Report

Equiangular Lines in Rd Moshe Rosenfeld University of Washington Shanghai Jiao Tong University July 1, 2013 How wide and how even can you spread your chop-sticks? Equiangular Lines • Definition: A set of lines through the origin in Rd is equiangular if the angle between any pair of distinct lines is the same. • In R2, the maximum number of equiangular lines is 3. • The only possible angle among 3 equiangular lines in R2 is 60o. • The largest minimal angle among k lines through the origin in R2 is 180/k. • What is happening in higher dimensions? Study of equiangular lines involves: • Topology • Geometry • Linear Algebra • Graph Theory • Group Theory • Combinatorial Designs • Quantum Theory Equiangular Lines in R3 • The four diagonals of the regular cube in R3 form a set of four equi-angular lines. • Angle: arccos(1/3). More equiangular lines in R3 • The six diagonals of the icosahedron form a set of 6 equiangular lines in R3. • Angle ???? • Is it fair to ask a student in a final exam to calculate the angle? • The coordinates of the 12 vertices? • How many other angles are possible among 4 equiangular lines in R3 ? • How many equiangular lines are possible in R3 • How do you approach such a problem? • Matrices Eigenvalues Graphs Seidel’s adjacency matrix For a given graph G, define the S-adjacency matrix as follows: Si,i = 0 Si,j = -1 is (i,j) E(G) Si,j = 1 otherwise. Note: for a given graph G, S = J – 2A – I. 1. For the icosahedron, Select a unit vector on each diagonal. 2. The Gram-Schmidt matrix generated by them will be: 1 1 1 1 1 1 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 R3 with angle = cos Constructing the icosahedron from a graph 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 I 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 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. • Note: the last matrix is the S matrix of this graph. The S-matrix of this graph: 0 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 This is an orthogonal matrix. So A2 = 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! 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 • Graphs on 4 vertices fall into 3 switching classes: • 1. K4 4 diagonals of the cube • 2. C4 degenerate case • 3. K4 – K2 (any four diagonals of the icosahedron). • Conclusion: there are only two possible angles among 4 – equiangular lines in R3 • For each d, there are finitely many angles possible for d+1 -equiangular lines in Rd How wide can you spread your chopsticks? • For any set Ln of n lines through the origin in R3 let (Ln) = smallest angle among all distinct pairs of lines from Ln • Let (n) = max (Ln) (4) = arccos(1/3) (5) = (6) = arccos(1/5) (7) = ? How wide can you spread your chopsticks? • For any set of n > d lines through the origin in Rd we have: cos( ) nd d ( n 1) Equality holds only if the lines are equiangular. For n = d+1 we get cos 1/d. For d n = 2d we get : cos 1/ 2 − 1 Maximum number of equiangular lines in Rd (Aart Blokhuis) • Let {u1,…,un} be unit vectors on n distinct αequiangular lines in Rd. • Define: fi(x1,…,xd) = <ui,x>2 - α2<x,x>. • fi(uk,1,…,uk,d) = 0 if k ≠ i and 1 - α2 if k = i • Hence {fi(x1,…,xd)} are linearly independent. • fi(x1,…,xd) span {x12, …,xd2, xixk} • So: n ≤ d 1 2 In perpetual progress.. • (Seidel) For a given dimension n, what is the maximum number of equiangular lines in Rn? 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 Rd. • Which angles appear in infinitely many dimensions? 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 Rd. • Example: arccos(1/(2k+1)) • Others? (Babai) • Arccos(1/(1+2k) In perpetual progress.. • There are angles for which Rd contains d+1 equiangular lines for infinitely many dimensions. • For example: arccos 1/2k (k 4). • If there are d+1 equiangular lines in Rd 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). It is easy to be odd… • • • • • • • • • What about arccos(1/2k)? Exist in Rd only if d 2k mod 4. Only d + 1 lines. For d = 4 we have: (with Brendan McKay) P4 + i.K2 + (45 – 6i)K1 (0 ≤ i ≤ 7) C8 + i.K2 + (21 – 6i)K1 (0 ≤ i ≤ 3) C6 + i.K2 + (21 – 6i)K1 (0 ≤ i ≤ 2) C7 + i.K2 + (18 – 6i)K1 (0 ≤ i ≤ 3) K1,3 + i.K2 + (21 – 6i)K1 (0 ≤ i ≤ 3) Arccos(1/4) • The S-matrix of every G(4n + 1, 2n + 2) has –4 as an eigenvalue. • Examples: C6 + C3 • (3-cube) 5K1 (9 lines in R8) (13 lines in R12) • Petersen 7K1 (17 lines in R16) • Heawood 11K1 (25 lines in R24) • K7,7,7 – 7K3 (21 lines in R20) Some constructions 1. (-3,-3,1,1,1,1,1,1) : 28 equiangular lines in R7 (= the upper bound for d = 7). 2. Petersen: 10 equiangular lines in R5 3. Clebsch: 16 equiangular lines in R6 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 -22 – 1 is the smallest eigenvalue of its S-matrix. Clebsch’s graph Petersen {1,2,3,4,5} {} {(x, {x,y})} {(,x)} 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. Best known upper bound: • Dom de Caen • 2/9(d + 1)2 equiangular lines in Rd for d = 3.22t – 1 – 1 • Using association schemes and quadratic forms over GF(2). • Produce graphs with only four eigenvalues. Equiangular lines in R4 (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. Equiangular lines in R4 (a simple demonstration). • 5K1 • K3 • C4 cos = 1 (degenerate case) cos = cos = 2 33 1 2 17 1 2 • K2 cos = • C5 cos = 1/5 • 2K2 cos = 1/3 • K5 cos = 1/4 33 1 Construction of “many” equiangular lines: sample. • • • • • • • • • Petersen: A2 = J – A +2I Eigenvalues: {3, 1{5}, (-2){4}} S-matrix: {3{5}, (-3){5}}. Yielding 10 arccos(1/3) – equiangular lines in R5 Clebsch: A2 = 2(J – A) + 3I. Eigenvalues: {5, 1{10}, (-3)5} S-{10, 5{5}, (-3){10}} Yielding 16 arccos(1/3) – equiangular lines in R6 Higman-Sims G(100, 22, 0, 6) S-{55, 15{22} , (-5){77}} Yielding 100 arccos(1/5) equiangular lines in R23. (-3,-3, 1,1,1,1,1,1): 28 equiangular lines in R7 Graph Designs • Equiangular lines partition K2n into two graphs Gi(2n, n – 1) and a perfect matching. K6 = C6 + 2K3 + 3K2 Decomposing K6 • K6 has 6 vertices and 15 edges. It can be decomposed into two: 6-cycles and a perfect matching. K6 = 2C6 + 3K2 Decomposing K6 (continued) Can all edges be treated equal? Can you decompose the 30 edges of 2K6 into five 6-cycles? Is: 2K6 = 5C6 ? • 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 2K6 = 5C6 Decomposing K6 (continued) • Can you decompose 2K6 into five copies of 2K3? Is: 2K6 = 5(2K3) ? 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. 1 1 2 5 4 3 2 1 2 1 2 1 2 6 We now need to place three 3’s in the last three triangles. Which is not possible. 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. 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] Graph Designs: (2n-1)G(2n, n – 1) = (n -1)K2n • Can we cover the edges of K2n by (2n – 1) copies of G(2n,2n – 1) so that every edge of K2n appears in exactly n – 1 of the copies. • Example 1: (2n – 1)(Kn + Kn) = (n – 1)K2n ? • True iff there is a Hadamard matrix of order 2n. • Example 2: G(2n, n – 1) = Kn,n – nK2 • If (2n – 1)(Kn,n – nK1) = (n – 1)K2n 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. A “matchable” Hadamard Matrix 1. 2. 3. 4. 5. 6. 7. 8. 6 + + – – + + – – + – – + + – – + + + + + – – – – + – + – – + – + + + – – – – + + + – – + – + + – 1 The other six cubes are constructed similarly 7 3 8 2 + – + – + – + – 5 • We interpret the 4n – 1 columns of the Hadamard matrix of order 4n as labeled copies of Kn,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. • • • • A := {1,2,…,2n} Let X = {x1, … ,xn} and Y = {y1, … ,yn} X & Y are matchable if: {(xi – ys(i)) mod 2n + 1} = {1,2,…,2n} for some permutation s. – Example: n = 10. – X = { 1, 3, 4, 9, 5} – Y = {10,8, 7, 2, 6} {1, 2, 5, 8, 10} {3, 4, 6, 7, 9} • Let GF*(q) = {x1,…,xq-1} • QR(q) = Quadratic Residues, • Theorem: {QR(q), NR(q)} are matchable for q > 9. • J. Kratochvil, J. Nesetril, M.R. and T. Szony NR(q) = GF*(q) \ QR(q) 1, 2, 4 Example: 6, 5, 3 All differences (ai – bi) mod 7 are distinct. • Hadamard matrices derived by the Paley construction are matchable. • Conjecture: all hadamard matrices are matchable. • A matchable hadamard matrix yields a solution to (2n – 1)(Kn,n – nK2) = (n – 1)K2n • 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. Final simple conjecture • A := {1,2,…,2n} • Let X = {x1, … ,xn} and Y = {y1, … ,yn} • Is it true that there is a permutation s such that: {(xi – ys(i)) mod 2n + 1} = 2n? • The answer in general is NO! • For which integers n, any partition (X,Y) is matchable? Xia Xia