Symmetric structures
Péter Csikvári
This note follows the treatment of the book A course in combinatorics by
J. H. van Lint and R. M. Wilson.
1. Block designs
Definition 1.1. Let P = {P1 , . . . , Pv } be set of points and B = {B1 , . . . , Bb }
be set of blocks. We say that S = (P, B) is a block design of parameters
t − (v, k, λ) if Bi ⊆ P, |Bi | = k and for every T ⊆ P with |T | = t there are
exactly λ blocks Bi such that T ⊆ Bi .
Before we start proving anything let’s see some examples.
First of all, there are very boring examples like one set B containing all
points or if we choose all sets of size k of a v-element set. Let’s see a bit more
interesting example.
The Fano-plane is a 2 − (7, 3, 1) block design: let P = {1, 2, 3, 4, 5, 6, 7},
and let B consist of the following sets: {1, 2, 4}, {2, 3, 5}, {3, 4, 6}, {4, 5, 7},
{5, 6, 1}, {6, 7, 2}, {7, 1, 3}. One can easily check that any two points are in
exactly 1 set. So it is indeed a block design of parameters 2 − (7, 3, 1).
There are at least two possible ways to generalize the Fano-plane. If we
keep the condition that every block has size 3 and every pairs is in exactly
one block then we arrive to the so-called Steiner triple system (STS): this is a
block design of parameters 2 − (n, 3, 1). We can imagine it as a decomposition
of Kn into triangles such that every edge is in exactly
one triangle. There are
two necessary conditions for the parameters: 3 | n2 and that 2 | n − 1 since
the degree of every vertex is n − 1 and a triangle covers 2 edges at a given
vertex. The two conditions together give that n = 6k + 1 or n = 6k + 3. This
condition turns out to be sufficient. For n = 6k + 3 there is a very simple
construction: take three copies of the cyclic group Z2k+1 , then the vertices of
Kn can be denoted by (x)i , where x is an element of Z2k+1 and i = 1, 2, 3.
Let us consider the following triangles: ((x)i , (y)i , x+y
), where i + 1 is
2
i+1
meant by mod 3 and x 6= y, and ((x)1 , (x)2 , (x)3 ). It is easy to check that this
construction gives a decomposition of Kn into triangles.
There is another way to generalize the Fano-plane, and this leads us to the
concept of finite projective plane. If we call the blocks of the Fano-plane lines,
then in the Fano-plane any two points determine exactly one line, and any
two lines intersect in exactly one point. So Fano-plane mimics the projective
plane. So let us require these two conditions: (i) any any two points determine
1
2
exactly one line, and (ii) any two lines intersect in exactly one point. It turns
out that there are two boring examples satisfying these conditions: one line
containing all points; and one line containing all except one point and the extra
point not on the line is incident to lines of size 2 determined by the points on
the line and the extra point. If we exclude these two boring examples we get
the concept of finite projective planes. It turns out that it is equivalent with
the design 2 − (n2 + n + 1, n + 1, 1). In this design every point is on exactly
n+1 lines, and as the second parameter shows, every line contains n+1 points.
The Fano-plane corresponds to the case n = 2.
There is a funny way to construct finite projective planes: consider a regular
(n2 + n + 1)-gon, and choose n + 1 points such a way that all possible distances
appear among the n + 1 vertices. Such an (n + 1)-gon is called totally irregular
polygon. Let us rotate this totally irregular polygon every possible ways then
we get n2 + n + 1 lines of a projective plane. For instance, if we take the vertex
1, 2, 4 in a regular 7-gon, then we get back that the Fano -plane. If we consider
the sets (0 + k, 1 + k, 3 + k, 9 + k) in the cyclic group Z13 with k ∈ Z13 then we
get the design 2 − (13, 4, 1), the finite projective plane of rank 3. If n = 4 then
we can take the vertices 3, 6, 7, 12, 14 to get a totally irregular polygon in the
regular 21-gon, this leads to the (unique) design of parameters 2 − (21, 5, 1),
the finite projective plane of rank 4. The projective planes obtained by such
a construction is called a cyclic projective plane.
There is another much more standard way to construct finite projective
planes from finite fields. Let q be a prime power and Fq be the finite field
with q elements. In F3q \ {(0, 0, 0)} let us introduce the equivalence relation
(x, y, z) ' (x0 , y 0 , z 0 ) if there exists some λ 6= 0 such that x0 = λx, y 0 = λy, z 0 =
λz. It is indeed an equivalence relation and the number of equivalence classes
3 −1
is qq−1
= q 2 + q + 1. In our projective plane an equivalence class [x, y, z] will
correspond to each point, and also an equivalence class [a, b, c] will correspond
to each line. A point is incident to a line if ax + by + cz = 0. Note that it is
independent of which elements of the equivalence classes of the point or line
we consider. It is easy to check that every line contains exactly q + 1 points,
any any two points determine exactly one line, and any two lines intersect in
exactly one point. Surprisingly, the finite projective planes obtained this way
can be obtained as cyclic projective planes.
2. Lower bounds on the number of blocks
2.1. Double counting. The conditions of a block design t − (v, k, λ) is so
strong that it gives a lot of possibility to carry out double counting arguments.
3
Theorem 2.1. Let S = (P, B) be a block design of parameters t − (v, k, λ).
Let I be a subset of P with i points, where i ≤ t. Let bi be the number of blocks
containing I. Then
bi = λ
v−i
t−i
.
k−i
t−i
Proof. Let us consider the set
A = {(T, B) | I ⊆ T ⊆ B, |T | = t, B ∈ B}.
Then we can count |A| in two different ways. First we can choose T in
ways and having T we can choose B in λ ways. Hence
v−i
|A| = λ
.
t−i
v−i
t−i
Another way to count |A| is as follows. We choose B in bI ways and then from
B we can choose t − i elements from B \ I in k−i
ways. Hence
t−i
k−i
|A| = bI
.
t−i
Hence
bI = λ
v−i
t−i
k−i
t−i
depends only on the size of I so we can call it bi .
Corollary 2.2. Let S = (P, B) be a block design of parameters t − (v, k, λ),
then it is a block desing for all i ≤ t.
Corollary 2.3. Let S = (P, B) be a block design of parameters t − (v, k, λ).
Let I be a subset of P with i points, where i ≤ t. Let us consider the system
SI = (PI , BI ), where PI = P \ I and
BI = {B \ I | B ∈ B and
I ⊆ B}.
Then SI = (PI , BI ) is a block design of parameters (t − i) − (v − i, k − i, λ).
This is called a derived block design.
We will also need a dual result of Theorem 2.1.
Theorem 2.4. Let S = (P, B) be a block design of parameters t − (v, k, λ).
Let J be a subset of P with j points, where j ≤ t. Let bj be the number of
blocks disjoint from J. Then
v−j
j
k b = λ v−t .
k−t
4
Naturally, j is just a superscript in bj , not a power.
Proof. We will use the inclusion-exclusion principle. For sake of simplicity, let
J = {1, 2, . . . , j}. Let Ai ⊆ B be defined as follows:
Ai = {B ∈ B | i ∈ B}.
Then clearly we would like to determine the cardinality of B \ ∪jm=1 Am . By
inclusion-exclusion principle we have
j
X
j
B \ ∪ Ai = |B| −
|Ai | +
i=1
X
|Ai1 ∩ Ai2 | + . . .
1≤i1 <i2 ≤j
i=1
Clearly, the set Ai1 ∩ · · · ∩ Air is the set of blocks which contains the set
I = {i1 , . . . , ir }. By Theorem 2.1, its cardinality is
v−r
bI = λ
t−r
.
k−r
t−r
Hence the number bJ of blocks avoiding J is
j
X
J
i j
b =
(−1)
λ
i
i=0
v−i
t−i
.
k−i
t−i
Now we can use various identities for proving our result or we can use the
following shortcut. From the above formula we can see that the number bJ
only depends on |J| = j, so we can use the notation bj for it. So let us count
the cardinality of the set
A = {(J, B) | |J| = j, B ∈ B, J ∩ B = ∅}.
First of all,
v−k
|A| = b
,
j
where b = b0 is the number of blocks:
v
t
.
k
t
b = b0 = λ
On the other hand,
v j
|A| =
b.
j
5
Hence
j
v−k
j
b =
(vt)
λ k
()
t =λ
v
j
v−k v
j
t
k v
t
j
=λ
(v−k)!
v!
j!(v−k−j)! t!(v−t)!
k!
v!
t!(k−t)! j!(v−j)!
=λ
(v − k)!(v − j)!(k − t)!
.
(v − k − j)!(v − t)!k!
Note that
v−j
k v−t
k−t
λ
=
(v−j)!
k!(v−k−j)!
λ (v−t)!
(v−k)!(k−t)!
=λ
Hence
(v − j)!(v − k)!(k − t)!
.
(v − t)!k!(v − k − j)!
v−j
k v−t .
k−t
j
b =λ
Corollary 2.5. Let S = (P, B) be a block design of parameters t − (v, k, λ).
Let I and J be subsets of P with i and j points, where i + j ≤ t. Let bji be the
number of blocks which contains I and are disjoint from J. Then
v−i−j
bji = λ
k−i
v−t
k−t
.
Proof. Let us consider the derived block design (S)I introduced in Corollary 2.3
and apply Theorem 2.4. Then
(v−i)−j
v−i−j
bji = λ
k−i
(v−i)−(t−i)
(k−i)−(t−i)
=λ
k−i
v−t
k−t
.
2.2. Lower bounds on the number of blocks.
Theorem 2.6 (Fisher’s inequality). Let b the number of blocks of a block
design 2 − (v, k, λ). If v > k (in other words, no block contains all elements)
then b ≥ v.
Proof. Let N be the v × b incidence matrix: Ni,B = 1 if i ∈ B, otherwise
Ni,B = 0. Let us consider the matrix M = N N T . Then Mij simply counts
the number of blocks which contain both i and j, by definition it is λ if i 6= j.
On the other hand, if i = j, then it is the number of blocks containing i, let
r denote this number. Note that r is independent of i by Theorem 2.1 and
bigger than λ since we can choose an element i contained in a block B, and
element not contained in B as v > k. Hence
N N T = (r − λ)I + J.
6
Note that the rank of (r − λ)I + J is v. Indeed, its eigenvalues are (r − λ) + vλ
and r − λ with multiplicity v − 1. Hence the rank of N is at least v. On the
other hand, it is at most min(v, b). Hence b ≥ v.
Next we give a generalization of the previous theorem where we really use
the results from the previous subsection.
Theorem 2.7 (Ray-Chaudhuri and Wilson). Let S be a block design of parameters t − (v, k, λ) with b blocks. Assume
that the positive integer s satisfies
that t ≥ 2s and v ≥ k + s. Then b ≥ vs .
Proof.
For an i ≤ t let us consider i-th incidence matrix Ni : it is matrix of size
v
×
b
whose elements indexed by pairs (I, B), where |I| = i and B ∈ B, and
i
NI,B = 1 if I ⊆ B, and NI,B = 0 otherwise.
Let us also introduce the matrix Wij of size vi × vj whose elements indexed
by pairs (I, J), where |I| = i and |J| = j, and NI,J = 1 if I ⊆ J, and NI,J = 0
otherwise.
The main claim is that
s
X
Ns NsT =
bi2s−i WisT Wis .
i=0
Let E and F be two sets of size s such that |E ∩ F | = µ. Then
(Ns NsT )E,F = |{B ∈ B | E, F ⊆ B}| = |{B ∈ B | E ∪ F ⊆ B}| = b2s−µ
since |E ∪ F | = 2s − µ. Note that
(WisT Wis )E,F
µ
= |{I | I ⊆ E, F, |I| = i}| = |{I | I ⊆ E ∩ F, |I| = i}| =
.
i
So what we need to prove is that
b2s−µ =
s
X
bi2s−i
i=0
µ
.
i
Recall that
v−2s+µ
t−2s+µ
.
k−2s+µ
t−2s+µ
b2s−µ = λ
On the other hand,
X
s
s
X
µ
i
b2s−i
=
i
i=0
i=0
v−(2s−i)−i
k−(2s−i)
v−t
k−t
µ
=
i
s X
λ
v−t
k−t
i=0
v − 2s
k − (2s − i)
µ
µ−i
=
7
v − 2s + µ
= v−t
k − 2s + µ
k−t
λ
v−2s+µ
v−k
v−t
k−t
=λ
Let us introduce the notation r = 2s − µ. Then
(v−r)!
v−2s+µ
v−r
λ
t−2s+µ
k−2s+µ
t−2s+µ
=λ
t−r
k−r
t−r
=λ
(v−t)!(t−r)!
(k−r)!
(k−t)!(t−r)!
=λ
.
(v − r)!(k − t)!
.
(v − t)!(k − r)!
On the other hand,
v−2s+µ
v−k
v−t
k−t
λ
v−r
v−k
v−t
k−t
=λ
=
(v−r)!
(v−k)!(k−r)!
λ (v−t)!
(k−t)!(v−k)!
Hence
Ns NsT
=
s
X
=λ
(v − r)!(k − t)!
.
(v − t)!(k − r)!
bi2s−i WisT Wis .
i=0
Note that the matrices WisT Wis are positive semidefinite for all i, and bi2s−i
T
is non-negative. On the other hand, Wss
Wss = I · I = I and bss > 0 since
v ≥ k + s. Hence
s
X
bi2s−i WisT Wis
i=0
is positive definite. So its rank is vs . Hence the rank of Ns is also
the other hand, it is at most min(b, vs ). Hence b ≥ vs .
v
s
. On
Remark
2.8. For t ≥ 4, not many examples known where v > k + s and
b = vs . In fact, the only known example is the block design with parameters
4 − (23, 7, 1) and its complement. Note that for this block design we have
23
23 · 22 · 21 · 20
23 · 22
23
4
=
=
.
b = 7 =
7·6·5·4
2
2
4
If we apply derivation two times to this block design we get a design of parameters 2 − (21, 5, 1), the projective finite plane of rank 4. Surprisingly one can
extend this design of parameters 4 − (23, 7, 1) to get a design of parameters
5 − (24, 8, 1). This new design is called Witt-design and we will construct it
later.
3. Extended Golay-code
We will study certain codes in this section. I will use standard notations,
see for instance Miniature 5 of Matousek’s book.
The main theorem of this section is the following.
8
Theorem 3.1. Let C be a code of parameters (24, 212 , 8)2 such that 0 ∈ C.
Then C is unique up to permutations of the coordinates.
Proof. Let us delete one of the coordinate of C, and let C 0 be the obtained code.
Then C 0 has parameters (23, 212 , 7)2 . Now for every c ∈ C 0 let us consider the
ball
B(c) = {u ∈ {0, 1}23 | dH (u, c) ≤ 3},
where dH (x, y) is the Hamming-distance, the number of coordinates where x
and y differ. Note that for c1 , c2 ∈ C 0 we have B(c1 ) ∩ B(c2 ) = ∅ since if
u ∈ B(c1 ) ∩ B(c2 ) then dH (c1 , c2 ) ≤ dH (c1 , u) + dH (u, c2 ) ≤ 6 contradicts the
fact that the minimal distance in C 0 is at least 7. Note that
23
23
23
|B(c)| = 1 +
+
+
= 211 .
1
2
3
Hence
| ∪c∈C 0 B(c)| =
X
|B(c)| = 212 · 211 = 223 .
c∈C 0
So every u ∈ {0, 1}23 is exactly in one ball B(c).
Let Ai be the number of codewords c for which the weight w(c) = dH (c, 0) =
i. Now let us count
the number of vectors in {0, 1}23 which has weight i. Of
23
course, it is i . On the other since every u ∈ {0, 1}23 is exactly in one ball
B(c), we can count that how many ways we can get a word of weight i from a
codeword c. For instance
if c has weight i + 3, we have to delete 3 1’s from c
which we can do i+3
ways.
If we c contains i 1’s then we either do nothing
3
or we can change one 0 to 1 and one 1 to a 0, which we can do in 1 + i(23 − i)
ways. Similar argument shows that
23
= βi−3 Ai−3 +βi−2 Ai−2 +βi−1 Ai−1 +βi Ai +βi+1 Ai+1 +βi+2 Ai+2 +βi+3 Ai+3 ,
i
where
26 − i
25 − i
26 − i
βi−3 =
, βi−2 =
, βi−1 = (24 − i) + (i − 1)
,
3
2
2
i+1
i+2
i+3
βi = 1+i(23−i), βi+1 = (i+1)+(22−i)
, βi+2 =
, βi+3 =
.
2
2
3
Note that A0 = 1 as 0 ∈ C 0 , and A1 = A2 = A3 = A4 = A5 = A6 = 0 as
the minimal distance is 7 and 0 ∈ C 0 . It is clear that the equations above determines the number Ai ’s recursively. Tedious, but trivial computation shows
that A0 = A23 = 1, A7 = A16 = 253, A8 = A15 = 506, A11 = A12 = 1288.
9
Now let Ai be the number of codewords c ∈ C with weight i. We claim
that A0 = A23 = 1, A8 = A23 = 759, A12 = 2576. Indeed, if for instance there
were a c ∈ C such that w(c) = 7 then deleting a coordinate where c is 1 we
would get a code C 0 which contains a codeword of weight 6 which is impossible.
Similar arguments shows that Ai = 0 if i ∈
/ {0, 8, 12, 16, 24}.
Next we show that (c1 , c2 ) = 0, where we compute in F2 . For a vector u, let
supp(u) be the set of places where u is equal to 1. Note that |supp(u)| = w(u).
Then (u1 , u2 ) is the parity of |supp(u1 ) ∩ supp(u2 )|. Note that
1
(c1 , c2 ) ≡ |supp(c1 )∩supp(c2 )| = (|supp(c1 )| + |supp(c1 )| − |supp(c1 + c2 )|) =
2
1
= (w(c1 ) + w(c2 ) − w(c1 + c2 )) (mod 2).
2
We know that 4 | w(c1 ) and 4 | w(c2 ). Now we claim that 4 | w(c1 + c2 ).
Currently we don’t know that c1 + c2 ∈ C or not, but we can apply a little
trick. Let us consider the code C + c1 = {c + c1 | c ∈ C}. Note that 0 ∈ C + c1
and C + c1 has parameters (24, 212 , 8)2 . So everything which we proved so far
for C is also true for C + c1 . In particular, every codeword of C + c1 has weight
divisible by 4. In particular, 4 | w(c1 + c2 ). Hence (c1 , c2 ) = 0.
Now we are ready to show that C is a linear code. Let D be the linear code
generated by C, and let D⊥ its dual. Then dim D ≥ 12 as |C| = 212 . Note that
since any two codeword of C are orthogonal to each other we have C ⊆ D⊥
too. This means that dim D⊥ ≥ 12 is also true. Note that for any subspace V
and V ⊥ we have dim V + dim V ⊥ = 24. So dim D = dim D⊥ = 12 and C = D,
which means that C is a linear code.
Our next goal is to find a nice basis of C. First let’s choose a vector c1
with weight 12 and permute the coordinates such a way that the 2, 3, . . . , 13
coordinates in c1 are 0, the other coordinates are 1. Let us project C to
the coordinates 2, 3, . . . , 13. Then clearly c and c + c1 will have the same
projection. We show that there is no other codeword which has the same
projection as c and c + c1 . Since C is a linear code, it is enough to show
that if c2 projects to the 0 vector then it is 0 or c1 . If c2 projects to 0 and
c2 6= 0, c1 then 0 < w(c2 ) < 12. So w(c2 ) = 8 but then w(c1 + c2 ) = 4 which
is impossible. Finally observe that every projection contain even number of
1’s since (c, c1 ) = 0 for all c ∈ C. Hence the projection is the unique code
C 00 of parameters [12, 11, 2]2 which contains all words with even number of 1’s.
In particular, we can choose c2 , c3 , . . . , c12 such that i-th and 13-th coordinate
are 1 in ci and all other coordinates between 2, 3, . . . , 12 are 0. If necessary by
choosing ci + c1 instead of ci we can also ensure that the first coordinate of ci
10
are 0 for i = 2, . . . , 12. Hence the matrix whose rows are c1 , . . . , c12 has the
following form: (I A), where A is a 12 × 12 matrix whose first row and column
only contains 1’s except a11 = 0.
Let N be the 11 × 11 matrix which we get if we delete the first row and
column of A. Assume that ni is the i − 1-th row vector of N , this way ni
is a part of the vector ci for i = 2, . . . , 12. Since 2 ≤ w(ci ) ≤ 14 we have
w(ci ) = 8 or 12. On the other hand, if w(ci ) = 12 then w(c1 + ci ) = 4 which is
impossible. Hence w(ci ) = 8 and w(ni ) = 6. Let Bi = supp(ni ), then |Bi | = 6.
If |Bi ∩ Bj | = x then w(ci + cj ) = 2 + (6 − x) + (6 − x) = 14 − 2x so x = 1 or
x = 3. If x = 1 then w(ci + cj ) = 12, but again w(c1 + ci + cj ) = 4 as ci + cj
only contains two 1’s in the first 12 coordinates. Hence x = 3. So the sets Bi
has cardinalities 6 and |Bi ∩ Bj | = 3 for i 6= j.
We will show that there is a unique system of sets with the above properties.
There is a construction for these sets: Bi+1 = {i, i + 1, i + 3, i + 4, i + 5, i +
9} ⊆ F11 for i = 1, 2, . . . , 11. Note that if we write up all differences between
{0, 1, 3, 4, 5, 9} we will get every non-zero element exactly 3 times. This shows
that |Bi ∩ Bj | = 3 for all i 6= j. This way we can give a generator matrix of
our code C.



















1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
0
0
1
1
1
0
1
0
1
1
0
1
0
0
0
1
1
1
1
1
0
1
1
0
1
0
0
0
1
1
1
1
1
0
1
1
0
1
0
0
0
1
1
1
1
1
0
1
1
0
1
0
0
0
1
0
1
1
1
0
1
1
0
1
0
0
1
0
0
1
1
1
0
1
1
0
1
0
1
0
0
0
1
1
1
0
1
1
0
1
1
1
0
0
0
1
1
1
0
1
1
0
1
0
1
0
0
0
1
1
1
0
1
1
1
1
0
1
0
0
0
1
1
1
0
1



















Now we have two things left to do: (i) to prove that the this generator
matrix generates a code of parameters [24, 12, 8]2 , (ii) to prove the uniqueness
of our system of sets B2 , . . . , B12 .
Our next goal is to show that this code has indeed parameters [24, 12, 8]2 .
First of all, let’s show that for any u1 , u2 ∈ C we have (u1 , u2 ) = 0. In
other words, C is self-orthogonal. Clearly, (c1 , ci ) = 0 and (ci , cj ) = 0 for
i, j ∈ {2, . . . , 12}, so the generators are pairwise orthogonal, but then it is also
11
true for any u1 , u2 ∈ C. This means that not only G = (I A) is a generator
matrix for C, but also G0 = (−AT I) = (AT I) is also generator matrix for
C = C ⊥ . Next we show that every codeword has weight divisible by 4. This
is again true for the c1 , . . . , c12 , and if it is true for u1 and u2 then it is true
for u1 + u2 :
w(u1 + u2 ) = w(u1 ) + w(u2 ) − 2|supp(u1 ) ∩ supp(u2 )|,
but |supp(u1 ) ∩ supp(u2 )| ≡ (u1 , u2 ) ≡ 0 ( mod 2), so 4 | w(u1 + u2 ). So for all
vectors u ∈ C we have 4 | w(u). In order to show that the minimal distance is
8, it is enough to show that there is no vector with weight 4. Now assume that
u = (a, b) ∈ C has weight 4, where a and b are vectors of length 12. Clearly,
we have w(u) = w(a) + w(b). Note if u is generated by G then u = aG, and
similarly u = bG0 . Now if w(a) = 0 then u = 0, so w(u) = 0. If w(a) = 1 then
u = ci for some i and w(u) = 12 or 8. If w(a) = 2 then u = ci + cj , but then
w(u) = 8 again by the properties of the sets Bi . If w(a) = 3 then w(b) = 1
and u is row vector of G0 and so w(u) = 12 or 8 again. Finally, if w(a) = 4
then w(b) = 0 and u = 0, a contradiction again. Hence C does not contain
any codeword of weight 4. This means that the minimal distance is 8.
Finally, we have to show that the system of sets B2 , . . . , B12 is unique. So
assume that B2 , . . . , B12 ⊆ {1, 2, . . . , 11} such that |Bi | = 6 and |Bi ∩ Bj | = 3
if i 6= j. Let Ai be the complement of Bi then |Ai | = 5 and |Ai ∩ Aj | = 2 if
i 6= j (why?). First of all, we show that any two points p1 , p2 ∈ {1, 2, . . . , 11}
are contained in exactly 2 sets Ai . (There is a very general claim which immediately implies it, but we don’t refer to anything here, so we do it very
elementarily.) First we show that there are no three sets Ai , Aj , Ak all containing p1 , p2 . Indeed, if Ai , Aj , Ak all contains p1 , p2 then since the pairwise
intersection of them has 2 elements, we get that Ai ∪ Aj ∪ Ak = {1, 2, . . . , 11}.
But then if Am is a fourth set which contains k elements from p1 , p2 then
|Am | = k + 3(2 − k) = 6 − 2k 6= 5. So every two points are contained in at
most 2 sets. On the other hand, let us consider the set
S = {(pi , pj , Ak ) | pi , pj ∈ Ak }.
5
Then |S| = 11 · 2 since if we fix Ak (11 ways) we can choose pi , pj in 52 ways.
On the other hand, we can choose
pi , pj first in 11
ways
2
and then Ak in at
11
5
11
most 2 ways. Hence |S| ≤ 2 · 2. Since 11 · 2 = 2 · 2 we get that every
two points are contained in exactly two sets.
Next we define some graphs on the set A2 . For every p ∈
/ A2 let us consider
the following graph Γ(p): for a pair x, y ∈ A2 let (x, y) ∈ E(Γ(p)) if the
other set going through x, y contains p. Note that there are two sets, A2
and another set contain x and y. This way we get six graphs, these graphs
12
are called Hussein-graphs. We will show that for every p, the graph Γ(p) is
2-regular and if p 6= p0 then Γ(p) ∩ Γ(p0 ) contains exactly two edges. The
first claim follows from the observation that the neighbors of x in Γ(p) are
nothing else than the the intersection of the two sets with A2 wich contain
x and p. The second claim follows from the fact that Γ(p) ∩ Γ(p0 ) are the
intersection of A2 and the two sets containing p and p0 . Note that one can
get back the sets from the Hussein-graphs: every x, y ∈ A2 determine the set
{x, y, p1 , p2 , p3 }, where (x, y) ∈ E(Γ(pi )), where i = 1, 2, 3. So it is enough
to show that the system of Hussein-graphs is unique. This is indeed true:
every Hussein-graph is 2–regular on 5 vertices, so it must be a 5–cycle. Let
Γ(p1 ) be the cycle containing the edges (1, 2), (2, 3), (3, 4), (4, 5), (5, 1). The
other graphs contain exactly two of these edges. These two edges cannot be
adjacent, because then the 5–cycle have to contain the opposite edge too.
So then they can only contain two non-adjacent edges, but this completely
determine the graph: {(1, 3), (3, 2), (2, 5), (5, 4), (4, 1)} and its rotated versions
will be the 5 other graphs. This shows the uniqueness of the block system and
the Golay-code!
Theorem 3.2. Let 1, 2, . . . , 24 be the points of a block design and the blocks
are supp(c), where c is a codeword of the extended Golay-code of weight 8.
Then every 5 points are contained in exactly one block. Hence this is a block
design of parameters 5 − (24, 8, 1).
Proof. Note that every set T of size 5 can be contained in at most one block
since if T ⊆ supp(c1 ), supp(c2 ) then w(c1 + c2 ) ≤ 6 and this is impossible by
the definition of the extended Golay-code. Now let us consider the set
A = {(T, c) | |T | = 5, c ∈ C, w(c) = 8, T ⊆ supp(c)}.
Then |A| = 759 · 85 since there are 759 codewords of weight 8in C, and all of
them contain 85 sets of size 5. On the other hand, |A| ≤ 24
, since every T
5
is contained in at most one support of c ∈ C of weight 8. Note that
8
24
759 ·
=
,
5
5
so every T must be contained in exactly one block.
Remark 3.3. In the next section we will prove the uniqueness of a block design
of parameters 5 − (24, 8, 1) by showing that such a block design generates the
Golay-code which is unique.
Next we show another construction for the extended Golay-code.
13
Theorem 3.4. Let N be the adjacency matrix of the icosahedron considered as
a graph. Let J be the all 1 matrix of size 12 × 12. Then the matrix (I J − N )
generates the extended Golay-code.
Proof. Let us consider the icosahedron as a graph: it has 12 vertices, 5–regular
and two verticies have 2 or 0 common neighbors. So in every row of the matrix
H = (I J − N ) there are exactly 8 1’s and any two row vector are orthogonal
to each other. Hence this is true for any two codewords generated by H. Then
every codeword has weight divisible by 4: this is true for the row vectors of H
and if it is true for c1 and c2 then it is true for c1 + c2 :
w(c1 + c2 ) = w(c1 ) + w(c2 ) − 2|supp(c1 ) ∩ supp(c2 )|,
and |supp(c1 ) ∩ supp(c2 )| is even since (c1 , c2 ) = 0. So for any c ∈ C then
4 | w(c).
It is clear that dim C = 12. Since C is self-orthogonal, it implies that
C = C ⊥ . If C is generated by the matrix H = (I A) then C ⊥ is generated by
H ∗ = (−AT I) = (A I) since in our case −1 = 1 and J − N is a symmetric
matrix. So C is generated by the matrix H ∗ = (J − N I).
Now we are ready to prove that C does not contain a codeword c with weight
4. Assume for contradiction that c ∈ C such that w(c) = 4. Let us write c into
the form (a, b), where a and b have length 2. Then c = a(I J −N ) = b(J −N I).
Since w(a) + w(b) ≤ 4 we have w(a) ≤ 2 or w(b) ≤ 2. We can assume that
w(a) ≤ 2 since the case w(b) ≤ 2 is analogous. If w(a) = 0 then a = 0 and
c = 0 by c = aH. If w(a) = 1 then c is one of the row vector of H hence
w(c) = 8. If w(a) = 2 then c is the sum of two row vectors of H, but then
w(c) = 8 or 12 according to the corresponding vertices in the icosahedron
have 0 or 2 common neighbors. Hence w(c) = 4 is impossible. Since every
codeword have weight divisible by 4 we get that the minimal distance is at least
8. This means that C is a [24, 12, 8]2 code, but then it must be the extended
Golay-code by the uniqueness of this code.
Remark 3.5. There is a surprising construction for the Golay-code. Put all
0 − 1 vectors into lexicographic order and choose greedily vectors which do not
violate the condition of minimal distance 8. Then the obtained code is the
extended Golay-code.
4. Uniqueness of the Witt-design
Theorem 4.1. There exists a unique block design of parameters 5 − (24, 8, 1)
up to permutations of the ground set.
Proof. We have already seen the existence of this block design, we only need to
prove its uniqueness. The strategy will be the following: we will show that such
14
a block design generates the extended Golay-code and since we have proved
the uniqueness of the Golay-code this will immediately implies the uniqueness
of our block design.
So let B be a block design of parameters 5 − (24, 8, 1). This means that for
any T of size 5 there is a unique set B ∈ B such that T ⊆ B. In particular, if
B, B 0 ∈ B then |B ∩ B 0 | ≤ 4 unless B = B 0 .
We will decompose the proof into several lemmas.
Lemma 4.2. If B, B 0 ∈ B then |B ∩ B 0 | =
6 3.
Before we start to prove this lemma we introduce an interesting substructure
called sunflower. Let A be a set of size 4 then for every p ∈
/ A there exists
a unique B ∈ B such that A ∪ {p} ⊆ B, now let us do it for every p ∈
/ A.
Then we will get 5 sets B1 , B2 , B3 , B4 and B5 such that A = Bi ∩ Bj =
B1 ∩ B2 ∩ B3 ∩ B4 ∩ B5 . The petals of the sunflower are Bi \ A, while its center
is A. Note that the union of B1 , B2 , B3 , B4 and B5 is the whole ground set.
Now we can start the proof of the lemma.
Proof. Assume that B ∩ B 0 = A0 such that |A0 | = 3. Then let A such that
A0 ⊆ A ⊆ B and |A| = 4. Let us create the sunflower with center A and sets
B1 = B, B2 , B3 , B4 , B5 . Since A0 ⊆ A ⊆ Bi and |B 0 ∩ Bi | ≤ 4 we get that
|B 0 ∩ (Bi \ A)| ≤ 1 for i = 2, 3, 4, 5 and |B 0 ∩ (Bi \ A)| = 0, but since B1 , . . . , B5
covers all points we get that
0
0
|B | ≤ |A | +
5
X
|B 0 ∩ (Bi \ A)| ≤ 7,
i=2
0
a contradiction. Hence |B ∩ B | cannot be 3.
Lemma 4.3. If B, B 0 ∈ B such that |B ∩ B 0 | = 4 then (B \ B 0 ) ∪ (B 0 \ B) ∈ B.
Proof. The proof of this lemma requires some case analysis. We can assume
that B = {1, 2, 3, 4, 5, 6, 7, 8} and B 0 = {1, 2, 3, 4, 9, 10, 11, 12}. Assume for
contradiction that U = {5, 6, 7, 8, 9, 10, 11, 12} is not a an element of B. There
is a unique Sx,y ∈ B such that {x, y, 9, 10, 11}, where {x, y} ⊆ {5, 6, 7, 8}. Note
that |Sx,y ∩ B 0 | ≥ 3, but from the previous lemma we know that it cannot be
exactly 3, and since it must be at most 4, we have |Sx,y ∩ B 0 | = 4. Now we
distinguish two cases according to Sx,y ∩ B 0 contains 12 or not for some {x, y}.
Case 1: There exists some x, y such that for S = Sx,y we have 12 ∈
/ S ∩ B0,
0
we can assume that {x, y} = {5, 6}. Hence {1, 2, 3, 4} ∩ (S ∩ B ) 6= ∅, we can
assume that 1 ∈ S ∩ B 0 . Now {1, 5, 6} ⊆ S ∩ B, so there must be a fourth
15
element in the intersection. This fourth element cannot be in {2, 3, 4}, because
then |S ∩ B 0 | ≥ 5. Hence 7 or 8 must be in S ∩ B 0 , we can assume that it is 7.
Note that 8 ∈
/ S since then |S ∩ B| ≥ 5.
Now let us consider the unique block R ∈ B which contains the set {9, 10, 11, 5, 8}.
Clearly, S 6= R because of the element 8. Note that if 12 ∈
/ R then the argument in the previous paragraphs gives that 6 or 7 is in R, but then |S ∩ R| ≥ 5
which is not possible. So 12 ∈ R, and 6, 7 ∈
/ R.
Now consider the unique block T ∈ B which contains the set {9, 10, 11, 7, 8}.
If it contains 12 then |T ∩R| ≥ 5 which is not possible. If 12 ∈
/ T then according
to the first paragraph 5 or 6 in T and so |S ∩ T | ≥ 5, a contradiction.
Case 2: For all {x, y} ⊆ {5, 6, 7, 8} we have 12 ∈ Sx,y ∩ B 0 . But then |S5,6 ∩
S5,7 | ≥ 5, a contradiction.
Lemma 4.4. For any B ∈ B there exist B 0 , B 00 ∈ B such that B, B 0 , B 00 are
pairwise disjoint.
Proof. Let us create a sunflower with set B1 = B, B2 , B3 , B4 , B5 . Then by the
previous lemma B 0 = (B2 \B3 )∪(B3 \B2 ) ∈ B and B 00 = (B4 \B5 )∪(B5 \B4 ) ∈
B, and clearly B, B 0 , B 00 are pairwise disjoint.
Lemma 4.5. If B, B 0 ∈ B then |B ∩ B 0 | =
6 1.
Proof. Assume for contradiction that for B, B 0 ∈ B we have |B ∩ B 0 | =
6 1.
Let us use the previous lemma to B. Then there are B ∗ , B ∗∗ ∈ B such that
B, B ∗ , B ∗∗ are pairwise disjoint. Since |B 0 ∩ B ∗ |, |B 0 ∩ B ∗∗ | ≤ 4, and
8 = |B 0 | = |B 0 ∩ B| + |B 0 ∩ B ∗ | + |B 0 ∩ B ∗∗ | = 1 + |B 0 ∩ B ∗ | + |B 0 ∩ B ∗∗ |,
we have |B 0 ∩ B ∗ | or |B 0 ∩ B ∗∗ | is equal to 3. But this is not possible by
the very first lemma. This contradiction shows that for B, B 0 ∈ B we have
|B ∩ B 0 | =
6 1.
Note that the first and last lemma together gives that for B, B 0 ∈ B we have
|B ∩ B 0 | even.
Now let
V = hb | b is a characteristic vector of a block B ∈ Bi.
We will show that V is the extended Golay-code. It is clear that if v 1 , v 2 ∈ V
then (v 1 , v 2 ) = 0 since it is true for the generating vectors. It is also true that
the weight of every vector is divisible by 4 since this is true for the generating
vectors and if it is true for v 1 and v 2 then it is true for v 1 + v 2 :
w(v 1 + v 2 ) = w(v 1 ) + w(c2 ) − 2|supp(v 1 ) ∩ supp(v 2 )|,
16
and |supp(v 1 ) ∩ supp(v 2 )| is even since (v 1 , v 2 ) = 0. So for any v ∈ V then
4 | w(v).
Lemma 4.6. There is no vector a of weight 4 in V .
Proof. Assume for contradiction that the a of weight 4 is in V . Let a be the
characteristic vector of A. Let us complete A to sunflower B1 , . . . , B5 . Let us
choose one-one point from all Bi \ A. Then there exists a block B ∈ B on this
five points. The sets B and A have intersection of size even since b and a are
orthogonal. This also means that B and Bi \ A have intersection of size even,
and since it is at least one we get that |B ∩ (Bi \ A)| ≥ 2. But it means that
8 = |B| ≥ 5 · 2 = 10, a contradiction. Hence there is no vector a of weight 4
in V .
Lemma 4.7. If v ∈ V has weight 8 then it is a characteristic vector of a block
in B.
Proof. Let A = supp(v). Let us choose 5 points from A. Then there is block
B with characteristic vector b containing these 5 points. So |A ∩ B| ≥ 5. And
since it is even, it must be either 6 or 8. If it is 6 then w(v + b) = 4 which
is impossible according to the previous lemma. So it must be 8 which means
that v = b.
Let us call a set D a dozen if it can be obtained as D = (B1 \ B2 ) ∪ (B2 \ B1 ),
where B1 , B2 ∈ B and |B1 ∩ B2 | = 2.
Lemma 4.8. If v ∈ V has weight 12 then it is a characteristic vector of a
dozen.
Proof. Let A = supp(v). Let us choose 5 points from A. Then there is block
B with characteristic vector b containing these 5 points. So |A ∩ B| ≥ 5. And
since it is even, it must be either 6 or 8. If it is 8 then w(v + b) = 4 which is
impossible according to the previous lemma. So it must be 6 which means that
v + b has weight 8 so it is a characteristic vector a block B 0 by the previous
lemma. Then A = (B \ B 0 ) ∪ (B 0 \ B), hence A is a dozen.
Lemma 4.9. The all 1 vector 1 is in V . Consequently, for a v ∈ V the weight
w(v) can be 0, 8, 12, 16, 24. If w(v) = 16 then it is the characteristic vector of
the complement of a block.
Proof. Adding together the characteristic vectors of a sunflower we get that 1
is in V . The other statements trivially follow from this observation.
Lemma 4.10. Let B ∈ B. Let Ui = |{B 0 ∈ B | |B ∩ B 0 | = i}|. Then
(i) There are exactly 759 blocks in B.
17
(ii) Every point is contained in 253 blocks.
(iii) U8 = 1, U1 = U3 = U5 = U6 = U7 = 0, U0 = 30, U2 = 448 and U4 = 280.
(iv) The number of dozens is 2576.
Proof. (i) Let us consider the set
Then |S| =
24
5
S = {(T, B) | |T | = 5, B ∈ B, T ⊆ B}.
= |B| 85 , hence |B| = 759.
(ii) Let us fix a point p, and let Bp be the set of blocks containing p. Let us
consider the set Let us consider the set
S 0 = {(T, B) | |T | = 5, B ∈ B, T ⊆ B, p ∈ T }.
Then |S 0 | = 23
= |Bp | 74 , hence |Bp | = 253.
4
(iii) It is clear that U8 = 1 and U1 = U3 = U5 = U6 = U7 = 0.
Next we show
8
that U4 = 280. Indeed, we can choose 4 elements of B in 4 = 70 ways and
each 4-element set is contained in 4 other blocks beside B, the other petals of
a sunflower. Hence U4 = 70 · 4 = 280. Next let us count the number of pairs
(B 0 , p), where p ∈ B∩B 0 . This is 8·253, because every point p of B is contained
in 253 sets including B itself. On the other hand, this is 2U2 +4U4 +8U8 . From
this we get that U2 = 448. Finally U0 = 759 − U2 − U4 − U8 = 30.
(iv) Let us consider the set
S 00 = {(B1 , B2 ) | B1 , B2 ∈ B, |B1 ∩ B2 | = 2}.
Then |S 00 | = |B| · U2 = 759 · 448. On the other hand, if D denotes the set of
dozens then we claim that |S 00 | = |D|· 61 12
. Indeed, if D = (B1 \B2 )∪(B2 \B1 ),
5
then |D ∩ B1 | = 6, and if |D ∩ B1 | = 6 then it determines B2 since b2 = d + b1
has weight 8. Now for a fixed D let us consider the set
R = {(T, B1 ) | T ⊆ D, |T | = 5, T ⊆ D ∩ B1 }.
First we can choose T in 12
ways. Then there exists a unique block B1
5
containing T . Let us mention that each such B1 intersects D in 6 points since
it must be even, at least 5 and it cannot be 8, because then d + b1 would have
weight 4. So |R| = 6nD , where nD is the number of blocks B1 intersecting D
in 6 points (the coefficient 6 comes
fact that T can be chosen
6 ways
from the
1 12
1 12
00
from D ∩ B1 ). Hence nD = 6 5 . Then |S | = |D|nD = |D| · 6 5 from which
|D| = 2576.
Theorem 4.11. V is the extended Golay-code.
18
Proof. Indeed, |V | = 2 · 1 + 2 · 759 + 2576 = 4096, so dim V = 12 and its
minimal distance is 8. Hence V has parameters [24, 12, 8]2 . Since the extended
Golay-code is unique, V must be the extended Golay-code.
Hence B is the Witt-design!
5. Derived symmetric structures
5.1. Strongly regular graphs.
Theorem 5.1. There exists a strongly regular graph with parameters (77, 16, 0, 4).
Proof. We only do the construction, to check the details is left as an exercise.
Consider the block design 3 − (22, 6, 1) which we get from 5 − (24, 8, 1) if we
fix two elements p, q ∈ {1, . . . , 24}, and consider those blocks which contain p
and q, and from all these block we remove p and q.
By double counting we get that there are
22
22 · 21 · 20
3
=
= 77
6
6
·
5
·
4
3
blocks. In the Witt-design two blocks can have intersection of size 0, 2, 4, 8, so
in this new block design the intersection can be 0, 2 or 6.
Now let us define the graph G as follows: its vertex set is the blocks in the
design of parameters 3 − (22, 6, 1), and two vertices are adjacent if the corresponding blocks are disjoint. This is a strongly regular graph of parameters
(77, 16, 0, 4).
Theorem 5.2. There exists a strongly regular graph with parameters (100, 22, 0, 6).
(This graph is called the Higman-Sims graph.)
Proof. We only do the construction, to check the details is left as an exercise.
Let us decompose this graph as follows: let v be its vertices, N (v) be its
neighbors and N2 (v) be the vertices of distance 2 from v. Clearly, |N (v)| = 22
and |N2 (v)| = 77. This suggests the following construction: to the 77 vertices
we put the strongly regular graph with parameters (77, 16, 0, 4). If w ∈ N2 (v)
then v and w should have 6 common neighbors: oh, the blocks have size 6,
so it is a natural guess to connect a block of size 6 with its elements. This is
indeed a good construction.
Remark 5.3. When we learned the Hoffman-Singleton theorem we have seen
that there exists a strongly regular graph with parameters (50, 7, 0, 1), the
Hoffman-Singleton graph. It turns out that it is an induced subgraph of the
19
Higman-Sims graph, and if we delete a copy of it from the Higman-Sims graph
then the remaining graph on 50 vertices is again a Hoffman-Singleton graph.
5.2. Equiangular lines.
Theorem 5.4. There exists
24
2
= 276 equiangular lines in R23 .
Proof. We need 276 unit vectors `1 , . . . , `276 such that (`i , `j ) = ± cos(α) for
some α.
It is a natural idea to start with the block design of parameters 4 − (23, 7, 1).
It has only
23
23 · 22 · 21 · 20
4
=
= 253
7
7
·
6
·
5
·
4
4
blocks. So the point-block incidence matrix N has size 23 × 253. Since 276 −
253 = 23, it is a natural guess that we will also need the identity matrix I.
Let nj be column vector of N , ej be a column vector of I.
We try to find the vectors `j in the form c(nj − b1), and d(ej − a1). Note
that
(ni − b1, nj − b1) = (ni , nj ) − 14b + 23b2 ,
where (ni , nj ) = 1 or 3 if i 6= j since in the Witt-design the intersection of
two different blocks can have size 0, 2 or 4, so in the punctured block design it
can be 1 or 3. The crucial observation is that two possible values of the scalar
product, 1 − 14b + 23b2 and 3 − 14b + 23b2 , have to be additive inverses to each
other, because (`i , `j ) = ± cos(α). Hence
−(1 − 14b + 23b2 ) = 3 − 14b + 23b2 .
Since their difference is 2 we have 3 − 14b + 23b2 = 1. Note that
(ni − b1, ni − b1) = (ni , ni ) − 14b + 23b2 = 7 − 14b + 23b2 = 4 + 1 = 5.
So even without computing b we get that c should be
(`i , `j ) =
Anyway,
√1 ,
5
and
±1
.
5
√
7± 3
b=
,
23
we can choose the plus sign.
Now let us consider the scalar product
(ni − b1, ej − a1) = (ni , ej ) − b − 7a + 23ab,
20
where (ni , ej ) is 1 if the corresponding block Bi contains the element j, otherwise it is 0. As before, the two possible values have to be additive invereses of
each other, so
1
−b − 7a + 23ba = − .
2
Then
√
b − 1/2
2−3 3
a=
=
.
23b − 7
46
One can check that
1
(ei − a1, ej − a1) = δi,j − 2a + 23a2 = δi,j + .
4
So d = √25 . Then for i 6= j we have
(d(ei − a1), d(ej − a1)) =
41
1
= ,
54
5
and
2
1
1
1
(c(nj − b1), d(ei − a1) = √ · √ · ±
=± .
2
5
5
5
1
Hence for any `i , `j we have (`i , `j ) = ± 5 .
5.3. Lattices. In this section we give the construction of the Leech-lattice. As
a warm up we construct the E8 lattice. Let us consider the Hamming-code with
parameters [7, 4, 3]2 , we can extend it uniquely to a code of parameters [8, 4, 4]2 .
In this code all words will have weight 0, 4 or 8. Now let us consider the
following lattice: (x1 , . . . , x8 ) ∈ Z8 if the vector (xi mod 2) is in the extended
Hamming-code. Then the minimal vectors are of the form (±2, 0, . . . , 0) or
(±1, ±1, ±1, ±1, 0, 0, 0, 0), where the four 1’s are restricted to the positions
given by the extended Golay-code. Hence there
8 · 2 + 14 · 24 = 240
minimal vectors, all of length 2.
Let us consider the following lattice: (x1 , . . . , x24 ) ∈ Z24 if the following
three conditions hold true for m = 0 or 1:
(i) xi ≡ m (mod 2),
(ii) xi −m
mod 2 is a word in the Golay-code,
P2 24
(iii) i=1 ≡ 4m (mod 8).
This lattice is called the Leech-lattice.
21
Proposition 5.5. √
There are 196560 minimal vectors in the Leech-lattice, all
of them has length 32.
Proof. Let us call the vector xi −m
mod 2 the reduced vector.
2
If m = 0 then we distinguish two cases according to the reduced vector is 0
or not. If it is 0 then all xi are divisible by 4, and by the third condition the
23
vector can’t
at least two ±4. The length of such a
√ be (±4, 0√ ), so there are 24
2
2
vector is 4 + 4 = 32. There are 2 22 such vectors.
If the reduced vector is non-zero then we have at least 8 xi ’s of form 4k + 2,
so the √
minimal vector
contains 8 ±2’s and 16 0’s, the length of such a vector is
√
2
again 8 · 2 = 32. From the 28 signings only half of them is good, because
of (iii). So there 759 · 27 such vectors.
If m = 1 then all xi ’s are odd. Observe that a vector of form (±1, . . . , ±1)
cannot be in the lattice, because the reduced vector being in the Golay-code
means that the number of −1’s is divisible by 4, but then (iii) cannot be
satisfied. So there should be at least ±3 and 23 ±1. Since 33 + 23 · 12 = 32 this
would give a minimal vector. Note that 1 and −3 reduces to 0, and −1 and
3 reduces to 1 in the reduced vector, since there are 212 words in the Golaycode we have 212 possibilites for the signing, so there are 24 · 212 such minimal
vectors.
Altogether we have
24 2
2 + 759 · 27 + 24 · 212 = 196560
2
√
minimal vectors, and their length is 32.
Remark 5.6. If we pack spheres into R8 , and to R24 such that their centers are
in the E8 lattice and the Leech-lattice, respectively and neighboring spheres
touch each other then these are the densest sphere packings in R8 and R24 ,
respectively. For d = 8 this was shown by Maryna Viazovski very recently,
and for d = 24 by Henry Cohn, Abhibnav Kumar, Stephen Miller, Danylo
Radchenko and Maryna Viazovska one week after Maryna’s result for d = 8.