VIGRE TALK: THE MIRACLE OCTAD GENERATOR

EMILY JENNINGS

1 Nov 2012

1. R. T. Curtis’s MOG

In 1974, Curtis used the Miracle Octad Generator to determinethe unique octad containing any 5 given points in a Steiner SystemS(5, 8, 24). He begins with a proof of a Steiner System S(5, 8, 24) usingTuryn’s construction. A sketch of this proof is below. But first we notethat if a such a Steiner system exists, there must be

(248

)/(85

)= 759

such 8-element subsets of the 24-element set.

Theorem 1. There exists a Steiner system S(5, 8, 24).

Proof. Let Ω be a 24-element set. Consider the power set P(Ω) as a24-dimensional vector space over F2, where the sum of two elements isdefined to be their symmetric difference. Let Λ ⊂ Ω be an 8-elementset. Let P and L be any two 3-dimensional subspaces of P(Λ) whosemembers are all 4-element subsets of Λ and whose intersection is theempty set. Here’s an example of two such subspaces:

Figure 1. Subspace P = span A,B,D of Λ.

Figure 2. Subspace L = span a, b, c of Λ.1

Let t be any member of L. Then t defines a 2-dimensional subspaceof P by lt = X ∈ P : |t ∩X| = 2. For ease of notation, if X ∈ lt letus say that X ∈ t. The following figure illustrates the correspondencebetween an element t and its subspace (line) lt.

Figure 3. 2-dimensional subspaces of P for each t ∈ L.

Note that X ∈ t ⇒ |X + t| = 4 and X /∈ t ⇒ |X + t| = 2 or 6. Sothen any even subset of Λ can be expressed as X + t or X ′ + t for someX ∈ P, t ∈ L where X ′ is the complement of X in Λ, i.e. X +X ′ = Λ.

Take 3 copies of Λ and define C to be the space of all sets of theform

(X or X ′) + t (Y or Y ′) + t (Z or Z ′) + tin Λ1 in Λ2 in Λ3

where X + Y + Z = 0.

Let [X ′Y Z]t denote X ′ + t ∈ Λ1, Y + t ∈ Λ2, Z + t ∈ Λ3. Then bylooking at the possible sizes of the C -sets, we see that the smallestnonzero C -sets are of size 8.

Suppose two octads have 5 or more points in common. Then theirsymmetric difference would have 16 − 10 = 6 or fewer points in it.Since the smallest C -sets have 8 elements, this cannot happen. Thusthe C -sets of size 8 form a Steiner system S(5, 8, 24). These ”octads”

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can be formed in the following ways:

Shape modulo complementation in Λi Number of distinct octads[000]o 3 distinct octads

[XX0]o 7 · 3 · 2 · 2 = 84 distinct octads[XX0]t 7 · 4 · 3 · 2 = 168 distinct octads[XY Z]t 7 · 3 · 3 · 2 · 2 · 2 = 504 distinct octads

for a total of 759 octads.

Example 2.

a) The 3 octads of shape [000]o are simply Λ1,Λ2, and Λ3 with therest of the 16 points empty.

b) To construct an octad of shape [XX0]o, there are

• 7 choices for X 6= 0• 2 choices for X or X ′ in Λi

• 2 choices for X or X ′ in Λj

• 1 choices for Λk (Λk = 0), 3 choices for k (k = 1, 2, or 3).

If the octad is not of shape [000]o (i.e. not Λ1,Λ2, or Λ3), then theoctad intersects Λ1,Λ2, or Λ3 in exactly 4 points. We will call thisΛi the ”heavy tetrad.” We will call the remaining Λj,Λk together the”square tetrad,” as there are 4 points remaining to complete the octad.

The ”heavy tetrad” may be one of any of the(84

)= 70 tetrads in

Λi.These 70 tetrads can be divided into 35 groups of tetrad couples,where each couple is given by (tetrad,Λi − tetrad).

Because P and L do not intersect, the ”square tetrad” must be oneof 140 special tetrads that intersect all the rows of the square with thesame parity, and all the columns of the square with the same parity.These 140 tetrads can be divided into 35 groups of 4, where each groupis given by (tetrad1,Λj − tetrad1, tetrad2,Λk − tetrad2).

There is a correspondence between these 2 systems of 35 groups: the6 tetrads should be a sextet, i.e. the union of any 2 of these tetradsforms an octad. This correspondence is shown in the Miracle OctadGenerator. Then using the MOG, one can read off the unique octad inS(5, 8, 24) given any 5 of its points.

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Figure 4. Curtis’s MOG

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Example 3. Find the octad containing the points 0, ∞, 19, 20, and10.From the labeling picture below (left), we can see that 3 points of these5 lie in Λ3 so Λ3 is the heavy tetrad and has shape below (right).

This could be one of five tetrads in the MOG. Looking at the corre-sponding square tetrads for each, only one contains the points 20 and10, namely that in the 5th column of the last row of Figure 4. Theremaining points are read off from the MOG as shown below (left).Hence the octad containing the points 22, 1, 12, 6, and 8 correspondsto the picture below (right), namely 0,∞, 19, 6, 20, 10, 14, 17.

2. The Hexacode

2.1. Definition and Properties of the hexacode.

Define F4 = 0, 1, ω, ω with the relations

1 + ω = ω, 1 + ω = ω, ω + ω = ωω = 1, ω2 = ω, ω2 = ω, ω3 = 1.

Definition 4. The [6, 3, 4] hexacode C6 is a code over F4 with generatormatrix 0 0 1 1 1 1

0 1 0 1 ω ω1 0 0 1 ω ω

Note that C6 has 43 = 64 codewords.

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C6 has the following symmetries:

(1) scalar multiplication by a power of ω (namely 1, ω, or ω),(2) switching the 2 digits in any two of the three couples,(3) any permutation of the three couples.

And, in fact, every hexacode word is an image under these symmetriesof one of the following words:

01 01 ωω, ωω ωω ωω, 00 11 11, 11 ωω ωω, 00 00 0036 images 12 images 9 images 6 images 1 image

Note that 36+12+9+6+1 = 64 = |C6|. Then we can check if a givenword is in the hexacode by inspecting its shape and its sign.

Lemma 5. A given word x1y1 x2y2 x3y3 belongs to C6 if and only if:

(1) up to permutations and flips of the couples, it has one of thefollowing shapes:

0a 0a bc, bc bc bc, 00 aa aa, aa bb cc, 00 00 00,

where a, b, c are distinct elements of the set 1, ω, ω, and(2) Π3

i=1sign(xiyi) > 0 or sign(xiyi) = 0 for i = 1, 2, 3, wheresign(xiyi) is defined as follows:

+ − 00a a0 001ω 1ω 11ωω ωω ωωω1 ω1 ωω

2.2. The hexacode and C24.

We can obtain codewords of C24 from hexacode words by replacingeach digit in the codeword by a column of length 4. (We now have 6columns of length 4.) We do so in one of two ways:

• Replace each digit by an odd interpretation, in any way suchthat the count of the top row is odd (i.e. the top row containsan odd number of non-blanks).• Replace each digit by an even interpretation, in any way such

that the count of the top row is even (i.e. the top row containsan even number of non-blanks).

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Figure 5. Interpretations of the hexacode digits. Oddinterpretations are to the left; even interpretations are tothe right.

Example 6. A single hexacode word corresponds to multiple Golaycodewords.

Figure 6. Some interpretations of the codeword 01 01 ωω.

Recall that octads are the Golay codewords of weight 8 and comprisethe Steiner system S(5, 8, 24). We have seen how to complete an octadgiven 5 points using Curtis’s MOG. Now let’s see how to complete anoctad given 5 of its points using Conway’s hexacode.

Example 7. Complete the octad containing

We know that we must either use odd interpretations such that thetop row becomes odd or even interpretations such that the top rowbecomes even.

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If odd: Since we can only add 3 points, columns 2, 3, and 5 mustbe left unchanged. (They are already odd.) We must add one point toeach of the columns 1, 4, and 6. Since 2, 3, and 5 are unchanged, wealready know the letters they correspond to in F4, shown below (left).Due to shape constraints, the hexacode word must be of the form 00aa aa, hence columns 4 and 5 must have an odd interpretation of 1that uses only one point, as shown below (right). We now need to add1 point to column 1 to obtain an odd interpretation of 0. No suchinterpretation exists.

If even: Since we can only add 3 points, columns 1, 4, and 6 mustbe left unchanged and we must add one point to each of the columns2, 3, and 5. Since columns 1, 4, and 6 are unchanged, we know theycorresponds to 1, 0, 0 respectively in F4, shown below (left). Dueto shape constraints, the hexacode word must be of the form 0a 0abc. So we have two cases: (1) columns 3 and 5 correspond to ω andcolumn 2 corresponds to ω, shown below (middle) or (2) columns 3and 5 correspond to ω and column 2 corresponds to ω, shown below(right).

Referring to Lemma 5, we see that the couples of 1ω ω0 ω0 have signs− − −, and hence this string does not correspond to a hexacode word.The couples of 1ω ω0 ω0 have signs + − −. Hence this string is a validhexacode word, and the octad containing the given 5 points is thatshown in the right of the figure above. In fact, we already knew thiswas the octad containing the given points from using Curtis’s MOGearlier.

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3. MOG The Projective Subgroup L2(23), A MaximalSubgroup of M24

Let us label the 24 letters of M24 ∞, 0, 1, ..., 22. Then PSL2(23) orL2(23) acts on these letters by linear fractional transformations,z 7→ (az + b)/(cz + d) with ad − bd a nonzero square mod 23. We

can represent these by matrices

[a bc d

]with determinant a nonzero

square modulo 23.Due to a theorem by Gleason and Prange, the full automorphism groupof an extended quadratic residue code of length n contains the groupPSL2(n).

Definition 8. For p and n primes such that x2 ≡ p mod n, the qua-dratic reside code of length n over Fp is the cyclic code whose generatorpolynomial has roots αi : i 6= 0 is a square mod n. The code has di-mension n+1

2.

Remark 9. Note that for p = 2, p is a quadratic residue mod n ifand only if n ≡ ±1 mod 8. So if p = 2, then n ≡ ±1 mod 8 for aquadratic residue code to exist.The [23,12,7] binary Golay code (C23) over F2 with dimension d = 12.C23 is the quadratic residue code with p = 2 and n = 2d−1 = 23 ≡ −1mod 8. Recall that we can obtain C24 (the extended binary Golay code)by appending a parity bit to words in C23.

C24 = span (Q), where Q = 0 ∪ quadratic residues mod 23 =0, 1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18.

Let the dodecad below correspond to Q and insert the numbers 0, 1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11 across the dodecad replacing the ∗ placesand negate the non-squares.

Figure 7. The dodecad used in construction of the MOG.

Fill in the remaining points of the dodecad using the linear fractionaltransformation γ : z 7→ −1/z, which correspond to the permutation in-dicated in the left half of Fig. 8. (Note this map acts by transpositions:(0 ∞)(1 22)(2 11)(3 15)(4 17)(6 19)(8 20)(9 5)(12 21)(13 7)(16 10)(18 14)).

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The right half of Fig. 8 is the end result: the standard MOG labeling.

Figure 8. The action of γ and the standard MOG labeling.

Then any map in L2(23) can be expressed as a permutation of M24.

4. The Tetracode, and MINIMOG

Let F3 = o,+,− with the usual relations.The tetracode C4 consists of the following 9 words over F3:

o ooo o + ++ o −−−+ o +− + +−o + − o+− o−+ − + o− − −+o

The MINIMOG is 4 × 3 array with rows are labeled o, +,−. Thecolumns of MINIMOG correspond to the digits of the tetracode.

The hexads for which

(1) the odd-men-out form part or all of a tetracode word, and(2) the column distribution is not 3 2 1 0

comprise all 132 hexads of the Steiner system S(5, 6, 12).

We will illustrate the process of completing a hexad from 5 pointsthrough examples.

Example 10. Odd-men-out form all of a tetracode word.

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Example 11. Odd-men-out form part of a tetracode word.

Hexads in S(5, 6, 12) can be extended to octads in S(5, 8, 24) by in-serting the MINIMOG into the MOG and completing the octad.

Example 12. Using the hexad from Example 10, we can construct anoctad in S(5, 8, 24).

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