New bounds for spherical two-distance sets and equiangular lines

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1 New bounds for spherical two-distance sets and equiangular lines Michigan State University Oct 8-31, 016 Anhui University

2 Definition If X = {x 1, x,, x N } S n 1 (unit sphere in R n ) and x i, x j = a or b for all i j, then we call X is a spherical two-distance set. Q : What is the maximum cardinality of a spherical two-distance set?

3 Figure: The maximum spherical two-distance set in R : Pentagon. Figure: The maximum spherical two-distance set in R 3 : Octahedron.

4 Known results Let g(n) denote the maximum size of spherical two-distance set in R n 1 Let e 1,..., e n+1 be the standard basis in R n+1. The points e i + e j, i j form a spherical two-distance set in the plane x x n+1 =, g(n) n(n + 1), n. (1)

5 Known results Let g(n) denote the maximum size of spherical two-distance set in R n 1 Let e 1,..., e n+1 be the standard basis in R n+1. The points e i + e j, i j form a spherical two-distance set in the plane x x n+1 =, g(n) n(n + 1), n. (1) Delsarte, Goethals, and Seidel in 1977 proved so-called harmonic bound : n(n + 3) g(n). () They also showed that this bound is tight for n =, 6,.

6 Known results Let g(n) denote the maximum size of spherical two-distance set in R n 1 Let e 1,..., e n+1 be the standard basis in R n+1. The points e i + e j, i j form a spherical two-distance set in the plane x x n+1 =, g(n) n(n + 1), n. (1) Delsarte, Goethals, and Seidel in 1977 proved so-called harmonic bound : n(n + 3) g(n). () They also showed that this bound is tight for n =, 6,. 3 Therefore, n(n+1) g(n) n(n+3).

7 Musin s result Musin used Delsarte s linear programming method to prove that g(n) = and g(3) = 76 or 77. n(n + 1) if 7 n 39, n, 3

8 Barg and Yu 014 We use the semidefinite programming (SDP) method showing that g(n) = In particular, g(3) = 76. n(n + 1), 7 n 93, n, 46, 78.

9 Maximum spherical two-distance sets Theorem (Yu 016+) n(n + 1) g(n) =, 7 n 417, n, 46, 78, 118, 166,, 86, 358 which are (k + 1) 3 for k =,, 9. For a + b 0, Oleg Musin proved the upper bounds are n(n+1). For a + b < 0, the upper bounds can be obtained from the bounds of equiangular lines in one higher dimension. Lemma If there are at most n(n 1) equiangular lines in R n, then maximum size of a spherical two-distance set in R n 1 is n(n 1). Musin Conjecture: g(n) = n(n + 1) n N, except n + 3 = (k + 1), k N.

10 Maximum spherical two-distance sets Theorem (Glazyrin and Yu 016+) g(n) = n(n + 1) n N, except n = (k + 1) 3, k N.

11 Definition A set of lines in R n is called equiangular if the angle between each pair of lines is the same. An equiangular line set can be defined as an unit vectors set S = {x i } M i=1 such that x i, x j = c, 1 i < j M for some c > 0. A equiangular line set can be defined as a spherical two-distance set with inner product value c and c. Question : What is the maximum cardinality of an equiangular line set in R n?

12 Figure: Maximum equiangular lines in R : 3 lines through opposite vertices of a regular hexagon. Figure: Maximum equiangular lines in R 3 : 6 lines through opposite vertices of the icosahedron.

13 Known results Let M(n) denote the maximum size of an equiangular line set in R n Hanntjes found M(n) for n = 3 and 4 in Van Lint and Seidel found the largest number of equiangular lines for 5 n 7 in Lemmens and Seidel used linear-algebraic methods to determine M(n) for most values of n in the region 8 n 3 in n M(n) 1/α n M(n) 1/α ; n ; n Table: Known bounds on M(n) in small dimensions

14 Our results Theorem (Barg and Yu 014) We use the semidefinite programming (SDP) method to show that M(n) = 76 for 4 n 41 and M(43) = 344. n M(n) SDP bound n M(n) SDP bound n n Table: Bounds on M(n) including new results Remark : Recently, we can show no 76 equiangular lines in R 19.

15 Gegenbauer polynomials Let G (n) k (t), k = 0, 1,... denote the Gegenbauer polynomials of degree k. They are defined recursively as follows: G (n) 0 1, G (n) 1 (t) = t, and G (n) k (k + n 4)tG(n) (t) = k 1 (t) (k 1)G(n) k + n 3 k (t), k.

16 Define a matrix Yk n (u, v, t), k 0 ( (Yk n (u, v, t)) ij = u i v j ((1 u )(1 v )) k/ G (n 1) t uv ) k (1 u )(1 v ) and a matrix Sk n (u, v, t) by setting S n k (u, v, t) = 1 6 σ S 3 Y n k (σ(u, v, t)), (3) I.J. Schoenberg (194) proved that if C is the finite set in S n 1, then (x,y) C G n k( x, y ) 0. (x,y,z) C 3 S n k (x y, x z, y z) 0.

17 Semidefinite programming min c T x m subject to F 0 + F i x i 0 where c, x R m and F i is an n by n symmetric matrix i. The sign means that the matrix is positive semidefinite. i=1 CVX (MATLAB toolkit) can solve an SDP in a second.

18 Theorem (Gerzon, absolute bounds) If there are M equiangular lines in R n, then M n(n+1). Gerzon bounds are known to be attained only for n =, 3, 7, and 3. Theorem (Neumann) If there are M equiangular lines in R n with angle arccos α and M > n, then 1/α is an odd integer. Theorem (Lemmens and Seidel) M 1/3 (n) = (n 1) for n 16, where M α (n) is the maximum size of an equiangular line set when the value of the angle is arccos α. Theorem (Relative bounds) M α (n) n(1 α ) 1 nα (4) valid for all α such that the denominator is positive.

19 SDP bound Theorem Let C be an equiangular line set with inner product values either a or a. Let p be a positive integer. The cardinality C is bounded above by the solution of the following semi-definite programming problem : subject to max(x 1 + x ) (5) ( 1 0 ) + 1 ( 0 1 ) ( 0 0 ) (x x ) + (x x 4 + x 5 + x 6 ) 0 (6) S n k (1, 1, 1) + S n k (a, a, 1)x 1 + S n k ( a, a, 1)x + S n k (a, a, a)x 3 + S n k (a, a, a)x 4 + S n k (a, a, a)x 5 + S n k ( a, a, a)x 6 0 (7) 3 + G (n) k (a)x 1 + G (n) k ( a)x 0, (8) where k = 0, 1,, p and x j 0, j = 1,, 6.

20 SDP bound table n 1/5 1/7 1/9 1/11 1/13 1/15 max Gerzon angle / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /7

21 SDP bound table (Cont.) n 1/5 1/7 1/9 1/11 1/13 1/15 max Gerzon angle / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /9

22 SDP bound table (Cont.) n 1/5 1/7 1/9 1/11 1/13 1/15 max Gerzon angle / / / / / / / / / / / / / / / / / / / / / / / / / / / / /5 * /5 * /5 * /5

23 Our results Theorem (Barg and Yu) We use the semidefinite programming method to show that M(n) = 76 for 4 n 41 and M(43) = 344 and we get tighter upper bounds for M(n) when n 136.

24 Observation angle dim bounds = = = = 11 Theorem (Yu 016+) We prove M a (n) 1 ( 1 a 1)( 1 a ) for all a N and for all 1 a n 3 a 16

25 Relaxation semidefinite programming problems Theorem Let C be an equiangular line set with inner product values either a or a. Let p be a positive integer. The cardinality C is bounded above by the solution of the following semi-definite programming problem : subject to max(x 1 + x ) (9) ( 1 0 ) + 1 ( 0 1 ) ( 0 0 ) (x x ) + (x x 4 + x 5 + x 6 ) 0 (10) S n k (1, 1, 1) + S n k (a, a, 1)x 1 + S n k ( a, a, 1)x + S n k (a, a, a)x 3 + S n k (a, a, a)x 4 + S n k (a, a, a)x 5 + S n k ( a, a, a)x 6 0 (11) 3 + G (n) k (a)x 1 + G (n) k ( a)x 0, (1) where k = 0, 1,, p and x j 0, j = 1,, 6. We find that above constraints only S n 1 0, S n 3 0 and (9) are crucial.

26 Symbolic semidefinite programming problems Theorem The solution of following optimization problem is 1 ( 1 a )( 1 a 1). max(1 + A) subject to a 4 (3a + 1) A + (6a 1)(a + 1) 3 ) B + a 4 (3a 1) (6a 1)(a 1) 3 ) C 0 (13) A + a 1 + a B + a a 1 C 0 (14) A(A 1) B + C (15) proof: We choose suitable t, where t = 16a 6 (6a 1)(a+1) (a 1) such that t a 1+a + a4 (3a+1) (6a 1)(a+1) = t a 3 a 1 + If we calculate t(14) + (13), we will get a4 (3a 1) (6a 1)(a 1) 3 a (t + 1)A + (t 1 + a + a 4 (3a + 1) (6a )(B + C) 0 1)(a + 1) 3 10a6 + 13a 4 8a + 1 (6a 1)(a 1) (a + 1) A a 4 (5a 1) (6a 1)(a 1) (B + C) 0 (a + 1)

27 notice that a 4 (5a 1) (6a 1)(a 1) 0 and we plug in (15). (a + 1) 10a6 + 13a 4 8a + 1 (6a 1)(a 1) (a + 1) A a 4 (5a 1) (6a 1)(a 1) A(A 1) 0 (a + 1) 10a6 + 13a 4 8a + 1 (6a 1)(a 1) (a + 1) a 4 (5a 1) (6a 1)(a 1) (A 1) (a + 1) 1 3a a 4 a 4 A 1 A 1 3a a 4 Then, it is not hard to see that A ( 1 a )( 1 a 1).

28 New bounds for equiangular lines Theorem (Glazyrin and Yu 016+) M 1 (n) n( a 3 a ) +, for all a 3 and for all n N. Theorem If n 359, then M(n) (a )(a 1), where a is the unique positive odd integer such that a n (a + ) 3. Notice that if n = a, then we just obtain Gerzon bound M(n) n(n+1). For other cases, we can prove that M(n) n(n 1).

29 Maximum spherical two-distance sets Theorem (Glazyrin and Yu 016+) g(n) = n(n + 1) n N, except n = (k + 1) 3, k N.

30 Future work 1 M(14) = 8 or 9. M(16) = 40 or 41. Can we determine them? The constructions and upper bounds for complex equiangular lines. 3 The maximum spherical 3-distance sets in R n. (only n =, 3, 8 and are known) 4 How much do we know the Maximum Separation Codes on sphere? If we have M points on S n 1, then max x i, x j M n n(m 1). = attained iff equiangular tight frames.

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