Complex Hadamard matrices and 3-class association schemes

Transcription

1 Complex Hadamard matrices and 3-class association schemes Akihiro Munemasa 1 1 Graduate School of Information Sciences Tohoku University (joint work with Takuya Ikuta) June 26, 2013 The 30th Algebraic Combinatorics Symposium Shizuoka University

2 Hadamard matrices and generalizations A (real) Hadamard matrix of order n is an n n matrix H with entries ±1, satisfying HH = ni. Conjecture Hadamard matrix of order n whenever n 0 (mod 4).

3 Hadamard matrices and generalizations A (real) Hadamard matrix of order n is an n n matrix H with entries ±1, satisfying HH = ni. Conjecture Hadamard matrix of order n whenever n 0 (mod 4). A complex Hadamard matrix of order n is an n n matrix H with entries in {ζ C ζ = 1}, satisfying HH = ni, where means the conjugate transpose.

4 Hadamard matrices and generalizations A (real) Hadamard matrix of order n is an n n matrix H with entries ±1, satisfying HH = ni. Conjecture Hadamard matrix of order n whenever n 0 (mod 4). A complex Hadamard matrix of order n is an n n matrix H with entries in {ζ C ζ = 1}, satisfying HH = ni, where means the conjugate transpose. We propose a strategy to construct infinite families of complex Hadamard matrices using association schemes, and demonstrate a successful case.

5 Circulant (complex) Hadamard matrices a 0 a 1 a n 1 a n 1 a 0 a 1 H = = n 1 a i A i, A = a 1 a Instead of n 2 entries, there are only n entries to determine.

6 Circulant (complex) Hadamard matrices a 0 a 1 a n 1 a n 1 a 0 a 1 H = = n 1 a i A i, A = a 1 a Instead of n 2 entries, there are only n entries to determine. Björck Fröberg ( ) circulant Hadamard, n 8 Faugère (2001), (2004) circulant Hadamard, n = 9, 10

7 Circulant (complex) Hadamard matrices a 0 a 1 a n 1 a n 1 a 0 a 1 H = = n 1 a i A i, A = a 1 a Instead of n 2 entries, there are only n entries to determine. Björck Fröberg ( ) circulant Hadamard, n 8 Faugère (2001), (2004) circulant Hadamard, n = 9, 10 On the other hand, it is conjectured that no circulant real Hadamard matrix of order > 4 exists.

8 Goethals Seidel (1970) symmetric regular (real) Hadamard matrix necessarily comes from a strongly regular graph (SRG) on 4s 2 vertices de la Harpe Jones (1990) SRG n: prime 1 (mod 4) symmetric circulant complex Hadamard

9 Goethals Seidel (1970) symmetric regular (real) Hadamard matrix necessarily comes from a strongly regular graph (SRG) on 4s 2 vertices de la Harpe Jones (1990) SRG n: prime 1 (mod 4) symmetric circulant complex Hadamard M. Watatani (1992) DRT n: prime 3 (mod 4) non-symmetric circulant complex Hadamard

10 Goethals Seidel (1970) symmetric regular (real) Hadamard matrix necessarily comes from a strongly regular graph (SRG) on 4s 2 vertices de la Harpe Jones (1990) SRG n: prime 1 (mod 4) symmetric circulant complex Hadamard M. Watatani (1992) DRT n: prime 3 (mod 4) non-symmetric circulant complex Hadamard H = α 0 A 0 + α 1 A 1 + α 2 A 2, A 0 = I

11 Goethals Seidel (1970) symmetric regular (real) Hadamard matrix necessarily comes from a strongly regular graph (SRG) on 4s 2 vertices de la Harpe Jones (1990) SRG n: prime 1 (mod 4) symmetric circulant complex Hadamard M. Watatani (1992) DRT n: prime 3 (mod 4) non-symmetric circulant complex Hadamard H = α 0 A 0 + α 1 A 1 + α 2 A 2, A 0 = I A 1 = A 1, A 2 = A 2 (G. J., de la H. J.) A 1 = A 2, (M. W.)

12 Goethals Seidel (1970) symmetric regular (real) Hadamard matrix necessarily comes from a strongly regular graph (SRG) on 4s 2 vertices de la Harpe Jones (1990) SRG n: prime 1 (mod 4) symmetric circulant complex Hadamard M. Watatani (1992) DRT n: prime 3 (mod 4) non-symmetric circulant complex Hadamard H = α 0 A 0 + α 1 A 1 + α 2 A 2, A 0 = I A 1 = A 1, A 2 = A 2 (G. J., de la H. J.) A 1 = A 2, (M. W.) Unifying principle: association schemes. (strongly regular graphs is a special case)

13 Godsil Chan (2010), and Chan (2011) classified complex Hadamard matrices of the form: H = α 0 I + α 1 A 1 + α 2 A 2 (we may assume α 0 = 1) where α 1 = α 2 = 1, and A 1 = adjacency matrix of a SRG Γ, A 2 = adjacency matrix of Γ.

14 Godsil Chan (2010), and Chan (2011) classified complex Hadamard matrices of the form: H = α 0 I + α 1 A 1 + α 2 A 2 (we may assume α 0 = 1) where α 1 = α 2 = 1, and A 1 = adjacency matrix of a SRG Γ, A 2 = adjacency matrix of Γ.

15 Godsil Chan (2010), and Chan (2011) classified complex Hadamard matrices of the form: H = α 0 I + α 1 A 1 + α 2 A 2 (we may assume α 0 = 1) where α 1 = α 2 = 1, and A 1 = adjacency matrix of a SRG Γ, A 2 = adjacency matrix of Γ. also found a complex Hadamard matrix of the form H = I + α 1 A 1 + α 2 A 2 + α 3 A 3 of order 15 from the line graph L(O 3 ) of the Petersen graph O 3.

16 The Bose Mesner algebra of a symmetric association scheme of class d A 0, A 1,..., A d = E 0, E 1,..., E d is a commutative semisimple algebra with primitive idempotents E 0, E 1,..., E d. A i = J, E i = I, A 0 = I, ne 0 = J. A j A k = δ jk A j, E j E k = δ jk E j

17 The Bose Mesner algebra of a symmetric association scheme of class d A 0, A 1,..., A d = E 0, E 1,..., E d is a commutative semisimple algebra with primitive idempotents E 0, E 1,..., E d. A i = J, E i = I, A 0 = I, ne 0 = J. A j A k = δ jk A j, E j E k = δ jk E j A j = p ij E i, A i = A i ( i), = p ji R.

18 The Bose Mesner algebra of a symmetric association scheme of class d A 0, A 1,..., A d = E 0, E 1,..., E d is a commutative semisimple algebra with primitive idempotents E 0, E 1,..., E d. A i = J, E i = I, A 0 = I, ne 0 = J. A j A k = δ jk A j, E j E k = δ jk E j A j = p ij E i, A i = A i ( i), = p ji R.

19 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard HH = ni

20 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard HH = ni ( p ki E k )( α j k=0 j=0 k=0 p kj E k ) = ni

21 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard HH = ni ( p ki E k )( α j k=0 j=0 k=0 p kj E k ) = ni

22 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( p ki E k )( α j k=0 j=0 k=0 p kj E k ) = ni

23 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( p ki E k )( k=0 k=0 j=0 α j j=0 k=0 α j p ki p kj E k = ni p kj E k ) = ni

24 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( p ki E k )( k=0 α j j=0 k=0 p kj E k ) = ni α j p ki p kj E k = ni = n k=0 j=0 k=0 E k

25 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( k=0 j=0 p ki E k )( k=0 α j j=0 k=0 p kj E k ) = ni α j p ki p kj E k = ni = n j=0 k=0 α j p ki p kj = n ( k) E k

26 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( k=0 j=0 p ki E k )( k=0 j=0 α j p kj E k ) = ni k=0 α j p ki p kj E k = ni = n j=0 α j p ki p kj = n ( k) k=0 E k

27 A i = When is H = matrix? p ki E k, E j E k = δ jk E j, p ki R. k=0 A i with = 1 a complex Hadamard H = ni ( k=0 j=0 0 i<j d p ki E k )( k=0 j=0 α j p kj E k ) = ni k=0 α j p ki p kj E k = ni = n j=0 p ki p kj = n ( k) α j ( αi + α ) j p ki p kj = n α j k=0 p 2 ki E k ( k)

28 a ij = α j + α j (0 i < j d). (1) 0 i<j d a ij p ki p kj = n p 2 ki ( k) (2)

29 a ij = α j + α j (0 i < j d). (1) 0 i<j d a ij p ki p kj = n p 2 ki ( k) (2) Step 1 Solve the system of linear equations (2) in {a ij } Step 2 Find { } from {a ij } by (1).

30 a ij = α j + α j (0 i < j d). (1) 0 i<j d a ij p ki p kj = n p 2 ki ( k) (2) Step 1 Solve the system of linear equations (2) in {a ij } Step 2 Find { } from {a ij } by (1). Problem: Given {a ij }, when { } satisfying (1)?

31 a ij = α j + α j (0 i < j d). (1) 0 i<j d a ij p ki p kj = n p 2 ki ( k) (2) Step 1 Solve the system of linear equations (2) in {a ij } Step 2 Find { } from {a ij } by (1). Problem: Given {a ij }, when { } satisfying (1)? f : (S 1 ) d+1 R d(d+1)/2, { } d { α j + α j } 0 i<j<d. where S 1 = {ζ C ζ = 1}.

32 a ij = α j + α j (0 i < j d). (1) 0 i<j d a ij p ki p kj = n p 2 ki ( k) (2) Step 1 Solve the system of linear equations (2) in {a ij } Step 2 Find { } from {a ij } by (1). Problem: Given {a ij }, when { } satisfying (1)? f : (S 1 ) d+1 R d(d+1)/2, { } d { α j + α j } 0 i<j<d. where S 1 = {ζ C ζ = 1}. Describe image of f

33 Instead of considering f : (S 1 ) d+1 R d(d+1)/2, consider f : (C ) d+1 C d(d+1)/2, { } d { α j + α j } 0 i<j<d. Describe the image of f. For example, for d = 2:

34 Instead of considering f : (S 1 ) d+1 R d(d+1)/2, consider f : (C ) d+1 C d(d+1)/2, { } d { α j + α j } 0 i<j<d. Describe the image of f. For example, for d = 2: f : (C ) 3 C 3, (x, y, z) ( x + y, x + z, y + z ) y x z x z y

35 Instead of considering f : (S 1 ) d+1 R d(d+1)/2, consider f : (C ) d+1 C d(d+1)/2, { } d { α j + α j } 0 i<j<d. Describe the image of f. For example, for d = 2: Not surjective. f : (C ) 3 C 3, (x, y, z) ( x + y, x + z, y + z ) y x z x z y

36 Instead of considering f : (S 1 ) d+1 R d(d+1)/2, consider f : (C ) d+1 C d(d+1)/2, { } d { α j + α j } 0 i<j<d. Describe the image of f. For example, for d = 2: f : (C ) 3 C 3, (x, y, z) ( x + y, x + z, y + z ) y x z x z y Not surjective. g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g( x y + y x, x z + z x, y z + z y ) = 0

37 Instead of considering f : (S 1 ) d+1 R d(d+1)/2, consider f : (C ) d+1 C d(d+1)/2, { } d { α j + α j } 0 i<j<d. Describe the image of f. For example, for d = 2: f : (C ) 3 C 3, (x, y, z) ( x + y, x + z, y + z ) y x z x z y Not surjective. g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g( x y + y x, x z + z x, y z + z y ) = 0 Indeed, image of f = zeros of g.

38 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3

39 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0

40 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0 image of f? = zeros of {g i,j,k }.

41 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0 image of f zeros of {g i,j,k }.

42 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0 image of f zeros of {g i,j,k }. Need h = (X )X 12 X 03 (X 01 X 23 + X 02 X 13 ) + 2(X 01 X 02 + X 13 X 23 ) (X ij = x i x j + x j x i ) (and similar polynomials obtained by permuting indices)

43 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0 image of f zeros of {g i,j,k }. Need h = (X )X 12 X 03 (X 01 X 23 + X 02 X 13 ) + 2(X 01 X 02 + X 13 X 23 ) (X ij = x i x j + x j x i ) (and similar polynomials obtained by permuting indices) image of f = zeros of g, h.

44 f : (C ) 4 C 6, (x 0, x 1, x 2, x 3 ) ( x i x j + x j x i ) 0 i<j 3 g(x, Y, Z) = X 2 + Y 2 + Z 2 XY Z 4. g i,j,k = g( x i x j + x j x i, x i x k + x k x i, x j x k + x k x j ) = 0 image of f zeros of {g i,j,k }. Need h = (X )X 12 X 03 (X 01 X 23 + X 02 X 13 ) + 2(X 01 X 02 + X 13 X 23 ) (X ij = x i x j + x j x i ) (and similar polynomials obtained by permuting indices) image of f = zeros of g, h. The same is true for m 4.

45 Theorem f : (C ) d+1 C d(d+1)/2, (α 0, α 1,..., α d ) ( α j + α j ) 0 i<j d The image of f coincides with the set of zeros of the ideal I in the polynomial ring C[X ij : 0 i < j d] generated by g(x ij, X ik, X jk ) h(x ij, X ik, X il, X jk, X jl, X kl ) where i, j, k, l are distinct, X ij = X ji, and g = X 2 + Y 2 + Z 2 XY Z 4, h = (Z 2 4)U Z(XW + Y V ) + 2(XY + V W ).

46 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1)

47 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1) How do we find ( ), and when does ( ) (S 1 ) d+1 hold?

48 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1) How do we find ( ), and when does ( ) (S 1 ) d+1 hold? Observe, for α C, α = 1 2 α + 1 α 2.

49 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1) How do we find ( ), and when does ( ) (S 1 ) d+1 hold? Observe, for α C, α = 1 2 α + 1 α 2.

50 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1) How do we find ( ), and when does ( ) (S 1 ) d+1 hold? Observe, for α C, So we need 2 a ij 2. α = 1 2 α + 1 α 2.

51 Given a zero (a ij ) of the ideal I, we know that there exists ( ) (C ) d+1 such that a ij = α j + α j (0 i < j d). (1) How do we find ( ), and when does ( ) (S 1 ) d+1 hold? Observe, for α C, α = 1 2 α + 1 α 2. So we need 2 a ij 2. Moreover, if a ij {±2} for all i, j, then = ±α j so the resulting matrix is a scalar multiple of a real Hadamard matrix Goethals Seidel (1970).

52 Theorem f : (C ) d+1 C d(d+1)/2, (α 0, α 1,..., α d ) ( α j + α j ) 0 i<j d Suppose (a ij ) the image of f, a ij R, and there exists 0 i 0 < i 1 d such that 2 < a i0,i 1 < 2. Let 0, 1 be a i0,i 1 = Define (0 i n, i i 0, i 1 ) by = 0 (a i0,i ) a i1,i1 a i0,i0. Then = α j and + α j = a ij α j (0 i < j d). (1) and every ( ) satisfying (1) is obtained in this way.

53 Theorem f : (C ) d+1 C d(d+1)/2, (α 0, α 1,..., α d ) ( α j + α j ) 0 i<j d Suppose (a ij ) the image of f, a ij R, and there exists 0 i 0 < i 1 d such that 2 < a i0,i 1 < 2. Let 0, 1 be a i0,i 1 = = Define (0 i n, i i 0, i 1 ) by ( ) 1 αi0 1 = 0 (a i0,i ) a i1,i1 a i0,i0. Then = α j and + α j = a ij α j (0 i < j d). (1) and every ( ) satisfying (1) is obtained in this way.

54 The procedure Step 1 Solve the system of equations g(x ij, X ik, X jk ) = 0, h(x ij, X ik, X il, X jk, X jl, X kl ) = 0, X ij p ki p kj = n 0 i<j d p 2 ki Step 2 List all solutions a ij with 2 a ij 2. Step 3 Find ( ) by where a i0,i 1 ±2, = 0 (a i0,i ) a i1,i1 a i0,i = a i0,i 1.

55 Step 1 Solve the system of equations Step 2 List all solutions a ij with 2 a ij 2, Step 3 Find ( )

56 Step 1 Solve the system of equations Step 2 List all solutions a ij with 2 a ij 2, Step 3 Find ( ) In many known examples of association schemes with d = 3, Step 2 failed.

57 Step 1 Solve the system of equations Step 2 List all solutions a ij with 2 a ij 2, Step 3 Find ( ) In many known examples of association schemes with d = 3, Step 2 failed. Theorem (Chan, arxiv: v1) There are only finitely many antipodal distance-regular graphs of diameter 3 whose Bose Mesner algebra contains a complex Hadamard matrix.

58 Step 1 Solve the system of equations Step 2 List all solutions a ij with 2 a ij 2, Step 3 Find ( ) In many known examples of association schemes with d = 3, Step 2 failed. Theorem (Chan, arxiv: v1) There are only finitely many antipodal distance-regular graphs of diameter 3 whose Bose Mesner algebra contains a complex Hadamard matrix. But Chan did find an example. L(O 3 ): the line graph of the Petersen graph.

59 q: a power of 2, q 4, Ω = PG(2, q): the projective plane over F q, Q = {[a 0, a 1, a 2 ] Ω a a 1 a 2 = 0}: quadric, X = {[a 0, a 1, a 2 ] Ω \ Q [a 0, a 1, a 2 ] [1, 0, 0]}, X = q 2 1.

60 q: a power of 2, q 4, Ω = PG(2, q): the projective plane over F q, Q = {[a 0, a 1, a 2 ] Ω a a 1 a 2 = 0}: quadric, X = {[a 0, a 1, a 2 ] Ω \ Q [a 0, a 1, a 2 ] [1, 0, 0]}, X = q 2 1. For x, y X, denote by x + y the line through x, y. 1 if i = 0, x = y, 1 if i = 1, (x + y) Q = 2, (A i ) xy = 1 if i = 2, (x + y) Q = 0, 1 if i = 3, (x + y) Q = 1, 0 otherwise. a complex Hadamard matrix in its Bose Mesner algebra.

61 q: a power of 2, q 4, Ω = PG(2, q): the projective plane over F q, Q = {[a 0, a 1, a 2 ] Ω a a 1 a 2 = 0}: quadric, X = {[a 0, a 1, a 2 ] Ω \ Q [a 0, a 1, a 2 ] [1, 0, 0]}, X = q 2 1. For x, y X, denote by x + y the line through x, y. 1 if i = 0, x = y, 1 if i = 1, (x + y) Q = 2, (A i ) xy = 1 if i = 2, (x + y) Q = 0, 1 if i = 3, (x + y) Q = 1, 0 otherwise. a complex Hadamard matrix in its Bose Mesner algebra.

62 q: a power of 2, q 4 q = 4, Ω = PG(2, q): the projective plane over F q, Q = {[a 0, a 1, a 2 ] Ω a a 1 a 2 = 0}: quadric, X = {[a 0, a 1, a 2 ] Ω \ Q [a 0, a 1, a 2 ] [1, 0, 0]}, X = q For x, y X, denote by x + y the line through x, y. 1 if i = 0, x = y, 1 if i = 1, (x + y) Q = 2, (A i ) xy = 1 if i = 2, (x + y) Q = 0, 1 if i = 3, (x + y) Q = 1, 0 otherwise. a complex Hadamard matrix in L(O 3 ).

63 Theorem The matrix H = I + α 1 A 1 + α 2 A 2 + α 3 A 3 is a complex Hadamard matrix if and only if (i) H belongs to the subalgebra forming the Bose Mesner algebra of a strongly regular graph (precise description omitted, already done by Chan Godsil), (ii) α α 1 = 2 q, α 2 = 1 α 1, α 3 = 1, (iii) α α 1 = (q 1)(q 2) (q + 2)r, q where r = (q 1)(17q 1) > 0. The case (ii) with q = 4 was found by Chan.