Complex Hadamard matrices contained in a Bose-Mesner algebra

A complex Hadamard matrix is a square matrix H with complex entries of absolute value 1 satisfying $HH^*= nI$, where $*$ stands for the Hermitian transpose and I is the identity matrix of order $n$. In this paper, we first determine the image of a certain rational map from the $d$-dimensional complex projective space to $\mathbb{C}^{d(d+1)/2}$. Applying this result with $d=3$, we give constructions of complex Hadamard matrices, and more generally, type-II matrices, in the Bose-Mesner algebra of a certain 3-class symmetric association scheme. In particular, we recover the complex Hadamard matrices of order 15 found by Ada Chan. We compute the Haagerup sets to show inequivalence of resulting type-II matrices, and determine the Nomura algebras to show that the resulting matrices are not decomposable into generalized tensor products.


Introduction
A complex Hadamard matrix is a square matrix H with complex entries of absolute value 1 satisfying HH * = nI, where * stands for the Hermitian transpose and I is the identity matrix of order n. They are the natural generalization of real Hadamard matrices. Complex Hadamard matrices appear frequently in various branches of mathematics and quantum physics.
A type-II matrix, or an inverse orthogonal matrix, is a square matrix W with nonzero complex entries satisfying W W (−) ⊤ = nI. Obviously, a complex Hadamard matrix is a type-II matrix.
A complete classification of complex Hadamard matrices, and that of type-II matrices are only available up to order n = 5 (see [7,14,10]).
Although it is shown by Craigen [7] that there are uncountably many equivalence classes of complex Hadamard matrices of order n whenever n is a composite number, some type-II matrices are more closely related to combinatorial objects than the others. Szollosi [16] used design theoretical methods to construct complex Hadamard matrices. Strongly regular graphs were used to construct type-II matrices in [5,6]. See [15] for a generalization. In this paper, we construct type-II matrices and complex Hadamard matrices in the Bose-Mesner algebra of a certain 3-class symmetric association scheme. In particular, we recover the complex Hadamard matrices of order 15 found by [4].
The method of finding a complex Hadamard matrices in the the Bose-Mesner algebra of a symmetric association scheme generalizes the classical work of Goethals and Seidel [9]. By assuming that the association scheme to be symmetric, the resulting complex Hadamard matrices are symmetric. It turns out that this assumption enables us to consider only the real parts of the entries of a complex Hadamard matrices, since the orthogonality can be expressed in terms of the real parts. Extending this reduction to type-II matrices, we are led to consider a rational map whose inverse is explicitly given in Section 2. In Section 3, we explain why only real parts come into play when we construct complex Hadamard matrices in the Bose-Mesner algebra of a symmetric association scheme. In Section 4, we take a particular family of a 3-class association scheme. This family was found after extensive computer experiment on the list of 3-class association schemes up to 100 vertices given in [8]. Surprisingly, most other association schemes up to 100 vertices, with the exceptions of amorphic or pseudocyclic schemes, do not admit a complex Hadamard matrix in their Bose-Mesner algebras. In Section 5, we compute the Haagerup set to show inequivalence of type-II matrices constructed in Section 4. In Section 6, we show that the Nomura algebra of each of the type-II matrices constructed in Section 4 has dimension 2. This implies that our matrices are not equivalent to a generalized tensor product by [11].
All the computer calculations in this paper are performed by magma [2].

The image of a rational map
We define a polynomial in three indeterminates X, Y, Z as follows: g(X, Y, Z) = X 2 + Y 2 + Z 2 − XY Z − 4.
For a finite set N and a positive integer k, we denote by N k the collection of all k-element subsets of N. Lemma 4. Let N = {0,1,. . . ,d},N 3 = N 3 and N 4 = N 4 . Let a i,j (0 ≤ i, j ≤ d, i = j) be complex numbers satisfying g(a i,j , a j,k , a i,k ) = 0 ({i, j, k} ∈ N 3 ), h(a i,j , a i,k , a i,ℓ , a j,k , a j,ℓ , a k,ℓ ) = 0 ({i, j, k, ℓ} ∈ N 4 ). Assume Let w i 0 , w i 1 be nonzero complex numbers satisfying Then Conversely, if complex numbers {w i } d i=0 satisfy (10), then (9) holds. Moreover, if a i,j (0 ≤ i < j ≤ d) are all real and Proof. Without loss of generality, we may assume i 0 = 0 and i 1 = 1.
Then the image of φ coincides with the zeros of the ideal generated by the polynomials h(X i,j , X i,k , X i,ℓ , X j,k , X j,ℓ , X k,ℓ ) = 0 ({i, j, k, ℓ} ∈ N 4 ), where X i,j = X j,i .
Proof. Let I denote the ideal of the polynomial ring C[X i,j | 0 ≤ i < j ≤ d] generated by (19) and (20). By Lemmas 1 and 3, the image of φ is contained in the set of zeros of I. Conversely, let a = (a i,j ) 0≤i<j≤d be a zero of I. If there exists i 0 , i 1 with 0 ≤ i 0 < i 1 ≤ d and a i 0 ,i 1 = ±2, then Lemma 4 implies that a is in the image of φ. Next suppose that a i,j ∈ {±2} for 0 ≤ i < j ≤ d. Then a 0,1 = a 0,1 2 + 2 a 0,1 .
The following lemma will be used in the proof of Theorem 2.
Lemma 5. In the rational function field with three indeterminates X 1 , X 2 , X 3 , set where X 0 = 1. Then Proof. Straightforward.

Type-II matrices contained in a Bose-Mesner algebra
Throughout this section, we let A denote a symmetric Bose-Mesner algebra with adjacency matrices A 0 = I, A 1 , . . . , A d . Let n be the size of the matrices A i , and we denote by P = (P i,j ) 0≤i≤d 0≤j≤d the first eigenmatrix of A. Then the adjacency matrices are expressed as holds. Since QP = nI and Q i,0 = P i,0 = 1 for i = 0, 1, . . . , d, we have Lemma 6. Let w 0 , w 1 , . . . , w d be nonzero complex numbers, and set Then the following are equivalent.
Proof. Set Clearly, W is type-II matrix if and only if β k β ′ k = n for 0 ≤ k ≤ d. In particular, (i) implies (ii).
To prove the converse, it suffices to show β 0 β ′ 0 = n provided β k β ′ k = n for 1 ≤ k ≤ d. Since W is symmetric, the diagonal entries of W W (−1) are all n. Thus If the matrix W given by (22) is a type-II matrix which is not equivalent to an ordinary Hadamard matrix, then the complex numbers a i,j defined by (10) are common zeros of the polynomials e k (1 ≤ k ≤ d) and satisfy (4)- (7).
are common zeros of the polynomials e k (1 ≤ k ≤ d) and satisfy (4)- (7), then there exist complex numbers w 0 , w 1 , . . . , w d satisfying (10) such that the matrix W is a type-II matrix which is not equivalent to an ordinary Hadamard matrix.
Moreover, the matrix W is a scalar multiple of a complex Hadamard matrix which is not equivalent to an ordinary Hadamard matrix if and only if a i,j defined by (10) are common real zeros of the polynomials e k (1 ≤ k ≤ d), satisfy (4)- (7) and (11).
Suppose first that W is a type-II matrix which is not equivalent to an ordinary Hadamard matrix. Then Lemma 6 implies that (23) holds. By (25), this implies that the complex numbers a i,j defined by (10) are common zeros of the polynomials e k (1 ≤ k ≤ d). Since W is not equivalent to an ordinary Hadamard matrix, (7) holds. Clearly (4) holds. From Lemma 1 and 3, we see that a i,j satisfy (5) and (6), respectively. Conversely, suppose that a i,j are common zeros of the polynomials e k (1 ≤ k ≤ d) and satisfy (4)- (7). From Lemma 4, there are complex numbers w 0 , w 1 , . . . , w d satisfying (10). Then by (25), we see that (23) holds, and hence W is a type-II matrix by Lemma 6, and W is not equivalent to an ordinary Hadamard matrix by (7).
Assume that W is a scalar multiple of a complex Hadamard matrix which is not equivalent to an ordinary Hadamard matrix. Then a i,j are common real zeros of the polynomials e k (1 ≤ k ≤ d) and (11) holds.
Conversely, suppose that the real numbers a i,j are common zeros of the polynomials e k (1 ≤ k ≤ d), satisfy (4)- (7) and (11). Then by Lemma 4, w 0 , w 1 , . . . , w d have the same absolute value. Therefore W is a scalar multiple of a complex Hadamard matrix.

Infinite families of complex Hadamard matrices
Let q ≥ 4 be an integer, and n = q 2 − 1. We consider a three-class association scheme X = (X, {R i } 3 i=0 ) with the first eigenmatrix: For q = 2 s with an integer s ≥ 2, there exists a 3-class association scheme with the first eigenmatix (26) (see [3, 12.1.1]).
Then, X has two non-trivial fusion schemes. One is an imprimitive scheme X 1 = (X, {R 0 , R 1 ∪ R 2 , R 3 }) with the first eigenmatrix: Another is a primitive scheme X 2 = (X, {R 0 , R 1 ∪ R 3 , R 2 }) with the first eigenmatrix: Theorem 2. Let w 1 , w 2 , w 3 be nonzero complex numbers. The matrix is a type-II matrix if and only if one of the following holds: Proof. Let K = Q(r), and consider the polynomial ring We assume that W is a type-II matrix. For i, j ∈ {0, 1, 2, 3}, define a i,j by (10), where w 0 = 1. We write a = (a 0,1 , a 0,2 , a 0,3 , a 1,2 , a 1,3 , a 2,3 ) for brevity. Since q 2 −1 is odd, W is not an ordinary Hadamard matrix. Then by Lemma 7, the complex numbers a i,j are common zeros of the polynomials e k (1 ≤ k ≤ 3) defined in (24), and satisfy (5)- (6). This implies that (a 0,1 , a 0,2 , . . . , a 2,3 ) is a common zero of the polynomials Let I be the ideal of R generated by these polynomials. Then we can verify that I contains the polynomial First assume a 1,2 is a zero of the polynomial b 1 . This implies w 1 = w 2 . Let I 1 denote the ideal generated by I and b 1 . Then we can verify that I 1 contains the polynomials X 0,3 + q 2 − 3 and c 1 c 2 , where In particular, If a 0,1 is a zero of the polynomial c 1 , then let I ′ 1 denote the ideal generated by I 1 and c 1 . We can verify that I ′ 1 contains the polynomial X 1,3 − 2. Hence w 1 = w 2 = w 3 , and we have Case (i).
If a 0,1 is a zero of the polynomial c 2 , then Let I ′′ 1 denote the ideal generated by I 1 and c 2 . We can verify that Therefore, by (9), (30), (31), (32), we obtain This gives Case (ii). Next assume a 1,2 is a zero of the polynomial b 2 . Let I 2 denote the ideal generated by I and b 2 . Then we can verify that I 2 contains the polynomials X 0,2 + 2, X 1,3 − 2, and (q 2 − 4)X 0,1 − 2(q 2 − 6). Hence Next assume a 1,2 is a zero of the polynomial b 3 . Let I 3 denote the ideal generated by I and b 3 . Then we can verify that I 3 contains the polynomials X 0,3 − 2 and c 4 c 5 , where In particular, w 3 = 1.
If a 0,2 is a zero of the polynomial c 4 , then Let I ′ 3 denote the ideal generated by I 3 and c 4 . We can verify that I ′ 3 contains the polynomial X 0,1 − 2. Hence w 1 = w 3 = 1, w 2 + 1/w 2 + 2(q 2 − 2)/q 2 = 0, and we have Case (iv). If a 0,2 is a zero of the polynomial c 5 , then Let I ′′ 3 denote the ideal generated by I and c 5 . We can verify that I ′′ 3 contains the polynomials qX 0,1 + 2 and q 2 X 1,2 + 2(q 2 − 2). Thus Since Next assume a 1,2 is a zero of the polynomial b + 4 b − 4 . Without loss of generality, we may assume a 1,2 is a zero of the polynomial b + 4 . Let I 4 denote the ideal generated by I and b + 4 . Then we can verify that I 4 contains the following polynomials: Therefore, we have Case (vi) by (9). Moreover, we can verify that I 4 contains the polynomial X 0, by Lemma 5. Since the ideal generated by I 4 and the polynomial is trivial, we conclude w 1 w 2 + w 3 = 0. Conversely, assume that w 1 , w 2 , and w 3 are given in Theorem 2. Then, we show that the matrix W given in (29) is a type-II matrix. To do this, form Lemma 7 we only check that a defined by (10) is a zero of the polynomials (24), (4), (5), and (6).
Next assume w 1 , w 2 , and w 3 are as in Case (iii). Then, by (10) we have and this is a zero of the polynomials (24), (4), (5), and (6). Hence W is a type-II matrix. Next assume w 1 , w 2 , and w 3 are as in Case (iv). Then, by (10) we have and this is a zero of the polynomials (24), (4), (5), and (6). Hence W is a type-II matrix.
Next assume w 1 , w 2 , and w 3 are as in Case (v). Then, by (10) we have and this is a zero of the polynomials (24), (4), (5), and (6). Hence W is a type-II matrix.
Lastly assume w 1 , w 2 , and w 3 are as in Case (vi). Then a is a zero of the polynomials (24), (4), (5), and (6). Hence W is a type-II matrix.

Corollary 1. Let W be a type-II matrix in Theorem 2. Then, W is a complex Hadamard matrix if and only if W is given in
Proof. Since q ≥ 4, we have n ≥ 15. Hence, |w 3 | = 1 for Cases (i) and (ii). For Cases (iii), (iv) and (v), it is easy to see that Therefore, by Lemma 4 the Cases (iii), (iv) and (v) give complex Hadamard matrices. Lastly, we consider the Case (vi). If r > 0, then we have 0 < a 0,1 < 2. In fact, since r > 3q, we have a 0,1 > 0. Also, Thus, we have 0 < a 0,1 < 2. By Lemma 7, we have |w i | = 1 for i = 1, 2, 3. Therefore, W is a complex Hadamard matrix.
If r < 0, then This implies |w 1 | = 1. Therefore, W is not a complex Hadamard matrix.
Chan [4], found three complex Hadamard matrices on the line graph of the Petersen graph. This is the 3-class association scheme with the first eigenmatrix (26), where q = 1, and the three matrices can be described as the matrix W in (29) with w 1 , w 2 , w 3 given as follows.
The complex Hadamard matrix of order 15 constructed in Theorem 2 (vi) seems to be new. This is obtained by by setting q = 4 and r = √ 201, and has coefficients w 1 , w 2 , w 3 = −w 1 w 2 , where We have verified using the span condition [13,Proposition 4.1] that, this matrix, as well as the one given by (41) are isolated, while the two matrices given by (42) and (43) do not satisfy the span condition.
Remark 1. None of the complex Hadamard matrices given in Corollary 1 is a Butson-Hadamard matrix, that is, entries are roots of unity. This follows from the fact that a i,j is not an algebraic integer for some i, j. Indeed, for the Case (iii), is not an integer.
As for the Case (v), a 0,1 = −2 q is not an integer.
Finally, for the Case (vi), first suppose that r is irrational. If a 0,1 is an algebraic integer, then so is But then a 0,1 + a ′ 0,1 = −1 + 4q + 2 q(q + 1) is an integer, which is absurd.

Equivalence
For a type-II matrix W of order n, the Haagerup set H(W ) (see [10]) is defined as We also define It should be remarked that, although H(W ) is an invariant, none of H i (W ) (i = 2, 3, 4) is.
where |{x 1 , x 2 , y 1 , y 2 }| = 3. If x 1 = x 2 , then w = 1. If x 1 = y 1 , then w = (w i w j /w k ) −1 , where (x 2 , x 1 ) ∈ R i , (x 1 , y 2 ) ∈ R j , (x 2 , y 2 ) ∈ R k , and 1 ≤ i, j, k ≤ d. All other cases can be treated in a similar manner, and we conclude that H 3 (W ) is contained in the right-hand side. Arguing in the opposite way, we obtain the reverse containment.

Lemma 10. Suppose that there exists
Proof. Clearly, H 4 (W ) \ {1} is contained in the right-hand side, so we show the reverse containment. Setting ∆ = {1, . . . , d − 1} in Lemma 9, we have It remains to show that Pick (x 1 , x 2 ) ∈ R i . Then by the assumption, there exists y 1 such that (x 1 , y 1 ) ∈ R i 1 and (x 2 , y 1 ) ∈ R j 1 . Similarly, there exists y 2 such that (x 1 , y 2 ) ∈ R j and (x 2 , y 2 ) ∈ R d . Now Replacing i 1 by d in the above argument gives Below, we determine the Haagerup set of the type-II matrices given in Theorem 2. In what follows, let X = (X, {R i } 3 i=0 ) be an association scheme with the first eigenmatrix (26), where q is an even positive integer with q ≥ 4. The intersection numbers of X are given by where B h has (i, j)-entry p j

a type-II matrix belonging to the Bose-Mesner algebra of X. Then
Proof. Note that H 3 (W ) is determined by Lemma 8. We can also easily see that the assumptions of Lemma 10 are satisfied by setting i = 2. This gives H 4 (W ) \ {1}. Since H 1 (W ) = {1} ⊂ H 4 (W ), we have the desired expression for H(W ).
Using Lemma 11, we can determine the Haagerup set H(W ) for each type-II matrix given in Theorem 2. Note that the description of H(W ) in Table 1 is valid for all even q ≥ 4, even though p 3 11 = 0 for q = 4. The elements of H(W ) given in Table 1 can be found as follows: As for the Case (i), K(W ) has two elements As for (ii), setting This implies 1 The Cases (iii) and (iv) are immediate. Finally, it is clear that K(W ) contains − 2 q and −2, in the Cases (v) and (vi), respectively. We do not need the remaining elements of K(W ) to prove the following propositions. Proposition 1. Let W 1 , . . . , W 6 be type-II matrices given in (i)-(vi) of Theorem 2, respectively. Then W 1 , . . . , W 6 are pairwise inequivalent.
Proposition 2. Let W + and W − be type-II matrices given in Theorem 2 (vi) with r > 0 and r < 0, respectively. Then W + and W − are inequivalent.
Proof. By Corollary 1, W + is a complex Hadamard matrix. Thus every element of H(W + ) has absolute value 1, and hence K(W + ) is contained in the interval [−2, 2]. On the other hand, the element does not belong to the interval [−2, 2]. Thus K(W + ) = K(W − ).
We were able to use the Haagerup set to distinguish some of the complex Hadamard matrices in Theorem 2. This is because the Haagerup set can be described by the intersection numbers of the association scheme, and is independent of the isomorphism class. In general, if q ≥ 8 is a power of 2, there may be many non-isomorphic association schemes with the eigenmatrix (26). We do not know whether complex Hadamard matrices having the same coefficients are equivalent if they belong to Bose-Mesner algebras of non-isomorphic association schemes.
Note that there are two type-II matrices described in Theorem 2(i), since w 1 = w 2 = w 3 is any one of the two zeros of a quadratic equation. Similarly, there are two type-II matrices in each of (ii)-(v) in Theorem 2. Moreover, there are four type-II matrices in (vi), since a 2 0,1 − 4 = 0. The following lemma shows that the two type-II matrices in Theorem 2(i) are inequivalent, and so are those in Theorem 2(ii).
Lemma 12. Let W and W ′ be type-II matrices belonging to the Bose-Mesner algebra of an association scheme X = (X, {R i } d i=0 ). Suppose W and W ′ have d + 1 distinct entries, the valencies of X are pairwise distinct, and min{p i 11 | 0 < i ≤ d} > |X| 2 . If W and W ′ are type-II equivalent, then W is a scalar multiple of W ′ .
Comparing (x 1 , y)-and (x 2 , y)-entries of (48), we find Since W has d + 1 distinct entries, we have Since W ′ has d + 1 distinct entries and the valencies are pairwise distinct, we obtain dd ′ w i = w ′ i . Hence dd ′ W = W ′ . For a matrix W with nonzero complex entries, we denote its entrywise inverse by W (−) .

Proposition 3. Let W be a type-II matrix given in (i) of Theorem 2.
Then W and W (−) are inequivalent. The same conclusion holds if W is a type-II matrix given in (ii) of Theorem 2.
Proof. First, let W be a type-II matrix given in (i) of Theorem 2. Then the matrices W and W (−) belong to the Bose-Mesner algebra of the association scheme (X, {R 0 , R 1 ∪ R 2 ∪ R 3 }). Since W has two distinct entries and this association scheme satisfies the hypothesis of Lemma 12, are inequivalent.
If W is a type-II matrix given in (ii) of Theorem 2, then W and W (−) belong to the Bose-Mesner algebra of the association scheme (X, {R 0 , R 1 ∪ R 2 , R 3 }). Since W has three distinct entries and this association scheme satisfies the hypothesis of Lemma 12, are inequivalent.
We do not know whether the two type-II matrices in each of (iii)-(v) in Theorem 2 are equivalent or not, and whether the two type-II matrices in Theorem 2(vi) with a given sign for r are equivalent or not.

Nomura algebras
Since q 2 − 1 is a composite, there are uncountably many inequivalent complex Hadamard matrices of order q 2 − 1 by [7]. Indeed, such matrices can be constructed using generalized tensor product [11]. We show that none of our complex Hadamard matrices is equivalent to a generalized tensor product. This is done by showing that the Nomura algebra of any of our complex Hadamard matrices has dimension 2. According to [11], the Nomura algebra of the generalized tensor product of type-II matrices is imprimitive, and this is never the case when it has dimension 2.
For a type-II matrix W ∈ M X (C) and a, b ∈ X, we define column vectors Y ab by setting The Nomura algebra N(W ) of W is the algebra of matrices in M n (C) such that Y ab is an eigenvector for all a, b ∈ X. It is shown in [12,Theorem 1] that the Nomura algebra is a Bose-Mesner algebra. We first show that the Nomura algebra consists of symmetric matrices. Throughout this section, we let N to denote the Nomura algebra of any of the type-II matrices given in Theorem 2.
Lemma 13. The algebra N is symmetric.
Proof. Suppose that N is not symmetric. Then by [12,Proposition 6(i) for some i ∈ {1, 2, 3}. Using the notation (10), we have It can be verified by computer that (49) is nonzero for each of the cases (i)-(vi) in Theorem 2.
Since N is symmetric, the adjacency matrices of N are the (0, 1)matrices representing the connected components of the Jones graph defined as follows (see [12,Sect. 3.3]). The Jones graph of a type-II matrix W ∈ M X (C) is the graph with vertex set X 2 such that two distinct vertices (a, b) and (c, d) are adjacent whenever Y ab , Y cd = 0, where , = 0 denotes the ordinary (not hermitian) scalar product. Proposition 4. The algebra N coincides with the linear span of I and J. In particular, the none of type-II matrices given in Theorem 2 is equivalent to a nontrivial generalized tensor product.
Proof. We aim to show that {(x, y) | x, y ∈ X, x = y} is a connected component of the Jones graph. Fix x, y, z with (x, y), (y, z), (z, x) ∈ R 3 . Then in the Jones graph, (x, y) and (y, x) belong to the same connected component, and (x, z) and (z, x) belong to the same connected component, by Lemma 13. We claim that (x, y) and (x, z) belong to the same connected component. Indeed, if (x, y) and (x, z) belong to different connected components, then (y, x) and (z, x) belong to different connected components. In particular, Then for j, k ∈ {1, 2}, c 1,j,k + c 2,j,k = c j,1,k + c j,2,k = c j,k,1 + c j,k,2 = p 3 jk , It can be verified by computer that these conditions give rise to a polynomial equation in q which has no solution in even positive integers. Therefore, we have proved the claim. This implies that, for each equivalence class C of the equivalence relation R 0 ∪ R 3 , (C × C) ∩ R 3 is a clique in the Jones graph.
It can be verified by computer that this leads to a polynomial equation in q which has no solution in even positive integers.
Similarly, for (x, z) ∈ R 2 , suppose Y xy , Y xz = 0 for all y ∈ R 3 (x). Then 3 i,j,k=0 and again this leads to a contradiction. Therefore, for any (x, z) ∈ R 1 ∪ R 2 , there exists y ∈ R 3 (x) such that Y xy , Y xz = 0. Since R 1 ∪ R 2 is connected, we see that R 1 ∪ R 2 ∪ R 3 is a connected component of the Jones graph. Therefore, dim N = 2.