Interpolated family of non group-like simple integral fusion rings of Lie type
aa r X i v : . [ m a t h . QA ] F e b INTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGSOF LIE TYPE
ZHENGWEI LIU, SEBASTIEN PALCOUX, AND YUNXIANG REN
Abstract.
This paper computes the generic fusion rules of the Grothendieck ring of Rep(PSL(2 , q )), q prime-power,by applying the Schur orthogonality relations on the generic character table. It then proves that this family of fusionrings interpolates to all integers q >
1, providing (when q is not prime-power) the first example of infinite family of nongroup-like simple integral fusion rings. Furthermore, they pass all the known criteria of (unitary) categorification.This provides infinitely many serious candidates for solving the famous open problem of whether there exists anintegral fusion category which is not weakly group-theoretical. A braiding criterion is finally discussed. Introduction
The original motivations of this work are in subfactor theory [17]. A subfactor encodes a Galois-like quantumgeneralization of the notion of symmetry [12, 19], analogous to a field extension, where its planar algebra [18] andits fusion category [8] are the analogous of the Galois group and its representation category, respectively. The indexof a subfactor is multiplicative with its intermediates, so a subfactor without proper intermediate (called a maximalsubfactor [1]) can be seen as a quantum analogous of the notion of prime number (about a quantum analogous of thenotion of natural number, see [32–34]). A group is a notion of classical symmetry. A finite group subfactor planaralgebra is in the bigger class of irreducible finite index depth 2 subfactor planar algebras, which is exactly the classof finite quantum group subfactor planar algebras [3, 4, 25, 36], encoding the finite dimensional Hopf C ∗ -algebras(also called Kac algebras [15]). The set of maximal ones in this class can be seen as a very close extension of theset of prime numbers, because the only known ones are those coming from groups, which then must be cyclic ofprime order, and it is an open problem whether there are other ones [32, Problem 4.12]. The intermediates of afinite quantum group subfactor correspond to the left coideal ∗ -subalgebras of the Kac algebra K [16], and thenormal left coideal ∗ -subalgebras correspond to the fusion subcategories of Rep( K ) [2]. A fusion category withoutnontrivial proper fusion subcategory is called simple . So a necessary condition for a subfactor planar algebra ofthis class to be maximal is that its fusion category must be simple. Moreover, Rep( K ) is a unitary integral fusioncategory. The Grothendieck ring of a fusion category is a fusion ring [27]. So we are looking for simple integralfusion rings, not coming from a group, but with a unitary categorification. More generally, a fusion category whichcan be “cooked up” from group theory is called weakly-group theoretical [11], which means that (up to equivalence)it has the same Drinfeld center as a sequence of group extensions. But a weakly group-theoretical simple integralfusion category (over the complex field) is precisely Grothendieck equivalent to Rep( G ) with G a finite simplegroup [11, Proposition 9.11]. So we are looking for integral fusion categories which are not weakly group-theoretical;their existence is one of the most famous open problem of the fusion category theory [11, Question 2]. Finally,Kaplansky’s 6th conjecture [20] states that every Kac algebra should be of Frobenius type (i.e. the dimension of itsirreducible complex representations divides its dimension), moreover its extension to every complex fusion categoryis open [11, Question 1].The paper [24] classifies all the simple integral fusion rings of Frobenius type up to rank 8, FPdim less than 4080(and = p a q b , pqr , by [11]). There are 4 group-like ones given by PSL(2 , q ) with q ∈ { , , , } , 28 ones excludedfrom unitary categorification by Schur product criterion (see Theorem 5.1), and exactly 2 other ones, denoted F and F (indexed by their FPdim). The second one is excluded from any categorification by the zero spectrumcriterion (see Theorem 5.12) and also the one spectrum criterion (see Theorem 5.16), whereas the first one passesall the criteria known to us (listed in Section 5), and we will now see the conceptual reason why. The family of finitesimple groups of Lie type PSL(2 , q ), with q prime-power, admits a generic character table depending on whether q is even, q ≡ − q >
1. The use of the Schur orthogonality relations (Theorem 2.3) on these interpolated tablesprovide rules (Theorems 4.5, 4.19 and 4.41), which turn out to be those of fusion rings and which have these formaltables as eigentables (Definition 2.1), by reconstruction Theorem 2.2. These interpolated fusion rings are integral,simple for q ≥
4, and non group-like when q is not a prime-power. They inherit all the good arithmetical properties of the group-like ones, more precisely Section 5 proves that they pass all the known criteria. It turns out that F is precisely given by the smallest non prime-power q = 6.Here are observations on the fusion matrices found in Section 4: • q even: all self-adjoint, and some non-trivial ones are of multiplicity 1, otherwise 2, • q ≡ − • q ≡ q = 5).This led to an interesting group-theoretical result: let G be a non-abelian finite simple group, then the Grothendieckring of Rep( G ) is of multiplicity ≤ G is isomorphic to some PSL(2 , q ) with q prime-power. Moreover,the family (PSL(2 , q )) is the only one, among the usual infinite families of non-abelian finite simple groups, wherethat fusion multiplicity is bounded above (because it is the only family of Lie type of rank 1). This makes thisfamily quite special, in the sense of fusion friendly. Question 1.1.
Is there a non prime-power q > for which the interpolated fusion ring admit a categorification? P. Etingof observed [7] that it is still premature to believe in a positive answer of above question, by pointingout that an interpolated family also exists from Rep( F q ⋊ F ⋆q ), whereas by [9, Corollary 7.4] there is a complexcategorification (if and) only if q is a prime-power. Section 7.1 proves that this family also pass all the criteria.The paper [22] already excluded the case q = 6 ( F ) from a categorification over a field of characteristic 0;otherwise, the positive characteristic p case (assuming spherical and pseudo-unitary) reduces to p = 41 only, andlooks promising. So the problem is open for q = 6 in characteristic 41, and for every non prime-power q = 6 inevery characteristic. To solve this problem we can try to generalize the localization strategy initiated in [22] and/orcompute the generic F-symbols of the symmetric fusion category Rep(PSL(2 , q )), q prime-power, and see how andwhere they can be interpolated in the non prime-power case (non-symmetric, so some adjustments may be required).Finally, Section 6 shows that, for q <
34 non prime-power, a complex categorification would not be braided, andask whether it should be the same for all q non prime-power.In this paper, we discuss the interpolated fusion rings for Rep(PSL(2 , q )), but it is natural to expect that, forevery family of Lie type, the representation rings admit such interpolation (as suggested by the data in [26]). Acknowledgements.
Thanks to Pavel Etingof, Dmitri Nikshych, Victor Ostrik and Arthur Jaffe for their encour-aging interest on this work. Thanks to Pavel for his moderation. Thanks to Jean Michel and Frank L¨ubeck, the firstled us to the second who helped us with [GAP]. The first author is supported by Grant 100301004 from TsinghuaUniversity and 2020YFA0713000 from NKPs; the third author, by Grant TRT 0159 and W911NF-20-1-0082.
Contents
1. Introduction 12. Preliminaries 33. Generic character table of PSL(2 , q ) 43.1. The case q even 43.2. The case q ≡ − q ≡ q even 54.2. The case q ≡ − q ≡ NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 3 Preliminaries
In this section, we show that the data of a commutative fusion ring is equivalent to the data of a formal tablesatisfying some assumptions.
Definition 2.1.
Let F be a commutative fusion ring. Let M , . . . , M r be its fusion matrices, and let D i = diag ( λ i,j ) , i = 1 , . . . , r be their simultaneous diagonalization. The eigentable of F is the table given by ( λ i,j ) . Note that the eigentable of the Grothendieck ring of Rep( G ), with G finite group, is the character table of G . Theorem 2.2.
Let ( λ i,j ) be a formal r × r table. Consider the space of functions from { , . . . , r } to C with someinner product h f, g i . Consider the functions ( λ i ) defined by λ i ( j ) = λ i,j , and assume that h λ i , λ j i = δ i,j . Considerthe pointwise multiplication ( f g )( i ) = f ( i ) g ( i ) , and the multiplication operator M f : g f g . Consider M i := M λ i ,and assume that for all i there is j (automatically unique, denoted i ∗ ) such that M ∗ i = M j . Assume that M isthe identity. Assume that for all i, j, k , n ki,j := h λ i λ j , λ k i is a nonnegative integer. Then ( n ki,j ) are the structureconstants of a commutative fusion ring and ( λ i,j ) is its eigentable. Moreover, every eigentable of a commutativefusion ring satisfies all the assumptions above.Proof. By assumption, the set { λ i | i = 1 , . . . , r } is an orthonormal basis of the r -dimensional space it generates,so this space must be the r -dimensional space of functions from { , . . . , r } to C . It follows that for all i, j , λ i λ j decomposes into a linear sum of λ k , and so by construction λ i λ j = X k n ki,j λ k . We can now show that ( n ki,j ) are the structure constants of a commutative fusion ring: • (commutative) λ i λ j = λ j λ i because the multiplication is pointwise and C is commutative, • (associative) ( λ i λ j ) λ k = λ i ( λ j λ k ) because the multiplication is pointwise and C is associative, • (Frobenius reciprocity) n ki,j = h M i λ j , λ k i = h λ j , M ∗ i λ k i = h M ∗ i λ k , λ j i = n ji ∗ ,k = n ji ∗ ,k = · · · = n ik,j ∗ , • (neutral) n k ,i = n ki, = h λ i λ , λ k i = h λ i , λ k i = δ i,k , • (adjoint) n i,j = n ji ∗ , = δ i ∗ ,j .Finally consider the function δ i : j δ i,j . Then M i δ j = λ i,j δ j . It follows that ( diag ( λ i,j )) i =1 ,...,r is a si-multaneous diagonalization of ( M i ) i =1 ,...,r . So ( λ i,j ) is an eigentable of the commutative fusion ring defined by( n ki,j ).Reciprocally, let ( λ i,j ) be the eigentable of a commutative fusion ring. Define ( λ i ) as above and the inner productby h λ i , λ j i = δ i,j . Then all the assumptions in the statement of the theorem follows easily. (cid:3) Theorem 2.3 (Schur orthogonality relations) . Let F be a commutative fusion ring, and ( λ i,j ) be its eigentable. Let c j = X i | λ i,j | (usually called the formal codegrees). Then X i λ i,j λ i,j ′ = δ j,j ′ c j and X j c j λ i,j λ i ′ ,j = δ i,i ′ . Proof.
The first relation follows from [30, Lemma 2.3] (which uses [28, Proposition 19.2(b)]). Next, let U be thematrix ( √ c j λ i,j ). The first relation means that U ∗ U = id , i.e. U is an isometry. But in the finite dimensional case,an isometry is unitary, so U U ∗ = id also, which means that: X j λ i,j √ c j λ i ′ ,j √ c j = δ i,i ′ . The second relation follows. (cid:3)
Note that for the character table of a finite group G , then ( | G | / c j ) are exactly the class sizes. Corollary 2.4.
The inner product in Theorem 2.2 can always be taken of the form: h f, g i := X j c j f ( j ) g ( j ) ZHENGWEI LIU, SEBASTIEN PALCOUX, AND YUNXIANG REN Generic character table of
PSL(2 , q )The generic character table for PSL(2 , q ) requires to consider three cases depeding on q mod 4. We will use thedata provided by [GAP] as a reference for these tables, but for more usual references, see (for example) [13, § § The case q even.Theorem 3.1. Here is the generic character table of
PSL(2 , q ) with q even: ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤ charparam c classparam k { } { } { , . . . , q − } { , . . . , q }{ } { , . . . , q } q − − − πkcq +1 ) { } q − { , . . . , q − } q + 1 1 2 cos( πkcq − ) 0 class size q − q ( q + 1) q ( q − Proof.
Here is the [GAP] code providing these data (recall that for q even, PSL(2 , q ) = SL(2 , q )): gap> Print(CharacterTableFromLibrary("SL2even")); (cid:3) The case q ≡ − .Theorem 3.2. Here is the generic character table of
PSL(2 , q ) with q ≡ − : ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤ charparam c classparam k { } { , } { , . . . , q − } { , . . . , q − } { q +14 }{ } { , } q − − i ( − k + c √ q − k +1 ( − k +1 { , . . . , q − } q − − − πkcq +1 ) − − c { } q − − { , . . . , q − } q + 1 1 2 cos( πkcq − ) 0 0 class size q − q ( q + 1) q ( q − q ( q − Proof.
Here is the [GAP] code providing these data: gap> Print(CharacterTableFromLibrary("PSL2odd")); (cid:3)
The case q ≡ .Theorem 3.3. Here is the generic character table of
PSL(2 , q ) with q ≡ : ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤ charparam c classparam k { } { , } { , . . . , q − } { q − } { , . . . , q − }{ } { , } q +12 1+( − k + c √ q ( − k ( − k { , . . . , q − } q − − − πkcq +1 ) { } q − { , . . . , q − } q + 1 1 2 cos( πkcq − ) 2( − c class size q − q ( q + 1) q ( q +1)2 q ( q − Proof.
Here is the [GAP] code providing these data: gap> Print(CharacterTableFromLibrary("PSL2even")); (cid:3)
NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 5 Interpolated fusion rings
The generic character table of PSL(2 , q ) as provided by [GAP] can be interpolated to q non prime-power. In thissection, we show that this interpolated formal table is in fact the character table of an interpolated “PSL(2 , q )”,more precisely, we will prove that the application of the Schur orthogonality relations to the formal table provides afamily of fusion rings (depending on the integer q ) which are exactly the Grothendieck rings of Rep(PSL(2 , q )) when q is a prime-power, and otherwise, new non group-like simple integral fusion rings, whose eigentables are exactlythese formal tables. Theorem 4.1.
The generic character table of
PSL(2 , q ) , interpolated to every integer q ≥ , is the eigentable ofan integral fusion ring, called the interpolated fusion ring . Moreover, if q ≥ then it is simple, and if q is not aprime-power then it is not group-like.Proof. The result follows by Theorems 4.5, 4.19 and 4.41. To show it is simple if q ≥
4, just observe that every non-trivial simple object generates all, so that there is no proper non-trivial fusion subring. Finally, it is not group-likewhen q is not a prime-power by Lemma 4.3. (cid:3) The multiplicity of a fusion ring given by ( n ki,j ) is max i,j,k ( n ki,j ). Let the fusion multiplicity m ( G ) of a finitegroup G be the multiplicity of the Grothendieck ring of Rep( G ). The computation of the generic fusion rules madein this section leads to the following interesting result in finite group theory. Theorem 4.2.
Let G be a non-abelian finite simple group. Then m ( G ) ≤ if and only if G ≃ PSL(2 , q ) for someprime-power q . More precisely, m ( G ) = (cid:26) iff q even or q = 5 , iff q odd and q = 5 . Proof.
One way follows from Theorems 4.5, 4.19 and 4.41. Next, let d + ( G ) be the maximum among the degrees ofcomplex irreducible characters of G , and let Σ( G ) be their sum. By the contributions of Mikko Korhonen, GeoffRobinson, Denis Chaperon de Lauzi`eres and Frank L¨ubeck in [35], if G is non-abelian simple and non-isomorphicto some PSL(2 , q ) then d ( G ) > G ); but m ( G ) ≥ d / Σ( G ). (cid:3) Lemma 4.3.
For q non prime-power, the interpolated fusion ring is not the Grothendieck ring of Rep( G ) with G some finite group.Proof. Assume that for q non prime-power (and so ≥ G (obviously non-abelian).Then the interpolated fusion ring is simple, thus so is the Grothendick of Rep( G ), and G itself. But the interpolatedfusion rings are of multiplicity ≤
3, so by Theorem 4.2, G must be isomorphic to PSL(2 , q ′ ) for some prime-power q ′ , but by identifying their type, we see that [ q,
1] must be identified to [ q ′ , (cid:3) Definition 4.4.
Let x d,c denote the c -th simple object of FPdim d of a fusion ring. The case q even.Theorem 4.5. If q is even, the interpolated fusion ring is of rank q + 1 , FPdim q ( q − , type [[1 , , [ q − , q ] , [ q, , [ q + 1 , q − ]] , and given by the following rules: x q − ,c x q − ,c = δ c ,c x , + X c such that c + c + c = q +1 and c ,c ,c ) x q − ,c + (1 − δ c ,c ) x q, + X c x q +1 ,c ,x q − ,c x q, = X c (1 − δ c ,c ) x q − ,c + x q, + X c x q +1 ,c ,x q − ,c x q +1 ,c = X c x q − ,c + x q, + X c x q +1 ,c ,x q, x q, = x , + X c x q − ,c + x q, + X c x q +1 ,c ,x q, x q +1 ,c = X c x q − ,c + x q, + X c (1 + δ c ,c ) x q +1 ,c ,x q +1 ,c x q +1 ,c = δ c ,c x , + X c x q − ,c + (1 + δ c ,c ) x q, + X c such that c + c + c = q − and c ,c ,c ) x q +1 ,c + X c such that c + c + c = q − or c ,c ,c ) x q +1 ,c . The other rules follows by commutativity.
ZHENGWEI LIU, SEBASTIEN PALCOUX, AND YUNXIANG REN
Proof.
The rules are given by Theorem 3.1 and Lemmas 4.7 to 4.17. They satisfy all the axioms of a fusion ring byapplying Theorem 2.2 together with Proposition 4.18 (with the inner product as in Corollary 2.4). (cid:3)
The smallest q non prime-power in this case is q = 6. Here are the fusion matrices: , , , , , , In general, we deduce the following unexpected combinatorial result:
Corollary 4.6.
For ≤ c i ≤ q/ , the following map is invariant by permutation. ( c , c , c , c ) δ c ,c δ c ,c − {| q + 1 − | x || , x ∈ { c + c , c + c , c − c , c − c }} Proof.
Consider the following equality given by associativity h ( x q − ,c x q − ,c ) x q − ,c , x q − ,c i = h x q − ,c ( x q − ,c x q − ,c ) , x q − ,c i , then compute the LHS and RHS independently and directly by the fusion rules. The equality of the outcome ofthese two computations provide the result after straighforward reformulations. (cid:3) Note that other such combinatorial results can be given by considering other associativity equalities.The following lemmas compute every possible h x d ,c x d ,c , x d ,c i for every integer q even, with inner product asin Corollary 2.4. We will make the computation in full details for the first lemma (the other ones will be similar). Lemma 4.7. h x q − ,c x q − ,c , x q − ,c i = (cid:26) if c + c + c = q + 1 or c , c , c )1 elseProof. By Theorem 3.1: h x q − ,c x q − ,c , x q − ,c i = ( q − q ( q − − q − q + 1 q/ X k =1 πkc q + 1 )2 cos( 2 πkc q + 1 )2 cos( 2 πkc q + 1 )= ( q − q ( q + 1) − q − q + 1 q/ X k =1 ( ζ kc q +1 + ζ − kc q +1 )( ζ kc q +1 + ζ − kc q +1 )( ζ kc q +1 + ζ − kc q +1 )= ( q − q ( q + 1) − q − q + 1 q/ X k =1 X ǫ i = ± ( ζ P i =1 ǫ i c i q +1 ) k = ( q − q ( q + 1) − q − q + 1 12 X ǫ i = ± q X k =1 ( ζ P i =1 ǫ i c i q +1 ) k For fixed ( ǫ i ), let d = gcd( q + 1 , P i =1 ǫ i c i ), and let d ′ = ( q + 1) /d • if d < q + 1, then q X k =1 ( ζ P i =1 ǫ i c i q +1 ) k = d X r =1 rd ′ X k =( r − d ′ +1 ( ζ P i =1 ǫ i c i q +1 ) k − d X r =1 ! − − , • if d = q + 1, then q + 1 divides P i =1 ǫ i c i and so q X k =1 ( ζ P i =1 ǫ i c i q +1 ) k = q X k =1 q Now, 1 ≤ c i ≤ q/
2, so the only way for q + 1 to divide P i =1 ǫ i c i is to have c + c + c = q + 1 or 2max( c , c , c ),and then there are exactly two possible ( ǫ i ) for the divisibility to occur. Then • if P i =1 c i = q + 1 or 2max( c , c , c ) then h x q − ,c x q − ,c , x q − ,c i = ( q − q ( q + 1) − q − q + 1 12 (2 q − q − q + 1) − ( q + 1) − q ( q − q ( q + 1) = 0 . NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 7 • else h x q − ,c x q − ,c , x q − ,c i = ( q − q ( q + 1) − q − q + 1 12 ( − q − q + 1) − ( q + 1) + 4 qq ( q + 1) = q + qq ( q + 1) = 1 . (cid:3) Lemma 4.8. h x q − ,c x q − ,c , x q, i = 1 − δ c ,c .Proof. As for the proof of Lemma 4.7: h x q − ,c x q − ,c , x q, i = q ( q − q ( q − − q + 1 q/ X k =1 πkc q + 1 )2 cos( 2 πkc q + 1 )= q − q + 1 − q + 1 12 (2 q −
2) = 0 if c = c , = q − q + 1 − q + 1 12 ( −
4) = 1 else. (cid:3)
Lemma 4.9. h x q − ,c x q − ,c , x q +1 ,c i = 1 .Proof. As for the proof of Lemma 4.7: h x q − ,c x q − ,c , x q +1 ,c i = ( q − ( q + 1) q ( q −
1) + 1 q = q − q + 1 q = 1 . (cid:3) Lemma 4.10. h x q, x q, , x q − ,c i = 1 .Proof. As for the proof of Lemma 4.7: h x q, x q, , x q − ,c i = q ( q − q ( q − − q + 1 q/ X k =1 πkcq + 1 ) = qq + 1 − q + 1 12 ( −
2) = 1 . (cid:3) Lemma 4.11. h x q, x q, , x q, i = 1 .Proof. As for the proof of Lemma 4.7: h x q, x q, , x q, i = q q ( q −
1) + ( q − / q − − q/ q + 1 = q + ( q − q + 1) / − q ( q − / q − . (cid:3) Lemma 4.12. h x q, x q, , x q +1 ,c i = 1 .Proof. As for the proof of Lemma 4.7: h x q, x q, , x q, i = q ( q + 1) q ( q −
1) + 1 q − ( q − / X k =1 πkcq − qq − q − −
2) = 1 . (cid:3) Lemma 4.13. h x q +1 ,c x q +1 ,c , x q − ,c i = 1 .Proof. As for the proof of Lemma 4.7: h x q +1 ,c x q +1 ,c , x q − ,c i = ( q − q + 1) q ( q − − q = q + 1 q − q = 1 . (cid:3) Lemma 4.14. h x q +1 ,c x q +1 ,c , x q, i = 1 .Proof. As for the proof of Lemma 4.7: h x q +1 ,c x q +1 ,c , x q, i = q ( q + 1) q ( q −
1) + 1 q − ( q − / X k =1 πkc q − πkc q − q + 1 q − q − −
4) = 1 . (cid:3) Lemma 4.15. h x q +1 ,c x q +1 ,c , x q, i = 1 + δ c ,c . ZHENGWEI LIU, SEBASTIEN PALCOUX, AND YUNXIANG REN
Proof.
As for the proof of Lemma 4.7: h x q +1 ,c x q +1 ,c , x q, i = q ( q + 1) q ( q −
1) + 1 q − ( q − / X k =1 πkc q − πkc q − q + 1 q − q − q − −
2) = 2 if c = c , = q + 1 q − q − −
4) = 1 else. (cid:3)
Lemma 4.16. h x q +1 ,c x q +1 ,c , x q +1 ,c i = (cid:26) if c + c + c = q − or c , c , c )1 else .Proof. As for the proof of Lemma 4.7: h x q +1 ,c x q +1 ,c , x q +1 ,c i = ( q + 1) q ( q −
1) + 1 q + 1 q − ( q − / X k =1 πkc q − πkc q − πkc q − q + 1) q ( q −
1) + 1 q + 1 q − X ǫ i = ± q − X k =1 ( ζ P i =1 ǫ i c i q − ) k For fixed ( ǫ i ), let d = gcd( q − , P i =1 ǫ i c i ). • if d < q −
1, then P q − k =1 ( ζ P i =1 ǫ i c i q − ) k = − , • if d = q −
1, then P q − k =1 ( ζ P i =1 ǫ i c i q − ) k = q − ≤ c i ≤ ( q − /
2, so the only way for q − P i =1 ǫ i c i is to have c + c + c = q − c , c , c ), and then there are exactly two possible ( ǫ i ) for the divisibility to occur. Then • if P i =1 c i = q + 1 or 2max( c , c , c ) then h x q +1 ,c x q +1 ,c , x q +1 ,c i = ( q + 1) q ( q −
1) + 1 q + 1 q − q − − q + 2 q + 1) + ( q −
1) + ( q − q ) q ( q −
1) = 2 q − qq ( q −
1) = 2 • else h x q +1 ,c x q +1 ,c , x q +1 ,c i = ( q + 1) q ( q −
1) + 1 q + 1 q − − q + 2 q + 1) + ( q − − qq ( q −
1) = q − qq ( q −
1) = 1 . (cid:3) Lemma 4.17. h x q − ,c x q, , x q +1 ,c i = 1 .Proof. As for the proof of Lemma 4.7: h x q − ,c x q, , x q +1 ,c i = ( q − q ( q +1) q ( q − = 1 . (cid:3) Now, let prove that the orthogonality relation in the interpolated case:
Proposition 4.18 (Orthogonality Relation) . For all d i , c i , h x d ,c , x d ,c i = δ ( d ,c ) , ( d ,c ) . Proof.
As for the proof of Lemma 4.7: • h x , , x , i = q ( q − + q + q − q − + q q +1) = 1, • h x , , x q − ,c i = q − q ( q − − q − q +1 P k πkcq +1 ) = q − q ( q − − q − q +1 12 ( −
2) = 0, • h x , , x q, i = qq ( q − + q − q − − q q +1) = 0, • h x , , x q +1 ,c i = q +1 q ( q − + q + q − P k πkcq − ) = q +1 q ( q − + q + q − ( −
2) = 0, • h x q − ,c , x q − ,c i = ( q − q ( q − + q + q +1 12 (2 qδ c ,c − (2 + 2(1 − δ c ,c )) = · · · = δ c ,c , • h x q − ,c , x q, i = ( q − qq ( q − + q +1 12 ( −
2) = 0, • h x q − ,c , x q +1 ,c i = ( q − q +1) q ( q − − q = 0, • h x q, , x q, i = q q ( q − + ( q − / q − + q/ q +1 = · · · = 1 , • h x q, , x q +1 ,c i = q ( q +1) q ( q − + q − ( −
2) = 0.
NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 9 • h x q +1 ,c , x q +1 ,c i = ( q +1) q ( q − + q + q − (2( q − δ c ,c − (2 + 2(1 − δ c ,c )) = · · · = δ c ,c . (cid:3) The case q ≡ − .Theorem 4.19. If q ≡ − , the interpolated fusion ring is of rank ( q + 5) / , FPdim q ( q − / , type [[1 , , [ q − , , [ q − , q − ] , [ q, , [ q + 1 , q − ]] , and given by the following rules: x q − ,c x q − ,c = (1 − δ c ,c ) x , + δ c ,c X c (1 − δ c ,c ) x q − ,c + δ c ,c X c x q − ,c + (1 − δ c ,c ) X c x q +1 ,c ,x q − ,c x q − ,c = X c (1 − δ c ,c ) x q − ,c + X c (1 − δ c + c , q +14 ) x q − ,c + x q, + X c x q +1 ,c ,x q − ,c x q, = X c x q − ,c + x q, + X c x q +1 ,c ,x q − ,c x q +1 ,c = X c δ c ,c x q − ,c + X c x q − ,c + x q, + X c x q +1 ,c ,x q − ,c x q − ,c = δ c ,c x , + (1 − δ c + c , q +14 ) X c x q − ,c + X c such that c + c + c = q +12 or c ,c ,c ) x q − ,c + 2 X c such that c + c + c = q +12 and c ,c ,c ) x q − ,c + (2 − δ c ,c ) x q, + 2 X c x q +1 ,c ,x q − ,c x q, = X c x q − ,c + X c (2 − δ c ,c ) x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q − ,c x q +1 ,c = X c x q − ,c + 2 X c x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q, x q, = x , + X c x q − ,c + 2 X c x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q, x q +1 ,c = X c x q − ,c + 2 X c x q − ,c + 2 x q, + X c (2 + δ c ,c ) x q +1 ,c ,x q +1 ,c x q +1 ,c = δ c ,c x , + X c x q − ,c + 2 X c x q − ,c + (2 + δ c ,c ) x q, + 2 X c such that c + c + c = q − and c ,c ,c ) x q +1 ,c + 3 X c such that c + c + c = q − or c ,c ,c ) x q +1 ,c . Proof.
The rules are given by Theorem 3.2 and Lemmas 4.20 to 4.39. They satisfy all the axioms of a fusion ringby applying Theorem 2.2 together with Proposition 4.40 (with the inner product as in Corollary 2.4). (cid:3)
The smallest q non prime-power in this case is q = 15. Here are the fusion matrices: , , , , , , , , , Lemma 4.20. h x q − ,c x q − ,c , x q − ,c i = δ c ,c (1 − δ c ,c ) .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q − i ( − k + c √ q − − i ( − k + c √ q q + 1 q − X k =1 ( − k +1) + ( − q +14 +1) q + 1 = ( q − q ( q + 1) + 18 q X k =1 ( − i √ q ( − k (( − c + ( − c − ( − c )+ q ( − k (( − c + c − ( − c + c − ( − c + c )+ iq √ q ( − k ( − c + c + c )+ 1 q + 1= ( q − q ( q + 1) − q + 14 (( − c + c − ( − c + c − ( − c + c ) + 1 q + 1Observe that ( − c + c − ( − c + c − ( − c + c = (cid:26) − c , c , c ) = (1 , , , (2 , , h x q − ,c x q − ,c , x q − ,c i = (cid:26) c , c , c ) = (1 , , , (2 , , δ c ,c (1 − δ c ,c ). (cid:3) Lemma 4.21. h x q − ,c x q − ,c , x q − ,c i = δ c ,c .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q − i ( − k + c √ q − q + 1 q − X k =1 ( − k +1) ( − πkc q + 1 )) + ( − q +14 +1) ( − − c ) q + 1= ( q − q ( q + 1) + 14 q X k =1 ( − i √ q ( − k (( − c + ( − c ) + q ( − k ( − c + c ) − q + 1 X ǫ = ± q − X k =1 ( ζ c q +12 ) ǫk − − c q + 1Let S be { , , . . . , q − } . Then − S ∪ S = { , , . . . , q +12 } \ { q +14 , q +12 } mod q +12 . So: h x q − ,c x q − ,c , x q − ,c i = ( q − q ( q + 1) + 12 q ( − q ( − c + c ) + 2 q + 1 (( − c + 1) − − c q + 1= q − q + 1 − q − q + ( − c + c ( q + q )2 q ( q + 1) = (cid:26) c + c odd1 else = δ c ,c , because c , c ∈ { , } , so c + c is odd if and only if c = c . (cid:3) Lemma 4.22. h x q − ,c x q − ,c , x q, i = 0 .Proof. h x q − ,c x q − ,c , x q,c i = 2 q ( q − ) q ( q −
1) + 2 q + 1 q − X k =1 ( − k +1) ( −
1) + ( − q +14 +1) ( − q + 1= q − q + 1) − q + 1 q − − q + 1 = 0 . (cid:3) Lemma 4.23. h x q − ,c x q − ,c , x q +1 ,c i = 1 − δ c ,c .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q + 1)( q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q − i ( − k + c √ q q − q − q ( − q ( − c + c ) = 1 − δ c ,c . (cid:3) NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 11
Lemma 4.24. h x q − ,c x q − ,c , x q − ,c i = 1 − δ c , q +14 − c .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − ( q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q − − q + 1 q − X k =1 ( − k +1 ( − πkc q + 1 ))( − πkc q + 1 ))+ ( − q +14 +1 ( − − c )( − − c ) q + 1= ( q − q ( q + 1) − q − q + 1 X ǫ ,ǫ = ± q − X k =1 ( ζ ǫ c + ǫ c + q +14 q +12 ) k + 4 q + 1 ( − q +14 +1+ c + c Note that ζ q +14 q +12 = ζ − q +14 q +12 , so: h x q − ,c x q − ,c , x q − ,c i = ( q − q ( q + 1) − q + 4 q + 1 ( − q +14 +1+ c + c − q + 1 X ǫ ,ǫ = ± X ǫ = ± q − X k =1 ( ζ ǫ c + ǫ c + q +14 q +12 ) ǫk Now if c + c q +14 mod q +12 : X ǫ ,ǫ = ± X ǫ = ± q − X k =1 ( ζ ǫ c + ǫ c + q +14 q +12 ) ǫk = X ǫ ,ǫ = ± (( − ǫ c + ǫ c + q +14 −
1) = 4( − c + c + q +14 − X ǫ ,ǫ = ± X ǫ = ± q − X k =1 ( ζ ǫ c + ǫ c + q +14 q +12 ) ǫk = q − X ǫ ,ǫ = ± ǫ = ǫ (( − ǫ c + ǫ c + q +14 −
1) = q − − c + c + q +14 − c + c q +14 mod q +12 then c + c + q +14 is even. So: h x q − ,c x q − ,c , x q − ,c i = (cid:26) c + c q +14 mod q +12 − δ c , q +14 − c , because 1 ≤ c i ≤ q − , for i = 2 , (cid:3) Lemma 4.25. h x q − ,c x q − ,c , x q, i = 1 .Proof. h x q − ,c x q − ,c , x q, i = 2( q − q ( q − ) q ( q −
1) + 2 q + 1 q − X k =1 ( − k +1 ( − πkc q + 1 ))( −
1) + ( − q +14 +1 ( − − c )( − q + 1= q − q + 1 − q + 1 q − X k =1 X ǫ = ± ( ζ c + q +14 q +12 ) ǫk − − c + q +14 q + 1 = q − q + 1 + 2 q + 1 = 1 . (cid:3) Lemma 4.26. h x q − ,c x q − ,c , x q +1 ,c i = 1 .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − q + 1)( q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q − = q − q + 1 q = 1 . (cid:3) Lemma 4.27. h x q − ,c x q, , x q, i = 1 .Proof. h x q − ,c x q, , x q, i = 2 q ( q − ) q ( q −
1) + 2 q + 1 q − X k =1 ( − ( k +1) ( − + ( − ( q +14 +1) ( − q + 1= qq + 1 + 1 q + 1 = 1 . (cid:3) Lemma 4.28. h x q − ,c x q, , x q +1 ,c i = 1 .Proof. h x q − ,c x q, , x q +1 ,c i = 2 q ( q + 1)( q − ) q ( q −
1) = 1 . (cid:3) Lemma 4.29. h x q − ,c x q +1 ,c , x q +1 ,c i = 1 .Proof. h x q − ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) ( q − ) q ( q −
1) + 1 q X k =1 ( − i ( − k + c √ q = q + 1 q − q = 1 . (cid:3) Lemma 4.30. h x q − ,c x q − ,c , x q − ,c i = (cid:26) if c + c + c = q +12 or c , c , c )2 else .Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q −
1) + 1 q X k =1 ( − + 2 q + 1 q − X k =1 3 Y i =1 ( − πkc i q + 1 )) + 1 q + 1 Y i =1 ( − − c i )= 2( q − q ( q + 1) − q − q + 1 X ǫ i = ± q − X k =1 ( ζ P i =1 ǫ i c iq +12 ) k − q + 1 ( − P i =1 c i As for the proof of Lemmas 4.7 and 4.21, if P i =1 c i = q +12 and 2max( c , c , c ), then: h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q + 1) − q + 8 q + 1 (1 + ( − P i =1 c i ) − q + 1 ( − P i =1 c i = 2 . Else P i =1 c i = q +12 or 2max( c , c , c ), in particular P i =1 c i is even, and: h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q + 1) − q + 1 q + 1 (6(1 + ( − P i =1 c i ) − ( q − − q + 1 ( − P i =1 c i = 2( q − q ( q + 1) − q + 4 − ( q − q + 1 = 1 . (cid:3) Lemma 4.31. h x q − ,c x q − ,c , x q, i = 2 − δ c ,c .Proof. If c = c : h x q − ,c x q − ,c , x q − ,c i = 2 q ( q − q ( q −
1) + 2 q + 1 q − X k =1 ( − Y i =1 ( − πkc i q + 1 )) + 1 q + 1 ( − Y i =1 ( − − c i )= 2( q − q + 1 + 4 q + 1 (1 + ( − c + c ) − q + 1 ( − c + c = 2( q − q + 1 + 4 q + 1 = 2 . Else, c = c , and: h x q − ,c x q − ,c , x q − ,c i = 2( q − q + 1 + 2 q + 1 (1 + ( − c − q −
32 ) − q + 1 ( − c = 1 . (cid:3) Lemma 4.32. h x q − ,c x q − ,c , x q +1 ,c i = 2 . NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 13
Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q + 1)( q − q ( q −
1) + 1 q X k =1 ( − (1) = 2( q − q + 2 q = 2 . (cid:3) Lemma 4.33. h x q − ,c x q, , x q, i = 2 .Proof. h x q − ,c x q, , x q, i = 2 q ( q − q ( q −
1) + 2 q + 1 q − X k =1 ( − πkc q + 1 ))( − + − − c ( − q + 1= 2 qq + 1 + 2 q + 1 (1 + ( − c ) − q + 1 ( − c b = 2 qq + 1 + 2 q + 1 = 2 . (cid:3) Lemma 4.34. h x q − ,c x q, , x q +1 ,c i = 2 .Proof. h x q − ,c x q, , x q +1 ,c i = 2 q ( q + 1)( q − q ( q −
1) = 2 . (cid:3) Lemma 4.35. h x q − ,c x q +1 ,c , x q +1 ,c i = 2 .Proof. h x q − ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) ( q − q ( q −
1) + 1 q X k =1 ( − = 2( q + 1) q − q = 2 . (cid:3) Lemma 4.36. h x q, x q, , x q, i = 2 .Proof. h x q − ,c x q, , x q, i = 2 q q ( q −
1) + 2 q − q − X k =1 (1) + 2 q + 1 q − X k =1 ( − + ( − q + 1= 2 q q − q − q − − q + 1 q − − q + 1 = 2 . (cid:3) Lemma 4.37. h x q, x q, , x q +1 ,c i = 2 .Proof. h x q, x q, , x q +1 ,c i = 2( q + 1) q q ( q −
1) + 2 q − q − X k =1 (1) (2 cos( 4 πkc q − . Let S be { , , . . . , q − } . Then − S ∪ S = { , , . . . , q − } \ { q − } mod q − . So (as for Lemma 4.21): h x q, x q, , x q +1 ,c i = 2 qq − q − −
1) = 2 . (cid:3) Lemma 4.38. h x q, x q +1 ,c , x q +1 ,c i = 2 + δ c ,c .Proof. If c = c : h x q, x q +1 ,c , x q +1 ,c i = 2( q + 1) qq ( q −
1) + 2 q − q − X k =1 (1)(2 cos( 4 πkc q − πkc q − q + 1) q − q − −
1) = 2 . Else c = c , and: h x q, x q +1 ,c , x q +1 ,c i = 2( q + 1) q − q − − q −
32 ) = 3 . (cid:3) Lemma 4.39. h x q +1 ,c x q +1 ,c , x q +1 ,c i = (cid:26) if c + c + c = q − or c , c , c )2 else Proof. h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 1 q X k =1 (1) + 2 q − q − X k =1 3 Y i =1 (2 cos( 4 πkc i q − q + 1) q ( q −
1) + 2 q + 2 q − X ǫ i = ± q − X k =1 ( ζ P i =1 ǫ i c iq − ) k If P i =1 c i = q − and 2max( c , c , c ), then (as for some previous lemmas): h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 2 q + 8 q − −
1) = 2 . Else P i =1 c i = q − or 2max( c , c , c ), and: h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 2 q + 1 q − q − −
6) = 3 . (cid:3) Proposition 4.40 (Orthogonality Relation) . For all d i , c i , h x d ,c , x d ,c i = δ ( d ,c ) , ( d ,c ) . Proof.
As for the proof of previous lemmas: • h x , , x , i = q ( q − + q + q − q − + q +1 q − + q +1 = q − q ( q +1)( q − q ( q − q − q ( q − q ( q − = 1, • h x , , x q − ,c i = q − q ( q − + q P k =1 − − i ( − k + c √ q + q +1 P q − k =1 ( − k + q +1 ( − q − = q ( q +1) − q + q +1 = 0, • h x , , x q − ,c i = q − q ( q − − q + q +1 P q − k =1 ( − πkcq +1 )) + q +1 ( − − c ) = − q +1 + − c q +1 − − c q +1 = 0, • h x , , x q, i = qq ( q − + q − q − − q +1 q − − q +1 = q − + q − q − − q +1 = 0, • h x , , x q +1 ,c i = q +1) q ( q − + q + q − P q − k =1 (2 cos( πkcq − )) = q ( q − + q − q − = 0, • h x q − ,c , x q − ,c i = ( q − q ( q − + q P k =1 ( − i ( − k + c √ q − − i ( − k + c √ q )+ q +1 P q − k =1 ( − k +1) + q +1 ( − q − = q − q ( q +1) + q ( − c c q + q − q +1) + q +1 = q − q +1)(1+ q ( − c c )+ q ( q − q ( q +1) = ( q +1) q (1+( − c c )2 q ( q +1) = δ c ,c , • h x q − ,c , x q − ,c i = ( q − q ( q − − q P k =1 ( − i ( − k + c √ q )+ q +1 P q − k =1 ( − k (2 cos( πkc q +1 ))+ q +1 ( − q +14 ( − c = ( q − q ( q +1) + q − q +1 (1 + ( − c + q +14 ) + q +1 ( − c + q +14 = q − q +1 − qq ( q +1) = 0, • h x q − ,c , x q, i = ( q − qq ( q − + q +1 P q − k =1 ( − k + q +1 ( − q +14 = q +1 − q +1 = 0, • h x q − ,c , x q +1 ,c i = ( q − q +1) q ( q − + q P k =1 ( − i ( − k + c √ q ) = q − q = 0, • h x q − ,c , x q − ,c i = q − q ( q − + q + q +1 P q − k =1 (2 cos( πkc q +1 ))(2 cos( πkc q +1 )) + q +1 ( − c + c = q − q ( q +1) + q − q +1 (2(1 − δ c ,c )(1 + ( − c + c ) + δ c ,c (1 + ( − c + c − q − )) + q +1 ( − c + c = δ c ,c , • h x q − ,c , x q, i = q − qq ( q − + q +1 P q − k =1 (2 cos( πkc q +1 )) + q +1 ( − c = q +1 − q +1 (1 + ( − c ) + q +1 ( − c = 0, • h x q − ,c , x q +1 ,c i = q − q +1) q ( q − + q P k =1 ( −
1) = q − q = 0, • h x q, , x q, i = q q ( q − + q − P q − k =1 (1) + q +1 P q − k =1 (1) + q +1 (1) = qq − + q − q − + q +1 q − + q +1 = 1, • h x q, , x q +1 ,c i = q ( q +1) q ( q − + q − P q − k =1 (2 cos( πkc q − )) = q − − q − = 0, • h x q +1 ,c , x q +1 ,c i = q +1) q ( q − + q P k =1 (1) + q − P q − k =1 (2 cos( πkc q − ))(2 cos( πkc q − )) = q +1) q ( q − + q + q − ((1 − δ c ,c )( −
2) + δ c ,c ( − q − )) = δ c ,c . (cid:3) The case q ≡ .Theorem 4.41. If q ≡ , the interpolated fusion ring is of rank ( q + 5) / , FPdim q ( q − / , type [[1 , , [ q +12 , , [ q − , q − ] , [ q, , [ q + 1 , q − ]] , and given by the following rules: x q +12 ,c x q +12 ,c = δ c ,c x , + δ c ,c X c δ c ,c x q +12 ,c + (1 − δ c ,c ) X c x q − ,c + x q, + δ c ,c X c x q +1 ,c ,x q +12 ,c x q − ,c = X c (1 − δ c ,c ) x q +12 ,c + X c x q − ,c + x q, + X c x q +1 ,c , NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 15 x q +12 ,c x q, = X c x q +12 ,c + X c x q − ,c + x q, + X c x q +1 ,c ,x q +12 ,c x q +1 ,c = X c δ c ,c x q +12 ,c + X c x q − ,c + x q, + X c (1 + δ c + c , q − ) x q +1 ,c ,x q − ,c x q − ,c = δ c ,c x , + X c x q +12 ,c + X c such that c + c + c = q +12 or c ,c ,c ) x q − ,c + 2 X c such that c + c + c = q +12 and c ,c ,c ) x q − ,c + (2 − δ c ,c ) x q, + 2 X c x q +1 ,c ,x q − ,c x q, = X c x q +12 ,c + X c (2 − δ c ,c ) x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q − ,c x q +1 ,c = X c x q +12 ,c + 2 X c x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q, x q, = x , + X c x q +12 ,c + 2 X c x q − ,c + 2 x q, + 2 X c x q +1 ,c ,x q, x q +1 ,c = X c x q +12 ,c + 2 X c x q − ,c + 2 x q, + X c (2 + δ c ,c ) x q +1 ,c ,x q +1 ,c x q +1 ,c = δ c ,c x , + (1 + δ c + c , q − ) X c x q +12 ,c + 2 X c x q − ,c + (2 + δ c ,c ) x q, + 3 X c such that c + c + c = q − or c ,c ,c ) x q +1 ,c + 2 X c such that c + c + c = q − and c ,c ,c ) x q +1 ,c . Proof.
The rules are given by Theorem 3.3 and Lemmas 4.42 to 4.61. They satisfy all the axioms of a fusion ringby applying Theorem 2.2 together with Proposition 4.62 (with the inner product as in Corollary 2.4). (cid:3)
The smallest q non prime-power in this case is q = 21. Here are the fusion matrices: , , , , , , , , , , , , Lemma 4.42. h x q +12 ,c x q +12 ,c , x q +12 ,c i = δ c ,c δ c ,c .Proof. h x q +12 ,c x q +12 ,c , x q +12 ,c i = 2( q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q − k + c √ q − k + c √ q q − q − X k =1 ( − k + ( − q − q − = ( q + 1) q ( q −
1) + 18 q X k =1 (1 + √ q ( − k (( − c + ( − c + ( − c )+ q ( − k (( − c + c + ( − c + c + ( − c + c )+ q √ q ( − k ( − c + c + c ) − q −
1= ( q + 1) q ( q −
1) + 14 q + 14 (( − c + c + ( − c + c + ( − c + c ) − q − − c + c + ( − c + c + ( − c + c = (cid:26) c = c = c − h x q +12 ,c x q +12 ,c , x q +12 ,c i = (cid:26) c = c = c δ c ,c δ c ,c . (cid:3) Lemma 4.43. h x q +12 ,c x q +12 ,c , x q − ,c i = 1 − δ c ,c .Proof. h x q +12 ,c x q +12 ,c , x q − ,c i = 2( q − q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q − k + c √ q − q + 12 q − q X k =1 (1 + √ q ( − k (( − c + ( − c ) + q ( − k ( − c + c )= q + 12 q − q ( − c + c q = 1 − δ c ,c because c , c ∈ { , } , so that c + c is even if and only if c = c . (cid:3) Lemma 4.44. h x q +12 ,c x q +12 ,c , x q, i = 1 .Proof. h x q +12 ,c x q +12 ,c , x q, i = 2 q ( q +12 ) q ( q −
1) + 2 q − q − X k =1 ( − k + ( − q − q − q + 12( q −
1) + 2 q − q −
54 + 1 q − . (cid:3) Lemma 4.45. h x q +12 ,c x q +12 ,c , x q +1 ,c i = δ c ,c .Proof. h x q +12 ,c x q +12 ,c , x q +1 ,c i = 2( q + 1)( q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q − k + c √ q q − q − X k =1 ( − k πkc q − − q − − c q −
1= ( q + 1) q ( q −
1) + 1 + q ( − c + c q − q − − c ) + 2 q − − c = 12 q ( q −
1) ( q + 2 q + 1 + q + q ( − c + c − − q ( − c + c − q ) = δ c ,c . (cid:3) Lemma 4.46. h x q +12 ,c x q − ,c , x q − ,c i = 1 .Proof. h x q +12 ,c x q − ,c , x q − ,c i = 2( q − ( q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q − − NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 17 = q − q + 12 q X k =1 (1 + ( − k + c √ q )= q − q + 1 q = 1 . (cid:3) Lemma 4.47. h x q +12 ,c x q − ,c , x q, i = 1 .Proof. h x q +12 ,c x q − ,c , x q − ,c i = 2 q ( q − q +12 ) q ( q −
1) = 1 . (cid:3) Lemma 4.48. h x q +12 ,c x q − ,c , x q +1 ,c i = 1 .Proof. h x q +12 ,c x q − ,c , x q +1 ,c i = 2( q − q + 1)( q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q − q + 1 q − q = 1 . (cid:3) Lemma 4.49. h x q +12 ,c x q, , x q, i = 1 .Proof. h x q +12 ,c x q, , x q, i = 2 q ( q +12 ) q ( q −
1) + 2 q − q − X k =1 ( − k + ( − q − q − qq − − q − . (cid:3) Lemma 4.50. h x q +12 ,c x q, , x q +1 ,c i = 1 .Proof. h x q +12 ,c x q, , x q +1 ,c i = 2 q ( q + 1)( q +12 ) q ( q −
1) + 2 q − q − X k =1 ( − k πkc q − − q − − c q − q + 1 q − − q − − q − + c ) + 2( − q − + c q − . (cid:3) Lemma 4.51. h x q +12 ,c x q +1 ,c , x q +1 ,c i = 1 + δ c + c , q − .Proof. h x q +12 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) ( q +12 ) q ( q −
1) + 1 q X k =1 ( 1 + ( − k + c √ q q − q − X k =1 ( − k πkc q − πkc q − − q − − c − c q −
1= ( q + 1) q ( q −
1) + 1 q + 1 q − δ c + c , q − ( q − − − q − + c + c )) − (1 − δ c + c , q − )4(1 + ( − q − + c + c )) + 4( − q − + c + c q −
1= ( q + 1) q ( q −
1) + 1 q + δ c + c , q − ( q − − (1 − δ c + c , q − )4 q − δ c + c , q − + (1 − δ c + c , q − ) = 1 + δ c + c , q − . (cid:3) Lemma 4.52. h x q − ,c x q − ,c , x q − ,c i = (cid:26) if c + c + c = q +12 or c , c , c )2 else . Proof. h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q −
1) + 1 q X k =1 ( − + 2 q + 1 q − X k =1 3 Y i =1 ( − πkc i q + 1 ))= 2( q − q ( q + 1) − q − q + 1 X ǫ i = ± q − X k =1 ( ζ P i =1 ǫ i c iq +12 ) k If P i =1 c i = q +12 and 2max( c , c , c ), then (as for some previous lemmas): h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q + 1) − q − q + 1 ( −
1) = 2 . Else P i =1 c i = q +12 or 2max( c , c , c ), and then: h x q − ,c x q − ,c , x q − ,c i = 2( q − q ( q + 1) − q − q + 1 ( q − −
6) = 1 . (cid:3) Lemma 4.53. h x q − ,c x q − ,c , x q, i = 2 − δ c ,c .Proof. h x q − ,c x q − ,c , x q, i = 2 q ( q − q ( q − − q + 1 q − X k =1 2 Y i =1 ( − πkc i q + 1 ))= 2( q − q + 1) − q + 1 ( δ c ,c ( q − −
2) + (1 − δ c ,c )( − − δ c ,c . (cid:3) Lemma 4.54. h x q − ,c x q − ,c , x q +1 ,c i = 2 .Proof. h x q − ,c x q − ,c , x q +1 ,c i = 2( q + 1)( q − q ( q −
1) + 1 q X k =1 ( − = 2( q − q + 2 q = 2 . (cid:3) Lemma 4.55. h x q − ,c x q, , x q, i = 2 .Proof. h x q − ,c x q, , x q, i = 2 q ( q − q ( q −
1) + 2 q + 1 q − X k =1 ( − πkc q + 1 ))= 2 q ( q + 1) − q + 1 ( −
1) = 2 . (cid:3) Lemma 4.56. h x q − ,c x q, , x q +1 ,c i = 2 .Proof. h x q − ,c x q, , x q +1 ,c i = 2 q ( q + 1)( q − q ( q −
1) = 2 . (cid:3) Lemma 4.57. h x q − ,c x q +1 ,c , x q +1 ,c i = 2 .Proof. h x q − ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) ( q − q ( q −
1) + 1 q X k =1 ( −
1) = 2( q + 1) q − q = 2 . (cid:3) Lemma 4.58. h x q, x q, , x q, i = 2 .Proof. h x q, x q, , x q, i = 2 q q ( q −
1) + 2 q − q − X k =1 (1) + 1 q − + 2 q + 1 q − X k =1 ( − = 2 q q − q − q −
54 + 1 q − − q + 1 q −
14 = 2 . (cid:3) NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 19
Lemma 4.59. h x q, x q, , x q +1 ,c i = 2 .Proof. h x q, x q, , x q +1 ,c i = 2( q + 1) q q ( q −
1) + 2 q − q − X k =1 (2 cos( 4 πkc q − q − − c = 2 qq − − q − − c ) + 2( − c q − . (cid:3) Lemma 4.60. h x q, x q +1 ,c , x q +1 ,c i = 2 + δ c ,c .Proof. h x q, x q +1 ,c , x q +1 ,c i = 2( q + 1) qq ( q −
1) + 2 q − q − X k =1 (2 cos( 4 πkc q − πkc q − q − − c − c = 2( q + 1) q − − − δ c ,c )2(1 + ( − c + c ) + δ c ,c (1 + ( − c + c − q − )) q − − c + c q − δ c ,c . (cid:3) Lemma 4.61. h x q +1 ,c x q +1 ,c , x q +1 ,c i = (cid:26) if c + c + c = q − or c , c , c )2 else .Proof. h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 1 q X k =1 (1) + 2 q − q − X k =1 3 Y i =1 (2 cos( 4 πkc i q − q − − P i =1 c i If P i =1 c i = q − and 2max( c , c , c ), then (as for some previous lemmas): h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 2 q − − P i =1 c i ) q − − P i =1 c i q − q + 2 q + 1 + q − − qq ( q −
1) = 2 . Else P i =1 c i = q − or 2max( c , c , c ), in particular P i =1 c i is even, and then: h x q +1 ,c x q +1 ,c , x q +1 ,c i = 2( q + 1) q ( q −
1) + 2 q + q − − − P i =1 c i ) q − − P i =1 c i q −
1= 2 q + 4 q + 2 + 2 q − q − qq ( q −
1) = 3 . (cid:3) Proposition 4.62 (Orthogonality Relation) . For all d i , c i , h x d ,c , x d ,c i = δ ( d ,c ) , ( d ,c ) . Proof.
As for the proof of previous lemmas: • h x , , x , i = q ( q − + q + q − q − + q − + q +1 q − = q − q ( q +1)( q − q ( q − q ( q − = 1, • h x , , x q − ,c i = q ( q − + q P k =1 ( − k + c √ q ) + q − P q − k =1 ( − k + q − ( − q − = q ( q − + q − q − = 0, • h x , , x q − ,c i = q ( q +1) + q P k =1 ( − − q +1 P q − k =1 (2 cos( πkcq +1 )) = q ( q +1) − q − q +1 ( −
1) = 0, • h x , , x q, i = q − + q − q − + q − − q +1 q − = q +1)( q − − ( q − q − = 0, • h x , , x q +1 ,c i = q ( q − + q + q − P q − k =1 (2 cos( πkcq − )) + − c q − = q ( q − + q − q − (1 + ( − c ) + − c q − = 0, • h x q +12 ,c , x q +12 ,c i = q +12 q ( q − + q P k =1 ( − k + c √ q )( − k + c √ q ) + q − P q − k =1 ( − k + q − ( − q − = q +12 q ( q − + q (1 + ( − c + c q ) + q − q − + q − = q +12 q ( q − + q + ( − c + c + q − q − = δ c ,c , • h x q +12 ,c , x q − ,c i = ( q +1)( q − q ( q − − q P k =1 ( − k + c √ q ) = q − q = 0, • h x q +12 ,c , x q, i = q ( q +1) q ( q − + q − P q − k =1 ( − k + q − ( − q − = q − − q − = 0, • h x q +12 ,c , x q +1 ,c i = ( q +1) q ( q − + q P k =1 ( − k + c √ q ) + q − P q − k =1 ( − k πkc q − ) + q − ( − q − − c = q +1 q ( q − + q − q − (1 + ( − q − + c ) + q − ( − q − + c = 0, • h x q − ,c , x q − ,c i = q − q ( q − + q P k =1 ( − + q +1 P q − k =1 ( − πkc q +1 ))( − πkc q +1 )) = q − q ( q +1) + q + q +1 ( δ c ,c ( q + 1) −
4) = q − q +2+ δ c ,c ( q + q ) − qq ( q +1) = δ c ,c ( q + q ) q ( q +1) = δ c ,c , • h x q − ,c , x q, i = q ( q − q ( q − + q +1 P q − k =1 (2 cos( πkc q +1 )) = q +1 − q +1 = 0, • h x q − ,c , x q +1 ,c i = q − q +1) q ( q − + q P k =1 ( −
1) = q − q = 0, • h x q, , x q, i = q q ( q − + q − q − + q − + q +1 q − = qq − + q − q − + q − + q − q +1) = q +( q − q +1)+( q − q − = 1, • h x q, , x q +1 ,c i = q ( q +1) q ( q − + q − P q − k =1 (2 cos( πkcq − )) + q − (2( − c ) = q − − q − (1 + ( − c ) + − c q − = 0, • h x q +1 ,c , x q +1 ,c i = q +1) q ( q − + q P k =1 (1) + q − P q − k =1 (2 cos( πkc q − ))(2 cos( πkc q − )) + q − (2( − c )(2( − c ) =2 + δ c ,c + q +1) q ( q − − q +1) q − + q = δ c ,c + q − q +2 q +2 − q − q +2 q − q ( q − = δ c ,c . (cid:3) Categorification criteria checking
In this section we check seven categorification criteria on the fusion rings of Section 4.5.1.
Schur product criterion.
Let F be a commutative fusion ring. Let Λ = ( λ i,j ) be its eigentable, λ i, =max j ( | λ i,j | ). Here is the commutative Schur product criterion: Theorem 5.1 ( [24] Corollary 8.5) . If F admits a unitary categorification then for all triple ( j , j , j ) we have X i λ i,j λ i,j λ i,j λ i, ≥ . Note that Theorem 5.1 is the corollary of a (less tractable) noncommutative version [24, Proposition 8.3], whichis an application of the quantum Schur product theorem on subfactors [21, Theorem 4.1].
Lemma 5.2.
Let m be a positive integer. The supremum on the positive roots of all the polynomials with integercoefficients in [ − m, m ] is m + 1 .Proof. If P ni =0 a i x i = 0 with x >
0, then | a n | x n ≤ P n − i =0 | a i | x i . So we only need to consider X n − m P n − i =0 X i . (cid:3) Proposition 5.3.
All the fusion rings of the interpolated family pass Schur product criterion.Proof.
According to the computations of Section 4 and the fact that the variables k and c play the same role inthe tables of Section 3, it is clear that we are reduced to consider rational functions and to show that they arenon-negative. Let first compute one example of such rational funtion, say for q even and j , j , j given by the lastcolumn in Theorem 3.1 (we will then see a general argument). As for the proof of Lemma 4.7:1 + 1 q − q/ X c =1 3 Y i =1 ( − πk i cq + 1 )) − q = 1 − q − q + 1 12 X ǫ i = ± q X c =1 ( ζ P i =1 ǫ i k i q +1 ) c = ( − q − q +1 12 ( −
1) = q + q − q ( q +1) if gcd( q + 1 , P i =1 ǫ i k i ) < q + 11 − q − q +1 12 ( q ) = q − q ( q +1) elseIn general, we already know that these rational functions are non-negative for q prime-power (because if G is afinite group then Rep( G ) is a unitary fusion category), so they must be non-negative for q large enough. It remainsto show that they are also non-negative for q small. Let P/Q be such a rational function, and assume the existenceof q small such that ( P/Q )( q ) <
0. We know that for q large enough ( P/Q )( q ) >
0, so it must exist x such that q < x < q and P ( x ) = 0. Now, it is clear according to the computations of Section 4 that the polynomial P canbe chosen with integer coefficents in [ − m, m ] with m equal to (say) 50; so by Lemma 5.2, x ≤
51, but we checkedby computer that the criterion is passed for all q ≤
70. The result follows. (cid:3)
Ostrik criterion.
Here is the commutative version of a criterion by V. Ostrik, using the formal codegrees( c i ). Theorem 5.4 ( [31] Theorem 2.21) . Let F be a commutative fusion ring. If it admits a pseudo-unitary complexcategorification, then P j c j ≤ c . Proposition 5.5.
All the fusion rings of the interpolated family pass Ostrik criterion.
NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 21
Proof. X j / c j = q ( q − + q + q −
22 1( q − + q q +1) = q − q − q +2 q ( q − if q even q ( q − + q + q − q − + q +1) q − + q +1) = q − q − q +6 q ( q − if q ≡ − q ( q − + q + q − q − + q +1) q − + q − = q − q − q − q +6 q ( q − if q ≡ < <
12 (1 + 1 c ) . (cid:3) Drinfeld center criterion.
This criterion is also a commutative version of a criterion proved by V. Ostrik,which also involves the formal codegrees ( c i ). Theorem 5.6 ( [31] Corollary 2.14) . Let F be a commutative fusion ring. If it admits a complex categorificationthen for all j , c / c j is an algebraic integer. If F admits a complex spherical categorification C , then by [31, Theorem 2.13] ( c / c j ) are exactly the FPdim ofthe simple objects of the Drinfeld center which contains the trivial object in C under the forgetful functor. Proposition 5.7.
All the fusion rings of the interpolated family pass Drinfeld center criterion.Proof.
The numbers ( c / c j ) are precisely the class sizes mentioned in the tables of Section 3, they are clearlyintegers. (cid:3) Extended cyclotomic criterion.
The following theorem is a slight extension of the usual cyclotomic criterion(on the simple object FPdims) of a fusion ring to all the entries of its eigentable, in the commutative case.
Theorem 5.8.
Let F be a commutative fusion ring, and let ( λ i,j ) be its eigentable. If F admits a complexcategorification then λ i,j is a cyclotomic integer, for all i, j .Proof. It follows from [10, Theorem 8.51] (see [23] for more details). (cid:3)
Proposition 5.9.
All the fusion rings of the interpolated family pass the extended cyclotomic criterion.Proof.
All the entries in the tables of Section 3 are clearly cyclotomic integers (for the ones with a square root,consider the quadratic Gauss sums). (cid:3)
Conjectural Isaacs criterion.
Let F be a commutative fusion ring and let ( λ i,j ) be its eigentable. Theorem5.6 admits the following conjectural stronger version extending Theorem 3.7 of Isaacs’ book [14]. Conjecture 5.10. If F admits a complex categorification, then λ i,j c λ i, c j is an algebraic integer for all i, j . As observed by P. Etingof [7] (see the details in [23]), if Conjecture 5.10 is true then every complex abelian fusioncategory is of
Frobenius type (i.e. c λ i, is an algebraic integer, for all i ), which is the abelian case of a well-knownopen problem generalizing Kaplansky’s 6th conjecture. Proposition 5.11.
All the fusion rings of the interpolated family pass the conjectural Isaacs criterion.Proof.
The case j = 1 or λ i,j = 0 are obvious. If j = 1 and λ i,j = 0, then observe on the tables that c λ i, c j is aninteger, whereas λ i,j is a cyclotomic integer (by Theorem 5.8). (cid:3) Zero spectrum criterion.
Here is a general categorification obstruction over every field. It corresponds tothe existence of a pentagon equation of the form xy = 0, with x, y = 0 (see example as F in [22]). Theorem 5.12 ( [22]) . For a fusion ring R , if there are indices i , . . . , i , such that n i i ,i , n i i ,i , n i i ,i , n i i ,i , n i i ,i , n i i ,i = 0 , (1) X k n ki ,i n ki ∗ ,i n ki ,i ∗ = 0 , (2) n i i ,i = 1 , (3) X k n ki ,i n ki ,i ∗ = 1 or X k n ki ,i ∗ n ki ,i ∗ = 1 or X k n ki ∗ ,i n ki ,i ∗ = 1 , (4) X k n ki ,i n ki ,i ∗ = 1 or X k n ki ,i ∗ n ki ,i ∗ = 1 or X k n ki ∗ ,i n ki ,i ∗ = 1 . (5) then R cannot be categorified, i.e. R is not the Grothendieck ring of a fusion category, over any field K . Let take the index 1 for the trivial object.
Lemma 5.13.
Let F be a fusion ring, and assume the existence of indices i , . . . , i such that (1) and (2) hold.Then i j = 1 , for all j ∈ { , . . . , } .Proof. Assume that there is j ∈ { , . . . , } such that i j = 1. We will reach a contradiction. Let write the detailsfor j = 4 (the other cases are similar): assume that i = 1, by (1) and the neutral axiom of fusion ring, i = i and i = i ; in addition, n k ,i = δ i ,k . Then, by (2), n i i ∗ ,i n i i ,i ∗ = 0. Thus, using Frobenius reciprocity, n i i ,i = 0 or n i i ,i = 0, contradiction with (1). (cid:3) Lemma 5.14.
Let F be a fusion ring such that there is k with n k i,j = 0 for all i, j = 1 . Then F passes the zerospectrum criterion.Proof. Assume that F does not pass the zero spectrum criterion. Then there are indices i , . . . , i such that (1)and (2) hold. By Lemma 5.13, i j = 1, for all j ∈ { , . . . , } . Thus X k n ki ,i n ki ∗ ,i n ki ,i ∗ ≥ n k i ,i n k i ∗ ,i n k i ,i ∗ = 0by assumption, which contradicts (2). (cid:3) Proposition 5.15.
All the fusion rings of the interpolated family pass the zero spectrum criterion.Proof.
For q even or q ≡ x q +1 , and x q, respectively, fromTheorems 4.5 and 4.41. Now for q ≡ − P k n ki ,i n ki ∗ ,i n ki ,i ∗ does not involve x q − ,c x q − ,c with c = c (respectively with c = c ),then we can use the argument as for the proof of Lemma 5.14 with x q +1 , (respectively with x q − ,c and appropriate c ∈ { , . . . , q − } , according to the second line in Theorem 4.19, which is ok because we can take q ≥
15, so that q − ≥ x q − ,c x q − ,c and x q − ,c x q − ,c , with c = c and c = c ,where c i ∈ { , } . We will reach a contradiction. All the possible cases are similar, let write the details when ( i , i )and ( i ∗ , i ) are given by ( x q − ,c , x q − ,c ) and ( x q − ,c , x q − ,c ) respectively: by (1) and Frobenius reciprocity, n i i ,i n i i ,i ∗ = n i i ,i n i i ,i = 0 , so i must be given by a simple component of x q − ,c x q − ,c and x q − ,c x ∗ q − ,c , but there is no one by the firstline of Theorem 4.19, contradiction. (cid:3) One spectrum criterion.
Here is a general categorification obstruction over every field. It corresponds tothe existence of a pentagon equation of the form 0 = xyz , with x, y, z = 0 (see example as F in [22]). Theorem 5.16 ( [22]) . For a fusion ring R , if there are indices i , i , . . . , i such that n i i ,i , n i i ,i , n i i ,i , n i i ,i , n i i ,i , n i i ,i = 0 , (6) X k n ki ,i n ki ∗ ,i n ki ,i ∗ = 1 , (7) n i i ,i = n i i ∗ ,i = n i i ,i ∗ = 1 , (8) n i i ,i = 0 , (9) X k n ki ,i n ki ,i ∗ = 1 or X k n ki ,i ∗ n ki ,i ∗ = 1 or X k n ki ∗ ,i n ki ,i ∗ = 1 , (10) X k n ki ,i n ki ,i ∗ = 1 or X k n ki ,i ∗ n ki ,i ∗ = 1 or X k n ki ∗ ,i n ki ,i ∗ = 1 , (11) X k n ki ,i n ki ,i ∗ = 1 or X k n ki ,i ∗ n ki ,i ∗ = 1 or X k n ki ∗ ,i n ki ,i ∗ = 1 . (12) then R cannot be categorified, i.e. R is not the Grothendieck ring of a fusion category. Lemma 5.17.
Let F be a fusion ring, and assume the existence of indices i , . . . , i such that (6) and (9) hold.Then i j = 1 , for all j ∈ { , . . . , } .Proof. Assume that there is j ∈ { , . . . , } such that i j = 1. We will reach a contradiction. Let write the detailsfor j = 4 (the other cases are similar): assume that i = 1, by (6), i = i and i = i . So, n i i ,i = n i i ,i = 0 by(6), but n i i ,i = 0 by (9), contradiction. (cid:3) NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 23
Lemma 5.18.
Let F be a fusion ring such that there are k , k with k = k , n k i,j , n k i,j = 0 for all i, j = 1 . Then F passes the one spectrum criterion.Proof. Assume that F does not pass the one spectrum criterion. Then there are indices i , . . . , i such that (6), (7)and (9) hold. By Lemma 5.17, i j = 1, for all j ∈ { , . . . , } . Thus X k n ki ,i n ki ∗ ,i n ki ,i ∗ ≥ n k i ,i n k i ∗ ,i n k i ,i ∗ + n k i ,i n k i ∗ ,i n k i ,i ∗ ≥ (cid:3) Proposition 5.19.
All the fusion rings of the interpolated family pass the one spectrum criterion.Proof.
For q even: we can directly apply Lemma 5.18, with x q +1 ,c and c ∈ { , . . . , q − } (from Theorem 4.5) becausewe can assume q ≥
6, so that q − ≥ q ≡ n ki ,i n ki ∗ ,i n ki ,i ∗ = 0 if and only if k = i given by x q, . Thus P k = i n ki ,i n ki ∗ ,i n ki ,i ∗ = 0 , and as for the argument of Proposition 5.19, it implies that the above summust involve x q − ,c x q − ,c and x q − ,c x q − ,c , with c = c and c = c , which also reach to a contradiction.Finally, for q ≡ − x q − ,c x q − ,c then it must be at least 2by considering x q, and x q +1 , , ok. Else, if the sum (7) involves exactly one such product, and if c = c (respectively c = c ) then consider x q +1 ,c with c ∈ { , . . . , q − } (respectively x q − ,c with c ∈ { , . . . , q − } and c = q +14 − c ),which is also ok because we can assume q ≥
15, so that q − ≥
3. It remains that the sum (7) involves two suchproducts, x q − ,c x q − ,c and x q − ,c x q − ,c . If c = c and c = c (or both different) then applies previous step,else there is one equality and one non-equality, and by Theorem 4.41, the sum (7) must be zero. (cid:3) Braiding criterion
Once all the known criteria checked, we wonder whether the fusion rings of the interpolated family (in the nonprime-power case) can be categorified. We could naively expect that the F-symbols can be interpolated too, whichis not clear (at least without modification)... What is true is that the symmetric structure cannot be interpolated,so something else could be broken somewhere. What about the braiding structure? In this section we providea sufficient condition on the interpolated fusion rings to have no braided categorification. We refer to [8] for thenotions of degenerated, braided or symmetric fusion category.
Theorem 6.1 ( [29], Corollary 6.16) . Let C be a non-degenerate integral braided fusion category (over C ). If thereis a simple object of prime-power FPdim then there is a nontrivial symmetric subcategory.
Corollary 6.2.
For q non prime-power, if the interpolated fusion ring admits a simple object of prime-power FPdim then it has no braided categorification.Proof.
Assume it admits a braided categorification C . Then it can be non-degenerated (i.e. C ′ = V ec ) or degener-ated.If it is non-degenerated then, by Theorem 6.1, simplicity and Lemma 4.3, there is no nontrivial simple object ofprime-power FPdim.Else it is degenerated, i.e.
V ec = C ′ , so by simplicity C ′ = C , so C is symmetric, and by [5] and simplicity, C ≃
Rep( G ) as fusion category (without considering the symmetric structure), with G a finite simple group,contradiction with Lemma 4.3. (cid:3) Here is the list of q <
200 such that the associated fusion ring has no nontrivial simple object of prime-powerFPdim, so that Corollary 6.2 cannot be applied, and the existence of a braided categorification is open: • q even: 34, 56, 76, 86, 92, 94, 116, 118, 134, 142, 144, 146, 154, 160, 176, 184, 186, 188, • q ≡ − • q ≡ Question 6.3.
Can we exclude the existence of a braided categorification for every q non prime-power? Appendix
Etingof ’s moderation.
The previous sections shows that the Grothendieck rings of the family of fusioncategories Rep(PSL(2 , q )), with q prime-power, interpolates to a family of fusion rings for every integer q ≥ q for which the corresponding interpolated fusion ring admits a complex categorification; because, as we will see in this section, the family of fusion categories Rep( F q ⋊ F ⋆q ), with q prime-power, interpolates to a family of fusion rings for every integer q ≥
2, which also pass all the known criteria, butaccording to [9, Corollary 7.4], the ones which admit a complex categorification are exactly those with q prime-power.The Grothendieck ring of Rep( F q ⋊ F ⋆q ) is of rank q , FPdim q ( q − , q − , [ q − , x ,c x ,c = x ,c + c mod q − ,x ,c x q − , = x q − , x ,c = x q − , ,x q − , x q − , = P c x ,c + ( q − x q − , . Then, the generic character table of F q ⋊ F ⋆q is ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤ charparam c classparam k { } { , . . . , q − } { }{ , . . . , q − } ζ kcq − { } q − − q q − q prime-power interpolates into a family of fusion rings for all integer q ≥
2, with eigentable as above. Let see why they all pass every categorification criterion mentioned in Section 5: • Schur product criterion: if ( j , j , j ) from Theorem 5.1 involves the middle column of the eigentable thenthe presence of 0 reduces the checking to the one for the cyclic group C q − , which is true. It remains onlyfour cases, whose computation give q ( q − , , q, q ( q − q − respectively, and all of them are non-negative. • Ostrik criterion: 2 P j c j = 2( q ( q − + q − q − + q ) = q − q − q +2) q ( q − ≤ < c . • Drinfeld center criterion: ( c / c j ) are the class sizes mentioned in the eigentable, they are all integers. • Conjectural Isaacs criterion: observe that c / c j λ i, is always an integer when λ i,j = 0. • Extended cyclotomic criterion: all the entries of the eigentable are cyclotomic integers. • Zero spectrum criterion: see Lemma 7.1. • One spectrum criterion: see Lemma 7.2.
Lemma 7.1.
All the fusion rings of above family pass the zero spectrum criterion.Proof.
Consider i , . . . , i of Theorem 5.12. If none of them is given by x q − , then it reduces to the cyclic group C q − . Else, one of them is given by x q − , , and then it is easy to see by the fusion rules together with (1), that atleast two among i , . . . , i are given by x q − , : • if there are exactly two, and if they are in the (bottom) indices of the same coefficient in (2), say i and i ,then i j , with j = 5 , , ,
9, are given by some x ,c j , with c − c c − c mod q −
1, contradiction with c + c ≡ c + c given by n i i ,i , n i i ,i = 0 at (1); else they are in the indices of different coefficients in (2),say i and i , and then i must be given by x q − , because n i i ,i = 0, contradiction with n i i ,i = 0. • if there are exactly three, and if they are in the indices of three different coefficients in (2) then this sumcannot be zero, so they must be in the indices of exactly two different coefficients in (2), say i , i , i , thenwe also get a contradiction using n i i ,i , n i i ,i = 0. • if there are strictly more than three, then (as above), they must be in the indices of exactly two differentcoefficients in (2), which also prevents the sum to be zero. (cid:3) Lemma 7.2.
All the fusion rings of above family pass the one spectrum criterion.Proof.
As for the proof of Lemma 7.1, we can assume that at least two among i , . . . , i are given by x q − , . If i is given by x q − , then the indices of each coefficent of (7) must involve at least one x q − , : • if it is exactly one each, say i , i , i , then by (6) and using the notations of the proof of Lemma 7.1, c ≡ c + c , c ≡ c − c and c ≡ c + c , so that c + c ≡ c mod q −
1, contradiction with (9). • if there is one coefficient for two ones, and the two others with one, say i , i , i , i , then by n i i ,i , n i i ,i = 0in (6), i and i must be given by x q − , too, contradiction with (9). • if there is two coefficients for two ones, and the other with one, say i , i , i , i , i , then by n i i ,i , n i i ,i = 0in (6), i and i must be given by x q − , too, contradiction with (9). • else it is exactly two each, the sum in (7) cannot be equal to one.Then i is not given by x q − , , and so the bottom indices of each coefficient in (7) must be both x q − , or both not: • if it is “both not” for all, we reduce to the cyclic group C q − , NTERPOLATED FAMILY OF NON GROUP-LIKE SIMPLE INTEGRAL FUSION RINGS OF LIE TYPE 25 • if it is “both” for exactly one, say i and i , then by n i i ,i , n i i ,i = 0 in (6), i and i must be given by x q − , too, contradiction with (9), • if it is “both” for exactly two, say i , i , i , i , then by n i i ,i , n i i ,i = 0 in (6), i and i must be given by x q − , too, contradiction with (9), • else it is “both” for all, then the sum in (7) cannot be equal to one. (cid:3) L¨ubeck’s trick.
Here is a trick (given by Frank L¨ubeck) to display the character table directly on [GAP] for q not necessarily a prime-power, by removing the restriction to prime-power as follows: gap> gt02 := CharacterTableFromLibrary("SL2even");;gap> gt02.domain := function(q) return q mod 2 = 0; end;;gap> gt1 := CharacterTableFromLibrary("PSL2even");;gap> gt1.domain := function(q) return q mod 4 = 1; end;;gap> gt3 := CharacterTableFromLibrary("PSL2odd");;gap> gt3.domain := function(q) return q mod 4 = 3; end;; Then we can display as usual for every integer q using the following command: gap> Display(CharacterTable("PSL",2,q)); In addition, we can get the fusion ring for every given integer q by applying the Schur orthogonality relationswith the following function: gap> FusionPSL2:=function(q)> Ch:=CharacterTableWithSortedCharacters(CharacterTable("PSL",2,q));;> irr:=Irr(Ch);; n:=Size(irr);;> return List([1..n],i->List([1..n],j->List([1..n],k->ScalarProduct(irr[i]*irr[j],irr[k]))));> end;; Remark 7.3.
Theorems 4.5, 4.19 and 4.41 provide the generic fusion rules, whereas above function compute themfor every given integer. We used it to double-checked (for q small) that these theorems contain no mistake. References [1]
D. Bisch , A note on intermediate subfactors . Pacific J. Math. 163 (1994), no. 2, 201–216.[2]
S. Burciu , Kernels of representations and coideal subalgebras of Hopf algebras . Glasg. Math. J. 54 (2012), no. 1, 107–119.[3]
M.C. David , Paragroupe d’Adrian Ocneanu et alg`ebre de Kac , Pacific J. Math. 172 (1996), no. 2, 331–363 (French)[4]
P. Das and V. Kodiyalam , Planar algebras and the Ocneanu-Szymanski theorem , Proc. Amer. Math. Soc. 133 (2005), no. 9,2751–2759[5]
P. Deligne , Cat´egories tannakiennes , (French) [Tannakian categories] The Grothendieck Festschrift, Vol. II, 111–195, Progr. Math.,87, (1990).[6]
F. Digne, J. Michel , Representations of finite groups of Lie type , London Mathematical Society Student Texts, 95. CambridgeUniversity Press, Cambridge, 2020. viii+258 pp.[7]
P. Etingof , private communication (2020).[8]
P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik , Tensor Categories , American Mathematical Society, (2015). MathematicalSurveys and Monographs Volume 205.[9]
P. Etingof, S. Gelaki, and V. Ostrik , Classification of fusion categories of dimension pq . Int. Math. Res. Not. (2004), no. 57,3041–3056.[10]
P. Etingof, D. Nikshych, and V. Ostrik , On fusion categories , Annals of Mathematics, 162, pp. 581–642 (2005).[11]
P. Etingof, D. Nikshych, and V. Ostrik , Weakly group-theoretical and solvable fusion categories , Adv. Math., 226 (2011),pp. 176–205.[12]
D.E. Evans, Y. Kawahigashi , Quantum symmetries on operator algebras . Oxford Mathematical Monographs (1998) xvi+829 pp.[13]
W. Fulton, J. Harris , Representation theory. A first course , Graduate Texts in Mathematics, 129. Readings in Mathematics.Springer-Verlag, New York, 1991. xvi+551 pp.[GAP] The GAP Group,
GAP – Groups, Algorithms, and Programming, Version 4.11.0 ; 2020, .[14]
I.M. Isaacs , Character theory of finite groups , Corrected reprint of the 1976 original [Academic Press], AMS Chelsea Publishing,xii+310 (2006).[15]
M. Izumi, H. Kosaki , Kac algebras arising from composition of subfactors: general theory and classification . Mem. Amer. Math.Soc. 158 (2002), no. 750, 198 pp.[16]
M. Izumi, R. Longo, S. Popa , A Galois correspondence for compact groups of automorphisms of von Neumann algebras with ageneralization to Kac algebras.
J. Funct. Anal. 155 (1998), no. 1, 25–63.[17]
V. Jones, V.S. Sunder , Introduction to subfactors . London Mathematical Society Lecture Note Series, 234. Cambridge UniversityPress, Cambridge, 1997. xii+162 pp.[18]
V.F.R. Jones , Planar algebras, I , (1999),.[19]
V. Jones , On the origin and development of subfactors and quantum topology , Bull. Amer. Math. Soc. (N.S.) 46 (2009), no. 2,309–326.[20]
I. Kaplansky , Bialgebras . Lecture Notes in Mathematics. Department of Mathematics, University of Chicago (1975) iv+57 pp.[21]
Z. Liu
Exchange relation planar algebras of small rank , Trans. Amer. Math. Soc., 368 (2016), pp. 8303–8348. [22]
Z. Liu, S. Palcoux and Y. Ren , Triangular Prism Equations and Categorification , in preparation.[23]
Z. Liu, S. Palcoux and Y. Ren , Classification of Grothendieck rings of complex fusion categories of multiplicity one up to ranksix , arXiv:2010.10264, (2020).[24]
Z. Liu, S. Palcoux and J. Wu , Fusion Bialgebras and Fourier Analysis , arXiv: 1910.12059, (2020).[25]
R. Longo , A duality for Hopf algebras and for subfactors. I , Comm. Math. Phys. 159 (1994), no. 1, 133–150.[26]
F. L¨ubeck , Character Degrees and their Multiplicities for some Groups of Lie Type of Rank < , [27] G. Lusztig , Leading coefficients of character values of Hecke algebras , Proc. Symp. in Pure Math., 47, pp. 235–262 (1987).[28]
G. Lusztig , Hecke algebras with unequal parameters . CRM Monograph Series, 18 (2003) vi+136 pp.[29]
D. Nikshych , Morita equivalence methods in classification of fusion categories , Hopf algebras and tensor categories, 289–325,Contemp. Math., 585, Amer. Math. Soc. (2013).[30]
V. Ostrik , On formal codegrees of fusion categories . Math. Res. Lett. 16 (2009), no. 5, 895–901.[31]
V. Ostrik , Pivotal fusion categories of rank 3 , Mosc. Math. J., 15, pp. 373–396, 405 (2015).[32]
S. Palcoux , Ore’s theorem for cyclic subfactor planar algebras and beyond , Pacific J. Math. 292-1 (2018)[33]
S. Palcoux , Euler totient of subfactor planar algebras . Proc. Amer. Math. Soc. 146 (2018), no. 11, 4775–4786[34]
S. Palcoux , Ore’s theorem on subfactor planar algebras , Quantum Topology, 11 (2020), no. 3, 525–543[35]
S. Palcoux , Characterization of the family of simple groups PSL(2,q) by tensor multiplicity , https://mathoverflow.net/q/382380 .[36] W. Szymanski , Finite index subfactors and Hopf algebra crossed products , Proc. Amer. Math. Soc. 120 (1994), no. 2, 519–528.
Z. Liu, Yau Mathematical Sciences Center, Department of Mathematics, Tsinghua University, and Beijing Institute ofMathematical Sciences and Applications, Beijing, 100084, China
Email address : [email protected] S. Palcoux, Beijing Institute of Mathematical Sciences and Applications, Huairou District, Beijing, China
Email address : [email protected] URL : https://sites.google.com/view/sebastienpalcoux Y. Ren, Department of Physics, Harvard University, Cambridge, 02138, USA
Email address ::