1 Introduction

The concept of equivalence—or isomorphism, depending on the setting—is central in the study of mathematical structures. Equivalence is essentially about dealing with ambient symmetries and leads to a partitioning of structures into equivalence classes. For example, the aim of a classification of structures is to find a transversal of the equivalence classes. The equivalence mappings from a structure onto itself give the symmetries of the structure (and form a group under composition).

The definition of equivalence depends on the studied structures, whose main properties should be respected. For some structures, there is one obvious definition, but for others the situation may be more involved. In any case, it is generally desirable to have definitions that lead to as large equivalence classes as possible.

Hadamard and related matrices are considered in this paper; for more information about such structures, [2, 4, 7] can be consulted. See also [8, 9, 11,12,13,14] for specific classification studies. Two types of equivalence are commonly considered for Hadamard and related matrices: monomial equivalence and Hadamard equivalence. It is not obvious whether the set of Butson-type Hadamard matrices is closed under Hadamard equivalence. We shall show in this paper that this is indeed the case.

Equivalence of Hadamard and related matrices is considered in Sect. 2, and the main proof regarding Hadamard equivalence of Butson-type matrices is given in Sect. 3. Throughout the paper, the group operation is identified with multiplication.

2 Equivalence of Hadamard matrices

A (real) Hadamard matrix of order n is an \(n\times n\) matrix \(H = (h_{i,j})\) with entries from the group \((\{-1,1\},\cdot )\) such that for any distinct \(1 \le a,b \le n\), the multiset \(\{h_{a,k}h_{b,k}^{-1} : 1 \le k \le n\}\) contains \(-1\) and 1 equally often (n/2 times each). (Here, \(h = h^{-1}\) for both elements of the group, but the definition is given in the particular form for the sake of a later generalization.) It is a classical result that a necessary condition for a Hadamard matrix of order n to exist is that \(n=1,2\) or n is divisible by 4. It is further conjectured that this is also a sufficient condition; proving this conjecture is the main open question in the theory of Hadamard and related matrices.

A generalized Hadamard matrix of order \(m\lambda \) and index \(\lambda \) is an \(m\lambda \times m\lambda \) matrix \(H = (h_{i,j})\) with entries from a group G of order m such that for any distinct \(1 \le a,b \le m\lambda \), the multiset

$$\begin{aligned} \{h_{a,k}h_{b,k}^{-1} : 1 \le k \le m\lambda \} \end{aligned}$$
(1)

contains each element of G equally often (\(\lambda \) times each). A comparison with the definition of real Hadamard matrices given earlier shows that real Hadamard matrices are generalized Hadamard matrices of order n and index n/2.

To consider equivalence of Hadamard and related matrices, we define the monomial group \({\mathcal {G}}_n\) as the matrix group that is the subgroup of \(\text{ GL}_n({\mathbb {Z}}G)\) whose elements are the matrices in which each row and column contains exactly one element of G. We further let the group \({\mathcal {G}}_n \times {\mathcal {G}}_n\) act on an \(n \times n\) matrix M by \((X,Y)\cdot M = XMY^{\mathrm {T}}\).

Definition 1

Two \(n\times n\) matrices with entries from a group G, \(H_1\) and \(H_2\), are said to be monomially equivalent, denoted by \(H_1\cong H_2\), if \(H_1 = XH_2Y^{\mathrm {T}}\) for some \((X,Y) \in {\mathcal {G}}_n \times {\mathcal {G}}_n\).

In other words, two matrices are monomially equivalent if one can be obtained from the other using row and column permutations and, respectively, left- and right-multiplication of rows and columns with elements from G. Such left- and right-multiplications permute the values in entries of individual rows and columns, respectively, and those permutions are called local equivalence operations in [5, Sect. 4.4.1]. If the group G is nonabelian, we get two different sets of permutations, but if the group G is abelian, the order of multiplication does not matter, and we get only one.

The set of generalized Hadamard matrices is closed under monomial equivalence; see [6, Theorem 1.16] for a proof.

In addition to local equivalence operations, one may also consider global equivalence operations [5, Sect. 4.4.3], which are permutations of the elements of the group (in all entries) under which a given class of matrices is closed. The term ’global equivalence operations’ is used in [5] only for those permutations that are not local equivalence operations, but the author of the current paper finds it more natural to let the local equivalence operations form a subset of the global equivalence operations. As shown in [5, Theorem 4.4.10], for many classes of Hadamard and related matrices and with an abelian group G, the global equivalence operations are generated by the local equivalence operations and the group automorphisms, \(\mathrm {Aut}(G)\).

We write f(H) for the matrix obtained by applying the function f to each entry of H.

Definition 2

Two \(n\times n\) matrices with entries from a group G, \(H_1\) and \(H_2\), are said to be Hadamard equivalent if \(H_1 \cong \phi (H_2)\) for some \(\phi \in \mathrm {Aut}(G)\), that is, \(H_1 = \phi (XH_2Y^{\mathrm {T}})\) where \((X,Y) \in {\mathcal {G}}_n \times {\mathcal {G}}_n\).

Definition 2 is [7, Definition 4.12]. For some matrix classes, it is easy to show that they are closed under the mapping \(\phi (H)\). For example, for generalized Hadamard matrices, the elements of (1) become

$$\begin{aligned} \phi (h_{a,k})\phi (h_{b,k})^{-1} = \phi (h_{a,k})\phi (h_{b,k}^{-1}) = \phi (h_{a,k}h_{b,k}^{-1}), \end{aligned}$$
(2)

and because \(\phi \) is a bijection, the multiset remains unchanged. In the real case, \(|\mathrm {Aut}(G)| = 1\), and monomial and Hadamard equivalence coincide. Hence, there is no conflict between our definitions and the fact that both terms—and just equivalence—have been used for real Hadamard matrices in the literature.

Another generalization of real Hadamard matrices are complex Hadamard matrices. A complex Hadamard matrix of order n is an \(n\times n\) matrix \(H = (h_{i,j})\) with entries that are unit complex numbers, that is, lie on the unit circle in the complex plane, such that

$$\begin{aligned} HH^*=nI_n, \end{aligned}$$
(3)

where \(H^*\) denotes the conjugate transpose of H, \(I_n\) is the \(n\times n\) identity matrix, and operations are carried out in the field \({\mathbb {C}}\). This is the same as saying that for any distinct \(1 \le a,b \le n\),

$$\begin{aligned} \sum _{k=1}^n h_{a,k}h_{b,k}^{-1} = 0, \end{aligned}$$
(4)

which makes a comparison with (1) easier.

The set of entries of complex Hadamard matrices forms an infinite group that is isomorphic to the circle group \(S^1\). General complex Hadamard matrices are of interest in physics [19], whereas in discrete mathematics, the main focus has been on matrices with entries from a finite subgroup. In fact, every finite subgroup of \({\mathbb {C}}^*\) is a (cyclic) group of roots of unity. A Butson-type Hadamard matrix of order n over mth roots of unity is a complex Hadamard matrix of order n with \(h^m=1\) for every entry h.

It is an easy exercise to show that the sets of complex and Butson-type Hadamard matrices are closed under monomial equivalence but what about Hadamard equivalence?

A basic fact for the circle group is that \(\mathrm {Aut}(S^1)\) is isomorphic to the unique group of order 2 (cf. [1, Ch. 4, Problem 25]), and the nontrivial element of \(\mathrm {Aut}(S^1)\) corresponds to complex conjugation here. Complex conjugation of H is denoted by \({\overline{H}}\). To verify that the set of complex Hadamard matrices is closed under conjugation, (3) can be used to get

$$\begin{aligned} {\overline{H}}\;{\overline{H}}^*={\overline{H}}\;\overline{H^*}=\overline{HH^*}=\overline{nI_n}=nI_n. \end{aligned}$$

So the only difference between monomial equivalence and Hadamard equivalence for complex Hadamard matrices is the inclusion of complex conjugation in the latter case. Complex conjugation indeed plays a central role in the study of complex Hadamard matrices [18].

For Butson-type Hadamard matrices over mth roots of unity, on the other hand, the entries are from a cyclic group of order m, that is, a group isomorphic to \(C_m\). By letting \(C_m = \langle g : g^m =e \rangle \), the elements of \(\mathrm {Aut}(C_m)\) are given by the functions

$$\begin{aligned} f(x) = x^k,\quad \gcd (k,m)=1, 1 \le k \le m. \end{aligned}$$
(5)

Hence, \(|\mathrm {Aut}(C_m)| = \varphi (m)\), where \(\varphi (m)\) is Euler’s totient function.

In the literature, studies of equivalence of Butson-type Hadamard matrices over mth roots of unity have mainly considered the case \(m=4\); see [13, 14]. We have \(|\mathrm {Aut}(C_4)|=2\), and the only nontrivial element corresponds to complex conjugation. Butson-type Hadamard matrices form a subset of complex Hadamard matrices, and we have seen that the set of such matrices is closed under complex conjugation.

However, there are studies of equivalence of Butson-type Hadamard matrices over mth roots of unity for larger values of m. One such example is [12], co-authored by the current author, where Hadamard equivalence is used without justifications. Those missing justifications will now be provided.

3 Hadamard equivalence of Butson-type matrices

Everything would be straightforward if (5) would be automorphisms of \({\mathbb {C}}\) for all admissible values of k, but this is not the case as only \(k=1\) (identity) and \(k=m-1\) (conjugation) give such automorphisms. However, considering cyclotomic fields \({\mathbb {Q}}(\zeta )\), where \(\zeta \) is a primitive mth root of unity, rather than \({\mathbb {C}}\) makes it possible to prove the main theorem in a direct manner.

Theorem 1

The set of Butson-type Hadamard matrices is closed under Hadamard equivalence.

Proof

Because the set of Butson-type Hadamard matrices is closed under monomial equivalence, it remains to consider the case of applying a group automorphism to each entry of the matrix. Let \(H = (h_{i,j})\) be a Butson-type Hadamard matrix of order n over mth roots of unity. By (2), we are done if we are able to show that \(\sum _{k=1}^n h_{a,k}h_{b,k}^{-1} = 0\) implies that \(\sum _{k=1}^n \phi (h_{a,k}h_{b,k}^{-1}) = 0\), for any \(\phi \) given by (5).

Let \(\zeta \) be a primitive mth root of unity. It is well known that \(\text{ Gal }({\mathbb {Q}}(\zeta )/{\mathbb {Q}})\) has order \(\varphi (m)\) and that the automorphisms map the mth roots of unity by a function (5); see, for example, [20, Theorem 2.5]. A function \(\phi \) given by (5) and whose domain is the mth roots of unity determines the entire automorphism \(\phi '\) of \({\mathbb {Q}}(\zeta )\). So in \({\mathbb {Q}}(\zeta )\),

$$\begin{aligned} \sum _{k=1}^n\phi (h_{a,k}h_{b,k}^{-1}) = \sum _{k=1}^n\phi '(h_{a,k}h_{b,k}^{-1}) = \phi '\left( \sum _{k=1}^n h_{a,k}h_{b,k}^{-1}\right) = \phi '(0) = 0. \end{aligned}$$

\(\square \)

We shall next give another, longer but more constructive, proof of the same result. We will need a few lemmas and theorems, some of which—especially Lemma 1—are well known but proofs are included for completeness.

Lemma 1

Let \(m \ge 2\). If \(\zeta \) is a primitive mth root of unity, then \(\sum _{i=0}^{m-1} \zeta ^i = 0\).

Proof

As \(\zeta \) is an mth root of unity, \(\zeta ^m=1\), that is, \(\zeta ^m-1\) = 0. We may now write

$$\begin{aligned} 0 = \zeta ^m-1 = (\zeta -1)(1+\zeta +\zeta ^2+\cdots + \zeta ^{m-1}), \end{aligned}$$

and the result follows as 1 is not a primitive mth root of unity, so \(\zeta \ne 1\). \(\square \)

Lemma 2

The function (5) maps a qth root of unity to a qth root of unity, for any q.

Proof

If \(z^q = 1\), then \((z^k)^q = (z^q)^k = 1^k = 1\). \(\square \)

Lemma 3

Let G be the group of mth roots of unity (\(m \ge 2\)), let \(\phi \in \mathrm {Aut}(G)\), let p be a prime factor of m, and let \(\zeta \) be a primitive mth root of unity. Then, \(\sum _{i=0}^{p-1}\phi (\zeta ^{im/p+j}) = 0\) for any integer j.

Proof

We get

$$\begin{aligned} \begin{aligned} \sum _{i=0}^{p-1}\phi (\zeta ^{im/p+j})&= \sum _{i=0}^{p-1}\phi (\zeta ^j\zeta ^{im/p}) = \sum _{i=0}^{p-1}\phi (\zeta ^j)\phi (\zeta ^{im/p})\\&=\phi (\zeta ^j)\sum _{i=0}^{p-1}\phi (\zeta ^{im/p}) = \phi (\zeta ^j)\sum _{i=0}^{p-1}\phi ((\zeta ^{m/p})^i)\\&=\phi (\zeta ^j)\sum _{i=0}^{p-1}\phi ((\zeta ^r)^i) \end{aligned} \end{aligned}$$

by substituting \(r = m/p\). Hence, \(\zeta ^r\) is a pth root of unity, which must be primitive as p is a prime (and \(\zeta ^r \ne 1\) as \(\zeta \) is a primitive mth root of unity). Then, \((\zeta ^r)^i\), \(0 \le i \le p-1\) are precisely the p pth roots of unity. By Lemma 2 and the fact that automorphisms are bijections, the values of \(\phi ((\zeta ^r)^i)\) for \(0 \le i \le p-1\) are also the pth roots of unity. Therefore, \(\sum _{i=0}^{p-1}\phi ((\zeta ^r)^i) = 0\) by Lemma 1. \(\square \)

We need one more result, which has been proved in several different ways in the literature [3, 10, 15,16,17].

Theorem 2

Let \(\zeta \) be a primitive mth root of unity, and, for \(1\le i \le n\), let \(z_i\) be an mth root of unity. Every relation of the form \(\sum _{i=1}^{n}a_iz_i = 0\) can be obtained as a \({\mathbb {Z}}\)-linear combination of relations, where each relation is of the form \(\sum _{i=0}^{p-1} \zeta ^{im/p+j} = 0\) for some prime factor p of m and integer j.

We are now ready for an alternative proof of the main theorem.

Proof of Theorem 1

As in the earlier proof, we show that \(\sum _{k=1}^n h_{a,k}h_{b,k}^{-1} = 0\) implies that \(\sum _{k=1}^n \phi (h_{a,k}h_{b,k}^{-1}) = 0\), for any \(\phi \) given by (5).

By Theorem 2, we may write

$$\begin{aligned} 0 = \sum _{k=1}^n h_{a,k}h_{b,k}^{-1} = \sum _{s=1}^N \left( T_s\sum _{i=0}^{p_s-1} \zeta ^{im/p_s+j_s} \right) , \end{aligned}$$

where \(\zeta \) is a primitive mth root of unity and, for \(1 \le s \le N\), \(T_s\) are integers, \(p_s\) are (not necessarily distinct) prime factors of m, and \(j_s\) are integers. Now consider the linear combination obtained by substituting \(h_{a,k}h_{b,k}^{-1}\) with \(\phi (h_{a,k}h_{b,k}^{-1})\) for all \(1 \le k \le n\):

$$\begin{aligned} \sum _{k=1}^n\phi (h_{a,k}h_{b,k}^{-1}) = \sum _{s=1}^N \left( T_s\sum _{i=0}^{p_s-1} \phi (\zeta ^{im/p_s+j_s}) \right) = \sum _{s=1}^N (T_s\cdot 0) = 0, \end{aligned}$$

utilizing Lemma 3. \(\square \)