Additional notes on equiangular lines
Péter Csikvári
This is some additional text to Miniature 8 of Matousek’s Thirty-three
miniatures. We follow the treatment of the book Algebraic graph theory by
C. Godsil and G. Royle.
1. Equiangular lines
We say that distinct lines `1 , `2 , . . . , `n of Rd going through the origin are
equiangular if there exists some angle α such that for any i 6= j, the angle of
`i and `j is α. Clearly, if we chose a vector xi of length 1 on the line `i , then
xTi xj = (xi , xj ) = ± cos α. (Note that we consider xi ’s as column vectors.) For
sake of simplicity let us introduce the notation t = cos α.
It will be convenient to introduce the matrices Xi = xi xTi . Note that If i 6= j
then we have
T r(Xi Xj ) = T r(xi xTi xj xTj ) = T r(xi (xTi xj )xTj ) = T r((xTi xj )xTj xi ) = t2
and
T r(Xi Xi ) = T r(xi xTi xi xTi ) = T r(xi (xTi xi )xTi ) = T r(xi xTi ) = T r(xTi xi ) = 1.
We can see the same way that T r(Xi ) = T r(xi xTi ) = T r(xti xi ) = 1.
In particular if for some matrix Y we have
n
X
Y =
ci X i
i=1
then

T r(Y 2 ) = T r 
n
X
!2 
ci X i
i=1
=
n
X
c2i +2
i=1
X
ci cj t2 = (1−t2 )
n
X
c2i +t2
i=1
1≤i<j≤n
n
X
i=1
This will be one of our main equations, so we repeat it:

!2 
!2
n
n
n
X
X
X
(1)
T r(Y 2 ) = T r 
ci Xi  = (1 − t2 )
c2i + t2
ci .
i=1
i=1
i=1
Via this equation we will give another proof of the absolute bound given in
Miniature 8 of Matousek’s Thirty-three miniatures.
d
Theorem 1.1. (Absolute bound.) If distinct lines
`1 , `2 , . . . , `n of R going
d+1
through the origin are equiangular then n ≤ 2 .
1
!2
ci
.
2
Proof. Note that the matrices Xi = xi xTi are symmetric matrices of size d ×
d. We will show that X1 , . . . , Xn are linearly independent matrices in the
vectorspace of symmetric
matrices of size d × d. Since the dimension of this
d+1
vectorspace is 2 , this would immediately prove our result.
Now if X1 , . . . , Xn were not linearly independent then for some c1 , . . . , cn
not all 0, we have
n
X
0d =
ci X i ,
i=1
where 0d denotes the all 0 matrix of size d × d. Then by Equation (1) we have
!2
n
n
X
X
ci .
0 = T r(02d ) = (1 − t2 )
c2i + t2
i=1
i=1
P
But the latter expression is clearly positive as t2 < 1 and ni=1 c2i > 0. This
contradiction
proves that X1 , . . . , Xn are linearly independent, and so n ≤
d+1
.
2
Let us introduce the matrix A which has columns x1 , x2 , . . . , xn . Then
T
AA =
n
X
xi xTi
=
n
X
i=1
Xi .
i=1
On the other hand,
AT A = I + tS,
where S is ±1 matrix, except in the diagonal where it is 0, since xTi xj =
sij cos α = sij t. The matrix S is called the Seidel-matrix.
The next theorem will be essential when we start to study the case of equality
in the absolute bound.
Theorem 1.2. Assume that
I=
n
X
ci X i .
i=1
Then ci = nd , and
d(1 − t2 )
.
1 − dt2
Furthermore, the eigenvalues of S are n−d
with multiplicity d, and −1
with
dt
t
−1
multiplicity n − d. If n 6= 2d then t is an integer. Moreover, if n 6= 2d and
d ≤ n − 2 then −1
is an odd integer.
t
n=
3
Proof. Since I =
Pn
i=1 ci Xi ,
by Equation (1) we have
d = T r(I 2 ) = (1 − t2 )
n
X
c2i + t2
i=1
Note that
Xk =
n
X
n
X
!2
ci
.
i=1
ci Xi Xk ,
i=1
so
1 = T rXk =
n
X
2
ci T r(Xi Xk ) = ck + t
i=1
X
2
2
ci = (1 − t )ck + t
i6=k
n
X
ci .
i=1
This means that
P
1 − t2 ni=1 ci
ck =
,
1 − t2
which means that c1 = c2 = · · · = cn = c. Then
n
X
d = T rI = c
T rXi = nc.
i=1
Hence we have c =
d
.
n
If we write it back to the previous equation we get that
P
1 − t2 ni=1 ci
d
1 − t2 d
=
=
,
n
1 − t2
1 − t2
which means that
n=
d(1 − t2 )
.
1 − dt2
P
We have seen that AAT = ni=1 Xi , this means that AAT = nd I. Note that
AAT and AT A have the same eigenvalues except the 0’s. This means that the
eigenvalues of AAT = I + tS are nd with multiplicity d and 0 with multiplicity
n − d. In other words, the eigenvalues of S are n−d
with multiplicity d, and
dt
−1
with multiplicity n − d.
t
Since the minimal polynomial of −1
divides the characteristic polynomial of
t
S, and a minimal polynomial with coefficients in Q cannot have multiple zeros1
we get that the minimal polynomial of −1
has degree 1 or 2. (Alternative
t
way: the minimal polynomial of S has degree 2 since it is symmetric. The
minimal polynomial of −1
divides the minimal polynomial of S.) If the minimal
t
−1
polynomial of t is of degree 2 then −1
and n−d
are algebraic conjugates
t
dt
1This
is true for any field of characteristic 0 (Why? Hint: consider P 0 if P is the minimal
polynomial.), and even for many fields with characteristic p including the finite fields Fq .
4
and the characteristic polynomial of S is a power of their common minimal
polynomial, which would mean that n − d = d. So if n − d 6= d then −1
is an
t
integer since it is algebraic integer and rational.
Finally, if n 6= 2d and d ≤ n − 2 then let us consider the matrix
1
M = (J − I + S),
2
where J is the all 1 matrix. This is a 0 − 1 matrix with 1’s at exactly at the
places where S has 1’s. The eigenspace of J belonging to the eigenvalue 0
has dimension n − 1, The eigenspace of S belonging to the eigenvalue −1
has
t
dimension n − d. If d ≤ n − 2 then these two eigenspaces have to intersect
non-trivially, i. e. there is an eigenvector x such that Jx = 0 and Sx = −1
x
t
and so
1
1
−1
−1 +
x.
M x = (J − I + S)x =
2
2
t
This means that 12 −1 + −1
is an integer,
is an algebraic integer, and since −1
t
t
it must be odd.
Now we are ready to study the case of equality in the absolute bound.
Theorem 1.3. If distinct lines `1 , `2 , . . . , `n of Rd going through the origin are
equiangular and n = d+1
then d = 2, 3 or d + 2 is an odd perfect square.
2
Proof. If n − d = d and n = d+1
then d = 3. Let us assume that d 6= 3, and
2
so n − d 6= d. We have seen that X1 , . . . , Xn are linearly independent in the
vectorspace of symmetric matrices of size d × d. Then they form a basis, and
so there are c1 , . . . , cn such that
n
X
I=
ci X i .
i=1
Then by the previous theorem
d+1
d(1 − t2 )
=n=
2
1 − dt2
which gives that t12 = d + 2. If n − d 6= d then −1
is an integer, so d + 2 must
t
be a perfect square.
Finally, if n 6= 2d and d > 2 and then d ≤ d+1
− 2 = n − 2 so 1t is an odd
2
integer, so d + 2 must be an odd perfect square.
For d = 3, the six main diagonals of an icosahedron form a system of equiangular lines. For d = 7, we can consider the following 28 vectors in R8 : xij is
5
the vector where the i and j-th coordinates are √324 , the other coordiantes are
−1, for instance
1
x35 = √ (−1, −1, 3, −1, 3, −1, −1, −1).
24
Then ||xij || = 1 and xTij xkm = ± 13 . Seemingly, all vectors are in R8 , but since
all xij are orthogonal to the all 1 vector, we can consider them as vectors in
R7 . So this 82 = 28 lines form a system of equiangular lines in R7 . For d = 23 we will construct a system of equiangular lines of size 24
= 276.
2
Before we do it, we will introduce the so-called Witt-design.