Order-Six CHMs Containing Exactly Three Distinct Elements

Yanzu Huang; Mengfan Liang; Lin Chen

doi:10.2478/qic-2026-0001

Intrduction

Mutually unbiased bases (MUBs) are a significant concept in quantum physics. In general, MUBs in Hilbert space ℂ⁶ are orthogonal bases such that the inner product of any two vectors from different bases has a modulus of $\frac{1}{\sqrt{n}}$ {1 \over {\sqrt n }}. When the number of MUBs reaches n + 1, they are referred to as complete MUBs. Complete MUBs exist in ℂⁿ when n is a prime power [1]. The problem of finding complete MUBs in ℂ⁶ is an unsolved case and a well-known open problem in quantum information. Various approaches have been used to study the MUB problem. For instance, paper [2] explored the average distance between four bases in six dimensions, providing strong evidence against the existence of four mutually unbiased bases in ℂ⁶. Paper [3] introduced an infinite family of MUB triplets in dimension 6, demonstrating that this family cannot be extended to complete MUBs. Paper [4] showed that if complete MUBs in dimension 6 exist, they cannot include more than one product basis. Paper [5] examined the number of product vectors in a set of four MUBs in dimension 6, showing that each of the remaining three MUBs contains at most two product vectors. Further research on this topic can be found in [6–15].

The complex Hadamard matrix (CHM) is also an important concept since it is usually related to MUB problem. An n × n matrix H whose elements all have modulus one is called a CHM if HH^† = nI. If a set of four MUBs in ℂ⁶ exists and contains the identity matrix, then any other matrix U in the set satisfies that $\sqrt{6} U$ \sqrt 6 U is a 6 × 6 CHM, and we refer to the set as an MUB trio. The final target for us is to find whether there exists a MUB trio.

The complete classification of 6 × 6 CHMs is also a longstanding open problem. Paper [16] characterized CHMs whose all elements are roots of unity, such as Fourier matrices. Paper [17] introduced the Tao matrix, consisting only of $1, e^{\frac{2 π i}{3}}, e^{\frac{4 π i}{3}}$ 1,{{\rm{e}}^{{{2\pi {\rm{i}}} \over 3}}},{{\rm{e}}^{{{4\pi {\rm{i}}} \over 3}}}, which does not belong to any parameterized family of matrices. Paper [18] presented a method to apply faster algorithms for the homogeneous case to the inhomogeneous case, discovering a family of 6 × 6 CHMs not included in the Butson matrices. In 2011, Karlsson introduced a three-parameter family of CHMs in ℂ⁶[19], termed ”the H₂-reducible matrices.” The most known 6 × 6 CHMs, such as the Haagerup matrix [20], belong to this family, except for the Tao matrix. Paper [21] proposed a four-parameter 6 × 6 CHM family, though its analytic form remains unknown. Additional studies on the classification problem are available in [22–28].

In this paper, we present a novelty approach for studying the classification problem, that is, the investigation of the CHM of order 6 containing only three distinct elements. Prior classifications mainly focus on the amount of parameters, so this paper can complete the prior classifications from a new perspective. We begin with several theorems and lemmas in preliminaries. Then in Theorem 1, we claim that the CHM containing only {1, –1, a} is complex equivalent to Diţă matrix D₀. In Corollary 1, we show that an order-six CHM containing only elements {1, –1, i, –i} is complex equivalent to D₀. We extend Lemma 8 to finish our classification of all H₂-reducible CHM containing only three elements, and the result is that H₂-reducible CHM containing only three elements is complex equivalent to D₀. Next, in Theorem 2, we claim that CHM containing only {1, a, ā} is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)}, the Tao matrix. In Lemma 9 we introduce a method to roughly but efficiently classify all the cases. The core idea of that is to preliminarily examine the real part of the inner product of two rows or columns, and that method will be used throughout all subsequent discussions. In Theorem 3, we claim that CHM containing only {1, a, –ā} does not exist. In Lemma 10 we extend our method to do a brief classification, we define the modified pending terms, which is somewhat different from several terms in the inner product. Theorem 4 shows the main result in this paper, that is, CHM containing only three distinct elements is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)} or D₀. We prove this by the method in Lemma 9 and the skills of group theory. We also define the ”outward-pointing” inner product, which provides the orientation of all inner products. Finally, we examine the two matrices $S_{6}^{(0)}$ S_6^{(0)} and D₀, with the results in [29,30], we claim that CHM containing only three distinct elements does not belong to a MUB trio.

The rest of this paper is organized as follows. In Section 2, we introduce the definitions and facts used in this paper. Then we introduce our main results on CHM containing only three distinct elements and whether such CHM belongs to MUB trios in Section 3. Finally we concluded in Section 4.

Preliminaries

We follow the definition in [31] to define the equivalence and complex equivalence.

Definition 1.

(i) Let the monomial unitary matrix be a unitary matrix each of whose rows and columns has exactly one nonzero entry. The entry has modulus one. Let ℳ_n be the set of n × n monomial unitary matrices.

(ii) Two n × n matrices U and V are complex equivalent when U = PVQ where P, Q ∈ ℳ_n. If P, Q are both permutation matrices then we say that U, V are equivalent.

Remark 1.

The complex equivalence relations defined in Definition 1 cannot keep the condition ”CHMs have exactly three distinct matrix elements” invariant, but the equivalence relations can. We use complex equivalence relations to link our results to the well-known matrices, such as Diţă matrix and Tao matrix.

Remark 2.

We introduce several notations used in this paper. The ”ω” refers to $e^{\frac{2 π i}{3}}$ {e^{{{2\pi i} \over 3}}}. An overline above a letter (representing a complex number) indicates the conjugate of that, for example, $\bar{x}$ {\bar x} is the conjugate of x for x ∈ ℂ.

Lemma 1.

Suppose S is a CHM containing exactly three distinct elements, and all elements of the first row of S are one. Then S is complex equivalent to the Tao matrix, denoted by $S_{6}^{(0)}$ S_6^{(0)}, or the matrix in an equivalent form, also called Butson matrices.1 $S_{6}^{(0)} = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & ω & ω & ω^{2} & ω^{2} \\ 1 & ω & 1 & ω^{2} & ω^{2} & ω \\ 1 & ω & ω^{2} & 1 & ω & ω^{2} \\ 1 & ω^{2} & ω^{2} & ω & 1 & ω \\ 1 & ω^{2} & ω & ω^{2} & ω & 1 \end{matrix}] .$ S_6^{(0)} = \left[ {\matrix{ 1 & 1 & 1 & 1 & 1 & 1 \cr 1 & 1 & \omega & \omega & {{\omega ^2}} & {{\omega ^2}} \cr 1 & \omega & 1 & {{\omega ^2}} & {{\omega ^2}} & \omega \cr 1 & \omega & {{\omega ^2}} & 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^2}} & \omega & 1 & \omega \cr 1 & {{\omega ^2}} & \omega & {{\omega ^2}} & \omega & 1 \cr } } \right].

The above result is from [29]. We want to generalize this result.

Lemma 2.

Suppose S is a CHM containing only {1, ω, ω²}, then S is complex equivalent to the Tao matrix.

Proof.

We can take A ∈ ℳ₆ which contains only {0, 1, ω, ω²} to let HA be a CHM with the property that all elements in row 1 are one. Moreover, HA also contains only {1, ω, ω²}, so Lemma 1 shows that such matrix is complex equivalent to the Tao matrix.

Lemma 3.

If a 6 × 6 CHM X contains a 2 × 3 submatrix with rank one, then X is complex equivalent to the matrix from the following two-parameter family 2 $H (α, β) = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & - 1 & - 1 & - 1 \\ 1 & ω & ω^{2} & α & α ω & α ω^{2} \\ 1 & ω & ω^{2} & - α & - α ω & - α ω^{2} \\ 1 & ω^{2} & ω & β & β ω^{2} & β ω \\ 1 & ω^{2} & ω & - β & - β ω^{2} & - β ω \end{matrix}] .$ H(\alpha ,\beta ) = \left[ {\matrix{ 1 & 1 & 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & { - 1} & { - 1} & { - 1} \cr 1 & \omega & {{\omega ^2}} & \alpha & {\alpha \omega } & {\alpha {\omega ^2}} \cr 1 & \omega & {{\omega ^2}} & { - \alpha } & { - \alpha \omega } & { - \alpha {\omega ^2}} \cr 1 & {{\omega ^2}} & \omega & \beta & {\beta {\omega ^2}} & {\beta \omega } \cr 1 & {{\omega ^2}} & \omega & { - \beta } & { - \beta {\omega ^2}} & { - \beta \omega } \cr } } \right].

Lemma 4.

Suppose a 6 × 6 CHM H contains a submatrix $[\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & ω & ω & ω^{2} & ω^{2} \end{matrix}]$ \left[ {\matrix{ 1 & 1 & 1 & 1 & 1 & 1 \cr 1 & 1 & \omega & \omega & {{\omega ^2}} & {{\omega ^2}} \cr } } \right] or $[\begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & ω \\ 1 & ω \\ 1 & ω^{2} \\ 1 & ω^{2} \end{matrix}]$ \left[ {\matrix{ 1 & 1 \cr 1 & 1 \cr 1 & \omega \cr 1 & \omega \cr 1 & {{\omega ^2}} \cr 1 & {{\omega ^2}} \cr } } \right] where $ω = e^{\frac{2 π i}{3}}$ \omega = {e^{{{2\pi i} \over 3}}}. Then H is complex equivalent to the Tao matrix or the matrix from the two-parameter family H(α, β) in Lemma 3.

Lemmas 3 and 4 were obtained in [29].

Lemma 5.

Suppose |a| = |b| = 1. We have

(i)
a + b is real if and only if $a = \bar{b}$ a = \bar b or a = –b;
(ii)
ab is real if and only if $a = \bar{b}$ a = \bar b or $a = - \bar{b}$ a = - \bar b;
(iii)
a + b + ab is real if and only if $a = \bar{b}$ a = \bar b or a = –1 or b = –1;
(iv)
$a + b + a \bar{b}$ a + b + a\bar b is real if and only if a = –b or a = 1 or b = –1;
(v)
$a + b + \bar{a} \bar{b}$ a + b + \bar a\bar b is real if and only if $a = \bar{b}$ a = \bar b or a = 1 or b = 1.
(vi)
if also |c| = |d| = 1 and a + b = c + d, then we have a = c or a = d.

Proof.

All these results can be proven easily by assuming a = cos θ₁ + i sin θ₁, b = cos θ₂ + i sin θ₂ and considering the coefficient of i. Moreover, we can let c = –a, d = –b, so a + b + ab = –c – d + cd, now c + d – cd is real if and only if $a = \bar{b}$ a = \bar b or a = –1 or b = –1, that is, $c = \bar{d}$ c = \overline d or c = 1 or d = 1. The same skill can show us the condition to let $a + b - a \bar{b}$ a + b - a\bar b, $a + b + \bar{a} \bar{b}$ a + b + \bar a\bar b be real.

Finally we introduce a lemma which shows some properties of MUB trios. This lemma is the result from [29].

Lemma 6.

An MUB trio does not contain a CHM which has a 3 × 3 Hadamard submatrix.

Results

In this section, we will take a complete classification of CHMs containing only three distinct elements. We will first classify the following three cases clearly and use the results to take a complete classification of CHMs containing only three distinct elements. Here the three cases are CHMs containing three distinct elements {1, – 1, a}, {1, a, ā} or {1, a, –ā}. The results of these cases play a significant role in the classification of CHMs containing only {1, a, b}, which is the general case. The connections of them can be found at the beginning of Appendix E. Before that we will give a useful lemma to simplify our discussion.

Lemma 7.

There does not exist a CHM H of order six containing only {1, a, b} which is complex equivalent to H (α, β) in Lemma 3.

Proof.

If we do multiplications by diagonal matrices in ℳ₆, it is easy to find that for any two rows, the inner product of them before and after the multiplications will be the same, in the sense of multiplying by a complex number of modulus 1.

If b = – 1, since the inner product of the columns 4-6 of rows 1,2 in H(α, β) is -3, so that each column in columns 4-6 of the first two rows of H must be the same. We shall assume the columns 4-6 of row 1 to be [1, 1, 1]. And the inner product of the columns 4-6 of rows 1,3 is α (1 + ω + ω²) = 0, which shows the columns 3-6 of row of that CHM contains three distinct elements. But that CHM contains only {1, –1, a}, and 1 –1 + ā ≠ 0 shows a contradiction. We can do similar discussion for a = – 1 or a = –b. In the latter case we can consider āH which contains only {1, – 1, ā}.

If a, b ≠ – 1 and a ≠ –b, the only possible H₂ matrices containing only {1, a, b} are 3 $[\begin{array}{l} 1 & a \\ b & 1 \end{array}], [\begin{array}{l} a & 1 \\ b & a \end{array}], [\begin{array}{l} b & 1 \\ a & b \end{array}] .$ \left[ {\matrix{ 1 \hfill & a \hfill \cr b \hfill & 1 \hfill \cr } } \right],\left[ {\matrix{ a \hfill & 1 \hfill \cr b \hfill & a \hfill \cr } } \right],\left[ {\matrix{ b \hfill & 1 \hfill \cr a \hfill & b \hfill \cr } } \right]. in the sense of exchanging their rows and columns. We can obtain three relations of a, b from the three matrices, they are b = –a, b = –a², a = –b². Since the first two rows of H(α, β) contains exactly 3 disjoint H₂ matrices, if they are all the same, since the inner product of the columns 4-6 of rows 1,2 in H(α, β) is -3, then we shall assume the columns 4-6 of rows 1,2 of H to be [1, a]. Consider the inner product of columns 4-6 of rows 2,3, we have 1 + a + b = 0. But the equation has no common solutions with one of the three relations above, so we deduce the contradiction. If the three H₂ submatrices contain no less than two of the three H₂ matrices listed above, we can solve {a, b} = {ω, –ω²} or {–ω, ω²}. By taking conjugate of H, we only need to consider {a, b} = {ω, –ω²}. We shall assume a ≠ –1, and the inner product of rows 1,3 of H(α, β) is (1 + a)(1 + ω + ω²) = 0 (if a = –1 we can consider the inner product of rows 1,4) and for x, y ∈ {1, ω, –ω²} we have xȳ ∈ {1, ±ω, ±ω²}, so the rows 1,3 of matrix containing only {1, ω, –ω²} must be 4 $[\begin{matrix} 1 & 1 & 1 & 1 & ω & ω \\ 1 & 1 & ω & ω & 1 & 1 \end{matrix}] .$ \left[ {\matrix{ 1 & 1 & 1 & 1 & \omega & \omega \cr 1 & 1 & \omega & \omega & 1 & 1 \cr } } \right]. in the sense of exchanging columns. However, there are three disjoint H₂ matrices in rows 1,2 of the matrix, and they are all from the matrices in (3). Since the first row of such three H₂ matrices must all contain two distinct elements, but the first row of the matrix contains 1 four times, hence we deduce the contradiction.

The result of Lemma 7 can be used in the proof of the following theorem, which gives the complete classification of CHMs containing only {1, – 1, a}.

3.1.

CHM Containing Only {1, –1, a}

For {1, –1, a}, we prove a brief conclusion.

Theorem 1.

Suppose H is a CHM of order 6 containing only {1, – 1, a}. Then H is complex equivalent to 5 $H^{(1)} = [\begin{matrix} i & 1 & 1 & 1 & 1 & 1 \\ 1 & i & 1 & 1 & - 1 & - 1 \\ 1 & 1 & i & - 1 & 1 & - 1 \\ 1 & 1 & - 1 & i & - 1 & 1 \\ 1 & - 1 & 1 & - 1 & i & 1 \\ 1 & - 1 & - 1 & 1 & 1 & i \end{matrix}] .$ {H^{(1)}} = \left[ {\matrix{ i & 1 & 1 & 1 & 1 & 1 \cr 1 & i & 1 & 1 & { - 1} & { - 1} \cr 1 & 1 & i & { - 1} & 1 & { - 1} \cr 1 & 1 & { - 1} & i & { - 1} & 1 \cr 1 & { - 1} & 1 & { - 1} & i & 1 \cr 1 & { - 1} & { - 1} & 1 & 1 & i \cr } } \right].

Here a = i. Moreover, H⁽¹⁾ is complex equivalent to Diţă matrix 6 $D_{0} = [\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & - 1 & i & - i & - i & i \\ 1 & i & - 1 & i & - i & - i \\ 1 & - i & i & - 1 & i & - i \\ 1 & - i & - i & i & - 1 & i \\ 1 & i & - i & - i & i & - 1 \end{matrix}]$ {D_0} = \left[ {\matrix{ 1 & 1 & 1 & 1 & 1 & 1 \cr 1 & { - 1} & i & { - i} & { - i} & i \cr 1 & i & { - 1} & i & { - i} & { - i} \cr 1 & { - i} & i & { - 1} & i & { - i} \cr 1 & { - i} & { - i} & i & { - 1} & i \cr 1 & i & { - i} & { - i} & i & { - 1} \cr } } \right]

Notice that for permutation matrix 7 $P = [\begin{matrix} - i & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{matrix}], Q = [\begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & i \\ 0 & 0 & 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 & i & 0 \end{matrix}],$ P = \left[ {\matrix{ { - i} & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr } } \right],\quad Q = \left[ {\matrix{ 1 & 0 & 0 & 0 & 0 & 0 \cr 0 & i & 0 & 0 & 0 & 0 \cr 0 & 0 & i & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & i \cr 0 & 0 & 0 & i & 0 & 0 \cr 0 & 0 & 0 & 0 & i & 0 \cr } } \right], we have P · H⁽¹⁾ Q = D₀, then H⁽¹⁾ is indeed complex equivalent to D₀. Since H⁽¹⁾ contains only {1, –1, i}, while D₀ contains four distinct elements, we prefer to choose H⁽¹⁾ as the representative in the following.

More details of the proof will be provided in Appendix A.

Corollary 1.

Suppose H is an order-six CHM containing only {1, – 1, i, – i}. Then H is complex equivalent to H⁽¹⁾ in (5).

Proof.

Since we can do the right multiplication to let all the elements in row 1 in H be 1, and the elements in other rows are in {1, –1, i, – i}, that is just the case we have discussed before.

Corollary 2.

Suppose H is a 6 × 6 CHM containing only {1, a, –a}, then H is complex equivalent to H⁽¹⁾ in (5).

Proof.

Noting that a⁻¹ H is a CHM containing only {1, –1, a⁻¹}, by Theorem 1 we have finished our proof.

Based on the previous results, we introduce a lemma which can immensely simplify our discussion in the latter subsections.

Lemma 8.

If an H₂-reducible CHM of order 6 H containing only three distinct elements {1, a, b}, then it is complex equivalent to H⁽¹⁾ in (5). Moreover, a = –b or –1 ∈ {a, b}.

The proof will be provided in Appendix B. This lemma indicates that all H₂-reducible CHMs of order 6 containing only three distinct elements are complex equivalent to H⁽¹⁾ due to Theorem 1 and Corollary 2.

3.2.

CHM Containing Only {1, a, ā}

In this subsection we prove the following theorem.

Theorem 2.

Suppose H is a CHM of order 6 containing only {1, a, ā}, then H is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)} or H⁽¹⁾ in Lemma 2 and Theorem 1.

Proof.

We will prove that by analyzing several special cases and using a valid method to deal with the general cases.

Noting that if x, y ∈ {1, a, ā}, then xȳ ∈ {1, a, ā, a², ā² }. Then for the inner product of two rows, we denote the times 1, a, ā, a², ā² appear in the six items in this inner product by n₁, n₂, n₃, n₄, n₅ respectively. Then we must have 8 $n_{1} + n_{2} a + n_{3} \bar{a} + n_{4} a^{2} + n_{5} {\bar{a}}^{2} = 0.$ {n_1} + {n_2}a + {n_3}\bar a + {n_4}{a^2} + {n_5}{{\bar a}^2} = 0.

We claim that 0 ≤ n_i < 3(i = 1, …, 6), otherwise the matrix H must have a 2 × 3 submatrix with rank one. Then Lemmas 3 and 7 show a contradiction.

3.2.1.

Several Special Cases

We first pay attention to two special cases, that is, a = ω or a = –ω. For a = ω, then {1, a, ā} = {1, ω, ω²}, Lemma 2 shows that the matrix H is complex equivalent to the Tao matrix. For a = –ω, then by doing multiplications with some A ∈ ℳ₆, we find that H is equivalent to a matrix with the property that all the elements of the first row are 1. This matrix contains only {1, ω, ω², –ω, –ω²}. In the last paragraph of the proof of Lemma 8, we know such CHM is complex equivalent to the Tao matrix. In fact, one example of a CHM containing only {1, –ω, –ω²} is the following form, denoted by $S_{6}^{(1)}$ S_6^{(1)}.9 $[\begin{matrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & - ω & - ω & - ω^{2} & - ω^{2} \\ 1 & - ω & 1 & - ω^{2} & - ω^{2} & - ω \\ 1 & - ω & - ω^{2} & 1 & - ω & - ω^{2} \\ 1 & - ω^{2} & - ω^{2} & - ω & 1 & - ω \\ 1 & - ω^{2} & - ω & - ω^{2} & - ω & 1 \end{matrix}] .$ \left[ {\matrix{ 1 & 1 & 1 & 1 & 1 & 1 \cr 1 & 1 & { - \omega } & { - \omega } & { - {\omega ^2}} & { - {\omega ^2}} \cr 1 & { - \omega } & 1 & { - {\omega ^2}} & { - {\omega ^2}} & { - \omega } \cr 1 & { - \omega } & { - {\omega ^2}} & 1 & { - \omega } & { - {\omega ^2}} \cr 1 & { - {\omega ^2}} & { - {\omega ^2}} & { - \omega } & 1 & { - \omega } \cr 1 & { - {\omega ^2}} & { - \omega } & { - {\omega ^2}} & { - \omega } & 1 \cr } } \right].

So far we have done a complete analysis of these cases. Now we move on to the general cases.

3.2.2.

General Cases

We introduce a method to roughly but efficiently classify all possible cases. That is, for such an array [n₁,…, n₅], we try to analyze just whether the left-hand side of equation (8) is real. For example, for [1,1,1,1,2], which means that the left-hand side of (8) is 1 + a + ā + a² + 2ā², we can easily find that 1, a + ā, and a² + ā² are always real, that is to say, to let 1 + a + ā + a² + 2ā² be real, ā² must also be real. And from this we conclude that ā² = ±1, then a = ±1 or ±i, and all these cases are discussed before. This method gives us an easy way to classify all cases, let alone the solutions of original equations. We can remove some terms of the equation that are obviously real, like a + ā. The remaining terms of the equation are what we can not select a part of the remaining terms that must be real, and we call these remaining terms ”pending terms”, also, we call the amount of the remaining terms ”the amount of pending terms”, here the amount is calculated with multiplicity.

One important thing is that the solutions obtained from letting the pending terms be real, we call it “the solutions from the pending terms” in the following pages, are not always the solutions of the original equation, while the solutions of the original equation are always the solutions from the pending terms. That is to say, the solutions of the original equation are included in the solutions from the pending terms. Now we use this method to prove our claim below.

Lemma 9.

Suppose H is a 6 × 6 CHM containing only {1, a, ā} then for any two distinct arrays [n₁,…, n₅] and $[n_{1}^{'}, \dots, n_{5}^{'}] (0 \leq n_{i}, n_{i}^{'} < 3, i = 1, \dots, 5)$ \left[ {n_1^\prime , \ldots ,n_5^\prime } \right]\left( {0 \le {n_i},\;n_i^\prime < 3,\;i = 1, \ldots ,5} \right), we call the equations with coefficient from the two arrays their original equations, if their original equations are not the same and not be conjugate to each other, then the two original equations either have common but simple solutions or have no common solutions. Here simple solutions {1, a, ā} are included in {1, ω, ω²}, {1, –ω, –ω²}, {1,a, –a}, {1, –1,a}.

The proof of Lemma 9 will be present in Appendix C.

Lemma 9 inspires us that for a CHM H containing only {1, a, ā}, if two inner products of different rows are neither the same nor conjugate to each other, then we claim that H is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)} or H⁽¹⁾. Also, if the original equation is not Eq.1-4 or Eq.1′-4′, our claim also holds. So the remaining case is that all the original equations derived from the inner product of two distinct rows are the same or conjugate to each other and included in Eq.1-4 and Eq.1′-4′.

We use the group theory to simplify this problem. For an additive group 〈Z₅, +〉 acting on a set A = {1, a, ā, a², ā²}, we try to let such action representing the inner product of two elements. To realize that, we try to set up an isomorphism between A and Z₅. We define a mapping f with f (1) = 0, f(a) = 1, f (ā) = 4, f (a²) = 2, f (ā²) = 3, although the mapping f is not a isomorphism, we have f (a · ā) = f (a) + f(ā) = 0,f (ā² · a²) = 0, f(a · a) = f(a) + f(a) = 2 = f(a²), f(ā²) = 2f(ā). We ignore the value of f(a · a²) and f(ā · ā²) because these values make no difference in analysing our problem. We specify that when computing the inverse mapping f⁻¹, we must have f⁻¹ [0,1,2,3,4] = [1, a, a², ā², ā].

Now we can make the group action on a set represent the inner product of two elements, moreover, for two rows [x₁, …, x₆], [y₁, …, y₆], x_i, y_i ∈ A(i = 1, …, 6), the inner product of the two rows can be computed in the following steps: first, compute f [x₁, …, x₆] and – f [y₁, …, y₆]; Then compute f [x₁, …, x₆] – f [y₁, …, y₆] and denote that expression by [z₁, …, z₆]; Finally, we compute $\sum_{i = 1}^{6} f^{- 1} (z_{i})$ \sum\limits_{i = 1}^6 {{f^{ - 1}}} \left( {{z_i}} \right) and this is the inner product of the two rows. We can easily verify that claim by the equation $f (x_{1}) - f (y_{1}) \equiv f (x_{1}) + f (\bar{y_{1}}) \equiv f (x_{1} \bar{y_{1}})$ f(\bar y) = 5 - f(y) mod 5. In particular, we claim that $f (\bar{y}) = 5 - f (y)$ f\left( {{x_1}} \right) - f\left( {{y_1}} \right) \equiv f\left( {{x_1}} \right) + f\left( {\overline {{y_1}} } \right) \equiv f\left( {{x_1}\overline {{y_1}} } \right) if y ≠ 1, $f (\bar{y}) = f (y) = 0$ f(\bar y) = f(y) = 0 if y = 1.

Hence when we consider a CHM H with all elements in row 1 are one and containing only elements from A, we can also consider f (H), which contains only elements from Z5 and all elements in the first row of f(H) are 0.

Now consider the first three rows, by exchanging rows we shall assume them to be rows 1-3. Then we shall assume the first three rows of f (H) to be: 10 $[\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ x_{1} & x_{2} & x_{3} & x_{4} & x_{5} & x_{6} \\ y_{1} & y_{2} & y_{3} & y_{4} & y_{5} & y_{6} \end{matrix}] .$ \left[ {\matrix{ 0 & 0 & 0 & 0 & 0 & 0 \cr {{x_1}} & {{x_2}} & {{x_3}} & {{x_4}} & {{x_5}} & {{x_6}} \cr {{y_1}} & {{y_2}} & {{y_3}} & {{y_4}} & {{y_5}} & {{y_6}} \cr } } \right].

The mapping f maps the left side of Eq.1-4 and Eq.1’-4’ to 15,15,8,9 and 15,15,12,11. We shall assume that the inner products of row 1 and rows 2,3 are equal according to inclusion-exclusion principle, then $\sum_{i = 1}^{6} x_{i} = \sum_{i = 1}^{6} y_{i}$ \sum\limits_{i = 1}^6 {{x_i}} = \sum\limits_{i = 1}^6 {{y_i}} . Consider the inner product of rows 2,3, we have 11 $\sum_{i = 1}^{6} f (f^{- 1} (x_{i}) \bar{f^{- 1} (y_{i})}) \equiv 30 + \sum_{i = 1}^{6} (x_{i} - y_{i}) = 30 \equiv 0 mod 5.$ \sum\limits_{i = 1}^6 f \left( {{f^{ - 1}}\left( {{x_i}} \right)\overline {{f^{ - 1}}\left( {{y_i}} \right)} } \right) \equiv 30 + \sum\limits_{i = 1}^6 {\left( {{x_i} - {y_i}} \right)} = 30 \equiv 0\,\bmod \,5.

But the inner product of rows 2,3 must have the same form as other inner products, which means $\sum_{i = 1}^{6} x_{i} \equiv \sum_{i = 1}^{6} y_{i} \equiv 0$ \sum\limits_{i = 1}^6 {{x_i}} \equiv \sum\limits_{i = 1}^6 {{y_i}} \equiv 0 mod 5. Hence $\sum_{i = 1}^{6} x_{i}$ \sum\limits_{i = 1}^6 {{x_i}} must be 15, corresponding to [1,2,2,3,3,4] or [1,1,2,3,4,4] respectively.

If the second row is [1,2,2,3,3,4], then consider that the inner product of rows 2,3 has the same form and row 3 is a permutation of [1,2,2,3,3,4], we can solve the third row, that is [3,4,3,2,1,2]. Since such claim holds for any row in rows 3-6, we can never solve the fourth row then, hence we deduce a contradiction. The contradiction can be similarly deduced for the case when the second row is [1,1,2,3,4,4]. Hence we have finished our proof of Theorem 2.

Corollary 3.

Suppose H is a CHM of order 6 containing only {1, a, a²}, then H is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)} or H⁽¹⁾ in Lemma 2 and Theorem 1.

Proof.

Noting that a⁻¹H is a CHM containing only {1, a, ā}, by Theorem 2 we have finished the proof.

In Lemma 9 we have proposed a set of simple solutions, at that time the set contains only {1, ω, ω²}, {1, –ω, –ω}, {1,a, –a}, {1, –1, a}, now we add {1, a, a}, {1, a, a²} to the set due to Theorem 2 and Corollary 3. In particular, the simple solutions {1, ω, ω²}, {1, –ω, –ω²} are included in {1, a, ā}. The meaning of this set is that, in the subsequent text, when we encounter these cases, we can directly use the existing results to handle them.

3.3.

CHM Containing Only {1, a, –ā}

In this subsection we prove a main result about CHM containing only {1, a, –ā}.

Theorem 3.

6 × 6 CHM containing only {1, a, –ā} does not exist.

We will prove that by analyzing several special cases and using a similar method in the previous subsection to deal with the general cases.

Noting that if x, y ∈ {1, a, –ā}, then xȳ ∈ {1, a, –a, ā, –ā, –a², –ā²}. Then for the inner product of two rows, we denote the times 1, a, –a, ā, –ā, –a², –ā² appear in the six items in this inner product by n₁, n₂, n₃, n₄, n₅, n₆, n₇ respectively. Then we must have 12 $n_{1} + (n_{2} - n_{3}) a + (n_{4} - n_{5}) \bar{a} - n_{6} a^{2} - n_{7} {\bar{a}}^{2} = 0.$ {n_1} + \left( {{n_2} - {n_3}} \right)a + \left( {{n_4} - {n_5}} \right)\bar a - {n_6}{a^2} - {n_7}{{\bar a}^2} = 0.

Lemmas 3,7 shows that 0 ≤ n_i < 3(i = 1,…, 6).

3.3.1.

Several Special Cases

We first pay attention to the special cases, that is, a = ±ω, since we can take conjugate to H, so we only need to discuss the case a = ω. The CHM H containing only {1, ω, –ω²}, by the proof of Lemma 8, we know that such CHM never exists. So we have finished our discussion of the special cases.

3.3.2.

General Cases

We introduce a lemma which is similar to Lemma 9 and use the method by analyzing the pending terms to prove it.

Lemma 10.

Suppose H is a 6×6 CHM containing only {1, a, −ā}, then for any two distinct arrays [n₁, …, n₇] and $[n_{1}^{'}, \dots, n_{7}^{'}] (0 \leq n_{i}, n_{i}^{'} < 3, i = 1, \dots, 5)$ \left[ {n_1^\prime , \ldots ,n_7^\prime } \right]\left( {0 \le {n_i},n_i^\prime < 3,\;i = 1, \ldots ,5} \right), if their original equations are neither the same nor conjugate to each other, then the two original equations either have common but simple solutions or have no common solutions. Now the simple solutions are a = ±1, ±i, ±ω, ±ω².

The proof of Lemma 10 will be provided in Appendix D.

From the proof of Lemma 10 we find that only when the original equation is ±(a+ ā) – 2(a² + ā²) = 0, ±2(a + ā) – (a² + ā²) = 0 or 2 ± (a + ā) – (a² + ā²) = 0, it has no simple solutions. Then we denote A₁ = {1, a, ā, −a², −ā²}, A₂ = {1, –a, –ā, –a², –ā²}, and define f_i : A_i → Z₅ by f₁ [1, a, ā, –a², –ā²] = [0,1,4,3,2] and f₂ [1, –a, –ā, –a², –ā²] = [0,1,4,3,2]. For n₃ = n₅ we will consider the mapping f₁ and for n₂ = n₄ we will consider the mapping f₂, so we can do similar discussions as in the last subsection. Since the group actions are similar, so the result in the last subsection still holds, which means if the inner products of distinct rows or columns are the same or conjugate to each other and the solutions of the original equations are not simple, then such CHM does not exist. Recall that all simple solutions a = ±1, a = ±i, a = ±ω, a = ±ω² correspond to CHM containing only {1, ω, –ω²} or {1, –ω, ω²}, hence we have proven Theorem 3.

Corollary 4.

The 6 × 6 CHM containing only {1, a, –a² } does not exist.

Proof.

Noting that a⁻¹H is a CHM containing only {1, –a, ā}, if we let b = –a, then by Theorem 3 we have finished the proof.

Now the set of simple solutions contains {1, −1, a}, {1, a, −a}, {1, a, ā}, {1, a, a²}, {1, a, −ā}, {1, a,−a²}, we have classified all these cases before.

3.4.

CHM Containing Only {1, a, b}

Now we give a complete classification to all CHM containing only {1, a, b}. Here is our main result of this paper.

Theorem 4.

Suppose H is a 6 × 6 CHM containing only {1, a, b}, then H is complex equivalent to $S_{6}^{(0)}$ S_6^{(0)} or H⁽¹⁾ in Lemma 2 and Theorem 1. Moreover, any CHM containing only three distinct elements is not a member of MUB trio.

The proof of Theorem 4 will be provided in Appendix E.

Conclusion

We have taken a complete classification of CHM containing only three distinct elements. The surprising and brief result is that all such CHM can only be complex equivalent to an H₂-reducible matrix H⁽¹⁾ in Theorem 1 or a non-H₂-reducible matrix $S_{6}^{(0)}$ S_6^{(0)} in Lemma 2. The result gives us a deeper understanding of CHMs of order 6 since the least amount of distinct elements one CHM can contain is just three. And by analyzing CHM containing a small amount of distinct elements we shall obtain some special matrices which can give us a new angle to find more non-H₂-reducible CHM and understand their property. Our method in proving the main result can also help us to analyze CHM of order 6 whose elements in the first row are all 1. For example, if such matrix contains exactly 7 distinct elements, and there are three pairs of pairwise conjugated elements, our method works. We will focus on such matrices containing more distinct elements and try to do a complete classification of them, in this way we may classify all the CHM of order 6 step by step.

Order-Six CHMs Containing Exactly Three Distinct Elements

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