Gauged fermionic matrix quantum mechanics
GGauged fermionic matrix quantum mechanics
David Berenstein a and Robert de Mello Koch b,c, a Department of Physics, University of California,Santa Barbara, CA 93106, USA b School of Physics and Telecommunication Engineering , South China Normal University, Guangzhou 510006, China c National Institute for Theoretical Physics,School of Physics and Mandelstam Institute for Theoretical Physics,University of the Witwatersrand, Wits, 2050,South Africa
ABSTRACT
We consider the gauged free fermionic matrix model, for a single fermionic matrix. In thelarge N limit this system describes a c = 1 / [email protected] [email protected] a r X i v : . [ h e p - t h ] M a r ontents χ FR ( σ ) for the hook R
85 Fermion Schur polynomials are traces 126 Conclusion 16A Young’s Orthogonal Representation 17
The discovery of the gauge/gravity duality [1] has made precise how a theory of physicalstrings in higher dimensions arises from the large N limit of gauge theories, as originallysuggested by ’t Hooft [2]. Such a string theory (with appropriate boundary conditions) is infact now considered to be equivalent to gauge theory. This progress, in principle, has givenus a fully consistent description of quantum gravity.Unfortunately, many very interesting and difficult problems in quantum gravity are stilllargely inaccessible in the dual gauge theory dynamics: the dual gauge theory needs to besolved in the strong coupling regime.It is often useful to study simpler, even exactly solvable, large N models in order to tryto understand better the emergence of the additional dimensions in string theory. Manytimes such models arise as (possibly protected) sectors of a larger theory, or they can be astarting point to do perturbation theory in some coupling constant.This strategy has been very successful in the case of the half BPS sector in N = 4 SYMtheory. This sector is actually protected by supersymmetry and is generated by traces of asingle scalar field tr( Z k ), where Z is a highest weight state for SO (6) in the N = 4 gaugemultiplet. The first important result was the full exact diagonalization of the two pointfunction [3]. This was done using combinatorial techniques to express a complete basis ofstates in terms of of Young diagrams. The naive basis of traces is not orthogonal: there arenon-trivial overlaps generated at order 1 /N . These overlaps generate complications whenanalyzing anomalous dimensions in other setups.The main statement of that paper is that the states built this way are orthogonal. It was1hen realized that such a system could be thought of in terms of a 2D fermion description,similar to the integer quantum hall effect [4]. This description it made clear that stringscould be interpreted as chiral edge excitations of a droplet, and also that it is possible toidentify a class of D-branes (giant gravitons and dual giant gravitons [5, 6, 7], see also [8])in terms of single fermion and hole states in the droplet dynamics.A big surprise is that the fermion droplet prescription also describes the supergravitysolutions exactly [9] and for each droplet configuration in the gauge theory one can find asolution of supergravity that describes it. This sector alone has led to numerous additionalinsights in the theory of quantum gravity. Making the combinatorial problem of relating thetraces and Schur functions more precise, it has been noted that the topology of spacetimecan be changed by superposing states of a fixed topology [10, 11].In this paper we study the fermion counterpart of this dynamics: a single fermionicgauged matrix model. The states arising here can be thought of as a special class of states ofthe SU (1 |
1) sector of N = 4 SYM. These states are built from products of traces of a singleWeyl fermion (spin up) with tr( ψ k + ), but they are not protected by symmetry: a non-trivialanomalous dimension is generated at higher loop orders in perturbation theory. This sectorhas been analyzed in some detail in [12].Just like in the half BPS sector, the fermionic matrix model can be studied in its ownright. There is a basis of traces and another basis based on Young diagrams (Schur functions).This paper studies in detail the relationship between these two and we find that surprisingly,they are the same basis, although they have different normalizations. In contrast to thebosonic case, the Schur functions for the fermions involve a twist due to Fermi statistics.This has consequences for the map between basis states, which produce non-trivial factorsthat are square roots of integers. These arise when the representations of the symmetricgroup are treated explicitly using the Young orthogonal representation.The rest of the paper is organized as follows. In the next section we introduce the gaugedfermionic matrix model and review relevant background from the corresponding bosonicmodel. This section also develops a precise statement of the conjectured relation betweenthe trace and Schur function bases. Section 3 reviews the construction of the Schur functionsfor fermions. The novel ingredient in the construction is a twisted character, consideredin detail in section 4. This discussion is enough to prove a special case of the generalconjecture. In section 5 we give a complete proof of our conjecture. The proof uses elementsfrom the representation theory of both the symmetric and unitary groups, combinatorics,orthogonality at infinite N and the ring structure of multiplying by traces, to develop aninduction argument. We draw conclusions and suggest some avenues for further study insection 6. 2 Preliminaries
The gauged fermionic matrix model is defined by the following first order action S = (cid:90) dt tr (cid:2) ¯ ψiD t ψ − m ¯ ψψ (cid:3) (1)where D is a covariant derivative and ψ transforms in the adjoint of U ( N ). If we choosethe gauge A = 0 the dynamics is free with ψ acting as raising operators and ¯ ψ as loweringoperators giving rise to a fermionic Fock space of states.The vacuum is gauge invariant (a singlet of U ( N )). Any state in the dynamics can beaccessed by raising operators acting on the vacuum. However, we need to impose the Gauss’law constraint. A fermion ψ ab has an upper and lower U ( N ) index. These need to be fullycontracted to form a singlet. These contractions are in the form of traces. For example, asingle trace state acting on the vacuum is given by | k (cid:105) = tr( ψ k ) | (cid:105) (2)Using the cyclic property of the trace and the fermionic character of ψ it can be easily shownthat tr( ψ k ) = ( − k − tr( ψ k ) (3)so that only traces with an odd number of fields ψ are allowed. Each of these traces hasfermi statistics and has energy k = 2 s + 1, where s = 0 , , . . . . At infinite N each trace issupposed to correspond to a different ‘particle’.If we normalize the energies in units of 1 /
2, we get a single particle state for each halfinteger ( s + 1 / c = 1 / χ ( θ + 2 π ) = − χ ( θ )).This suggests that this is a matrix model for a c = 1 / c = 1 chiral boson in 1 + 1 dimensions. Themain insight of this map to the chiral boson is that in the bosonic matrix model one canreduce the dynamics to the eigenvalues of the matrix X . When the Gauss’ law is implementedthe eigenvalues act as fermions. This dynamics for bosons is usually best described in termsof a first order dynamics S = (cid:90) dt tr (cid:2) ¯ ZiD t Z − m ¯ ZZ (cid:3) (4)where Z = X + iP and ¯ Z are complex and the Gauss’ law constraint requires them tocommute. The corresponding fermions are in the phase space of X and the ground statecan be described by a droplet in phase space. This is a familiar story of the quantum halleffect. The eigenvalue dynamics is the effective field theory of free electrons in the lowestLandau level in 2 + 1 dimensions, which has been slightly deformed by a binding potential3roportional to x + y . The traces tr( Z k ) are collective excitations of the droplet withangular momentum k on the edge.Unlike the bosonic matrix model, in the fermionic setup we can not choose a gauge wherewe diagonalize the fermionic degrees of freedom. The interpretation of an edge dynamics isharder to do and will not be pursued here.Because the bosonic system can also be interpreted in terms of first quantized fermionsin a bulk 2 + 1 system, we can also write the wave functions in terms of Slater determinantsof single particle states. These wave functions are governed by Schur polynomials of thematrices. These are obtained by traces in irreducible representations of U ( N ), which arelabeled by Young diagrams. The map from multi-traces to Schur functions in the bosonicmodel is non-trivial [14]. It is also obtained from character expansions of the symmetricgroup. Edges of strongly coupled quantum systems can exhibit c = 1 / U ( N ) invariance. A free system actually has a larger symmetry U L ( N ) × U R ( N ) where the first U L ( N ) rotates only the upper indices of the fermions (as afundamental) and the second U R ( N ) rotates only the lower indices (as an antifundamental).If we take a k particle state in the Fock space, it will have k upper indices and k lower indices.We want to decompose the state into irreducible representation of U L ( N ). Since the state isin a tensor product of k fundamentals, it will be decomposed into irreducible representationsthat are labeled by a Young diagram: one needs to symmetrize or antisymmetrize the tensorindices. These are distinct irreducibles, they have different Casimir’s and because the actionof U L ( N ) is unitary, any pair of states labeled by two different diagrams are orthogonal toeach other. The same analysis can be done with the lower indices. Fermi statistics guaranteesthat symmetrizing in the upper indices corresponds to antisymmetrizing on lower indices.Now, we need to gauge the diagonal embedding U ( N ) → U ( N ) L × U R ( N ). We thus need theYoung diagram representation of the upper indices to be the same as the one for the lowerindices, but these are mirrors of each other. Thus only diagrams that are self-conjugate areallowed. There is a unique singlet for such self-conjugate representation. This will be calledthe Schur function. To properly define the Schur function requires building the map morecarefully, which will be described in the next section.Now, we want to motivate how the Schur functions for fermions and the traces basisshould be the same. 4o motivate this equality, consider the following identity for the bosonic matrix modeltr( Z k ) = (cid:88) single hooks ( − s − . . . ... (5)where the sum is over Young diagrams given by hooks with k boxes, and s is the number ofrows of the diagram. A precise accounting of how traces act on the full Schur basis can befound in [11]. Acting with a trace adds skew hooks of length k to a given Young diagram inall possible ways with a sign that is − naively substitute fermions in equation (5), basically arguing by analogy, we getquite a few restrictions because the only allowed Young tableaux are those that are equal totheir reflection about the diagonal. Basically, if a similar equation holds for fermions, thereis only one hook that appears in the sum. In this sense, a natural guess is that each trace isequal to a hook tr( ψ s +1 ) = α s . . . ... (6)where the single row and single column each has s + 1 boxes. Here we allow the possibilityof a non-trivial normalization factor. The origin of this factor is that at leading order in N we have that (cid:104) | tr( ¯ ψ s +1 ) tr( ψ s +1 ) | (cid:105) (cid:39) (2 s + 1) N s +1 (7)whereas for Young tableaux with k boxes we usually have that (cid:104) | Y T ( k ) Y T ( k ) | (cid:105) (cid:39) N k (8)In these equations we use fields with canonical normalization. Other normalizations arepossible so that the right hand side has no powers of N , and these are useful for taking thestrict N → ∞ limit.This suggests that α s = √ s + 1. This is very different to the bosonic matrix modelwhere the coefficients in the translation are all ± s + 1 just comes fromthe number of Young diagrams that contribute.Also, if we consider the general action of the product of a trace on a given Schur function,which is by adding skew hooks of length 2 s + 1 in all possible ways, the condition of reflectionsymmetry of the allowed Young tableaux means that there is only one place where the skewhook can be attached: it must be attached symmetrically with respect to the diagonal. Thehook can only be attached if the diagram does not already contain a hook of the givenlength. This is indicative of the Fermi statistics of the traces where one does not allowdouble occupation of a state. 5ow, let us describe the conjecture we will prove in this paper. Consider a trace structuretr( ψ s +1 ) . . . tr( ψ s k +1 ) | (cid:105) (9)describing a state in the field theory. We will show that this state is (up to a normalizationcoefficient) equal to the state given by the tableau . . . . . .
21 ... . . .... 21 (10)where there are exactly 2 s + 1 ones (the largest hook on the diagonal has 2 s + 1 boxes),there are exactly 2 s + 1 boxes with a label 2 etc, where the label just indicates how weassociate different traces to different hooks. The bold-face numbers are on the diagonal andthey label the hooks. We will also show that the normalization coefficient is ± (cid:81) i √ s i + 1. The Schur polynomial basis constructed in [3] for a single adjoint scalar, diagonalizes the freefield two point function and manifestly accounts for the trace relations that appear at finite N . In this section we will review the analogous construction, for a single adjoint fermion,given in [17].Consider a single fermion ψ ij transforming in the adjoint of the gauge group U ( N ). Thetwo point function is (cid:104) ψ ij ( ψ † ) kl (cid:105) = δ il δ kj (11)Since fermionic fields anticommute, it is important to spell out how products of the fermionfields are ordered. With the convention( ψ ⊗ n ) IJ = ψ i j ψ i j · · · ψ i n j n ( ψ †⊗ n ) KL = ψ † k n l n · · · ψ † k l ψ † k l (12)for ordering, the two point function is given by (cid:104) ( ψ ⊗ n ) IJ ( ψ †⊗ n ) KL (cid:105) = (cid:88) σ ∈ S n sgn( σ ) σ IL ( σ − ) KJ (13)where sgn( σ ) is the sign of permutation σ . The sign of the permutation is given by sgn( σ ) =( − m where m is the number of transpositions in the product. The ordering in (12) isadopted to ensure that there are no n dependent phases in (13).6ased on experience with the bosonic case, we expect the Schur polynomials are a linearcombination of traces (cid:88) σ ∈ S n C σ Tr V ⊗ n ( σψ ⊗ n ) (14)The anti-commuting nature of the fields must be reflected in the above sum. To see how thishappens, consider changing summation variable from σ to γ − σγ . The permutation γ swapsfields inside the trace. Since we are swapping fermions, we get − (cid:88) σ ∈ S n C σ Tr V ⊗ n ( σψ ⊗ n ) = (cid:88) σ ∈ S n C γ − σγ Tr V ⊗ n ( σγψ ⊗ n γ − )= (cid:88) σ ∈ S n C γ − σγ sgn( γ )Tr V ⊗ n ( σψ ⊗ n ) (15)Swapping fields must be a symmetry of the basis so that the coefficients must obey C γ − σγ = sgn( γ ) C σ (16)This is enough to determine the coefficients C σ . To show this we will make use of theClebsch-Gordan coefficient for R × R to couple to the antisymmetric irrep [1 n ], denoted by S [1 n ] R Rm m (cid:48) . These Clebsch-Gordan coefficients obey (this is obtained by specializing formula7-186 of [19]) Γ
Rij ( σ )Γ Rkl ( σ ) S [1 n ] R Rj l = sgn( σ ) S [1 n ] R Ri k (17)Assume without loss of generality that we have an orthogonal representation, so that S [1 n ] R Rm l Γ Rlk ( σ ) = sgn( σ )Γ Rmi ( σ ) S [1 n ] R Ri k ⇒ Γ S ( σ ) O = sgn( σ ) O Γ S ( σ ) (18)where O m m (cid:48) = S [1 n ] R Rm m (cid:48) . OO T commutes with every element of the group and hence isproportional to the identity matrix. Thus, after suitable normalization, we have OO T = (19) O can only be non-zero for self conjugate irreps because S [1 n ] R Rm m (cid:48) is only non-zero for selfconjugate irreps. Recall that given a Young diagram, the conjugate (or transposed) diagramis obtained by exchanging the roles of the rows and columns. A self-conjugate irrep is labeledby a Young diagram which coincides with its conjugate diagram. Note that C γ − σγ = Tr (cid:0) O Γ R ( γ − σγ ) (cid:1) = sgn( γ )Tr (cid:0) Γ R ( γ − ) O Γ R ( σ )Γ R ( γ ) (cid:1) γ ) C σ (20)which proves that the coefficients of our polynomials do indeed obey (16). Using thesecoefficients we immediately obtain the Schur polynomials for fermions. Spelling out indexstructures, our conventions are χ R ( ψ ) = 1 n ! (cid:88) σ ∈ S n χ FR ( σ ) ψ i i σ (1) · · · ψ i n i σ ( n ) χ † R ( ψ ) = 1 n ! (cid:88) σ ∈ S n χ FR ( σ ) ψ † i n i σ ( n ) · · · ψ † i i σ (1) (21)where we have introduced the twisted character χ FR ( σ ) ≡ Tr (cid:0) O Γ R ( α ) (cid:1) (22)The two point function is easily evaluated, to give (cid:104) χ R χ † S (cid:105) = δ RS f R (23)where f R is the product of factors, one for each box, of Young diagram R . Recall that thebox in row i and column j has factor N − i + j . χ FR ( σ ) for the hook R In this section we will evaluate the twisted character when R is the hook representation. Inthis case R has a single row of length > R is self conjugate.This implies that the number rows of length 1 in R is equal to the number of columns oflength 1. Our argument uses Young’s orthogonal representation for the symmetric group,which is reviewed in Appendix A.Before turning to the evaluation it is useful to review the explicit formula given in [19]for the Clebsch-Gordan coefficient S [1 n ] R Rm m (cid:48) = O mm (cid:48) . This requires that we know somethingabout how to label states in a given symmetric group irrep R . Towards this end, recall thata Young diagram with n boxes can be filled with a unique integer 1 , , · · · , n in each box.A tableau is called standard if the entries in each row and each column are increasing. Forevery standard tableau there is a unique state in the vector space carrying representation R and the dimension of symmetric group irrep R is given by the number of standard tableauthat can be obtained by filling R . For the self conjugate Young diagrams introduce thenotation | i (cid:105) = 1 2 43 56 | i T (cid:105) = 1 3 62 54 (24)which implies that there is a natural pairing of the states belonging to a self conjugate irrep.Consider the standard tableau | i = 1 (cid:105) = 1 2 34 56 (25)8ny other standard tableau is given a sign depending on how many swaps are needed to getit to match | (cid:105) . For an even number of swaps the sign is +1 and for an odd number it is − i , denote this sign by Λ i . Then formula (7-211a) of [19] says O = d R (cid:88) i =1 Λ i | i (cid:105)(cid:104) i T | (26)with d R the dimension of irrep R . Evaluating the character is computing the sum χ R ( Oσ ) = d R (cid:88) i =1 Λ i (cid:104) i T | Γ R ( σ ) | i (cid:105) (27)Notice that O has no diagonal elements. There is only a non-zero contribution to thetwisted character when σ can turn | i (cid:105) into | i T (cid:105) . In addition, because of (18) only elementswith sgn( σ ) = 1 can have a non-zero twisted character. Note also that elements in the sameconjugacy class have the same character, up to a sign χ FR ( σ ) = Tr (cid:0) O Γ R ( α ) (cid:1) = Tr (cid:0) O Γ R ( ρ )Γ R ( ρ − )Γ R ( α ) (cid:1) = sgn( ρ )Tr (cid:0) O Γ R ( ρ − )Γ R ( α )Γ R ( ρ ) (cid:1) = sgn( ρ ) χ FR ( ρ − σρ ) (28)Thus, to prove the character of a given permutation vanishes we can study any permutationin the conjugacy class.With these observations in hand, we will now argue that only a single conjugacy class hasa non-vanishing twisted character when R is a hook. If a hook representation is to be selfconjugate, the corresponding Young diagram must have an odd number of boxes. Considera hook with 2 k + 1 boxes. Flipping the standard tableau implies that all of the labels in thepattern greater than 1 change position. From the structure of Young’s orthogonal represen-tation we know that a permutation, after it is written in terms of adjacent transpositions,only swaps the labels of boxes that are named in the permutation. Thus, to satisfy the factthat all labels greater than 1 change position we know that all labels greater than 1 mustappear. A 2 k + 1 cycle may give a non-zero result. One might expect a 2 k cycle will givea non-zero result. This is not the case. We can see this in two ways as follows: (i) the 2 k cycle is odd so we know its twisted character vanishes and (ii) we could use the 2 k cycle(1 , , , · · · , k − , k ) that leaves 2 k + 1 inert. This permutation never moves label 2 k + 1and hence never changes | i (cid:105) into | i T (cid:105) . This second observation shows that we need all 2 k + 1labels to appear in the permutation for a non-zero result.Now, use all the 2 k + 1 labels to form a permutation built from smaller disjoint cycles.At least one of these cycles must have an odd length. The labels of boxes can be shuffledbetween boxes named in a given cycle, but we only mix boxes named in the same cycle.Choose the cycle with odd length to have the form (2 k + 1 , k, k − , ..., i + 1 , i, i − | i T (cid:105) from | i (cid:105) by just shuffling the labels of these boxes.9onsequently we conclude that for the self conjugate hook representation with 2 k + 1 boxes,only the 2 k + 1 cycle gives a non-zero twisted character.We will now evaluate the only non-zero character χ FR ( σ ) which is for σ a 2 k + 1 cycle.The evaluation is a straightforward application of the rules of Appendix A. The importantaspects of the computation are the following1. The only states that contribute to the character have patterns such that 2 i + 1 and 2 i for i = 1 , , · · · , k appear in different arms (horizontal or vertical) of the hook. Thisimplies that a total of 2 k states contribute to the character.2. The character picks up a factor of √ (2 i +1)(2 i − i for i = 1 , , · · · , k , because 2 i + 1 and2 i are swapped.3. The labels 2 i and 2 i − ± ± i − ,for i = 1 , , · · · , k . In the end the signs conspire so that only the overall sign is notfixed.The result for the only non-zero twisted character is χ FR ( σ ) = ± k (cid:89) i =1 (cid:112) (2 i + 1)(2 i − i (cid:18) i − (cid:19) = ±√ k + 1 (29)This has an interesting and immediate consequence for the Schur polynomial χ R ( ψ ) when R is a self conjugate hook. Recall that χ R ( ψ ) = 1(2 k + 1)! (cid:88) σ ∈ S k +1 χ FR ( σ ) ψ i i σ (1) · · · ψ i k +1 i σ (2 k +1) (30)The only contribution to the sum is for σ a 2 k + 1 cycle, and there are (2 k )! such terms. Thesign of the character is correlated with the sign of the trace ψ i i σ (1) · · · ψ i k +1 i σ (2 k +1) = ± Tr ( ψ k +1 ),so that in the end we find χ R ( ψ ) = 1 √ k + 1 Tr ( ψ k +1 ) (31)which is a special case of the general result we prove in this paper.Before concluding this section, we note that given the above value of the twisted character,there is a straightforwards extension to representations R made up by stacking hooks. Asan example, stacking hooks of length 13, 7 and 3 produces (32)10abel the representation by the hook lengths of the stacked hooks. The above representa-tion is labeled (13 , , χ FR ( σ ) with R the representation (2 k + 1 , k +1 , · · · , k l + 1) and σ a permutation with cycle structure (2 k + 1)(2 k + 1) · · · (2 k l + 1). Wewill write σ = σ k +1 σ k +1 · · · σ k l +1 . We make 3 basic observations:1. Not all states contribute to the trace. In going from | i (cid:105) to | i T (cid:105) it is clear that boxesare not swapped between hooks. Thus the labels appearing in the 2 k q + 1-cycle of σ must all populate the hook of length 2 k q + 1. All states that don’t obey this conditioncan be dropped as they don’t contribute.2. As reviewed in Appendix A, the action of a given permutation is determined by thecontent of the Young diagram. This content is the same for the stacked or un stackedhooks. Thus, the action of the (2 k l + 1) cycle on the hook of length 2 k l + 1 is thesame whether or not it is stacked in R . An illustration of this rule for the content isas follows 0 1 2 3 4 5 − − − − − − − − − − − − − − R that participate to the fermionic character. Theycan be decomposed into labels, one for each hook, and the labels runs over all thestates of that hook.The above observations taken together imply that there is a tensor product structure tothe subspace of states in R that contribute to the character and further that the action ofthe permutation σ splits up so that each (disjoint) cycle in σ acts on a different hook. Thus χ FR ( σ ) = (cid:88) I (cid:104) R, I | σ k +1 σ k +1 · · · σ k l +1 | R, I (cid:105) = (cid:88) i ,i , ··· i l (cid:104) (2 k + 1) , i | ⊗ (cid:104) (2 k + 1) , i | ⊗ · · · ⊗ (cid:104) (2 k l + 1) , i l | σ k +1 σ k +1 · · · σ k l +1 | (2 k + 1) , i (cid:105) ⊗ | (2 k + 1) , i (cid:105) ⊗ | (2 k l + 1) , i l (cid:105) = (cid:88) i (cid:104) (2 k + 1) , i | σ k +1 | (2 k + 1) , i (cid:105) · · · (cid:88) i l (cid:104) (2 k l + 1) , i l | σ k l +1 | (2 k l + 1) , i l (cid:105) = χ (2 k +1) F ( σ k +1 ) · · · χ (2 k l +1) F ( σ k l +1 ) Here ( n ) denotes a cycle of length n . The permutation ( n )( m ) comprises a disjoint n -cycle and m -cycle. ± l (cid:89) i =1 (cid:112) k i + 1 (34)where in the last line we used the fermionic character for a hook.This is almost the proof we need. We still need to show that no other cycle structure hasa non-trivial character on a given tableau as above. The result obtained in (31) shows that the fermion Schur polynomial labeled by a hook isgiven by a single trace. In this section this result will be generalized to any self conjugaterepresentation R , where there is a single trace structure that contributes.Any such representation is obtained by stacking self conjugate hooks. The result we willprove shows that the Schur polynomial for a representation obtained by stacking k hooksis equal to a product of k traces, one for each hook. The number of fields inside the traceequals the number of boxes in the corresponding hook. k is equal to the number of boxes onthe diagonal in R .The expression (34) will give the normalization that relates the trace structure to thetableau. To finish the proof we employ an induction argument. We use the idea that singletraces can be treated as orthogonal particles at large N : each trace represents a differentcreation operator, so that when we multiply by a trace we should get a new element of theFock space of states with that particle present.We will assume that we have proved the result for all tableaux with at most k selfconjugate hooks. In the particle language, this is assuming that we have proved the resultfor all occupation numbers less than or equal to k . The one hook result is the one particleresult. In that case, the single trace is equal to a single hook times the square root of thenumber of fields in the trace.The idea is then to add an extra trace (particle) and to prove the result with the extratrace included, which should now have k + 1 hooks and to account for the new states thatare generated this way. Since we have already shown the result for k = 1, this will prove theresult by induction for all k .Consider a trace structure of the form tr( ψ s ) . . . tr( ψ s k ). Using cyclicity and the anticom-muting nature of ψ it is easy to see that each power s i i = 1 , · · · , k is odd for a non-zero trace.Therefore each trace is an anticommuting variable. This means that for a non-zero productall the s i must be distinct from each other. Order the s i in decreasing order s > s > . . . s k .Also, associate to this trace structure the following permutation in S L with L = (cid:80) ki =1 s i (1 , , . . . , s )( s + 1 , s + 2 , . . . , s + s ) . . . ( s + · · · + s k − + 1 , . . . , L ) (35)12his permutation is constructed by taking the numbers from 1 , . . . , L and doing an orderedcycle on the first s elements 1 , . . . , s , a cycle of order s on the next set of elements etc.This gives a unique element of the permutation group for each trace structure.Now let the above permutation act on a self conjugate representation labeled by a Youngdiagram with L boxes. To get a non-zero answer, when acting an a given standard tableauthe labeling must be reflected by the permutation that we have chosen. For example, theaction of the permutation must send1 2 4 63 8 9 115 10 137 12 −→ ψ s k +1 ) to the structure, with s k +1 < s k . Thatis, we want to multiply the fermion Schur polynomial by the single hook tableaux with s k +1 elements. We will now do induction on s k +1 to check what the structure of the producttableaux should be.The first thing to notice is that a single permutation ( a , a + 1 , . . . , a + 2 s ) can onlyperform reflections between a pair of consecutive a + i , a + i + 1. To see this, note that thecycle permutation can be written as( a, a + 1 , . . . , a + 2 s ) = ( a + 2 s − , a + 2 s ) ◦ · · · ◦ ( a + 1 , a + 2) ◦ ( a, a + 1) (37)and a + i appears in at most two places. These are the only times in which the permutationcan move a + i .Since the number of elements inside the permutation is odd, a reflection of the labelsmust have a fixed point. Such a fixed point needs to be a fixed point of the labeling of thestandard tableau. For the example discussed in (36) above, 1 , k traces needs to have at the least k fixed points. Now we will showthat it has exactly k such fixed points.To prove this consider the representation of U ( N ) associated to the trace structure westart with and tensor it with the single self conjugate hook of s k +1 = 2 n − . . . n (38)When we tensor a fixed representation T with this one (in the sense of representations of U ( N )), we need to sum over all tableaux where we have added exactly 2 n − T . We13hen distribute the labels in the boxes above in the boxes added to T respecting the followingrule (the Littlewood-Richardson rule). Read the numbers in reverse order from the right tothe left in each successive row. This will produce a pattern like 1 , , , , , . . . which we callthe semistandard tableau associated to the product. The rules for filling the semistandardtableau are that in each row, the order of the labels is non-increasing from right to left, andin each column the labels are strictly decreasing from bottom to top (increasing from top tobottom). The second rule (adapted to the present case) is that there is at least one label 1before the label 2, one 2 before a 3 etc in the word of the pattern. One can check that itis impossible to add boxes in a shape of a 2 × ⊗ × k ≤ k and for arbitrarily big tableaux. Now lets us do induction on the value of s k +1 itself. Namely,we do induction on the total particle number and a second induction on the energy of thelightest particle.Let us start with the smallest possible value s k +1 = 1. We can only add one self dual14orner, and we can only do it if all the other s k are higher. For example ⊗ = (40)is allowed, but ⊗ = 0 (41)because we can not add a hook of length one on the diagonal for the second term..This procedure shows that we have generated all possible diagrams with exactly one boxin the k + 1 hook: we had all possible diagrams with k hooks before (by hypothesis) andnow we have produced all states with k + 1 hooks where the last hook has length one.At large N each such selfdual diagram with M boxes has norm N M . Dividing by N M ,which only depends on the number of boxes and not the particular shape, we get an or-thonormal basis.Let us go to the next case: by multiplying by tr( ψ ), due to large N factorization, weshould obtain a state that is orthogonal to all configurations that do not have a tr( ψ ) in it.In particular, when we multiply it by previous known diagrams with k self-dual hooks, if thetrace splits into more than one skew hook in the product, it generates tableaux that haveexactly one box in the last diagonal. But we already generated all of these by states thathave a tr( ψ ), and by orthogonality of the tableaux states, the coefficient with which they aregenerated must be zero. Hence the three boxes of tr( ψ ) must all lie in the same hook. Thatis, when we multiply by tr( ψ ) the only option is not to divide the skew hook and add toadd it using the prescription implied by the equality of the Schur and trace basis. We do thesame for tr( ψ ): it is easy to show that we can not split the hook into various skew hooksbecause they lead to states that are already accounted for by states produced with tr( ψ ) ortr( ψ ). The induction then becomes straightforward: any new product must be orthogonalto all states where the last hook has lower length, which were already all generated. Thatis, the last hook is not divided and all the boxes of the hook we are multiplying by mustbelong to the same hook.Now, this actually completes the proof that the bases are proportional to each other. Theargument uses orthogonality at infinite N , but what we are studying is the ring structure ofmultiplying by traces which is independent of N . What we are seeing is how multiplying bytraces produces new states.The final check is that the norm of a product of traces is the product of the norms of theindividual traces, but this is already implicit in (34), where we computed the character andshowed that we have a product structure. 15 neat corollary of the result here is that when we take products of twisted Schurs, thetwisted Littlewood Richardson coefficients are all ± , In this paper we have exhibited a remarkable connection between the basis of traces and thebasis of Schur functions, constructed for a single fermionic gauged matrix model: they arethe same bases, albeit with different normalizations. To prove the equivalence we have hadto develop some new formulas for twisted characters. The results of this paper provide acomplete set of twisted character values for any permutation and any representation, whichis more than what is known explicitly for the usual symmetric group characters. The proof ofthe relation between the Schur and trace bases itself is performed by doing a double induction,both on the number of traces and in the number of fields in the last trace added. It usesrepresentation theory of both the symmetric and unitary group, large N factorization andthe ring structure of multiplying by traces and it needs all of these ingredients to work. Thetwisted Schurs form a ring and the structure constants of this ring are the twisted LittlewoodRichardson coefficients. Our results prove that the twisted Littlewood Richardson coefficientsare all ± , N we expect that expectation values of products of traces factorize, butthere should be mixing corrections of order 1 /N . For the fermionic matrix model we havestudied here, our results prove that there is a similar factorization, but in this case thefactorization is exact. There are still corrections to the norm of individual states that arepowers of 1 /N , but the system has a well defined notion of particle number at any value of N . It would be fascinating to properly explore the consequences of this factorization.Our results probably have immediate application to some operator-state problems inCFT. As an example, there are single fermion sectors in the free N = 4 SYM in the SU (1 | c = 1 / Acknowledgements
16e would like to thank Sanjaye Ramgoolam for useful discussions. The work of D.B. issupported in part by the Department of Energy under grant DE-SC 0011702. The work ofRdMK is supported by the South African Research Chairs Initiative of the Department ofScience and Technology and National Research Foundation as well as funds received fromthe National Institute for Theoretical Physics (NITheP).
A Young’s Orthogonal Representation
This representation is specified by giving the action of the “adjacent transpositions” whichare swaps of the form ( i, i + 1). A box in row i and column j has content j − i . Here is anexample of a Young diagram with the content of each box displayed0 1 2 3 4 5 − − − − − − − a in the standard tableau have content c a . The state | ST ( a ↔ a + 1) (cid:105) is labeled by the tableau obtained by swapping a and a + 1 in | ST (cid:105) . Young’s orthogonalrepresentation is defined by( i, i + 1) | ST (cid:105) = 1 c i +1 − c i | ST (cid:105) + (cid:115) − c i +1 − c i ) | ST ( a ↔ a + 1) (cid:105) (43)This defines the irrep because any element of the group can be written as a product ofadjacent permutations. The following example is obtained using (43)(3 , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:43) = 13 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:43) + √ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:43) (44) References [1] J. M. Maldacena, “The Large N limit of superconformal field theories and supergrav-ity,” Int. J. Theor. Phys. , 1113 (1999) [Adv. Theor. Math. Phys. , 231 (1998)]doi:10.1023/A:1026654312961, 10.4310/ATMP.1998.v2.n2.a1 [hep-th/9711200].[2] G. ’t Hooft, “A Planar Diagram Theory for Strong Interactions,” Nucl. Phys. B , 461(1974). doi:10.1016/0550-3213(74)90154-0 173] S. Corley, A. Jevicki and S. Ramgoolam, “Exact correlators of giant gravi-tons from dual N=4 SYM theory,” Adv. Theor. Math. Phys. , 809 (2002)doi:10.4310/ATMP.2001.v5.n4.a6 [hep-th/0111222].[4] D. Berenstein, “A Toy model for the AdS / CFT correspondence,” JHEP , 018(2004) doi:10.1088/1126-6708/2004/07/018 [hep-th/0403110].[5] J. McGreevy, L. Susskind and N. Toumbas, “Invasion of the giant gravitons fromAnti-de Sitter space,” JHEP , 008 (2000) doi:10.1088/1126-6708/2000/06/008 [hep-th/0003075].[6] M. T. Grisaru, R. C. Myers and O. Tafjord, “SUSY and goliath,” JHEP , 040 (2000)doi:10.1088/1126-6708/2000/08/040 [hep-th/0008015].[7] A. Hashimoto, S. Hirano and N. Itzhaki, “Large branes in AdS and their field theorydual,” JHEP , 051 (2000) doi:10.1088/1126-6708/2000/08/051 [hep-th/0008016].[8] V. Balasubramanian, M. Berkooz, A. Naqvi and M. J. Strassler, “Giant gravitons inconformal field theory,” JHEP , 034 (2002) doi:10.1088/1126-6708/2002/04/034 [hep-th/0107119].[9] H. Lin, O. Lunin and J. M. Maldacena, “Bubbling AdS space and 1/2 BPS geometries,”JHEP , 025 (2004) doi:10.1088/1126-6708/2004/10/025 [hep-th/0409174].[10] D. Berenstein and A. Miller, “Can Topology and Geometry be Measured by an Oper-ator Measurement in Quantum Gravity?,” Phys. Rev. Lett. , no. 26, 261601 (2017)doi:10.1103/PhysRevLett.118.261601 [arXiv:1605.06166 [hep-th]].[11] D. Berenstein and A. Miller, “Superposition induced topology changes in quantumgravity,” JHEP , 121 (2017) doi:10.1007/JHEP11(2017)121 [arXiv:1702.03011 [hep-th]].[12] M. Staudacher, “The Factorized S-matrix of CFT/AdS,” JHEP , 054 (2005)doi:10.1088/1126-6708/2005/05/054 [hep-th/0412188].[13] D. Berenstein, “A Matrix model for a quantum Hall droplet with manifest particle-hole symmetry,” Phys. Rev. D , 085001 (2005) doi:10.1103/PhysRevD.71.085001 [hep-th/0409115].[14] M. Stone, “Schur Functions, Chiral Bosons and the Quantum Hall Effect Edge States,”Phys. Rev. B , 8399 (1990). doi:10.1103/PhysRevB.42.8399[15] A. Kitaev, “Anyons in an exactly solved model and beyond,” Annals Phys. , no. 1,2 (2006). doi:10.1016/j.aop.2005.10.005 1816] Kasahara, Y. and Ohnishi, T. and Mizukami, Y. and Tanaka, O. and Ma, Sixiao andSugii, K. and Kurita, N. and Tanaka, H. and Nasu, J. and Motome, Y. and Shibauchi, T.and Matsuda, Y., “Majorana quantization and half-integer thermal quantum Hall effect ina Kitaev spin liquid”, Nature, vol 559, pp 227-231, 2018 doi = 10.1038/s41586-018-0274-0[arxiv:1805.05022[cond-mat.str-el]][17] R. de Mello Koch, P. Diaz and N. Nokwara, “Restricted Schur Polynomialsfor Fermions and integrability in the su(2—3) sector,” JHEP1303