Biangular Lines Darcy Best March 24, 2014 Joint work with: Hadi - PowerPoint PPT Presentation

Biangular Lines Darcy Best March 24, 2014 Joint work with: Hadi Kharaghani (University of Lethbridge)

Hadamard Matrices A Hadamard matrix, H , is a square matrix (of order n ) with entires in {± 1 } such that HH T = nI . Example of a Hadamard Matrix:   1 1 1 1 1 − 1 −     1 1 − −   1 − − 1 Note: − denotes − 1.

Constructing Hadamard Matrices We can easily construct a 2 × 2 Hadamard matrix as follows: � 1 � 1 H 2 = . 1 − We can use this matrix to generate H 4 (on the previous slide):   1 1 1 1 � H 2 � H 2 1 − 1 −   H 4 = =  .   H 2 − H 2 1 1 − −  1 − − 1

Constructing Hadamard Matrices Sylvester noticed that H 2 k could be created for any k ∈ N by continuing the same procedure. � 1 � H 2 k − 1 � � 1 H 2 k − 1 H 2 k = ⊗ H 2 k − 1 = 1 − H 2 k − 1 − H 2 k − 1 This observation can be extended to arbitrarily large matrices: If we have two Hadamard matrices of order p and q , then we can construct another Hadamard matrix of order pq in the following way: H pq = H p ⊗ H q

Hadamard Conjecture It is easily proven that if H is a Hadamard matrix of order n , then n = 1 , 2 or 4 k . The converse, though, is known as the Hadamard conjecture. Conjecture A Hadamard matrix exists for every order n = 4 k. (True for n < 668)

Unbiased Hadamard Matrices We call two Hadamard matrices (say H and K ) unbiased if HK T = √ nL where L is a Hadamard matrix. Example:     1 1 1 1 1 1 1 − 1 1 − − 1 1 − 1     H =  , K =     1 − 1 − 1 − 1 1    1 − − 1 − 1 1 1     2 2 2 2 1 1 1 1 2 2 − 2 − 2 1 1 − − HK T =      = 2     2 − 2 2 − 2 1 − 1 −    − 2 2 2 − 2 − 1 1 − � �� � L

Unbiased Hadamard Matrices If we have a set of Hadamard matrices that are pairwise unbiased, then we have a set of mutually unbiased Hadamard matrices.     1 1 1 1 1 1 1 − 1 1 − − 1 1 − 1     H =  , K =     1 − 1 − 1 − 1 1    1 − − 1 − 1 1 1 Can we find any more Hadamard matrices that are unbiased with both H and K ? No.

Unbiased Hadamard Matrices There are larger sets of mutually unbiased Hadamard matrices of order 4 if we allow complex entries. Two differences between the real and complex case: ◮ Entries must lie on the unit circle. ◮ Use the conjugate transpose, H ∗ , instead of the transpose.     1 1 1 1 1 − j j 1 1 − − 1 − i i     H 1 =  , H 2 =  ,     1 − 1 − 1 1 i j   1 − − 1 1 1 j i     1 j j − 1 j − j 1 j i i 1 j 1 i     H 3 =  , H 4 =     1 i i 1 1 i − i    1 i j j 1 i 1 j Note: j denotes − i .

Deconstruct the Hadamard Matrices Take the rows of the Hadamard matrices and place them into a set.         1 1 1 1 1 − j j 1 j j − 1 j − j 1 1 −− 1 − i i 1 j i i 1 j 1 i          ,  ,  ,         1 − 1 − 1 1 i j 1 i i 1 1 i − i      1 −− 1 1 1 j i 1 i j j 1 i 1 j ↓   � � � � � � � �   1 , 1 , 1 , 1 1 1 1 1 1 1 −− 1 − 1 − 1 −− 1 ,     2 2 2 2         � � � � � � � �   1 , 1 , 1 , 1  1 − j j 1 − i i 1 1 i j 1 1 j i ,    2 2 2 2 V = � � � � � � � � 1 , 1 , 1 , 1  1 j j − 1 j i i 1 i i 1 1 i j j ,      2 2 2 2         � � � � � � � � 1 , 1 , 1 , 1   1 j − j 1 j 1 i 1 i − i 1 i 1 j     2 2 2 2

Deconstruct the Hadamard Matrices   � � � � � � � �   1 , 1 , 1 , 1 1 1 1 1 1 1 −− 1 − 1 − 1 −− 1 ,     2 2 2 2         � � � � � � � �   1 , 1 , 1 , 1  1 − j j 1 − i i 1 1 i j 1 1 j i ,    2 2 2 2 V = � � � � � � � � 1 , 1 , 1 , 1   1 j j − 1 j i i 1 i i 1 1 i j j ,    2 2 2 2         � � � � � � � �  1 , 1 , 1 , 1   1 j − j 1 j 1 i 1 i − i 1 i 1 j     2 2 2 2 Properties: ◮ Take any two vectors from the same matrix, inner product is 0. ◮ Take any two vectors from different matrices, the inner product has absolute value 1 2 . � 1 � � 1 � ∗ � √ = 1 4 HK ∗ = 1 � = 1 2 H 2 K 4 L 2 L 4

Biangular Lines Definition A set of unit vectors V is called biangular if for any (distinct) pair of vectors, u , v ∈ V , |� u , v �| ∈ { 0 , α } , where 0 < α < 1. Question Given n (the dimension) and α , what is the maximum size of a set of biangular vectors?

Bounds on the Size of Sets Theorem Let V ⊂ R n be a set of unit vectors. If |� v , w �| ∈ { 0 , α } for all v , w ∈ V , v � = w, where 0 < α < 1 , then � n + 2 � | V | ≤ . 3 Proof We can show that A = { X v := v ⊗ v ⊗ v | v ∈ V } ⊂ S 3 ( R n ) is a set of linearly independent vectors in S 3 ( R n ) which has dimension � n +2 � . 3

Bounds on the Size of Sets Theorem Let V ⊂ R n be a set of unit vectors. If |� v , w �| ∈ { 0 , α } for all v , w ∈ V , v � = w, where 0 < α < 1 , then � n + 2 � | V | ≤ . 3 Theorem Let V ⊂ C n be a set of unit vectors. If |� v , w �| ∈ { 0 , α } for all v , w ∈ V , v � = w, where 0 < α < 1 , then � n + 1 � | V | ≤ n . 2

Bounds on the Size of Sets – Now with α ! Theorem Let V ⊂ R n be a set of unit vectors. If |� v , w �| ∈ { 0 , α } for all v , w ∈ V , v � = w, where 0 < α < 1 , then | V | ≤ n ( n + 2)(1 − α 2 ) 3 − ( n + 2) α 2 if the denominator is positive. Theorem Let V ⊂ C n be a set of unit vectors. If |� v , w �| ∈ { 0 , α } for all v , w ∈ V , v � = w, where 0 < α < 1 , then | V | ≤ n ( n + 1)(1 − α 2 ) 2 − ( n + 1) α 2 if the denominator is positive.

Comparisons ◮ AUB (Absolute Upper Bound) – Bound without α ◮ SUB (Special Upper Bound) – Bound with α R n C n n α AUB SUB AUB SUB 2 1 / 2 4 3 6 18/5 3 1 / 2 10 45/7 18 9 4 1 / 2 20 12 40 20 5 1 / 2 35 21 75 45 6 1 / 2 56 36 126 126 7 1 / 2 84 63 196 N/A 8 1 / 2 120 120 288 N/A 9 1 / 2 165 297 405 N/A 10 1 / 2 220 N/A 550 N/A

Old Example   � � � � � � � �  1 , 1 , 1 , 1  1 1 1 1 1 1 −− 1 − 1 − 1 −− 1 ,     2 2 2 2         � � � � � � � �   1 , 1 , 1 , 1 1 − j j 1 − i i 1 1 i j 1 1 j i ,     2 2 2 2 V = � � � � � � � � 1 , 1 , 1 , 1  1 j j − 1 j i i 1 i i 1 1 i j j ,     2 2 2 2         � � � � � � � �  1 , 1 , 1 , 1   1 j − j 1 j 1 i 1 i − i 1 i 1 j     2 2 2 2 V has 16 vectors, but the special upper bound is 20, can we attain this amount? Yes. � � V ∪ (1 0 0 0) , (0 1 0 0) , (0 0 1 0) , (0 0 0 1)

Unbiased Hadamard Matrices When you have m mutually unbiased Hadamard matrices (in C ) of order n , then we can deconstruct the matrices into mn biangular 1 vectors with α = √ n . We can always add the rows of the identity to the set of vectors and preserve biangularity. ( m + 1) n = mn + n ≤ n ( n + 1)(1 − α 2 ) = ( n + 1) n 2 − ( n + 1) α 2 So... m ≤ n . When you have m mutually unbiased Hadamard matrices (in R ), then m ≤ n 2 .

Unbiased Hadamard Matrices When n = 4, this bound is attained (seen above). When else is the bound attained? Theorem If n is a prime power, then there are n mutually unbiased Hadamard matrices. (Constructive proof) Conjecture No other value of n attains this upper bound. In the first non-prime power case ( n = 6), the upper bound is conjectured to be only 2. (Strong computational evidence to support)

Orthogonality Between Matrices So far, the examples we have given always have an inner product of zero within a matrix, and nonzero between different matrices. We will loosen that slightly to allow the inner product between matrices to be zero as well.     1 1 1 1 1 1 1 1 1 1 1 − 1 − 1 1 1 1 − 1 −− 1 − 1 − 1 1 1 1 1 −         1 −− 1 1 −− 1 1 −− 1 −− 1 1         1 −−−− 1 1 1 1 1 − 1 1 −−−     H = , K =  1 1 −− 1 1 −−   1 1 −−− 1 1 −       1 1 1 −−−− 1   1 − 1 −−−−−      1 − 1 1 − 1 −− 1 −−− 1 1 − 1     1 − 1 − 1 − 1 − 1 1 1 1 − 1 − 1 The inner product between vectors from H and K are in { 0 , ± 4 } .