ( N, k ) ’t Hooft-like limit (3.62). Due to the behav-ior of N in the representation (0; Λ − ), the final eigenvalues for the representation (Λ + ; Λ − ) (inthe product of (Λ + ; 0) and (0; Λ − )) come from the contribution from (Λ + ; 0). Furthermore,for the representation (Λ + ; Λ + ), the eigenvalues are given by the multiple of the fundamentalquantity which is given by the eigenvalue for the representation ( f ; f ) (or ( f ; f )). When therepresentation Λ − appears in the branching of Λ + , the eigenvalue leads to the representation( | Λ + | − | Λ − | ; 0) where | Λ ± | denotes the number of boxes.that the Γ µ term does not contribute to the eigenvalue for the representation (Λ + ; 0) also.Then one obtains the following identity h (Λ + ; Λ − ) = h (Λ + ; Λ − ) + 1( N + k + 2) 1 N ( N + 2) ˆ u (Λ + ; Λ − ) . Among four eigenvalues studied in sections 2 and 3, the eigenvalues for the spin 2 currentare different from the ones in sections 6, 7 and 8. The quantum numbers l ± is the samebecause the Γ µ dependent terms in the adjoint spin 1 current (6.1) do not contribute to thefinal eigenvalue equations. Therefore, we present only the eigenvalue h (Λ + ; Λ − ) with theeigenvalues for the higher spin 3 current in next sections. , , currents in the SU ( N +2) SU ( N ) coset The higher spin 1 current in the linear version is the same as the previous higher spin 1current in (4.1). The higher spin 2 current in the linear version behaves as the one in thenonlinear version exactly as long as the eigenvalues are concerned (note that the higher spin currents contain the product of adjoint spin 1 current and adjoint spin current. See alsothe equation (4 . 7) of [22]). Therefore, one needs to have the eigenvalues for the higher spin95Λ + ; Λ − )0 N (1 − λ ) λ N (1 − λ ) λ N (1 − λ ) λ N (1 − λ ) λ λ ( λ + 4) 12 λ ( λ + 4) 12 λ ( λ + 4) 12 λ ( λ + 4)12 λ ( λ + 4) 12 λ ( λ + 4) 12 λ ( λ + 4) 12 λ ( λ + 4) N (1 − λ ) λ λ ( λ + 3) 32 λ ( λ + 3) 32 λ ( λ + 3)96 λ N (1 − λ ) λ λ λ λ ( λ + 3) 32 λ ( λ + 3) N (1 − λ ) λ λ ( λ + 3)96 λ λ λ N (1 − λ ) λ Table 5: The (continued) eigenvalue v + under the large ( N, k ) ’t Hooft-like limit (3.62). Dueto the behavior of N in the representation (0; Λ − ), the final eigenvalues for the representation(Λ + ; Λ − ) (in the product of (Λ + ; 0) and (0; Λ − )) come from the contribution from (Λ + ; 0).Furthermore, for the representation (Λ + ; Λ + ), the eigenvalues are given by the multiple ofthe fundamental quantity which is given by the eigenvalue for the representation ( f ; f ). Onthe other hand, for the representation (Λ + ; Λ − ) where the representation Λ − appears in thebranching of Λ + , the eigenvalues can be written in terms of the eigenvalue for ( f ; 0).3 current only. In next two sections, the previous eigenvalues φ (1)0 and v ± acting on (Λ + ; Λ − )described in sections 4 and 5 are still valid.Recall that the higher spin 3 current in the linear version [39] is Φ ( ) ( z ) = Φ (1)2 ( z ) − c LΦ ( ) ( z ) − c ∂ Φ ( ) ( z ) − c ∂ UΦ ( ) ( z ) − c U ∂ Φ ( ) ( z ) (7.1) − c UUΦ ( ) ( z ) − c ∂ Γ µ Φ ( ) ,µ ( z ) − c Γ µ ∂ Φ ( ) ,µ ( z ) − c ∂ Γ µ Γ µ Φ ( ) ( z ) → Φ (1)2 ( z ) − c LΦ ( ) ( z ) − c ∂ Φ ( ) ( z ) − c ∂ UΦ ( ) ( z ) − c U ∂ Φ ( ) ( z ) − c UUΦ ( ) ( z ) , where the c , c and c terms do not contribute to the eigenvalues for the representation(Λ + ; Λ − ). Recall that the higher spin currents Φ ( ) ,µ ( z ) contain the product of adjoint spin1 current and adjoint spin current. Here the ( N, k ) dependent coefficients are given by c = 16( k − N )(6 kN + 5 k + 5 N + 4) , c = − ( k − N )(6 kN + 29 k + 29 N + 52)3( k + N + 2)(6 kN + 5 k + 5 N + 4) ,c = − k + N + 2) , c = 4( k + N + 2) ,c = 16( k − N )( k + N + 2)(6 kN + 5 k + 5 N + 4) ,c = 6 i (2 + k + N ) , c = − i (2 + k + N ) ,c = − k − N )( k + N + 2)(6 kN + 5 k + 5 N + 4) . (7.2)96Λ + ; Λ − ) 00 0 12( λ − λ − 1) 12( λ − λ − N λ N λ λ − λ − N λ λ − λ − N λ N λ N λ λ − λ − N λ N λ λ − λ − N λ λ − λ − N λ N λ λ − λ − N λ Table 6: The eigenvalue v − under the large ( N, k ) ’t Hooft-like limit (3.62). Due to the be-havior of N in the representation (Λ + ; 0), the final eigenvalues for the representation (Λ + ; Λ − )(in the product of (Λ + ; 0) and (0; Λ − )) come from the contribution from (0; Λ − ). One can seethe N behavior in the representation (Λ + ; Λ + ).(Λ + ; Λ − )0 96(1 − λ ) 32( λ − λ − 1) 96(1 − λ ) 32( λ − λ − − λ ) 32( λ − λ − 1) 96(1 − λ ) 32( λ − λ − − λ ) 32( λ − λ − 1) 96(1 − λ ) 32( λ − λ − N λ λ − λ − 1) 96(1 − λ ) 32( λ − λ − − λ ) N λ − λ ) 32( λ − λ − − λ ) 32( λ − λ − N λ λ − λ − − λ ) 32( λ − λ − 1) 96(1 − λ ) N λ Table 7: The (continued) eigenvalue v − under the large ( N, k ) ’t Hooft-like limit (3.62). Dueto the behavior of N in the representation (Λ + ; 0), the final eigenvalues for the representation(Λ + ; Λ − ) (in the product of (Λ + ; 0) and (0; Λ − )) come from the contribution from (0; Λ − ).One can see the N behavior in the representation (Λ + ; Λ + ).In the primary basis [40], the following quantity can be constructed e Φ ( ) ( z ) ≡ Φ ( ) ( z ) + p ∂ Φ ( ) ( z ) + p LΦ ( ) ( z ) → Φ (1)2 ( z ) + ( p − c ) LΦ ( ) ( z ) + ( p − c ) ∂ Φ ( ) ( z ) − c ∂ UΦ ( ) ( z ) − c U ∂ Φ ( ) ( z ) − c UUΦ ( ) ( z ) (7.3)where the two quantities are introduced p ≡ − ( k − N )(3 kN + 16 k + 16 N + 29)3( k + N + 2)(3 kN + 4 k + 4 N + 5) , p ≡ k − N )(3 kN + 4 k + 4 N + 5) . (7.4) The relation (7.1) is used in (7.3). In the remaining parts of this paper, the higher spin 3current is given by (7.3). We will examine the corresponding eigenvalues by considering both97 Λ + ; Λ − ) 00 0 φ (1)2 (0; ) − φ (1)2 (0; )= (1 − λ )(2 − λ ) = − (1 − λ )(2 − λ ) φ (1)2 ( ; 0) φ (1)2 ( ; ) φ (1)2 [( ; 0) − (0; )]= − λ ( λ + 1) = − N λ (2 λ − 1) = − ( λ − λ + 1) − φ (1)2 ( ; 0) − φ (1)2 [( ; 0) − (0; )] − φ (1)2 ( ; )= λ ( λ + 1) = ( λ − λ + 1) = N λ (2 λ − φ (1)2 ( ; 0) φ (1)2 ( ; 0) φ (1)2 [2( ; 0) − (0; )]= − λ ( λ + 1) = − λ ( λ + 1) = − (3 λ − λ + 2)2 φ (1)2 ( ; 0) φ (1)2 ( ; 0) φ (1)2 [2( ; 0) − (0; )]= − λ ( λ + 1) = − λ ( λ + 1) = − (3 λ − λ + 2) − φ (1)2 ( ; 0) − φ (1)2 [2( ; 0) − (0; )] − φ (1)2 ( ; 0)= λ ( λ + 1) = (3 λ − λ + 2) = λ ( λ + 1) − φ (1)2 ( ; 0) − φ (1)2 [2( ; 0) − (0; )] − φ (1)2 ( ; 0)= λ ( λ + 1) = (3 λ − λ + 2) = λ ( λ + 1) Table 8: The eigenvalue φ (1)2 under the large ( N, k ) ’t Hooft-like limit (3.62). The generalstructure for the eigenvalue with Λ + = Λ − in the product of (Λ + ; 0) and (0; Λ − ) is given bythe linear combinations of the one of (0; f ) (or (0; f )) and the one of ( f ; 0) (or ( f ; 0)). Theneach coefficient depends on the the number of boxes in Λ + and Λ − . When the representationΛ − appears in the branching of Λ + , the eigenvalue leads to the representation ( | Λ + | − | Λ − | ; 0)where | Λ ± | denotes the number of boxes. The eigenvalue with Λ + = Λ − can be written interms of the multiple of the eigenvalue of ( f ; f ) or ( f ; f ). The N behavior in this case iswritten explicitly in this Table and next one. We also present the eigenvalues in terms of the’t Hooft coupling constant λ .the higher spin 3 current in the nonlinear version and the other remaining parts separately inorder to see how they (Φ (1)2 ( z ) and e Φ ( ) ( z )) behave under the large ( N, k ) ’t Hooft like limit. (0; f ) and (0; f ) representations The relevant subsections are given by 2.1.1 and 4.1. The two eigenvalues associated with oneof the minimal representations can be summarized by h (0; ) = ( N k + 2 N + 1)2 N ( N + k + 2) ,φ ( ) (0; ) = 4 k ( N + 5 k N + + 24 kN + 16 kN − k + 4 N + 23 N + 18 N )3 N ( k + N + 2) (3 kN + 4 k + 4 N + 5) . (7.5) For the first eigenvalue, one should calculate the OPE between T ( z ) and Q ( w ) (in SU (5))and read off the second order pole. The coefficient of Q ( w ) in the right hand side of thisOPE is the corresponding eigenvalue. For the second eigenvalue, one computes the OPE98 φ (1)2 (0; ) 2 φ (1)2 (0; ) − φ (1)2 (0; ) − φ (1)2 (0; )= (1 − λ )(2 − λ ) = (1 − λ )(2 − λ ) = − (1 − λ )(2 − λ ) = − (1 − λ )(2 − λ ) φ (1)2 [( ; 0) + 2(0; )] φ (1)2 [( ; 0) + 2(0; )] φ (1)2 [( ; 0) − φ (1)2 [( ; 0) − ( λ − λ + 4) = ( λ − λ + 4) = − (3 λ − λ + 4) = − (3 λ − λ + 4) − φ (1)2 [( ; 0) − − φ (1)2 [( ; 0) − − φ (1)2 [( ; 0) + 2(0; )] − φ (1)2 [( ; 0) + 2(0; )]= (3 λ − λ + 4) = (3 λ − λ + 4) = − ( λ − λ + 4) = − ( λ − λ + 4)2 φ (1)2 ( ; ) 2 φ (1)2 [( ; 0) + (0; )] 2 φ (1)2 [( ; 0) − (0; )] 2 φ (1)2 [( ; 0) − (0; )]= − N λ (2 λ − 1) = − (2 λ − 1) = − ( λ − λ + 1) = − ( λ − λ + 1)2 φ (1)2 [( ; 0) + (0; )] 2 φ (1)2 ( ; ) 2 φ (1)2 [( ; 0) − (0; )] 2 φ (1)2 [( ; 0) − (0; )]= − (2 λ − 1) = − N λ (2 λ − 1) = − ( λ − λ + 1) = − ( λ − λ + 1) − φ (1)2 [( ; 0) − (0; )] − φ (1)2 [( ; 0) − (0; )] − φ (1)2 ( ; ) − φ (1)2 [( ; 0) + (0; )]= ( λ − λ + 1) = ( λ − λ + 1) = N λ (2 λ − 1) = (2 λ − − φ (1)2 [( ; 0) − (0; )] − φ (1)2 [( ; 0) − (0; )] − φ (1)2 [( ; 0) + (0; )] − φ (1)2 ( ; )= ( λ − λ + 1) = ( λ − λ + 1) = (2 λ − 1) = N λ (2 λ − Table 9: The (continued) eigenvalue φ (1)2 under the large ( N, k ) ’t Hooft-like limit (3.62).The general structure for the eigenvalue with Λ + = Λ − in the product of (Λ + ; 0) and (0; Λ − )is given by the linear combinations of the one of (0; f ) (or (0; f )) and the one of ( f ; 0) (or( f ; 0)). Then each coefficient depends on the the number of boxes in Λ + and Λ − . When therepresentation Λ − appears in the branching of Λ + , the eigenvalue leads to the representation( | Λ + | − | Λ − | ; 0) where | Λ ± | denotes the number of boxes. The eigenvalue with Λ + = Λ − canbe written in terms of the multiple of the eigenvalue of ( f ; f ) or ( f ; f ). The N behavior inthis case is written explicitly in this Table and previous one. We also present the eigenvaluesin terms of the ’t Hooft coupling constant λ .between e Φ ( ) ( z ) and Q ( w ) (in SU (5)) and read off the third order pole. Of course, all thehigher spin currents do not contain the spin-1 currents V a ( z ). By counting the highest powersof k or N (the sum of powers in k and N for the expressions containing both dependences)in the numerators and the denominators appearing in the above eigenvalues, one can observethe behaviors under the large ( N, k ) ’t Hooft like limit and these eigenvalues approach to thefinite values .Similarly, the other two eigenvalues can be also obtained from h (0; ) = ( N k + 2 N + 1)2 N ( N + k + 2) ,φ ( ) (0; ) = − k ( N + 5 k N + + 24 kN + 16 kN − k + 4 N + 23 N + 18 N )3 N ( k + N + 2) (3 kN + 4 k + 4 N + 5) . (7.6) The highest power terms of the eigenvalue for the higher spin 3 current are exactly the same as the onesin (4.2) although the denominators are little different from each other at finite ( N, k ). current Q ( w ). By reading off the correspondingcoefficients in the appropriate poles, the above eigenvalues can be determined. ( f ; 0) and ( f ; 0) representations The relevant subsections are given by 2.1.2 and 4 . 2. The two eigenvalues can be described as h ( ; 0) = ( N + 1)( N + 3)2( N + 2)( N + k + 2) ,φ ( ) ( ; 0) = − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × N ( N + 10 k N + 8 k + + 36 kN + 71 kN + 43 k + 5 N + 26 N + 50 N + 30) . (7.7)One obtains these eigenvalues by substituting the SU ( N + 2) generators T a ∗ into the zeromode of the spin 1 current V a in the corresponding “reduced” (higher) spin currents whereall the Q a ( z ) dependent terms are ignored. Then one has the unitary matrix acting on thecorresponding state and the diagonal elements of the last 2 × N ↔ k and 0 ↔ , theeigenvalues do not remain the same, compared to (7.5). However, the large ( N, k ) ’t Hooftlimit provide the coincident values with minus sign .When we consider the complex conjugated representation, the following results hold h ( ; 0) = ( N + 1)( N + 3)2( N + 2)( N + k + 2) ,φ ( ) ( ; 0) = 43( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × N ( N + 10 k N + 8 k + + 36 kN + 71 kN + 43 k + 5 N + 26 N + 50 N + 30) . (7.8)According to the previous analysis, the first eigenvalue remains the same and the secondeigenvalue has the extra minus signs, compared to the ones in (7.7). One can use h ( ; 0) + N + k +2) 1 N ( N +2) ˆ u ( ; 0) = (2 N +3)4( N + k +2) + ( − N ) ( N + k +2) 1 N ( N +2) to obtain the eigenvalue in(7.7). Eigenvalues for the higher representations with thehigher spin- , , currents in the SU ( N +2) SU ( N ) coset In this section, there are 22 subsections where we consider the explicit 22 higher represen-tations as done in section 5. The same analysis done in section 5 is applied to this section.Note that each subsection in this section corresponds to each section of section 5. For therepresentations where Λ − appears in the branching of Λ + , we write down the eigenvaluesfor the higher spin 3 current completely. On the other hand, for the representations wherethe product of (Λ + ; 0) and (0; Λ − ) occurs, we describe some relations for them rather thanpresenting the complete expressions because they are rather involved due to the presence ofimaginary i . ( f ; f ) representation In this case, when one takes the N × N subdiagonal unitary matrix inside of ( N + 2) × ( N + 2)unitary matrix, the corresponding diagonal elements for the higher spin currents provide thefollowing two eigenvalues h ( ; ) = ( N + 1) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )= − N − k )( + 10 kN + 8 kN + 4 N + 19 N + 22 N + 6)3 N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) . (8.1)The previous relation in (5.1) is used. The first eigenvalue can be interpreted as follows. Byrecalling that the stress energy tensor in the linear version has the extra contribution fromthe spin 1 current, one can write down the precise relation from (2.10) h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = 1( N + k + 2) + 1 N ( N + 2)( k + N + 2) , which is exactly the same as the first eigenvalue of (8.1). By power counting of N and k , onesees that the above eigenvalues behave as N dependence under the large ( N, k ) ’t Hooft likelimit.Let us introduce the difference between the higher spin 3 currents in the linear and non-linear version from (7.3) Φ ( z ) ≡ " ( p − c ) LΦ ( ) + ( p − c ) ∂ Φ ( ) − c ∂ UΦ ( ) − c U ∂ Φ ( ) − c UUΦ ( ) ( z ) . (8.2)101he coefficients are given by (7.2) and (7.4). Let us denote the eigenvalue of the zero modeof this field as ∆ and it turns out that the eigenvalue in the above higher representation isdescribed as∆( ; ) = − k − N )( − k + + 12 N + 7 N − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.3)This can be obtained as follows. Let us denote the zero mode eigenvalues for LΦ ( ) , ∂ Φ ( ) , ∂ UΦ ( ) , U ∂ Φ ( ) , and UUΦ ( ) as l φ ( ) , ∂ φ ( ) , ∂ u φ ( ) , u ∂φ ( ) , and uu φ ( ) . Then theseeigenvalues for the representation ( , ) are given by " kN + 2 kN + N + 5 N + 6 N + 1) N ( N + 2)( k + N + 2) , k + N + 2) , − i q N ( N + 2)( k + N + 2) , − i q N ( N + 2)( k + N + 2) , − N ( N + 2)( k + N + 2) . (8.4)Then by substituting the coefficients (7.2) and (7.4) into the relation (8.2) together with (8.4)one sees the final result in (8.3). Note that although the third and fourth eigenvalues in (8.4)contain the imaginary i with same ( N, k ) coefficients, due to the vanishing of ( c + c ) in (7.2),there is no contribution for these eigenvalues. It is easy to see that the simple power countingof ( N, k ) implies that (8.3) behaves as N under the large ( N, k ) ’t Hooft limit . In particular,the highest power of ( N, k ) in the eigenvalue for the higher spin 3 current in (8.1) plays therole of next leading order N and this will be the basic quantity for the arbitrary representation(Λ + ; Λ + ) because the eigenvalue will be the multiple of the eigenvalue for ( , ). ( f ; f ) representation The two eigenvalues corresponding to the zero modes of the (higher spin) currents of spins 2and 3 which act on the representation ( ; ) can be summarized by h ( ; ) = ( kN + 2 kN + N + 6 N + 8 N + 2)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) One has the following eigenvalue in the complex conjugated representation ( ; )) φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; ) = − φ (1)2 ( ; ) − ∆( ; )together with (5.1) and (8.3). See also the footnote 18. The extra minus sign comes from the spin 3 associatedwith the cubic generators of SU ( N + 2). ( N − k )( − N − k N − k + + 4 kN − kN − k + 2 N + 8 N + 20 N + 24) . (8.5) The eigenvalue for the higher spin 3 current in the nonlinear version can be found in (5.2).Note that the eigenvalue for the spin 2 current does not have any contribution from thecommutator [( T ) , Q ¯ A − ]. This provides only the eigenvalue for the representation (0; ). Alsothe term Q ¯ A − ( T ) acting on the representation ( ; 0) gives the eigenvalue h ( ; 0) with Q ¯ A − acting on the state | ( ; 0) > . By inserting the overall factor into this state, one has the finalstate associated with the representation ( ; ). Therefore, one arrives at the above eigenvaluefor the spin 2 current . As before, one can describe this eigenvalue in the context of thecorresponding eigenvalue nonlinear version. That is, h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = 12 + ( − N − N ( N + 2)( k + N + 2) One can calculate the eigenvalue equations [22] for each term in (8.2) l φ ( ) ( ; 0) = − N (cid:0) kN + 4 k + 3 N + 12 N + 11 (cid:1) N + 2)( k + N + 2) ( ; 0) = − l φ ( ) ( ; 0) ,∂ φ ( ) ( ; 0) = − N ( k + N + 2) ( ; 0) = − ∂ φ ( ) ( ; 0) ,∂ u φ ( ) ( ; 0) = − iN q NN +2 k + N + 2) ( ; 0) = − ∂ u φ ( ) ( ; 0) , u ∂φ ( ) ( ; 0) = − iN q NN +2 k + N + 2) ( ; 0) = − u ∂φ ( ) ( ; 0) , uu φ ( ) ( ; 0) = N N + 2)( k + N + 2) ( ; 0) = − uu φ ( ) ( ; 0) . (8.6)Similarly, one has the following eigenvalue equations l φ ( ) (0; ) = − k (3 kN + 2 N + 6 N + 1)2 N ( k + N + 2) (0; ) = − l φ ( ) (0; ) ,∂ φ ( ) (0; ) = − k ( k + N + 2) (0; ) = − ∂ φ ( ) (0; ) ,∂ u φ ( ) (0; ) = − ik q N +2 N k + N + 2) (0; ) = ∂ u φ ( ) (0; ) , u ∂φ ( ) (0; ) = − ik q N +2 N k + N + 2) (0; ) = u ∂φ ( ) (0; ) , uu φ ( ) (0; ) = k ( N + 2)4 N ( k + N + 2) (0; ) = − uu φ ( ) (0; ) . (8.7)In the third and fourth eigenvalues, there are no sign changes under the ↔ . h (0; ) + h ( ; 0) = ( kN + 2 N + 1)2 N ( k + N + 2) + ( N + 1)( N + 3)2( N + 2)( k + N + 2) . In the first line, the relations in (2.11) are used. In the second line, the previous relations in(7.6) and (7.7) are used.For the eigenvalue corresponding to the higher spin 3 current, there exists the contributionfrom the lower order poles appearing in the commutator [( Φ ( ) ) , Q ¯ A − ]. One describes thefollowing eigenvalues for each term appearing in (8.2) " − (2 kN + 2 kN − k + 6 N + 15 N + 6)2( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)2( k + N + 2) , − i q NN +2 (3 k + N + 2)2( k + N + 2) , ( − kN − k + 5 N + 16 N + 12)4( N + 2)( k + N + 2) . (8.8) Note that the eigenvalues in the third and the fourth are different from each other. This is thereason why there exists an imaginary i term in the eigenvalue in (8.5). Then by substitutingthe coefficients (7.2) and (7.4) into the relation (8.2) together with (8.8) one sees the finalresult in (8.5).Furthermore, there are ∆( ; 0) and ∆(0; ). They can be obtained explicitly. Once again,the large ( N, k ) ’t Hooft like limit for ∆( ; 0) + ∆(0; ) lead to the N behavior. Note thatthe eigenvalues appearing in (8.6) and (8.7) are finite but the coefficients in (7.2) and (7.4)behave as N . Similarly, one can easily see that due to the finiteness in (8.8), the final extraterms in (8.5) vanish under the large ( N, k ) ’t Hooft limit. Therefore, the eigenvalues for thehigher spin 3 current in this higher representation in both linear and nonlinear versions arethe same as each other . In the similar higher representation, one obtains h ( ; ) = ( kN + 2 kN + N + 6 N + 8 N + 2)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " − i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )(27 k N + 86 k N + 57 k + 9 kN + 64 kN + 119 kN + 48 k + 14 N + 44 N + 20 N − , where the last two terms in the eigenvalue of higher spin 3 current can be determined from the followingeigenvalues " − (4 kN + 10 kN + 3 k + 6 N + 15 N + 6)2( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)2( k + N + 2) , i q NN +2 (3 k + N + 2)2( k + N + 2) , .3 The ( f ; symm ) representation This higher representation can be obtained from the product of ( ; 0) and (0; ). It turnsout that the two eigenvalues are given by h ( ; ) = (2 kN + 4 kN + N + 6 N + 11 N + 8)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " − i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 146 k N + 105 k + 31 kN + 104 kN + 78 k − N − N − N − . (8.9) The previous relation (5.6) can be inserted. Note that the eigenvalue for the spin 2 currentdoes not have any contribution from the commutator [( T ) , Q − Q − ]. This provides onlythe eigenvalue for the representation (0; ). Also the term Q − Q − ( T ) acting on therepresentation ( ; 0) gives the eigenvalue h ( ; 0) with Q − Q − acting on the state | ( ; 0) > .By inserting the overall factor into this state, one has the final state associated with therepresentation ( ; ). Therefore, one arrives at the above eigenvalue for the spin 2 current.Furthermore, one can interpret the above eigenvalue of spin 2 current from the one in thenonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 4 k + 7)4( N + k + 2) + ( − N + N + 2) N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + N + 2) N ( k + N + 2) + ( N + 1)( N + 3)2( N + 2)( k + N + 2) . In the first line, the first result can be obtained from the conformal dimension formula inprevious sections and the ˆ u charge is the sum of ˆ u charges of (0; ) and ( ; 0). In the secondline the previous relation (7.7) is used and the relation (8.17) is used.Furthermore, there are ∆( ; 0) and ∆(0; ). The former can be obtained from therelation (8.6) while the latter is given by (8.18).For the second eigenvalue corresponding to the higher spin 3 current, there is the contri-bution from the lower order poles appearing in the commutator, [( Φ ( ) ) , Q − Q − ]. By using − (5 kN + 12 k − N − N − N + 2)( k + N + 2) . See also (7.8) and (7.5) for the eigenvalue of the spin 2 current. The eigenvalues ∆( ; 0) and ∆(0; ) can bedetermined from the previous relations (8.6) and (8.7). Furthermore, the footnote 20 is needed for the explicitexpression of φ (1)2 ( ; ). " − (4 kN + 10 kN + 3 k + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6) k + N + 2 , i q NN +2 (3 k + N + 2) k + N + 2 , − (11 kN + 24 k − N − N − N + 2)( k + N + 2) , one determines the last two terms in the eigenvalue of the higher spin 3 current in (8.9) . ( f ; symm ) representation This higher representation can be obtained from the product of ( ; 0) and (0; ). It turnsout that the two eigenvalues are given by h ( ; ) = (2 kN + 4 kN + N + 6 N + 11 N + 8)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 130 k N + 87 k − − kN + 16 kN + 42 k − N − N − N − . (8.10) The previous relation in (5.9) can be inserted. The eigenvalue for the spin 2 current does nothave any contribution from the commutator [( T ) , Q − Q − ]. This provides only the eigen-value for the representation (0; ). Also the term Q − Q − ( T ) acting on the representation( ; 0) gives the eigenvalue h ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By insertingthe overall factor into this state, one has the final state associated with the representation( ; ). Therefore, one arrives at the above eigenvalue for the spin 2 current.Furthermore, one can interpret the above eigenvalue of spin 2 current from the one in thenonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 4 k − N + k + 2) + ( − N − N − N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + N + 2) N ( k + N + 2) + ( N + 1)( N + 3)2( N + 2)( k + N + 2) . For the similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the last terms in (8.9) with an extra minus sign because the cubic generators can have an extra minussign under the complex conjugated representation. Note that the quantity φ (1)2 ( ; ) can be obtained in thefootnote 22. Of course, ∆( ; 0) can be determined from the previous relation (8.6). u charge is the sum of ˆ u charges of (0; ) and ( ; 0). In the secondline the previous relation (7.7) is used and the relation (8.17) is used.Furthermore, there are ∆( ; 0) and ∆(0; ). The former can be obtained from the relation(8.6) while the latter is given by (8.18) together with the footnote 56.For the eigenvalue corresponding to the higher spin 3 current (the corresponding eigenvalueof the higher spin 3 current in the nonlinear version appears in (5.9)), there is the contribu-tion from the lower order poles appearing in the commutator, [( Φ ( ) ) , Q − Q − ]. From thefollowing eigenvalues for each term in (8.2) " − (2 kN + 2 kN − k + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − (13 kN + 24 k − N − N − N + 2)( k + N + 2) , and the known results for the coefficients in (7.2) and (7.4), the above eigenvalue of (8.10)can be observed . ( f ; antisymm ) representation The two eigenvalues are given by h ( ; ) = (2 kN + 4 kN + N + 10 N + 19 N + 8)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " − i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 146 k N + 105 k + + 140 kN + 78 k + 11 N + 26 N − N − . (8.11) The relation in (5.12) can be inserted. Note that the eigenvalue for the spin 2 current does nothave any contribution from the commutator [( T ) , Q − Q − ]. This provides only the eigen-value for the representation (0; ). Also the term Q − Q − ( T ) acting on the representation One has the following eigenvalues for the similar higher representation φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the last terms in (8.10) with an extra minus sign. Note that the quantity φ (1)2 ( ; ) can be obtained inthe footnote 23. Of course, ∆( ; 0) can be determined from the previous relation (8.6). Moreover ∆(0; )can be determined from the footnote 56. 107 ; 0) gives the eigenvalue h ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By insertingthe overall factor into this state, one has the final state associated with the representation( ; ). Therefore, one arrives at the above eigenvalue for the spin 2 current.Furthermore, one can interpret the above eigenvalue of spin 2 current from the one in thenonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 4 k + 15)4( N + k + 2) + ( − N + N + 2) N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 3 N + 2) N ( k + N + 2) + ( N + 1)( N + 3)2( N + 2)( k + N + 2) . In the first line, the first result can be obtained from the conformal dimension formula inprevious sections and the ˆ u charge is the sum of ˆ u charges of (0; ) and ( ; 0). In the secondline the previous relation (7.7) is used and the relation (8.20) is used.Furthermore, there are ∆( ; 0) and ∆(0; ). The former can be obtained from the relation(8.6) while the latter is given by (8.21).For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutator [( Φ ( ) ) , Q − Q − ]. Bycombining the following eigenvalues with the various coefficients " − (4 kN + 10 kN + 3 k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)( k + N + 2) , i q NN +2 (3 k + N + 2)( k + N + 2) , − (11 kN + 24 k − N − N − N + 2)( k + N + 2) , the last two terms in eigenvalue of the higher spin 3 current of (8.11) can be determinedexplicitly . ( f ; antisymm ) representation The two eigenvalues are described by h ( ; ) = (2 kN + 4 kN + N + 10 N + 19 N + 8)2 N ( N + 2)( k + N + 2) , One has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the last terms in (8.11) with an extra minus sign. Note that the quantity φ (1)2 ( ; ) can be obtained inthe footnote 24. Of course, ∆( ; 0) can be determined from the previous relation (8.6). Moreover ∆(0; )can be determined from (8.21). ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 130 k N + 87 k − − kN + 52 kN + 42 k − N − N − N − . (8.12) One can insert the relation (5.15). Note that the eigenvalue for the spin 2 current does nothave any contribution from the commutator [( T ) , Q − Q − ]. This provides only the eigen-value for the representation (0; ). Also the term Q − Q − ( T ) acting on the representation( ; 0) gives the eigenvalue h ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By insertingthe overall factor into this state, one has the final state associated with the representation( ; ). Therefore, one arrives at the above eigenvalue for the spin 2 current.Furthermore, one can interpret the above eigenvalue of spin 2 current from the one in thenonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 4 k + 7)4( N + k + 2) + ( − N − N − N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 3 N + 2) N ( k + N + 2) + ( N + 1)( N + 3)2( N + 2)( k + N + 2) . In the first line, the first result can be obtained from the conformal dimension formula inprevious sections and the ˆ u charge is the sum of ˆ u charges of (0; ) and ( ; 0). In the secondline the previous relation (7.7) is used and the relation (8.20) is used.Furthermore, there are ∆( ; 0) and ∆(0; ). The former can be obtained from the relation(8.6) while the latter is given by (8.21) with the footnote 57.For the eigenvalues corresponding to the remaining higher spin currents, there are thecontributions from the lower order poles appearing in the commutator [( Φ ( ) ) , Q − Q − ]. Bycombining the following eigenvalues with the various coefficients " − (2 kN + 2 kN − k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − (13 kN + 24 k − N − N − N + 2)( k + N + 2) , the last two terms in eigenvalue of the higher spin 3 current of (8.12) can be determinedexplicitly One has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) .7 The ( symm ; 0) representation This higher representation can be obtained from the product of the minimal representation( f ; 0) and itself. The two eigenvalues with this representation can be described as h ( ; 0) = ( N + 1)( N + 4)( N + 2)( k + N + 2) ,φ ( ) ( ; 0) = φ (1)2 ( ; 0) + ∆( ; 0)= − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × N ( N + 10 k N + 8 k + + 39 kN + 89 kN + 58 k + 2 N + 8 N + 35 N + 30) . (8.13)The relation (5.18) is used. It is easy to see that the first eigenvalue is the twice of the one in(7.7) under the ( N, k ) ’t Hooft limit. One also realizes that the description from the nonlinearanalysis for the spin 2 current implies the following expression (with the help of (2.13)) h ( ; 0) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; 0) = ( N + 2)( N + k + 2) + N N ( N + 2)( k + N + 2) , which coincides with the above eigenvalue in (8.13). Furthermore, if one sees the last eigen-value closely, one observes that the highest power terms in the numerator (8.13) are givenby − N (3 k N + 6 kN ) which is the twice of the ones in (7.7). Note that the denominatorsin both expressions are the same at finite ( N, k ). This implies that the eigenvalue for thehigher spin 3 current in this higher representation can be interpreted as the additive quantumnumber and it is given by the sum of each eigenvalue for the higher representation in theminimal representation ( ; 0) under the large ( N, k ) ’t Hooft like limit.The difference between the nonlinear and linear cases which vanishes in the large N ’tHooft limit is given by∆( ; 0) = − N ( k − N )( − kN − k − − N − N − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.14)More explicitly, the eigenvalues for each term in (8.2) have the following form " − N ( kN + 2 k + 2 N + 9 N + 8)( N + 2)( k + N + 2) , − N ( k + N + 2) , − iN q NN +2 ( k + N + 2) , − iN q NN +2 ( k + N + 2) , N ( N + 2)( k + N + 2) . plus the last terms in (8.12) with an extra minus sign. Note that the quantity φ (1)2 ( ; ) can be obtained inthe footnote 25. Of course, ∆( ; 0) can be determined from the previous relation (8.6). Moreover ∆(0; )can be determined from (8.21) together with the footnote 57. i , theyare the same and their coefficients are opposite to each other. This leads to the absence ofimaginary i in (8.14). One can easily see that under the large ( N, k ) ’t Hooft like limit botheigenvalues for the higher spin 3 currents in the nonlinear and linear versions coincide witheach other because the expression in (8.14) behaves as N . ( antisymm ; 0) representation This higher representation can be obtained from the product of the minimal representation( f ; 0) and itself. The two eigenvalues with this representation can be described as h ( ; 0) = N ( N + 3)( N + 2)( k + N + 2) ,φ ( ) ( ; 0) = φ (1)2 ( ; 0) + ∆( ; 0)= − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × N ( N + 10 k N + 8 k + + 21 kN + 17 kN − k + 2 N − N − N − . (8.15)It is easy to see that the first eigenvalue is the twice of the one in (7.7) under the ( N, k ) ’tHooft limit. One also realizes that the description from the nonlinear analysis for the spin 2current implies the following expression (with the help of the subsection 2.2.4: in particular(2.14)) h ( ; 0) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; 0) = N ( N + k + 2) + N N ( N + 2)( k + N + 2) . which coincides with the above eigenvalue in (8.15). Furthermore, if one sees the last eigen-value closely (8.15), one observes that the highest power terms in the numerator are givenby − N (3 k N + 6 kN ) which is the twice of the ones in (7.7). Note that the denominatorsin both expressions are the same at finite ( N, k ). This implies that the eigenvalue for thehigher spin 3 current in this higher representation can be interpreted as the additive quantumnumber and it is given by the sum of each eigenvalue for the higher representation in theminimal representation ( ; 0) under the large ( N, k ) ’t Hooft like limit. One has φ ( ) ( ; 0) = φ (1)2 ( ; 0) + ∆( ; 0) . Note that here the quantity ∆( ; 0) can be obtained from the corresponding quantity ∆( ; 0) by changingthe sign. We need to have the footnote 26 for the quantity φ (1)2 ( ; 0). N ’tHooft limit is given by∆( ; 0) = − N ( k − N )( − k − − N − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.16)This can be determined by the following eigenvalues " − N ( kN + 2 k + 2 N + 7 N + 4)( N + 2)( k + N + 2) , − N ( k + N + 2) , − iN q NN +2 ( k + N + 2) , − iN q NN +2 ( k + N + 2) , N ( N + 2)( k + N + 2) , together with the coefficients. Note that the highest order terms in (8.16) are the same asthe ones in (8.14). One can easily see that the above quantity (8.16) vanishes under the large( N, k ) ’t Hooft limit . (0; symm ) representation This higher representation can be obtained from the product of the minimal representation(0; f ) and itself. The two eigenvalues with this representation can be described as h (0; ) = ( kN + N + 2) N ( k + N + 2) ,φ ( ) (0; ) = φ (1)2 (0; ) + ∆(0; ) (8.17)= 8 k ( N + 2 k N + + 9 kN − kN − k + 4 N − N − N )3 N ( k + N + 2) (3 kN + 4 k + 4 N + 5) . The relation (5.20) is inserted. It is easy to see that the first eigenvalue is the twice of theone in (7.5) under the large ( N, k ) ’t Hooft limit. One also realizes that the description fromthe nonlinear analysis for the spin 2 current implies the following expression (with the helpof (2.17)) h (0; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u (0; ) = k ( N + k + 2) + ( N + 2) N ( N + 2)( k + N + 2) , Similarly, one has φ ( ) ( ; 0) = φ (1)2 ( ; 0) + ∆( ; 0) . The quantity ∆( ; 0) can be obtained from the corresponding quantity ∆( ; 0) by changing the sign. Thefootnote 27 for the quantity φ (1)2 ( ; 0) can be used. k (6 k N + 3 kN ) which is the twice of theones in (7.5). This implies that the eigenvalue for the higher spin 3 current in this higherrepresentation can be interpreted as the additive quantum number and it is given by the sumof each eigenvalue for the higher representation in the minimal representation (0; ) underthe large ( N, k ) ’t Hooft like limit.The difference between the nonlinear and linear cases which vanishes in the large N ’tHooft limit is given by∆(0; ) = 16 k ( k − N )( N − − kN − k − N − N − N ( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.18)Here the eigenvalues for each term appearing in (8.2) are used " − k (2 kN + N + 3 N + 2) N ( k + N + 2) , − k ( k + N + 2) , − ik q N +2 N ( k + N + 2) , − ik q N +2 N ( k + N + 2) , k ( N + 2) N ( k + N + 2) . (8.19)Again, the large ( N, k ) ’t Hooft limit of the above eigenvalue (8.18) behaves as N and therefore,both eigenvalues for the higher spin 3 currents in the nonlinear and linear versions have thesame value . (0; antisymm ) representation This higher representation can be obtained from the product of the minimal representation(0; f ) and itself. The two eigenvalues with this representation can be described as h (0; ) = ( kN + 3 N + 2) N ( k + N + 2) ,φ ( ) (0; ) = φ (1)2 (0; ) + ∆(0; ) (8.20)= 8 k ( N + 2 k N + + 27 kN + 4 kN − k + 4 N + 35 N + 27 N )3 N ( k + N + 2) (3 kN + 4 k + 4 N + 5) . One has the following eigenvalues for similar higher representation φ ( ) (0; ) = φ (1)2 (0; ) + ∆(0; ) . The quantity φ (1)2 (0; ) can be read off from the footnote 28 while the quantity ∆(0; ) can be obtainedfrom the eigenvalues in (8.19) by changing the signs in the first, second, and last eigenvalues together withthe coefficients in (7.2) and (7.4). N, k ) ’t Hooft limit. One also realizes that the description fromthe nonlinear analysis for the spin 2 current implies the following expression (with the helpof (2.19)) h (0; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u (0; ) = ( k + 2)( N + k + 2) + ( N + 2) N ( N + 2)( k + N + 2) . Furthermore, one observes that the highest power terms in the numerator (8.20) are given by8 k (6 k N + 3 kN ) which is the twice of the ones in (7.5). This implies that the eigenvaluefor the higher spin 3 current in this higher representation can be interpreted as the additivequantum number and it is given by the sum of each eigenvalue for the higher representationin the minimal representation (0; ) under the large ( N, k ) ’t Hooft like limitThe difference between the nonlinear and linear cases which vanishes in the large N ’tHooft limit is given by∆(0; ) = 16 k ( k − N )( N − − kN − k + N − N − N ( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.21)Here the corresponding eigenvalues are given by " − k (2 kN + N + 5 N + 2) N ( k + N + 2) , − k ( k + N + 2) , − ik q N +2 N ( k + N + 2) , − ik q N +2 N ( k + N + 2) , k ( N + 2) N ( k + N + 2) . (8.22)Note that the highest order terms in (8.21) are the same as the ones in (8.18). Because thelarge ( N, k ) ’t Hooft limit of the above eigenvalue (8.21) behaves as N , both eigenvalues forthe higher spin 3 currents in the nonlinear and linear versions have the same value . ( symm ; symm ) representation The two eigenvalues are given by h ( ; ) = 2( N + 2 N + 2) N ( N + 2)( k + N + 2) , One has φ ( ) (0; ) = φ (1)2 (0; ) + ∆(0; ) . The quantity φ (1)2 (0; ) can be read off from the footnote 29 while the quantity ∆(0; ) can be obtained fromthe eigenvalues in (8.22) by changing the signs in the first, second, and last eigenvalues together with thecoefficients in (7.2) and (7.4). ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )= 163 N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × ( N + 10 k N + 8 k N − − kN − kN + 10 kN + 24 k − N − N − N − N ) . (8.23)Note that the eigenvalue φ (1)2 ( ; ) is given by (5.22). The highest power terms of ( N, k )in the eigenvalue of the higher spin 3 current (8.23) are twice of the terms in (8.1). Thedescription from the nonlinear analysis for the spin 2 current implies the following expression(with the help of (3.5)) h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = 2( N + k + 2) + 4 N ( N + 2)( k + N + 2) . By realizing that the following eigenvalues hold in this higher representation " kN + 2 kN + N + 6 N + 8 N + 4) N ( N + 2)( k + N + 2) , k + N + 2) , − i q N ( N + 2)( k + N + 2) , − i q N ( N + 2)( k + N + 2) , − N + 2 N ) ( k + N + 2) , (8.24)and putting together with the various known coefficients, the difference between the nonlinearand linear cases in the eigenvalue for the higher spin 3 current (which vanishes in the large( N, k ) ’t Hooft limit) can be summarized by ∆( ; ) = − k − N )( − kN − k + + 12 N + 2 N − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) . (8.25) By inserting this value and the eigenvalue for the higher spin 3 current in the nonlinear version,the final eigenvalue in the linear version can be written as the one in (8.23). Furthermore,the highest power terms of ( N, k ) in (8.25) are the four times of the ones in (8.3). ( symm ; f ) representation The two eigenvalues are summarized by h ( ; ) = ( N + 1)( N + 7 N + 2)2 N ( N + 2)( k + N + 2) , One has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; ) . The quantity φ (1)2 ( ; ) can be read off from the footnote 30 while the quantity ∆( ; ) can be obtainedfrom the expression in (8.25) by changing the sign. ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )= − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( N + 39 k N − k N − k N − k N + N + 405 k N + 1334 k N + 1493 k N + 271 k N − k N − k + 60 kN + 643 kN + 3823 kN + 2172 kN + 220 kN − k + 25 N + 280 N + 1248 N + 2012 N + 1304 N + 288 N + 2565 kN ) . (8.26) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.23). One observes that these eigenvalues(8.26) are twice of the ones in (7.7) respectively if one takes the highest order terms inthe numerators. It is easy to see this behavior in the spin 2 current. For the higher spin3 current, more explicitly, the corresponding terms in the second eigenvalue in (8.26) aregiven by (18 k N + 36 k N ) = 6 kN (3 k N + 6 kN ) and the denominator contains the fator N (6 kN + 5 k + 5 N + 4) → kN , compared to the one in (7.7). Therefore, one observes thatthe corresponding quantity N (3 k N + 6 kN ) appears in the eigenvalue in (7.7).The description from the nonlinear analysis for the spin 2 current implies the followingexpression (with the help of (3.6)) h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 11)4( N + k + 2) + ( − N + 1) N ( N + 2)( k + N + 2) . It is obvious that this is equal to the eigenvalue in (8.26).The difference between the nonlinear and linear cases which vanishes in the large ( N, k )’t Hooft limit is given by∆( ; ) = − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N − k N − k N + − kN − kN − k − − N − N + 108 N + 136 N − . (8.27)Here as before, the following eigenvalues, corresponding to the five terms in (8.2), are used inthis calculation " − (2 kN + 4 kN + 3 N + 10 N − N − N − N ( N + 2)( k + N + 2) , − (2 N − k + N + 2) , − i ( N − p N ( N + 2)( k + N + 2) , − i ( N − p N ( N + 2)( k + N + 2) , ( N − N ( N + 2)( k + N + 2) . Then the final eigenvalue for the higher spin 3 current in the linear version can be obtainedby combining the one in the nonlinear version and the (8.27) as in (8.26) . One can consider the following eigenvalues φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; ) , .13 The ( symm ; f ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + 2 N + 12 N + 13 N + 2)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( N − k )( − N − k N − k + + 23 kN − kN − k + 10 N + 34 N + 40 N + 24) . (8.28) The relation (5.24) is used. Note that the eigenvalue for the spin 2 current does not haveany contribution from the commutator [( T ) , Q ¯ A − ]. This provides only the eigenvalue for therepresentation (0; ). Also the term Q ¯ A − ( T ) acting on the representation ( ; 0) gives theeigenvalue h ( ; 0) with Q ¯ A − acting on the state | ( ; 0) > . By inserting the overall factorinto this state, one has the final state associated with the representation ( ; ). Therefore,one arrives at the above eigenvalue for the spin 2 current. As before, one can describe thiseigenvalue in the context of the corresponding eigenvalue nonlinear version. That is, h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (4 N + 2 k + 7)4( N + k + 2) + ( − N − N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 2 N + 1)2 N ( k + N + 2) + ( N + 1)( N + 4)( N + 2)( k + N + 2) . In the first line the previous relations in (3.7) and in the second line, the relations (7.6) and(8.13) are used.For the eigenvalue corresponding to the higher spin 3 current, there exists the contributionfrom the lower order poles appearing in the commutator [( Φ ( ) ) , Q ¯ A − ]. One describes thefollowing eigenvalues for each term appearing in (8.2) " − (2 kN + kN − k + 6 N + 15 N + 6)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , ( − kN − k + 7 N + 20 N + 12)2( N + 2)( k + N + 2) . (8.29) Note that the eigenvalues in the third and the fourth are different from each other. This is thereason why there exists an imaginary i term in the eigenvalue in (8.28). Then by substituting where the quantity φ (1)2 ( ; ) can be read off from the footnote 31 while the quantity ∆( ; ) can beobtained from the expression in (8.27) by changing the sign. . ( symm ; antisymm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 8 N + 12 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " − i q NN +2 ( k + N + 2) − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( N − k )( N + 145 k N + 108 k + + 68 kN + 160 kN + 84 k + 19 N + 52 N + 4 N − . (8.30) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.27). As before, one can describe thiseigenvalue in the context of the corresponding eigenvalue in nonlinear version. That is, h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = ( N + k + 6)( N + k + 2) + 4 N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 3 N + 2) N ( k + N + 2) + ( N + 1)( N + 4)( N + 2)( k + N + 2) . In the first line the previous relations in (3.9) and in the second line, the relations (8.20) and(8.13) are used. For similar higher representation, one obtains φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the contributions from the following eigenvalues together with the previous coefficients " (4 kN + 11 kN + 4 k + 6 N + 15 N + 6)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , (4 kN + 12 k + N − N − N + 2)( k + N + 2) . Note that the quantity φ (1)2 ( ; ) can be obtained from the footnote 32, the quantity ∆( ; 0) can bedetermined from (8.14) with an extra minus sign and the quantity ∆(0; ) can be read off from (8.7). " − kN + 11 kN + 4 k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)( k + N + 2) , i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + N + 8 N + 12)( N + 2)( k + N + 2) , (8.31) provide the last two terms of (8.30) . ( symm ; antisymm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 8 N + 12 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " i q NN +2 ( k + N + 2) − N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( N − k )( N + 125 k N + 108 k − − kN − kN + 12 k − N − N − N − . (8.32) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.30). The contributions from the followingeigenvalues together with the previous coefficients " kN + kN − k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) , It is straightforward to see that one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus other contributions from the following eigenvalues " kN + 11 kN + 4 k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − − kN − k + N + 8 N + 12)( N + 2)( k + N + 2) , together with the previous coefficients. The quantity φ (1)2 ( ; ) can be obtained from the footnote 33, thequantity ∆( ; 0) can be determined from (8.14) with an extra minus sign and the quantity ∆(0; ) can beread off from (8.21). . ( symm ; symm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 6 N + 8 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 131 k N + 84 k − − kN − kN + 36 k − N − N − N − . (8.33) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.33). Note that the eigenvalue for thespin 2 current does not have any contribution from the commutator [( T ) , Q − Q − ]. Thisprovides only the eigenvalue for the representation (0; ). Also the term Q − Q − ( T ) actingon the representation ( ; 0) gives the eigenvalue h ( ; 0) with Q − Q − acting on the state | ( ; 0) > . By inserting the overall factor into this state, one has the final state associatedwith the representation ( ; ). Therefore, one arrives at the above eigenvalue for the spin2 current.Furthermore, the above eigenvalue for the spin 2 current can be interpreted as the one inthe nonlinear version h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = ( N + k )( N + k + 2) + ( − N − N ( N + 2)( k + N + 2) For similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the other contributions which can be determined by the following eigenvalues " − kN + kN − k + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)( k + N + 2) , i q NN +2 (3 k + N + 2)( k + N + 2) , − − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) . The quantity φ (1)2 ( ; ) can be obtained from the footnote 35, the quantity ∆( ; 0) can be determinedfrom (8.14) with an extra minus sign and the quantity ∆(0; ) can be read off from (8.21) with an extra minussign. h (0; ) + h ( ; 0) = ( kN + N + 2) N ( k + N + 2) + ( N + 1)( N + 4)( N + 2)( k + N + 2) . In the first line, the relations in (3.11) are used and similarly in the second line, the relations(8.17) and (8.13) are used also.The ∆(0; ) can be obtained from ∆(0; ) by changing the sign. As before, the last twoterms of (8.33) can be obtained from the following eigenvalues " − kN + kN − k + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) , (8.34) with the previous coefficients . ( antisymm ; antisymm ) representation The two eigenvalues are h ( ; ) = 2( N + 2 N + 2) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )= 163 N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5) × ( N + 10 k N + 8 k N − + 3 kN + 47 kN + 58 kN + 24 k − N − N + 5 N + 6 N ) . (8.35)The highest power terms of the last term in the eigenvalue of the higher spin 3 current in(8.35) are twice of the ones in (8.1). Note that the eigenvalue φ (1)2 ( ; ) is given by (5.36). For similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus other contributions from the eigenvalues " kN + 11 kN + 4 k + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − − kN − k + N + 8 N + 12)( N + 2)( k + N + 2) . The quantity φ (1)2 ( ; ) can be obtained from the footnote 36, the quantity ∆( ; 0) can be determinedfrom (8.14) with an extra minus sign and the quantity ∆(0; ) can be read off from (8.18). h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = 2( N + k + 2) + 4 N ( N + 2)( k + N + 2) . The previous relations in (3.16) are used in this calculation and this is equal to the one in(8.35).The difference between the nonlinear and linear cases which vanishes in the large N ’tHooft limit is obtained as follows: ∆( ; ) = − k − N )( − kN − k + + 12 N + 2 N − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) , (8.36) which is equal to the one in (8.25). The following eigenvalues are used in this expression " kN + 2 kN + N + 6 N + 8 N + 4) N ( N + 2)( k + N + 2) , k + N + 2) , − i √ N + 2 N ( k + N + 2) , − i √ N + 2 N ( k + N + 2) , − N ( N + 2)( k + N + 2) , together with the previous coefficients. These eigenvalues are the same as the ones in (8.24) . ( antisymm ; f ) representation The two eigenvalues are described by h ( ; ) = ( N + 4 N + N + 2)2 N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )= − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( N + 39 k N − k N − k N − k N + N + 189 k N + 146 k N − k N − k N − k N − k + 60 kN + 247 kN + 177 kN − kN − kN − kN − k + 25 N + 100 N + 84 N − N − N − N ) . (8.37) One has, for similar higher representation, φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; ) . The quantity φ (1)2 ( ; ) can be obtained from the footnote 37 and the quantity ∆( ; ) can be determinedfrom (8.36) with an extra minus sign. φ (1)2 ( ; ) is given by (5.37). Note that the eigenvalue for the spin 2current can be also obtained the one in the nonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (2 N + 3)4( N + k + 2) + ( − N + 1) N ( N + 2)( k + N + 2) , where the relations in (3.17) are used.One observes that these eigenvalues are twice of the ones in (7.7) respectively if onetakes the highest order terms in the numerators. It is easy to see this behavior in the spin2 current. For the higher spin 3 current, more explicitly, the corresponding terms in thesecond eigenvalue in (8.37) are given by (18 k N + 36 k N ) = 6 kN (3 k N + 6 kN ) and thedenominator contains the factor N (6 kN + 5 k + 5 N + 4) → kN , compared to the one in(7.7). Therefore, one observes that the corresponding quantity N (3 k N + 6 kN ) appears inthe eigenvalue in (7.7).The difference between the nonlinear and linear cases which vanishes in the large N ’tHooft limit is given by∆( ; ) = − N ( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N − k N − k N + − kN − kN − kN − k − − N − N + 60 N + 40 N − . (8.38)Here the following eigenvalues corresponding to the various fields in (8.2) are used " − (2 kN + 4 kN + 3 N + 6 N − N − N − N ( N + 2)( k + N + 2) , − N − k + N + 2) , − i ( N − √ N + 2 N ( k + N + 2) , − i ( N − √ N + 2 N ( k + N + 2) , ( N − N ( N + 2)( k + N + 2) . One sees that the highest power terms in (8.38) are the same as the ones in (8.27) . ( antisymm ; f ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + 2 N + 8 N + 5 N + 2)2 N ( N + 2)( k + N + 2) , For the similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; )The qunatity φ (1)2 ( ; ) can be obtained from the footnote 38 and the quantity ∆( ; ) can be obtained from(8.38) with an extra minus sign. ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 49 " i q NN +2 ( k + N + 2)+ 18( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 77 k N + 48 k − − kN + 35 kN + 30 k − N − N − N − . (8.39) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.38). One sees that the highest power termsof ( N, k ) in the last term of the eigenvalue (8.39) for the higher spin 3 current also appear in(8.28).For the eigenvalue of spin 2 current, one obtains the following analysis from the nonlinearcase h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = (4 N + 2 k − N + k + 2) + ( − N − N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 2 N + 1)2 N ( k + N + 2) + N ( N + 3)( N + 2)( k + N + 2) . The relations in (3.18) are used in the first line and the previous relations (7.6) and (8.15)are also used in the second line.One uses the following eigenvalues " − (2 kN + 3 kN + 6 N + 15 N + 6)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , ( − kN − k + 7 N + 20 N + 12)2( N + 2)( k + N + 2) , (8.40)in order to obtain the last two terms in the eigenvalue of the higher spin 3 current in (8.39).The last eigenvalue of (8.40) is the same as the one in (8.29) . One obtains φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the contributions from the following eigenvalues with the coefficients " (4 kN + 9 kN + 6 N + 15 N + 6)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , (4 kN + 12 k + N − N − N + 2)( k + N + 2) . One can read off φ (1)2 ( ; ) from the footnote 39, ∆( ; 0) from (8.16) with minus sign, and ∆(0; ) from(8.7). .20 The ( antisymm ; symm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 4 N + 4 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " − i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 139 k N + 96 k + + 44 kN + 100 kN + 60 k + N − N − N − . (8.41) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.41). The eigenvalue for the spin 2 currentcan be also obtained from the one in the nonlinear version as follows: h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = 1 + 4 N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + N + 2) N ( k + N + 2) + N ( N + 3)( N + 2)( k + N + 2) . We use the relations in (3.20) in the first line and the relations in (8.17) and (8.15) in thesecond line.The highest power of ( N, k ) in the last term of the eigenvalue of higher spin 3 current(8.41) is the same as the one in (8.30). Furthermore, the following eigenvalues " − kN + 9 kN + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)( k + N + 2) , i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + N + 8 N + 12)( N + 2)( k + N + 2) (8.42) are used for the last two terms in (8.41). The last eigenvalue of (8.42) is the same as the onein (8.31) . In this case, also for the similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus other contributions from the following eigenvalues " kN + 9 kN + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , .21 The ( antisymm ; symm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 4 N + 4 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 137 k N + 96 k − − kN + 20 kN + 60 k − N − N − N − . (8.43) The highest power terms of ( N, k ) in the last term of the eigenvalue of higher spin 3 current(8.43) are exactly the same as the ones in (8.33). Note that the eigenvalue φ (1)2 ( ; ) in(8.43) is given by (5.44). It is easy to check that the above last two terms of (8.43) can bedetermined by using the following eigenvalues " − kN + 3 kN + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) , (8.44) together with the coefficients. The last eigenvalue of (8.44) is the same as the one in (8.34) . − − kN − k + N + 8 N + 12)( N + 2)( k + N + 2) , together with the previous coefficients. The eigenvalue φ (1)2 ( ; ) can be obtained from the footnote 40, theeigenvalue ∆( ; 0) can be determined from (8.16) with the extra minus sign, and the eigenvalue ∆(0; ) canbe read off from (8.18). One has following eigenvalues φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; )plus the contributions of the following eigenvalues " kN + 3 kN + 3 N + 12 N + 12)( N + 2)( k + N + 2) , , − i q NN +2 ( k + 3 N + 6)( k + N + 2) , i q NN +2 (3 k + N + 2)( k + N + 2) , − − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) . .22 The ( antisymm ; antisymm ) representation The two eigenvalues are given by h ( ; ) = ( kN + 2 kN + N + 6 N + 8 N + 4) N ( N + 2)( k + N + 2) ,φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) + 169 " i q NN +2 ( k + N + 2)+ 9( N + 2)( k + N + 2) (3 kN + 4 k + 4 N + 5)(6 kN + 5 k + 5 N + 4) × ( k − N )( N + 137 k N + 96 k − − kN + 56 kN + 60 k − N − N − N − . (8.45) Note that the eigenvalue φ (1)2 ( ; ) is given by (5.47). One sees that the eigenvalue for thespin 2 current can be interpreted as the one in the nonlinear version h ( ; ) + 1( N + k + 2) 1 N ( N + 2) ˆ u ( ; ) = ( k + N )( k + N + 2) + ( − N − N ( N + 2)( k + N + 2)= h (0; ) + h ( ; 0) = ( kN + 3 N + 2) N ( k + N + 2) + N ( N + 3)( N + 2)( k + N + 2) . In the first line, the conformal dimension formula and the additive property of ˆ u charge canbe used and in the second line, the relations (8.17) and (8.15) can be used.The eigenvalue ∆( ; 0) is given by (8.16) with minus sign and the eigenvalue ∆(0; ) canbe obtained from ∆(0; ) in (8.21) by changing the sign.Moreover, the last two terms in (8.45) can be obtained from the following eigenvalues " − kN + 3 kN + 9 N + 24 N + 12)( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − kN − k + 5 N + 16 N + 12)( N + 2)( k + N + 2) , (8.46) together with the coefficients. The last eigenvalue of (8.46) is the same as the one in (8.34).Note that the highest power terms of ( N, k ) in the last term of the eigenvalue of higher spin3 current (8.45) are exactly the same as the ones in (8.33) . The eigenvalue φ (1)2 ( ; ) is given by the footnote 42, the eigenvalue ∆( ; 0) is given by (8.16) with minussign, and the eigenvalue ∆(0; ) is given by (8.18) with an extra minus sign. For similar higher representation, one has φ ( ) ( ; ) = φ (1)2 ( ; ) + ∆( ; 0) + ∆(0; ) N, k ). It turns out that theyreproduce the ones in the nonlinear version described in section 5 under the large ( N, k ) ’tHooft like limit. We present the first two Tables where one can find the conformal dimensions for the spin 2current acting on the various representations in the coset under the large ( N, k ) ’t Hooft likelimit (up to three boxes). The other quantum numbers l ± and ˆ u can be made in Tables but wedid not do it. We have found other quantum numbers associated with the higher spin currentson some part of representations in the above Tables. The explicit large ( N, k ) behaviors ofthose eigenvalues for the higher spin 3 currents are summarized in the last two Tables. Onerealizes that although the conformal dimensions are equal to each other for different tworepresentations, the eigenvalues for the higher spin 3 current for those two representationsare different from each other under the large ( N, k ) ’t Hooft like limit. For example, forthe representations ( ; ) and ( ; ), the eigenvalues φ (1)2 are different but their conformaldimensions h are the same under the large ( N, k ) ’t Hooft like limit. For the eigenvalue h , thecorresponding conformal dimension for the spin 2 current is 2 while for the eigenvalue φ (1)2 ,the corresponding conformal dimension for the higher spin 3 current is 3. Under the complexconjugation, the former remains the same (even spin) but the latter changes the sign (oddspin).Let us present some related and open problems in the near future. • The other representationsOne can calculate the cases where the higher representations have Λ + = 0 and Λ − = , , and . It is straightforward to calculate them because one has the known higherspin currents for several N values and the OPEs between these currents and the product of plus the contributions from the following eigenvalues " (cid:0) kN + 9 kN + 9 N + 24 N + 12 (cid:1) ( N + 2)( k + N + 2) , , i q NN +2 ( k + 3 N + 6)( k + N + 2) , − i q NN +2 (3 k + N + 2)( k + N + 2) , − (cid:0) − kN − k + N + 8 N + 12 (cid:1) ( N + 2)( k + N + 2) . The eigenvalue φ (1)2 ( ; ) can be obtained from the footnote 43, the eigenvalue ∆( ; 0) is given by (8.16) withminus sign, and the eigenvalue ∆(0; ) is given by (8.18). fermions can be obtained as before.1) The (0; antisymm) representation with three boxesThe relevant subsection is given by 2 . . 7. The four eigenvalues are given by φ (1)0 (0; ) = − k ( N + k + 2) ,v + (0; ) = 216 k ( k + N + 2) ,v − (0; ) = 12 k (17 k + 2 N + 2)( k + N + 2) ,φ (1)2 (0; ) = 4 k ( N − k + + 51 kN − k + 5 N + 64 N + 12)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (9.1)One can see that the first eigenvalue is three times of the one in (4.2) even finite ( N, k ). Thelast eigenvalue is also three times of the one in (4.2) under the large ( N, k ) ’t Hooft-like limit.See also (5.21).2) The (0; antisymm) representation with three boxesThe four eigenvalues are φ (1)0 (0; ) = 3 k ( N + k + 2) ,v + (0; ) = 216 k ( k + N + 2) ,v − (0; ) = 12 k (17 k + 2 N + 2)( k + N + 2) ,φ (1)2 (0; ) = − k ( N − k + + 51 kN − k + 5 N + 64 N + 12)( k + N + 2) (6 kN + 5 k + 5 N + 4) . (9.2)The first eigenvalue is three times of the one in (4.3) even finite ( N, k ). The last eigenvaluealso is three times of the one in (4.3) under the large ( N, k ) ’t Hooft-like limit. See also thefootnote 29.3) The (0; mixed) representation with three boxesThe relevant subsection is given by 2 . . 8. The four eigenvalues are described by φ (1)0 (0; ) = − k ( N + k + 2) ,v + (0; ) = 216 k ( k + N + 2) , − (0; ) = 12 k (13 k + 12 N − k + N + 2) ,φ (1)2 (0; ) = 4 k ( N − k + + 15 kN − k + 5 N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . The first two eigenvalues are the same as the ones in (9.1) while the last eigenvalue is thesame as the one in (9.1) under the large ( N, k ) ’t Hooft-like limit.4) The (0; mixed) representation with three boxesThe four eigenvalues are summarized by φ (1)0 (0; ) = 3 k ( N + k + 2) ,v + (0; ) = 216 k ( k + N + 2) ,v − (0; ) = 12 k (13 k + 12 N − k + N + 2) ,φ (1)2 (0; ) = − k ( N − k + + 15 kN − k + 5 N − N − k + N + 2) (6 kN + 5 k + 5 N + 4) . The first two eigenvalues are the same as the ones in (9.2) while the last eigenvalue is thesame as the one in (9.2) under the large ( N, k ) ’t Hooft-like limit.Then one considers the following higher representationsΛ + = , , , , , , Λ − = , , , , which appears in the product of (Λ + ; 0) and (0; Λ − ) with Λ + = Λ − . One expects that theeigenvalues are summarized in the following two Tables. • The three-point functionsAs mentioned in the abstract, one can determine the three-point functions [41] of thehigher spin currents with two scalar operators at finite ( N, k ). From (5.18), < O ( ;0) O ( ;0) Φ (1)2 > = " − N ( N + 5 k + + 45 kN + 43 k − N − N + 12)3( k + N + 2) (6 kN + 5 k + 5 N + 4) < O ( ;0) O ( ;0) > −→ − λ ( λ + 1) < O ( ;0) O ( ;0) > . The large ( N, k ) ’t Hooft like limit (3.62) in the final expression is taken. It is straightforwardto write down all the three-point functions we have found in this paper. • The higher spin 3 current in terms of adjoint spin 1 and spin currents130 φ (1)0 (0; ) 3 φ (1)0 (0; ) − φ (1)0 (0; ) − φ (1)0 (0; ) φ (1)0 [( ; 0) + 3(0; )] φ (1)0 [( ; 0) + 3(0; )] φ (1)0 [( ; 0) − φ (1)0 [( ; 0) − φ (1)0 [ − ( ; 0) + 3(0; )] φ (1)0 [ − ( ; 0) + 3(0; )] φ (1)0 [ − ( ; 0) − φ (1)0 [ − ( ; 0) − φ (1)0 [2( ; 0) + 3(0; )] φ (1)0 [2( ; 0) + 3(0; )] φ (1)0 [2( ; 0) − φ (1)0 [2( ; 0) − φ (1)0 [2( ; 0) + 3(0; )] φ (1)0 [2( ; 0) + 3(0; )] φ (1)0 [2( ; 0) − φ (1)0 [2( ; 0) − φ (1)0 [ − 2( ; 0) + 3(0; )] φ (1)0 [ − 2( ; 0) + 3(0; )] φ (1)0 [ − 2( ; 0) − φ (1)0 [ − 2( ; 0) − φ (1)0 [ − 2( ; 0) + 3(0; )] φ (1)0 [ − 2( ; 0) + 3(0; )] φ (1)0 [ − 2( ; 0) − φ (1)0 [ − 2( ; 0) − Table 10: Expectation of the eigenvalue φ (1)0 at finite ( N, k ). Due the fact that this higherspin current has conformal dimension 1, there are no contributions from the commutatorbetween the zero mode of this higher spin current and the corresponding mode Q − Q − Q − , Q − Q − Q − (and its complex conjugated ones) from the subsections 2 . . . . 8. Theeigenvalues are given by the linear combinations of the one of (0; f ) (or (0; f )) and the one of( f ; 0) (or ( f ; 0)). Then each coefficient depends on the number of boxes in Λ + and Λ − . Thefirst row can be seen from the previous description. In order to express these eigenvalues interms of ( N, k ) dependence explicitly, the relations (4.2) and (4.4) are needed.In this paper, we used the higher spin currents for several N values which are written interms of adjoint spin 1 and spin currents. In principle, one can find the explicit expressionfor the higher spin 3 current (by hand) as described in section 4. Although it will be rathercomplicated to obtain this form because that all the calculations on the OPEs should bechecked step by step, it will be worthwhile to determine this full expression. Once this willbe found, then it will be an open problem to obtain the corresponding eigenvalues associatedwith any representations. • The spectrum for the higher spin 4 currentWhat happens for the higher spin current with different spin? For example, the next 16higher spin current contains the higher spin 4 current Φ ( s =2)2 ( z ). One expects that the behaviorof large ( N, k ) ’t Hooft-like limit in the eigenvalues on this higher spin 4 current looks similarto the results of this paper. The general structure for the eigenvalue with Λ + = Λ − in theproduct of (Λ + ; 0) and (0; Λ − ) is given by the linear combinations of the one of (0; ) (or(0; )) and the one of ( ; 0) (or ( ; 0)). Then each coefficient depends on the the number ofboxes in Λ + and Λ − . Also one has plus sign for the fundamental representation while minussign for the complex conjugated (anti fundamental) representation. The corresponding basic131 φ (1)2 (0; ) 3 φ (1)2 (0; ) − φ (1)2 (0; ) − φ (1)2 (0; ) φ (1)2 [( ; 0) + 3(0; )] φ (1)2 [( ; 0) + 3(0; )] φ (1)2 [( ; 0) − φ (1)2 [( ; 0) − φ (1)2 [ − ( ; 0) + 3(0; )] φ (1)2 [ − ( ; 0) + 3(0; )] φ (1)2 [ − ( ; 0) − φ (1)2 [ − ( ; 0) − φ (1)2 [2( ; 0) + 3(0; )] φ (1)2 [2( ; 0) + 3(0; )] φ (1)2 [2( ; 0) − φ (1)2 [2( ; 0) − φ (1)2 [2( ; 0) + 3(0; )] φ (1)2 [2( ; 0) + 3(0; )] φ (1)2 [2( ; 0) − φ (1)2 [2( ; 0) − φ (1)2 [ − 2( ; 0) + 3(0; )] φ (1)2 [ − 2( ; 0) + 3(0; )] φ (1)2 [ − 2( ; 0) − φ (1)2 [ − 2( ; 0) − φ (1)2 [ − 2( ; 0) + 3(0; )] φ (1)2 [ − 2( ; 0) + 3(0; )] φ (1)2 [ − 2( ; 0) − φ (1)2 [ − 2( ; 0) − Table 11: Expectation of the eigenvalue φ (1)2 under the large ( N, k ) ’t Hooft-like limit (3.62).The eigenvalues are given by the linear combinations of the one of (0; f ) (or (0; f )) and theone of ( f ; 0) (or ( f ; 0)). Then each coefficient depends on the the number of boxes in Λ + and Λ − . The first row can be seen from the previous description. In order to express theseeigenvalues in terms of λ dependence explicitly, the relations (4.2) and (4.4) with the large( N, k ) ’t Hooft-like limit (3.62) are needed.eigenvalues with an appropriate normalization are found in [22] φ (2)2 ( ; 0) = 125 λ (1 + λ )(2 + λ ) , φ (2)2 (0; ) = − 125 (1 − λ )(2 − λ )(3 − λ ) . Their complex conjugated ones remain the same because this higher spin current has theconformal spin 4. When the representation Λ − appears in the branching of Λ + , one expectsthat the eigenvalue leads to the representation ( | Λ + | − | Λ − | ; 0) where | Λ ± | denotes the numberof boxes. The eigenvalue with Λ + = Λ − can be written in terms of the multiple of theeigenvalue of ( ; ) or ( ; ). It would be interesting to observe whether these behaviorsoccur. Similarly, it is an open problem to obtain the eigenvalues for the higher spin 2 currentΦ ( s =2)0 ( z ). • The three boxes in Λ + In this paper, the boxes for Λ + in the eigenvalues of the higher spin currents are limitedto 2. One considers the case where Λ + contains the three boxes: , , and (andits conjugated ones). At least, one needs to have the SU ( N + 2) generators with N =3 , , , , 11 in these higher representations in order to extract the eigenvalues. It is knownthat the dimensions for the above higher representations are given by ( N + 2)( N + 3)( N + 4), ( N + 2)( N + 3)( N + 1), and ( N + 2)( N + 1) N respectively. In particular, for N = 11, thesebecome 455, 728, and 286. This implies that the 84 generators of SU (13) should be writtenin terms of 455 × 455 matrices, 728 × 728 matrices, and 286 × 286 matrices respectively. It israther difficult to obtain 84 generators in the mixed representation by using the methods in132ppendix D . It is an open problem to find out the systematic way to read off the complete84 generators which are 728 × 728 matrices by using the general formula described in thesubsection 3 . • The three-point functions from the decomposition of the four-point functions of scalaroperators with Virasoro conformal blocksRecently [42], using the decomposition of the scalar four-point functions by Virasoroconformal blocks, the three-point functions including N corrections in the two dimensional(bosonic) W N minimal model were obtained using the result of [43] (see also the works of[44, 45]). The N corrections for the conformal dimension 6 , , W N minimal model. As observed in [42], it is an open problem to obtain the three-pointfunctions from the decomposition of four-point functions in the large N = 4 holography. • The orthogonal Wolf space coset spectrumOne can ask what happens for the orthogonal Wolf space coset spectrum. The relevantprevious works are given by [46, 47]. It is an open problem to obtain the eigenvalues forthe higher spin currents in the higher representations. One should obtain the generators of SO ( N + 4) in various higher representations explicitly and obtain the higher spin currents(where the spins are 2, 3 or 4) for several N values. Acknowledgments This research was supported by Kyungpook National University Research Fund, 2017.CA acknowledges warm hospitality from C.N. Yang Institute for Theoretical Physics, StonyBrook University. CA would like to thank M.H. Kim for discussions and T. Kephart forpointing out the reference [29]. 133 The SU (5) generators in the symmetric 15 repre-sentation of SU (5) with two boxes In order to determine the eigenvalue equations for the zero modes of (higher spin) currentsin the higher representations, one should obtain the SU ( N + 2 = 5) generators in the higherrepresentations. The 24 generators are given by ( N + 2)( N + 3) × ( N + 2)( N + 3) = 15 × V ) for the fundamentalrepresentation as follows [48]:ˆ e = , ˆ e = , ˆ e = , ˆ e = , ˆ e = . For given 24 SU (5) generators which are 5 × T ˆ e = ˆ e , T ˆ e = ˆ e , T ˆ e = ˆ e , T ˆ e = ˆ e ,T ˆ e = ˆ e , T ˆ e = ˆ e , T ˆ e = ˆ e , T ˆ e = ˆ e ,T ˆ e = ˆ e , T ˆ e = ˆ e , T ˆ e = ( i √ e , T ˆ e = ( − i √ e ,T ˆ e = − √ e T ˆ e = ( i √ √ 10 )ˆ e , T ˆ e = ( i √ √ 10 )ˆ e ,T ˆ e = ( i √ √ 10 )ˆ e , T ˆ e = ( − i s 32 + 12 √ 10 )ˆ e ,T ˆ e = − s 25 ˆ e . (A.1)The remaining half of the generators can be obtained by taking the transpose and the complexconjugate on these generators. The coefficients of the right hand side of (A.1) give us thenonzero matrix elements of these generators. For example, the nonzero component of T isgiven by the (4 , 1) matrix element and the numerical value is equal to 1. Similarly, one ofnonzero components of T is given by the (1 , 1) matrix element and the numerical value isequal to ( i + √ ).Then the basis vectors of direct product space V ⊗ V can be obtained from E ij ≡ ˆ e i ⊗ ˆ e j where i, j = 1 , , · · · , 5. The basis for the symmetric tensors can be realized byˆ u = E , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = E , u = 1 √ E + E ) , ˆ u = E , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = 1 √ E + E ) , ˆ u = E , ˆ u = 1 √ E + E ) , ˆ u = E . Then it is straightforward to calculate the following quantities based on (A.1) T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u ,T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u ,T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = √ u , T ˆ u = ˆ u ,T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u ,T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u ,T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = √ u ,T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = √ u ,T ˆ u = √ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u , T ˆ u = ˆ u ,T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u ,T ˆ u = ˆ u , T ˆ u = √ u , T ˆ u = √ u , T ˆ u = ˆ u , T ˆ u = ˆ u ,T ˆ u = ( i + 1 √ u , T ˆ u = 1 √ u , T ˆ u = ( i − √ u ,T ˆ u = ( − i + 1 √ u , T ˆ u = ( − i − √ u , T ˆ u = − √ u ,T ˆ u = ( i √ u , T ˆ u = ( i √ u , T ˆ u = ( − i √ u ,T ˆ u = ( − i √ u , T ˆ u = − √ u , T ˆ u = − √ u ,T ˆ u = ( i √ √ 10 )ˆ u , T ˆ u = ( i √ √ 10 )ˆ u , T ˆ u = ( i √ √ 10 )ˆ u ,T ˆ u = ( i √ √ 10 )ˆ u , T ˆ u = ( i √ √ 10 )ˆ u , T ˆ u = ( i √ √ 10 )ˆ u ,T ˆ u = ( − i √ √ 10 )ˆ u , T ˆ u = ( 5 i √ − √ u , T ˆ u = ( − i √ √ 10 )ˆ u ,T ˆ u = ( 5 i √ − √ u , T ˆ u = ( − i √ √ 10 )ˆ u , T ˆ u = ( 5 i √ − √ u ,T ˆ u = ( − i r 32 + 1 √ 10 )ˆ u , T ˆ u = ( − i √ − √ u , T ˆ u = − r 25 ˆ u . (A.2) For example, one can write down T ˆ u = T ˆ e ⊗ ˆ e which can be written as T ˆ e ⊗ ˆ e +ˆ e ⊗ T ˆ e .By using (A.1), this can be written as ˆ e ⊗ ˆ e + ˆ e ⊗ ˆ e . Then this is equal to √ u as in(A.2). 135rom (A.2), the 24 generators in terms of 15 × 15 unitary matrix are given as follows: T = √ √ ,T = √ √ , = √ √ ,T = √ √ ,T = √ √ , = √ √ ,T = √ √ ,T = √ √ , = √ √ ,T = √ √ ,T = diag (cid:18) i + 1 √ , √ , 16 (3 i − √ , − i + 1 √ , 16 ( − i − √ , − √ , 16 (3 i + √ , 16 (3 i + √ , 16 ( − i + √ , 16 ( − i + √ , − √ , − √ , , , (cid:19) ,T = diag (cid:18) i √ √ , i √ √ , i √ √ , i √ √ , i √ √ , + i √ √ , − i √ √ , 160 (5 i √ − √ , − i √ √ , 160 (5 i √ − √ , − i √ √ , 160 (5 i √ − √ , − i r 32 + 1 √ , 120 ( − i √ − √ , − r ! . (A.3) For example, from the first line of (A.2), the nonzero components of the generator T aregiven by (7 , , , , 7) and (14 , 8) and their numerical values are √ 2, 1, 1, √ T a T a ∗ ) = ( N + 2 + 2) = 7 with139 = 3. The number 7 is the index l of the representation of SU (5) [27]. B The SU (5) generators in the antisymmetric 10representation of SU (5) with two boxes The 24 generators are given by ( N + 2)( N + 1) × ( N + 2)( N + 1) = 10 × 10 matrix for theantisymmetric representation .The basis for the antisymmetric tensors can be realized byˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) , ˆ u = 1 √ E − E ) . (B.1)After applying the 24 generators which are 5 × × 10 unitary matrix are given as follows as in Appendix A : T = − − , T = − ,T = , T = − − − , = − − , T = − ,T = , T = ,T = − , T = ,T = daig (cid:18) √ , 16 (3 i − √ , 16 ( − i − √ , 16 (3 i + √ , 16 (3 i + √ , + 16 ( − i + √ , 16 ( − i + √ , − √ , − √ , (cid:19) ,T = diag (cid:18) i √ √ , i √ √ , i √ √ , − i √ √ , 160 (5 i √ − √ , − i √ √ , 160 (5 i √ − √ , − i √ √ , 160 (5 i √ − √ , 120 ( − i √ − √ (cid:19) . Again the half of generators can be obtained from these by taking the transpose and complexconjugate. One has Tr( T a T a ∗ ) = ( N + 2 − 2) = 3 which is the index l of the representation of SU (5) [27]. For the generators of SU ( N + 2) with N = 5 , , , 11, the similar constructionscan be done. 141 The SU (5) generators in the symmetric 35 repre-sentation of SU (5) with three boxes The basis vectors of direct product space V ⊗ V ⊗ V can be obtained from E ijk ≡ ˆ e i ⊗ ˆ e j ⊗ ˆ e k where i, j, k = 1 , , · · · , 5. The basis for the symmetric ( ) tensors can be realized by ˆ u = E , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = E , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = E , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , In the notation of [27], their ˆ u charge is twice of the one of this paper. For example, the branching rule(Table A. 71 of [27]) of SU (5) → SU (3) × SU (2) × U (1) gives = = ( , ) +( , ) +( , ) − +( , ) − by taking the complex conjugation of . One should take the complex conjugation of SU (3) representationand take the minus sign of ˆ u charge. This can be compared to the relation in (3.24). For N ≥ 7, the threebox does not contain the complex conjugated notation. Note that the corresponding relation for the SU (11)branching in = is given by Table A. 77 of [27]. u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E + E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = E , ˆ u = 1 √ E + E + E ) , ˆ u = 1 √ E + E + E ) , ˆ u = E . (C.1)After applying the 24 generators which are 5 × × 35 unitary matrix are given as follows as in Appendices A and B . We presentonly the nonzero elements. T : (20 , , (22 , , (25 , , (27 , , (30 , , (34 , 19) has 1 , (17 , 11) has 2 , : (13 , , (15 , , (18 , , (24 , , (29 , , (33 , 18) has √ , (11 , , (32 , 17) has √ ,T : (11 , , (15 , , (18 , , (27 , , (30 , , (34 , 26) has 1 , (24 , 20) has 2 , : (13 , , (17 , , (22 , , (25 , , (29 , , (33 , 25) has √ , (20 , , (32 , 24) has √ ,T : (11 , , (13 , , (18 , , (20 , , (25 , , (34 , 31) has 1 , (29 , 27) has 2 , : (15 , , (17 , , (22 , , (24 , , (30 , , (33 , 30) has √ , (27 , , (32 , 29) has √ ,T : (21 , , (23 , , (25 , , (28 , , (30 , , (33 , 17) has 1 , (19 , 12) has 2 , : (14 , , (16 , , (18 , , (26 , , (31 , , (34 , 18) has √ , (12 , , (35 , 19) has √ ,T : (12 , , (16 , , (18 , , (28 , , (30 , , (33 , 24) has 1 , (26 , 21) has 2 , : (14 , , (19 , , (23 , , (25 , , (31 , , (34 , 25) has √ , (21 , , (35 , 26) has √ ,T : (12 , , (14 , , (18 , , (21 , , (25 , , (33 , 29) has 1 , (31 , 28) has 2 , : (16 , , (19 , , (23 , , (26 , , (30 , , (34 , 30) has √ , (28 , , (35 , 31) has √ ,T : (12 , , (14 , , (16 , , (21 , , (23 , , (28 , 27) has 1 , (34 , 33) has 2 , : (18 , , (19 , , (25 , , (26 , , (30 , , (31 , 30) has √ , (33 , , (35 , 34) has √ ,T : (9 , , (22 , , (23 , , (24 , , (25 , , (26 , 19) has 1 , (4 , 2) has 2 , : (5 , , (8 , , (13 , , (14 , , (20 , , (21 , 14) has √ , (2 , , (7 , 4) has √ ,T : (8 , , (22 , , (23 , , (29 , , (30 , , (31 , 19) has 1 , (6 , 3) has 2 , : (5 , , (9 , , (15 , , (16 , , (27 , , (28 , 16) has √ , (3 , , (10 , 6) has √ ,T : (3 , , (15 , , (16 , , (29 , , (30 , , (31 , 26) has 1 , (9 , 8) has 2 , : (5 , , (6 , , (22 , , (23 , , (27 , , (28 , 23) has √ , (8 , , (10 , 9) has √ ,T : (8 , 8) has − i, (3 , 3) has i, : (27 , , (28 , 28) has − √ , (29 , , (30 , , (31 , 31) has − √ , : (13 , , (14 , 14) has 1 √ , (10 , 10) has − √ , : (9 , 9) has 12 ( −√ − i ) , (6 , 6) has 12 ( −√ i ) , (22 , , (23 , 23) has 16 ( −√ − i ) , (15 , , (16 , 16) has 16 ( −√ i ) , : (4 , 4) has 12 ( √ − i ) , (2 , 2) has 12 ( √ i ) , : (25 , 25) has 16 ( √ − i ) , (20 , , (21 , 21) has 13 ( √ − i ) , : (7 , 7) has 12 ( √ − i ) , (18 , 18) has 16 ( √ i ) , : (11 , , (12 , 12) has 13 ( √ i ) , (1 , 1) has 12 ( √ i ) ,T : (35 , 35) has − r , (18 , , (25 , , (30 , 30) has − i (5 √ − i √ √ , : (14 , , (16 , , (23 , 23) has i (5 √ i √ √ , (34 , 34) has − i (5 √ − i √ √ , : (19 , , (26 , , (31 , 31) has i (5 √ i √ √ , (32 , 32) has 3( √ − i √ √ , : (33 , 33) has − ( √ i √ √ , (12 , , (21 , , (28 , 28) has − (3 √ − i √ √ , : (1 , , (2 , , (3 , , (4 , , (5 , , (6 , , (7 , , (8 , , (9 , , (10 , 10) has (3 √ i √ √ , : (11 , , (13 , , (15 , , (20 , , (22 , , (27 , 27) has (9 √ − i √ √ , : (17 , , (24 , , (29 , 29) has (9 √ − i √ √ . The remaining generators can be obtained from these by taking the transpose and the complexconjugation. One has Tr( T a T a ∗ ) = ( N + 2 + 2)( N + 2 + 3) = 28 which is the index l of therepresentation of SU (5) [27]. D The SU (5) generators in the mixed 40 represen-tation of SU (5) with three boxes The basis for the mixed ( ) tensors can be realized by ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E + E − E ) , The branching rule (Table A. 71 of [27]) of SU (5) → SU (3) × SU (2) × U (1) gives = = ( , ) +( , ) + ( , ) + ( , ) − + ( , ) − + ( , ) − which can be compared to the relation in (3.24). For N ≥ SU (11) branching in = is given by Table A. 77 of [27]. u = 1 √ E + E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ E − E − E ) , ˆ u = 1 √ E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 1 √ 12 ( − E + 2 E − E − E + 2 E − E ) , (D.1)ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 (2 E − E + 2 E − E − E − E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 ( − E − E − E + 2 E − E + 2 E ) , ˆ u = 12 ( E − E − E + E ) , ˆ u = 1 √ 12 ( − E − E − E + 2 E − E + 2 E ) , ˆ u = 12 ( E − E + E − E ) , ˆ u = 1 √ 12 ( − E − E − E + 2 E − E + 2 E ) , ˆ u = 12 ( E − E + E − E ) . It is not obvious to see these basis at first sight. There is other way to obtain these basis by usingthe roots and weights [49]. After applying the 24 generators which are 5 × × 40 unitary matrix are given as follows. We present onlythe nonzero elements. T : (10 , , (15 , , (34 , 26) has − , (35 , , (39 , , (40 , 30) has − , : (36 , 22) has 12 , (7 , , (20 , , (33 , 25) has 1 , (13 , , (17 , 28) has − r , (24 , , (28 , 2) has r , : (23 , , (27 , 2) has − √ , (13 , , (17 , 27) has 1 √ , : (19 , , (31 , 4) has √ , (40 , 29) has − √ , : (35 , , (36 , , (39 , 30) has √ ,T : (15 , 12) has − , (27 , , (39 , 37) has − , : (28 , , (40 , 38) has 12 , (3 , , (13 , , (20 , , (31 , , (32 , 26) has 1 , : (17 , 36) has − r , (7 , , (24 , , (36 , 9) has r , : (35 , 9) has − √ , (7 , , (17 , , (23 , 5) has 1 √ , : (19 , , (33 , 11) has √ , (27 , , (28 , , (39 , , (40 , 37) has √ ,T : (33 , 37) has − , (34 , 38) has 12 , : (3 , , (10 , , (17 , , (20 , , (23 , , (24 , , (31 , , (32 , 30) has 1 , : (19 , , (40 , 16) has − r , (7 , , (28 , 6) has r , : (19 , , (39 , 16) has − √ , (7 , , (27 , 6) has 1 √ , : (13 , , (35 , 12) has − √ , (33 , , (34 , 37) has − √ ,T : (11 , , (16 , , (19 , 7) has − , (33 , , (34 , , (37 , , (39 , 27) has − , : (38 , , (40 , 28) has 12 , (8 , 4) has 1 , : (14 , , (18 , , (20 , 32) has − r , (26 , , (30 , , (32 , 3) has r , : (25 , , (29 , , (31 , 3) has − √ , (14 , , (18 , , (20 , 31) has 1 √ , : (33 , 24) has − √ , (34 , , (37 , , (38 , , (39 , , (40 , 27) has √ ,T : (16 , , (19 , 13) has − , (29 , , (31 , , (39 , , (40 , 36) has − , : (30 , , (32 , 24) has 12 , (4 , , (14 , 11) has 1 , : (18 , , (20 , 34) has − r , (8 , , (26 , , (34 , , (38 , 9) has r , : (33 , , (37 , 9) has − √ , (8 , , (18 , , (20 , , (25 , 5) has 1 √ , (40 , 35) has − √ , (29 , , (30 , , (31 , , (32 , , (39 , 36) has √ ,T : (19 , 17) has − , (31 , , (33 , , (34 , 36) has − , : (32 , 28) has 12 , (4 , , (11 , , (18 , , (25 , , (26 , 22) has 1 , : (20 , 40) has − r , (8 , , (30 , , (40 , 15) has r , : (39 , 15) has − √ , (8 , , (20 , , (29 , 6) has 1 √ , : (14 , , (37 , 12) has − √ , (34 , 35) has − √ , : (31 , , (32 , , (33 , 36) has √ ,T : (4 , , (11 , , (16 , , (20 , , (25 , , (26 , , (29 , , (30 , , (37 , , (38 , 36) has 1 , : (8 , , (14 , , (32 , , (34 , 13) has r , (8 , , (14 , , (31 , , (33 , 13) has 1 √ , : (18 , , (39 , 17) has − √ ,T : (35 , , (37 , 29) has − , (36 , , (38 , 30) has 12 , : (5 , , (12 , , (13 , , (14 , , (33 , , (34 , 32) has 1 , : (9 , , (10 , , (11 , , (21 , , (23 , , (25 , 4) has √ , : (35 , , (36 , , (37 , , (38 , 29) has − √ ,T : (9 , 5) has − , (35 , , (36 , , (37 , , (38 , , (39 , 31) has − , : (40 , 32) has 12 , (6 , , (17 , , (18 , 8) has 1 , : (12 , 22) has − r , (22 , 1) has r , : (21 , 1) has − √ , (12 , 21) has 1 √ , : (15 , , (16 , , (27 , , (29 , 4) has √ , : (36 , , (38 , , (39 , , (40 , 31) has − √ , (35 , , (37 , 26) has √ ,T : (39 , 33) has − , (40 , 34) has 12 , : (2 , , (12 , , (17 , , (18 , , (27 , , (28 , , (29 , , (30 , 26) has 1 , : (15 , , (16 , , (36 , , (38 , 11) has − r , : (6 , , (22 , 5) has r , (6 , , (21 , 5) has 1 √ , : (15 , , (16 , , (35 , , (37 , 11) has − √ , (39 , , (40 , 33) has − √ ,T : (9 , 9) has − i, (2 , 2) has i, : (15 , , (16 , 16) has − √ , (17 , , (18 , , (39 , , (40 , 40) has − √ , : (23 , , (24 , , (25 , , (26 , 26) has 1 √ , : (12 , 12) has 12 ( −√ − i ) , (6 , 6) has 12 ( −√ i ) , : (35 , , (36 , , (37 , , (38 , 38) has 16 ( −√ − i ) , : (27 , , (28 , , (29 , , (30 , 30) has 16 (cid:16) −√ i (cid:17) , : (5 , 5) has 12 ( √ − i ) , (1 , 1) has 12 ( √ i ) , : (13 , , (14 , , (33 , , (34 , 34) has 16 ( √ − i ) , (10 , , (11 , 11) has 13 ( √ − i ) , : (7 , , (8 , , (31 , , (32 , 32) has 16 ( √ i ) , (3 , , (4 , 4) has 13 ( √ i ) ,T : (19 , 19) has − i (5 √ − i √ √ , : (31 , , (32 , , (33 , , (34 , , (39 , , (40 , 40) has − i (5 √ − i √ √ , : (4 , , (11 , , (16 , , (25 , , (26 , , (29 , , , (37 , , (38 , i (5 √ i √ √ , : (20 , 20) has − i (5 √ − i √ √ , (8 , , (14 , , (18 , 18) has i (5 √ i √ √ , : (1 , , (2 , , (5 , , (6 , , (9 , , (12 , , (21 , , (22 , 22) has (3 √ i √ √ , : (3 , , (10 , , (15 , , (23 , , (24 , , (27 , , (28 , , (35 , , (36 , √ − i √ √ , : (7 , , (13 , , (17 , 17) has (9 √ − i √ √ . The remaining generators can be obtained similarly. One has Tr( T a T a ∗ ) = ( N + 2) − l of the representation of SU (5) [27]. The SU (5) generators in the antisymmetric 10representation of SU (5) with three boxes The basis for the mixed ( ) tensors can be realized by ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) , ˆ u = 1 √ E − E + E − E + E − E ) . (E.1)After applying the 24 generators which are 5 × × 10 unitary matrix are given as follows as done in previous Appendices: T = − − , T = − − , The branching rule (Table A. 71 of [27]) of SU (5) → SU (3) × SU (2) × U (1) gives = = ( , ) +( , ) + ( , ) − which can be compared to the relation in (3.24). For N ≥ 7, the three box does not containthe complex conjugated notation. Note that the corresponding relation for the SU (11) branching in =is given by Table A. 77 of [27]. = , T = ,T = − − , T = − − ,T = , T = ,T = − − , T = ,T = diag (cid:18) , √ , √ , 16 ( −√ i ) , 16 ( −√ i ) , 16 ( √ i ) , 16 ( −√ − i ) , 16 ( −√ − i ) , 16 ( √ − i ) , − √ (cid:19) , = diag (cid:18) 120 (3 √ 10 + 5 i √ , 160 (9 √ − i √ , − √ 10 + i √ , 160 (9 √ − i √ , − √ 10 + i √ , − ( 1 √ 10 + i √ , 160 (9 √ − i √ , − √ 10 + i √ , − √ − i √ , − √ − i √ (cid:19) . The remaining generators can be obtained from these generators by taking the transpose and thecomplex conjugation. One has Tr( T a T a ∗ ) = ( N + 2 − N + 2 − 3) = 3 which is the index l ofthe representation of SU (5) [27]. For the generators of SU ( N + 2) with N = 5 , , , 11, thecorresponding ones in the higher representations more than three boxes are rather involved. References [1] M. R. Gaberdiel and R. 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