1. Equiangular Lines in R d Moshe Rosenfeld University of Washington Shanghai Jiao Tong University July 1, 2013

2. How wide and how even can you spread your chop-sticks?

3. Equiangular Lines Definition: A set of lines through the origin in R d is equiangular if the angle between any pair of distinct lines is the same. In R 2, the maximum number of equiangular lines is 3. The only possible angle among 3 equiangular lines in R 2 is 60 o. The largest minimal angle among k lines through the origin in R 2 is 180/k. What is happening in higher dimensions?

4. Study of equiangular lines involves: Topology Linear Algebra Group Theory Quantum Theory Geometry Graph Theory Combinatorial Designs

5. Equiangular Lines in R 3 The four diagonals of the regular cube in R 3 form a set of four equi-angular lines. Angle: arccos(1/3).

6. More equiangular lines in R 3 The six diagonals of the icosahedron form a set of 6 equiangular lines in R 3. Angle ???? Is it fair to ask a student in a final exam to calculate the angle? The coordinates of the 12 vertices?

7. How many other angles are possible among 4 equiangular lines in R 3 ? How many equiangular lines are possible in R 3 How do you approach such a problem? Matrices  Eigenvalues  Graphs

8. Seidel’s adjacency matrix For a given graph G, define the S-adjacency matrix as follows: S i,i = 0 S i,j = -1 is (i,j)  E(G) S i,j = 1 otherwise. Note: for a given graph G, S = J – 2A – I.

9. 1.For the icosahedron, Select a unit vector on each diagonal. 2.The Gram-Schmidt matrix generated by them will be: This is a positive semi-definite matrix of rank 3. Conversely, for every  for which this matrix is PSD with rank 3 we can form six equiangular lines in R 3 with angle  = cos 

10. Constructing the icosahedron from a graph The smallest eigenvalue of the S matrix is -  5 and its multiplicity is 3. Yielding 6 equiangular lines with angle arccos(1/  5 ) – the icosahedron.

11. Note: the last matrix is the S matrix of this graph.

12. The S-matrix of this graph: This is an orthogonal matrix. So A 2 = 5I and its eigenvalues are {  5 {3}, -  5 {3} }. So we found the angle: arccos(1/  5 ) and we can calculate the coordinates of the 12 vertices of the icosahedron!

13. Doesn’t the matrix depend on the choice of unit vectors? Switching: U  V(G). Switching G with respect to U means if u  U and x  V(G) \ U then remove the edge ux if ux  E(G) and add it otherwise. Switching is an equivalence relation. All graphs in a switching class will produce isometric equiangular lines. Switching: J.J.Seidel

14. Graphs on 4 vertices fall into 3 switching classes: 1. K 4  4 diagonals of the cube 2. C 4  degenerate case 3. K 4 – K 2 (any four diagonals of the icosahedron). Conclusion: there are only two possible angles among 4 – equiangular lines in R 3 For each d, there are finitely many angles  possible for d+1  -equiangular lines in R d

15. How wide can you spread your chopsticks? For any set L n of n lines through the origin in R 3 let  (L n ) = smallest angle among all distinct pairs of lines from L n Let  (n) = max  (L n )  (4) = arccos(1/3)  (5) =  (6) = arccos(1/  5)  (7) = ?

16. How wide can you spread your chopsticks? d

17. Maximum number of equiangular lines in R d (Aart Blokhuis) Let {u 1,…,u n } be unit vectors on n distinct α- equiangular lines in R d. Define: f i (x 1,…,x d ) = 2 - α 2. f i (u k,1,…,u k,d ) = 0 if k ≠ i and 1 - α 2 if k = i Hence {f i (x 1,…,x d )} are linearly independent. f i (x 1,…,x d )  span {x 1 2, …,x d 2, x i x k } So: n ≤

18. In perpetual progress.. (Seidel) For a given dimension n, what is the maximum number of equiangular lines in R n ? The maximum number of equiangular lines in dimensions 3 through 18 are: 6, 6, 10, 16, 28, …, 28 (d = 14), 36, 40, 48, 48,... For a given dimension d, there are finitely many angles  for which in every there are d+1 equiangular lines in R d. Which angles appear in infinitely many dimensions?

19. In progress.. Ubiquitous angles: (Barry Guidulli, M. R.)‏ There are angles  for which in every dimension d > d(  ) there are d+1 equiangular lines in R d. Example: arccos(1/(2k+1)) Others? (Babai) Arccos(1/(1+2  k)

20. In perpetual progress.. There are angles  for which R d contains d+1 equiangular lines for infinitely many dimensions. For example: arccos 1/2k (k  4). If there are d+1 equiangular lines in R d with angle arccos ¼ then d = 4k. Conjecture: for all integers k there are S matrices of order 4k+1 with smallest eigenvalue -4 (verified for k  13).

21. It is easy to be odd… What about arccos(1/2k)? Exist in R d only if d  2k mod 4. Only d + 1 lines. For d = 4 we have: (with Brendan McKay) P 4 + i. K 2 + (45 – 6i)K 1 (0 ≤ i ≤ 7) C 8 + i. K 2 + (21 – 6i)K 1 (0 ≤ i ≤ 3) C 6 + i. K 2 + (21 – 6i)K 1 (0 ≤ i ≤ 2) C 7 + i. K 2 + (18 – 6i)K 1 (0 ≤ i ≤ 3) K 1,3 + i. K 2 + (21 – 6i)K 1 (0 ≤ i ≤ 3)

22. Arccos(1/4) The S-matrix of every G(4n + 1, 2n + 2) has –4 as an eigenvalue. Examples: C 6 + C 3 (9 lines in R 8 ) (3-cube)  5K 1 (13 lines in R 12 ) Petersen  7K 1 (17 lines in R 16 ) Heawood  11K 1 (25 lines in R 24 ) K 7,7,7 – 7K 3 (21 lines in R 20 )

23. Some constructions 1.(-3,-3,1,1,1,1,1,1) : 28 equiangular lines in R 7 (= the upper bound for d = 7). 2.Petersen: 10 equiangular lines in R 5 3.Clebsch: 16 equiangular lines in R 6 4.The second largest eigenvalue of regular graphs and equiangular lines: 5.If G(n,r) is an r-regular graph and 2 is its second largest eigenvalue, then -2 2 – 1 is the smallest eigenvalue of its S-matrix.

24. Clebsch’s graph Petersen  {1,2,3,4,5}  {  }  {(x, {x,y})}  {( ,x)}

25. Clebsch graph Vertices: { {n,m} | 1  m < n  5; 1,2,3,4,5,  } Edges: {( , n), (n, {n,m}), Petersen} This graph is a strongly regular (5, 0, 2) graph. We can now calculate its eigenvalues and the eigenvalues of its Seidel matrix.

26. Best known upper bound: Dom de Caen 2/9(d + 1) 2 equiangular lines in R d for d = 3. 2 2t – 1 – 1 Using association schemes and quadratic forms over GF(2). Produce graphs with only four eigenvalues.

27. Equiangular lines in R 4 (a simple demonstration). Every switching class of graphs of order 2k+1 contains a unique Eulerian graph. These are the six switching classes of graphs of order 5.

28. Equiangular lines in R 4 (a simple demonstration). 5K 1 cos  = 1 (degenerate case) K 3 cos  = C 4 cos  = K 2 cos  = C 5 cos  = 1/  5 2K 2 cos  = 1/3 K 5 cos  = 1/4

29. Construction of “many” equiangular lines: sample. Petersen: A 2 = J – A +2I Eigenvalues: {3, 1 {5}, (-2) {4} } S-matrix: {3 {5}, (-3) {5} }. Yielding 10 arccos(1/3) – equiangular lines in R 5 Clebsch: A 2 = 2(J – A) + 3I. Eigenvalues: {5, 1 {10}, (-3) 5 } S-{10, 5 {5}, (-3) {10} } Yielding 16 arccos(1/3) – equiangular lines in R 6 Higman-Sims G(100, 22, 0, 6) S-{55, 15 {22}, (-5) {77} } Yielding 100 arccos(1/5) equiangular lines in R 23. (-3,-3, 1,1,1,1,1,1): 28 equiangular lines in R 7

30. Graph Designs Equiangular lines partition K 2n into two graphs G i (2n, n – 1) and a perfect matching. K 6 = C 6 + 2K 3 + 3K 2

31. Decomposing K 6 K 6 has 6 vertices and 15 edges. It can be decomposed into two: 6-cycles and a perfect matching. K 6 = 2C 6 + 3K 2

32. Decomposing K 6 (continued) Can all edges be treated equal? Can you decompose the 30 edges of 2K 6 into five 6-cycles? Is: 2K 6 = 5C 6 ?

33. Answer: YES (easy). Color the edges properly by 5 colors B, G, R, Y, P B-G, G-R, R-Y, Y-P,P-B 2K 6 = 5C 6

34. Decomposing K 6 (continued) Can you decompose 2K 6 into five copies of 2K 3? Is: 2K 6 = 5(2K 3 ) ?

35. We wish to number the six vertices of every pair of triangles by 1,2,…,6 so that every pair {j,k} will appear exactly twice.

36. 1 111 1 32 4 56 2 222 We now need to place three 3’s in the last three triangles. Which is not possible.

37. We wish to number the eight vertices of every tetrahedral pair by 1,2,…,8 so that every pair {j,k} will appear exactly 3 times.

38. 1.+ +++++++ 2.+ –+–+– +– 3.+ +–– ++–– 4.+ –– ++––+ 5.+ +++– – – – 6.+ –+–– +–+ 7.+ +–– ––++ 8.+ –– +–++– [1,3,5,7] [1,2,5,6] [1,4,5,8]………….. [1,4,6,7] [2,4,6,8] [3,4,7,8] [2,3,6,7]………….. [2,3,5,8]

39. Graph Designs: (2n-1)G(2n, n – 1) = (n -1)K 2n Can we cover the edges of K 2n by (2n – 1) copies of G(2n,2n – 1) so that every edge of K 2n appears in exactly n – 1 of the copies.

40. Example 1: (2n – 1)(K n + K n ) = (n – 1)K 2n ? True iff there is a Hadamard matrix of order 2n. Example 2: G(2n, n – 1) = K n,n – nK 2 If (2n – 1)(K n,n – nK 1 ) = (n – 1)K 2n then there is an (2n)x(2n – 1) matrix such that: a) Each column has n 1’s and n (-1)’s. b) The Hamming distance between any two rows is  n – 1. c) In each column we can select a “matching” so that by “ignoring” it the Hamming distance between any two rows will be exactly n – 1.

41. A “matchable” Hadamard Matrix 1. +++++++ 2. –+–+– +– 3. +–– ++–– 4. –– ++––+ 5. +++– – – – 6. –+–– +–+ 7. +–– ––++ 8. –– +–++– 1 3 5 7 6 8 2 The other six cubes are constructed similarly

42. We interpret the 4n – 1 columns of the Hadamard matrix of order 4n as labeled copies of K n,n In each copy we wish to select a perfect matching so that each pair {k,j} will be selected exactly once. If possible, we say that the Hadamard matrix is matchable.

43. A := {1,2,…,2n} Let X = {x 1, …,x n } and Y = {y 1, …,y n } X & Y are matchable if: {  (x i – y s(i) ) mod 2n + 1} = {1,2,…,2n} for some permutation s. –Example: n = 10. –X = { 1, 3, 4, 9, 5} {1, 2, 5, 8, 10} –Y = {10,8, 7, 2, 6}{3, 4, 6, 7, 9}

44. Let GF*(q) = {x 1,…,x q-1 } QR(q) = Quadratic Residues,NR(q) = GF*(q) \ QR(q) Theorem: {QR(q), NR(q)} are matchable for q > 9. J. Kratochvil, J. Nesetril, M.R. and T. Szony 1, 2, 4 Example: 6, 5, 3 All differences  (a i – b i ) mod 7 are distinct. Hadamard matrices derived by the Paley construction are matchable. (2n-1)(K n,n – nK 2 ) = (n – 1)K 2n for all n for which there exists a Payley Hadamard matrix of order 2n.

45. Conjecture: all hadamard matrices are matchable. A matchable hadamard matrix yields a solution to (2n – 1)(K n,n – nK 2 ) = (n – 1)K 2n So do matchable conference matrices. Simple conjecture:  n there is an (2n)x(2n – 1) matrix such that: a) Each column has n 1’s and n (-1)’s. b) The Hamming distance between any two rows is  n – 1.

46. Final simple conjecture A := {1,2,…,2n} Let X = {x 1, …,x n } and Y = {y 1, …,y n } Is it true that there is a permutation s such that: {  (x i – y s(i) ) mod 2n + 1} = 2n? The answer in general is NO! For which integers n, any partition (X,Y) is matchable?

47. Xia