Extremal lattices of minimum 8 related to the Mathieu group M_{22}.

Christine Bachoc ^{∗}  Gabriele Nebe ^{†}

dedicated to Martin Kneser

Abstract

In this paper, we construct three new extremal lattices of minimum 8; one is 3-modular and of dimension 40, the two others are unimodular of dimension 80. They are strongly connected to the 20dimensional lattice with automorphism group isomorphic to 2.M_{22}.2.

1 Introduction

An even, unimodular lattice can only exist in dimensions n which are multiples of 8, and one knows, from the modular properties of its theta series, that its minimum is bounded by 2([n/24] + 1) (see [CoS 88]). A lattice attaining this bound is said to be extremal. Extremal lattices are known up to dimension 64; the most wanted would be a 72-dimensional lattice of minimum 8, which is not yet proved to exist. In this paper, we construct two extremal lattices in dimension 80. One of them belongs to a series of three lattices of minimum 8, in dimension 20, 40, 80, which are modular of level 7, 3, respectively 1.

∗Recherche effectuée au sein de l’Unitée Mixte C.N.R.S.–Enseignement supérieur U.R.M. 9936

†supported by the DFG

1991 Mathematics Subject Classification: 11H31, 11H56, 94B05

An even lattice is said to be modular if it is similar to its dual. The norm of the similarity is then called the level of the lattice. In particular, its determinant is equal to the level to the power n/2. The modular properties of the theta series of such lattices, studied in [Que 95] lead to the notion of extremal lattices; if the level l is a prime number such that 1 + l divides 24, an extremal lattice has minimum 2([n/k] + 1), where k = 2 · 24/(1 + l). For the levels 7, 3, 1, we get k = 24, 12, 6, which means that the “jump dimensions” (which are the dimensions where the minimum may increase) for level 1 (resp. 3) is twice the “jump dimensions” for level 3 (resp. 7). In order to shift from one level to another, we can take the tensor product of a lattice with suitable elementary lattices. An example of such a situation is the following, and was communicated to the first author by B. Gross: let α satisfy the equation α^{2} − α + 2 = 0; then ℤ[α] is the ring of integers of the imaginary quadratic field of discriminant -7. This ring embeds into any maximal order 𝔐 of the quaternion field ℚ_{3,∞} over ℚ ramified at 3 and ∞, as follows. We set i^{2} = −1, j^{2} = −3, ij = k so that ℚ_{3,∞} = ℚ(i, j) and choose 𝔐 = ℤ[1, i, (1 + j)/2, (i + k)/2]. Then α = (1 + i + j + k)/2 ∈ 𝔐. It embeds also in a maximal order of the octonions over ℚ, as described in ([Gro 96],

§4). Then the hermitian matrix E = ( \begin{matrix}2 & α & −1 \\ \overline{α} & 2 & α \\ −1 & \overline{α} & 2 \end{matrix} ) defines over these

three rings a lattice for the scalar product x.y = trace(x^{t}E\overline{y}), where trace is the reduced trace on each algebra, which is isometric to the Barnes lattice P_{6}, the Coxeter-Todd lattice K_{12}, respectively the Leech lattice Λ_{24}. In other words, if P_{6} is the Barnes lattice considered with its hermitian structure over ℤ[α], then we have the isometries 𝔸_{2}^{2} ⊗_{ℤ[α]} P_{6} ∼ K_{12} and 𝔼_{8} ⊗_{ℤ[α]} P_{6} ∼ Λ_{24} (together with the scalar product trace(x\overline{y}), the previous maximal orders define respectively the root lattices 𝔸_{2}^{2} and 𝔼_{8}).

We point out the existence of a similar series of three extremal lattices of minimum 8. The first one is a 20-dimensional even lattice of level 7, which is a ℤ[α]-unimodular lattice; its automorphism group is the group 2.M_{22}.2, as shown in [PlN 95], [CCNPW 85]. Its structure as a ℤ[α]-lattice is explicitly given in [CCNPW 85]. Then L_{40} = 𝔸_{2}^{2} ⊗_{ℤ[α]} L_{20} and L_{80} = 𝔼_{8} ⊗_{ℤ[α]} L_{20} are of level 3 respectively unimodular; from this definition we determine their automorphism groups in §3. As shown in §2, one can easily get a Gram matrix for these lattices from a Gram matrix of L_{20}, following a general procedure that constructs modular Gram matrices (and we give several examples of such constructions); but this is of no use in order to compute the minimum of the 80-dimensional lattices, which might become lower than 8, since the algorithms designed to compute the minimum of a lattice are by this time limited to the dimensions around 60. In §4 we show that these lattices are special cases of a general construction using a (10, 5, 4)-binary code, which allows us to compute their minimum. It moreover gives two non-isometric lattices in dimension 80.

A few constructions were already proposed for an 80-dimensional unimodular extremal lattice. The most natural would perhaps be a lattice invariant under SL_{2}(79), in connection with the corresponding quadratic residue code; see [Sch 93]. In fact, there is a unique even unimodular SL_{2}(79)-invariant lattice of dimension 80 which remains a candidate to be extremal. In [Neb 97] resp. [BQS 95] three other candidates are constructed, which provide the groups SL_{2}(41)⊗S̃_{3} resp. C_{4} ×C_{41} : C_{40} as subgroups of their automorphism groups. None of them can overcome the problem of the determination of its minimum (although LLL-reduction doesn’t give any vector of norm 6).

It is worth noticing that the center density of an 80-dimensional unimodular lattice of minimum 8 is equal to 2^{40}, which is slightly less than the one of the lattice of the same dimension constructed independently by Elkies and Shioda which has center density 2^{40.14..} ([Shi 91]).

2 A convenient construction

In this section we give a convenient construction for modular lattices. In particular Gram matrices for the lattices L_{40} and L_{80} may easily be obtained from a Gram matrix of L_{20}, the latter being given in [PlN 95] (cf. also [CCNPW 85]).

Proposition 2.1 Let K be either ℚ or an imaginary quadratic number field and F ∈ K^{n×n} a hermitian positive definite matrix. Let a, d ∈ ℚ, b ∈ K, such that a, d, ad − bb̄ > 0. Then the matrix

Q(F, a, b, d) := ( \begin{matrix}aF & bI_{n} \\ b̄I_{n} & dF^{−1} \end{matrix} )

is a positive definite hermitian matrix of determinant (ad − bb̄)^{n}. The map

g → ( \begin{matrix}g & 0 \\ 0 & g^{−tr} \end{matrix} ) embeds Aut(F) into Aut(Q(F, a, b, d)).

If K = ℚ then Q(F, a, b, d) is ad − b^{2}-modular with respect to the sym-

plectic similarity ( \begin{matrix}0 & I_{n} \\ −I_{n} & 0 \end{matrix} ).

\underline{Proof:} Straightforward, using

(⋆) Q(F, a, b, d) = ( \begin{matrix}1 & 0 \\ 0 & F^{−1} \end{matrix} ) (( \begin{matrix}a & b \\ b̄ & d \end{matrix} ) ⊗ F) ( \begin{matrix}1 & 0 \\ 0 & F^{−1} \end{matrix} ) .

□

Remark: Let B be a basis of the lattice L in the hermitian space (K^{n}, ϕ_{0}) with Gram matrix F and B^{∗} the dual basis of B which is a lattice basis of the dual lattice L^{∗}. Then Q(F, a, b, d) is the Gram matrix of the lattice L ⊕ L^{∗} in the hermitian space (K^{n} ⊕ K^{n}, ϕ) with respect to the basis (B, B^{∗}), where the hermitian form ϕ : K^{n} ⊕ K^{n} → K is given by

ϕ(l_{1} + l′_{1}, l_{2} + l′_{2}) = aϕ_{0}(l_{1}, l_{2}) + bϕ_{0}(l_{1}, l′_{2}) + b̄ϕ_{0}(l′_{1}, l_{2}) + dϕ_{0}(l′_{1}, l′_{2})

for all l_{1}, l_{2}, l′_{1}, l′_{2} ∈ K^{n}.

Many well known lattices can be described as above:

For example the root lattices 𝔻_{4} = Q(𝔸_{2}, 1, 1, 3) and 𝔼_{8} = Q(𝔻_{4}, 1, 1, 2) = Q(𝔸_{4}, 1, 2, 5), but also the Coxeter Todd lattice K_{12} = Q(P_{6}, 1, 2, 7) can be constructed in this way. Examples: Some extremal 2-modular lattices:

(i) The Barnes-Wall lattice (cf. [CoS 88] p.129) BW_{16} = Q(𝔸_{2}⊗𝔻_{4}, 1, 2, 6).

(ii) The 2 known extremal 2-modular lattices of dimension 20 (cf. [PlN 95] p. 49) may be obtained from the Craig lattices (cf. [CoS 88] p.222) in dimension 10: [2.M_{12}.2]_{20} = Q(A_{10}^{(2)}, 1, 3, 11) and [SU_{5}(2)^{2(2)} ∘ SL_{2}(3)]_{20} = Q(A_{10}^{(3)}, 1, 3, 11).

(iii) The extremal 2-modular lattice of dimension 24 with maximal finite automorphism group ([Neb 95], [Neb 96]): [6.U_{4}(3).2 ^{2} □ × \sqrt{-3} SL_{2}(3)]_{24} = Q(K_{12}, 1, 1, 3).

(iv) Applying this construction to a Gram matrix F of the 6-modular lattices of minimum 8 belonging to the maximal finite subgroups 16 and 17 of GL_{24}(ℚ) (in [Neb 95] and [Neb 96]) one obtains extremal 2-modular lattices Q(F, 1, 2, 6) of dimension 48.

(v) If F is a Gram matrix of the 6-modular lattice of minimum 6 in dimension 16 with maximal finite automorphism group number 9 of [NeP 95] then Q(F, 1, 2, 6) is a non extremal 2-modular lattice of dimension 32. So this construction does i.g. not preserve the minimum.

It should be noted, that the construction Q only depends on the lattice and not on the choice of the Gram matrix. For unimodular F, the lattice

Q(F, a, b, d) is simply the tensor product ( \begin{matrix}a & b \\ b & d \end{matrix} )⊗F. For arbitrary integral

F, the construction Q is a natural way to obtain a (modular) overlattice of this tensor product, which is also invariant under the automorphism group Aut(F).

Let F be a Gram matrix of L_{20}. Then the corresponding bilinear form is of the form trace ∘ f, where L_{20} is a 10-dimensional ℤ[α]-module and f is a unimodular hermitian form L_{20} × L_{20} → ℤ[α].

Proposition 2.2 (i) Q(F, 1, 2, 7) ∈ ℤ^{40×40} is a Gram matrix of the ℤlattice L_{40}.

(ii) g := Q(f, 1, \frac{2}{\sqrt{−7}}, 1) ∈ ℤ[α]^{20×20} is a Gram matrix of the hermitian ℤ[α]-lattice L_{40}.

(iii) Q(g, 1, \frac{−α}{\sqrt{−7}}, 3) ∈ ℤ[α]^{40×40} is a Gram matrix of the hermitian ℤ[α]- lattice L_{80}.

\underline{Proof:} (ii) Let 𝔐 be the maximal order of ℚ_{3,∞}. As described in the introduction, 𝔐 is a hermitian ℤ[α]-lattice. With respect to the ℤ[α]-basis (\frac{i−k}{2}, 1), the form trace(x\overline{y}) on 𝔐 has the hermitian Gram matrix

h := ( \begin{matrix}1 & \frac{2}{\sqrt{-7}} \\ −\frac{2}{\sqrt{-7}} & 1 \end{matrix} ) .

The ℤ[α]-hermitian form on L_{40} = L_{20}⊗_{ℤ[α]} 𝔐 is given by the Gram matrix h ⊗ f. Since f is unimodular, one finds (ii) using (⋆).

(i) may be deduced from (ii) by blowing up the matrix g: Let (b_{1}, …, b_{10}) be a basis of the ℤ[α]-lattice L_{20} with Gram matrix f and (b_{1}^{∗}, …, b_{10}^{∗}) the corresponding dual basis (i.e. f(b_{i}, b_{j}^{∗}) = δ_{ij}). A ℤ-basis of L_{20} is given by B := (b_{1}, αb_{1}, …, b_{10}, αb_{10}). Let F be the corresponding Gram matrix of trace ∘ f. Then the dual basis is B^{∗} = (\frac{1}{7}(4 − α)b_{1}^{∗}, \frac{1}{7}(1 − 2α)b_{1}^{∗}, …, \frac{1}{7}(4 − α)b_{10}^{∗}, \frac{1}{7}(1−2α)b_{10}^{∗}) (with Gram matrix F^{−1}). To get integral scalar products we multiply the elements of B^{∗} by \sqrt{−7} and compute trace(g(b_{i}, \frac{4−α}{\sqrt{−7}}b_{j}^{∗})) = δ_{ij}(trace(−\frac{2}{\sqrt{−7}} \frac{4−ᾱ}{\sqrt{−7}})) = 2δ_{ij}, etc.. Therefore trace ∘ g = Q(F, 1, 2, 7).

(iii) The matrix

o := ( \begin{matrix}1 & \frac{2}{\sqrt{−7}} & \frac{−α}{\sqrt{−7}} & 0 \\ −\frac{2}{\sqrt{−7}} & 1 & 0 & \frac{−α}{\sqrt{−7}} \\ \frac{ᾱ}{\sqrt{−7}} & 0 & 1 & −\frac{2}{\sqrt{−7}} \\ 0 & \frac{ᾱ}{\sqrt{−7}} & \frac{2}{\sqrt{−7}} & 1 \end{matrix} )

is a Gram matrix of the ℤ[α]-lattice 𝔼_{8}. Hence (iii) follows from the equation

( \begin{matrix}I_{10} & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & f & 0 \\ 0 & 0 & 0 & I_{10} \end{matrix} ) Q(g, 1, \frac{−α}{\sqrt{−7}}, 3) ( \begin{matrix}I_{10} & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & f & 0 \\ 0 & 0 & 0 & I_{10} \end{matrix} ) = o ⊗ f.

□

3 The automorphism groups

In this section we deduce the automorphism groups Aut(L), i.e. the group of all orthogonal transformations mapping the lattice L into itself, of the two lattices L = L_{40} and L = L_{80}. Since L_{40} only has 360360 minimal vectors, it might be possible to compute Aut(L_{40}) with the computer program described in [PlS 97]. But L_{80} has 1250172000 vectors of length 8 (which moreover cannot be found directly by computer), so here this method will fail. We use the classification of finite simple groups to show that the automorphism groups of both lattices are the tensor products of Aut(L_{20}) = 2.M_{22}.2 with the automorphism groups of the hermitian lattices 𝔸_{2}^{2} respectively 𝔼_{8}.

Proposition 3.1 (i) The hermitian automorphism group of the ℤ[α]-lattice 𝔸_{2}^{2} is the unit group of the maximal order 𝔐 of ℚ_{3,∞}, isomorphic to S̃_{3}.

(ii) The hermitian automorphism group of the ℤ[α]-lattice 𝔼_{8} is 2.Alt_{7}.

(iii) Aut(L_{20}) = 2.M_{22}.2 is an extension of the double cover 2.M_{22} of the Mathieu group M_{22} by the outer automorphism of order 2. This group is also the hermitian automorphism group of the ℤ[α]-lattice L_{20}.

\underline{Proof:} (i) The automorphism group of the integral lattice 𝔸_{2}^{2} is D_{12} ≀ C_{2} the wreath product of the dihedral group of order 12 by C_{2}. The hermitian automorphism group U ≤ GL_{4}(ℤ[α]) of this lattice may be regarded as a subgroup of D_{12}≀C_{2} commuting with an element α ∈ GL_{8}(ℚ) with α^{2}−α+2 = 0. By construction, U contains the unit group S̃_{3} of 𝔐. Since 7 does not divide the order of U, the commuting algebra C_{ℚ^{8×8}}(U) of U in ℚ^{8×8} contains ℚ[α] properly. One concludes that C_{ℚ^{8×8}}(U) = C_{ℚ^{8×8}}(S̃_{3}) ≅ ℚ_{3,∞}. Therefore U embeds into ℚ_{3,∞} and hence U = S̃_{3}. (ii) Let U ≤ GL_{4}(ℤ[α]) be the hermitian automorphism group of 𝔼_{8}. Then U ≤ Aut(𝔼_{8}) = 2.O_{8}^{+}(2).2 is contained in the stabilizer of a maximal totally singular subspace T_{1} = α𝔼_{8}/2𝔼_{8}. By [CCNPW 85] this subspace is unique up to the action of 2.O_{8}^{+}(2).2 and its stabilizer is 2_{+}^{1+6}.Alt_{8}. One computes the centralizer of α in this group to be the subgroup 2.Alt_{7} of index 512. (iii) Follows from [CCNPW 85] p. 39 (cf. also [PlN 95] Chapter IX). □

Theorem 3.2 The automorphism group of L_{80} is G := Aut(L_{80}) = 2.Alt_{7}⊗\sqrt{−7} 2.M_{22}.2 and a maximal finite subgroup of GL_{80}(ℚ). The G-invariant lattices in ℚ^{1×80} are of the form cL_{80}, where 0 ≠ c ∈ C_{ℚ^{80×80}}(G) ≅ ℚ[α] is an invertible element in the commuting algebra of G (i.e. G is a GIR in the sense of [Gro 90]).

\underline{Proof:} Let U := 2.Alt_{7} ⊗\sqrt{−7} 2.M_{22}.2. By construction U is a subgroup of Aut(L_{80}) = G. The commuting algebra of U is isomorphic to ℚ[α] and U fixes up to isomorphism only one lattice (cf. [CCNPW 85] and [JLPW 95]). So we only have to show that G = U. Since L_{80} is orthogonally indecomposable, G is a maximal finite primitive subgroup of GL_{80}(ℚ).

Assume first, that 2.M_{22} ≤ G is a normal subgroup of G. Then C := C_{G}(2.M_{22}) embeds into GL_{4}(ℚ[α]) and contains 2.Alt_{7}. Since 2.Alt_{7} is the full automorphism group of its up to isomorphism unique lattice in ℚ[α]^{1×4} this implies that C = 2.Alt_{7} and G contains 2.Alt_{7} ⊗\sqrt{−7} 2.M_{22} of index ≤ 2 = |Out(2.M_{22})|. Hence G = U. The same conclusion holds under the assumption that 2.Alt_{7} is a normal subgroup of G, because then C := C_{G}(2.Alt_{7}) is 2.M_{22}.2, a maximal finite subgroup of GL_{20}(ℚ). Therefore 2.M_{22} ≤ G and G = U.

Hence we may assume that none of the two groups is a normal subgroup of G. Let F := F it(G) = ∏_{p||G|} O_{p}(G) be the Fitting group of G. The primitivity of G implies that all abelian characteristic subgroups of F are cyclic. A Theorem of P. Hall, which classifies those p-groups, all of which abelian characteristic subgroups are cyclic (cf. [Hup 67], p. 357), yields only rather restricted possibilities for F: If p ≠ 2, 5, then O_{p}(G) is cyclic, for p = 5, one additionally has the possibility O_{5}(G) ≅ 5_{+}^{1+2} and O_{2}(G) embeds into GL_{16}(ℚ), leaving many possibilities (cf. [NeP 95]). None of the possible groups F have an outer automorphism group the order of which is divisible by 11.

Therefore one concludes that S := G^{(∞)}/Z(G) is a finite simple group containing U^{(∞)}/Z(U). An inspection of the orders and character tables of the 26 sporadic finite simple groups in [CCNPW 85] shows that S is a finite Chevalley group. Bounds for the minimal degrees of the projective representation of the finite Chevalley groups are given in [SeZ 93] (cf. also [LaS 74]). Using the fact, that the minimal degree of a faithful projective modular representation of U^{(∞)}/Z(U) is 3 · 6 = 18 (cf. [JLPW 95]), one immediately excludes the case that S is a classical group. The bounds in [SeZ 93] now only leave the possibilities that S is one of F_{4}(2), G_{2}(2), G_{2}(3), G_{2}(4), Sz(8), Sz(32), ^{2}G_{2}(3), ^{2}G_{2}(27), ^{3}D_{4}(2), or ^{2}F_{4}(2). But none of these 10 groups is of order divisible by 11, contradicting the assumption that M_{22} ≤ S. □

Using the same arguments one shows the following

Theorem 3.3 The automorphism group of L_{40} is G := Aut(L_{40}) = S̃_{3} ⊗\sqrt{−7} 2.M_{22}.2 and a maximal finite subgroup of GL_{40}(ℚ). The G-invariant lattices in ℚ^{1×40} are of the form cL_{40} or cL_{40}^{#}, where 0 ≠ c ∈ C_{ℚ^{40×40}}(G) ≅ ℚ[α] is an invertible element in the commuting algebra of G.

4 The minimum

In order to prove that the minimum of L_{20}, L_{40}, L_{80} is 8, we shall reconstruct them using codes. This general construction, of which a special case is already described in [Bac 97] for the lattice of dimension 20, also shows the existence of a second unimodular lattice of minimum 8 in dimension 80.

The first step is a special case of [Que 84]. Let (L_{0}, b(x, y)) be an even, integral lattice of odd determinant and of dimension k; then the quotient (L_{0}/2L_{0}, \overline{b}) admits an hyperbolic decomposition L_{0}/2L_{0} = T_{1} ⊕T_{2} (i.e. T_{i}^{⊥} = T_{i} for the induced non degenerate form \overline{b}). For i = 1, 2, we denote by L_{i} the sublattice of L_{0} which is the preimage of T_{i}. We again denote by b, \overline{b} the forms over L_{0}^{n}, (L_{0}/2L_{0})^{n} deduced from the previous ones.

The Hamming weight of a word x ∈ 𝔽_{2}^{n} is the number wt(x) of its non zero coordinates. We denote by x.y the usual scalar product x.y = ∑_{i=1}^{n} x_{i}y_{i} in 𝔽_{2}^{n}. Let C be a binary linear code of length n (see [MWS 77]), and let

C = T_{1} ⊗_{𝔽_{2}} C ⊕ T_{2} ⊗_{𝔽_{2}} C^{⊥} ⊂ (L_{0}/2L_{0})^{n}.

For x ∈ L_{0} (respectively x ∈ L_{0}/2L_{0}), and u ∈ 𝔽_{2}^{n}, we denote by x ⊗ u ∈ L_{0}^{n} (respectively x ⊗ u ∈ (L_{0}/2L_{0})^{n}) the n-tuple which i-th coordinate is equal to x if i belongs to the support of u and is zero otherwise.

The n-tuple of 𝔽_{2}^{n} with all coordinates equal to 1 (resp. 0) is denoted by 1 (resp. 0).

Proposition 4.1 With the previous notations, we have

(i) C^{⊥} = C.

(ii) Let L_{C} be the preimage of C in L_{0}^{n}, endowed with the form x.y = \frac{1}{2}b(x, y). It is an integral lattice of determinant det(L_{C}) = det(L_{0})^{n}.

(iii) If moreover \frac{1}{\sqrt{2}}L_{i} is even, or if the codes C and C^{⊥} are even (i.e. the Hamming weight takes even values on them), then L_{C} is even.

\underline{Proof:}

(i) The 𝔽_{2}-vector space C is generated by elements of the form t ⊗ u, where either u belongs to C and t to T_{1}, or u belongs to C^{⊥} and t to T_{2}. Since \overline{b}(t ⊗ u, s ⊗ v) = (u.v)\overline{b}(t, s), this is zero in all cases. Hence C ⊂ C^{⊥}; equality holds from the equality of the dimensions. (ii) follows from (i) and [Que 84]. (iii) The lattice L_{C} is generated by (2L_{0})^{n} which is obviously even with respect to \frac{1}{2}b(x, y), and by the x ⊗ u, where u belongs to C (resp C^{⊥}) and x to L_{1} (resp to L_{2}). Since (x ⊗ u).(x ⊗ u) = wt(u)b(x, x)/2, the lattice L_{C} is even under the hypothesis of the proposition. □

In order to measure the norm of the lattice L_{C}, we define a weight w on L_{0}/2L_{0} by the following:

Definition 4.2 We set, for t_{1} ∈ T_{1} and t_{2} ∈ T_{2},

w(t_{1} + t_{2}) = \{ \begin{matrix}0 if t_{1} = t_{2} = 0 \\ 1 if \overline{b}(t_{1}, t_{2}) ≠ 0 \\ 2 if \overline{b}(t_{1}, t_{2}) = 0 and (t_{1}, t_{2}) ≠ (0, 0) \end{matrix}

We extend the function w to (L_{0}/2L_{0})^{n} by setting w(x) = ∑_{i=1}^{n} w(x_{i}). If C ⊂ (L_{′}/∈L_{′})^{\}, the weight of C is defined by

w(C) = min{w(t) | t ∈ C \ {0}}.

Lemma 4.3 If L_{0} is one of the lattices ℤ[α], 𝔸_{2}^{2}, 𝔼_{8}, we can choose a decomposition L_{0}/2L_{0} = T_{1} ⊕ T_{2} as above with L_{i} isometric to \sqrt{2}L_{0}; then

∀ t ∈ L_{0}/2L_{0}, w(t) = min{b(x, x)/2 | x ∈ t}.

\underline{Proof:} The fact that these three lattices are ℤ[α]-modules shows the existence of such a decomposition: one can take L_{1} = αL_{0}, L_{2} = \overline{α}L_{0} (see the Introduction). In all three cases it is well known that the classes of L_{0} modulo 2L_{0} are represented by the vectors of norm 2 and 4 (see [CoS 88], [Bac 97]) and that one class doesn’t contain elements of the two types. For x_{i} ∈ L_{i}, the equality b(x_{1} + x_{2}, x_{1} + x_{2}) = b(x_{1}, x_{1}) + b(x_{2}, x_{2}) + 2b(x_{1}, x_{2}) shows that

(⋆) 2w(t) ≡ b(x, x) (mod 4) ∀ t ∈ L_{0}/2L_{0}, ∀ x ∈ t

which concludes the proof. □

Now we apply the construction described in Proposition 4.1 to the three lattices L_{0} of Lemma 4.3, which show that, in those cases, the weight w is the right one to be considered in order to determine the minimum of the lattice L_{C}.

Example. If C is the length 3 code generated by 1, it is easy to see that the corresponding C has weight 4; from Lemma 4.3 and the fact that 2L_{0} has minimum 4, the lattices L_{C} have minimum 4. We recover B. Gross constructions of the Coxeter-Todd and Leech lattices described in the Introduction.

In ([Bac 97], Theorem 6.3), a code C × C^{⊥} over 𝔽_{2} × 𝔽_{2} of weight 8 and length 10 is defined; this is the case L_{0} = ℤ[α], where ℤ[α]/2ℤ[α] is identified with 𝔽_{2} × 𝔽_{2}. We prove here that the extension of this code to the two other quotients L_{0}/2L_{0} keeps the weight 8 (clearly C = L_{0}/2L_{0} ⊗_{𝔽_{2}×𝔽_{2}} (C × C^{⊥})). We start by the definition of the code C and some of its properties which will be of later use. We identify an element of 𝔽_{2}^{n} and the set of its non zero coordinates.

Lemma 4.4 Let C be the binary code of length 10 generated by the rows of the matrix

1 0 0 0 0 1 1 1 0 0
0 1 0 0 0 0 1 1 1 0
0 0 1 0 0 0 0 1 1 1
0 0 0 1 0 1 0 0 1 1
0 0 0 0 1 1 1 0 0 1

(i) It is a cocyclic code, equivalent to its dual; it contains 15 words of weight 4, 15 words of weight 6, and one word of weight 10.

(ii) C ∩ C^{⊥} = {0, 1}.

(iii) A word of weight 6 of C contains exactly three words of weight 4 of C^{⊥}, which are, up to permutation, in the following position:

1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 1 1 .

and this is a one to one correspondence between the words of weight 6 of C and the sets of three words of weight 4 of C^{⊥} in the previous position.

\underline{Proof:} Straightforward from the definition of C. □

Proposition 4.5 If C is the binary code defined in Lemma 4.4, in all three cases C has weight 8.

\underline{Proof:} We set (e_{1}, …, e_{k}) a basis of L_{0}/2L_{0} such that the k/2 first vectors generate T_{1}, while the k/2 last vectors generate T_{2} and satisfy \overline{b}(e_{i}, e_{k/2+j}) = δ_{i,j}. We write x = e_{1} ⊗ u_{1} + … + e_{k} ⊗ u_{k} a non zero element of C, with u_{1}, …, u_{k/2} ∈ C and u_{k/2+1}, …, u_{k} ∈ C^{⊥}. Up to a change of basis, we can assume that {u_{1}, …, u_{k/2}} are distinct. ¿From (⋆) and the fact that (L_{C}, \frac{1}{2}b(x, y)) is even (Proposition 4.1), the weight is even on C and we only need to prove that w(x) ≥ 7. Obviously w(x) ≥ card(∪u_{i}), so we can assume that card(∪u_{i}) ≤ 6. If at least one of the words u_{i} has weight 6, for example u_{1}, the others are contained in it. Since wt(C) = 4, u_{2}, …, u_{k/2} = 0. ¿From the definition of w, w(x) = 6 if and only if u_{k/2+1} = u_{1}, which is not possible since C ∩ C^{⊥} = {0, 1}.

The last case to consider is the case where all the non zero u_{i} have weight 4. Two distinct weight 4 words of C ∪ C^{⊥} intersect on zero, one, or two coordinates. Since we assume card(∪u_{i}) ≤ 6, we only need to consider the case where the intersections have cardinality 2. If at most one of the {u_{1}, …, u_{k/2}} and at most one of the {u_{k/2+1}, …, u_{k}} is non zero, then clearly at least four coordinates belong to T_{1} ∪ T_{2} \ {0} and have weight 2. If at least two of the {u_{1}, …, u_{k/2}} are weight 4 words, they are in the position of Lemma 4.4 and define a weight 6 word of C^{⊥}. Under our assumption, this means that u_{k/2+1}, …, u_{k} = 0, and the nonzero coordinates of x all have weight 2. □

We have the following estimate of the norm of the lattice L_{C}:

• It contains (2L_{0})^{n}, which has norm 4, attained on the vectors of the form (x, 0, …, 0) up to permutation, where x is a minimal vector of L_{0}. The next value of the norm on this lattice is 8.

• min(L_{C} \ (2L_{0})^{n}) = w(C).

The goal of the second step of the construction, in the case of the code C given in Lemma 4.4, is to get rid of the norm 4 vectors of L_{C}. Therefore, we define a sublattice L′ of L_{C} not containing them, and show the existence of an even overlattice Λ of L′, with the same index, and of minimum 8. This must be understood as a “neighbouring” procedure. The lattice L′ is easy to define:

L′ = {x = (x_{1}, …, x_{n}) ∈ L_{C} | ∑_{i=1}^{n} x_{i} ∈ 2L_{1}}.

We also set

L′′ = L_{C} + \{\frac{1}{2}(z, z, …, z) | z ∈ L_{1}\}.

Proposition 4.6 Assume that n ≡ 2 (mod 4), 1 ∈ C ∩ C^{⊥} and that w(C) = 8. Then

(i) min(L′) = 8.

(ii) The map ψ : x → ∑_{i=1}^{n} x_{i} induces an isometry of the 𝔽_{2}-quadratic spaces (L′′/L′, \frac{1}{2}b(x, x)) and (L_{1}/2L_{1}, \frac{1}{4}b(x, x)). Moreover ψ(L_{C}/L′) = 2L_{0}/2L_{1}.

\underline{Proof:}

(i) Clearly, the norm 4 vectors of L_{C}, previously defined, don’t belong to L′ because the minimum of L_{1} is 4. (ii) With the previous notations, a vector y of L_{C} can be written y = ∑ l_{1} ⊗ u + ∑ l_{2} ⊗ v + 2t where u ∈ C, v ∈ C^{⊥}, l_{i} ∈ L_{i} for i = 1, 2, and t ∈ L_{0}^{n}. Then ∑ y_{i} = ∑ wt(u)l_{1} + ∑ wt(v)l_{2} + 2 ∑ t_{i} ∈ 2L_{0}, and, if x = y + \frac{1}{2}(z, …, z), ψ(x) = ∑ x_{i} = ∑ y_{i} + \frac{n}{2}z ≡ ∑ y_{i} + z (mod 2L_{1}) belongs to 2L_{1}, if and only if x ∈ L′. Clearly, ψ(x) ∈ 2L_{0} if and only if x ∈ L_{C}, i.e. z ∈ 2L_{1}.

Since the lattice L_{C} is even for \frac{1}{2}b(x, y), one has for y ∈ L_{C}, b(y, y) ≡ 0 (mod 4). Hence b(x, x) = b(y + \frac{1}{2}(z, …, z), y + \frac{1}{2}(z, …, z)) ≡ b(∑ y_{i}, z) + 2(b(z, z)/4) (mod 4) is even and \frac{1}{2}b(ψ(x), ψ(x)) = \frac{1}{2}b(∑ y_{i} + z, ∑ y_{i} + z) ≡ b(∑ y_{i}, z) + \frac{1}{2}b(z, z) (mod 4) since ∑ y_{i} ∈ 2L_{0}, which shows that ψ is an isometry of the quadratic forms. □

The last step of the construction shows that certain overlattices of L′ keep the minimum 8.

Theorem 4.7 Let C be the binary code defined in Lemma 4.4, and let L_{C}, L′ be the lattices previously defined from it. Let B be a maximal totally singular subspace of the 𝔽_{2}-quadratic space L′′/L′, which is a complement of L_{C}/L′ and let Λ(B) be the sublattice of L′′ which is the preimage of B. Then the minimum of Λ(B) is 8.

\underline{Proof:} Let B be a fixed maximal totally singular subspace as in the Theorem and Λ := Λ(B). From the definition of Λ, Λ∩L_{C} = L′ which has minimum 8, and Λ ⊂ L_{C} +\{\frac{1}{2}(z, …, z) | z ∈ L_{1}\}. For y ∈ L_{0} and z ∈ L_{1} \2L_{0}, since L_{1} ∼ \sqrt{2}L_{0}, b(y +z/2, y +z/2) = b(y, y)+b(y, z)+b(z, z)/4 ∈ ℕ_{≥1}. Hence a vector X in L_{C} + \frac{1}{2}(z, …, z) satisfies X.X ≥ 5 (recall that X.X = b(X, X)/2 ∈ ℕ from Proposition 4.6(ii)). We shall prove that such a translate of L_{C} doesn’t contain any vector of norm 6, which is enough to prove that Λ has minimum 8. We fix z ∈ L_{1} \ 2L_{0}. For i = 1, 2, 3 we set

S_{i} = {y ∈ L_{0} | b(y + z/2, y + z/2) = i}.

Then a vector Y ∈ L_{C} such that X = Y + \frac{1}{2}(z, …, z) has norm 6 would have either 8 coordinates in S_{1} and 2 in S_{2}, or 9 coordinates in S_{1} and one in S_{3}. We need a precise description of these sets. The case L_{0} = ℤ[α] is treated in [Bac 97] (in this case, S_{3} = ∅), so we now concentrate on the two other cases.

Lemma 4.8 With the previous notations,

(i) If L_{0} = 𝔸_{2}^{2}, there are ϵ_{i} ∈ \frac{1}{2}L_{0}, 1 ≤ i ≤ 4, such that 2ϵ_{i} belongs to the class of z modulo 2L_{0}, and b(ϵ_{i}, ϵ_{i}) = 1 for i = 1, 2, b(ϵ_{i}, ϵ_{i}) = 3 for i = 3, 4, and b(ϵ_{i}, ϵ_{j}) = 0 for i ≠ j, such that

\begin{aligned}S_{1} & = \{−\frac{z}{2} ± ϵ_{i} | i = 1, 2\} \\ S_{2} & = \{−\frac{z}{2} + \frac{1}{2} ∑_{i=1}^{4} k_{i}ϵ_{i} | k_{i} = ±1, ∑_{i=1}^{4} k_{i} ≡ 0 (mod 4)\} \\ S_{3} & = \{−\frac{z}{2} ± ϵ_{i} | i = 3, 4\} \end{aligned}

(ii) If L_{0} = 𝔼_{8}, there are ϵ_{i} ∈ \frac{1}{2}L_{0}, 1 ≤ i ≤ 8, such that 2ϵ_{i} belongs to the class of z modulo 2L_{0}, and b(ϵ_{i}, ϵ_{j}) = δ_{i,j} for all i, j, such that

S_{1} = \{−\frac{z}{2} ± ϵ_{i} | 1 ≤ i ≤ 8\}
S_{2} = \{−\frac{z}{2} + \frac{1}{2} ∑_{i=1}^{8} k_{i}ϵ_{i} | k_{i} = ±1, ∑_{i=1}^{8} k_{i} ≡ 0 (mod 4)\}
S_{3} = \{−\frac{z}{2} ± ϵ_{i} ± ϵ_{j} ± ϵ_{l} | 1 ≤ i, j, l ≤ 8 pairwise distinct \}

\underline{Proof:} In the case L_{0} = 𝔼_{8}, z/2 is a deep hole and its class modulo 2L_{0} contains an orthogonal frame denoted by (2ϵ_{i})_{1≤i≤8} ([CoS 88], p.169). In the case L_{0} = 𝔸_{2}^{2}, it is easy to check that any class modulo 2L_{0} which doesn’t contain a minimal vector, contains such an orthogonal set.

In both cases, the vectors ϵ_{i} ± ϵ_{j} generate a sublattice of index 2 in L_{0} (isometric to 𝔻_{8} in the case L_{0} = 𝔼_{8}). So we may assume (up to a change of signs) that (ϵ_{1} + … + ϵ_{k})/2 ∈ L_{0}. We set y + x/2 = (∑_{i=1}^{8} k_{i}ϵ_{i})/2, where the k_{i} are integers with the same parity, and find the announced solutions. □

Now we need to study the decomposition of the elements of the sets S_{i} in L_{0}/2L_{0} = T_{1} ⊕ T_{2}. We set p_{i} the projection on T_{i}. Since T_{1} ⊕ T_{2} is a hyperbolic decomposition, for all x, y ∈ L_{0}/2L_{0}, \overline{b}(x, y) = \overline{b}(p_{1}(x), p_{2}(y)) + \overline{b}(p_{2}(x), p_{1}(y)). Moreover, we can choose z arbitrarily in its class modulo 2L_{0}, so we take z = −2ϵ_{1}. We fix a basis (e_{1}, …, e_{k/2}) of T_{1} such that e_{k/2} = 2ϵ_{i} (mod 2L_{0}) for all i, and choose for (e_{k/2+1}, …, e_{k}) its dual basis in T_{2}, defined by the conditions \overline{b}(e_{i}, e_{k/2+j}) = δ_{i,j}. If −z/2 ± ϵ_{i} and −z/2 ± ϵ_{j} are two distinct elements of S_{1}, p_{1}(−x/2 ± ϵ_{i}) − p_{1}(−x/2 ± ϵ_{j}) = p_{1}(±ϵ_{i} ± ϵ_{j}) ≠ 0, since, for i ≠ j, ±ϵ_{i} ± ϵ_{j} has norm 2, hence cannot belong to L_{1}, and, for i = j, p_{1}(2ϵ_{i}) = e_{k/2}. So, in both cases, the map p_{1} is a bijection from S_{1} to T_{1}. Since p_{2}(2ϵ_{i}) = 0, the image of S_{1} under p_{2} is half of T_{2}. More precisely, we have b(2ϵ_{1}, ϵ_{1} ± ϵ_{i}) ≡ 0 (mod 2), so \overline{b}(e_{k/2}, p_{2}(s)) = 0 for all s ∈ S_{1}, and p_{2}(S_{1}) = 𝔽_{2}e_{k/2+1} + … + 𝔽_{2}e_{k−1}.

Let now s belong to S_{2}. We have \overline{b}(e_{k/2}, p_{2}(s)) ≡ b(2ϵ_{1}, ϵ_{1}+∑_{i=1}^{k} k_{i}ϵ_{i})/2 ≡ 1 (mod 2), which proves that p_{2}(s) has a non zero coordinate on e_{k}. Then it is not possible for Y to have 8 coordinates in S_{1} and 2 in S_{2}, since the code C^{⊥} doesn’t contain any word of weight 2.

We now consider the case where Y has exactly one coordinate in S_{3}. If s ∈ S_{3}, \overline{b}(e_{k/2}, p_{2}(s)) = 0, and the coordinates of Y don’t have any component on e_{k}. In order to deal with this case, we need to know the exact decomposition of S_{1} on the basis {e_{1}, …, e_{k}}, i.e., since S_{1} is in a one-to-one correspondence with T_{1}, we need to know the map ϕ : 𝔽_{2}^{k/2} → 𝔽_{2}^{k/2−1} such that for all s ∈ S_{1}, and for all x ∈ 𝔽_{2}^{k/2}, p_{1}(s) = ∑_{i=1}^{k/2} x_{i}e_{i} ⟺ p_{2}(s) = ∑_{i=k/2+1}^{k−1}(ϕ(x))_{i−k/2}e_{i}. In general, this map is not linear. ¿From the previous remarks, we have ϕ(0) = 0 and ϕ(x + e_{k/2}) = ϕ(x); hence we only need to know ϕ|_{<e_{1},...,e_{k/2−1}>}, again denoted by ϕ. In the case L_{0} = 𝔸_{2}^{2}, k = 4, and ϕ is already determined by these conditions: the only possibility is ϕ(x_{1}e_{1}) = x_{1}e_{3}. In the case L_{0} = 𝔼_{8}, we need to take into account the mutual scalar products of the elements of S_{1}. Since, for 1 ≠ i ≠ j ≠ 1, \overline{b}(ϵ_{1} ± ϵ_{i}, ϵ_{1} ± ϵ_{j}) = 1, we have ∀x, x′ ∈ 𝔽_{2}^{k/2−1} \ {0} | x ≠ x′, x^{t}ϕ(x′) + x′^{t}ϕ(x) = 1. This set of affine conditions leaves eight solutions for ϕ, which are transitively permuted by changes of the base {e_{1}, …, e_{k/2−1}}. One of them is:

x \{ \begin{matrix}0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & e_{1} \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & e_{2} \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & e_{3} \end{matrix}
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ϕ(x) \{ \begin{matrix}0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & e_{5} \\ 0 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & e_{6} \\ 0 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & e_{7} \end{matrix}
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C_{i}  1 2 4 6 8 5 7 3

where each column contains x ∈ 𝔽_{2}^{k/2−1} and ϕ(x).

Lemma 4.9 For all t ∈ S_{3}, there is a unique s ∈ S_{1} such that t = s + l_{1}, where l_{1} ∈ L_{1} and p_{1}(l_{1}) ∉ 𝔽_{2}e_{k/2}.

\underline{Proof:} In both cases, since x ∈ L_{1}, the vectors 2ϵ_{i} belong to L_{1} for all i. Since b(x, y) ∈ 2ℤ for all x, y ∈ L_{1}, we have the inclusion L_{1} ⊂ ⊕_{i=1}^{k}ℤϵ_{i}. Then, from the isometries L_{1} ∼ \sqrt{2}L_{0}: If L_{0} = 𝔸_{2}^{2}, ∀ i = 3, 4, ∃! j(i) ∈ {1, 2} | ϵ_{i} ± ϵ_{j(i)} ∈ L_{1}. Clearly ϵ_{i} ± ϵ_{j(i)} ≢ 2ϵ_{i} (mod 2L_{0}). If L_{0} = 𝔼_{8}, ∀1 ≤ i ≠ j ≠ l ≤ 8, ∃! s = s(i, j, l) | ϵ_{i} ± ϵ_{j} ± ϵ_{l} ± ϵ_{s} ∈ L_{1}. (The unicity of s in this case comes from the fact that the words of the Hamming code form a 3-design ([MWS 77]).) Again, ϵ_{i} ±ϵ_{j} ±ϵ_{l} ±ϵ_{s} ≢ 2ϵ_{i} (mod 2L_{0}).

□

Now we conclude in the case L_{0} = 𝔸_{2}^{2}: if Y has nine coordinates in S_{1} and one in S_{3}, say Y_{i_{0}}, from Lemma 3.4, Y_{i_{0}} = s + l_{1}, with p_{1}(l_{1}) ∉ 𝔽_{2}e_{2}. Then Y = e_{1} ⊗ u_{1} + e_{2} ⊗ u_{2} + e_{3} ⊗ u_{3} (mod 2L_{0}) with u_{1}, u_{2} ∈ C and u_{3} ∈ C^{⊥}; the determination of ϕ shows that u_{1} = u_{3} up to the coordinate i_{0} which is different since p_{1}(l_{1}) ∉ 𝔽_{2}e_{2}. Hence they cannot be even at the same time.

Let L_{0} = 𝔼_{8} and Y = ∑_{i=1}^{7} e_{i} ⊗ u_{i}, with u_{1}, …, u_{4} ∈ C and u_{5}, … u_{7} ∈ C^{⊥}. We arrange the elements of 𝔽_{2}^{3} in the lexicographic order: 𝔽_{2}^{3} = {C_{1}, C_{2}, …, C_{8}}, C_{1} = (0, 0, 0), C_{2} = (0, 0, 1), C_{3} = (0, 1, 0),…,C_{8} = (1, 1, 1) and set, for 1 ≤ i ≤ 8, A_{i} = {j | (u_{5,j}, u_{6,j}, u_{7,j}) = C_{i}}. Moreover, we set a_{i} = card(A_{i}) (mod 2). Then, the determination of ϕ in this case shows that, up to the coordinate i_{0} where Y_{i_{0}} ∈ S_{3}, we have for i = 1, 2, 3, u_{i} = v_{i}, where:

\{ \begin{matrix}v_{1} & = A_{3} ∪ A_{5} ∪ A_{6} ∪ A_{8} \\ v_{2} & = A_{3} ∪ A_{4} ∪ A_{6} ∪ A_{7} \\ v_{3} & = A_{2} ∪ A_{3} ∪ A_{5} ∪ A_{7} \end{matrix}

More precisely, since C is even, u_{i} = v_{i} if and only if v_{i} is even. Case 1: u_{5}.u_{6} = 0. Then, from the parity of C^{⊥},

\{ \begin{matrix}a_{1} + a_{2} = 0 \\ a_{3} + a_{4} = 0 \\ a_{5} + a_{6} = 0 \\ a_{7} + a_{8} = 0 \\ a_{1} + a_{3} + a_{5} + a_{7} = 0 \end{matrix}

and wt(v_{3}) ≡ a_{2} + a_{3} + a_{5} + a_{7} = a_{1} + a_{2} = 0 (mod 2), which shows that the component of p_{1}(l_{1}) over e_{3} is zero and hence that v_{3} ∈ C. Then the scalar product of this word with u_{5}, u_{6}, u_{7} must be zero which leads to the additional conditions:

\{ \begin{matrix}a_{5} + a_{7} = 0 \\ a_{3} + a_{7} = 0 \\ a_{2} = 0 \end{matrix}

This implies that a_{i} = 0 for all 1 ≤ i ≤ 8, which means that p_{1}(l_{1}) ∈ 𝔽_{2}e_{4}, in contradiction with Lemma 3.4. Case 2: u_{5}.u_{6} = 1. Now we have the conditions

\{ \begin{matrix}a_{1} + a_{2} = 1 \\ a_{3} + a_{4} = 1 \\ a_{5} + a_{6} = 1 \\ a_{7} + a_{8} = 1 \\ a_{1} + a_{3} + a_{5} + a_{7} = 0 \end{matrix}

which imply

wt(v_{1}) ≡ a_{3} + a_{7} (mod 2)
wt(v_{2}) ≡ a_{5} + a_{7} (mod 2)
wt(v_{3}) ≡ a_{1} + a_{2} = 1 (mod 2)

If a_{3} +a_{7} = 0, then v_{1} ∈ C which is in contradiction with v_{1}.u_{6} = a_{3} +a_{8} = 1. If a_{5} +a_{7} = 0, then v_{2} ∈ C which is in contradiction with v_{2}.u_{5} = a_{6} +a_{7} = 1.

If a_{3} + a_{7} = a_{5} + a_{7} = 1, since v_{3}.u_{5} = a_{5} + a_{7} = 1 and v_{3}.u_{6} = a_{3} + a_{7} = 1, i_{0} ∈ A_{7} and v_{3}.u_{7} = a_{2} = 0. Then a_{2} = a_{3} = a_{5} = a_{8} = 0 and a_{1} = a_{4} = a_{6} = a_{7} = 1. But v_{1} + v_{2} ∈ C which is in contradiction with (v_{1} + v_{2}).u_{5} = a_{5} + a_{7} + a_{8} = 1. □

The group S := Stab_{Aut(L_{C})}(L′) acts on the set of maximal totally singular subspaces B of L′′/L′ which are complements of L_{C}/L′, as defined in Theorem 4.7. From Proposition 4.6(ii), L′′/L′ is a non-singular 𝔽_{2}-quadratic space of dimension k = dim(L_{0}) and Witt defect 0. Let k = 2m; there are 2^{m(m−1)/2} such subspaces B. If L_{0} is one of ℤ[α], 𝔸_{2}^{2}, or 𝔼_{8} we find using the notation of Theorem 4.7.

Proposition 4.10 (i) If L_{0} = ℤ[α], then B, hence Λ(B), is unique and Λ(B) is isometric to L_{20}.

(ii) If L_{0} = 𝔸_{2}^{2}, then the two subspaces B lie in one orbit under the action of S. Hence also in this case one gets a unique lattice Λ(B), which is isometric to L_{40}.

(iii) If L_{0} = 𝔼_{8}, the 64 subspaces B fall into 2 orbits under the action of S of length 8 and 56 with representatives say B and B′. The two lattices Λ(B) ≅ L_{80} and Λ(B′) are non isometric.

\underline{Proof:} The group S_{0} := Aut(L_{0})∩Aut(L_{1})∩Aut(L_{2}) acts on the set of maximal totally singular subspaces of L_{1}/2L_{1} which are complements of 2L_{0}/2L_{1}. They correspond bijectively to the set of subspaces B as in Theorem 4.7. by the isometry ψ : L′′/L′ → L_{1}/2L_{1} as defined in Proposition 4.6. (ii). The action of S_{0} is induced by the action of a subgroup of S on L′′/L′. One calculates S_{0} = {^{±}1}, D_{8} ^{|}∧ S_{3}, resp. 2.Alt_{8} and the decomposition of the set of ψ(B) into 1, 1, resp. 2 orbits under S_{0} of lengths 1, 2, resp. 8 and 56. The stabilizers of representatives of the last two orbits are the two maximal subgroups 2.Alt_{7} resp. (SL_{2}(5) × C_{3}).2 of 2.Alt_{8}. To show the proposition, we have to show that the lattices defined by the latter two subspaces B are non isometric and that the lattices L_{20}, L_{40}, and L_{80} as defined in the Introduction are of the form Λ(B).

The explicit description of the lattice L_{20} given in [CCNPW 85] shows that L_{20} is isometric to the lattice Λ of ([Bac 97], Theorem 6.4), which clearly satisfies the conditions of Theorem 4.7. (with the notations of [Bac 97], L′ = L_{C}^{e}). Let us denote by a superscript ^{(20)} the previous constructions when applied to the first case L_{0} = ℤ[α]. It was already noticed that, for the three cases L_{0} = ℤ[α], 𝔸_{2}^{2}, 𝔼_{8},

C = L_{0}/2L_{0} ⊗_{𝔽_{2}×𝔽_{2}} C^{(20)}.

Hence we also have

L_{C} ≃ L_{0} ⊗_{ℤ[α]} L_{C}^{(20)}

and clearly

L′ ≃ L_{0} ⊗_{ℤ[α]} L′^{(20)}.

Then, if L_{20} defines the maximal self-orthogonal subspace B^{(20)} of L′′^{(20)}/L′^{(20)}, the even lattice L_{0} ⊗_{ℤ[α]} L_{20} defines the maximal self-orthogonal subspace B = L_{0}/2L_{0} ⊗_{𝔽_{2}×𝔽_{2}} B^{(20)} ⊂ L_{0}/2L_{0} ⊗_{𝔽_{2}×𝔽_{2}} L′′^{(20)}/L′^{(20)} ≃ L′′/L′ (notice that this is the 2-torsion part of (L′)^{∗}/L′) which is a complement of L_{C}/L′. Therefore the lattices L_{20}, L_{40}, and L_{80} are isometric to Λ(B) in the respective cases.

It remains to consider the dimension 80. The lattice L_{80} corresponds to the orbit of length 8 with stabilizer 2.Alt_{7} = Aut_{ℤ[α]}(L_{0}). Since the stabilizer of a representative B′ of the orbit of length 56 has endomorphism ring ≅ ℤ[β := \frac{−1+\sqrt{-15}}{2}], it is natural to construct the lattice M_{80} := Λ(B′) replacing ℤ[α] by ℤ[β].

Let L_{C} be the ℤ[β]-lattice in ℚ[β]^{10} constructed as in Proposition 4.1. from L_{0} = ℤ[β], L_{0}/2L_{0} = ℘/(2)⊕\overline{℘}/(2), and the binary code of Lemma 4.4, where ℘ := (β, 2) is a prime ideal of ℤ[β] containing 2. Let

M_{20} := \{(x_{1}, …, x_{10}) ∈ L_{C} | ∑ x_{i} ∈ 2℘\} + \frac{℘^{2}}{2} (1, …, 1).

Lemma 4.11 (i) The lattice M_{20} is a ℤ[β]-unimodular lattice which as ℤlattice has minimum 8, 180 minimal vectors, and automorphism group Aut_{ℤ}(M_{20}) = Aut_{ℤ[β]}(M_{20}) = 2^{5} : S_{6} (a maximal subgroup of 2.M_{22}.2).

(ii) Let M_{80} := M_{20} ⊗_{ℤ[β]} 𝔼_{8}, where Aut_{ℤ[β]}(𝔼_{8}) = (SL_{2}(5)×C_{3}).2 as above. Then M_{80} is an even unimodular lattice not isometric to L_{80}.

\underline{Proof:} (i) May be calculated with the computer ([PlS 97]). Note that the minimal vectors of M_{20} are up to permutation of the coordinates the vectors (^{±}2,^{±}2, 0, …, 0) ∈ 2ℤ[β]^{10}. In particular, the automorphism group of M_{20} fixes L_{C}. This can be seen from the fact that the weight w to be defined on L_{0}/2L_{0} in order to satisfy the equality of Lemma 4.3. takes the values 0, 1, 4 (cf. Definition 4.2) since the ideal ℘ is not principal. Hence one can show that the minimum min(L_{C} \ 2ℤ[β]^{10}) is 10. (ii) M_{80} is clearly an even unimodular lattice. By construction its automorphism group contains the subgroup (SL_{2}(5) ⊗ C_{3}).2 ⊗ 2^{5}.S_{6}, which is not a subgroup of Aut(L_{80}) = 2.Alt_{7} ⊗_{ℤ[α]} 2.M_{22}.2 (cf. [CCNPW 85]). Therefore M_{80} is not isometric to L_{80}. □

As above one now concludes that M_{80} is of the form Λ(B′) and hence of minimum 8 by Theorem 4.7. Since M_{80} is not isometric to L_{80}, the subspace B′ ≤ L′′/L′ belongs to a second orbit (of length 56) under S. □

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