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(  )

nd
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/ 2𝑑 − 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+2k)
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…
•
•
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•
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•
•
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 -22 – 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.
•
•
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•
•
•
•
•
•
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.
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[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
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1
The other six cubes are constructed similarly
7
3
8
2
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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