ON HADAMARD MATRICES AT ROOTS OF UNITY Introduction A

ON HADAMARD MATRICES AT ROOTS OF UNITY TEODOR BANICA AND JEAN-MARC SCHLENKER Abstract. We study Hadamard matrices of order n, formed by l-th roots of unity. A main problem is to find the allowed values of (n, l), and we discuss here the following statement: for l = pa1 1 . . . pas s we must have n ∈ p1 +. . .+ps N. For s = 1 this is a previously known result, for s = 2, 3 this is a result that we prove in this paper, and for s ≥ 4 this is a conjecture that we raise. We present as well some remarks and comments regarding the other known obstructions.

Introduction A complex Hadamard matrix is a square matrix h ∈ Mn (C), whose entries are on the unit circle, and whose rows are pairwise orthogonal. The basic example is the Fourier matrix: Fn = wij with w = e2πi/n . The story goes back to work of Sylvester and Hadamard, who studied the real matrices. These have the numbers ±1 as entries. The order of such a matrix must be of the form n = 2 or n = 4k, and the existence problem for arbitrary values of k is still open. See Kharaghani and Tayfeh-Rezaie [15]. The matrices with entries at arbitrary roots of unity were first considered by Butson [9], as a natural generalization of the ±1 case. There are many questions about the existence of such matrices. See Turyn [25], Drake [10], de Launey [16], Brock [7], de Launey and Dawson [17], Winterhof [26]. The matrices with arbitrary complex entries appeared in the eighties. Popa discovered that such matrices produce orthogonal subalgebras of Mn (C), hence commuting squares [20], which are in turn related to the classification of subfactors [21]. Bj¨orck discovered a connection between circulant Hadamard matrices and cyclic n-roots [6]. The classification problem for small Hadamard matrices was investigated by Haagerup in [12], with complete results at n ≤ 5. The complex Hadamard matrices appear in several branches of coding theory ˙ and quantum physics. See Tadej and Zyczkowski [23]. Also, a quite unexpected application to a problem in harmonic analysis was found by Tao in [24]. 2000 Mathematics Subject Classification. 05B20. Key words and phrases. Complex Hadamard matrix. 1

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TEODOR BANICA AND JEAN-MARC SCHLENKER

The complex Hadamard matrices are also known for their highly non-trivial quantum algebraic invariants. The story here goes back to the papers of Popa [20] and Jones [13]. For a discussion of the main problems, see Jones [14]. The quantum algebraic problems can be formulated as well in terms of quantum permutation groups, thanks to a general result in [1]. The explicit construction is worked out in [4]. The whole subject was recently surveyed in [3]. Summarizing, the complex Hadamard matrices appear to be at the heart of several branches of combinatorics, quantum algebra, and mathematical physics in general. Most problems about them are open. The purpose of this paper is to make some advances on Hadamard matrices having as entries the roots of unity. This is probably the simplest case: (1) (2) (3) (4)

The basic example, Fn = wij with w = e2πi/n , is of this form. Regarding the other known examples, most of them are also of this form. The degree of the roots of unity is a useful complexity parameter. Some techniques from number theory can be used.

Another motivating remark is the fact that the Hadamard matrices at roots of unity should correspond to subfactors or planar algebras or quantum permutation groups “at roots of unity”, somehow in the spirit of Drinfeld’s paper [11]. The problem that we consider is the classical one, namely to find the connection between the order of roots l and the order of the matrix n. Not all the pairs (n, l) are allowed, as shown by the following results: (1) Sylvester obstruction: l = 2 implies n = 2 or 4|n. (2) Butson obstruction: l = pa implies p|n. (3) de Launey obstruction: coming from |det|2 = nn . The names here are assigned according to pioneering work. For the Butson and the de Launey obstructions, the current statements are due to Winterhof [26]. These obstructions are of quite different nature. However, as a very rough classification, the Sylvester and de Launey obstructions mainly concern the l p, we obtain the following estimate: n > (p − 1)(q − 1) + (p − 1) = (p − 1)q This shows that the numbers n, n − q, n − 2q, . . . , n − (p − 1)q are all positive. Now since p must divide one of them, we get n ∈ pN + qN, which is the desired result. In the other case p > r we get the estimate n > (r − 1)q, and by using the same argument we obtain n ∈ qN + rN, so we are done as well. 6. Bar systems Our next goal is to extend Proposition 5.2 to exponents l having 3 prime factors. In this section we present a geometric approach to the problem. The idea is as follows: (1) The relevant case is l = abc, with a, b, c prime. (2) We can identify Zl with a collection of abc copies of the unit cube. (3) With this picture, the fact that a, b, c are prime is actually irrelevant. These three steps will be explained in detail later on. For the moment, let us state the relevant problem in R3 . We regard R3 as being a union of copies of the unit cube: those having vertices at the points of Z3 . These copies of the unit cube, that we call “small cubes”, will be the building blocks for all the considerations in this section. For instance all the sets X ⊂ R3 to be considered will be unions of small cubes. The size of such a set is the number of small cubes the set is made of. Definition 6.1. Let a ≤ b ≤ c be three positive integers, and consider the (a, b, c)cube in R3 , viewed as a union of abc small cubes. (1) A bar is any of the ab + bc + ac cubular segments of length a, b, c. (2) A bar system is any collection of bars.

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(3) A virtual bar system is a difference of bar systems S 0 − S, with S ⊂ S 0 . (4) A ghost bar is a non-empty virtual bar system containing no bar. The relation with sums of roots is as follows. If l = abc with a < b < c prime, we have the following set-theoretic isomorphism: Zl = Zabc = Za × Zb × Zc = {1, . . . , a} × {1, . . . , b} × {1, . . . , c} Thus we can identify the elements of Zl with the small cubes of the (a, b, c)cube. With this identification, the above notions read as follows: (1) The bars are the basic sums of type Sar , Sbr , Scr . (2) The bar systems are the trivial sums. (3) The virtual bar systems are the vanishing sums (cf. Theorem 4.3). (4) The ghost bars are the vanishing sums containing no basic sums. Probably most illustrating here is the picture of the vanishing sum in Proposition 5.3. With our new notations a, b, c instead of p, q, r, this sum is: S =

Saab + Sa2ab + . . . + Sa(c−1)ab (a−1)bc +Sbbc + Sb2bc + . . . + Sb −Scbc − Sc2bc − . . . − Sc(a−1)bc

The corresponding picture inside the (a, b, c)-cube is as follows: take an ac-face face minus an a-bar, then an ab-face minus a b-bar, and remove the ac-face minus c-bar which appears. We get in this way a ghost bar, corresponding to S. We recall from the previous section that the size of this particular ghost bar is (a − 1)(b − 1) + (c − 1). We prove now that this number is optimal. Lemma 6.2. The minimal size of a ghost bar is (a − 1)(b − 1) + (c − 1). Proof. Let X be a ghost bar, chosen to be of minimal size. We can decompose X into slices X1 , . . . , Xc , by following the c direction. Each slice is an (a, b)-rectangle, or matrix, that we denote as follows:   k k X11 . . . X1b Xk =  . . . . . . . . .  k k Xa1 . . . Xab That is, Xijk ∈ N is the multiplicity of the (i, j, k) small sube inside X. The difference between two given slices Xk and Xl comes from certain a-bars and b-bars, having integer coefficients. That is, we have:   A1 + B1 . . . Ab + B1 ... ...  Xk = Xl +  . . . A1 + Ba . . . Ab + Ba

ON HADAMARD MATRICES AT ROOTS OF UNITY

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Here the coefficients are integers: Ai , Bj ∈ Z. Case 1. Assume first that X has an empty slice:   0 ... 0 Xl = . . . . . . . . . 0 ... 0 Thus any slice is of the following form:   A1 + B1 . . . Ab + B1 ... ...  Xk =  . . . A1 + Ba . . . Ab + Ba We can permute rows and columns as to have A1 ≤ . . . ≤ Ab and B1 ≤ . . . ≤ Ba . Since Xk contains no a-bar and no b-bar, we must have Ai = Bj = 0 for any i, j. Thus all the slices of X are empty, contradiction. Case 2. Assume now that X has a slice containing one element. By permuting rows and columns we can assume that this element appears on top left:   1 0 ... 0  0 0 ... 0   Xl =  . . . . . . . . . . . . 0 0 ... 0 Thus any slice is of the following form:  A1 + B1 + 1 A2 + B1  A1 + B2 A2 + B2 Xk =   ... ... A1 + Ba A2 + Ba

 . . . Ab + B1 . . . Ab + B2   ... ...  . . . Ab + Ba

Once again by rearranging Ai ’s and Bj ’s and by using the fact that Xk contains no a-bar and no b-bar, we get that either Xk = Xl or Xk = Xl0 , where:   0 0 ... 0  0 1 ... 1   Xl0 =  . . . . . . . . . . . . 0 1 ... 1 Indeed if Xk 6= Xl then one of the Ai or one of the Bj must be negative, we suppose for instance that one of the Ai is negative. Let i0 be such that Ai0 is minimal among the Ai , and suppose that i0 6= 0, then, since the entries of Xk are non-negative, Bj ≥ −Ai0 for all j. Since Xk contains no bar, it follows that all Ai are equal and all Bj are equal to −Ai0 , so that Xk = Xl . If i0 = 1, the same argument can be used and shows that Xk = Xl0 .

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Now since X contains no c-bar, there should be at least one slice of type Xl0 , and this gives the desired estimate on the size of X: n ≥ (a − 1)(b − 1) + (c − 1) Observe that the estimate is sharp, because of the example discussed before stating the present lemma. A careful examination of the picture shows that this example consists indeed of c − 1 slices of type Xl , and a slice of type Xl0 . Case 3. Assume now that all slices have at least 2 elements. We choose one slice Xl having minimal number of elements, say m ≥ 2. Now since X contains no c-bar, there should be at least one slice different from Xl . We let Xk be such a slice, chosen as to have minimal size. We know that Xk can be obtained from Xl by adding or removing a number of a-bars and b-bars. Since Xk contains no a-bar or b-bar, at least one of these bars has to come with a negative coefficient. We assume that this bar is an a-bar, coming with coefficient −1. The other situations, which would actually lead to bigger lower bounds for n, will be discussed at the end of the proof. By permuting rows and colums we may assume that the removed a-bar is the first one, and that the nonzero entries in the first row of Xl appear on top:   k1 . . . . . . . . . . . . . . . . . . . . .    k . . . . . . . . . Xl =  v   0 . . . . . . . . . . . . . . . . . . . . . 0 ... ... ... Here v ≤ m is a certain number, k1 , . . . , kv ≥ 1 are some other numbers, and the dots have a vertical meaning. The right part of the matrix, starting from the second column, is irrelevant for the rest of the proof. Now recall that Xk is obtained from Xl by removing an a-bar from the first row, then by adding or removing some other a-bars and b-bars. Since removing the first a-bar makes a number of −1 signs appear, we must compensate with b-bars, one for each of these −1 signs. Thus Xk must be at least as big as:     k1 . . . . . . . . . −1 0 . . . 0 . . . . . . . . . . . .  . . . . . . . . . . . .      k . . . . . . . . . −1 0 . . . 0  Xl0 =  v +  1 ... 1   0 . . . . . . . . .  0 . . . . . . . . . . . .  . . . . . . . . . . . . 0 ... ... ... 0 1 ... 1 In other words, the size of Xk satisfies: nk = m + ((a − v)(b − 1) − v)


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≥ m + ((a − m)(b − 1) − m) ≥ (a − m)(b − 1) This gives the desired inequality on the size of X: since Xl is the c-slice with the minimal number of small cubes, n ≥ ≥ > >

(a − m)(b − 1) + (c − 1)m a(b − 1) + m(c − b) a(b − 1) + (c − b) (a − 1)(b − 1) + (c − 1)

Finally, it remains to justify the assumption that we made, namely that Xk comes from Xl by removing one a-bar, then by adding or removing some other a-bars and b-bars. But this follows from the following two observations: (1) when interchanging a, b in the above constructions the size of Xl0 gets bigger, because a ≤ b, and (2) when removing 2 or more a-bars instead of 1, the size of Xl0 gets bigger as well, because the entries 1 are replaced by entries ≥ 2. Theorem 6.3. Any virtual bar system has the size of a bar system. Proof. First, the size of a bar system can be any element of aN + bN + cN. The virtual bar system consists of bars, having size a, b, c, and of a ghost bar, having size n ≥ (a − 1)(b − 1) + (c − 1). Thus the size of the ghost bar satisfies n ≥ (a − 1)b, and we can conclude as in the proof of Proposition 5.3. 7. The Butson obstruction We have now all ingredients nedeed for generalizing the Butson obstruction. We begin with a statement coming from results in the previous section, which generalizes at the same time Proposition 5.2 and Proposition 5.3. Theorem 7.1. For l = pa11 . . . pas s with s ≤ 3, the following are equivalent: (1) There is a n-sum of l-roots which vanishes. (2) n ∈ p1 N + . . . + ps N. Proof. For s ≤ 2 this follows from Proposition 5.2. So, assume that we are in the case s = 3, and let S be a vanishing sum of l-roots, having n terms. Since S is Z-trivial, we can write is as a sum of basic sums Spri , with ± coefficients. These basic sums can intersect or not, depending on the values of the rotation parameters r. Now when two parameters r, r0 belong to distinct classes modulo p1 p2 p3 , the corresponding basic sums cannot intersect. Thus we can decompose S as a disjoint union of vanishing sums, one of each class. Since our problem is additive in the size of the sum, we can assume that S consists only of one component. Moreover, by performing a suitable rotation, we

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can assume that this component is the one corresponding to the class of r = 0. This is the same as assuming that we have l = p1 p2 p3 . Now we can regard S as a virtual bar system, as explained in the previous section, and the result follows from Theorem 6.3. We can reformulate the above result, as an Hadamard matrix obstruction: Theorem 7.2. Assume that we have l = pa11 . . . pas s , with s ≤ 3. Then, in order for Hn (l) to be non-empty, we must have n ∈ p1 N + . . . + ps N. Proof. This follows indeed from Theorem 7.1, by using the fact that the scalar product between the first two rows of any element h ∈ Hn (l) can be regarded as a vanishing n-sum of l-roots. Observe that at s = 1 we have the statement “l = pa implies p|n”, which is Winterhof’s generalization of the Butson obstruction [26]. Now regarding numeric applications, here is a table of impossible values of n, for exponents l ≤ 30, composite and squarefree: l 6 10 14 15 21 22 26 30 n 3 3,5 2,4,7 2,4,5,8,11 3,5,7,9 3,5,7,9,11,13 This table can be continued up to l = 2·3·5·7 = 210, which is the first exponent not covered by our result. However, we believe of course that the statement holds in general, and we have the following conjecture. Conjecture 7.3. The above result holds for any s ∈ N. Now back to applications, it is probably useful to have as well a criterion excluding l’s once n is given. We have here the following algorithm: (1) Find all possible decompositions n = p1 m1 + . . . + ps ms , with pi prime. (2) K is the set of products k = p1 . . . ps , with p1 , . . . , ps as above. (3) D ⊂ K is obtained by erasing k ∈ K if d|k with d ∈ K, d < k. (4) The obstruction is d|l for some d ∈ D. According to the above discussion, this is in general a conjectural algorithm. The excluded values of l are so far those having at most 3 prime factors. This gives the following alternative table of applications of Theorem 7.2: n 2 3 4 5 6 7 8 9 10 11 d|l 2 3 2 5,6 2,3 6,7,10 2,15 3,10 2,5,21 6,10,14,15 As an example of application, we have H7 (15) = ∅, a previously unknown result. Indeed, the n = 7 box contains the numbers d = 6, 7, 10, whose mission is to divide l = 15. But this is impossible.


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8. Global obstructions Consider an Hadamard matrix h ∈ Hn (l). With w = e2πi/l , the entries of h are powers of w, so the determinant of h is an element of the ring Z[w]. On the other hand the Hadamard matrix condition is hh∗ = nIn , where h = ¯ ji ) is the adjoint matrix. By appying determinants we get: (h | det(h)|2 = nn The fact that this condition produces obstructions on (n, l) was pointed out by de Launey in [16]. For general statements in this sense, see Winterhof [26]. In this section we will be concerned with exponents l ≤ 6. The situation here is as follows. At l = 2, 4 there is no obstruction coming from the determinant, and so seems to be the case at l = 5. At l = 3, 6 we can use the following well-known result: Proposition 8.1. For an odd number N , we have N = |d|2 for some d ∈ Z[e2πi/3 ] if and only if N = m2 p1 . . . ps with pi 6= 5 (mod 6). Summarizing, we have the following result, coming by putting together the various obstructions on (n, l), for small values of l. Theorem 8.2. We have the following obstructions for Hn (l) to be non-empty. (1) (2) (3) (4) (5)

l=2 l=3 l=4 l=5 l=6

implies implies implies implies implies

n = 2 or 4|n. 6|n, or n = 3m2 p1 . . . ps is odd with pi 6= 5 (mod 6). 2|n. 5|n. 2|n, or n = m2 p1 . . . ps is odd with pi 6= 5 (mod 6).

At l = 2, 4 it is conjectured that there are no other obstructions: for l = 2 this corresponds to a well-known problem, for l = 4 see Turyn [25]. At l = 3, 5, 6 we conjecture that there are no other obstructions either. In other words, we have the following conjecture. Conjecture 8.3. The above obstructions are the only ones at l ≤ 6. A first piece of evidence for the conjecture comes from the fact that the above result captures indeed all one can get from the known obstructions. The other piece of evidence comes from numeric verifications. At l = 3 we have the matrices F3 , T , and various tensor products between them, so the first missing matrix is at n = 12. At l = 5 we have the matrix F5 , so the first missing matrix is at n = 10. Theorem 8.4. The sets H10 (5) and H12 (3) are non-empty.

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Proof. Here are the two needed examples, written in k standing for e2πik/l :  0 0 0 0 0 0 0 0  0 0 1 1 2 2 3 3   0 1 0 3 2 4 1 4   0 1 3 4 3 1 0 2   0 2 3 0 1 3 4 1 5 X10 =   0 2 4 2 0 1 3 4  0 3 1 2 4 0 4 2   0 3 2 4 1 4 2 3   0 4 2 1 4 3 1 0 0 4 4 3 3 2 2 1 

3 X12

         =        

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 1 1 1 2 2 2 2

0 0 0 1 0 2 2 2 1 1 1 2

0 0 1 2 2 0 1 2 0 1 2 1

0 1 0 2 2 1 2 0 2 0 1 1

0 1 2 0 1 2 0 2 0 2 1 1

0 1 2 1 2 0 0 1 2 1 0 2

0 1 2 2 0 2 1 1 1 0 2 0

0 2 1 0 2 0 2 1 1 2 1 0

logarithmic form, i.e. with 0 4 2 4 2 3 1 0 3 1

0 4 3 2 4 1 3 1 2 0



0 2 1 1 0 2 1 0 2 2 0 1

0 2 1 2 1 1 0 2 1 0 0 2

0 2 2 1 1 1 2 0 0 1 2 0

             

                  

These matrices were obtained by using a computer program.

In the case l = 6, we have the various matrices from the l = 2, 3 cases, and the tensor products between them. The first problem appears at n = 7, where no such tensor product is available. But we have here the Petrescu matrix:   1 1 1 1 1 1 1 1 w w4 w5 w3 w3 w    1 w 4 w w 3 w 5 w 3 w    P 1 = 1 w 5 w 3 w w 4 w w 3  1 w 3 w 5 w 4 w w w 3    1 w 3 w 3 w w w 4 w 5  1 w w w3 w3 w5 w4 This matrix was found as well by using a computer program [19].


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9. Extensions of the Sylvester obstruction We present now two obstructions to the existence of Hadamard matrices for some values of n and l. Both are extensions of the Sylvester obstruction, which holds for n = 2 and l = 4k + 2, k ≥ 1. Theorem 9.1. If p ≥ 3 is prime and a ≥ 1, then Hp+2 (2pa ) = ∅. Proof. Let M ∈ Hn (l), written in logarithmic form (so that its entries are elements of Z/lZ). Theorem 5.1 shows that each row of M contains one 2-cycle (two numbers differing by l/2) and one p-cycle. The two elements of the 2-cycle have opposite parities, while the elements of the p-cycle have the same parity. Therefore, each row of M has either exactly one odd entry or exactly one even entry. The same applies to the difference between two row, since rows correspond to pairwise orthogonal vectors. Let L1 , L2 be two rows of M , and let L2 − L1 be their difference. (1) If L1 and L2 both have exactly one even entry, then L2 − L1 has either no odd entry (if the even entries of L1 and L2 are at the same position) or exactly two odd entries (if these even entries are at different positions). (2) The same holds if L1 and L2 both have exactly one odd entry. (3) If L1 has exactly one even entry, and L2 has exactly one odd entry, then L2 − L1 has either no even entry (if the positions correspond) or exactly two even entries (if the positions are different). We can see that in all the three cases, L2 − L1 cannot have either exactly one odd entry or exactly one even entry, a contradiction. Theorem 9.2. If q > p ≥ 3 are prime and a, b ∈ N, then H2p (2a q b ) = ∅. Proof. Let M ∈ H2p (2a q b ), written in logarithmic form. Theorem 5.1 shows that each row of M is a union of 2-cycles and of q-cycles. Since q > p, there can be no q-cycle, since one q-cycle would leave an odd number of elements which can not be grouped in 2-cycles. So, each row of M is a union of 2-cycles. The same argument shows that the difference between two rows is also a union of 2-cycles. This proves that the reduction of M modulo 2 is a real Hadamard matrix (written in logarithmic form). The usual Sylvester obstruction can therefore be used, and shows that there can be no such matrix, since p is odd.

10. Summary The results presented in this paper can be illustrated by an extension of the table at the end of section 2, extended up to n = 12 and l = 15.

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n\l 2 3 4 5 6 7 8 9 10 11 12


2 F2 × F2,2 × ×s × L × ×s × M

3 × F3 × × T × × F3,3 × × 3 X12

4 F2 × F2,2 × H × L × 4 B10 × M

5 × × × F5 × × × × 5 X10 × ×

6 F2 F3 F2,2 ×l T P1 L F3,3

3 X12

7 × × × × × F7 × × × × ×

8 F2 × F2,2 × H × L × 4 B10 × M

9 × F3 × × T × × F3,3 × × 3 X12

10 F2 × F2,2 F5 ×gs ×gs L B9 F10 M

11 × × × × × × × × × F11 ×

12 F2 F3 F2,2 ×h T P1 L F3,3 4 B10 3 X12

13 × × × × × × × × × × ×

14 F2 × F2,2 × ×gs F7 L ×gs ×

15 × F3 × F5 T × F3,3 5 X10

×gs

3 X12

In this table, we use the notation L = F2,2,2 . The matrix M in the last row denotes one of the 12 × 12 real Hadamard matrices, well-known to exist. The symbol ×h for n = 5, l = 12 describes the obstruction coming from the Haagerup classification of Hadamard matrices for n = 5, while the symbol ×gs for l = 10 and n = 6, 7 as well as for n = 6, l = 14 and n = 9, l = 14 corresponds to the generalized Sylvester obstruction in Theorem 9.1 and Theorem 9.2. The matrix B9 appearing for n = 9 and l = 10 is a known example, see [8]. 4 The matrix B10 on the last row is another known example, see [8]. The empty cells, for several values of l for n = 11 and for n = 8, l = 15 and n = 10, l = 6, means that we do not know whether Hn (l) is empty or not. References [1] T. Banica, Representations of compact quantum groups and subfactors, J. Reine Angew. Math. 509 (1999), 167–198. [2] T. Banica, J. Bichon and G. Chenevier, Graphs having no quantum symmetry, Ann. Inst. Fourier 57 (2007), 955–971. [3] T. Banica, J. Bichon and B. Collins, Quantum permutation groups: a survey, Banach Center Publ., to appear. [4] T. Banica and R. Nicoara, Quantum groups and Hadamard matrices, Panamer. Math. J. 17 (2007), 1–24. [5] K. Beauchamp and R. Nicoara, Orthogonal maximal abelian ∗-subalgebras of the 6 × 6 matrices, arxiv:math/0609076. [6] G. Bj¨ orck, Functions of modulus 1 on Zn whose Fourier transforms have constant modulus, and cyclic n-roots, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci. 315 (1990), 131–140. [7] B.W. Brock, Hermitian congruence and the existence and completion of generalized Hadamard matrices, J. Combin. Theory Ser. A 49 (1988), 233–261. [8] W.T. Bruzda, Complex Hadamard matrices and set of basis of optimal quantum measurement, Master thesis, Krakow (2006). [9] A.T. Butson, Generalized Hadamard matrices, Proc. Amer. Math. Soc. 13 (1962), 894–898.


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