On Classification of Modular Tensor Categories

Transcription

1 Commun. Math. Phys. 9, (009) Digital Object Identifier (DOI) 0.007/s z Communications in Mathematical Physics On Classification of Modular Tensor Categories Eric Rowell,, Richard Stong,, Zhenghan Wang 3, Department of Mathematics, Texas A&M University, College Station, TX 77843, U.S.A. Center for Communications Research, 430 Westerra Court, San Diego, CA 9-969, U.S.A. 3 Microsoft Station Q, University of California, CNSI Bldg, Rm 37, Santa Barbara, CA , U.S.A. zhenghwa@microsoft.com Received: 6 December 007 / Accepted: 3 June 009 Published online: 5 August 009 The Author(s) 009. This article is published with open access at Springerlink.com Abstract: We classify all unitary modular tensor categories (UMTCs) of rank 4. There are a total of 35 UMTCs of rank 4 up to ribbon tensor equivalence. Since the distinction between the modular S-matrix S and S has both topological and physical significance, so in our convention there are a total of 70 UMTCs of rank 4. In particular, there are two trivial UMTCs with S = (±). Each such UMTC can be obtained from 0 non-trivial prime UMTCs by direct product, and some symmetry operations. Explicit data of the 0 non-trivial prime UMTCs are given in Sect. 5. Relevance of UMTCs to topological quantum computation and various conjectures are given in Sect. 6.. Introduction A modular tensor category (MTC) in the sense of V. Turaev determines uniquely a (+)-topological quantum field theory (TQFT) [Tu] (a seemingly different definition appeared in [MS].) The classification of MTCs is motivated by the application of MTCs to topological quantum computing [F,Ki,FKW,FLW,FKLW,P], and by the use of MTCs in developing a physical theory of topological phases of matter [Wil,MR, FNTW,Ki,Wa,LWe,DFNSS]. G. Moore and N. Seiberg articulated the viewpoint that rational conformal field theory (RCFT) should be treated as a generalization of group theory [MS]. The algebraic content of both RCFTs and TQFTs is encoded by MTCs. Although two seemingly different definitions of MTCs were used in the two contexts [MS,Tu], the two notions are essentially equivalent: an MTC in [MS] consists of essentially the basic data of a TQFT in [Wal]. The theory of MTCs encompasses the most salient feature of quantum mechanics in the tensor product: superposition. Therefore, even without any applications in mind, the classification of MTCs could be pursued as a quantum generalization of the classification of finite groups. The first author is partially supported by NSA grant H The second and third authors are partially supported by NSF FRG grant DMS

2 344 E. Rowell, R. Stong, Z. Wang There are two natural ways to organize MTCs: one by fixing a pair (G,λ), where G is a compact Lie group, and λ a cohomology class H 4 (BG; Z); and the other by fixing the rank of an MTC, i.e. the number of isomorphism classes of simple objects. If a conjecture of E. Witten were true, then every MTC would come from a Chern-Simons-Witten (CSW) TQFT labelled by a pair (G,λ) [Witt,MS,HRW]. Classification by fixing a compact Lie group G has been carried out successfully for G=finite groups [DW,FQ], G = T n torus [Ma,BM], and G = A, B, C, D simple Lie groups [FK,KW,TW]. In this paper, we will pursue the classification by fixing the rank. This approach is inspired by the study of topological phases of matter and topological quantum computing. Another reason is that we have evidence that there might be exotic (+)-TQFTs other than CSW theories [HRW]. Topological phases of matter are like artificial elements. The only known topological phases of matter are fractional quantum Hall liquids: electron systems confined on a disk immersed in a strong perpendicular magnetic field at extremely low temperatures [Wil,DFNSS]. Electrons in the disk, pictured classically as orbiting inside concentric annuli around the origin, organize themselves into some topological order [Wen,WW,WW]. Therefore, the classification of topological phases of matter resembles the periodic table of elements. The periodic table does not go on forever, and simpler elements are easier to find. The topological quantum computing project is to find MTCs in Nature, in particular those with non-abelian anyons. Therefore, it is important that we know the simplest MTCs in a certain sense because the chance for their existence is better. There is a hierarchy of structures on a tensor category: rigidity, pivotality, sphericity. We will always assume that our category is a fusion category: a rigid, semi-simple, C-linear monoidal category with finitely many isomorphism classes of simple objects, and the trivial object is simple. It has been conjectured that every fusion category has a pivotal structure [ENO]. Actually, it might be true that every fusion category is spherical. Another important structure on a tensor category is braiding. A tensor category with compatible pivotal and braiding structures is called ribbon. In our case a ribbon category is always pre-modular since we assume it is a fusion category. For each structure, we may study the classification problem. The classification of fusion categories by fixing the rank has been pursued in [O,O]. Since an MTC has considerably more structures than a fusion category, the classification is potentially easier, and we will see that this is indeed the case in Sects. 3 and 4. The advantage in the MTC classification is that we can work with the modular S matrix and T matrix to determine the possible fusion rules without first solving the pentagon and hexagon equations. For the classification of MTCs of a given rank, we could start with the infinitely many possible fusion rules, and then try to rule out most of the fusion rules by showing the pentagon equations have no solutions. However, pentagon equations are notoriously hard to solve, and we have no theories to practically determine when a solution exists for a particular set of fusion rules (Tarski s theorem on the decidability of the first-order theory of real numbers provides a logical solution). So being able to determine all possible fusion rules without solving the pentagon equations greatly simplifies the classification for MTCs. As shown in [HH], all structures on an MTC can be formulated as polynomial equations over Z. Hence the classification of MTC is the same as counting points on certain algebraic varieties up to equivalence. But all the data of an MTC can be presented over a certain finite degree Galois extension of Q, probably over an abelian Galois extension of Q if normalized appropriately. Therefore, the classification problem is closer to number theory than to algebraic geometry. The argument in Sects. 3 and 4 is basically Galois theory

3 On Classification of Modular Categories 345 plus elementary yet complicated number theory. To complete the classification, we need to solve the pentagons and hexagons given the fusion rules. A significant complication comes from the choices of bases of the Hom spaces when solving the pentagon equations. The choices of basis make the normalization of 6 j symbols into an art: so far no computer programs are available to solve pentagons with a fusion coefficient >, but one set of such fusion rules is solved completely [HH]. Currently, there are no theories to count the number of solutions of pentagon equations for a given set of fusion rules without solving the pentagons. For unitary MTCs, there is tension between two desirable normalizations for 6 j symbols: to make the F matrix unitary, or to present all data of the theory in an abelian Galois extension of Q. For the Fibonacci theory, unitarity of the F matrix and abelianess of the Galois extension of Q cannot be achieved simultaneously, but with different F matrices, each can be obtained separately [FW]. This is the reason that we will only define the Galois group of a modular fusion rule and a modular data, but not the Galois group of an MTC. The main result of this paper is the classification of MTCs with rank=, 3 and unitary MTCs of rank=4. The authors had obtained the classification of all unitary MTCs of rank 4 in 004 [Wa]. The delay is related to the open finiteness conjecture: There are only finitely many equivalence classes of MTCs for any given rank. By Ocneanu rigidity the conjecture is equivalent to: There are only finitely many sets of fusion rules for MTCs of a given rank. Our classification of MTCs of rank 4 supports the conjecture. We also listed all quantum group MTCs up to rank in Sect. 5. Two well-known constructs of MTCs are the quantum group method, and the quantum double of spherical tensor categories or the Drinfeld center. The quantum double is natural for MTCs from subfactor theory using Ocneanu s asymptotic inclusions [EK]. It seems that this method might produce exotic MTCs in the sense of [HRW]. Our main technique is Galois theory. Galois theory was introduced into the study of RCFT by J. de Boer and J. Goeree [dbg], who considered the Galois extension K of Q by adjoining all the eigenvalues of the fusion matrices. They made the deep observation that the Galois group of the extension K over Q is always abelian. This result was extended by A. Coste and T. Gannon who used their extension to study the classification of RCFTs [CG]. Fusion rules of an MTC are determined by the modular S-matrix through Verlinde formulas. It follows that the Galois extension K is the same as adjoining to Q all entries of the modular S matrix. When a Galois group element applies to the S matrix entry-wise, this action is a multiplication of S by a signed permutation matrix, which first appeared in [CG]. It follows that the entries of the S matrix are the same up to signs if they are in the same orbit of a Galois group element. For a given rank 4, this allows us to determine all possible S-matrices, therefore, all possible fusion rules. Note that the Galois group of a modular data does not change the fusion matrices, but it can change a unitary theory into a non-unitary theory. For example, the Galois conjugate of the Fibonacci theory is the Yang-Lee theory, which is non-unitary. We might expect that for each modular data, one of its Galois conjugates would be realized by a unitary ( MTC. ) This is actually ( ) false. For example, take a rank= modular data with 0 S =, and T =. No Galois actions can change the S matrix, hence 0 i the quantum dimension of the non-trivial simple object from to, though the same fusion rules can be realized by a unitary theory: the semion theory. Reference [Ro] contains a set of fusion rules which has non-unitary MTC realizations, but has no unitary realizations at all.

4 346 E. Rowell, R. Stong, Z. Wang A Table. Unitary MTCs of rank 4 Z (A, 3) A 4 N 4 U Z 3 (A, ) (A, 5) A 4 N 6 N 4 U A 0 A 8 N 8 N 4 N 6 Z Z Z 4 (A, 3) (A, 7) Fib Fib U U U The paper is organized as follows. In Sect., we study the implications of the Verlinde formulas using Galois theory. In Sects. 3 and 4, we determine all self-dual modular S matrices of modular symbols of rank=, 3, and unitary ones for rank=4. Rank= is known to experts, and rank=3 fusion rules have been previously classified [CP]. For modular data, Theorems 3. and 3. can also be deduced from [O,O]. In Sect. 5, we determine all UMTCs of rank 4. In Sect. 6, we discuss some open questions about the structure and application of MTCs. In the Appendix, together with S. Belinschi, we determine all non-self dual unitary modular data of rank 4. We summarize the classification of all rank 4 unitary MTCs into Table. There are a total of 70 unitary MTCs of rank 4 (a total of 35 up to ribbon tensor equivalence). The count is done in Sect Each such UMTC can be obtained from 0 non-trivial prime UMTCs by direct product, and some symmetry operations. The 0 non-trivial prime UMTCs are the semion MTC, the Fibonacci MTC or (A, 3),the Z 3 MTC, the Ising MTC, the (A, ) MTC, the even half of an SU() MTC at level 5or(A, 5),theZ 4 MTC, the toric code MTC, the (D 4, ) MTC, and the even half of an SU() MTC at level 7 or (A, 7). Their explicit data are listed in Sect Out of the 0 non-trivial prime UMTCs, 9 are quantum group categories for a simple Lie group: the semion=su(), the Fibonacci=(G ),thez 3 =SU(3), the Ising=complex conjugate of (E 8 ),the(a, )=SU(),theZ 4 = SU(4), the toric code= Spin(6), the (D 4, ) = Spin(8), and the (A, 7) =complex conjugate of (G ). The Ising MTC and the SU() MTC have the same fusion rules, but the Frobenius-Schur indicators of the non-abelian anyon σ are +,, respectively. The toric code MTC and the Spin(8) MTC have the same fusion rules, but the twists are {,,, }, and {,,, }, respectively. We choose q = e πi l in the quantum group construction. In the Ising case, it is the q = e πi l theory for E 8 at level=. For notation and more details, see Sect We do not know how to construct (A, 5) by cosets of quantum group categories. The information for each rank is contained in one row of Table. Each box contains information of the MTCs with the same fusion rules. The center entry in a box denotes the realization of the fusion by a quantum group category or their products. We also use Fib to denote the Fibonacci category (A, 3). The upper left corner has either A or N, where A means that all anyons are abelian, and N that at least one type of anyons is

5 On Classification of Modular Categories 347 non-abelian. The right upper corner has a number which is the number of different unitary theories with that fusion rule. If the lower right corner has a U, it means that at least one type of anyons has universal braiding statistics for topological quantum computation. The detailed information about which anyon is abelian or non-abelian, universal or non-universal is given in Sect It is worth noticing that the list of all fusion rules up to rank=4 agrees with the computer search for RCFTs in [GK]. We believe this continues to be true for rank=5. The rank= 6 list in [GK] is not complete. Finally, we comment on the physical realization of UMTCs. The existence of abelian anyons in ν = 3 FQH liquids is established theoretically with experimental support, while non-abelian anyons are believed to exist at the ν = 5 and ν = 5 plateaus (see [DFNSS] and the references therein). Current experimental effort is focused on FQH liquids at ν = 5. But the fermionic nature of electrons complicates direct application of MTCs to FQH liquids because only anyonic properties of bosonic systems can be described fully by MTCs. In other words, we need a refined theory, e.g. a spin MTC, to describe a fermionic system [DW,BM].. Galois Theory of Fusion Rules In this section, we study the implication of Verlinde formulas for fusion rules of MTCs. For more related discussion, see the beautiful survey [G]. Definition.. () A rank=n label set is a finite set L of n elements with a distinguished element, denoted by 0, and an involution ˆ: L L such that ˆ0 = 0. A label i L is self dual if î = i. The charge conjugation matrix is the n nmatrixc = (δ i ĵ ). Note that C is symmetric and C = I n,then n identity matrix. () A rank=n modular fusion rule is a pair (N ; S), where N is a set of n n nmatrices N i = (ni, k j ) 0 j,k n, indexed by a rank=n label set L, with ni, k j Q, and S = ( s ij ) 0 i, j n is an n n matrix satisfying the following: (a) s 00 =, s = s i, ĵ i, j, and all s i,0 s are non-zero; n (b) If we let D = i=0 s i,0, then S = S D is a symmetric, unitary matrix. Furthermore, the matrices N i in N and S are related by the following: N i S = S i (.) for all i L, where i = (δ ab λ ia ) n n is diagonal, and λ ia = s ia s 0a. The identities (.) or equivalently the Verlinde formulas (.3) below imply many symmetries among ni, k j :nk 0, j = δ jk, ni, k j = nk j,i = n ˆk = n ĵ. î, ĵ i,ˆk The matrix N i will be called the i th fusion matrix. From identities (.), the diagonal entries in i are the eigenvalues of N i, and the columns of S are the corresponding eigenvectors. The non-zero number D will be called the total quantum order, d i = s i0 the quantum dimension of the i th label, and D = n i=0 d i the global quantum dimension. (3) A rank=n modular symbol consists of a triple (N ; S, T ). The pair (N ; s S 00 ) is a rank=n modular fusion rule with all ni, k j N ={0,,, }(here s 00 is the (0,0)- entry of the unitary matrix S = (s ij ) 0 i, j n ), and the n nmatrixt = (δ ab θ a ) n n is diagonal, and θ 0 =. Furthermore, S and T satisfy

6 348 E. Rowell, R. Stong, Z. Wang (i) (ST) 3 = (D + s 00 )S ; (ii) S = C; (iii) θ i U() and θî = θ i for each i, where D ± = n. The following identity can be deduced: i=0 θ i ± di D + D = D. (.) The complex number θ i will be called the twist of the i th label. Note that s 00 might be D. A modular symbol is called unitary if each quantum dimension d i is the Frobenius-Perron eigenvalue of the corresponding fusion matrix N i. In particular, the quantum dimensions d i s are positive real numbers. (4) A modular symbol (N ; S, T ) is called a modular data if there is an MTC whose fusion rules, modular S-matrix, and T -matrix are given by N, s S 00, T of the modular symbol. (5) Let ={λ ij } i, j L for a rank=n modular fusion rule, and let K = Q(λ ij ), i, j L be the Galois extension of Q. Then the Galois group G of the Galois field K over Q is called the Galois group of the modular fusion rule. We are interested in searching for n + tuples (N 0,...,N n ; S, T ) related in the correct fashion. We will index the rows and columns of matrices by 0,,...,n. Since N i S = S i, the columns of S must be eigenvectors of N i with eigenvalues λ i,0, λ i,,..., and λ i,n, respectively. Looking at the first entries of these columns and of N i S, and using the only non-zero of the first row of N i, we see that λ i,0 = d i, and d j λ i, j = s i, j. It follows that K is the same as Q( s ij ), i, j L. Since S is symmetric, we see that for i = j we have d j λ i, j = d i λ j,i, and s i, j = d i λ j,i = d j λ i, j for all i and j.letni, k j denote the ( j, k) entry of N i. Since N i = we compute for 0 j, k n, n ni, k j = m=0 λ i,0 0 0 D S 0 λ i, λ i,n s i,m s j,m s k,m n D = d m m=0 S, λ i,m λ j,m λ k,m d m D. (.3) The fusion matrices can also be described equivalently by fusion algebras. For a rank=n fusion rule, each label i is associated with a variable X i. Then the fusion ring R is the free abelian ring Z[X 0,...,X n ] generated by X i s modulo relations (called fusion rules) X i X j = n i=0 nk i, j X k. The fusion algebra will be F = R Z K, where K is the Galois field of the fusion rules above. We may replace K by C. If the modular fusion rule is realized by an MTC, then X i is an equivalence class of simple objects, and the multiplication X i X j is just the tensor product. There are modular symbols that are not modular data. Example.. Take the following: S = 0,

7 On Classification of Modular Categories 349 and T = Diag(,θ, ). The fusion matrices N i are determined by the formulas (.3), hence are independent of θ. They are the same as those of the Ising MTC in Sect Therefore, for any θ U(), we get a modular symbol. But only when θ is a 6 th root of unity, do we have modular data. Very likely the modular symbol of an MTC determines the MTC, and we do not know when a modular symbol becomes a modular data. Proposition.3. If (N ; S, T ) is a modular data, then we have: () θ i θ j s ij = k nk s k0θ k. îj () j θ A 4 ij j = θ 3 j A ij i, where A ij = n j iî ni ij + n j ii ni. jî (3) Let ν k = D i, j L ni k, j d θi id j, then ν θ k is 0 if k = ˆk, and is ± if k = ˆk. ν k is j called the Frobenius-Schur indicator of k. (4) D + s 00 = e πic 4 for some c Q. The rational number c mod 8 is called the topological central charge of the modular data. Proof. For (), see [BK, Eq. (3..)] on p. 47. For (), it is [BK, Theorem 3..9] found on p. 57. Formula (3) from [Ba] for RCFTs can be generalized to MTCs. (4) follows from Theorem.5. Proposition.3 () implies that the θ i are actually roots of unity of finite order, which is often referred to as Vafa s Theorem. But from Example., we know that this is not true for general modular symbols, in particular Q(θ i ) might not be algebraic for modular symbols. This leads to: Definition.4. Given a modular data (N ; S, T ), letk N be the Galois field Q( s ij, D,θ i ), i, j L. Then the Galois group of K N over Q will be called the Galois group of the modular data. Theorem.5. () (de Boer-Goeree theorem): The Galois group of a modular fusion rule is abelian. () The Galois group of a modular data is abelian. By the Kronecker-Weber theorem, there is an integer m such that K N Q(ζ m ), where ζ m = e πi m. The smallest such m for K N is called the conductor of K N, and the order of T always divides N (we intentionally build N into the notation K N ). The Galois group of Q(ζ N ) is the cyclic group of units l such that gcd(l, N) =. Each l acts on K N as the Frobenius map σ l : ζ N ζn l. Consequently, σ l(t ) = T l and σ l (S) = S P σ, where the signed permutation matrix P σ corresponds to the Galois element σ in the Galois group of the modular fusion rule. It is known that the fusion algebra of a rank=n MTC is isomorphic to the function algebra of n points. A Galois group element σ of the associated modular fusion rule induces an isomorphism of the fusion algebra. It follows that σ determines a permutation of the label set. When we have only a modular fusion rule, the two algebra structures on the fusion algebra a priori might not be isomorphic to each other. But still a Galois group element of the modular fusion determines a permutation of the label set and the

8 350 E. Rowell, R. Stong, Z. Wang de Boer-Goeree theorem holds. Actually what we are using in this paper are identities among modular S entries up to some parity signs ɛ i,σ =±associated to each Galois element σ. Such parity signs first appeared in [CG] for Galois automorphisms of Q(λ i, j, D). First we note the following easy, but very useful fact that the ordered set of eigenvalues of N i determines the label i uniquely. Proposition.6. There do not exist indices j = k such that λ i, j = λ i,k for all i for any modular fusion rule (N ; S). Proof. If there were such indices, then the dot product of rows j and k of S would be D = n i=0 s i, j > 0, a contradiction. Except (5), the following theorem is contained in [CG]. Theorem.7. Let G be the Galois group of a rank=n modular fusion rule (N ; S). Then () the simultaneous action of the Galois group G on the set ={λ ij } gives an injective group homomorphism ι : G S n, where S n is the permutation group of n letters; for σ G, ι(σ)(i) is the associated element in S n. () For any σ G, the matrix P σ = d σ(0) S σ( S) is a signed permutation matrix; furthermore, the map σ P σ gives a group homomorphism from G to the signed permutation matrices modulo ± which lifts ι. (3) For each σ G, there are ɛ i,σ =±such that Moreover, and σ( s j,k ) = d σ(0) ɛ σ(k),σ s j,σ (k). (.4) s j,k = ɛ σ(j),σ ɛ k,σ s σ(j),σ (k), (.5) ɛ σ (k),σ = ɛ σ(0),σ ɛ 0,σ ɛ k,σ. (.6) (4) The Galois group G is abelian. (5) If n is even, then n i=0 ɛ i,σ = ( ) σ D. If n is odd, then D K, and σ(d) = ɛ σ d σ(0), where ɛ σ =±. We have n i=0 ɛ i,σ = ɛ σ ( ) σ. We are going to use σ for both the element of the Galois group G and its associated element of S n. When σ G applies to a matrix, σ applies entry-wise. Proof. Let K = Q[{λ i, j } 0 i, j n ] be the Galois extension of Q generated by the eigenvalues of all the N i and let G be the associated Galois group as above. The action of G on the eigenvalues gives an injection G S n S n S n, where there are n factors. Note that we have not assumed the N i have distinct eigenvalues, therefore this map is not necessarily unique and is not necessarily a group homomorphism. This is not a problem as we will resolve the ambiguity shortly. Just fix one such map for now. Let (σ,σ,...,σ n ) denote the image of σ G under this injection. Note that a priori,

9 On Classification of Modular Categories 35 there is no relationship between the σ i.let i be the diagonal matrix with diagonal entries λ i, j,so N i = S i S. Let P σi = (δ i=σi ( j)) 0 i, j n be the permutation matrix corresponding to σ i. Since σ(λ i, j ) = λ i,σi ( j), wehaveσ( i ) = Pσ i i P σi. Since N i is rational we have S i S = N i = σ(n i ) = σ( S)P σ i i P σi σ( S). Rewriting this gives i [ S σ( S)P σ i ]=[ S σ( S)P σ i ] i. Hence B i,σ = S σ( S)P σ i commutes with i. It follows that B i,σ is block diagonal, with blocks corresponding to the equal eigenvalues of N i. In formulas, if the ( j, k) entry of B i,σ is nonzero, then λ i, j = λ i,k.let S σ( S) = B i,σ P σi = C σ. Note two facts, if the ( j, k) entry of C σ is nonzero, then the ( j,σ i (k)) entry of B i,σ is nonzero and hence λ i, j = λ i,σi (k). The second fact is that C σ (as the notation suggests) does not depend on i, only on σ. Suppose C σ has nonzero entries in column k, saythe( j, k) and (l, k) entries. Then λ i, j = λ i,σi (k) = λ i,l for all i, contradicting Proposition.6 above. If a row or column of C σ is all zeroes, then det(c σ ) = 0, a contradiction. Hence C σ has exactly one nonzero entry in every row and in every column. Thus there is a unique permutation σ S n and a diagonal matrix B σ such that C σ = B σ P σ. Note that we are now using σ for both the element of the Galois group and its associated element of S n. Note that C σσ = S σσ ( S) = S σ( S)σ ( S σ ( S)) = C σ σ(c σ ), from which it follows that the map G S n is a group homomorphism. Thus we have proved that the simultaneous action of the Galois group G on the eigenvalues λ i, j of N i for all i gives an injective group homomorphism G S n. Note that the squared length of column zero of S is D = n i=0 d i, which must be equal to the squared length of column σ(0). Hence ( n n ) D = dσ(0) λ i,σ (0) = d σ(0) σ = dσ(0) σ(d ). i=0 Rewriting gives ( ) σ D = d σ(0) D. It follows that G acts in the same way on the quantities {d j /D }. The Verlinde formulas (.3) encode the symmetry of the N i matrices, and give us the complete symmetry under interchanging the last n N i and simultaneously reordering the last n rows and columns of all matrices. Thus ni, k j is invariant under G and hence is necessarily rational if we define it first to be only in R. Transposing the identity S σ( S) = C σ and inverting this identity gives the two equations σ( S) S = Cσ T and i=0 λ i,0 C σ = σ( S) S = D σ(d ) σ( S) S = d σ(0) C T σ.

10 35 E. Rowell, R. Stong, Z. Wang Hence the matrices d σ(0) C σ and d σ(0) B σ are orthogonal. Since B σ is diagonal it follows that B σ = d σ(0) ɛ 0,σ ɛ,σ 0 0 ɛ n,σ for some choices of ɛ i,σ =±. The map σ d σ(0) C σ gives a group homomorphism from G to the signed permutation matrices modulo ± which lifts the homomorphism ι of (). Rewrite the definition of C σ as σ( S) = SB σ P σ. Picking out the ( j, k) entry, we have σ( s j,k ) = d σ(0) ɛ σ(k),σ s j,σ (k). Moreover, since the left-hand side is symmetric we get SB σ P σ = Pσ B σ S. In coordinates this condition becomes s j,k = ɛ k ɛ σ(j) s σ(j),σ (k). Consider the action of G on pairs ( j, k) defined by σ ( j, k) (σ ( j), σ (k)). Then we see that s j,k is constant on orbits of this action. To see identity (.6), we apply σ to identity (.4) and compare with identity (.5). Note that s σ (0),σ (0) = ɛ σ(0),σ ɛ 0,σ by identity (.5). Given σ,σ G, consider first σ σ ( s j,k ) = σ ( d ɛ σ σ (0) (k),σ s j,σ (k)) = σ ( d ɛ σ σ (0) (k),σ s σ (k), j) = d σ (0)λ ɛ σ σ (0),σ (0) (k),σ ɛ σ ( j),σ s σ (k),σ ( j). Then consider ( ) σ σ ( s j,k ) = σ σ ( s k, j ) = σ ɛ σ ( j),σ d s k,σ ( j) σ (0) ( ) = σ ɛ σ ( j),σ d s σ ( j),k = ɛ σ ( j),σ σ (0) d ɛ σ (k),σ s σ ( j),σ (k). σ (0)λ σ (0),σ (0) Hence σ σ = σ σ using d i λ j,i = d j λ i, j, i.e. G is abelian. Suppose now that the rank n = r is even. Then det( S) = D n, hence det( S) = ±D r. Since the determinant is a polynomial in the entries of the matrix det(σ ( S)) = ±σ(d ) r, with the same sign as det( S). Hence det( S σ( S)) = d n σ(0). Since det(c σ ) = d n σ(0) ( )σ n j=0 ɛ j,σ, we conclude n j=0 ɛ j,σ = ( ) σ. For odd rank n = r +, det( S) =±D r+, hence D K. Hence σ(d) = ɛ σ D/d σ(0),ɛ σ =±and one gets the formula n j=0 ɛ j,σ = ɛ σ ( ) σ. Note that the resulting Eqs. (.5) for the entries s j,k are unchanged if we replace B σ with B σ. We will use this to assume ɛ 0 = below. Next we will use the fact that the θ i U() to produce a series of twist inequalities on the entries of S. Theorem.8. Given a modular symbol (N ; S, T ) and S is a real matrix, then () max i s i, j D s jj + D for any j. () If j = k, then D n s j,k i=0 s i, j s i,k. (3) n ɛ σ(j) s j,σ ( j) j=0 θ j θ σ(j) = D i: σ(i)=i θ iɛ σ(i).

11 On Classification of Modular Categories 353 Proof. Rewrite the twist equation as T ST ST = D + S. Then taking the ( j, k) entry of this formula gives n θ j θ k θ i s i, j s i,k = D + s j,k. i=0 Since D + =D and θ i =, the largest of the n+ numbers s i, j s i,k,0 i n, and D s j,k must be at most the sum of the other n. If j = k, then i s i, j = D > D s jj. Hence this inequality is trivial unless the largest is one of the first n and we get max i s i, j D s n jj + i=0 If j = k then i s i, j s i,k = 0 and the nontrivial case is D n s i, j s i,k. s j,k i=0 We will refer to these as the twist inequalities. Suppose σ G corresponds to signs ɛ i as above. We drop σ for notational easiness. Multiply the identity above by ɛ σ(j) /(θ j θ σ(j) ),setk = σ(j), and sum over j. The result is n n ɛ σ(j) j=0 i=0 n θ i s i, j s i,σ ( j) = D + j=0 s i, j. ɛ σ(j) s j,σ ( j) θ j θ σ(j). Interchanging the sums and using the fact that s i,σ ( j) = ɛ σ(j) ɛ σ(i) s σ(i), j gives n n θ i ɛ σ(i) i=0 j=0 n s i, j s σ(i), j = D + j=0 ɛ σ(j) s j,σ ( j) θ j θ σ(j). By orthogonality of the rows of S, the innermost sum on the left is zero if i = σ(i) and D = D + D if i = σ(i). Hence n j=0 ɛ σ(j) s j,σ ( j) θ j θ σ(j) = D i: σ(i)=i If σ is fixed point free, then n ɛ σ(j) s j,σ ( j) j=0 θ j θ σ(j) = 0. θ i ɛ σ(i). 3. Rank= and 3 Modular S Matrices In this section, we determine all possible modular S matrices for rank= and 3 modular symbols. The rank=3 case first appeared in [CP], but our proof is new. Theorem 3.. The only possible rank= modular S matrices of some modular symbols are

12 354 E. Rowell, R. Stong, Z. Wang () () where ɛ = ; where ϕ = +ϕ. ( ) ɛ, ɛ ( ) ϕ, ϕ ( Proof. ) Since all labels are self-dual, S is a symmetric ( ) real unitary matrix of the form d 0. The fusion matrix N d is of the form, so we have d m = +md. Simplifying D + D = D leads to θ + θ = d = m d. Since θ U(),so md. If d > 0, then d = m+ m +4, hence m = 0,. If d < 0, then d = m m +4, hence [Q(θ + θ ) : Q]. It follows that θ = e pπi q for some (p, q) =, and q is one of {,, 3, 4, 5, 6}. Direct computation shows there are no integral solutions p, q for cos( pπ q ) = m m m +4 except for q =, 5 and m = 0,. Theorem 3.. Then the only possible rank=3 modular S matrices of some modular symbols up to permutations are () ɛ ɛ ɛ ω ω, ɛ ω ω () where ɛ =, and ω 3 =,ω =. d d 0 d, d (3) where d =. d d d d, d d where d is a real root of x 3 x x +and d = d /(d ) which is a root of x 3 x x +. The largest d = cos(π/7) cos(π/7) = ,and d = cos(π/7) =

13 On Classification of Modular Categories 355 Proof. The non-self dual case is given in the Appendix. Hence we assume all fusion rules are self-dual, so S is a real, symmetric, unitary matrix up to the scalar D. Itfollows that the fusion matrices N i s are commutative, symmetric, integral matrices. One approach to proving the theorem is to analyze case by case for the Galois groups of fusion rules G =, Z, Z 3. This strategy will be fully exploited in the rank=4 case in the next section. Instead we will argue directly from the S-matrix in this section. The fusion matrices N, N are symmetric, and N N = N N. Therefore, they can be written as N = 0 0 m k 0 k l and N = k l l n such that There characteristic polynomials are and +ml + kn = k + l. p (x) = x 3 (l + m)x + (ml k )x + l = 0 p (x) = x 3 (k + n)x + (nk l )x + k = 0, respectively. Next we turn to the S matrix, which is of the following form: S = d d d s s. d s s Orthogonality of the columns of the S matrix translates into the equations d + d s + d s = 0, d + d s + d s = 0, d d + s ( s + s ) = 0. The first two equations give s = d s /d and s = d s /d. Plugging these into the third equation gives hence (d + d ) s +d d s d d = 0, s = d d ± D.

14 356 E. Rowell, R. Stong, Z. Wang Thus and s = d ± D, Thus the eigenvalues of N are s = d ± D. d, b = s = d d ± D, and c = s = d d d d ( ± D) and the eigenvalues of N are We compute d, e = s = d d ± D, and f = d d d ( ± D). d b + d f = d c + d e = bc + ef =. Since d bc = l and d ef = k, these are equivalent to l c + k e = l b + k f = l d + k d =. Also note that d e = d b. Let s deal with the case where l = 0 first. Then we have k = kn +. Hence k = and n = 0. Thus the eigenvalues of N are,, and - and the eigenvalues of N are (m + m +8)/, 0, and (m m +8)/. Since N has eigenvalues d, b, c, and d = 0, hence c = 0 which implies m = 0. This gives (k,l,m, n) = (, 0, 0, 0) and S = d d 0 d, d where d =. The case k = 0 gives essentially the same solution, so we will henceforth assume l and k are positive. Since p (l) = k l 0 and p (0) = l 0, we see that the largest root of p is >l, one of the remaining roots is in (0,l) and the other root is negative. Similarly the largest root of p is > k and the other roots are in (0, k) and (, 0).

15 On Classification of Modular Categories 357 Case. The polynomial p (x) is reducible. Since d >l, d cannot be an integer. Thus p must split into a linear and an irreducible quadratic. Thus Q[d, D] is a quadratic extension of Q. Hence Q[d, d, D] has degree or 4 over Q. Thus p is also reducible and also splits into a linear and an irreducible quadratic. Since the l/b + k/f = l/c + k/e =, the integral roots must be either b and f or c and e. Without loss, we may assume the integer roots are b and f.let d = α + β s and e = α β s for rational (in fact integer or half-integer) α and β and integer s. Then since d e = d b and c is the conjugate of d we have d = b α + β s α β s and c = b α β s α + β s. Hence l = d bc = b 3. Since f = k/(d e) = k/(α β s) and b = l/b = k/f = α + β s. Therefore solving = k/d + l/d for k gives k = d l d = α + β s + b 3 e d b = α + β s (α + β s)(α β s) = α(α β s) + β[α + β s] s. Since k is an integer, this forces α β s =, hence b =. Since l>0, this means l = and b =. Also from the equations above we get k = α, d = α + α +, e = α α +, f = α, d = α ++α α += d /, c = α + α α +, and D = α ++α α +. Thus and p (x) = x 3 (α +)x (α +)x + p (x) = x 3 3αx + (α )x +α. Thus (k,l,m, n) = (α,, α,α)and α ++α α + α + α + S = α ++α α + α α + α + α + α α + α + α. α + Note that n = α must be a non-negative integer. Setting α = 0 gives the example found above again. Thus we may assume α. Since d = d /, the equation for the θ s is +θ d + (/4)θ d 4 =D = +(/)d. If a solution did exist, then we would have + (/)d (/4)d4 d, hence 7 (d 3) or d 3+ 7 = Since d + 3 = , this cannot occur. Thus these S matrices, for positive α, do not give a modular symbol.

16 358 E. Rowell, R. Stong, Z. Wang Case. The polynomial p (x) is an irreducible cubic. By Case, we see that p (x) is also an irreducible cubic. Then there must be a Galois symmetry σ with σ(d ) = b, σ(b) = c and σ(c) = d. Hence σ(d ) = f, σ(f ) = e, and σ(e) = d since these roots of p pair with the corresponding roots of p. Applying σ to the identity d e = d b, gives d b = fc. Thus we must have fc= d d /( ± D). Since fc= d d ( ± D) + d + d d d ( ± D) + = d d d d ( ± D) + D ± D d d ( ± D), we compute and hence Since D >, we get that ± D = ( ± D) d d fc= ± ( ) ± D =. d d D( ± D), d d s = d d ± D =±, and hence b =±/d and e =±/d. Thus d and d are units in the ring of algebraic integers. Hence k = l = and hence m + n =. Without loss we may assume m = and n = 0. Then p (x) = x 3 x x + and p (x) = x 3 p (/x) = x 3 x x +. Then one computes d = cos(π/7) = , cos(π/7) b = /d, c = /(d ), d = d /(d ) = , e = /d, and f = d and S = d d d d. d d 4. Rank = 4 Modular S Matrices First we introduce the following notation. For an integer m, define φ m = m + m +4, that is, φ m is the unique positive root of x mx = 0. Note that any algebraic number φ whose only conjugate is /φ must be φ m for some integer m. Also note the only rational φ m is φ 0 =.

17 On Classification of Modular Categories 359 Theorem 4.. The only possible rank=4 modular S matrices of unitary modular symbols up to permutations are () ω ω, ω ω () (3) (4) where ω =±i; ; ; ϕ ϕ ϕ ϕ ϕ ϕ, ϕ ϕ (5) (6) where ϕ = + 5 is the golden ratio; ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ; ϕ ϕ ϕ d d + d d 0 d + d d + d + d, d d d where d is the largest real root of x 3 3x. Proof. The non-self dual case is treated in the Appendix, so we will assume that S is real in the following. Since the fusion coefficients ni, k j are totally symmetric in i, j and k for self-dual categories, we will instead write n i, j,k in what follows. For notational easiness, when the Galois group element σ is clear from the context, we simply write ɛ i,σ as ɛ i. Identities for ɛ i and s jk that are not referenced are all from Theorem.7. All the twist inequalities are from Theorem.8.

18 360 E. Rowell, R. Stong, Z. Wang Case. G contains a 4-cycle. By symmetry we may assume σ = (0 3) G.The conditions s j,k = ɛ k ɛ σ(j) s σ(j),σ (k) and ɛ ɛ ɛ 3 = give d d d 3 d S = ɛ ɛ d ɛ d 3 ɛ d ɛ d 3 ɛ d. d 3 ɛ ɛ d ɛ ɛ d By symmetry under interchanging N and N 3, we may assume ɛ = +. Note that σ (d ) = λ, = /d = d. Hence the characteristic polynomial p of N is irreducible. Since σ (d ) = d 3 /d < 0. Hence σ (d ) = d. Thus p is irreducible. Since ɛ /d 3 is a root of p, it follows that p 3 is also irreducible. We see that λ, = ɛ d /d, λ, = d 3 /d, and λ,3 = ɛ /d 3. In particular d λ, λ, λ,3 =. Orthogonality of the rows of S is equivalent to d + ɛ d d d d 3 + ɛ d 3 = 0, or + λ,3 = λ, +. d λ, Write p (x) = x 4 c x 3 +c x +c 3 x. Then p 4 (x) = x 4 ɛ c 3 x 3 c x +ɛ c x. Note that c = Trace(N ) 0 and ɛ c 3 = Trace(N 3 ) 0. Multiplying together the orthogonality condition above and five of its formal conjugates gives 8 + (c 3 c ) 6c c 3 +(c 3 c )c +3c = 0. This equation forces c and c 3 to be even. Let = c c 3 and = c +c 3 (hence and are even and congruent mod 4). Then solving the quadratic equation above for c we see that we must have ( 3)( +3) to be a square and c = 3 ± ( 3)( +3) 6. It follows that 6. If and are multiples of 4, then we see they are multiples of 8 and either sign gives an integral c.if and are both mod 4, then there is a unique choice of the sign for which c is integral. The Galois group of p must be Z/4Z, otherwise it would contain the (0 )( 3). Applying this to the orthogonality identity above gives /λ,3 + d = λ, +/λ,. Multiplying this by the original identity gives d λ,3 + d λ,3 = λ, λ, + λ, λ,. Hence d λ,3 = (λ, λ, ) ±, either of which contradicts the product of all four roots being. In particular, p cannot have complex roots, since complex conjugation would give a transposition in the Galois group. Applying σ to the orthogonality identity gives d +/λ, = λ, +/λ,3.

19 On Classification of Modular Categories 36 We know from the preliminary discussion that all three of the resulting N i matrices will be rational. Define P = 6c 3 =± 3, then we compute n,, = 5c 3c 3 c c 3 P, 8 8 n,, = ɛ (P ), n,,3 = ɛ ( c + c 3 8 n,, = c + c 3 4 n,,3 = P, n,3,3 = c + c 3 ( 8 n,, = ɛ c c c c 3 4 c c 3 8 c c 3 + c 4 P, ( c + c 3 n,,3 = ɛ + c c n,3,3 = ɛ (P +), and n 3,3,3 = ɛ ( 5c3 3c 8 ) P, P, ) P, ) P, c c 3 8 ) P. Recall that the n i, j,k must be nonnegative integers. This restricts the c i. First looking at n,,3, we see that P must be a positive integer. Hence c must be given by the upper sign. This condition in fact guarantees integrality of all the n i, j,k. (The additional factors of in the denominator cancel out if c is integral and can be ignored.) Integrality of P severely restricts, since it requires all odd prime factors of + 3 to be congruent to mod 8 (since and are both squares mod any such prime). In particular either = 0or 6. Since P = 0, we see that + 64, hence >. Thus c 3 must be positive and >0. Since we saw above ɛ c 3 0, we see that ɛ = +. Thus rewriting the orthogonality relation gives d /d 3 = (d )/(d +). The twist inequality coming from the (0, 3) entry reads D ( + d d 3 ) (+d ). Plugging in the preceding identity simplifies this to D d. Rearranging gives 3d d + d 3 +> d + d 3 and plugging in the identity d = d 3 (d )/(d +) yields ( ) d3 < 3d d + (d +) < 3. To see why this is helpful, expand the equations Trace(N i ) = c i for i =, 3 and use the identity above to eliminate d. The result is c = d d d 3 + d +d, and d (d +) d d 3 c 3 = d +d d (d +) d 3 d d. d d 3

20 36 E. Rowell, R. Stong, Z. Wang Subtracting these gives = c c 3 = 4 d 3 d + +d +. d d 3 Hence > 4 d 3 d + > 4 3/ > 4.9. However, we saw above that either = 0 or 6. It follows that 0. Since 6 and c 3 /6, it follows that c > = c c 3. Thus p 3 () = c c c 3 < 0. Thus d 3 >. Hence we see <(d +)/(d d 3 )<4. It follows that = 0, i.e., c = c 3. Since = 0, = c is a multiple of 8 and ( ) P ( c ) =. 3 4 In this case the characteristic polynomials become p (x) = x 4 c x 3 + (c /4) x + c x, p (x) = x 4 4 (c /4) x 3 6x +4 (c /4) x +, and p 3 (x) = x 4 c x 3 (c /4) x + c x. In particular p (x) >0forx c. Hence d < c.wehave p (x) = (x t x )(x t x ), where t > 0 > t are the two roots of t 4 (c /4) t 4 = 0. Since the larger root of x tx is an increasing function of t, d must correspond to t. Hence d = t + > t = 4 (c /4) d + 4 > 4 (c /4) t. In particular, d > 4 since the square root above is integral. Finally the twist inequality coming from the (0, ) entry reads ( D + d ) d 3. d Squaring and using the identity d 3 = d (d +)/(d ) to eliminate d 3 gives the inequality 4(d +) d 4 d (d 3 4d + d +4)d d (d ) (d 3) 0. Dividing through by 4(d +) d and rearranging gives d d (d +) (d3 4d + d +4) + d (d ) (d 3) 4(d +) d. The right-hand side of this inequality is an increasing function of d for d > 4 and a decreasing function of d, hence we may replace d by its upper bound c and d by the lower bound above. The result is c + 64c4 8c6 + (608 76c +3c4 c ) 6 0.

21 On Classification of Modular Categories 363 Since the coefficient of the square root is nonnegative it follows that c + 64c4 8c6 + (608 76c +3c4 )c 0. This polynomial in c is negative for c > 6, hence c 8. The only multiple of 4 in the range 3 c 8 for which the Pell equation above is satisfiable is c = 4. This gives c = c 3 = 4, c =, P = 3. However plugging in shows that the twist inequality D ( +d d 3 /d ) does not actually hold in this case. Thus there are no solutions in this case. Case. G is the Klein 4-group. Let σ, σ, and σ 3 be the elements of G which correspond to (0 )( 3), (0 )( 3), and (0 3)( ), respectively. Let B σ correspond to signs ɛ i with ɛ ɛ ɛ 3 = and B σ correspond to signs δ i with δ δ δ 3 =. Then using the usual identities gives d d d 3 d d d 3 d S = ɛ ɛ d 3 ɛ ɛ d d d ɛ d 3 s, s = s, δ d 3 s,3,3 d δ d 3 δ δ δ d. d 3 ɛ ɛ d s,3 ɛ s, d 3 s,3 δ δ d δ s, Comparing these we see ɛ = δ, s, = ɛ, and s, = δ, hence d d d 3 d S = d ɛ ɛ d 3 ɛ d 3 δ ɛ ɛ d ɛ δ d. d 3 ɛ ɛ d ɛ δ d ɛ δ Orthogonality of the rows of S gives the three conditions (+ɛ )(d + ɛ d d 3 ) = (+δ )(d + ɛ d d 3 ) = (+ɛ δ )(d 3 + ɛ ɛ d d ) = 0. Suppose ɛ = +, then we see d = ɛ d d 3, hence ɛ = and the remaining orthogonality relations become d (+δ )( d 3 ) = d 3 (+δ )( d ). We cannot have d = d 3 =, since this would make d =, hence δ =. This gives d d 3 d d 3 d S = d 3 d 3 d d d 3 d d. 3 d 3 d d d 3 The eigenvalues of N are d and /d each with multiplicity. Hence d = φ m for some integer m. The eigenvalues of N 3 are d 3 and /d 3 each with multiplicity, hence d 3 = φ n for some integer n. So φ m φ n φ m φ n φ S = m φ n φ n φ m φ m φ n φ m φ. n φ n φ m φ m φ n

22 364 E. Rowell, R. Stong, Z. Wang The resulting N i matrices are necessarily rational, but in this case they are all integral, namely, N = mn m n N N 3 = N 3 N = 0 m 0, N = 0 n m 0 0 n 0 0 m 0, and N 3 = n Note that nonnegativity of the entries forces m, n 0 and hence φ m,φ n. The strongest twist inequalities are the (0, ) and (, 3) cases which give D 4or(φm + )(φn +) 6. This gives, up to symmetry, the solutions (m, n) = (0, 0), (0, ), (0, ), or (, ). These are all excluded since the resulting Galois group G is at most Z/Z. (These examples will return when we look at smaller Galois groups.) Case 3. G contains a 3-cycle. Since we can exclude Cases and above, the image of G in S 4 cannot be transitive. It follows that G must fix the point j not on the 3-cycle. Thus λ i, j is rational (hence integral) for every i. Up to symmetry there are two cases for the 3-cycle. We could have σ = ( 3) or σ = (0 ). Ifσ = ( 3), then the d i are integral. The identities s j,k = ɛ σ(j) ɛ k s σ(j),σ (k) and ɛ ɛ ɛ 3 = giveɛ i = for all i (since d i = s 0,i = ɛ i s 0,i+ = ɛ i d i+ for i ) and d d d d S = d s, s 3,3 s 3,3 s, s, s., d s, s, s 3,3 Orthogonality of the columns of S gives s, + s, + s 3,3 = and s, s, + s, s 3,3 + s 3,3 s, = d. The first of these gives /d = λ, + λ, + λ,3 fromwhichwesee /d is an algebraic integer. Hence d = and λ,i <. The second equation gives λ, λ, + λ, λ,3 + λ,3 λ, =. Hence λ,, λ,, and λ,3 are the three roots of g(x) = x 3 + x x + n for some integer n. This cubic must be irreducible and have three real roots all less than. Irreducibility excludes n = 0 and n =. For the roots of g to be less than, we must have g() >0orn +> 0. Hence n. However, this results in complex roots. Thus this case gives no solutions. Thus we must have σ = (0 ) and λ i,3 is integral for all i. The identities for s j,k give d d d 3 d S = d ɛ ɛ d ɛ ɛ ɛ d ɛ d 3 ɛ ɛ d. 3 d 3 ɛ d 3 ɛ ɛ d 3 s 3,3 Since σ(d 3 ) = λ 3, = ɛ d 3 /d and σ (d 3 ) = ɛ ɛ d 3 /d,wemusthaveσ(d 3 ) = d 3. (Otherwise ɛ = ɛ = d = d = which fails.) Thus d 3 is a root of an irreducible cubic g(x) = x 3 c x + c x c 3 and ɛ d and ɛ ɛ d are ratios of roots of g.ifg had

23 On Classification of Modular Categories 365 Galois group S 3, then the ratios of the roots of g would be roots of an irreducible sextic. Thus g has Galois group Z/3Z and G ={,σ,σ }. Note that c = d 3 + λ 3, + λ 3, = d 3 d d (d d + ɛ d + ɛ ɛ d ), c = d 3 λ 3, + λ 3, λ 3, + λ 3, d 3 = d 3 d d (ɛ d + ɛ + ɛ ɛ d ), and c 3 = d 3 λ 3, λ 3, = ɛ d 3 3 d d. Orthogonality of the columns of S gives c = ɛ d + ɛ ɛ d + ɛ d d c 3 d3 =, and c = +ɛ d + ɛ ɛ d = s 3,3 = λ 3,3 Z. c 3 d 3 d 3 Thus g(x) = x 3 cx + ncx + c for integer n, c. Since g has Galois group Z/3Z, δ = c discr(g) = (n +4)c n(n +9)c 7 must be a square. The resulting n i, j,k are n,, = ɛ (δ nc ) ɛ δ n +3, n,, = ɛ ɛ (n +3) ( δ + nc n +3), n,,3 = (n +3) ( nδ + (n +)c +3n), ɛ n,, = (n +3) (δ + nc n +3), n,,3 = ɛ (3n c), n +3 n,3,3 = ɛ (n +3) (δ nc +n 3), n,, = ɛ ɛ (δ + nc +) + ɛ ɛ δ n +3, n,,3 = (n +3) (nδ + (n +)c +3n), n,3,3 = ɛ ɛ (n +3) ( δ nc +n 3), n 3,3,3 = c + n3 n +3. Integrality of n,,3 requires c 3n (mod n + 3). If we write c = 3n + a(n +3) for integer a, then we compute δ = (n +3) (a (n +4) +an 3). Hence δ = (n +3)β, where β is integral and β = a (n +4) +an 3. Note that in particular this forces

24 366 E. Rowell, R. Stong, Z. Wang a = 0. Rewriting it as β = (an +) +4a 4, we see that β an + (mod ). Thus we compute n,, = ɛ ((n +)β 3n an(n +3) ), n,, = ɛ ɛ ( β + an +), n,,3 = ( nβ +3n + a(n +)), n,, = ɛ (β + an +), n,,3 = ɛ a, n,3,3 = ɛ (β an ), n,, = ɛ ɛ ((n +)β +3n + an(n +3) +), n,,3 = (nβ +3n + a(n +)), n,3,3 = ɛ ɛ ( β an ), n 3,3,3 = a + n, and these are all integral. Nonnegativity of these entries gives further restrictions on the parameters. Looking at n,, + n,3,3 = ɛ β, we see that ɛ is the sign of β (or β = 0, but this gives a =±, n = a, c = a and g(x) = (x a)(x ) which is reducible). Looking at n,, + n,3,3 = ɛ ɛ β, we see that ɛ =. Looking at n,,3 we see that a > 0. Nonnegativity provides additional constraints on the parameters, but instead we look at the twist inequalities. We saw above that a > 0, hence c = 3n + a(n +3) n +3n +3 > 0. Thus two of the roots d 3, λ 3, and λ 3, of g must be positive. By symmetry, we may assume d 3 >λ 3, > 0 >λ 3,. Then ɛ = and we have S = d 3 /λ 3, d 3 /λ 3, d 3 d 3 /λ 3, d 3 /λ 3, d 3 d 3 /λ 3, d 3 /λ 3, d 3 d 3 d 3 d 3 nd 3. Let M = max(/λ 3,, / λ 3, ) so that Md 3 = max(d, d ). Since D = (n +3)d 3 and d 3 + λ + 3, λ = n +, 3, the diagonal twist inequality coming from the (0, 0) entry gives n M n d 3 This inequality allows only finitely many choices of the parameters.

25 On Classification of Modular Categories 367 If n > 0, then g( /φ n ) = φn 3 < 0 and therefore λ 3, > /φ n. Thus M >φ n and we get φ n < n +3+ Further since d 3 λ 3, = Mc and d 3 >λ 3,,wehave n +3 d 3. d 3 > φ n c φ n (n +3n +3). For n, we get a contradiction by noting that φ n > n, hence these equations force n < n +3+ n, a contradiction. For n =, plugging in gives a contradiction. If n = 0, then β = 4a 3, hence a = and g(x) = x 3 3x +3. Since s 3,3 = 0, the (3, 3) entry of the twist equation gives θ 3 d 3 (+θ + θ ) = 0, hence θ = θ = e ±πi/3. The (0, 3) and (, 3) entries give θ 3 (+d θ d θ ) = D + = θ θ 3 (d d θ + θ ). This case is realized by (A, 7). If n < 0, then /M = λ 3,. One easily checks that M and d 3 are increasing functions of c, therefore it suffices to check that the inequality M n +3+(n +3) / d3 fails for a = and hence c = n +3n +3. The inequality fails for n. (To see this simply compute both sides for n =. For n 3, note that g( /n) = ((n +)/n) 3 < 0 and g((n +) ) = (n +) 4 + n +< 0. Therefore M < n and d 3 >(n +). Hence we have M > n > n + 4 and (n +3) / /d 3 <, but these combine to contradict the inequality.) For n =, a =, the inequality holds, but the resulting polynomial g(x) = x 3 x x + is reducible. Moving up to the next case n =, a = 3, the inequality fails. Thus there are no solutions in this case. With the cases above completed, we consider G which is not transitive and contains no 3-cycle. Up to symmetry, it follows that G must be a subgroup of Z/Z Z/Z = (0 ), ( 3). Case 4. G contains the transposition σ = ( 3). In this case the parity condition gives ɛ ɛ ɛ 3 =. Three instances of the usual identity give d = s 0, = ɛ s 0,3 = ɛ d 3, d 3 = s 0,3 = ɛ 3 s 0, = ɛ 3 d, and d = s 0, = ɛ s 0, = ɛ d. Since the d i are positive we conclude ɛ = ɛ = ɛ 3 =, a contradiction. Thus we are left with only three possibilities. Either G = Z/Z = (0 )( 3), G = Z/Z = (0 ), org is trivial. Case 5. G contains σ = (0 )( 3). Using the identities s j,k = ɛ σ(j) ɛ k s σ(j),σ (k) and ɛ ɛ ɛ 3 = gives d d d 3 d S = d ɛ ɛ d 3 ɛ d 3 s, ɛ ɛ d s.,3 d 3 ɛ ɛ d s,3 ɛ s,