ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.369865 J. Algebra Comb. Discrete Appl. 5(2) • 51–63 Received: 6 May 2017 Accepted: 27 October 2017 Journal of Algebra Combinatorics Discrete Structures and Applications Fourier matrices of small rank Research Article Gurmail Singh Abstract: Modular data is an important topic of study in rational conformal field theory. Cuntz, using a computer, classified the Fourier matrices associated to modular data with rational entries up to rank 12, see [3]. Here we use the properties of C-algebras arising from Fourier matrices to classify complex Fourier matrices under certain conditions up to rank 5. Also, we establish some results that are helpful in recognizing C-algebras that not arising from Fourier matrices by just looking at the first row of their character tables. 2010 MSC: 05E30, 05E99, 81R05 Keywords: Fourier matrices, Modular data, Fusion rings, C-algebras 1. Introduction Fourier matrices are a fundamental ingredient of modular data. Modular data is a basic component of rational conformal field theory, see [5]. Further, rational conformal field theory has important applications in physics, see [4] and [8]. In particular, it has nice applications to string theory, statistical mechanics, and condensed matter physics, see [10] and [13]. Modular data give rise to fusion rings, C-algebras and C∗-algebras, see [3] and [11]. These rings and algebras are interesting topics of study in their own right. A unitary and symmetric matrix whose first column has positive real entries is called a Fourier matrix if its columns under entrywise multiplication produce integral structure constants. The set of columns of a Fourier matrix under entrywise multiplication and usual addition generate a fusion algebra, see [3]. But a two-step rescaling on Fourier matrices gives rise to self-dual C-algebras. Cuntz, using a computer, classified the Fourier matrices with rational entries up to rank 12, see [3]. But rational Fourier matrices do not include some other important matrices, see sections 4, 5 and 6. Here we use C-algebra perspective to classify the complex Fourier matrices up to rank 5 under certain conditions. Also, we establish some results that are helpful in recognizing the C-algebras that are not arising from Fourier matrices by mere looking at the first row of their character tables. Gurmail Singh; Department of Mathematics and Statistics, University of Regina, Canada, S4S 0A2 (email: gurmail.singh@uregina.ca). 51 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 In Section 2, we collect the definitions and introduce a two-step rescaling of Fourier matrices. In Section 3, we summarize the results that are useful to recognize the C-algebras that are not arising from Fourier matrices. In Section 4, we classify Fourier matrices of rank 2 and 3. In sections 5 and 6, we classify non-homogeneous Fourier matrices of rank 4 and rank 5, respectively. For the classification of homogenous Fourier matrices see [11, Theorem 13]. 2. C-algebras arising from Fourier matrices A scaling of the rows of a Fourier matrix gives the basis of a Fusion algebra that contains the identity element. But a two-step rescaling of a Fourier matrix gives the standard basis of C-algebra. Definition 2.1. Let A be a finite dimensional and commutative algebra over C with distinguished basis B = {b0 := 1A,b1, . . . ,br−1}, and an R-linear and C-conjugate linear involution ∗ : A → A. Let δ : A → C be an algebra homomorphism. Then the triple (A,B,δ) is called a C-algebra if it satisfies the following properties: 1. for all bi ∈ B, (bi)∗ = bi∗ ∈ B, 2. for all bi,bj ∈ B, we have bibj = ∑ bk∈B λijkbk, for some λijk ∈ R, 3. for all bi,bj ∈ B,λij0 6= 0 ⇐⇒ j = i∗, 4. for all bi ∈ B,λii∗0 = λi∗i0 > 0. 5. for all bi ∈ B, δ(bi) = δ(bi∗ ) > 0. The algebra homomorphism δ is called a degree map, and the values δ(bi), for all bi ∈ B, are called the degrees of A. For i 6= 0, δ(bi) is called a nontrivial degree. If δ(bi) = λii∗0, for all bi ∈ B, we say that B is a standard basis. The order of a C-algebra is denfined as δ(B+) := ∑r−1 i=0 δ(bi). A C-algebra is called symmetric if bi∗ = bi, for all i. A C-algebra with rational structure constants is called a rational C-algebra. The readers interested in C-algebras are directed to [1], [2] and [7]. To keep the generality, in the following definition of a Fourier matrix we assume the structure constants to be integers instead of nonnegative integers, see [3, Definition 2.2]. Definition 2.2. Let r ∈ Z+ and I an r ×r identity matrix. Then S is called a Fourier matrix if 1. S is a unitary and symmetric matrix, that is, SS̄T = I,S = ST , 2. Si0 > 0, for 0 ≤ i ≤ r − 1, where S is indexed by {0, 1, 2, . . . ,r − 1}, 3. Nijk = ∑ l SliSljS̄lkS −1 l0 ∈ Z, for all 0 ≤ i,j,k ≤ r − 1. Let S be a Fourier matrix. Let s = [sij] be a matrix with entries sij = Sij/Si0, for all i,j, we call it an s-matrix associated to S (briefly, s-matrix). Since S is a unitary matrix, ss̄T =diag(d0,d1, . . . ,dr−1) is a diagonal matrix, where di = ∑ j sijs̄ij. The numbers di are called norms of s-matrix. The relation sij = Sij/Si0 implies the structure constants Nijk = ∑ l slisljslkd −1 l , for all i,j,k. Since the structure constants Nijk are integers, the numbers sij are algebraic integers, see [3, Section 3]. Therefore, if S has only rational entries then entries of s-matrix are rational integers, and such s-matrices are known as integral Fourier matrices, see [3, Definition 3.1]. Cuntz classified the integral Fourier matrices up to rank 12 by using a computer, see [3]. In this paper, we consider the broader class of s-matrices that have algebraic integer entries. There is an interesting row-and-column operation (two-step rescaling) procedure that can be applied to a Fourier matrix S that results in the first eigenmatrix, the character table, of a self-dual C-algebra. The steps of the procedure are reversed to obtain the Fourier matrix S from the first eigenmatrix. The 52 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 explanation of the procedure is as follows. Let S = [Sij] be a Fourier matrix indexed with {0, 1, . . . ,r−1}. We divide each row of S with its first entry and obtain the s-matrix. The multiplication of each column of the s-matrix with its first entry gives the P-matrix associated to S (briefly, P-matrix), the first eigenmatrix of a self-dual C-algebra. That is, sij = SijS −1 i0 and pij = sijs0j, for all i,j, where pij denotes the (i,j)-entry of the P-matrix. Conversely, to obtain the s-matrix from a P-matrix, divide each column of the P-matrix with the squareroot of its first entry. Further, the Fourier matrix S is obtained from the s-matrix by dividing the ith row of s-matrix by √ di, where di = ∑ j |sij| 2. That is, sij = pij/ √ p0j, and Sij = sij/ √ di, for all i,j. Since the entries of an s-matrix are algebraic integers, the entries of a P-matrix are also algebraic integers. Remark 2.3. Throughout this paper, unless mentioned explicitly, the sets of columns of a P-matrix and an s-matrix are denoted by B = {b0,b1, . . . ,br−1} and B̃ = {b̃0, b̃1, . . . , b̃r−1}, respectively. The structure constants generated by the columns, with entrywise multiplication, of a P-matrix and an s-matrix are denoted by λijk and Nijk, respectively. MT denotes the transpose of a matrix M. Let S be a Fourier matrix and A := CB, a C-span of B. Define a map ∗ : A −→ A by ( ∑ j ajbj) ∗ =∑ j ājbj∗ = ∑ j āj[p̄0j, p̄1j, . . . , p̄r−1,j] T . This map ∗ is an involution on A, and the map δ : A −→ C defined as δ( ∑ j ajb̃j) = ∑ j ājs0j, that is, δ(bi) = δ(s0ib̃i) = s 2 0i for all i, is a positive degree map of A. Since bi = s0ib̃i, the structure constants generated by the basis B are given by λijk = Nijks0is0js −1 0k , for all i,j,k. S is a unitary matrix, therefore, Nij0 = ∑ l SliSljS̄l0S −1 l0 6= 0 ⇐⇒ j = i ∗ and Nii∗0 = 1 > 0, for all i,j. Thus λij0 6= 0 ⇐⇒ j = i∗ and λii∗0 > 0, for all i,j. Therefore, the vector space A := CB is a C-algebra of order d0, B is the standard basis of A, and P-matrix is the first eigenmatrix of A, see [11, Theorem 4], and we say (A,B,δ) is a C-algebra arising from a Fourier matrix S. Note that, entries of the first eigenmatrix P are the entries of the character table A. Thus at some places we consider the P-matrix of A as the character table of A and the ith row of P-matrix as the ith irreducible character of A. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S. Since S is a symmetric matrix, A is a self-dual C-algebra and d0 = djδ(bj), for all j. The entries of an s-matrix and the associated P-matrix are algebraic integers. Therefore, if A has rational degrees then both the degrees and norms are rational integers and both divide the order of A, see [11, Proposition 5]. Note that, a C-algebra arising from a Fourier matrix S is a symmetric C-algebra if and only if S is a real matrix. A C-algebra that has at least two different nontrivial degrees is called a non-homogeneous C-algebra and we call the associated Fourier matrix (s-matrix) a non-homogeneous Fourier matrix (non-homogeneous s-matrix, respectively). Every self-dual C-algebra not necessarily have rational degrees. For example, a self-dual C-algebra of rank 2 with basis {1,x}, and the structure constants given by the equation x2 = 1 + x does not have rational degrees. We remark that this C-algebra does not arise from a Fourier matrix. But a rational C-algebra arising from a Fourier matrix has integral degrees, see [11, Proposition 5]. Lemma 2.4. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S of rank r. 1. If A has rational order then the order of A is an integer. 2. If A has nonnegative structure constants then degrees of A are greater or equal to 1. 3. If A has rational order and all the degrees of A different from 1 are all equal then the degrees of A are integers. (Note: the algebra A need not be homogeneous.) Proof. (i). Since degrees of A are algebraic integers and order of A is the sum of degrees of A, the order of A is a rational integer. (ii). For any i, δ(bi) is the first entry of column vector (b̃i)2 and Nii0 = 1. Note that, Nijk ≥ 0 for all i,j,k, because A has nonnegative structure constants. Therefore, δ(bi) = 1 + m, where m is a nonnegative algebraic integer. (iii). Let all the degrees of A different from 1 be a positive real number t. Therefore, d0 = m + nt, where m is the number of degrees equal to 1 and n is the number of degrees equal to t. Since the order d0 is an integer, t is an integer. 53 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 3. Recognition of C-algebras arising from Fourier matrices The following results are useful for recognizing C-algebras that are not arising from Fourier matrices by mere looking at the degrees of C-algebras, that is, the first row of the character tables. All the character tables of the association schemes used here are produced by Hanaki and Miyamoto, see [6]. Lemma 3.1. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S with nonnegative structure constants. Let L(B) = {b ∈ B : δ(b) = 1}. Then L(B) is an abelian group. Proof. Since δ(b0) = 1, b0 ∈ L. Let bi,bj ∈ L(B). Therefore, bi = b̃i and bj = b̃j. Thus bibj = b̃ib̃j =∑ k Nijkb̃k implies 1 = δ(bibj) = ∑ k Nijkδ(b̃k). By Lemma 2.4, δ(bl) ≥ 1, thus δ(b̃l) = √ δ(bl) ≥ 1, for all l. Therefore, bibj = b̃k, for some bk ∈ B. Also, for all bi ∈ L(B), bibi∗ = b0 + ∑ j λii∗jbj and λii∗j ≥ 0 imply bibi∗ = b0, that is, b −1 i = bi∗ ∈ L(B). Hence L(B) is an abelian group. Proposition 3.2. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S with nonnegative structure constants. Let S be a real Fourier matrix. 1. Let the order of A be a rational number. If all the degrees of A different from 1 are equal to t then t might be a power of 2. 2. If rank of A is an even integer then A cannot have only one degree different from 1. 3. Let the order of A be a rational number. If the rank of A is greater than 3 then A cannot have only one degree greater or equal to r and all other degrees equal to 1. Proof. (i). Since S is a real Fourier matrix, by Lemma 3.1, the elements of B with degree 1 form an elementary abelian group. Thus the number of elements of B with degree 1 is a power of 2. By Lemma 2.4, t is an integer. The result follows from the fact that t divides the order of the algebra, see [11, Proposition 5 (ii)]. (ii). If rank of A is 2 then both degrees are equal to 1, see Section 4. By Lemma 3.1, the set of elements of B with degree 1 form an elementary abelian group. Therefore, the order of the group is a power of 2, say 2m, where m is a nonnegative integer. Since A has only one degree different from 1, 2m = r − 1, a contradiction to the fact that r is an even integer. (iii). Suppose A has only one degree k that is greater or equal to r and the remaining degrees are equal to 1. Without loss of generality, let the first row of the character table be [1, 1, . . . , 1,k], where k ≥ r. Since δ(b0) = . . . = δ(br−2), d0 = . . . = dr−2. The structure constants are nonnegative, therefore, |pij| ≤ p0j for all i,j, see [12, Proposition 4.1]. Therefore, the only possible entries of row 2, . . ., row r − 1 of P-matrix are [1, 1, . . . , 1,−k], which is not possible as P nonsingular. The adjacency algebras of the association schemes as12(9), as14(4), as16(10), as16(20), as16(21) and as16(62) have the degree patterns that violate the above result, see [6]. Therefore, they are not arising from Fourier matrices. The character table of the association scheme as4(2) [6] is an example where part (iii) of the above proposition fails for the rank 3. The above proposition also helps to sieve out a lot of C-algebras even if their self-duality is not known. The next proposition helps to recognize the C-algebras not arising from Fourier matrices. Proposition 3.3. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S of rational order. Let the number of i’s such that δ(bi) = 1 be t and the remaining r − t degrees are equal to k. Then the possible values of k are the divisors of t. Proof. By Lemma 2.4, the order and degrees of A are integers. Since entries of s-matrix are algebraic integers, all the norms are also integer, that is, d0k−1 ∈ Z. Therefore, d0 = t+ (d0−t)k implies tk−1 ∈ Z. Hence k is in the subset of the divisors of t. 54 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 The above proposition illustrates that the adjacency algebras of the association schemes as9(3), as9(8), as10(6), as16(20), as16(21) and as16(62) [6] are not arising from Fourier matrices. In the next proposition we examine the possible number of occurrences of a degree if it is one of the degrees and satisfy a certain criteria. Proposition 3.4. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S with integral degrees. 1. If for a given j, δ(bj) (6= 1) is a smallest nontrivial degree that divides all the nontrivial degrees δ(bl) (6= 1) then the number of degrees equal to 1 is a multiple of δ(bj). 2. Let δ(bt) be a degree divisible by all the smaller degrees and divides all the bigger degrees. Let δ(bs) be the largest degree among all the degrees strictly less than δ(bt). Let the sum of the degrees less than δ(bs) be β1δ(bs) and the number of degrees equal to δ(bs) be β2. Then β1δ(bs) + β2δ(bs) is divisible by δ(bs). (Note: β1δ(bs) is not equal to zero only if δ(bs) > 1.) 3. Suppose A has nonnegative structure constants. If for all i < j, δ(bi) divides δ(bj) then A has integral structure constants. Proof. (i). Since degrees of A are integers, the norms are integers. Thus, d0δ(bj)−1 ∈ Z. Therefore, ( 1 + r−1∑ i=1 δ(bi) ) δ(bj) −1 = ( ∑ δ(bi)=1 δ(bi) + ∑ δ(bi)≥δ(bj) δ(bi) ) δ(bj) −1 = ( ∑ δ(bi)=1 δ(bi) ) δ(bj) −1 + α ∈ Z, where α ∈ Z. Hence the number of degrees equal to 1 are multiple of δ(bj). (ii). The degrees and norms of A are integers. Therefore, d0δ(bt) −1 = (r−1∑ i=0 δ(bi) ) δ(bt) −1 = ( ∑ δ(bi)<δ(bs) δ(bi) + ∑ δ(bi)=δ(bs) δ(bi) + ∑ δ(bi)>δ(bs) δ(bi) ) δ(bt) −1 = ( β1δ(bs) + ∑ δ(bi)=δ(bs) δ(bi) ) δ(bt) −1 + γ ∈ Z, where γ ∈ Z. Thus β1δ(bs) + β2δ(bs) is divisible by δ(bt). (iii). Since λijk are nonnegative, Nijk are nonnegative, because λijk = Nijks0is0js −1 0k , for all i,j,k. Let b̃ib̃j = ∑ k Nijkb̃k. On comparing the first entry of both sides, we conclude that b̃k cannot occur with nonzero coefficient whenever s0k > s0is0j, that is, √ δ(bk) > √ δ(bi)δ(bj) implies Nijk = 0. Hence the assertion follows from the relation between λijk and Nijk. For example, the adjacency algebras of association schemes as7(2), as8(5), as8(6), as9(8), as9(9) and adjacency algebras of homogenous schemes have the degree patterns that violate the above proposition, see [6]. Therefore, they are not arising from Fourier matrices. Lemma 3.5. Let (A,B,δ) be a C-algebra arising from a rational Fourier matrix S of odd rank and odd order. Let the odd degree among all the degrees of A be maximum. Then the rank of A must be at least 11. Proof. Let d0 = δ(bi)ai. By [3, Lemma 3.7], d0 is an odd square, thus ai is a square. Let δ(b1) be an odd integer and δ(b1) ≥ δ(bi) for each i. Therefore, d0 ≥ 9δ(b1), and d0 ≤ 1 + (r − 1)δ(b1). Thus 9δ(b1) ≤ 1 + (r − 1)δ(b1) implies δ(b1)(9 − (r − 1)) ≤ 1 implies δ(b1) ∈ Z+ only if r − 1 ≥ 9. 55 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 4. Fourier matrices of rank 2 and 3 In this section we classify Fourier matrices of rank 2 and 3. In fact, we find the P-matrices of C-algebras arising from Fourier matrices of rank 2 and 3. But the associated Fourier matrix S can be recovered easily from the P-matrix as described in Section 2. Since the row sum of a character table is zero, the character table of a C-algebra of rank 2 with standard basis B = {b0,bi} is given by P = [ 1 n 1 −1 ] , and the structure constants are given by b21 = nb0 + (n − 1)b1. Therefore, the structure constant N111 is integer only for n = 1, and the associated Fourier matrix S = 1 √ 2 [ 1 n 1 −1 ] . Let P be the character table for a symmetric C-algebra arising from a Fourier matrix of rank 3 with standard basis B = {b0,b1,b2}. Let b1b2 = ub1 + vb2. Then P =   1 k1 k21 φ1 φ2 1 ψ1 ψ2   , where φ1 = (v −u− 1 + √ (u−v − 1)2 + 4u)/2, φ2 = (u−v − 1 − √ (u−v − 1)2 + 4u)/2, and ψ1 = (v −u− 1 − √ (u−v − 1)2 + 4u)/2, ψ2 = (u−v − 1 + √ (u−v − 1)2 + 4u)/2. Therefore, d0 = 1 + k1 + k2, d1 = 1 + |φ1|2 k1 + |φ2|2 k2 , d2 = 1 + |ψ1|2 k1 + |ψ2|2 k2 . Lemma 4.1. There is no symmetric homogenous C-algebra of rank 3 arising from a Fourier matrix S. Proof. Suppose (A,B,δ) is a symmetric homogenous C-algebra arising from a Fourier matrix. Since ST = S, φ2k = ψ1l implies u = v. The structure constant N210 = 0 implies k = 2u. Thus φ1 = (−1 + √ 1 + 2k)/2 and φ2 = (−1 − √ 1 + 2k)/2. A homogenous C-algebra arising from a Fourier matrix has all degrees equal to 1, see [11, Proposition 12]. But the structure constants N112 = 1 (2k + 1) √ k [k2 + k 2 ] and N222 = 1 (2k + 1) √ k [k2 − 1 − 3 2 k] are not integers for k = 1, a contradiction. Theorem 4.2. Let (A,B,δ) be a symmetric non-homogeneous C-algebra of rank 3 arising from a Fourier matrix S with integral degrees. Then the corresponding matrices P, s and S are as follows. P =   1 1 21 1 −2 1 −1 0   , s =   1 1 √ 2 1 1 − √ 2 1 −1 0   and S =   1/2 1/2 1/ √ 2 1/2 1/2 −1/ √ 2 1/ √ 2 −1/ √ 2 0   . Proof. Let δ(bi) = ki, for all i. Since the integral degrees of A divide the order, k1 divides 1 + k2, and k2 divides 1 + k1. Therefore, the only possible degree pattern of A are [1, 1, 2] and [1, 2, 3], up to the permutations. Since N012 = 0, 1− v k1 − u k2 = 0. Therefore, the degree patterns [1, 1, 2] and [1, 2, 3] imply v = 1 − u 2 and v = 2 − 2u 3 , respectively. Case 1. Let the degree pattern be [1, 1, 2], that is, k1 = 1 and k2 = 2. Therefore, v = 1 − u 2 . Since N011 = 1, u3(u− 1) = 0. Hence (u,v) = (0, 1), or (u,v) = (1, 1/2). Subcase 1. Let (u,v) = (0, 1). 56 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Therefore, P =   1 1 21 1 −2 1 −1 0   , s =   1 1 √ 2 1 1 − √ 2 1 −1 0   and S =   1/2 1/2 1/ √ 2 1/2 1/2 −1/ √ 2 1/ √ 2 −1/ √ 2 0   . Subcase 2. Let (u,v) = (1, 1 2 ). Therefore, P =   1 1 2 1 −3 + √ 17 4 −1 − √ 17 4 1 −3 − √ 17 4 −1 + √ 17 4   . Note that δ(b1) = δ(b0), but d1 6= d0, a contradiction. Hence u = 1 and v = 1 2 is not a possible case. Case 2. Let the degree pattern be [1, 2, 3], that is, k1 = 2 and k2 = 3. Therefore, v = 2 − 2u 3 . Since N011 = 1, 625u4 − 1850u3 + 2520u2 − 1296u + 243 = 0. But it has no real roots, see [9]. Thus we rule out [1, 2, 3] degree pattern, because an s-matrix associated with a symmetric C-algebra might be a real matrix. Remark 4.3. The above P-matrix is given by the character table of the adjacency algebra of an asso- ciation scheme as4(2), see [6]. In the next theorem we prove that there is only one asymmetric C-algebra of rank 3 arising from a Fourier matrix. Moreover, the following theorem shows that for rank 3 it is not necessary to assume |sij| ≤ s0j to prove that the homogeneous C-algebra arising from a Fourier matrix is a group algebra, see [11, Theorem 13]. Theorem 4.4. Let (A,B,δ) be an asymmetric C-algebra arising from a Fourier matrix S of rank 3. Then the P-matrix is the first eigenmatrix of the group algebra of a group of order 3. Proof. The P-matrix of an asymmetric C-algebra of rank 3 is as follows, P =   1 k k1 α ᾱ 1 ᾱ α   , where α = (−1 + i √ 1 + 2k)/2. Since A is homogenous, k = 1, see [11, Proposition 12]. Therefore, P(= s) =   1 1 11 ζ3 ζ23 1 ζ23 ζ3   and S = 1√ 3   1 1 11 ζ3 ζ23 1 ζ23 ζ3   . 5. Fourier matrices of rank 4 In this section we classify the Fourier matrices under certain conditions, and we show that there is no non-homogeneous integral Fourier matrix of rank 4. For homogenous Fourier matrices see [11]. 57 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Lemma 5.1. Let (A,B,δ) be a C-algebra arising from a Fourier matrix S of rank r. Let |sij| ≤ s0j, for all j. Let δ(bj) = kj for all j, and ki = 1 for some i > 0. 1. Then |pij| = kj for all j. 2. If s is a real matrix then pij = ±kj for all j. Proof. (i). Since δ(bi) = 1, di = d0. Therefore, the row 1 and row i of the s-matrix are [1, 1, √ k2, . . . ,√ kr−1] and [1,pi1,pi2/ √ k2, . . . ,pi,r−1/ √ kr−1], respectively. Since d0 = di, |pij|/ √ kj = √ kj, for all j. Hence |pij| = kj, for all j. (ii). By Part (i), |pij| = kj for all j. Since s is a real matrix, pij = ±kj for all j. The next proposition classify the non-homogeneous Fourier matrices of rank 4 with one nontrivial degree equal to 1. Proposition 5.2. Let (A,B,δ) be a non-homogeneous rational C-algebra arising from a Fourier matrix S of rank 4. Let |sij| ≤ s0j, for all j. Let δ(bj) = kj, for all j, and ki = 1, for some i > 0. Then the associated P-matrix is   1 1 4 6 1 1 4 −6 1 1 −2 0 1 −1 0 0   . Proof. Since A is a rational C-algebra, the degrees of A are integers, see [11, Proposition 5]. Let δ(bi) = ki, for all i. Without loss of generality, suppose k1 = 1. Therefore, we have P =   1 1 k2 k3 1 p11 p12 p13 1 p21 p22 p23 1 p31 p32 p33   . Case 1. If p11,p12 and p13 are not rational integers. Since d0 = d1, |p11| = 1, |p12| = k2 and |p13| = k3. Thus p11,p12 and p13 cannot be irrational real numbers. Therefore, they can be non-real algebraic integers. Since the structure constants are rational numbers, a complex conjugate of an irreducible character of A is an irreducible character of A. Without loss of generality, assume that the third irreducible character is a complex conjugate of the second character. Thus k2 = 1. But A is non-homogenous and d0 = diδ(bi), therefore, k3 = 3 and d3 = 2. Since S is a symmetric matrix, |p31| = |p32| = 1, thus d3 > 3, a contradiction. Case 2. If p11,p12 and p13 are rational integers. By Lemma 5.1, p11 = ±1,p12 = ±k2 and p13 = ±k3. Since the row sum of P-matrix is zero, the second row of P-matrix is either [1,−1,k2,−k2] or [1, 1,k2,−(k2 + 2)]. Subcase 1. Let the second row of P be [1,−1,k2,−k2]. Then, the first row of P-matrix is [1, 1,k2,k2]. Since k2 divides d0 and A is non-homogenous, k2 = 2. Thus, by the symmetry of Fourier matrix S and orthogonality of characters, we have P =   1 1 2 2 1 −1 2 −2 1 1 −1 −1 1 −1 −1 1   But the s-matrix associated to the above P-matrix does not have integral structure constants. Subcase 2. Let the second row of P be [1, 1,k2,−(k2 + 2)]. 58 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Then, the first row of P-matrix is [1, 1,k2,k2 + 2]. Since k2 divides d0 and A is non-homogenous, k2 = 2 or 4. Thus, by the symmetry of Fourier matrix S and orthogonality of characters, we have P =   1 1 2 4 1 1 2 −4 1 1 −2 0 1 −1 0 0   or P =   1 1 4 6 1 1 4 −6 1 1 −2 0 1 −1 0 0   . But the s-matrix associated to the first P-matrix does not have integral structure constants. Remark 5.3. The above P-matrix is the first eigenmatrix of the adjacency algebra of an association scheme as12(8), see [6]. Cuntz, with a computer, shows that there is no non-homogenous rational Fourier matrix of rank 4, see [3]. In the next theorem, we use C-algebra perspective to show that there is no non-homogeneous rational Fourier matrix of rank 4, that is, there is no non-homogeneous s-matrix with integral entries. Unlike the above proposition, we do not assume any nontrivial degree equal to 1. Theorem 5.4. There is no non-homogenous rational Fourier matrix S of rank 4. Proof. Let (A,B,δ) be a C-algebra arising from a rational Fourier matrix S of rank 4. Let δ(bi) = ki, for all i > 0. Since s-matrix is integral, k1,k2 and k3 are square integers, see [11, Proposition 5 (iii)]. As d0 = 1 + k1 + k2 + k3 and k1,k2 and k3 divide d0, therefore, k2 + k3 ≡−1 mod k1, k1 + k3 ≡−1 mod k2 and k1 + k2 ≡−1 mod k3. Claim: k1 = k2 = k3 = 1. Without loss of generality, suppose k1 is an even integer. Since k1,k2 and k3 are squares, k1 ≡ 0 mod 4 and d0 6≡ 0 mod 4, a contradiction to the fact that k1 divides d0. Therefore k1,k2 and k3 are odd integer. Suppose k1 ≥ k2,k3 and k1 > 1. Now, if all k1,k2 and k3 are odd integers then k1,k2,k3 ≡ 1 mod 4. But d0 ≡ 0 mod 4 implies d0 = k1a, a ≥ 4. Therefore, k1(a− 1) = 1 + k2 + k3 implies 3k1 ≤ 1 + k2 + k3. Thus k2 or k3 > k1, again a contradiction. 6. Fourier matrices of rank 5 In this section we prove that there is no non-homogenous s-matrix with integral entries (integral Fourier matrix) of rank 5. But the following proposition shows, under certain conditions, that there are three s-matrices of rank 5 with algebraic integer entries. Recall that, a Fourier matrix S with rational entries has associated integral s-matrix, and a complex Fourier matrix S has associated s-matrix with algebraic integer entries. Proposition 6.1. Let (A,B,δ) be a non-homogeneous rational C-algebra arising from a Fourier matrix of rank 5. If |sij| ≤ s0j for all j. If δ(bi) = 1 for one i > 0 and δ(bj) = kj for all j 6= i. Then up to simultaneous row and column permutations the P-matrices are as follows,  1 1 2 2 2 1 1 2 −2 −2 1 1 −2 0 0 1 −1 0 √ 2 − √ 2 1 −1 0 − √ 2 √ 2   ,   1 1 2 4 8 1 1 2 4 −8 1 1 2 −4 0 1 1 −2 0 0 1 −1 0 0 0   and   1 1 4 3 3 1 1 4 −3 −3 1 1 −2 0 0 1 −1 0 √ 3 − √ 3 1 −1 0 − √ 3 √ 3   . Proof. Since A is a rational C-algebra, the degrees of A are integers. Let δ(bi) = ki, for all i. Without loss of generality, let k1 = 1. Therefore, first eigenmatrix of A is given by P =   1 1 k2 k3 k4 1 p11 p12 p13 p14 1 p21 p22 p23 −(1 + p21 + p22 + p23) 1 p31 p32 p33 −(1 + p31 + p32 + p33) 1 p41 p42 p43 −(1 + p41 + p42 + p43)   . 59 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Case 1. If p11,p12,p13 and p14 are not all rational integers. Since d0 = d1, |p11| = 1, |p12| = k2, |p13| = k3 and |p14| = k4. Since the row sum is zero, at least two of these p11,p12,p13 and p14 can be non-real algebraic integers. Since the structure constants are rational numbers, a complex conjugate of an irreducible character is an irreducible character. Without loss of generality, we assume that the third irreducible character is a complex conjugate of the second character, thus k2 = 1. Without loss of generality, let k3 ≤ k4. Therefore, d0 = dimi implies (k3,k4) ∈ {(1, 2), (1, 4), (2, 5), (3, 6), (6, 9)}. If k3 = 1 then d3 = d0 = d1 = d2 and |sij| ≤ s0j imply all the entries of the rows 1, 2, 3 and 4 are nonzero. Since k3 6= k4, the entries of the fifth row are rational integers because Galois conjugate of an irreducible character is an irreducible character, and rows of s-matrix corresponding to the conjugate characters have equal norm. But k3 = 1 implies d4 ≤ 3. Thus there might be at least two zero entries in the fifth row of P-matrix. But S is a symmetric matrix, we get a contradiction. For (k3,k4) ∈{(2, 5), (3, 6), (6, 9)}, k3 6= k4, thus the entries of the row 5 are rational integers because k2 = k1 = k0 = 1 and the Galois conjugate of an irreducible character is an irreducible character. Each entry of the row 1, 2 and 3 of P-matrix is non-zero. But for each of the above pair there are exactly 3 zeros in the fifth row. Since S is a symmetric matrix, we get a contradiction. Case 2. If p11,p12,p13 and p14 are all rational integers. By Lemma 5.1, the only possible degree patterns are: [1,−1,k2,k3,−(k2 + k3)], [1, 1,k2,k3,−(k2 + k3 + 2)], [1, 1, k2, −k3, − (k2 −k3 + 2)], [1, − 1, k2, −k3, − (k2 −k3)], [1, 1, −k2, −k3, k2 + k3 − 2], [1, − 1, −k2, −k3, k2 + k3]. Subcase 1. Let the second row of P be [1,−1,k2,k3,−(k2 + k3)]. Then the first row of the character table is [1, 1,k2,k3,k2 + k3] and (k2 + k3) ∣∣(2 + k2 + k3). Thus (k2,k3) = (1, 1). Therefore, by the orthogonality of characters, we have P =   1 1 1 1 2 1 −1 1 1 −2 1 1 p22 p23 −1 1 1 p32 p33 −1 1 −1 p42 p43 1   . Since k2 = k3 = 1, d2 = d3 = 6. But each of |p22|, |p23|, |p32| and |p33| can be at most 1. Thus both d2 and d3 are strictly less than 6, a contradiction. Hence this case is not possible. Subcase 2. Let the second row of P be [1, 1,k2,k3,−(k2 + k3 + 2)]. Then the first row of the character table is [1, 1,k2,k3,k2 + k3 + 2]. Therefore, by the orthogonality of the irreducible characters and symmetry of the matrix S, we have P =   1 1 k2 k3 k2 + k3 + 2 1 1 k2 k3 −(k2 + k3 + 2) 1 1 p22 −2 −p22 0 1 1 p32 −2 −p32 0 1 −1 0 0 0   . Without loss of generality, let k2 ≤ k3. Since k2 ∣∣(2k3 + 4) and k3∣∣(2k2 + 4), we have (k2,k3) ∈{(1, 2), (1, 3), (1, 6), (2, 4), (2, 8), (3, 10), (4, 6), (4, 12), (6, 16), (8, 10), (12, 28)}. Note that k2 6= k3, thus d3 6= d4. The structure constants are rational. Therefore, if p22 or p33 is not rational then row 3 and 4 of P-matrix should be Galois conjugates. But the rows of s-matrix corresponding to conjugate irreducible characters should have equal norm. Thus p22 and p32 are rational integers. Therefore, det(P) ∈ Z and (detP)2 = n5. Thus n = 2(k2 + k3 + 2) need to be a square. But the only two pairs (2, 4), (4, 12) do not fail this test. For (k2,k3) = (2, 4), d2 = 8 = 1 + 1 + (p22√ 2 )2 + (−2 −p22√ 4 )2 . 60 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Since the entries of P-matrix are algebraic integers, we have p22 = 2. Similarly, d3 = 4 implies p32 = −2. Therefore, P =   1 1 2 4 8 1 1 2 4 −8 1 1 2 −4 0 1 1 −2 0 0 1 −1 0 0 0   . For (k2,k3) = (4, 12), d2 = 9 and d3 = 3. Therefore, we have p22 = 4,−5 and p32 = 1,−2. Since S is a symmetric matrix, p22 = 4 and p32 = −2 is the only possibility. But for p22 = 4 and p32 = −2, the associated s-matrix does not have integral structure constants. Subcase 3. Let the second row of P be [1, 1, k2, −k3, − (k2 −k3 + 2)]. Then, the first row of the character table is [1, 1,k2,k3,k2 −k3 + 2]. Therefore, by the orthogonality of characters, symmetry of the Fourier matrix S and PP̄ = nI, we have P =   1 1 k2 k3 k2 −k3 + 2 1 1 k2 −k3 −(k2 −k3 + 2) 1 1 −2 0 0 1 −1 0 p33 −p33 1 −1 0 p43 −p43   . Therefore, k2 ∣∣4, k3∣∣2k2 + 4, (k2 − k3 + 2)∣∣(k2 + k3 + 2) and k2 − k3 + 2 > 0. Hence (k2,k3) ∈ {(1, 1), (1, 2), (2, 2), (4, 2), (4, 3), (4, 4)}. Since |sij| ≤ s0j, (k2,k3) 6∈ {(1, 1), (1, 2)}. For (k2,k3) = (2, 2), d3 = d4 = 4. Thus p33p̄33 = 2, p43p̄43 = 2. But the integrality of the structure constants of s-matrix and orthogonality of characters forces p33 = ± √ 2 and p43 = ∓ √ 2. Therefore, up to simultaneous permutation of row 4 and row 5, and column 4 and column 5, we have P =   1 1 2 2 2 1 1 2 −2 −2 1 1 −2 0 0 1 −1 0 √ 2 − √ 2 1 −1 0 − √ 2 √ 2   . For (k2,k3) = (4, 2), d3 = 6. Thus |p33| = 4 √ 3 > 2, a contradiction. For (k2,k3) = (4, 3), k4 = 3, d3 = 4 and d4 = 4. Thus |p33| = √ 3 and |p43| = √ 3. But the integrality of structure constants and orthogonality of characters forces p33 = ± √ 3 and p43 = ∓ √ 3. Therefore, up to simultaneous permutation of row 4 and row 5, and column 4 and column 5, we have P =   1 1 4 3 3 1 1 4 −3 −3 1 1 −2 0 0 1 −1 0 √ 3 − √ 3 1 −1 0 − √ 3 √ 3   . Although the structure constants are not all integers, for example λ342 = 3/2, but the associated s-matrix has integral structure constants. Subcase 4. Let the second row of P be [1, − 1, k2, −k3, − (k2 −k3)]. Then, the first row of the character table is [1, 1, k2, k3, k2 −k3]. Therefore, by the orthogonality of the characters and symmetry of the matrix S, we have P =   1 1 k2 k3 k2 −k3 1 −1 k2 −k3 −(k2 −k3) 1 1 0 p23 −(2 + p23) 1 −1 p32 p33 −(p32 + p33) 1 −1 p42 p43 −(p42 + p43)   . 61 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Since PP̄ = nI, from row 1, 2 and column 3, we get k2 = 0, a contradiction. Subcase 5. Let the second row of P be [1, 1, −k2, −k3, k2 + k3 − 2]. Then the first row of the character table is [1, 1, k2, k3,k2 + k3 − 2]. Therefore, by the symmetry of the matrix S and orthogonality of characters, we have P =   1 1 k2 k3 k2 + k3 − 2 1 1 −k2 −k3 k2 + k3 − 2 1 −1 p22 −p22 0 1 −1 p32 −p32 0 1 1 0 0 −2   . Therefore k2 ∣∣2k3, k3∣∣2k2, (k2 + k3 − 2)∣∣(k2 + k3 + 2) and k2 + k3 − 2 > 0. Without loss of generality, let k2 ≤ k3. Hence (k2,k3) ∈ {(1, 2), (2, 2)}. Since |sij| ≤ s0j, (k2,k3) 6= (1, 2). For (k2,k3) = (2, 2), d2 = d3 = 4. Thus p22p̄22 = 2, p32p̄32 = 2. But the integrality of structure constants and orthogonality of characters forces p22 = ± √ 2 and p32 = ∓ √ 2. Therefore, up to simultaneous permutation of rows and columns, we have P =   1 1 2 2 2 1 1 −2 −2 2 1 −1 √ 2 − √ 2 0 1 −1 − √ 2 √ 2 0 1 1 0 0 −2   . Subcase 6. Let the second row of P be [1, − 1, −k2, −k3, k2 + k3]. Then the first row of the character table is [1, 1, k2, k3, k2 + k3]. Therefore, by the symmetry of the matrix S and orthogonality of the characters, we have P =   1 1 k2 k3 k2 + k3 1 −1 −k2 −k3 k2 + k3 1 −1 p22 p23 −1 1 −1 p32 p33 −1 1 1 p42 p43 0   . Similar to the Subcase 1, k2 = k3 = 1 implies d2 = d3 = 6, and we get a contradiction. Remark 6.2. We note that the associated s-matrices to the P-matrices in the above proposition are not integral Fourier matrices. The first two matrices are the character tables of as08(10), as16(24), respectively, see [6]. The third matrix is not a first eigenmatrix of an adjacency algebra of any association scheme because the structure constants generated by the columns of P-matrix are not all nonnegative integers. In the next theorem, by using the properties of C-algebras, we show that there is no non-homogeneous rational Fourier matrix of rank 5. Theorem 6.3. There is no non-homogeneous rational Fourier matrix S of rank 5. Proof. Let (A,B,δ) be a non-homogenous C-algebra arising from a rational Fourier matrix S of rank 5. Let δ(bi) = ki, for all i. Since s-matrix is integral, ki are perfect square integers, see [11, Proposition 5]. By [3, Lemma 3.7], d0 is a square. Therefore, d0 ≡ 0, 1 mod 4. Let d0 = k4a,d0 = k3b,d0 = k2c,d0 = k1d. Then each of a,b,c and d is greater than 1 and a square integer because A is non-homogeneous. Case 1. If each of k1,k2,k3,k4 is an odd integer, then d0 is odd. Also, the fact that k1,k2,k3,k4 are odd implies a,b,c,d are odd and greater than or equal to 9. Without loss of generality, let k4 ≥ k1,k2,k3. Therefore, d0 ≥ 9k4,d0 = 1 + k1 + k2 + k3 + k4 ≤ 1 + 4k4, a contradiction. 62 http://orcid.org/0000-0002-0819-8221 G. Singh / J. Algebra Comb. Discrete Appl. 5(2) (2018) 51–63 Case 2. If three of k1,k2,k3,k4 are odd and one is even, then d0 is even. Without loss of generality, suppose k4 is even. Thus d0 = k4a implies a ≥ 4. Subcase 1. If k4 > k1,k2,k3, then d0 ≥ 4k4, d0 ≤ 1 + (k4 −1) + (k4 −1) + (k4 −1) + k4 = 4k4 −2, a contradiction. Subcase 2. If k4 < one of k1,k2,k3, say k3, so k1,k2 ≤ k3, then d0 ≥ 4k3 because b is an even square. Thus d0 ≤ 1 + k3 + k3 + k3 + (k3 − 1) = 4k3 implies k1 = k2 = k3 and k4 = k3 − 1, d0 = 4k3. Now d0 = 4x 2 because d0 is a square and an even integer. Hence k1 = k2 = k3 = x2 and k4 = x2 − 1. Since x2 − 1 divides 4x2 and x is an odd integer, x2 − 1 divides 4, we get a contradiction. Case 3. If two of k1,k2,k3,k4 are odd and two are even, then d0 ≡ 3 mod 4, a contradiction. Case 4. If one of k1,k2,k3,k4 is odd and three are even, then d0 ≡ 2 mod 4, a contradiction. Acknowledgment: The author would like to thank Professor Allen Herman whose valuable sug- gestions helped him to improve this paper. References [1] Z. Arad, E. Fisman, M. Muzychuk, Generalized table algebras, Israel J. Math. 114(1) (1999) 29–60. [2] H. I. Blau, Table algebras, European J. Combin. 30(6) (2009) 1426–1455. [3] M. Cuntz, Integral modular data and congruences, J. Algebraic Combin. 29(3) (2009) 357–387. [4] P. Francesco, P. Mathieu, D. Sénéchal, Conformal Field Theory, Springer–Verlag, New York, 1997. [5] T. Gannon, Modular data: The algebraic combinatorics of conformal field theory, J. Algebraic Combin. 22(2) (2005) 211–250. [6] A. Hanaki, I. Miyamoto, Classification of association schemes with small vertices, 2017, available at: math.shinshu-u.ac.jp/ hanaki/as/. [7] D. G. Higman, Coherent algebras, Linear Algebra Appl. 93 (1987) 209–239. [8] J. D. Qualls, Lectures on Conformal Field Theory, arXiv:1511.04074 [hep-th]. [9] E. L. Rees, Graphical Discussion of the Roots of a Quartic Equation, Amer. Math. Monthly 29(2) (1922) 51–55. [10] M. Schottenloher, A Mathematical Introduction to Conformal Field Theory, Springer–Verlag, Berlin, Heidelberg, 2nd edition, 2008. [11] G. Singh, Classification of homogeneous Fourier matrices, arXiv:1610.05353 [math.RA]. [12] B. Xu, Characters of table algebras and applications to association schemes, J. Combin. Theory Ser. A 115(8) (2008) 1358–1373. [13] A. Zahabi, Applications of Conformal Field Theory and String Theory in Statistical Systems, Ph.D. dissertation, University of Helsinki, Helsinki, Finland, 2013. 63 http://orcid.org/0000-0002-0819-8221 https://doi.org/10.1007/BF02785571 https://doi.org/10.1016/j.ejc.2008.11.008 https://doi.org/10.1007/s10801-008-0139-y http://www.springer.com/us/book/9780387947853#otherversion=9781461274759 https://doi.org/10.1007/s10801-005-2514-2 https://doi.org/10.1007/s10801-005-2514-2 http://math.shinshu-u.ac.jp/~hanaki/as/ http://math.shinshu-u.ac.jp/~hanaki/as/ https://doi.org/10.1016/S0024-3795(87)90326-0 https://arxiv.org/abs/1511.04074 https://doi.org/10.2307/2972804 https://doi.org/10.2307/2972804 http://www.springer.com/gp/book/9783540686255 http://www.springer.com/gp/book/9783540686255 https://arxiv.org/abs/1610.05353 https://doi.org/10.1016/j.jcta.2008.02.005 https://doi.org/10.1016/j.jcta.2008.02.005 https://helda.helsinki.fi/bitstream/handle/10138/40728/zahabi_dissertation.pdf?...1 https://helda.helsinki.fi/bitstream/handle/10138/40728/zahabi_dissertation.pdf?...1 Introduction C-algebras arising from Fourier matrices Recognition of C-algebras arising from Fourier matrices Fourier matrices of rank 2 and 3 Fourier matrices of rank 4 Fourier matrices of rank 5 References