Complex Hadamard matrices of order 6: a four-parameter family
aa r X i v : . [ m a t h . OA ] A ug COMPLEX HADAMARD MATRICES OF ORDER :A FOUR-PARAMETER FAMILY FERENC SZ ¨OLL ˝OSI
Dedicated to Professor Uffe Haagerup on the occasion of his th birthday Abstract.
In this paper we construct a new, previously unknown four-parameter familyof complex Hadamard matrices of order 6, the entries of which are described by algebraicfunctions of roots of various sextic polynomials. We conjecture that the new, generic family G (4)6 together with Karlsson’s degenerate family K (3)6 and Tao’s spectral matrix S (0)6 form anexhaustive list of complex Hadamard matrices of order 6. Such a complete characterizationmight finally lead to the solution of the famous MUB-6 problem. Primary 05B20, secondary 46L10.
Keywords and phrases.
Complex Hadamard matrices, mutually unbiased bases, MUB Introduction
Complex Hadamard matrices form an important family of orthogonal arrays with theadditional unimodularity constraint imposed on their entries. These matrices obey the alge-braic identity HH ∗ = nI n where ∗ stands for the Hermitean transpose, and I n is the identitymatrix of order n . They appear in various branches of mathematics frequently, includinglinear algebra [6], coding- and operator theory [7, 15] and harmonic analysis [13, 19]. Theyplay an important rˆole in quantum optics, high-energy physics, and they are one of the keyingredients to quantum teleportation- and dense coding schemes [20] and mutually unbiasedbases (MUBs) [2]. An example of complex Hadamard matrices is the Fourier matrix F n ,well-known to exists for all n . It is natural to ask how does a “typical” complex Hadamardmatrix of order n look like, and a satisfying answer to this question can be given providedwe have a complete characterization of Hadamard matrices of order n at our disposal. Thesetypes of problems, however, are notoriously difficult even for small n . Naturally, one is inter-ested in the essentially different matrices only, and we identify two matrices H and K andsay that they are equivalent if H = P D KD P for some unitary diagonal matrices D , D and permutational matrices P , P . Recall that a complex Hadamard matrix is dephased,if all entries in its first row and column are equal to 1. While studying and classifying realHadamard matrices is naturally a discrete, finite problem which can be handled by deepalgebraic methods and sophisticated computer programs to some extent, the complex case,however, behaves essentially different. In particular, due to the appearance of various para-metric families one cannot hope for a finite list of inequivalent matrices, but rather for afinite list of constructions, each of them leading to an infinite family of complex Hadamardmatrices.The complete classification of complex Hadamard matrices is available up to order n = 5only. It is trivial that F n is the unique complex Hadamard matrix for orders n ≤
3. The case
Date : August, 2010., Preprint.This work was supported by the Hungarian National Research Fund OTKA K-77748. n = 4 is still elementary, and it was shown by Craigen that all complex Hadamard matricesof order 4 belong to an infinite, continuous one-parameter family [4]. In order 5 we haveuniqueness again, a result which is absolutely non-trivial already. In particular, Lov´asz wasthe first who showed [12] that F is the only circulant complex Hadamard matrix in thisorder, and a decade later Haagerup managed to prove the uniqueness of F by discoveringan algebraic identity (cf. formula (17)) relating the matrix entries in a surprising way [7].In order 6 various one- [1, 5, 14, 21], two- [11, 17] and three-parameter families [9, 10] havebeen constructed recently and it is conjectured that these are part of a more general, four-parameter family of complex Hadamard matrices, yet to be discovered [2]. This conjectureis supported by overwhelming numerical evidence [16], however so far only a fairly smallsubset of it was described by closed analytic formulæ, including an isolated matrix S (0)6 anda three-parameter matrix K (3)6 [9, 10].The reason why the 6 × C d , B and B are unbiased if for every e ∈ B , f ∈ B we have |h e, f i| = 1 /d . A family of orthonormal bases is said to be mutually unbiased ifevery two of them are unbiased. The famous MUB-6 problem asks for the maximal numberof mutually unbiased bases in C . On the one hand this number is at least 3, as thereexists various infinite families of triplets of MUBs in this order [8, 17, 21], on the otherhand it is well-known that it cannot be larger than 7 (cf. the references of [2]). In fact,it is conjectured that a triplet is the best one can come up with in dimension 6 [21]. Theconnection between MUBs and Hadamard matrices of order 6 has been exploited in [8] veryrecently, where a discretization scheme was offered to attack the problem and it was provedby means of computers, however, in a mathematically rigorous way, that the members of thetwo-parameter Fourier family F (2)6 ( a, b ) and its transpose cannot belong to a configuration of7 MUBs containing the standard basis in dimension 6. One reasonable hope to finally settlethe MUB-6 problem is to give a complete characterization of complex Hadamard matricesof order 6 and apply the same technique to them.The goal of this paper is to propose a general framework towards the complete classificationof complex Hadamard matrices of order 6. In particular, by characterizing the orthogonaltriplets of rows in complex Hadamard matrices we generalize an observation of Haagerup [7]to obtain a new algebraic identity relating the matrix entries in an unexpected way. This isan essentially new tool to study complex Hadamard matrices of small orders, and one of themain achievements of this paper. We apply this result to obtain complex Hadamard matrices,moreover we conjecture that the the construction we present here reflects the true natureof complex Hadamard matrices of order 6. It has the following three features: Firstly, it isgeneral in contrast with the earlier attempts where always some additional extra structurewas imposed on the matrices including self-adjointness [1], symmetry [14], circulant blockstructure [17] or H -reducibility [9]. Secondly, it has 4 degrees of freedom and thirdly all theentries of the obtained matrices can be described by algebraic functions of roots of varioussextic polynomials. This suggests on the one hand the existence of a four-parameter familyof complex Hadamard matrices of order 6 and reminds us on the other hand the fact thatthe desired algebraical description where the entries are expressed by radicals might not bepossible at all. However, from the applicational point of view, and in particular, to utilize thecomputer-aided attack of [8] to the MUB-6 problem we shall need these matrices numericallyanyway. OMPLEX HADAMARD MATRICES OF ORDER 6 3
The outline of the paper is as follows. In Section 2 we briefly discuss the main ideasof the construction to motivate the various auxiliary results we prove there. The excitedreader might want to skip this section at first, and jump right ahead to Section 3, where theconstruction of the new family is presented from a high-level perspective. In Section 4 weanalyze the construction thoroughly.2.
Preliminary results
In this section we present the ingredients necessary to construct our new family of complexHadamard matrices of order 6, including a characterization of the mutual orthogonality ofthree rows in Hadamard matrices (cf. Theorem 2.4). First, however, we would like to motivatethese efforts by describing the main ideas of the construction.We start with a submatrix(1) E ( a, b, c, d ) := a b c d and attempt to embed it into a complex Hadamard matrix of order 6(2) G (4)6 ( a, b, c, d ) := a b e s s c d f s s g h ∗ ∗ ∗ t t ∗ ∗ ∗ t t ∗ ∗ ∗ ≡ (cid:20) E BC D (cid:21) . with 3 × E, B, C and D in two steps, as follows. First we construct the submatrices B and C featuring unimodular entries to obtain three orthogonal rows and columns of G .Secondly we find the unique lower right submatrix D to get a unitary matrix. Should theentries of this matrix become unimodular, we have found a complex Hadamard matrix. Weconjecture that the submatrix E can be chosen, up to equivalence, in a way that there willbe only finitely many candidates for the blocks B and C and therefore we can ultimatelydecide whether the submatrix E can be embedded into a complex Hadamard matrix. Theresulting matrix G can be thought as the “Hadamard dilation” of the operator E .We shall heavily use the following through the paper without any further comment: theconjugate of a complex number of modulus 1 is its reciprocal, and hence the conjugate ofa multivariate polynomial with real coefficients depending on indeterminates of modulus 1is just the polynomial formed by entrywise reciprocal of the aforementioned indeterminates.We computed various Gr¨obner bases [3] in this paper with the aid of Mathematica. Thereader is advised to use a computer algebra system for bookkeeping purposes and consult[18] for the standard notations for well-known complex Hadamard matrices such as S (0)6 , K (3)6 ,etc.We begin with recalling two elementary results from the existing literature. Lemma 2.1.
Suppose that we have a partial row (1 , a, b, e, ∗ , ∗ ) composed from unimodularentries. Then one can specify some unimodular numbers s and s in place of the unknownnumbers ∗ to make this row orthogonal to (1 , , , , , if and only if (3) | a + b + e | ≤ . FERENC SZ ¨OLL ˝OSI
Proof.
To ensure orthogonality, we need to have 1 + a + b + e + s + s = 0 from which itfollows that | a + b + e | = | s + s | ≤
2. It is easily seen geometrically, that in this casewe can define the unimodular numbers required. (cid:3)
The missing coordinates featuring in Lemma 2.1, s and s , can be obtained algebraicallythrough the well-known Lemma 2.2 ((Decomposition formula, [14])) . Suppose that the rows (1 , , , , , and (1 , a, b, e, s , s ) containing unimodular entries are orthogonal. Let us denote by Σ := 1 + a + b + e , and suppose that < | Σ | ≤ . Then (4) s , = − Σ2 ± i Σ | Σ | r − | Σ | . If Σ = 0 then s is independent from a, b, e but s = − s .Proof. Clearly s and s are the unimodular numbers with s + s = − Σ. (cid:3) Now we proceed by investigating the orthogonality of triplets of rows. In order to do this,the following is a crucial
Definition 2.3 ((Haagerup polynomial)) . The Haagerup polynomial H associated to therows (1 , , , , , , (1 , a, b, e, ∗ , ∗ ) and (1 , c, d, f, ∗ , ∗ ) of a complex Hadamard matrix read H ( a, b, c, d, e, f ) := (1 + a + b + e )(1 + c + d + f )(1 + ca + db + f e ) . The first result of ours is the following
Theorem 2.4.
Suppose that we have the partial rows (1 , a, b, e, ∗ , ∗ ) and (1 , c, d, f, ∗ , ∗ ) ,composed from unimodular entries. Then one can specify some unimodular numbers s , s , s and s in place of the unknown numbers ∗ to make these rows together with (1 , , , , , mutually orthogonal if and only if (5) H ( a, b, c, d, e, f ) = 4 − | a + b + e | − | c + d + f | − | ca + db + f e | with (6) |H ( a, b, c, d, e, f ) | ≤ . Proof.
First we start by proving that (5) holds. To do this, we utilize Haagerup’s idea [7] asfollows: by pairwise orthogonality, we find that1 + a + b + e = − s − s c + d + f = − s − s ca + db + f e = − s s − s s Now, by multiplying these three equations together we find that (6) follows and further H = − ( s + s )( s + s )( s s + s s ) = − ( s s + s s + s s + s s )( s s + s s ) = −| s s + s s | − ( s s + s s )( s s + s s ) = −| s s + s s | − ℜ ( s s + s s ) . To conclude the proof, we need to show that2 ℜ ( s s + s s ) = | a + b + e | + | c + d + f | − OMPLEX HADAMARD MATRICES OF ORDER 6 5
To see the converse direction, we need to show that (5) essentially encodes orthogonality.Let us use the notations Σ := 1 + a + b + e , ∆ := 1 + c + d + f , Ψ := 1 + ca + db + f e . Withthis notation condition (5) boils down to(7) H = Σ∆Ψ = 4 − | Σ | − | ∆ | − | Ψ | . Clearly, if | Σ | ≤ | ∆ | ≤ s , s , s and s to ensure orthogonality to row (1 , , , , , , a, b, e, s , s ) and (1 , c, d, f, s , s ) reads(8) Ψ + s s + s s = 0 . Suppose first that we have the trivial case Σ = ∆ = 0. Then, by the decomposition formulawe have s = − s and s = − s , and (7) implies that | Ψ | = 2. Therefore, if we set theunimodular number s := − Ψ s / = 0. Then we have s = − s , and from (7)it follows that | Σ | ≤
2, and in particular(9) | Ψ | = p − | Σ | . Now we can use the Decomposition formula to find out the values of s and s and theorthogonality equation (8) becomes(10) Ψ + s − i Σ | Σ | r − | Σ | ! = 0 . This holds, independently of s , if | Σ | = 2, as by (9) Ψ = 0 follows. Otherwise, set theunimodular number s := − i ΣΨ | Σ || Ψ | to ensure the orthogonality through (10).Finally, let us suppose that Σ = 0 and ∆ = 0. Now observe that in this case the value ofΨ needed for formula (8) can be calculated through (7). The other ingredient, namely thevalue of s s + s s can be established through the Decomposition formula, once we derivethe required bounds | Σ | ≤ | ∆ | ≤
2. Depending on the value of H , we treat severalcases differently.CASE 1: Suppose that −| Ψ | ≤ H . This implies, by formula (7), that | Σ | + | ∆ | ≤ | Σ | ≤ | ∆ | ≤ | Ψ | from (7).Suppose first that H ≥
0. Hence, after taking absolute values, (7) becomes | Ψ | + | Σ || ∆ || Ψ | + | Σ | + | ∆ | − , and the only non-negative root is(11) | Ψ | = −| Σ || ∆ | + p (4 − | Σ | )(4 − | ∆ | )2 . Now suppose that −| Ψ | ≤ H <
0. Hence, after taking absolute values, (7) becomes(12) | Ψ | − | Σ || ∆ || Ψ | + | Σ | + | ∆ | − , and we find that the only non-negative root we have under the assumption | Σ | + | ∆ | ≤ | Ψ | = | Σ || ∆ | + p (4 − | Σ | )(4 − | ∆ | )2 . FERENC SZ ¨OLL ˝OSI
CASE 2: Suppose now that H < −| Ψ | . This implies that | Σ | + | ∆ | >
4, and we do nothave a priori the bounds | Σ | ≤ | ∆ | ≤
2. Nevertheless, we derive equation (12) again, andwe find that the values of | Ψ | can be any of(14) | Ψ | , = | Σ || ∆ | ± p (4 − | Σ | )(4 − | ∆ | )2 , provided that the roots are real, namely we have either | Σ | > | ∆ | > | Σ | ≤ | ∆ | ≤
2. The first case is, however, not possible, as it would imply | Ψ | > | Σ | ≤ | ∆ | ≤ | Ψ | has beenfound, we are free to use the Decomposition formula to obtain the values of s , s , s and s . Clearly, we can set s with the + sign, while s with the − sign as in formula (4), up toequivalence. However, we do not know a priori how to distribute the signs amongst s and s , and to simplify the notations we define(15) s = − ∆2 ± i ∆ | ∆ | r − | ∆ | , s = − ∆2 ∓ i ∆ | ∆ | r − | ∆ | . In particular, by using (7) and (15) we find that the orthogonality equation (8) becomes(16) 4 − | Σ | − | ∆ | − | Ψ | Σ∆ + Σ∆2 ± | Σ || ∆ | r − | Σ | r − | ∆ | , where the ± sign agrees with the definition of s . To conclude the theorem plug in all of thepossible values of | Ψ | as described in (11), (13), (14) into (16) to verify that for some choiceof the sign it holds identically. (cid:3) Remark . The two possible signs described by formula (14) can be realized. In particular,there are two different orthogonal triplet of rows composed of sixth roots of unity where | Σ | = | ∆ | = √ | Ψ | = 1 while | Ψ | = 2 in theother. Corollary 2.6 ((Haagerup’s trick, [7])) . Suppose that the rows (1 , , , , , , (1 , a, b, e, s , s ) and (1 , c, d, f, s , s ) composed of unimodular entries are mutually orthogonal. Then (17) H ( a, b, c, d, e, f ) ∈ R . Haagerup used the property (17) to give a complete characterization of complex Hadamardmatrices of order 5, or equivalently, describe the orthogonal maximal abelian ∗ -subalgebrasof the 5 × B and C of G . Once we have three orthogonal rows and columns we readily fill out the remaininglower right submatrix D . This is explained by the following two lemmata, the first of whichbeing a special case of a more general matrix inversion OMPLEX HADAMARD MATRICES OF ORDER 6 7
Lemma 2.7. If U and V are n × n matrices then ( I n + U V ) − = I n − U ( I n + V U ) − V provided that one of the matrices I n + U V or I n + V U is nonsingular.Proof.
By symmetry, we can suppose that the matrix I n + V U is nonsingular. Then, we have( I n + U V )( I n − U ( I n + V U ) − V ) = I n + U V − U ( I n + V U )( I n + V U ) − V = I n . (cid:3) Lemma 2.8.
Suppose that we have a × partial complex Hadamard matrix consisting ofthree orthogonal rows and columns, containing no vanishing × minor. Then there is aunique way to construct a unitary matrix containing these rows and columns as a submatrix.Proof. Let U be a 6 × × A, B, C and D , as the following: U = (cid:20) A BC D (cid:21) . By the orthogonality of the first three rows and columns and using the fact that the entriesare unimodular, we have(18) AA ∗ + BB ∗ = 6 I (19) A ∗ A + C ∗ C = 6 I To ensure orthogonality in-between the first three and the last three rows we need to have AC ∗ + BD ∗ = O , the all 0 matrix. As B is nonsingular by our assumptions we can define(20) D := − CA ∗ ( B − ) ∗ . Now we need to show that the last three rows are mutually orthogonal as well. Indeed, byusing (20) and (18) we have CC ∗ + DD ∗ = C (cid:0) I + A ∗ ( BB ∗ ) − A (cid:1) C ∗ = C (cid:0) I + A ∗ (6 I − AA ∗ ) − A (cid:1) C ∗ , which, by Lemma 2.7 and (19) is C I + 16 A ∗ (cid:18) I − AA ∗ (cid:19) − A ! C ∗ = C (cid:18) I − A ∗ A (cid:19) − C ∗ = 6 C ( C ∗ C ) − C ∗ = 6 I . (cid:3) We do not state that the obtained unitary matrix U is Hadamard, which is not true ingeneral. Recall that our goal is to embed the submatrix E into the matrix G (cf. (1)-(2)).We have the following trivial Lemma 2.9.
Suppose that a submatrix E can be embedded into a complex Hadamard matrix G of order in which the upper right submatrix B is invertible. Then for some unimodularsubmatrix C for which the first three columns of G are orthogonal the lower right submatrix D = − CE ∗ ( B − ) ∗ is unimodular.Proof. Indeed, this is exactly what embedding means. (cid:3)
In particular, if the submatrices B and C are chosen carefully, the unimodular propertyof D follows for free. FERENC SZ ¨OLL ˝OSI
Corollary 2.10.
Start from a submatrix E and suppose that there are only finitely many(invertible) candidate matrices B ∈ SOL B and C ∈ SOL C such that the first three rows andcolumns of the matrix G are orthogonal. Then E can be embedded into a complex Hadamardmatrix of order if and only if there is some B ∈ SOL B and C ∈ SOL C such that the matrix D = − CE ∗ ( B − ) ∗ is unimodular. Note that due to the finiteness condition in Corollary 2.10 once we have all (but finitelymany) candidate matrices B and C we can decide algorithmically whether the submatrix E can be embedded into a complex Hadamard matrix.The next step is to characterize 6 × × Lemma 2.11.
Suppose that in a dephased × complex Hadamard matrix there exist anoninitial row (or column) containing three identical entries x . Then x = ± and this row(or column) reads (1 , , , − , − , − , up to permutations.Proof. Suppose to the contrary, that there are three nonreal numbers x in a row (column).Then the sum of these numbers x together with the leading 1 read | x | = 10+6 ℜ ( x ) > − x and hence the last part of the statement follows. (cid:3) Recall that the core of a dephased complex Hadamard matrix of order n is its lower right( n − × ( n −
1) submatrix. A vanishing sum of order k is a k -term sum adding up to 0.The following breakthrough result was obtained very recently. Theorem 2.12 ((Karlsson, [9, 10])) . Let H be a dephased complex Hadamard matrix oforder . Then the following are equivalent:(a) H belongs to the three-parameter degenerate family K (3)6 ;(b) H is H -reducible;(c) The core of H contains a − ;(d) Some row or column of H contains a vanishing sum of order ;(e) Some row or column of H contains a vanishing sum of order . In particular, Theorem 2.12 gives a characterization of complex Hadamard matrices oforder 6 containing F as a submatrix. The term H -reducibility refers to the beautifulstructure of these matrices: they have a canonical form in which all 9 of their 2 × K (3)6 , and we shall heavily use these conditions through ourpaper. Part (e) tells us that once we have four entries in a row or column of a matrix whichlies outside the family K (3)6 the Decomposition formula readily derives the unique remainingtwo values through (4). As the family K (3)6 forms a three-parameter subset, the matrices itcontains are atypical, hence the adjective “degenerate”. Corollary 2.13.
Suppose that in a dephased × complex Hadamard matrix H there exist anoninitial row (or column) containing three identical entries. Then H belongs to the family K (3)6 . Lemma 2.14.
Suppose that a × complex Hadamard matrix H has a vanishing × minor. Then H belongs to the family K (3)6 . OMPLEX HADAMARD MATRICES OF ORDER 6 9
Proof.
Suppose that H has a vanishing 3 × E ( a, b, c, d ),as in formula (1). Such an assumption can be made, up to equivalence. As det( E ) = b + c − a − d + ad − bc = 0, we find that if any of the indeterminates a, b, c, d is equal to 1then E contains a noninitial row (or column) containing three 1s, and therefore by Corollary2.13 we conclude that this matrix belongs to the family K (3)6 . Otherwise, we can supposethat none of a, b, c, d is equal to 1 and hence we find that d = ( a + bc − b − c ) / ( a − dd − b = a or c = a should hold, but then we have either d = c or d = b as well. In particular, wefind that E has two identical rows (or columns). After enphasing the matrix, again, we findthat there is a full column (or row) of entries 1 and a reference to Corollary 2.13 concludesthe lemma. (cid:3) Therefore to investigate those matrices which lie outside the family K (3)6 we can safely useLemma 2.8 and in particular the inversion formula (20).It turns out, that the isolated matrix S (0)6 (cf. [18]) requires a special treatment as well. Itis featured in the following Lemma 2.15.
Suppose that in a × dephased complex Hadamard matrix H there is anoninitial row and column composed of cubic roots of unity. Then H is either equivalent to S (0)6 or belongs to the family K (3)6 . Let us denote by ω the principal cubic root of unity once and for all, that is ω := e π i / . Proof.
First suppose that the cubic row and column meet in a common 1. Then our matrixlooks like as the matrix H on the left below, up to equivalence: H = ω ω ω ω ω a b c d ω ∗ ∗ ∗ ∗ ω ∗ ∗ ∗ ∗ ω ∗ ∗ ∗ ∗ , H ′ = ω ω ω ω a b c d ∗ ω ∗ ∗ ∗ ∗ ω ∗ ∗ ∗ ∗ ω ∗ ∗ ∗ . Now by orthogonality of the first three rows we find that a + b = − ω and c + d = −
1. Hence,we can assume, up to equivalence, that a = 1 , b = − ω , and c = ω , d = ω . But then, wecan fill out the fourth row, and the third and fourth column as well. We conclude that theobtained matrix is equivalent to S (0)6 .Secondly, let us suppose that the cubic row and column meet in a common ω . This matrix H ′ is depicted on the right above. Then, by calculating the orthogonality equations, we findthat either a = −
1, and hence the matrix (if it can be completed to a Hadamard at all)belongs to the family K (3)6 , or a = ω , b = ω , and, up to equivalence c = ω , d = ω . Thisimplies that the third row and third column feature cubic entries only meeting in a common1 therefore reducing the situation to the first case. The third case, namely when the cubicrow and column meet in a common ω can be treated similarly. (cid:3) Now we turn to the presetting of the submatrix E ( a, b, c, d ) (see (1)). In order to avoidthe case when the system of equations (5)–(17) is linearly dependent we need to excludevarious input quadruples ( a, b, c, d ). However, it shall turn out that we are free to do such restrictions, up to equivalence. Before proceeding further, let us define the following two-variable function mapping T to C as follows: E ( x, y ) := x + y + x + y + xy + x y. We say that y is an elliptical pair of x , if E ( x, y ) = 0. Observe that for a given x = − y + y = − (1 + x ) / (1 + x ). The following is a strictlytechnical Proposition 2.16 ((Canonical transformation)) . Suppose that we have a complex Hadamardmatrix H inequivalent from S (0)6 and any of the members of the family K (3)6 . Then H has a × submatrix E ( a, b, c, d ) as in formula (1) , up to equivalence, satisfying (21) ( b − c − b − d )( c − d )( b − c )( bc − d ) E ( b, d ) E ( c, d ) = 0 . Proof.
The strategy of the proof is the following: first we pick a “central element” d fromthe core of the matrix and then we show that b and c can be set satisfying (21). Recall, thatby Lemma 2.15 there is no a noninitial row and column composed from cubic roots of unityin the matrix.First let us assume that there is a 1 in the core somewhere. Suppose that there is a, tosay, row full of cubics. In this case set d = ω , and c = ω . Now in the column containing d there is a non-cubic entry γ , and we are free to set b = γ . If there is neither a full row nor acolumn of cubics in the matrix, then set d = 1, and choose a non-cubic c from its row. Notethat there cannot be a further noninitial 1 in the row or column of d by Corollary 2.13. Nowobserve that as the elliptical pairs of 1 are ω and ω , we can choose a suitable b from thecolumn of d unless all entries there are members of the set { ω, ω , c, c } . Note that ω togetherwith ω cannot be in the column of d at the same time, and from this it is easily seen thatwe cannot define the value of b only if the column of d is one of the following four cases, upto permutations: (1 , , ω, c, c, c ), (1 , , ω, c, c, c ), (1 , , ω , c, c, c ), (1 , , ω , c, c, c ). However,by normalization and by orthogonality, the sum of the entries in this column should addup to 0, and we find in all cases that the unimodular solution to c is a cubic root of unity,contradicting the choice of c . Therefore one of the entries in the column of d is different from ω, ω , c, c which will be chosen as b .Secondly, let us suppose, that there is no 1 in the core. In particular, all entries in thecore are different rowwise and columnwise.Pick any d from the core of the matrix. Let us denote by c and c the elliptical pairsof d (maybe c = c , or they are undefined). Now we have several cases depending on theappearance of these values in the row and column containing d .CASE 1: c and c present in both the row and column containing d . Hence, in the rowand column containing d there are the entries 1, d , c and c already, the remaining two, α and β are uniquely determined by the Decomposition formula. Note that α = d , asotherwise we would have β = − (2 d + d + d ) / (1 + d ) (as the sum of all entries in a noninitialrow add up to 0, and the sum of the elliptical pairs is known), which is unimodular if andonly if d = ± i or d = ω or d = ω . In the first case we have β = ∓ i and we are dealing witha member of the family K (3)6 . The second and third case imply that we have a full row and afull column of cubics, a contradiction. If β = d/α then by picking c = α from the row we canset b = β in the column. Otherwise, in case we have β = d/α , then reset the central element d to the α which is in the same row as d . Now observe that after this exchange we shouldmet the requirements of Case 1 again (otherwise we are done here), and hence the elliptical OMPLEX HADAMARD MATRICES OF ORDER 6 11 pairs of α , α and α , should present in the row and column of α , which therefore containexactly the same entries. However α , = d , as otherwise orthogonality with the condition E ( α, d ) = 0 would imply d = ±
1, a contradiction; α , = d/α , as otherwise d = ω or d = ω would follow from the same argument implying that the row containing α has a noninitial1, a contradiction. Therefore, the only option left is that α , = c , . But then we can set c = d = α and b = d/α = α , and we are done.CASE 2: c and c present in (to say) the row containing d , but only one of these values(say c ) is present in the column of d . Let us denote by α and β the two further entries inthis row which, again, are different from d . In the column of d there is already 1, d and c , and observe that the remaining three entries cannot be ( d , α, β ) as this would imply d = c , contradicting our case-assumption. Therefore one of the three unspecified entries γ is different from d , α and β . Now if γ = d/α then set c = α , b = γ otherwise set c = β , b = γ . We are done.CASE 3: Only the value c is in (to say) the row containing d and in the column of d aswell. This is a tricky case, as it might happen that the undetermined triplet in both the d -th row and column is precisely ( d , α, d/α ), and therefore we cannot ensure condition (21).However, from the orthogonality equation 1 + d + d + c + α + d/α = 0, from its conjugate,and the elliptical condition E ( c , d ) = 0 we can form a system of equations, the solution ofwhich can be found by computing a Gr¨obner basis. By investigating the results, we findthat either α = − α + d = 0 leading us to the family K (3)6 or c = d . In the last case,however, we can calculate the values of d and α explicitly. In particular, we find that thevalues of d and α are given by some of the unimodular roots of the following polynomials1 + 2 d + 2 d + d = 0 and 1 + 4 α − α − α − α − α − α + 4 α + α = 0. Now reset the“central” entry d to α , and observe that the elliptical pairs of α are not present in its row,and therefore the conditions of this subcase are no longer met. Otherwise we can supposethat the undetermined triplet in the column of d is not ( d , α, d/α ). Set c = α = d . Pick γ from the column which is different from d , α , d/α , set b = γ and we are done.CASE 4: Only the value c is in (to say) the row containing d and in the column of d there is the other elliptical value c = c . We can suppose that two of the undeterminedentries in the row of d satisfy α = d and β = d and set c = α . Now if in the column theundefined triplet is precisely ( α , d/α , d ), then observe that the same triplet cannot appearin the row, as otherwise c = c would follow. Therefore we can reset c to a value differentfrom α, d/α, d , and set b = α . Otherwise there is an entry in the column which can be setto b , we are done.CASE 5: In the column of d there is no elliptical value at all. Pick any c = α = d fromthe row. Now in the column there are four unspecified entries. Clearly, one of them, say γ will be different from d , α and d/α . Set b = γ . We are done. (cid:3) Not every submatrix E can be embedded into a complex Hadamard matrix of order 6.To offer a necessary condition, let us recall first that an operator A is called a contraction,if k A k ≤
1, where k . k denotes both the Euclidean norm on C and the induced operatornorm on the space of 6 × Lemma 2.17. If A is any × submatrix of a complex Hadamard matrix H of order then A/ √ is a contraction. Proof.
Clearly, we can assume that this submatrix A is the upper left of the matrix H , whichwe will write in block form, as follows: H = (cid:20) A BC D (cid:21) . Now suppose, to the contrary that there is some vector s , such that k As k > √ k s k andconsider the block vector s ′ := ( s, T ∈ C . We have k Hs ′ k = (cid:13)(cid:13) ( As, Cs ) T (cid:13)(cid:13) ≥ (cid:13)(cid:13) ( As, T (cid:13)(cid:13) = k As k > √ k s k = √ k s ′ k = k Hs ′ k , where in the last step we used that the matrix H/ √ (cid:3) In particular, we have the following
Corollary 2.18.
If the submatrix E can be embedded into a complex Hadamard matrix oforder , then every eigenvalue λ of the matrix E ∗ E satisfy λ ≤ . Corollary 2.18 is a useful criterion to show that a matrix E cannot be embedded intoa complex Hadamard matrix, however, it is unclear how to utilize it for our purposes. Inparticular, we do not know how to characterize those 3 × E to be embedded. The answer to this question mightdepend on the dimension, as it is easily seen that while every 2 × × The construction: A high-level perspective
Here we describe the generic family G (4)6 from a high-level perspective. In particular, weoutline the main steps only, and do not discuss some degenerate cases which might come upduring the construction. The next section is dedicated to investigate the process in details.The main result of this paper is the following Construction 3.1 ((The Dilation Algorithm)) . Do the following step by step to obtaincomplex Hadamard matrices of order . INPUT : the quadruple ( a, b, c, d ), forming the upper left 3 × E ( a, b, c, d ), asin formula (1). G (4)6 (see (2)) to obtain a quadraticequation to f :(22) F + F f + F f = 0 , where the coefficients F , F and F depend on the parameters a, b, c, d and the inde-terminate e , and derive the following linearization formula from it:(23) f = − F F − F F f. OMPLEX HADAMARD MATRICES OF ORDER 6 13 f :(24) G + G f + G f = 0 , where, again, the coefficients G , G and G depend on the parameters a, b, c, d and theindeterminate e . Plug the linearization formula (23) into (24) and rearrange to obtainthe companion value of e f = F ( e ), where(25) F ( e ) := − F G − F G F G − F G . | f | = 1 should hold, calculate the sextic polynomial G ( e ) coming from the equation F ( e ) F ( e ) − e . G find all unimodular triplets ( e, s , s ) satisfying e + s + s = − − a − b , calculate the companion values f = F ( e ), s = F ( s ) and s = F ( s )through formula (25) and store all sextuples ( e, s , s , f, s , s ) in a solution set called SOL B . SOL C . SOL B and SOL C construct the submatrices B and C , check if the first three rows and columns are mutually orthogonal and finally useLemma 2.8 to compute the lower right submatrix D through formula (20). OUTPUT: all unimodular matrices found in step
The construction: The nasty details
Here we investigate the steps of Construction 3.1 in details.STEP a, b, c, d ) in compliance with the Canonical Transformationdescribed by Proposition 2.16 as the
INPUT , and form the submatrix E . Check if it meetsthe requirements of Corollary 2.18; if yes, then proceed, otherwise conclude that it cannot beembedded into a complex Hadamard matrix of order 6. Experimental results show that oncethree out of the four parameters are fixed the last one can be easily set to a value such thatthe quadruple ( a, b, c, d ) leads to a complex Hadamard matrix. Heuristically this means thatthere is nothing “mystical” in the choice of the initial quadruple and hence the parametersshould be independent from each other.STEP F
0, independently of e . Indeed,suppose otherwise, which means that the following system of equations (where the last twoare the conjugate of the first two, up to some irrelevant constant factors) abc + a bc + abd + ab d + a bcd + ab cd ≡ ,b c + ab c + a d + a bd + acd + bcd ≡ ,a + b + ac + abc + bd + abd ≡ ,a b + ab + bc + b c + ad + a d ≡ , are fulfilled. We compute a Gr¨obner basis and find that the polynomial bc (1 + c )( c − d ) d (1 + d )( c + d )(1 + d + d ) is a member of it. After substituting back into the original equations we find that there iseither a vanishing sum of order 2 in E or a = b = 1 and therefore the whole family is amember of K (3)6 , or we have E = F or E = F ∗ but these matrices have b = c which howeveris not allowed by the Canonical Transformation.It might happen that F e making it vanish, which cannotbe anything else, but(26) e = a b + a d + ab + ad + b c + bcabc + abd + ac + a + bd + b . Nevertheless, we can suppose that in case of one of the pairs ( e, f ), ( s , s ) and ( s , s ) we donot set e as above, otherwise we would have e = s = s which, by Lemma 2.11 would imply e = s = s = −
1, obtaining some member of the family K (3)6 by Corollary 2.13. Hence,we can suppose that e is different than the value described by formula (26) above, and weconclude that F = 0.STEP f from e in general, as formula(25) might suggests. Indeed, there are complex Hadamard matrices in which s = e , but s = f . The reason for this phenomenon is that formulas (22) and (24) might be linearlydependent. After plugging (23) into (24) we obtain the expression F G − F G + ( F G − F G ) f = 0 , which can lead us to one of the following three cases:CASE 1: Both the polynomials F G − F G and F G − F G vanish identically, indepen-dently of e , meaning that in this case we do not have another condition on f . Luckily thiscan never happen, as by calculating a Gr¨obner basis (again, to speed up the computationswe have added the conjugates of the equations as well) we find that the polynomial( b − b ( b − c ) c ( bc − d )(1 + c + d )( b − d ) E ( c, d )is a member of the basis. Therefore we have either c + d = − c + d = − { c, d } consists of nontrivial cubic roots only. By directly solving the corresponding equations, wefind that the quadruples (1 , , ω, ω ), ( ω , ω, ω, ω ) and (1 , , ω , ω ), ( ω, ω , ω , ω ) can vanishboth polynomials, however these cases were excluded by the Canonical Transformation.CASE 2: Both the polynomials F G − F G and F G − F G vanish for some | e | = 1,meaning that in this case we do not have another condition on f . However, having thesenumbers e at our disposal we can recover the two possible values of f from (23). Once wehave the candidate pairs ( e, f ) and ( e, f ) we readily calculate the remaining pairs ( s , s )and ( s , s ) through the Decomposition Formula. Store all suitable sextics ( e, s , s , f, s , s )in the solution set SOL B . Proceed to step e to make these twopolynomial vanish at the same time. Hence we can derive formula (25) for F ( e ). Proceed tostep f is of modulus one. To do this, we calculate thefundamental polynomial P a,b,c,d ( e ) ≡ |F G − F G | − |F G − F G | . After some calculations, it will be apparent that P has the following remarkable structure: P ≡ a b c d P + a b c d P (2 + 2 a + 2 b ) e + a b c d Qe + Re + Qe + P (2 + 2 a + 2 b ) e + P e OMPLEX HADAMARD MATRICES OF ORDER 6 15 where the coefficients
P, Q and R depend on the quadruple ( a, b, c, d ) only. If P ≡ P ( e ) of modulus one.STEP e, f ) , ( s , s ) and ( s , s ) is symmetric, and from all of the roots of P ofmodulus one we select all possible triplets ( e, s , s ) satisfying e + s + s = − − a − b , whichis needed to ensure orthogonality of the first two rows. From (25) we compute the uniquecompanion values f = F ( e ) , s = F ( s ) and s = F ( s ).Otherwise, should the number on the right hand side of (26) is of modulus one, then forevery root e of P we calculate its unique companion value f = F ( e ), and then we use theDecomposition formula to determine the pairs ( s , s ) and ( s , s ).If no unimodular roots are found at this point then the matrix E ( a, b, c, d ) cannot beembedded into a complex Hadamard matrix of order 6, and we do not proceed any further.At the end of this step we store all obtained sextics ( e, s , s , f, s , s ) in the solution set SOL B . Typically two sextics are found.STEP SOL C in a similar way.STEP SOL B and SOL C we check if the first threerows and columns are orthogonal, disregard those cases in which the submatrix B is singularand finally use Lemma 2.8 to obtain the lower right submatrix D . Note that by Lemma 2.14we disregard members of the family K (3)6 only. We check if D is composed of unimodularentries.STEP OUTPUT all unimodular matrices found during the process. We remarkhere that by Corollary 2.10 if no unimodular matrices were found then the submatrix E cannot be embedded into any complex Hadamard matrices of order 6. If unimodular matricesare found, then typically we find two matrices, as the solution set SOL B and SOL C containstwo suitable sextics each, however, experimental results show that for each sextic in SOL B there is a unique sextic in SOL C making D unimodular as required.We have finished the discussion of Construction 3.1. The results are summarized in thefollowing Theorem 4.1.
Start from a submatrix E as in (1) and suppose that there are only finitelymany (invertible) candidate submatrices B and C such that the first three rows and columnsof the matrix G (see (2) ) are orthogonal. Then Construction 3.1 gives an exhaustive list ofall complex Hadamard matrices of order , up to equivalence, containing E as a submatrix. The interested reader might want to see an example of generic Hadamard matrices whichcan be described by closed analytic formulæ, that is for which the fundamental polynomials P a,b,c,d and P a,c,b,d are both solvable. Such a matrix can be obtained when we choose the inputquadruple ( a, a, c, a ) where the real part of a is the unique real solution of 4 ℜ [ a ] − ℜ [ a ]+1 =0 and c = ( − a + a + a + 1) / ( a + a + a − a ). It is easily seen that the matrices we obtainstarting from the submatrix E ( a, a, c, a ) are inequivalent from S (0)6 and do not belong to thefamily K (3)6 . Remark . When
P ≡ B . In this case we have the trivial restriction (3) on e , while the companion value f coming from (25) is unimodular unconditionally. Although in principle we can find three orthogonal rows through the Decomposition formula for everysuitable e , we do not know which one to favourize in order to obtain a unimodular submatrix D via formula (20). Also, it might happen that the polynomial P a,c,b,d obtained during step P a,b,c,d P a,c,b,d
0, then we have a finitelymany choices for the submatrices B and C and we can use Corollary 2.10 to conclude theconstruction. Remark . The polynomial P formally can vanish when we have F G − F G ≡ F G −F G ≡
0, however this is excluded by the Canonical Transformation and explained in detailsin Case 1 of step P canbe calculated by means of Gr¨obner bases, but as these coefficients are rather complicatedobtaining such a basis turned out to be a task beyond our capabilities. Nevertheless, weconjecture that the case P ≡ S (0)6 and K (3)6 , can be recoveredfrom Construction 3.1.It is reasonable to think that every complex Hadamard matrix of order 6 has some 3 × E leading to nonvanishing fundamental polynomials. In particular, we do notexpect any complex Hadamard matrices of order 6 (except maybe S (0)6 and K (3)6 ) whichcannot be recovered from Construction 3.1. Therefore we formulate the following Conjecture 4.4.
The list of complex Hadamard matrices of order is as follows: the isolatedmatrix S (0)6 , the three-parameter degenerate family K (3)6 and the four-parameter generic family G (4)6 as described above. It would be nice to understand the structure of G (4)6 more thoroughly and express theentries of these matrices by some well-chosen trigonometric functions in a similar fashion as K (3)6 is described, however, as we have encountered sextic polynomials already the appearanceof such formulas is somewhat unexpected. Also, it is natural to ask which matrices satisfy theconditions of Corollary 2.18. An algebraic characterization of these matrices might lead to adeeper understanding of the generic family G (4)6 and hopefully to the desired full classificationof complex Hadamard matrices of order 6. References [1]
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