Equiangular Lines in R - PowerPoint Presentation

Slide1

Equiangular Lines in Rd

Moshe Rosenfeld

University of Washington

Shanghai Jiao Tong University

July 1, 2013Slide2

How wide and how even can you spread your chop-sticks?Slide3

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?Slide4

Study of equiangular lines involves:

Topology

Linear Algebra

Group Theory

Quantum Theory

Geometry

Graph Theory

Combinatorial DesignsSlide5

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).Slide6

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?Slide7

How many other angles are possible among 4 equiangular lines in R

3

?

How many equiangular lines are possible in R3

How do you approach such a problem?Matrices  Eigenvalues  GraphsSlide8

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.Slide9

For the icosahedron, Select

a unit vector on each

diagonal.

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

Slide10

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

.Slide11

Note: the last matrix is the S matrix of this graph.Slide12

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!Slide13

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.SeidelSlide14

Graphs on 4 vertices fall into 3 switching classes:

1. K

4

 4 diagonals of the cube2. C4

 degenerate case3. 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 RdSlide15

How wide can you spread your chopsticks?

For any set

L

n

of n lines through the origin in R3 let (Ln) = smallest angle among all distinct pairs of lines from Ln

Let (n

) = max



(

L

n) (4) = arccos(1/3)(5) = (6) = arccos(1/5)(7) = ?Slide16

How wide can you spread your chopsticks?

For any set of n > d lines through the origin in R

d

we have

:

Equality

holds only if the lines are equiangular

.

For n = d+1 we get

cos

  1/d.

For n = 2d we get :

cos

 

1/

dSlide17

Maximum number of equiangular lines in

R

d

(Aart

Blokhuis)Let {u1,…,u

n} be unit vectors on n distinct α-equiangular lines in Rd.

Define:

f

i

(x

1,…,xd) = <ui,x

>

2

-

α2<x,

x>.

f

i(uk,1

,…,uk,d) = 0 if k ≠ i and 1 - α2 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 ≤ Slide18

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?Slide19

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+2



k)Slide20

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 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).Slide21

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.K2 + (45 – 6i)K1 (0 ≤ i ≤ 7)C8 + i.K

2

+ (21 – 6i)K

1 (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)Slide22

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 R8)

(3-cube)



5K

1

(13 lines in R12)Petersen  7K1 (17 lines in R16)Heawood  11K1 (25 lines in R24)

K

7,7,7

– 7K

3 (21 lines in R20) Slide23

Some constructions

(-3,-3,1,1,1,1,1,1) :

28 equiangular lines in R7 (= the upper bound for d = 7).

Petersen: 10 equiangular lines in R5Clebsch

: 16 equiangular lines in R6The second largest eigenvalue of regular graphs and equiangular lines: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.Slide24

Clebsch’s graph

Petersen



{1,2,3,4,5}



{



}



{(x, {x,y})}



{(



,x)}Slide25

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.

Slide26

Best known upper bound:

Dom de Caen

2/9(d + 1)

2

equiangular lines in R

d for d = 3.22t – 1 – 1

Using association schemes and quadratic forms over GF(2).

Produce graphs with only four eigenvalues.Slide27

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.Slide28

Equiangular lines in R

4

(a simple demonstration).

5K1 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



= 1/4Slide29

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: 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 R7Slide30

Graph Designs

Equiangular lines

partition

K

2n into two graphs Gi(2n, n – 1) and a perfect matching.

K

6

= C

6

+ 2K

3

+ 3K

2Slide31

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

2Slide32

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

?Slide33

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

6Slide34

Decomposing K

6

(continued)

Can you decompose 2K

6 into five copies of 2K3?

Is: 2K

6

= 5(2K

3

) ?Slide35

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.Slide36

1

1

1

1

1

3

2

4

5

6

2

2

2

2

We now need to place three 3’s in the last three triangles. Which is not possible.Slide37

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.Slide38

+ + + +

+ + + +

+ – + –

+ – + –

+ + – –

+ + – –

+ – – +

+ – – +

+ + + +

– – – –

+ – + – – + – ++ + – – – – + +

+ – – +

– + + –

[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]Slide39

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 K2n appears in exactly n – 1 of the copies.Slide40

Example 1: (2n – 1)(K

n

+ K

n

) = (n – 1)K2n ? True iff there is a Hadamard matrix of order 2n.

Example 2: G(2n, n – 1) = K

n,n

– nK

2

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.Slide41

A “matchable” Hadamard Matrix

+ +

+

+ + + +

–

+

– + – + – + –

–

+ + – –

– – +

+ – – +

+

+ + –

– –

–

–

+

–

–

+

–

+

+

–

–

–

–

+

+

–

–

+

–

+

+

–

1

3

5

7

6

8

2

The other six cubes are constructed similarlySlide42

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.Slide43

A := {1,2,…,2n}

Let X = {x

1

, … ,

xn} and Y = {y1, … ,yn}

X & Y are matchable if:

{



(x

i

– ys(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}Slide44

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.

Slide45

Conjecture: all hadamard matrices are matchable.

A matchable hadamard matrix yields a solution to (2n – 1)(K

n,n

– nK

2) = (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.Slide46

Final simple conjecture

A := {1,2,…,2n}

Let X = {x

1

, … ,x

n} and Y = {y1, … ,yn}

Is it true that there is a permutation s such that:

{



(x

i – ys(i)) mod 2n + 1} = 2n?The answer in general is NO!For which integers n, any partition (X,Y) is matchable?Slide47

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