EE_{8}-lattices and dihedral groups

Robert L. Griess Jr. Department of Mathematics University of Michigan Ann Arbor, MI 48109 USA

Ching Hung Lam Department of Mathematics and National Center for Theoretical Sciences National Cheng Kung University Tainan, Taiwan 701

Abstract

We classify integral rootless lattices which are sums of pairs of EE_{8}-lattices (lattices isometric to \sqrt{2} times the E_{8}-lattice) and which define dihedral groups of orders less than or equal to 12. Most of these may be seen in the Leech lattice. Our classification may help understand Miyamoto involutions on lattice type vertex operator algebras and give a context for the dihedral groups which occur in the Glauberman-Norton moonshine theory.

Contents

1 Introduction  2
2 Background and notations  5
3 Tensor products  7
4 Uniqueness  9
5 DIH_{4} and DIH_{8} theories  10
5.1 DIH_{4}: When is M + N rootless? . . . . . . . . . . . . . . . . . . . . 10
5.2 DIH_{8} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2.1 DIH_{8}: What is F? . . . . . . . . . . . . . . . . . . . . . . . . 12
5.2.2 DIH_{8}: Given that F = 0, what is J? . . . . . . . . . . . . . . 14
5.2.3 DIH_{8}: Given that F ≅ AA_{1} ⊥ AA_{1}, what is J? . . . . . . . . 15
5.2.4 DIH_{8}: Given that F_{M} ≅ F_{N} ≅ DD_{4}, what is J? . . . . . . . 18

6 DIH_{6} and DIH_{12} theories  25
6.1 DIH_{6} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
6.1.1 DIH_{6}: Explicit glueing . . . . . . . . . . . . . . . . . . . . . 30
6.2 DIH_{12} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6.2.1 DIH_{12}: Study of J_{M} and J_{N} . . . . . . . . . . . . . . . . . . 32
6.2.2 DIH_{12}: the structure of F . . . . . . . . . . . . . . . . . . . 32
6.2.3 DIH_{12}: A comparison of eigenlattices . . . . . . . . . . . . . 33
6.3 s = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
6.4 s ∈ {2, 4} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
6.5 Uniqueness of the case DIH_{12}(16) . . . . . . . . . . . . . . . . . . . 35
7 DIH_{10} theory  35
7.1 DIH_{10}: Which ones are rootless? . . . . . . . . . . . . . . . . . . . . 36
7.2 DIH_{10}: An orthogonal direct sum . . . . . . . . . . . . . . . . . . . 37
7.3 DIH_{10}: From AA_{4}^{4} to L . . . . . . . . . . . . . . . . . . . . . . . . . 39
7.4 DIH_{10}: Explicit glueing and tensor products . . . . . . . . . . . . . 39
A General results about lattices  44
B Characterizations of lattices of small rank  49
C Nonexistence of particular lattices  53
D Properties of particular lattices  54
E Values of the Hermite function  66
F Embeddings of NREE8 pairs in the Leech lattice  66
F.1 Leech lattice and its isometry group . . . . . . . . . . . . . . . . . . 67
F.2 Standard EE_{8}s in the Leech lattice . . . . . . . . . . . . . . . . . . . 68
F.2.1 EE_{8} corresponding to octads in different frames . . . . . . . 68
F.2.2 EE_{8} associated to an even permutation in octad stabilizer . . 69
F.3 EE_{8} pairs in the Leech lattice . . . . . . . . . . . . . . . . . . . . . . 70
F.3.1 |g| = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
F.3.2 |g| = 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
F.3.3 |g| = 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
F.3.4 |g| = 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
F.3.5 |g| = 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

1 Introduction

By lattice, we mean a finite rank free abelian group with rational valued, positive definite symmetric bilinear form. A root in an integral lattice is a norm 2 vector. An integral lattice is rootless if it has no roots. The notation EE_{8} means \sqrt{2} times the famous E_{8} lattice.

In this article, we classify pairs of EE_{8}-lattices which span an integral and rootless lattice and whose associated involutions (isometries of order 2) generate a dihedral group of order at most 12. Examples of such pairs are easy to find within familiar lattices, such as the Barnes-Wall lattices of ranks 16 and 32 and the Leech lattice, which has rank 24.

Our main theorem is as follows. These results were announced in [GL].

Main Theorem 1.1. Let M, N ≅ EE_{8} be sublattices in a Euclidean space such that L = M + N is integral and rootless. Suppose that the involutions associated to M and N (2.4) generate a dihedral group of order less than or equal to 12. Then the possibilities for L are listed in Table 1 and all these possibilities exist. The lattices in Table 1 are uniquely determined (up to isometry of pairs M, N) by the notation in column 1 (see Table 3). Except for DIH_{4}(15), all of them embed as sublattices of the Leech lattice.

Table 1: NREE8SUMs: integral rootless lattices which are sums of EE_{8}s

Name  ⟨t_{M}, t_{N}⟩ Isometry type of L (contains) D(L) In Leech?
DIH_{4}(12)  Dih_{4} ≥ DD_{4}^{⊥3}  1^{4}2^{6}4^{2}  Yes
DIH_{4}(14)  Dih_{4} ≥ AA_{1}^{⊥2} ⊥ DD_{6}^{⊥2}  1^{4}2^{8}4^{2}  Yes
DIH_{4}(15)  Dih_{4} ≥ AA_{1} ⊥ EE_{7}^{⊥2}  1^{2}2^{14}  No
DIH_{4}(16)  Dih_{4}  ≅ EE_{8} ⊥ EE_{8}  2^{16}  Yes
DIH_{6}(14)  Dih_{6} ≥ AA_{2} ⊥ A_{2} ⊗ E_{6}  1^{7}3^{3}6^{2}  Yes
DIH_{6}(16)  Dih_{6}  ≅ A_{2} ⊗ E_{8}  1^{8}3^{8}  Yes
DIH_{8}(15)  Dih_{8} ≥ AA_{1}^{⊥7} ⊥ EE_{8}  1^{10}4^{5}  Yes
DIH_{8}(16, DD_{4}) Dih_{8} ≥ DD_{4}^{⊥2} ⊥ EE_{8}  1^{8}2^{4}4^{4}  Yes
DIH_{8}(16, 0)  Dih_{8}  ≅ BW_{16}  1^{8}2^{8}  Yes
DIH_{10}(16)  Dih_{10} ≥ A_{4} ⊗ A_{4}  1^{12}5^{4}  Yes
DIH_{12}(16)  Dih_{12} ≥ AA_{2} ⊥ AA_{2}
⊥ A_{2} ⊗ E_{6}  1^{12}6^{4}  Yes

X^{⊥n} denotes the orthogonal sum of n copies of the lattice X.

Table 2: Containments of NREE8SUM

Name  Sublattices
DIH_{8}(15)  DIH_{4}(12)
DIH_{8}(16, DD_{4})  DIH_{4}(12)
DIH_{8}(16, 0)  DIH_{4}(16)
DIH_{12}(16)  DIH_{4}(12), DIH_{6}(14)

Our methods are probably good enough to determine all the cases where M +N is integral, but such a work would be quite long.

This work may be considered purely as a study of positive definite integral lattices. Our real motivation, however, is the evolving theory of vertex operator algebras (VOA) and their automorphism groups, as we shall now explain.

The primary connection between the Monster and vertex operator algebras was established in [FLM]. Miyamoto showed [Mi1] that there is a bijection between the conjugacy class of 2A involutions in the Monster simple group and conformal vectors of central charge \frac{1}{2} in the moonshine vertex operator algebra V^{♮}. The bijection between the 2A-involutions and conformal vectors offers an opportunity to study, in a VOA context, the McKay observations linking the extended E_{8}-diagram and pairs of 2A-involutions [LYY]. This McKay theory was originally described in purely finite group theory terms.

Conformal vectors of central charge \frac{1}{2} define automorphisms of order 1 or 2 on the VOA, called Miyamoto involutions when they have order 2. They were originally defined in [Mi]; see also [Mi1]. Such conformal vectors are not found in most VOAs but are common in many VOAs of great interest, mainly lattice type VOAs and twisted versions [DMZ, DLMN]. Unfortunately, there are few general, explicit formulas for such conformal vectors in lattice type VOAs. We know of two. The first such formula (see [DMZ]) is based on a norm 4 vector in a lattice. The second such formula (see [DLMN], [GrO+]) is based on a sublattice which is isometric to EE_{8}. This latter formula indicates special interest in EE_{8} sublattices for the study of VOAs.

We call the dihedral group generated by a pair of Miyamoto involutions a Miyamoto dihedral group. Our assumed upper bound of 12 on the order of a Miyamoto dihedral group is motivated by the fact that in the Monster, a pair of 2A involutions generates a dihedral group of order at most 12 [GMS]. Recently, Sakuma [Sa] announced that 12 is an upper bound for the order of a Miyamoto dihedral group in an OZVOA (= CFT type with zero degree 1 part)[GNAVOA1] with a positive definite invariant form. This broad class of VOAs contains all lattice type VOAs V_{L}^{+} such that the even lattice L is rootless, and the Moonshine VOA V^{♮}. If a VOA has nontrivial degree 1 part, the order of a Miyamoto dihedral group may not be bounded in general (for instance, a Miyamoto involution can invert a nontrivial torus under conjugation). See [GNAVOA1].

If L is rootless, it is conjectured [LSY] that the above two kinds of conformal vectors will exhaust all the conformal vectors of central charge 1/2 in V_{L}^{+}. This conjecture was proved when L is a \sqrt{2} times a root lattice or the Leech lattice [LSY, LS] but it is still open if L is a general rootless lattice. The results of this paper could help settle this conjecture, as well as provide techniques for more work on the Glauberman-Norton theory [GlNo].

The first author acknowledges financial support from the National Science Foundation, National Cheng Kung University where the first author was a visiting distinguished professor, and the University of Michigan. The second author acknowledges financial support from the National Science Council of Taiwan (Grant No. 95-2115-M-006-013-MY2).

2 Background and notations

Convention 2.1. Lattices in this article shall be rational and positive definite. Groups and linear transformations will generally act on the right and n-tuples will be row vectors.

Definition 2.2. Let X be an integral lattice. For any positive integer n, let X_{n} = {x ∈ L| (x, x) = n} be the set of all norm n elements in X.

Definition 2.3. If L is a lattice, the summand of L determined by the subset S of L is the intersection of L with the ℚ-span of S.

Notation 2.4. Let X be a subset of Euclidean space. Define t_{X} to be the orthogonal transformation which is −1 on X and is 1 on X^{⊥}.

Definition 2.5. A sublattice M of an integral lattice L is RSSD (relatively semiselfdual) if and only if 2L ≤ M + ann_{L}(M). This implies that t_{M} maps L to L and is equivalent to this property when M is a direct summand.

The property that 2M^{∗} ≤ M is called SSD (semiselfdual). It implies the RSSD property, but the RSSD property is often more useful. For example, if M is RSSD in L and M ≤ J ≤ L, then M is RSSD in J, whence the involution t_{M} leaves J invariant.

Example 2.6. An example of a SSD sublattice is \sqrt{2}U, where U is a unimodular lattice. Another is the family of Barnes-Wall lattices.

Lemma 2.7. If the sublattice M is a direct summand of the integral lattice L and (det(L), det(M)) = 1, then SSD and RSSD are equivalent properties for M.

Proof. It suffices to assume that M is RSSD in M and prove that it is SSD.

Let V be the ambient real vector space for L and define A := ann_{V} (M). Since (det(L), det(M)) = 1, the natural image of L in D(M) is D(M), i.e., L+A = M^{∗}+A (A.13). We have 2(L + A) = 2(M^{∗} + A), or 2L + A = 2M^{∗} + A = M ⊥ A. The left side is contained in M + A, by the RSSD property. So, 2M^{∗} ≤ M + A. If we intersect both sides with ann_{V} (A), we get 2M^{∗} ≤ M. This is the SSD property. □

Lemma 2.8. Suppose that L is an integral lattice and N ≤ M ≤ L and both M and N are RSSD in L. Assume that M is a direct summand of L. Then ann_{M}(N) is an RSSD sublattice of L.

Proof. This is easy to see on the level of involutions. Let t, u be the involutions associated to M, N. They are in O(L) and they commute since u is the identity on ann_{L}(M), where t acts as the scalar 1, and since u leaves invariant M = ann_{L}(ann_{L}(M)), where t acts as the scalar −1. Therefore, s := tu is an involution. Its negated sublattice L^{−}(s) is RSSD (2.4), and this is ann_{M}(N). □

Definition 2.9. An IEE8 pair is a pair of sublattices M, N ≅ EE_{8} in a Euclidean space such that M +N is an integral lattice. If M +N has no roots, then the pair is called an NREE8 pair. An IEESUM is the sum of an IEE8 pair and an NREE8SUM is the sum of an NREE8 pair.

Table 3: Notation and Terminology

Notation  Explanation  Examples in text
Φ_{A_{1}}, ⋯, Φ_{E_{8}}  root system of the indicated type  Table 1, 2
A_{1}, ⋯, E_{8}  root lattice for root system Φ_{A_{1}}, …, Φ_{E_{8}}  Table 1, 2
AA_{1}, ⋯, EE_{8}  lattice isometric to \sqrt{2} times the lattice  Table 1
A_{1}, ⋯, E_{8}
ann_{L}(S) {v ∈ L| (v, s) = 0 for all s ∈ S} (2.4), (6.21)
BRW^{+}(2^{d})  the Bolt-Room-Wall group, a subgroup of

O(2^{d}, ℚ)of shape 2_{+}^{1+2d}Ω^{+}(2d, 2)

BW_{2^{n}}  the Barnes-Wall lattice of rank 2^{n}  Table 1, 6, (5.2.2)
DIH_{n}(r)  an NREE8SUM M, N such that the SSD  Table 1, Sec. F.3

involutions t_{M}, t_{N} generate a dihedral group of order n and M + N is of rank r

DIH_{8}(16, X) an NREE8SUM DIH_{8}(16) such that Table 1, Sec. F.3

X ≅ ann_{M}(N) ≅ ann_{N}(M)

DIH_{n}-theory  the theories for DIH_{n}(r) for all r  Sec. 5.2, 6.2
D(L)  discriminant group of integral lattice L: L^{∗}/L  (A.3), (D.17)
HS_{n} or D_{n}^{+}  the half spin lattice of rank n, i.e.,  (6.22)

the lattice generated by D_{n} and \frac{1}{2}(11 ⋯ 1)

HHS_{n}^{+} or DD_{n}^{+}  \sqrt{2} times the half spin lattice HS_{n}^{+}  (F.4)
IEE8 pair  a pair of EE_{8} lattices whose sum is integral  (2.9), Sec. F.3
IEE8SUM  the sum of an IEE8 pair
NREE8 pair  an IEE8 pair whose sum has no roots  (2.9), Sec. F.3
NREE8SUM  the sum of an NREE8 pair  Table 1
L^{∗}  the dual of the rational lattice L, i.e., those  (2.4), (A.3)

elements u of ℚ ⊗ L which satisfy (u, L) ≤ ℤ

Λ the Leech lattice Sec. F.3
L^{+}(t), L^{−}(t)  the eigenlattices for the action of t
on the lattice L: L^{ε}(t) := {x ∈ L|xt = εx}  (6.22)
T el(L, t) total eigenlattice for action of t (6.30)
on L; L^{+}(t) ⊥ L^{−}(t)
T el(L, D) total eigenlattice for action of an elementary (A.2)

abelian 2-group D on L; T el(L, D) := ∑ L^{λ},

where λ ∈ Hom(D, {±1}) and

L^{λ} = {α ∈ L| αg = λ(g)α for all γ ∈ D}

m^{n} the homocyclic group ℤ_{m}^{n} = ℤ_{m} × ⋯ × ℤ_{m}, (6.9), (D.12)
n times
|g|, |G|  order of a group element, order of a group  (D.6), Sec. F.3
O(X) or Aut X  the isometry group of the lattice X  (5.3)

Table 4: Notation and Terminology (continued)

Notation  Explanation  Examples in text
O_{p}(G)  the maximal normal p-subgroup of G  (A.11), (A.13)
O_{p′}(G)  the maximal normal p′-subgroup of G,  (A.14)
p-rank  the rank of the maximal elementary  (A.3)

abelian p-subgroup of an abelian group

root  a vector of norm 2  Sec. 1, (3.5)
rectangular  a lattice with an orthogonal basis  (B.2)

lattice

square lattice  a lattice isometric to some \sqrt{m} ℤ^{n}  (B.2)
W eyl(E_{8}), W eyl(F_{4})  the Weyl group of type E_{8}, F_{4}, etc  (D.2)
X_{n}  the set of elements of norm n  (7.10), (7.11)

in the lattice X

X^{⊥n} the orthogonal sum of n copies Table 1, 5, 8

of the lattice X

ξ an isometry of Leech lattice (F.5), Sec. F.3

(see Notation F.5)

ℤ^{n}  rank n lattice with an orthonormal basis  (B.3)

Lemma 2.10. We use Definition 2.4. Let M and N be RSSD in an integral lattice L = M + N. A vector in L fixed by both t_{M} and t_{N} is 0.

Proof. We use L = M + N. If we tensor L with ℚ, we have complete reducibility for the action of ⟨t_{M}, t_{N}⟩. Let U be the fixed point space for ⟨t_{M}, t_{N}⟩ on ℚ ⊗ L. The images of M and N in U are 0, whence U = 0. □

3 Tensor products

Definition 3.1. Let A and B be integral lattices with the inner products ( , )_{A} and ( , )_{B}, respectively. The tensor product of the lattices A and B is defined to be the integral lattice which is isomorphic to A ⊗_{ℤ} B as a ℤ-module and has the inner product given by

(α ⊗ β, α′ ⊗ β′) = (α, α′)_{A} · (β, β′)_{B}, for any α, α′ ∈ A, and β, β′ ∈ B.

We simply denote the tensor product of the lattices A and B by A ⊗ B.

Lemma 3.2. Let D := ⟨t, g⟩ be a dihedral group of order 6, generated by an involution t and element g of order 3. Let R be a rational lattice on which D acts such that g acts fixed point freely. Suppose that A is a sublattice of R which satisfies at = −a for all a ∈ A. Then

(i) A ∩ Ag = 0; so A + Ag = A ⊕ Ag as an abelian group.

(ii) A + Ag is isometric to A ⊗ B, where B has Gram matrix (\begin{matrix}1 & −\frac{1}{2} \\ −\frac{1}{2} & 1 \end{matrix}).

(iii) Furthermore ann_{A+Ag}(A) = A(g − g^{2}) ≅ \sqrt{3}A.

Proof. (i) Take a ∈ A and suppose a = a′g. Since at = −a, we have −a = at = a′gt = a′tg^{2} = −a′g^{2}. That means a = a′g^{2} and a = ag. Thus a = 0 since g acts fixed point freely on R.

For any x, y ∈ R, we have 0 = (x, 0) = (x, y+yg+yg^{2}) = (x, y)+(x, yg)+(x, yg^{2}). Now, take x, y ∈ A. We have (x, yg) = (xt, ygt) = (−x, ytg^{2}) = (−x, −yg^{2}) = (x, yg^{2}). We conclude that (x, yg) = (x, yg^{2}) = −\frac{1}{2}(x, y).

Let bars denote images under the quotient ℤ⟨g⟩ → ℤ⟨g⟩/(1 + g + g^{2})

We use the linear monomorphism A ⊗ \overline{g^{i}} → R where ℤ⟨g⟩/(1 + g + g^{2}) has the bilinear form which take value 1 on a pair \overline{g^{i}}, \overline{g^{i}} and value −\frac{1}{2} on a pair \overline{g^{i}}, \overline{g^{j}} where j = i ± 1. This proves (ii)

For (iii), note that ψ : x ↦ xg − xg^{2} for x ∈ A is a scaled isometry and Im ψ is a direct summand of Ag ⊕ Ag^{2}. Note also that A ⊕ Ag = Ag ⊕ Ag^{2} = Im ψ ⊕ Ag. Thus we have

Im ψ ≤ ann_{A+Ag}(A) ≤ Im ψ ⊕ Ag.

By Dedekind law, ann_{A+Ag}(A) = Im ψ + (ann_{A+Ag}(A) ∩ Ag). Since (x, yg) = −\frac{1}{2}(x, y), ann_{A+Ag}(A) ∩ Ag = 0 and we have Im ψ = ann_{A+Ag}(A) as desired. □

Lemma 3.3. Suppose that A, B are lattices, where A ≅ A_{2}. The minimal vectors of A ⊗ B are just u ⊗ z, where u is a minimal vector of A and z is a minimal vector of B.

Proof. Let u be a minimal vector of A. The minimal vectors of ℤu ⊗ B have the above shape. Let u′ span ann_{A}(u). Then (u′, u′) = 6 and |A : ℤu + ℤu′| = 2. The minimal vectors of (ℤu ⊥ ℤu′) ⊗ B have the above shape. Now take a vector w in A ⊗ B \ (ℤu ⊥ ℤu′) ⊗ B. It has the form pu ⊗ x + qu′ ⊗ y, where p, q ∈ \frac{1}{2} + ℤ and p + q ∈ ℤ. The norm of this vector is therefore 2p^{2}(x, x) + 6q^{2}(y, y). A necessary condition that w be a minimal vector in A ⊗ B is that each of x, y be minimal in B and p, q ∈ \{±\frac{1}{2}\}.

Define v := \frac{1}{2}u + \frac{1}{2}u′. Then u′, v forms a basis for A. We have w = v ⊗ x + \frac{1}{2}u′ ⊗ (y − x). Since w ∈ A ⊗ B, y − x ∈ 2B. Suppose y − x = 2b. If b = 0, done, so assume that b ≠ 0. In case x, y are minimal, (y − x, y − x) = 4(b, b) ≥ 4(x, x) and thus −2(x, y) ≥ 2(x, x). This implies x = −y and then w = (v − u′) ⊗ x as required. □

Notation 3.4. For a lattice L, let MinV ec(L) be the set of minimal vectors.

Lemma 3.5. We use the notations of (3.3). If B is a root lattice of an indecomposable root system and rank(B) ≥ 3, the only sublattices of A⊗B which are isometric to \sqrt{2}B are the u ⊗ B, for u a minimal vector of A.

Proof. Let S be a sublattice of A ⊗ B so that S ≅ \sqrt{2}B. Then S is spanned by MinV ec(S), which by (3.3) equals M_{u} ∪ M_{v} ∪ M_{w}, where u, v, w are pairwise nonproportional minimal vectors of A which sum to 0 and where M_{t} := (t ⊗ B) ∩ MinV ec(S), for t = u, v, w. Note that (u, v) = (v, w) = (w, u) = −1.

We suppose that M_{u} and M_{v} are nonempty and seek a contradiction. Take b, b′ ∈ B so that u⊗b ∈ M_{u}, v⊗b′ ∈ M_{v}. Then (u⊗b, v⊗b′) = (u, v)(b, b′) = −(b, b′). Since S is doubly even, all such (b, b′) are even.

We claim that all such (b, b′) are 0.

Assume that some such (b, b′) ≠ 0. Then, since b, b′ are roots, (b, b′) is ±2 and b = ±b′. Then u ⊗ b, v ⊗ b ∈ S, whence w ⊗ b ∈ S. In other words, A ⊗ b ≤ S.

Since rank(S) = rank(B) ≥ 3, S properly contains A ⊗ b. Since S is generated by its minimal vectors and the root system for B is connected, S contains some t ⊗ d where d ∈ MinV ec(B) and (d, b) ≠ 0. It follows that (d, b) = ±1. Take t′ ∈ MinV ec(A) so that (t, t′) = ±1. Then (t ⊗ d, t′ ⊗ b) = ±1, whereas S is doubly even, a contradiction. The claim follows.

The claim implies that M_{u} and M_{v} are orthogonal. Similarly, M_{u}, M_{v}, M_{w} are pairwise orthogonal, and at least two of these are nonempty. Since MinV ec(S) is the disjoint union of M_{u}, M_{v}, M_{w}, we have a contradiction to indecomposability of the root system for B. □

4 Uniqueness

Theorem 4.1. Suppose that L is a free abelian group and that L_{1} is a subgroup of finite index. Suppose that f : L_{1} × L_{1} → K is a K-valued bilinear form, where K is an abelian group so that multiplication by |L : L_{1}| is an invertible map on K. Then f extends uniquely to a K-valued bilinear form L × L → K.

Proof. Our statements about bilinear forms are equivalent to statements about linear maps on tensor products. We define A := L_{1} ⊗L_{1}, B := L⊗L and C := B/A. Then C is finite and is annihilated by |L : L_{1}|^{2}. From 0 → A → B → C → 0, we get the long exact sequence 0 → Hom(C, K) → Hom(B, K) → Hom(A, K) → Ext^{1}(C, K) → · · ·. Each of the terms Hom(C, K) and Ext^{1}(C, K) are 0 because they are annihilated by |C| and multiplication by |C| on K is an automorphism. It follows that the restriction map from B to A gives an isomorphism Hom(B, K) ≅ Hom(A, K). □

Remark 4.2. We shall apply (4.1) to L = M + N when we determine sufficient information about a pairing M_{1} × N_{1} → ℚ, where M_{1} is a finite index sublattice of M and N_{1} is a finite index sublattice of N. The pairings M × M → ℚ and N × N → ℚ are given by the hypotheses M ≅ N ≅ EE_{8}, so in the notation of (4.1) we take L_{1} = M_{1} + N_{1}.

Remark 4.3. We can determine all the lattices in the main theorem by explicit glueing. However, it is difficult to prove the rootless property in some of those cases. In Appendix F, we shall show that all the lattices in Table 1 can be embedded into the Leech lattice except DIH_{4}(15). The rootless property follows since the Leech lattice has no roots. The proof that DIH_{4}(15) is rootless will be included at the end of Subsection 5.1.

5 DIH_{4} and DIH_{8} theories

5.1 DIH_{4}: When is M + N rootless?

Notation 5.1. Let M, N be EE_{8} lattices such that the dihedral group D := ⟨t_{M}, t_{N}⟩ has order 4. Define F := M ∩ N, P := ann_{M}(F) and Q := ann_{N}(F).

Remark 5.2. Since t_{M} and t_{M} commute, D fixes each of F, M, N, ann_{M}(F), ann_{N}(F). Each of these may be interpreted as eigenlattices since t_{M} and t_{N} have common negated space F, zero common fixed space, and t_{M}, t_{N} are respectively −1, 1 on ann_{M}(F) and t_{M}, t_{N} are respectively 1, −1 on ann_{N}(F). Since L = M +N, D has only 0 as the fixed point sublattice (cf. (2.10)). Therefore, the elementary abelian group D has total eigenlattice F ⊥ ann_{M}(F) ⊥ ann_{N}(F). Each of these summands is RSSD as a sublattice of L, by (2.8). It follows that \frac{1}{\sqrt{2}} (M ∩ N) is an RSSD sublattice in \frac{1}{\sqrt{2}}M and in \frac{1}{\sqrt{2}}N. Since \frac{1}{\sqrt{2}}M ≅ \frac{1}{\sqrt{2}}N ≅ E_{8}, we have that \frac{1}{\sqrt{2}} (M ∩ N) is an SSD sublattice in \frac{1}{\sqrt{2}}M and in \frac{1}{\sqrt{2}}N (cf. (2.7)).

Proposition 5.3. If M + N is rootless, F is one of 0, AA_{1}, AA_{1} ⊥ AA_{1}, DD_{4}. Such sublattices in M are unique up to the action of O(M).

Proof. This can be decided by looking at cosets of P +Q+F in M+N. A glue vector will have nontrivial projection to two or three of span_{ℝ}(P), span_{ℝ}(Q), span_{ℝ}(F).

Since F is a direct summand of M by (A.10) and \frac{1}{\sqrt{2}}F is an SSD by (5.2), we have \frac{1}{\sqrt{2}}F ≅ 0, A_{1}, A_{1} ⊥ A_{1}, A_{1} ⊥ A_{1} ⊥ A_{1}, A_{1} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, D_{4}, D_{4} ⊥ A_{1}, D_{6}, E_{7} and E_{8} by (D.2). If \frac{1}{\sqrt{2}}F = E_{8}, then t_{M} = t_{N} and D := ⟨t_{M}, t_{N}⟩ is only a cyclic group of order 2. Hence, we can eliminate \frac{1}{\sqrt{2}}F = E_{8}.

Now we shall note that in each of these cases, F ⊥ P contains a sublattice A ≅ AA_{1}^{8} such that F ∩ A ≅ AA_{1}^{k} and A ∩ P ≅ AA_{1}^{8−k}, where k = rank F. We use an orthogonal basis of A to identify M/A with a code. Since M ≅ EE_{8}, this code is the Hamming [8, 4, 4] binary code H_{8}. Let φ : M/A → H_{8} be such an identification. Then φ((F ⊥ P)/A) is a linear subcode of H_{8}.

Next, we shall show that (\frac{1}{\sqrt{2}}F )^{∗} contains a vector v of norm 3/2 if \frac{1}{\sqrt{2}}F ≅ A_{1} ⊥ A_{1} ⊥ A_{1}, A_{1} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, D_{4} ⊥ A_{1}, D_{6} or E_{7}.

Recall that if A_{1} = ℤα, (α, α) = 2, then A_{1}^{∗} = \frac{1}{2}ℤα and \frac{1}{2}α ∈ A_{1}^{∗} has norm \frac{1}{2}. Since (A_{1}^{⊥k})^{∗} = (A_{1}^{∗})^{⊥k}, (A_{1}^{⊥k})^{∗} contains a vector of norm 3/2 if k ≥ 3.

We use the standard model

D_{n} = {(x_{1}, x_{2}, …, x_{n}) ∈ ℤ^{n}| x_{1} + ⋯ + x_{n} ≡ 0 mod 2}.

Then \frac{1}{2}(1, …, 1) ∈ D_{n}^{∗} and its norm is \frac{1}{4}n. Therefore, there exists vectors of norm 3/2 in (D_{4} ⊥ A_{1})^{∗} = D_{4}^{∗} ⊥ A_{1}^{∗} and D_{6}^{∗}. Finally, we recall that E_{7}^{∗}/E_{7} ≅ ℤ_{2} and the non-trivial coset is represented by a vector of norm 3/2.

Now suppose \frac{1}{\sqrt{2}}F ≅ A_{1} ⊥ A_{1} ⊥ A_{1}, A_{1} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, D_{4} ⊥ A_{1}, D_{6} or E_{7} and let γ ∈ 2F^{∗} be a vector of norm 3. Since F ∩ A ≅ AA_{1}^{k}, F ∩ A has a basis {α_{1}, …, α_{k}} such that (α_{i}, α_{j}) = 4δ_{i,j}. Then

2F^{∗} < 2(F ∩ A)^{∗} = span_{ℤ}\{\frac{1}{2} ∑_{i=1}^{k} a_{i}α_{i}| a_{i} ∈ ℤ\}.

Thus, by replacing some basis vectors by their negatives, we have γ = \frac{1}{2}(α_{i_{1}} + α_{i_{2}} + α_{i_{3}}) for some 1 ≤ i_{1}, i_{2}, i_{3} ≤ k.

Since the image of the natural map M → D(F) is 2D(F), there exists a vector u ∈ M such that the projection of u to span_{ℝ}(F) is γ.

Now consider the image of u+A in H_{8} and study the projection of the codeword φ(u + A) to the first k coordinates. Since γ = \frac{1}{2}(α_{i_{1}} + α_{i_{2}} + α_{i_{3}}), the projection of φ(u + A) to the first k coordinates has weight 3.

If k = rank F ≥ 4, then the projection of (1, ⋯, 1) to the first k coordinates has weight k ≥ 4. Thus, φ(u + A) ≠ (1, …, 1) and hence φ(u + A) has weight 4 since φ(u + A) ∈ H_{8} .

If k = 3, then F ≅ AA_{1}^{3} and P ≅ AA_{1} ⊥ DD_{4}. Let K ≅ DD_{4} be an orthogonal direct summand of P and let ℤα = ann_{P} (K) ≅ AA_{1}. Note that F ⊥ ℤα < ann_{M}(K) ≅ DD_{4} and F = ℤα_{1} ⊥ ℤα_{2} ⊥ ℤα_{3}. Thus, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α) = \frac{1}{2}(γ + α) ∈ ann_{M}(K) < M and it has norm 4. Therefore, we may assume φ(u + A) has weight 4 and u is a norm 4 vector.

Similarly, there exists a norm 4 vector w ∈ N such that the projection of w in span_{ℝ}(F) is also γ. Then u − w ∈ L = M + N but (u, w) = (γ, γ) = 3 and hence u − w ∈ L is a root, which contradicts the rootless property of L. Therefore, only the remaining cases occur, i.e., F ≅ 0, AA_{1}, AA_{1} ⊥ AA_{1}, DD_{4}. □

Table 5: DIH_{4}: Rootless cases

M ∩ N  P ≅ Q dim(M + N)  Isometry type of L
0  EE_{8}  16  ≅ EE_{8} ⊥ EE_{8}
DD_{4}  DD_{4}  12  ≥ DD_{4} ⊥ DD_{4} ⊥ DD_{4}
AA_{1}  EE_{7}  15  ≥ AA_{1} ⊥ EE_{7} ⊥ EE_{7}
AA_{1} ⊥ AA_{1} DD_{6}  14  ≥ AA_{1} ⊥ AA_{1} ⊥ DD_{6} ⊥ DD_{6}

Remark 5.4. Except for the case F = M ∩ N ≅ AA_{1}, we shall show in Appendix F that all cases in Proposition 5.3 occur inside the Leech lattice Λ . The rootless property of L = M + N then follows from the rootless property of Λ. The rootless property for the case F = M ∩ N ≅ AA_{1} will be shown in the next proposition.

Proposition 5.5. [Rootless property for DIH_{4}(15)] If F = M ∩ N ≅ AA_{1}, then P ≅ Q ≅ EE_{7} and L = M + N is rootless.

Proof. We shall use the standard model for the lattice E_{7}, i.e.,

E_{7} = \{(x_{1}, …, x_{8}) ∈ ℤ^{8} | all x_{i} ∈ ℤ or all x_{i} ∈ \frac{1}{2} + ℤ and x_{1} + ⋯ + x_{8} = 0 \}

The dual lattice is E_{7}^{∗} = E_{7} ∪ (γ + E_{7}), where γ = \frac{1}{4}(1, 1, 1, 1, 1, 1, −3, −3). Recall that the minimal weight of E_{7}^{∗} is 3/2 [CS, p.125].

If F = M ∩ N = AA_{1} = ℤα, then it is clear that P ≅ Q ≅ EE_{7}. In this case, M = span_{ℤ}\{F + P, \frac{1}{2}α + ξ_{M}\} and N = span_{ℤ}\{F + Q, \frac{1}{2}α + ξ_{N}\} for some ξ_{M} ∈ P^{∗} and ξ_{N} ∈ Q^{∗} with (ξ_{M}, ξ_{M}) = (ξ_{N}, ξ_{N}) = 3. Therefore,

L = M + N = span_{ℤ}\{F + P + Q, \frac{1}{2}α + ξ_{M}, \frac{1}{2}α + ξ_{N}\}.

Take β ∈ L = M + N. If β ∈ F + P + Q, then (β, β) ≥ 4. Otherwise, β will have nontrivial projection to two or three of span_{ℝ}(P), span_{ℝ}(Q), span_{ℝ}(F). Now note that the projection of L onto span_{ℝ}(P) is span_{ℤ}\{P, ξ_{M}\} ≅ \sqrt{2}E_{7}^{∗} and the projection of L onto span_{ℝ}(Q) is span_{ℤ}\{Q, ξ_{N}\} ≅ \sqrt{2}E_{7}^{∗}. Both of them have minimal norm 3. On the other hand, the projection of L onto span_{ℝ}(F) is ℤ\frac{1}{2}α, which has minimal norm 1. Therefore, (β, β) ≥ 1 + 3 = 4 and so L is rootless. □

5.2 DIH_{8}

Notation 5.6. Let t := t_{N}, u := t_{N}, and g := tu, which has order 4. Define z := g^{2}, t′ := tz and u′ := uz. We define F := Ker_{L}(z − 1) and J := Ker_{L}(z + 1).

By Lemma A.6, L/(F ⊥ J) is an elementary abelian 2-group of rank at most min{rank(F), rank(J)}. We have two systems (M, t, Mg, t′) and (N, u, Ng, u′) for which the DIH_{4} analysis applies.

Notation 5.7. If X is one of M, N, we denote by L_{X}, J_{X}, F_{X} the lattices L := X + Xg, J, F associated to the pair X, Xg, denoted “M” and “N” in the DIH_{4} section.

5.2.1 DIH_{8}: What is F?

We now determine F.

Remark 5.8. It will turn out that the two systems (M, t, Mg, t′) and (N, u, Ng, u′) have the same DIH_{4} types (cf. Table 5). Also, we shall prove that rank(F_{X}) determines F_{X}, hence also determines J_{X}, for X = M, N.

Lemma 5.9. Let f = g or g^{−1}. Then (i) As endomorphisms of J, f^{2} = −1, (f − 1)^{2} = −2f. For x, y ∈ J, (x(f − 1), y(f − 1)) = 2(x, y).

(ii) (M ∩ J, (M ∩ J)f) = 0 and (N ∩ J, (N ∩ J)f) = 0.

(iii) For x, y ∈ M or x, y ∈ N, (x, y(f − 1)) = −(x, y).

Proof. (i) As endomorphisms of J, (f − 1)^{2} = f^{2} − 2f + 1 = −2f.

(ii) We take x, y ∈ M ∩ J (the argument for x, y ∈ N ∩ J is similar).

We have (x, yf) = (xt, yft) = (−x, ytf^{−1}) = (−x, −yfz) = (−x, yf) = −(x, yf), whence (x, yf) = 0.

(iii) We have (x, y(f − 1)) = (x, yf) − (x, y) = −(x, y). □

Lemma 5.10. (i) In ℚ ⊗ End(J), (g^{−1} − 1)^{−1}t(g^{−1} − 1) = u.

(ii) (M ∩ J)(g^{−1} − 1) ≤ N ∩ J and (N ∩ J)2(g^{−1} − 1)^{−1} ≤ M ∩ J.

(iii) rank(M ∩ J) = rank(N ∩ J).

(iv) rank(F_{M}) = rank(F_{N}).

Proof. We use the property that g^{−1} acts as −g on ℚ ⊗ J. We also abuse notation by identifying elements of ℚ[D] with their images in End(ℚ ⊗ J). For example, (g^{−1} − 1) is not an invertible element of ℚ[D], though its image in End(ℚ ⊗ J) is invertible.

For (i), observe that (g^{−1} − 1)^{2} = −2g^{−1}, so that g^{−1} − 1 maps J to J and has zero kernel. Secondly, 2(g^{−1} − 1)^{−1} maps J to J and has zero kernel.

The equation (g^{−1}−1)^{−1}t(g^{−1}−1) = u in ℚ⊗End(J) is equivalent to t(g^{−1}−1) = (g^{−1}−1)u which is the same as (g−1)t = (g^{−1}−1)u or tut−t = −gu−u = −tuu−u = −t − u, which is true since tut = −u.

The statement (ii) follows since in a linear representation of a group, a group element which conjugates one element to a second one maps the eigenspaces of the two elements correspondingly. Here, this means g^{−1} − 1 conjugates t to u, so that g^{−1} − 1 maps ℚ ⊗ (M ∩ J) to ℚ ⊗ (N ∩ J). Since g^{−1} − 1 maps J into J (though not onto J), g^{−1} − 1 maps the direct summand M ∩ J into the direct summand N ∩ J.

For (iii), observe that we have monomorphisms M ∩ J → N ∩ J → M ∩ J and N ∩ J → M ∩ J → N ∩ J by use of g^{−1} − 1 and 2(g^{−1} − 1)^{−1}. Therefore, (iv) follows from (iii). □

Lemma 5.11. Suppose that det(J ∩ M)det(J ∩ N) is the square of an integer (equivalently, that det(F_{M})det(F_{N}) is the square of an integer). Then rank(J ∩ M) = rank(J ∩ N) is even.

Proof. Note that rank(M ∩ J) = rank(N ∩ J) by (5.10). Let d := det(J ∩ M) and e := det(J ∩ N) and let r be the common rank of M ∩ J and N ∩ J. First note that (M ∩ J)(g^{−1} − 1) has determinant 2^{r}d and second note that (M ∩ J)(g^{−1} − 1) has finite index, say k, in N ∩ J. It follows that 2^{r}d = k^{2}e. By hypothesis, de is a perfect square. Consequently, r is even. □

Corollary 5.12. rank(F_{M}) = rank(F_{N}) is even.

Proof. We have rank(F) + rank(M ∩ J) = rank(M) = 8 and similarly for N. Since rank F_{M} = rank F_{N}, we have F_{M} ≅ F_{N} by (5.3) and hence detF_{M}detF_{N} = (detF_{M})^{2} is a square. Now use (5.11). □

Proposition 5.13. If L = M + N is rootless, then F_{M} ≅ F_{N} ≅ 0, AA_{1} ⊥ AA_{1} or DD_{4}. Moreover, M ∩ J ≅ N ∩ J.

Proof. Since by (5.12), rank(F_{M}) = rank(F_{N}) is even, Proposition 5.3 implies that F_{M} ≅ F_{N} ≅ 0, AA_{1} ⊥ AA_{1} or DD_{4}. It is well-known that there is one orbit of O(E_{8}) on the family of sublattices which have a given one of the latter isometry types. It follows that M ∩ J = ann_{M}(F_{M}) ≅ ann_{N}(F_{N}) ≅ N ∩ J. □

5.2.2 DIH_{8}: Given that F = 0, what is J?

By Proposition 5.13, when L is rootless, F_{M} ≅ F_{N} ≅ 0, AA_{1} ⊥ AA_{1} or DD_{4}. We now consider each case for F_{M} and F_{N} and determine the possible pairs M, N. The conclusions are listed in Table 6.

Table 6: DIH_{8} which contains a rootless DIH_{4} lattice

F_{M} ≅ F_{N} F_{M} ∩ F_{N} rank(M + N)  M + N  Isometry type
integral? roots ?  if rootless
0  0  16  rootless  ≅ BW_{16}
AA_{1}^{⊥2}  0  16  non-integral
AA_{1}^{⊥2}  AA_{1}  15  non-integral
AA_{1}^{⊥2}  AA_{1}^{⊥2}  14  non-integral
AA_{1}^{⊥2}  2A_{1}  15  has roots
DD_{4}  0  16  rootless  ≥ DD_{4}^{⊥2} ⊥ EE_{8}
DD_{4}  AA_{1}  15  rootless  ≥ AA_{1}^{⊥7} ⊥ EE_{8}
DD_{4}  2A_{1}  15  non-integral
DD_{4}  AA_{1}^{⊥2}  14  has roots
DD_{4}  AA_{1} ⊥ 2A_{1}  14  non-integral
DD_{4}  AA_{1}^{⊥3}  13  has roots
DD_{4}  AA_{3}  13  non-integral
DD_{4}  DD_{4}  12  has roots

Proposition 5.14. If F_{M} = F_{N} = 0, then L = M + N is isometric to the Barnes-Wall lattice BW_{16}.

Proof. The sublattice M′ := Mt_{N} is the 1-eigenspace for t_{M} and so M +M′ = M ⊥ M′. Consider how N embeds in (M +M′)^{∗} = \frac{1}{2}(M +M′). Let x ∈ N \(M +M′) and let y ∈ \frac{1}{2}M, y′ ∈ \frac{1}{2}M′ so that x = y + y′. We may replace y, y′ by members of y + M and y′ + M′, respectively, which have least norm. Both y, y′ are nonzero. Their norms are therefore one of 1, 2, by a property of the E_{8}-lattice. Since (x, x) ≥ 4, y and y′ each has norm 2. It follows that the image of N in D(M) is totally singular in the sense that all norms of representing vectors in M^{∗} are integers. A similar thing is true for the image of N in D(M′). It follows that these images are elementary abelian groups which have ranks at most 4. On the other hand, diagonal elements of the orthogonal direct sum M ⊥ M′ have norms at least 8, which means that N ∩ (M + M′) contains no vectors in N of norm 4. Therefore, N/(N ∩ (M + M′)) is elementary abelian of rank at least 4. These two inequalities imply that the rank is 4. The action of t_{N} on this quotient is trivial. We may therefore use the uniqueness theorem of [GrBWY] to prove that M + N is isometric to the Barnes-Wall lattice BW_{16}. □

5.2.3 DIH_{8}: Given that F ≅ AA_{1} ⊥ AA_{1}, what is J?

Proposition 5.15. If F_{M} ≅ F_{N} ≅ AA_{1} ⊥ AA_{1}, M + N is non-integral or has a root.

Proof. If F_{M} ≅ F_{N} ≅ AA_{1} ⊥ AA_{1}, then M ∩ J ≅ N ∩ J ≅ DD_{6}. We shall first determine the structure of M ∩ J + N ∩ J.

Let h = g^{−1}. Then, by Lemma 5.10, we have (M ∩ J)(h − 1) ≤ N ∩ J. Since (M ∩ J, (M ∩ J)h) = 0, (M ∩ J)(h − 1) ≅ 2D_{6} and det((M ∩ J)(h − 1)) = 2^{8}. Therefore, |N ∩ J : (M ∩ J)(h − 1)| = (2^{rank N∩J})^{1/2} = 2^{3}.

Let K = (M ∩ J)(h − 1). Then, by (D.4), there exists a subset {η_{1}, …, η_{6}} ⊂ N ∩ J with (η_{i}, η_{j}) = 4δ_{i,j} such that

K = span_{ℤ} {(η_{i} ± η_{j}) | i, j = 1, 2, …, 6}

and

N ∩ J = span_{ℤ} \{η_{1}, η_{2}, η_{4}, η_{6}, \frac{1}{2}(−η_{1} + η_{2} − η_{3} + η_{4}), \frac{1}{2}(−η_{3} + η_{4} − η_{5} + η_{6})\} .

By computing the Gram matrix, it is easy to show that {η_{1} +η_{2}, −η_{1} +η_{2}, −η_{2} + η_{3}, −η_{3} + η_{4}, −η_{4} + η_{5}, −η_{5} + η_{6}} forms a basis of K = (M ∩ J)(h − 1) ≅ 2D_{6}. Now let

\begin{matrix}α_{1} = (η_{1} + η_{2})(h − 1)^{−1}, α_{2} = (−η_{1} + η_{2})(h − 1)^{−1}, α_{3} = (−η_{2} + η_{3})(h − 1)^{−1}, \\ α_{4} = (−η_{3} + η_{4})(h − 1)^{−1}, α_{5} = (−η_{4} + η_{5})(h − 1)^{−1}, α_{6} = (−η_{5} + η_{6})(h − 1)^{−1}. \end{matrix}

Then {α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}} is a basis of M ∩ J. Moreover, (α_{1}, α_{2}) = 0, (α_{1}, α_{3}) = −2, (α_{i}, α_{i+1}) = −2 for i = 2, …, 5.

By the definition, we have η_{2} = −\frac{1}{2}(α_{1} + α_{2})(h − 1), (−η_{1} + η_{2} − η_{3} + η_{4}) = (α_{2} + α_{4})(h − 1) and (−η_{3} + η_{4} − η_{5} + η_{6}) = (α_{4} + α_{6})(h − 1). Hence,

N ∩ J = span_{ℤ} \{K, \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{2} + α_{4})(h − 1), \frac{1}{2}(α_{4} + α_{6})(h − 1)\} .

Since M ≅ EE_{8}, det(M) = 2^{8} and |M/(F_{M} + M ∩ J)| = 2^{2}. Note that M, F_{M} and M ∩ J are all doubly even. Recall that D_{6}^{∗}/D_{6} ≅ ℤ_{2} × ℤ_{2}. Since M ∩ J is a direct summand of M, the natural map \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}} (M ∩ J)) is onto. Similarly, the natural map \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}} (F_{M})) is also onto.

Define H := F_{M} ∩ F_{N}. Let H_{X} := ann_{F_{X}} (H), for X = M, N. Since H is the negated sublattice of the involution t_{N} on F_{M}, H is isometric to either 0, AA_{1}, AA_{1} ⊥ AA_{1} or 2A_{1} since F_{M} and F_{N} are rectangular.

Let {α_{M}^{1}, α_{M}^{2}} and {α_{N}^{1}, α_{N}^{2}} be bases of F_{M} and F_{N} such that (α_{M}^{i}, α_{M}^{j}) = 4δ_{i,j} and (α_{N}^{i}, α_{N}^{j}) = 4δ_{i,j}. Since |M : F_{M} + M ∩ J| = 2^{2} and the natural map \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}}F_{M}) is onto, there exist β^{1} ∈ (M ∩ J)^{∗}, β^{2} ∈ (M ∩ J)^{∗} so that

ξ_{M} = \frac{1}{2}α_{M}^{1} + β^{1} and ζ_{M} =\frac{1}{2}α_{M}^{2} + β^{2}

are glue vectors and the cosets \frac{1}{\sqrt{2}} (β^{1} + (M ∩ J)) , \frac{1}{\sqrt{2}} (β^{2} + (M ∩ J)) generate the

abelian group (\frac{1}{\sqrt{2}} (M ∩ J))^{∗} /\frac{1}{\sqrt{2}} (M ∩ J) ≅ ℤ_{2} ⊕ℤ_{2}. Since M is spanned by norm 4 vectors, we may also assume that ξ_{M} and ζ_{M} both have norm 4 and thus β_{1} and β_{2} have norm 3.

Recall that a standard basis for the root lattice D_{6} is given by {(1, 1, 0, 0, 0, 0), (−1, 1, 0, 0, 0, 0), (0, −1, 1, 0, 0, 0), (0, 0, −1, 1, 0, 0), (0, 0, 0, −1, 1, 0), (0, 0, 0, 0, −1, 1)} and the elements of norm 3/2 in (D_{6})^{∗} have the form \frac{1}{2}(±1, …, ±1) with evenly many − signs or \frac{1}{2}(±1, …, ±1) with oddly many − signs (cf. [CS, Chapter 5]). They are contained in two distinct cosets of (D_{6})^{∗}/D_{6}. Note that (D_{6})^{∗}/D_{6} have 3 nontrivial cosets and their elements have norm 3/2, 3/2, and 1 modulo 2ℤ, respectively.

Now define ϕ : D_{6} → M ∩ J by

\begin{matrix}(1, 1, 0, 0, 0, 0) ↦ α_{1}, (−1, 1, 0, 0, 0, 0) ↦ α_{2}, (0, −1, 1, 0, 0, 0) ↦ α_{3} \\ (0, 0, −1, 1, 0, 0) ↦ α_{4}, (0, 0, 0, −1, 1, 0) ↦ α_{5}, (0, 0, 0, 0, −1, 1) ↦ α_{6}. \end{matrix}

A comparison of Gram matrices shows that ϕ is \sqrt{2} times an isometry. Since \frac{1}{2}(−1, 1, −1, 1, −1, 1) and \frac{1}{2}(1, 1, −1, 1, −1, 1) are the representatives of the two cosets of (D_{6})^{∗}/D_{6} represented by norm 3/2 vectors, by (A.5),

ϕ (\frac{1}{2}(−1, 1, −1, 1, −1, 1)) = \frac{1}{2}(α_{2} + α_{4} + α_{6})

and

ϕ (\frac{1}{2}(1, 1, −1, 1, −1, 1)) = \frac{1}{2}(α_{1} + α_{4} + α_{6})

are the representatives of the two cosets of 2(M ∩ J)^{∗}/(M ∩ J) represented by norm 3 vectors. Therefore, without loss, we may assume

\{β^{1}, β^{2}\} = \{\frac{1}{2}(α_{2} + α_{4} + α_{6}), \frac{1}{2}(α_{1} + α_{4} + α_{6})\}.

Similarly, there exist γ^{1}, γ^{2} ∈ (N ∩ J)^{∗} with (γ^{1}, γ^{1}) = (γ^{2}, γ^{2}) = 3 such that

ξ_{N} = \frac{1}{2}α_{N}^{1} + γ^{1}, and ζ_{N} = \frac{1}{2}α_{N}^{2} + γ^{2}

are glue vectors and N = span_{ℤ}\{F_{N} + N ∩ J, ξ_{N}, ζ_{N}\}. Moreover, \frac{1}{\sqrt{2}} (γ^{1} + N ∩ J),

\frac{1}{\sqrt{2}} (γ^{2} + N ∩ J) generate the group (\frac{1}{\sqrt{2}} (N ∩ J))^{∗} /\frac{1}{\sqrt{2}} (N ∩ J).

We shall prove that (β, γ) ≡ \frac{1}{2}(mod ℤ), resulting in a contradiction.

Define φ : D_{6} → N ∩ J by

(1, 1, 0, 0, 0, 0) ↦ η_{1},  (−1, 1, 0, 0, 0, 0) ↦ η_{2},
(0, −1, 1, 0, 0, 0) ↦ \frac{1}{2}(η_{1} + η_{2} + η_{3} + η_{4}),  (0, 0, −1, 1, 0, 0) ↦ η_{4},
(0, 0, 0, −1, 1, 0) ↦ \frac{1}{2}(−η_{3} + η_{4} − η_{5} + η_{6}),  (0, 0, 0, 0, −1, 1) ↦ η_{6}.

By comparing the Gram matrices, it is easy to show that φ is a \sqrt{2} times an isometry. Thus, we may choose γ^{1}, γ^{2} such that

\begin{aligned}\{γ^{1}, γ^{2}\} & = \{φ (\frac{1}{2}(−1, 1, −1, 1, −1, 1)) , φ (\frac{1}{2}(1, 1, −1, 1, −1, 1))\} \\ & = \{\frac{1}{2}(η_{2} + η_{4} + η_{6}), \frac{1}{2}(η_{1} + η_{4} + η_{6})\} . \end{aligned}

By the definition of α_{1}, …, α_{6}, we have

\begin{aligned}η_{1} & = \frac{1}{2}(α_{1} − α_{2})(h − 1), η_{2} = \frac{1}{2}(α_{1} + α_{2})(h − 1), \\ η_{4} & = [\frac{1}{2}(α_{1} + α_{2}) + (α_{3} + α_{4})](h − 1), \\ η_{6} & = [\frac{1}{2}(α_{1} + α_{2}) + (α_{3} + α_{4} + α_{5} + α_{6})](h − 1). \end{aligned}

Thus,

(α_{2} + α_{4} + α_{6}, η_{2} + η_{4} + η_{6}) = −6 and (α_{2} + α_{4} + α_{6}, η_{1} + η_{4} + η_{6}) = −2.

Therefore, (β^{1}, γ^{1}) ≡ 1/2 mod ℤ.

Subcase 1. H ≅ 0, AA_{1} or AA_{1} ⊥ AA_{1}. In this case, we may choose the bases {α_{M}^{1}, α_{M}^{2}} and {α_{N}^{1}, α_{N}^{2}} of F_{M} and F_{N} such that (α_{M}^{i}, α_{N}^{j}) ∈ {0, 4}, for all i, j. Hence,

(ξ_{M}, ξ_{N}) = (\frac{1}{2}α_{M}, \frac{1}{2}α_{N}) + (β^{1}, γ^{1}) ≡ 1/2 mod ℤ

and L = M + N is non-integral.

Subcase 2. H ≅ 2A_{1}. Then H_{M} ≅ H_{N} ≅ 2A_{1}, also. By replacing α_{M}^{i} by −α_{M}^{i} and α_{N}^{i} by −α_{N}^{i} for i = 1, 2 if necessary, α_{M}^{1} + α_{M}^{2} = α_{N}^{1} + α_{N}^{2} ∈ H. Write ρ := α_{M}^{1} + α_{M}^{2} = α_{N}^{1} + α_{N}^{2}. Then we calculate the difference of the glue vectors

\begin{aligned}η_{M} − ζ_{M} & = \frac{1}{2}(α_{M}^{1} − α_{M}^{2}) + \frac{1}{2}(α_{2} + α_{4} + α_{6}) − \frac{1}{2}(α_{1} + α_{4} + α_{6}) \\ & ≡ \frac{1}{2}ρ + \frac{1}{2}(−α_{1} + α_{2}) mod (F_{M} + M ∩ J). \end{aligned}

Similarly,

\begin{aligned}η_{N} − ζ_{N} & = \frac{1}{2}(α_{N}^{1} − α_{N}^{2}) + \frac{1}{2}(η_{2} + η_{4} + α_{6}) − \frac{1}{2}(η_{1} + η_{4} + η_{6}) \\ & ≡ \frac{1}{2}ρ − \frac{1}{2}(−η_{1} + η_{2}) mod (F_{N} + N ∩ J). \end{aligned}

Let ν_{M} = \frac{1}{2}ρ + \frac{1}{2}(−α_{1} + α_{2}) and ν_{N} = \frac{1}{2}ρ − \frac{1}{2}(−η_{1} + η_{2}). Then ν_{M} and ν_{N} are both norm 4 vectors in L. Recall that (−η_{1} + η_{2}) = α_{2}(h − 1). Since (α_{i}, α_{M}^{j}) = (α_{i}, α_{N}^{j}) = 0 for all i, j, we have (ρ, α_{i}) = 0 for all i.

\begin{matrix}(ν_{M}, ν_{N}) = & (\frac{1}{2}ρ + \frac{1}{2}(−α_{1} + α_{2}), \frac{1}{2}ρ − \frac{1}{2}(−η_{1} + η_{2})) \\ = & \frac{1}{4}[(ρ, ρ) − (−α_{1} + α_{2}, α_{2}(h − 1))] \end{matrix}

Recall that (α_{M}^{i}, α_{M}^{j}) = 4δ_{i,j} and (α_{i}, α_{j}) = 4δ_{i,j} for i, j = 1, 2. Moreover, (x, yh) = 0 for all x, y ∈ M ∩J by (ii) of (5.9). Thus, (ρ, ρ) = (α_{M}^{1} +α_{M}^{2}, α_{M}^{1} +α_{M}^{2}) = 4+4 = 8 and (−α_{1} + α_{2}, α_{2}(h − 1)) = (−α_{1} + α_{2}, −α_{2}) = −4.

Therefore, (ν_{M}, ν_{N}) = \frac{1}{4}(8 − (−4)) = 3 and hence ν_{M} − ν_{N} is a root in L. □

5.2.4 DIH_{8}: Given that F_{M} ≅ F_{N} ≅ DD_{4}, what is J?

Next we shall consider the case F_{M} ≅ F_{N} ≅ DD_{4}. In this case, M ∩ J ≅ N ∩ J ≅ DD_{4}.

Notation 5.16. Let {α_{1}, α_{2}, α_{3}, α_{4}} ⊂ M ∩ J such that (α_{i}, α_{j}) = 4δ_{i,j}, i, j = 1, 2, 3, 4. Then M ∩ J = span_{ℤ}\{α_{1}, α_{2}, α_{3}, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})\}. In this case, the norm 8 elements of M ∩ J are given by ±α_{i} ± α_{j} for i ≠ j.

Lemma 5.17. Let h = g^{−1}. By rearranging the subscripts if necessary, we have

N ∩ J = span_{ℤ}\{(M ∩ J)(h − 1), \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{1} + α_{3})(h − 1)\}.

Proof. Let K := (M ∩ J)(h − 1). Then by (ii) of Lemma 5.10, we have K ≤ N ∩ J. Since (M ∩ J, (M ∩ J)h) = 0 by (5.9), K = (M ∩ J)(h − 1) ≅ 2D_{4}. Therefore, by (D.5), we have K ≤ N ∩ J ≤ \frac{1}{2}K.

Note that, by determinants, |N ∩J : K| = \sqrt{2^{4}} = 2^{2}. Therefore, there exists two glue vectors β_{1}, β_{2} ∈ (N ∩ J) \ K such that N ∩ J = span_{ℤ}{(M ∩ J)(h − 1), β_{1}, β_{2}}.

Since K has minimal norm 8 and N ∩J is generated by norm 4 elements, we may choose β_{1}, β_{2} such that both are of norm 4. On the other hand, elements of norm 4 in N ∩J are given by \frac{1}{2}γ(h−1), where γ ∈ M ∩J with norm 8, i.e., γ = ±α_{i} ±α_{j} for some i ≠ j. Since α_{1}(h−1), α_{2}(h−1), α_{3}(h−1), α_{4}(h−1) ∈ (M ∩J)(h−1) ≤ N ∩J, we may assume

β_{1} = \frac{1}{2}(α_{i} + α_{j})(h − 1) and β_{2} = \frac{1}{2}(α_{k} + α_{ℓ})(h − 1)

for some i, j, k, ℓ ∈ {1, 2, 3, 4}. Note that |{i, j} ∩ {k, ℓ}| = 1 because β_{1} + β_{2} ∉ K. Therefore, by rearranging the indices if necessary, we may assume β_{1} = \frac{1}{2}(α_{1} + α_{2})(h − 1), β_{2} = \frac{1}{2}(α_{1} + α_{3})(h − 1) and

N ∩ J = span_{ℤ}\{(M ∩ J)(h − 1), \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{1} + α_{3})(h − 1)\}.

as desired. □

Proposition 5.18. If F_{M} ≅ F_{N} ≅ DD_{4}, then M ∩ J + N ∩ J ≅ EE_{8}.

Proof. First we shall note that (M ∩ J) + (M ∩ J)(h − 1) = (M ∩ J) ⊥ (M ∩ J)h ≅ DD_{4} ⊥ DD_{4}. Moreover, we have |N ∩ J + M ∩ J : (M ∩ J) + (M ∩ J)(h − 1)| = |N ∩ J : (M ∩ J)(h − 1)| = \sqrt{(2^{8} · 4)/(2^{4} · 4)} = 4, by determinants. Therefore, det(M ∩ J + N ∩ J) = (2^{4} · 4)^{2}/4^{2} = 2^{8}.

Now by (5.17), we have

N ∩ J = span_{ℤ}\{(M ∩ J)(h − 1), \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{1} + α_{3})(h − 1)\}.

Next we shall show that (M ∩ J, N ∩ J) ⊂ 2ℤ. Since (M ∩ J, (M ∩ J)h)) = 0 and M ∩ J is doubly even, it is clear that (M ∩ J, (M ∩ J)(h − 1)) ⊂ 2ℤ. Moreover, for any i, j ∈ {1, 2, 3, 4}, i ≠ j,

(α_{k}, \frac{1}{2}(α_{i} + α_{j})(h − 1)) = \begin{cases}0 & if k ∉ \{i, j\}, \\ −2 & if k ∈ \{i, j\}, \end{cases}

and

(\frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}), \frac{1}{2}(α_{i} + α_{j})(h − 1)) = −2.

Since M∩J is spanned by α_{1}, α_{2}, α_{3} and \frac{1}{2}(α_{1}+α_{2}+α_{3}+α_{4}) and N∩J = span_{ℤ}\{(M∩ J)(h − 1), \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{1} + α_{3})(h − 1)\}, we have (M ∩ J, N ∩ J) ⊂ 2ℤ as required. Therefore, \frac{1}{\sqrt{2}} (M ∩ J + N ∩ J) is an integral lattice and has determinant 1 and thus M ∩ J + N ∩ J ≅ EE_{8}, by the classification of unimodular even lattices of rank 8. □

Lemma 5.19. Let α_{1}, α_{2}, α_{3}, α_{4} ∈ M ∩ J be as in Notation 5.16. Then

(M ∩ J)^{∗} = \frac{1}{4}span_{ℤ} \{α_{1} − α_{2}, α_{1} + α_{2}, α_{1} + α_{3}, α_{1} + α_{4}\} .

and

(N ∩ J)^{∗} = \frac{1}{4}span_{ℤ} \{α_{1}(h − 1), α_{2}(h − 1), α_{3}(h − 1), \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1)\} .

Proof. Since (ℤα_{i})^{∗} = \frac{1}{4}ℤα_{i} and M ∩ J = span_{ℤ}\{α_{1}, α_{2}, α_{3}, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})\},

(M ∩ J)^{∗}

= \{β = \frac{1}{4}(a_{1}α_{1} + a_{2}α_{2} + a_{3}α_{3} + a_{4}α_{4})| a_{i} ∈ ℚ, (β, γ) ∈ ℤ for all γ ∈ M ∩ J\}

= \{\frac{1}{4}(a_{1}α_{1} + a_{2}α_{2} + a_{3}α_{3} + a_{4}α_{4}) | a_{i} ∈ ℤ and ∑_{i=1}^{4} a_{i} ∈ 2ℤ, i = 1, 2, 3, 4\}
=\frac{1}{4}span_{ℤ} \{α_{1} − α_{2}, α_{1} + α_{2}, α_{1} + α_{3}, α_{3} + α_{4}\} .

Now by (5.17), we have

N ∩ J = span_{ℤ} \{(M ∩ J)(h − 1), \frac{1}{2}(α_{1} + α_{2})(h − 1), \frac{1}{2}(α_{1} + α_{3})(h − 1)\} .

Let β_{1} = \frac{1}{2}(α_{1} + α_{2})(h − 1), β_{2} = \frac{1}{2}(α_{1} − α_{2})(h − 1), β_{3} = \frac{1}{2}(α_{3} + α_{4})(h − 1), and β_{4} = \frac{1}{2}(α_{3} − α_{4})(h − 1). Then \{β_{1}, β_{2}, β_{3}, β_{4}\} forms an orthogonal subset of N ∩ J with (β_{i}, β_{j}) = 4δ_{i,j}. Note that \frac{1}{2}(β_{1} + β_{2} + β_{3} + β_{4}) = \frac{1}{2}(α_{1} + α_{3})(h − 1) ∈ N ∩ J. Thus, N ∩ J = span_{ℤ}\{β_{1}, β_{2}, β_{3}, \frac{1}{2}(β_{1} + β_{2} + β_{3} + β_{4})\} since both of them are isomorphic to DD_{4}. Hence we have

(N ∩ J)^{∗} = \frac{1}{4}span_{ℤ} \{β_{1} − β_{2}, β_{1} + β_{2}, β_{1} + β_{3}, β_{3} + β_{4}\}

= \frac{1}{4}span_{ℤ} \{α_{1}(h − 1), α_{2}(h − 1), α_{3}(h − 1), \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1)\} .

as desired. □

Lemma 5.20. We shall use the same notation as in (5.16). Then the cosets of 2(M ∩ J)^{∗}/(M ∩ J) are represented by

0, \frac{1}{2}(α_{1} + α_{2}), \frac{1}{2}(α_{1} + α_{3}), \frac{1}{2}(α_{2} + α_{3}),

and the cosets of 2(N ∩ J)^{∗}/(N ∩ J) are represented by

0, \frac{1}{2}α_{1}(h − 1), \frac{1}{4}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1), \frac{1}{4}(α_{1} + α_{2} + α_{3} − α_{4})(h − 1).

Moreover, (2(M ∩ J)^{∗}, 2(N ∩ J)^{∗}) ⊂ ℤ.

Proof. Since X := M ∩ J ≅ DD_{4}, it is clear that 2X^{∗}/X is a four-group. The three nonzero vectors in the list

0, \frac{1}{2}(α_{1} + α_{2}), \frac{1}{2}(α_{1} + α_{3}), \frac{1}{2}(α_{2} + α_{3}),

have norms two, so all are in 2X^{∗} \ X. Since the difference of any two has norm 2, no two are congruent modulo X. A similar argument proves the second statement.

For the third statement, we calculate the following inner products.

For any i, j, k ∈ {1, 2, 3, 4} with i ≠ j,

(α_{i} ± α_{j}, α_{k}(h − 1)) = \begin{cases}0 & if k ∉ \{i, j\}, \\ ±4 & if k ∈ \{i, j\}, \end{cases}

and

(α_{i} ± α_{j}, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1)) = 0 or − 4.

Since (M ∩ J)^{∗} = \frac{1}{4}span_{ℤ} \{α_{1} − α_{2}, α_{1} + α_{2}, α_{1} + α_{3}, α_{1} + α_{4}\} and (N ∩ J)^{∗} = \frac{1}{4}span_{ℤ} \{α_{1}(h − 1), α_{2}(h − 1), α_{3}(h − 1), \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1)\} by (5.19), we have ((M ∩ J)^{∗}, (N ∩ J)^{∗}) ⊂ \frac{1}{4}ℤ and hence (2(M ∩ J)^{∗}, 2(N ∩ J)^{∗}) ⊂ ℤ as desired. □

Remark 5.21. Note that the lattice D_{4} is BW_{2^{2}}, so the involutions in its isometry group BRW^{+}(2^{2}) ≅ W eyl(F_{4}) may be deduced from the theory in [GrIBW1], especially Lemma 9.14 (with d = 2). The results are in Table (7).

Notation 5.22. Define H := F_{M} ∩ F_{N} and let H_{X} := ann_{F_{X}} (H) for X = M, N. Since H is the negated sublattice of the involution t_{N} on F_{M}, we have the possibilities listed in Table 7. We label the case for \frac{1}{\sqrt{2}}H by the corresponding involution 2A, ⋯, 2G. (i.e., the involution whose negated space is \frac{1}{\sqrt{2}}H)

We shall prove the main result of this section, Theorem 5.26 in several steps.

Lemma 5.23. Suppose F_{M} ≅ F_{N} ≅ DD_{4}. If \frac{1}{\sqrt{2}}H ≅ AA_{1}, A_{1} ⊥ AA_{1} or A_{3} (i.e., the cases for 2B, 2D and 2F), then the lattice L is non-integral.

Table 7: The seven conjugacy classes of involutions in BRW^{+}(2^{2}) ≅ W eyl(F_{4})

Involution Multiplicity of −1 Isometry type of

negated sublattice

2A  1  A_{1}
2B  1  AA_{1}
2C  2  A_{1} ⊥ A_{1}
2D  2  A_{1} ⊥ AA_{1}
2E  3  A_{1} ⊥ A_{1} ⊥ A_{1}
2F  3  A_{3}
2G  4  D_{4}

Proof. We shall divide the proof into 3 cases. Recall notations (5.22).

Case 2B. In this case, \frac{1}{\sqrt{2}}H ≅ AA_{1} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ A_{3}. Then F_{M} ≥ H ⊥ H_{M} and M ≥ H ⊥ H_{M} ⊥ M ∩ J.

Let α ∈ H with (α, α) = 8. Then H = ℤα and H^{∗} = \frac{1}{8}ℤα.

By (A.5), we have (\frac{1}{\sqrt{2}}F_{M})^{∗}/\frac{1}{\sqrt{2}}F_{M} ≅ 2(F_{M})^{∗}/F_{M}. Thus, by (D.26), the natural map 2(F_{M})^{∗} → 2(H^{∗}) = \frac{1}{4}ℤα is onto. Therefore, there exists δ_{M} ∈ H_{M}^{∗} with (δ_{M}, δ_{M}) = 3/2 such that \frac{1}{4}α + δ_{M} ∈ 2(F_{M})^{∗}. Note that the natural map \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}}F_{M}) is also onto since \frac{1}{\sqrt{2}}M is unimodular and F_{M} is a direct summand of M. Therefore, there exists γ_{M} ∈ 2(M ∩ J)^{∗} with (γ_{M}, γ_{M}) = 2 such that

ξ_{M} = \frac{1}{4}α + δ_{M} + γ_{M}

is a glue vector for H ⊥ H_{M} ⊥ M ∩ J in M. Similarly, there exists δ_{N} ∈ H_{N}^{∗} with (δ_{N}, δ_{N}) = 3/2 and γ_{N} ∈ 2(N ∩ J)^{∗} with (γ_{N}, γ_{N}) = 2 such that

ξ_{N} = \frac{1}{4}α + δ_{N} + γ_{N}

is a glue vector for H ⊥ N_{N} ⊥ N ∩ J in N.

Since (γ_{M}, γ_{N}) ∈ ℤ by (5.20) and H_{M} ⊥ H_{N},

(ξ_{M}, ξ_{N}) = \frac{1}{16}(α, α) + (δ_{M}, δ_{N}) + (γ_{M}, γ_{N}) ≡ \frac{1}{2} mod ℤ,

which is not an integer.

Case 2D. In this case, \frac{1}{\sqrt{2}}H ≅ A_{1} ⊥ AA_{1} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ A_{1} ⊥ AA_{1}, also. Take α, β ∈ H such that (α, α) = 8, (β, β) = 4 and (α, β) = 0. Similarly, there exist α_{M}, β_{M} ∈ H_{M} and α_{N}, β_{N} ∈ H_{N} such that (α_{M}, α_{M}) = 8, (β_{M}, β_{M}) = 4 and (α_{M}, β_{M}) = 0 and (α_{N}, α_{N}) = 8, (β_{N}, β_{N}) = 4 and (α_{N}, β_{N}) = 0.

Note that ℤβ ⊥ ℤβ_{M} ⊥ ℤα_{M} ≤ ann_{F_{M}} (α) ≅ AA_{3}. Set A := ann_{F_{M}} (α) ≅ AA_{3}. Then |A : ℤβ ⊥ ℤβ_{M} ⊥ ℤα_{M}| = \sqrt{(4 × 4 × 8)/(2^{3} × 4)} = 2. Therefore, there exists a μ ∈ (ℤβ ⊥ ℤβ_{M} ⊥ ℤα_{M})^{∗} = \frac{1}{4}ℤβ ⊥ \frac{1}{4}ℤβ_{M} ⊥ \frac{1}{8}ℤα_{M} such that A = span_{ℤ}{β, β_{M}, α_{M}, μ} and 2μ ∈ ℤβ ⊥ ℤβ_{M} ⊥ ℤα_{M}.

Since A is generated by norm 4 vectors, we may choose μ so that μ has norm 4. The only possibility is μ = \frac{1}{2}(±β±β_{M}±α_{M}). Therefore, A = span_{ℤ}\{β, β_{M}, α_{M}, \frac{1}{2}(β+ β_{M} +α_{M})\} and 2A^{∗}/A ≅ ℤ_{4} is generated by \frac{1}{2}β+ \frac{1}{4}α_{M} +A. Note also that \frac{1}{2}β+ \frac{1}{4}α_{M} has norm 3/2.

Now recall that ℤα ≅ 2A_{1} and A ≅ AA_{3}. Thus, by (D.26), the natural map 2(F_{M})^{∗} → 2(H^{∗}) = \frac{1}{4}ℤα is onto. Thus, there exists a δ ∈ 2A^{∗} with (δ, δ) = 3/2 such that \frac{1}{4}α + δ ∈ 2(F_{M})^{∗}. By the previous paragraph, we may assume δ = \frac{1}{2}β + \frac{1}{4}α_{M}. Since the natural map \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}}F_{M}) is onto, there exists γ_{M} ∈ 2(M ∩J)^{∗} such that

ξ_{M} = \frac{1}{4}α + \frac{1}{2}β + \frac{1}{4}α_{M} + γ_{M}

is a glue vector for H ⊥ H_{M} ⊥ M ∩ J in M. Similarly, there exists γ_{N} ∈ 2(N ∩ J)^{∗} such that

ξ_{N} = \frac{1}{4}α + \frac{1}{2}β + \frac{1}{4}α_{N} + γ_{N}

is a glue vector for H ⊥ H_{N} ⊥ N ∩ J in N. Then

(ξ_{M}, ξ_{N}) = \frac{1}{16}(α, α) + \frac{1}{4}(β, β) + \frac{1}{16}(α_{M}, α_{N}) + (γ_{M}, γ_{N}) ≡ 1/2 mod ℤ,

since (α_{M}, α_{N}) = 0 and (γ_{M}, γ_{N}) ∈ ℤ by (5.20). Therefore, L is not integral.

Case 2F. In this case, \frac{1}{\sqrt{2}}H ≅ A_{3} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ AA_{1}. Then F_{M} ≥ H ⊥ H_{M} and F_{N} ≥ H ⊥ H_{N}. Let δ ∈ H^{∗}, α_{M} ∈ H_{M} and α_{N} ∈ H_{N} such that (δ, δ) = 3/2, (α_{M}, α_{M}) = 8 and (α_{N}, α_{N}) = 8.

Recall in Case 2B that \frac{1}{\sqrt{2}}H ≅ AA_{1} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ A_{3}. Now by exchanging the role of H with H_{M} (or H_{N}) and using the same argument as in Case 2B, we may show that there exist γ_{M} ∈ 2(M ∩ J)^{∗} and γ_{N} ∈ 2(N ∩ J)^{∗} such that ξ_{M} = δ + \frac{1}{4}α_{M} + γ_{M} is a glue vector for H ⊥ H_{M} ⊥ M ∩ J in M and ξ_{N} = δ + \frac{1}{4}α_{N} + γ_{N} is a glue vector for H ⊥ H_{N} ⊥ (N ∩ J) in N. However,

(ξ_{M}, ξ_{N}) = (δ, δ) + (γ_{M}, γ_{N}) ≡ 1/2 mod ℤ,

since (γ_{M}, γ_{N}) ∈ ℤ by (5.20). Again, L is not integral. □

Lemma 5.24. Let γ_{M} be any norm 2 vector in 2(M ∩ J)^{∗}. Then for each non-zero coset γ_{N} +(N ∩J) in 2(N ∩J)^{∗}/(N ∩J), there exists a norm 2 vector γ ∈ γ_{N} +(N ∩J) such that (γ_{M}, γ) = −1.

Proof. Recall from (5.19) that

2(M ∩ J)^{∗} = \frac{1}{2}span_{ℤ} \{α_{1} − α_{2}, α_{1} + α_{2}, α_{1} + α_{3}, α_{3} + α_{4}\} .

Thus, all norm 2 vectors in 2(M ∩ J)^{∗} have the form \frac{1}{2}(±α_{i} ± α_{j}) for some i ≠ j. Without loss, we may assume γ_{M} = \frac{1}{2}(α_{i} + α_{j}) by replacing α_{i}, α_{j} by −α_{i}, −α_{j} if necessary.

Now by (5.20), the non-zero cosets of 2(N ∩ J)^{∗}/(N ∩ J) are represented by \frac{1}{2}α_{1}(h − 1), \frac{1}{4}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1) and \frac{1}{4}(α_{1} + α_{2} + α_{3} − α_{4})(h − 1). Moreover, by (5.17),

N ∩ J = \frac{1}{2}span_{ℤ}\{ (α_{i} ± α_{j})(h − 1) | 1 ≤ i < j ≤ 4\}.

If γ_{N} + (N ∩ J) = \frac{1}{2}α_{1}(h − 1) + (N ∩ J), we take

γ = \frac{1}{2}α_{i}(h − 1) = \frac{1}{2}α_{1}(h − 1) + \frac{1}{2}(−α_{1} + α_{i})(h − 1) ∈ \frac{1}{2}α_{1}(h − 1) + (N ∩ J).

Recall from (5.16) that {α_{1}, α_{2}, α_{3}, α_{4}} ∈ M ∩ J and (α_{i}, α_{j}) = 4δ_{i,j} for i, j = 1, …, 4. Moreover, (x, yh) = 0 for all x, y ∈ M ∩ J by (5.9).

Thus, (γ, γ) = (\frac{1}{2}α_{i}(h − 1), \frac{1}{2}α_{i}(h − 1)) = \frac{1}{4}[(α_{i}h, α_{i}h) + (α_{i}, α_{i})] = 2 and (γ_{M}^{1}, γ) = (\frac{1}{2}(α_{i} + α_{j}), \frac{1}{2}α_{i}(h − 1)) = −\frac{1}{4}(α_{i} + α_{j}, α_{i}) = −1.

If γ_{N} + (N ∩ J) = \frac{1}{4}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1) + (N ∩ J), we simply take γ = \frac{1}{4}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1). Then (γ, γ) = 2 and

\begin{matrix}(γ_{M}, γ) = & (\frac{1}{2}(α_{i} + α_{j}), \frac{1}{4}(α_{1} + α_{2} + α_{3} + α_{4})(h − 1)) \\ = & − \frac{1}{8}(α_{i} + α_{j}, α_{1} + α_{2} + α_{3} + α_{4}) \\ = & − \frac{1}{8}(4 + 4) = −1. \end{matrix}

Finally we consider the case γ_{N} +N ∩J = \frac{1}{4}(α_{1} +α_{2} +α_{3} −α_{4})(h−1)+(N ∩J). Let {k, ℓ} = {1, 2, 3, 4} − {i, j} and take

\begin{aligned}γ & = \frac{1}{4}(α_{i} + α_{j} + α_{k} − α_{ℓ})(h − 1) = \frac{1}{4}(α_{1} + α_{2} + α_{3} − α_{4})(h − 1) + \frac{1}{2}(α_{4} − α_{ℓ}) \\ & ∈ \frac{1}{4}(α_{1} + α_{2} + α_{3} − α_{4})(h − 1) + N ∩ J. \end{aligned}

Then (γ, γ) = 2 and

\begin{matrix}(γ_{M}, γ) = & (\frac{1}{2}(α_{i} + α_{j}), \frac{1}{4}(α_{i} + α_{j} + α_{k} − α_{ℓ})(h − 1)) \\ = & − \frac{1}{8}(α_{i} + α_{j}, α_{i} + α_{j} + α_{k} − α_{ℓ}) \\ = & − \frac{1}{8}(4 + 4) = −1 \end{matrix}

as desired. □

Lemma 5.25. If \frac{1}{\sqrt{2}}H ≅ A_{1} ⊥ A_{1}, A_{1} ⊥ A_{1} ⊥ A_{1} or D_{4} (i.e., the cases for 2C, 2E and 2G), then the lattice L has roots.

Proof. We continue to use the notations (5.22). First, we shall note that the natural maps \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}}F_{M}), \frac{1}{\sqrt{2}}M → D(\frac{1}{\sqrt{2}} (M ∩ J)) and \frac{1}{\sqrt{2}}N → D(\frac{1}{\sqrt{2}}F_{N}), and \frac{1}{\sqrt{2}}N → D(\frac{1}{\sqrt{2}} (N ∩ J)) are all onto.

Case 2C. In this case, \frac{1}{\sqrt{2}}H ≅ A_{1} ⊥ A_{1} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ A_{1} ⊥ A_{1}. Let μ^{1}, μ^{2} be a basis of H such that (μ^{i}, μ^{j}) = 4δ_{i,j}. Let μ_{M}^{1}, μ_{M}^{2} and μ_{N}^{1}, μ_{N}^{2} be bases of H_{M} and H_{N} which consist of norm 4 vectors. Then, F_{M} = span_{ℤ}\{μ^{1}, μ^{2}, μ_{M}^{1}, \frac{1}{2}(μ^{1}+ μ^{2} + μ_{M}^{1} + μ_{M}^{2})\} and F_{N} = span_{ℤ}\{μ^{1}, μ^{2}, μ_{N}^{1}, \frac{1}{2}(μ^{1} + μ^{2} + μ_{N}^{1} + μ_{N}^{2})\}. Therefore, by the same arguments as in Lemma 5.20, the cosets representatives of (2F_{M}^{∗})/F_{M} are given by

0, \frac{1}{2}(μ^{1} + μ^{2}), \frac{1}{2}(μ^{1} + μ_{M}^{1}), \frac{1}{2}(μ^{2} + μ_{M}^{1}),

and the cosets representatives of (2F_{N}^{∗})/F_{N} are given by

0, \frac{1}{2}(μ^{1} + μ^{2}), \frac{1}{2}(μ^{1} + μ_{N}^{1}), \frac{1}{2}(μ^{2} + μ_{N}^{1}).

Therefore, there exist γ_{M}^{1}, γ_{M}^{2} ∈ (M ∩ J)^{∗} so that

ξ_{M} = \frac{1}{2}(μ_{1} + μ_{2}) + γ_{M}^{1} and ζ_{M} = \frac{1}{2}(μ_{1} + μ_{M}^{1}) + γ_{M}^{2},

are glue vectors for F_{M} + (J ∩ M) in M and such that γ_{M}^{1} + (M ∩ J), γ_{M}^{2} + (M ∩ J) generate 2(M ∩ J)^{∗}/(M ∩ J).

Similarly, there exist γ_{N}^{1}, γ_{N}^{2} ∈ (N ∩ J)^{∗} so that

ξ_{N} = −\frac{1}{2}(μ_{1} + μ_{2}) + γ_{N}^{1} and ζ_{N} = −\frac{1}{2}(μ_{1} + μ_{N}^{1}) + γ_{N}^{2},

are glue vectors for F_{N} + N ∩ J in N, and such that γ_{N}^{1} + (N ∩ J), γ_{N}^{2} + (N ∩ J) generate 2(N ∩ J)^{∗}/(N ∩ J).

By Lemma (5.24), we may assume (γ_{N}^{1}, γ_{M}^{1}) = −1. Then

(ξ_{M}, ξ_{N}) = (\frac{1}{2}(μ_{1} + μ_{2}) + γ_{M}^{1}, −\frac{1}{2}(μ_{1} + μ_{2}) + γ_{N}^{1})

= −\frac{1}{4}((μ_{1}, μ_{1}) + (μ_{2}, μ_{2})) + (γ_{M}^{1}, γ_{N}^{1})

= −\frac{1}{4}(4 + 4) − 1 = −3

and hence ξ_{M} + ξ_{N} is a root.

Case 2E. In this case, \frac{1}{\sqrt{2}}H ≅ A_{1} ⊥ A_{1} ⊥ A_{1} and \frac{1}{\sqrt{2}}H_{M} ≅ \frac{1}{\sqrt{2}}H_{N} ≅ A_{1}.

Let μ_{1}, μ_{2}, μ_{3} ∈ H be such that (μ_{i}, μ_{j}) = 4δ_{i,j}. Let μ_{M} ∈ H_{M} and μ_{N} ∈ H_{N} be norm 4 vectors. Then H_{M} = ℤμ_{M} and H_{N} = ℤμ_{N}. Moreover,

F_{M} = span_{ℤ}\{μ_{1}, μ_{2}, μ_{3}, \frac{1}{2}(μ_{1} + μ_{2} + μ_{3} + μ_{M})\} ≅ DD_{4}

and

F_{N} = span_{ℤ}\{μ_{1}, μ_{2}, μ_{3}, \frac{1}{2}(μ_{1} + μ_{2} + μ_{3} + μ_{N})\} ≅ DD_{4}.

Then, by (5.20), \frac{1}{2}(μ_{1} + μ_{2}) is in both 2(F_{M})^{∗} and 2(F_{N})^{∗}. Therefore, there exist γ_{M} ∈ 2(M ∩ J)^{∗} and γ_{N} ∈ 2(N ∩ J)^{∗} such that

ξ_{M} = \frac{1}{2}(μ_{1} + μ_{2}) + γ_{M} ∈ M and ξ_{N} = −\frac{1}{2}(μ_{1} + μ_{2}) + γ_{N} ∈ N

are norm 4 glue vectors for F_{M} ⊥ M ∩ J in M and F_{N} ⊥ N ∩ J in N, respectively. By Lemma (5.24), we may assume (γ_{M}, γ_{N}) = −1. Then,

\begin{matrix}(ξ_{M}, ξ_{N}) = & (\frac{1}{2}(μ_{1} + μ_{2}) + γ_{M}, −\frac{1}{2}(μ_{1} + μ_{2}) + γ_{N}) \\ = & − \frac{1}{4}(μ_{1} + μ_{2}, μ_{1} + μ_{2}) + (γ_{M}, γ_{N}) = −2 − 1 = −3 \end{matrix}

and ξ_{M} + ξ_{N} is a root.

Case 2G. In this case, \frac{1}{\sqrt{2}}H ≅ D_{4} and \frac{1}{\sqrt{2}}H_{M} = \frac{1}{\sqrt{2}}H_{N} = 0. Recall that 2(DD_{4})^{∗}/DD_{4} ≅ (D_{4})^{∗}/D_{4} by (A.5) and all non-trivial cosets of (D_{4})^{∗}/D_{4} can be represented by norm 1 vectors [CS, p. 117]. Therefore, the non-trivial cosets of 2(DD_{4})^{∗}/DD_{4} can be represented by norm 2 vectors. Thus we can find vectors γ ∈ 2H^{∗} with (γ, γ) = 2 and γ_{M} ∈ 2(M ∩ J)^{∗}, γ_{N} ∈ 2(N ∩ J)^{∗} such that

ξ_{M} = γ + γ_{M} ∈ M and ξ_{N} = −γ + γ_{N} ∈ N

are norm 4 glue vectors for F_{M} ⊥ M ∩ J in M and F_{N} ⊥ N ∩ J in N, respectively. Again, we may assume (γ_{M}, γ_{N}) = −1 by (5.24) and thus (ξ_{M}, ξ_{N}) = −3 and there are roots. □

Theorem 5.26. Suppose F_{M} ≅ F_{N} ≅ DD_{4}. If L = M + N is integral and rootless, then H = F_{M} ∩ F_{N} = 0 or ≅ AA_{1}.

The proof of Theorem 5.26 now follows from Lemmas 5.23 and 5.25.

6 DIH_{6} and DIH_{12} theories

We shall study the cases when D = ⟨t_{M}, t_{N}⟩ ≅ Dih_{6} or Dih_{12}. The following is our main theorem in this section. We refer to the notation table (Table 3) for the definition of DIH_{6}(14), DIH_{6}(16) and DIH_{12}(16).

Theorem 6.1. Let L be a rootless integral lattice which is a sum of sublattices M and N isometric to EE_{8}. If the associated dihedral group has order 6 or 12, the possibilities for L + M + N, M, N are listed in Table (8).

Table 8: DIH_{6} and DIH_{12}: Rootless cases
Name  F ≅  L contains …  with index … D(L)
DIH_{6}(14)  AA_{2}  ≥ A_{2} ⊗ E_{6} ⊥ AA_{2}  3^{2}  1^{9}3^{3}6^{2}
DIH_{6}(16)  0  A_{2} ⊗ E_{8}  1  3^{8}
DIH_{12}(16) AA_{2} ⊥ AA_{2} ≥ A_{2} ⊗ E_{6} ⊥ AA_{2}^{⊥2}  3^{2}  1^{12}6^{4}

6.1 DIH_{6}

Notation 6.2. Define t := t_{M}, h := t_{M}t_{N}. We suppose h has order 3. Then, N = Mg, where g = h^{2}. The third lattice in the orbit of D := ⟨g, t⟩ is Mg^{2}, but we shall not refer to it explicitly henceforth. Define F := M ∩ N, J := ann_{L}(F). Note that F is the common negated lattice for t_{M} and t_{N} in L, so is the fixed point sublattice for g and is a direct summand of L (cf. (A.10)).

Lemma 6.3. Let X = M or N. Two of the sublattices {(J ∩ X)g^{i}| i ∈ ℤ} are equal or meet trivially.

Proof. We may assume X = M. Suppose that 0 ≠ U = (J ∩ M)g^{i} ∩ (J ∩ M)g^{j} for i, j not congruent modulo 3. Then U is negated by two distinct involutions t^{g^{i}} and t^{g^{j}}, hence is centralized by g, a contradiction. □

Lemma 6.4. If F = 0, J ≅ A_{2} ⊗ E_{8}.

Proof. Use (3.2). □

Hypothesis 6.5. We assume F ≠ 0 and define the integer s by 3^{s} := |L/(J + F)|.

Lemma 6.6. L/(J ⊥ F) is an elementary abelian group, of order 3^{s} where s ≤ \frac{1}{2}rank(J).

Proof. Note that g acts trivially on both F and L/J since L/J embeds in F^{∗}. Observe that g − 1 induces an embedding L/F → J. Furthermore, g − 1 induces an embedding L/(J + F) → J/J(g − 1), which is an elementary abelian 3-group whose rank is at most \frac{1}{2}rank(J) since (g − 1)^{2} induces the map −3g on J. □

Lemma 6.7. s ≤ rank(F) and s ∈ {1, 2, 3}.

Proof. If s were 0, L = J + F and M would be orthogonally decomposable, a contradiction. Therefore, s ≥ 1. The two natural maps L → D(F) and L → D(J) have common kernel J ⊥ F. Their images are therefore elementary abelian group of rank s at most rank(F) and at most rank(J). In (6.6), we observed the stronger statement that s ≤ \frac{1}{2}rank(J). Since rank(J) ≥ 1, 8 = rank(J) + rank(F) > rank(J) ≥ 2s implies that s ≤ 3. □

Lemma 6.8. M/((M ∩ J) + F) ≅ L/(J ⊥ F) is an elementary abelian 3-group of order 3^{s}.

Proof. The quotient L/(J + F) is elementary abelian by (6.6). Since L = M + N and N = Mg, M covers L/L(g − 1). Since L(g − 1) ≤ J, M + J = L Therefore, L/(J ⊥ F) ≅ (M + J)/(J + F) = (M + (J + F))/(J + F) ≅ M/(M ∩ (J + F)). The last denominator is (M ∩ J) + F since F ≤ M. □

Lemma 6.9. D(F) ≅ 3^{s} × 2^{rank(F)}.

Proof. Since \frac{1}{\sqrt{2}}M ≅ E_{8} and the natural map of \frac{1}{\sqrt{2}}M to D(\frac{1}{\sqrt{2}}F) is onto and has kernel \frac{1}{\sqrt{2}}(M ∩ J ⊥ F), D(\frac{1}{\sqrt{2}}F) ≅ 3^{s} is elementary abelian. □

Notation 6.10. Let X = M ∩ J, Y = N ∩ J and K = X + Y. Note that Y = Xg and thus by Lemma 3.2, we have K ≅ A_{2} ⊗ (\frac{1}{\sqrt{2}}X) .

Let {α, α′} be a set of fundamental roots for A_{2} and denote α′′ = −(α + α′). Let g′ be the isometry of A_{2} which is induced by the map α → α′ → α′′ → α.

By identifying K with A_{2}⊗(\frac{1}{\sqrt{2}}X), we may assume X = M ∩J = ℤα⊗(\frac{1}{\sqrt{2}}X).

Recall that (x, x′g) = −\frac{1}{2}(x, x′) for any x, x′ ∈ K (cf. (3.2)). Therefore, for any β ∈ \frac{1}{\sqrt{2}}X, we may identify (α ⊗ β)g with α′ ⊗ β = αg′ ⊗ β and identify Y = Xg

with αg′ ⊗ (\frac{1}{\sqrt{2}}X).

Lemma 6.11. We have J = L(g − 1) + K, where K = J ∩ M + J ∩ N as in (6.10). The map g − 1 takes L onto J and induces an isomorphism of L/(J + F) and J/K, as abelian groups. In particular, both quotients have order 3^{s}.

Proof. Part 1: The map g − 1 induces a monomorphism. Clearly, L(g − 1) ≤ J and g acts trivially on L/L(g − 1). Obviously, F (g − 1) = 0. We also have L ≥ J + F ≥ K + F. Since M ∩ J ≤ K, t acts trivially on J/K. Therefore, so does g, whence J(g − 1) ≤ K. Since g acts trivially on L/J, L(g − 1)^{2} ≤ K.

Furthermore, (g − 1)^{2} annihilates L/(F + K), which is a quotient of L/F, where the action of g has minimal polynomial x^{2}+x+1. Therefore L/(F +K) is annihilated by 3g, so is an elementary abelian 3-group. We have 3L ≤ F + K.

Let P := {x ∈ L|x(g − 1) ∈ K}. Then P is a sublattice and F + J ≤ P ≤ L. By coprimeness, there are sublattices P^{+}, P^{−} so that P^{+} ∩ P^{−} = F + K and P^{+} + P^{−} = P and t acts on P^{ε}/(F + K) as the scalar ε = ±1. We shall prove now that P^{−} = F + K and P^{+} = F + J. We already know that P^{−} ≥ F + K and P^{+} ≥ F + J.

Let v ∈ P^{−} and suppose that v(g−1) ∈ K. Then v(g^{2} −g) ∈ K and this element is fixed by t. Therefore, v(g^{2} − g) ∈ ann_{K}(M ∩ J). By (3.2), there is u ∈ M ∩ J so that u(g^{2} − g) = v(g^{2} − g). Then u − v ∈ L is fixed by g and so u − v ∈ F. Since u ∈ K, v ∈ F + K. We have proved that P^{−} = F + K.

Now let v ∈ P^{+}. Assume that v ∉ F + J. Since D acts on L/J such that g acts trivially, coprimeness of |L/J| and |D/⟨g⟩| implies that L has a quotient of order 3 on which t and u act trivially. Since L = M + N, this is not possible. We conclude that F + J = P^{+}.

We conclude that P = P^{−} + P^{+} = F + J and so g − 1 gives an embedding of L/(J + F) into J/K.

Part 2: The map g −1 induces an epimorphism. We know that L/(F +J) ≅ 3^{s} and this quotient injects into J/K. We now prove that J/K has order bounded by 3^{s}.

Consider the possibility that t negates a nontrivial element x + K of J/K. By (A.7), we may assume that xt = −x. But then x ∈ M ∩ J ≤ K, a contradiction. Therefore, t acts trivially on the quotient J/K. It follows that the quotient J/K is covered by J^{+}(t). Therefore J/K embeds in the discriminant group of K^{+}(t), which by (3.2) is isometric to \sqrt{3}(M ∩ J). Since J/K is an elementary abelian 3-group and D(M ∩ J) ≅ 3^{s} × 2^{rank(M∩J)}, the embedding takes J/K to the Sylow 3-group of D(M ∩ J), which is isomorphic to 3^{s} (see (6.9) and use (A.13), applied to \frac{1}{\sqrt{2}}M and the sublattices \frac{1}{\sqrt{2}}F and \frac{1}{\sqrt{2}}(M ∩ J). □

Proposition 6.12. If L is rootless and F ≠ 0, then s = 1 and F ≅ AA_{2}. Also, L/(J ⊥ F) ≅ 3.

Proof. We have s ≤ 3, so by Proposition D.8, F ≅ AA_{2}, EE_{6} or AA_{2} ⊥ AA_{2} and s = 1, 1 or 2, respectively. Note X = M ∩ J is the sublattice of M which is orthogonal to F. Since M ≅ EE_{8}, X ≅ EE_{6}, AA_{2} and AA_{2} ⊥ AA_{2} if F ≅ AA_{2}, EE_{6} and AA_{2} ⊥ AA_{2}, respectively.

We shall show that L has roots if F ≅ EE_{6} or AA_{2} ⊥ AA_{2}. The conclusion in the surviving case follows from (6.9).

Case 1: F = EE_{6} and s = 1. In this case, X ≅ Y ≅ AA_{2}. Hence K ≅ A_{2} ⊗ A_{2}. As in Notation 6.10, we shall identify X with ℤα ⊗ A_{2} and Y with ℤαg′ ⊗ A_{2}. Then F ⊥ X ≅ EE_{6} ⊥ ℤ(α ⊗ A_{2}).

In this case, |M/(F + X)| = 3 and there exist γ ∈ (EE_{6})^{∗} and γ′ ∈ (A_{2})^{∗} with (γ, γ) = 8/3 and (γ′, γ′) = 2/3 such that M = span_{ℤ}{F + X, γ + α ⊗ γ′}. Then N = Mg = span_{ℤ}{F + Y, γ + αg′ ⊗ γ′} and we have

L = M + N ≅ span_{ℤ}{EE_{6} ⊥ (A_{2} ⊗ A_{2}), γ + (α ⊗ γ′), γ + (αg′ ⊗ γ′)}

Let β := (γ + (α ⊗ γ′)) − (γ + (αg′ ⊗ γ′)) = (α − αg′) ⊗ γ′. Then (β, β) = (α − αg′, α − αg′) · (γ′, γ′) = 6 · 2/3 = 4.

Let α_{1} be a root of A_{2} such that (α_{1}, γ′) = −1. Then α ⊗ α_{1} ∈ A_{2} ⊗ A_{2}, where α is in the first tensor factor and α_{1} is in the second tensor factor. Then (β, α⊗α_{1}) = (α−αg′, α)·(γ′, α_{1}) = (2+1)·(−1) = −3 and the norm of β +(α⊗α_{1}) is given by

(β +(α⊗α_{1}), β +(α⊗α_{1})) = (β, β)+(α⊗α_{1}, α⊗α_{1})+2(β, α⊗α_{1}) = 4+4−6 = 2.

Thus, a_{1} = β + (α ⊗ α_{1}) is a root in J. So, L has roots. In fact, we can say more. If we take a_{2} = βg + αg′ ⊗ α_{1}, then a_{2} is also a root and

(a_{1}, a_{2}) = (β + α ⊗ α_{1}, βg + αg′ ⊗ α_{1})

= (β, βg) + (β, αg′ ⊗ α_{1}) + (α ⊗ α_{1}, βg) + (α ⊗ α_{1}, αg′ ⊗ α_{1})

= −\frac{1}{2}(4 − 3 − 3 + 4) = −1.

Thus, a_{1}, a_{2} spans a sublattice A isometric to A_{2}.

Case 2: F = AA_{2} ⊥ AA_{2} and s = 2. In this case, X ≅ Y ≅ AA_{2} ⊥ AA_{2}. Hence, K ≅ A_{2} ⊗ (A_{2} ⊥ A_{2}). Again, we shall identify X with ℤα ⊗ (A_{2} ⊥ A_{2}) and F + X ≅ AA_{2} ⊥ AA_{2} ⊥ ℤα ⊗ (A_{2} ⊥ A_{2}) ≅ AA_{2} ⊥ AA_{2} ⊥ AA_{2} ⊥ AA_{2}. For convenience, we shall use a 4-tuple (ξ_{1}, ξ_{2}, ξ_{3}, ξ_{4}) to denote an element in (F + K)^{∗} ≅ (AA_{2})^{∗} ⊥ (AA_{2})^{∗} ⊥ (A_{2} ⊗ A_{2})^{∗} ⊥ (A_{2} ⊗ A_{2})^{∗}, where ξ_{1}, ξ_{2} ∈ (AA_{2})^{∗} and ξ_{3}, ξ_{4} ∈ (A_{2} ⊗ A_{2})^{∗}.

Recall that |M/(F +X)| = 3^{2} and the cosets of M/(F +X) can be parametrized by the tetracode (cf. [CS, p. 200]) whose generating matrix is given by

(\begin{matrix}1 & 1 & 1 & 0 \\ 1 & −1 & 0 & 1 \end{matrix}) .

Hence, there exists a element γ ∈ (AA_{2})^{∗} with (γ, γ) = 4/3 and γ′ ∈ (A_{2})^{∗} with (γ′, γ′) = 2/3 such that

M = span_{ℤ} \left\{{AA_{2} ⊥ AA_{2} ⊥ ℤα ⊗ (A_{2} ⊥ A_{2}), \atop (γ, γ, α ⊗ γ′, 0), (γ, −γ, 0, α ⊗ γ′)}\right\} .

Therefore, we also have

N = Mg = span_{ℤ} \left\{{AA_{2} ⊥ AA_{2} ⊥ ℤαg′ ⊗ (A_{2} ⊥ A_{2}), \atop (γ, γ, αg′ ⊗ γ′, 0), (γ, −γ, 0, αg′ ⊗ γ′)}\right\}

and

\begin{matrix}L = M + N = \\ span_{ℤ} \left\{{AA_{2} ⊥ AA_{2} ⊥ A_{2} ⊗ (A_{2} ⊥ A_{2}), (γ, γ, α ⊗ γ′, 0), \atop (γ, −γ, 0, α ⊗ γ′), (γ, γ, αg′ ⊗ γ′, 0), (γ, −γ, 0, αg′ ⊗ γ′)}\right\} . \end{matrix}

Let β_{1} = (γ, γ, α ⊗ γ′, 0) − (γ, γ, αg′ ⊗ γ′, 0) and β_{2} = (γ, −γ, 0, α ⊗ γ′) − (γ, −γ, 0, αg′ ⊗ γ′). Then β_{1}, β_{2} ∈ L(g − 1) ≤ J and both have norm 4.

Let α_{1} and α_{2} be roots in A_{2} such that (α_{1}, γ′) = (α_{2}, γ′) = −1 and (α_{1}, α_{2}) = 1. Denote

a_{1}^{1} = β_{1} + (0, 0, α ⊗ α_{1}, 0), a_{2}^{1} = β_{1}g′ + (0, 0, αg′ ⊗ α_{1}, 0),
a_{1}^{2} = β_{1} + (0, 0, α ⊗ α_{2}, 0), a_{2}^{2} = β_{1}g′ + (0, 0, αg′ ⊗ α_{2}, 0),
a_{1}^{3} = β_{2} + (0, 0, 0, α ⊗ α_{1}), a_{2}^{3} = β_{2}g′ + (0, 0, 0, αg′ ⊗ α_{1}),
a_{1}^{4} = β_{2} + (0, 0, 0, α ⊗ α_{2}), a_{2}^{4} = β_{2}g′ + (0, 0, 0, αg′ ⊗ α_{2}).

Then similar to Case 1, we have the inner products

(a_{1}^{i}, a_{1}^{i}) = (a_{2}^{i}, a_{2}^{i}) = 2, and (a_{1}^{i}, a_{2}^{i}) = −1,

for any i = 1, 2, 3, 4. Thus, each pair {a_{1}^{i}, a_{2}^{i}}, for i = 1, 2, 3, 4, spans a sublattice isometric to A_{2}. Moreover, (a_{k}^{i}, a_{ℓ}^{j}) = 0 for any i ≠ j and k, ℓ ∈ {1, 2}. Therefore, J ≥ A_{2} ⊥ A_{2} ⊥ A_{2} ⊥ A_{2}. Moreover, |span_{ℤ}{a_{1}^{i}, a_{2}^{i}| i = 1, 2, 3, 4}/K| = 3^{2} and hence J ≅ A_{2} ⊥ A_{2} ⊥ A_{2} ⊥ A_{2}. Again, L has roots. □

Corollary 6.13. |J : M ∩ J + N ∩ J| = 3 and M ∩ J = ann_{M}(F) ≅ EE_{6}.

Proof. (6.11) and (6.12). □

Corollary 6.14. (i) M ∩ J + N ∩ J is isometric to A_{2} ⊗ E_{6}. (ii) L = M + N is unique up to isometry.

Proof. For (i), use (3.2) and for (ii), use (4.1). □

Lemma 6.15. If v = v_{1} + v_{2} with v_{1} ∈ J^{∗} and v_{2} ∈ F^{∗}, then v_{2} has norm \frac{4}{3} and v_{1} has norm in \frac{8}{3} + 2ℤ.

Proof. Since 3v_{2} ∈ F, we may assume that 3v_{2} has norm 12 by (D.6) so that v_{2} has norm \frac{4}{3}. It follows that v_{1} has norm in \frac{8}{3} + 2ℤ. □

6.1.1 DIH_{6}: Explicit glueing

In this subsection, we shall describe the explicit glueing from F + M ∩ J + N ∩ J to L. As in Notation 6.10, X = M ∩ J, Y = N ∩ J and K = X + Y. Since F ≅ AA_{2}, we have X ≅ Y ≅ EE_{6} and K ≅ A_{2} ⊗ E_{6}. We also identify X with ℤα ⊗ (\frac{1}{\sqrt{2}}X) and Y with ℤαg′ ⊗ (\frac{1}{\sqrt{2}}X), where α is a root of A_{2} and g′ is a fixed point free isometry of A_{2} such that αg′ ⊗ β = (α ⊗ β)g as described in (6.10). Then F ⊥ X ≅ AA_{2} ⊥ ℤα ⊗ E_{6} ≅ AA_{2} ⊥ EE_{6}.

Recall that (AA_{2})^{∗}/AA_{2} ≅ ℤ_{2}^{2} × ℤ_{3}. Therefore, 2(AA_{2})^{∗}/AA_{2} is the unique subgroup of order 3 in (AA_{2})^{∗}/AA_{2}. Similarly, 2(ℤα⊗E_{6})^{∗}/(ℤα⊗E_{6}) is the unique subgroup of order 3 in (ℤα ⊗ E_{6})^{∗}/(ℤα ⊗ E_{6}) ≅ ℤ_{2}^{6} × ℤ_{3}.

Notation 6.16. Since F ⊥ X ≤ M and |M : F ⊥ X| = 3, there exists an element μ ∈ F^{∗} ⊥ X^{∗} such that 3μ ∈ F + X and M = span_{ℤ}{F + X, μ}. Let γ ∈ (AA_{2})^{∗} be a representative of the generator of the order 3 subgroup in 2(AA_{2})^{∗}/AA_{2} and γ′ a representative of the generator of the order 3 subgroup in (E_{6})^{∗}/E_{6}. Without loss, we may choose γ and γ′ so that (γ, γ) = 4/3 and (γ′, γ′) = 4/3. Since the image of μ in M/(F ⊥ X) is of order 3, it is easy to see that

μ ≡ ±(γ + α ⊗ γ′) or μ ≡ ±(γ − α ⊗ γ′) modulo F ⊥ X.

By replacing μ by −μ and γ′ by −γ′ if necessary, we may assume μ = γ + α ⊗ γ′. Then ν := μg = γ + αg′ ⊗ γ′ and N = span_{ℤ}{F + Y, ν}.

Proposition 6.17. With the notation as in (6.16), L = M + N ≅ span_{ℤ}{AA_{2} ⊥ A_{2} ⊗ E_{6}, γ + α ⊗ γ′, γ + αg′ ⊗ γ′}.

Remark 6.18. Let β = (α − αg′) ⊗ γ′ = (γ + α ⊗ γ′) − (γ + αg′ ⊗ γ′). Then β ∈ L(g − 1) = J = ann_{L}(F) but β = (α − αg′) ⊗ γ′ ∉ K ≅ A_{2} ⊗ E_{6}. Hence J = span_{ℤ}{β, K} as |J : K| = 3. Note also that (β, β) = 6 · 4/3 = 8.

Lemma 6.19. J^{+}(t) = ann_{J}(M ∩ J) ≅ \sqrt{6}E_{6}^{∗}.

Proof. By Remark (6.18), we have J = span_{ℤ}{β, K}, where β = (α − αg′) ⊗ γ′ and K = M∩J+N∩J ≅ A_{2}⊗E_{6}. Recall that M∩J is identified with ℤα⊗E_{6} and N∩J is identified with ℤαg′⊗E_{6}. Thus, by (3.2), ann_{K}(M∩J) = ℤ(αg′−αg^{′2})⊗E_{6} ≅ \sqrt{6}E_{6}. Since (α, αg′ − αg^{′2}) = 0, βg = (αg′ − αg^{′2}) ⊗ γ′ also annihilates M ∩ J. Therefore, J^{+}(t) = ann_{J}(M ∩ J) ≥ span_{ℤ}{ann_{K}(M ∩ J), βg}. Since γ′ + E_{6} is a generator of E_{6}^{∗}/E_{6}, we have span_{ℤ}{E_{6}, γ′} = E_{6}^{∗} and hence

span_{ℤ}\{ann_{K}(M ∩ J), βg\} = ℤ(αg′ − αg^{′2}) ⊗ span_{ℤ}\{E_{6}, γ′\} ≅ \sqrt{6}E_{6}^{∗}.

Note that (αg′ −αg^{′2}) has norm 6. Now by the index formula, we have det(J^{+}(t)) = 2^{6} × 3^{5} = det(\sqrt{6}E_{6}^{∗}) and thus, we have J^{+}(t) = ann_{J}(M ∩ J) ≅ \sqrt{6}E_{6}^{∗}. □

Corollary 6.20. J is isometric to the Coxeter-Todd lattice. Each of these is not properly contained in an integral, rootless lattice.

Proof. This is an extension of the result (D.35). Embed J in J′, a lattice satisfying (D.34) and embed the Coxeter-Todd lattice P in a lattice Q satisfying (D.34). Then both J′ and Q satisfy the hypotheses of (D.34), so are isometric. Since det(J) = det(P) = 3^{6}, J ≅ P. □

6.2 DIH_{12}

Notation 6.21. Let M, N be lattices isometric to EE_{8} such that their respective associated involutions t_{M}, t_{N} generate D ≅ Dih_{12}. Let h := t_{M}t_{N} and g := h^{2}. Let z := h^{3}, the central involution of D. We shall make use of the DIH_{6} results by working with the pair of distinct subgroups D_{M} := D_{t_{M}} := ⟨t_{M}, t_{M}^{g}⟩ and D_{N} := D_{t_{N}} := ⟨t_{N}, t_{N}^{g}⟩. Note that each of these groups is normal in D since each has index 2. Define M̃ := Mg, Ñ := Ng. If X is one of M, N, we denote by L_{X}, J_{X}, F_{X} the lattices L := X + X̃, J, F associated to the pair X, X̃, denoted “M” and “N” in the DIH_{6} section. We define K_{X} := (X ∩ J) + (X ∩ J)g, a D_{X}-submodule of J_{X}. Finally, we define F := F_{D} to be {x ∈ L|xg = g} and J := J_{D} := ann_{L}(F). We assume L is rootless.

Lemma 6.22. An element of order 3 in D has commutator space of dimension 12.

Proof. The analysis of DIH_{6} shows just two possibilities in case of no roots (cf. Lemma (6.4) and Proposition (6.12)). We suppose that g − 1 has rank 16, then derive a contradiction.

From (6.4), M + Mg ≅ A_{2} ⊗ E_{8}. From (3.5), there are only three involutions in O(A_{2} ⊗ E_{8}) which have negated space isometric to EE_{8}. Therefore, L > M + Mg. Since D(M + Mg) ≅ 3^{8}, L/(M + Mg) is an elementary abelian 3-group of rank at most 4, which is totally singular in the natural \frac{1}{3}ℤ/ℤ-valued bilinear form. Note also that L is invariant under the isometry of order 3 on A_{2} ⊗ E_{8} coming from the natural action of O(A_{2})×{1} on the tensor product. We now obtain a contradiction from (A.18) since L is rootless. □

Corollary 6.23. J_{M} ≅ J_{N} has rank 12 and F_{M} ≅ F_{N} ≅ AA_{2}.

Proof. Use (6.12) and (6.22). □

Corollary 6.24. The lattice J has rank 12 and contains each of J_{M} and J_{N} with finite index. The lattice F = F_{D} has rank 2, 3 or 4 and F/(F_{M} + F_{N}) is an elementary abelian 2-group.

Proof. Use (6.23) and (A.2), which implies that F/(F_{M} +F_{N}) is elementary abelian. □

Notation 6.25. Set t := t_{M} and u := t_{N}.

Lemma 6.26. The involutions t^{g} and u commute, and in fact t^{g}u = z.

Proof. This is a calculation in the dihedral group of order 12. See (6.21), (6.25). We have h = tu and h^{3} = z, so t^{g} = t^{h^{2}} = utut · t · tutu = utututu = uh^{3} = uz. □

We now study how t_{N} acts on the lattice J.

Lemma 6.27. For X = M or N, J_{X} and K_{X} are D-submodules.

Proof. Clearly, t^{g} fixes L_{M} = M +Mg, F_{M}, J_{M} and K_{M} = (M ∩J_{M})+(M ∩J_{M})g, the D_{M}-submodule of J generated by the negated spaces of all involutions of D_{M}. Since t_{N} = u = t^{g}z, it suffices to show that the central involution z fixes all these sublattices, but that is trivial. □

Lemma 6.28. The action of t on J/J_{M} is trivial.

Proof. Use (A.7) and the fact that M ∩ J ≤ J_{M}. □

6.2.1 DIH_{12}: Study of J_{M} and J_{N}

We work out some general points about F_{M}, F_{N}, J_{M}, J_{N}, K_{M} and K_{N}. We continue to use the hypothesis that L has no roots.

Lemma 6.29. As in (6.25), t = t_{M}, u = t_{N}.

(i) J(g − 1) ≤ J_{M} ∩ J_{N} and J/(J_{M} ∩ J_{N}) is an elementary abelian 3-group of rank at most \frac{1}{2}rank(J) = 6. Also, J/(J_{M} ∩ J_{N}) is a trivial D-module.

(ii) J = J_{M} + J_{N} = L(g − 1).

Proof. (i) Observe that g acts on J/J_{M} and that t acts trivially on this quotient (6.28). Since g is inverted by t, g acts trivially on J/J_{M}. A similar argument with u proves g acts trivially on J/J_{N}. Therefore, J(g − 1) ≤ J_{M} ∩ J_{N}. Since (g − 1)^{2} acts on J as −3g, (i) follows.

(ii) We observe that L_{X}(g − 1) ≤ J_{X}, for X = M, N. Since L = M + N, L(g − 1) ≤ J_{M} + J_{N}. The right side is contained in J = ann_{L}(F). Suppose that L(g − 1) ⪇ J. Then J/L(g − 1) is a nonzero 3-group which is a trivial module for D. It follows that L/(F + L(g − 1)) is an elementary abelian 3-group which has a quotient of order 3 and is a trivial D-module. This is impossible since L = M + N. So, J = J_{M} + J_{N} = L(g − 1). □

Lemma 6.30. F = F_{M} + F_{N}.

Proof. Since F is the sublattice of fixed points for g. Then F is a direct summand of L and is D-invariant. Also, D acts on F as a four-group and T el(F, D) has finite index in F. If E is any D-invariant 1-space in F, t or u negates E (because L = M +N). Therefore, F_{M} +F_{N} has finite index in F and is in fact 2-coelementary abelian (A.2).

Consider the possibility that F_{M} + F_{N} ⪇ F, i.e., that F_{M} + F_{N} is not a direct summand. Since F_{M} and F_{N} are direct summands, there are α ∈ F_{M}, β ∈ F_{N} so that \frac{1}{2}(α+β) ∈ L but \frac{1}{2}α and \frac{1}{2}β are not in L. Since by (6.23) F_{M} ≅ F_{N} ≅ AA_{2}, we may assume that α, β each have norm 4. Then by Cauchy-Schwartz, \frac{1}{2}(α + β) has norm at most 4 and, if equal to 4, α and β are equal. But then, \frac{1}{2}(α + β) = α ∈ L, a contradiction. □

6.2.2 DIH_{12}: the structure of F

Lemma 6.31. Suppose that F_{M} ≠ F_{N}. Then F_{M} ∩ F_{N} is 0 or has rank 1 and is spanned by a vector of norm 4 or norm 12.

Proof. Since F_{M} and F_{N} are summands, F_{M} ≠ F_{N} implies rank(F_{M} + F_{N}) ≥ 3, so F_{M} + F_{N} has rank 3 or 4. Assume that F_{M} ∩ F_{N} has neither rank 0 or rank 2. Since O(F_{M}) ≅ Sym_{3} × 2 and F_{M} ∩ F_{N} is an eigenlattice in F_{M} for the involution t_{N}, it must be spanned by a norm 4 vector or is the annihilator of a norm 4 vector. □

Lemma 6.32. Suppose that the distinct vectors u, v have norm 4 and u, v are in L \ (J + F). We suppose that u ∈ L_{M} \ (J + F) and v ∈ L_{N} \ (J + F). Write u = u_{1} + u_{2} and v = v_{1} + v_{2}, where u_{1}, v_{1} ∈ span_{ℝ}(J) and u_{2}, v_{2} ∈ span_{ℝ}(F). We suppose that each of u_{1}, u_{2}, v_{1}, v_{2} are nonzero. Then

(i) u_{1} and v_{1} have norm \frac{8}{3};

(ii) u_{2} and v_{2} have norm \frac{4}{3}.

Proof. Since 4 = (u, u) = (u_{1}, u_{1}) + (u_{2}, u_{2}), (i) follows from (ii). To prove (ii), use the fact that L/(J ⊥ F) is a 3-group, a rescaled version of (D.6) and (u_{2}, u_{2}) < (u, u). □

Lemma 6.33. (i) Suppose that F_{M} ∩ F_{N} = ℤu, where u ≠ 0. Let v span ann_{F_{M}} (u) and let w span ann_{F_{N}} (u). Then F = span\{u, v, w, \frac{1}{2}(u+v), \frac{1}{2}(u+w), \frac{1}{2}(v +w)\} and (u, u) = 4, (v, v) = 12 = (w, w).

(ii) If the rank of O_{3}(D(F)) is at least 2, then

(a) F = F_{M} ⊥ F_{N} (rank 4); or

(b) F has rank 3 and the number of nontrivial cosets of O_{3}(D(F)) ≅ 3 × 3 which have representing elements whose norms lie in \frac{4}{3} + 2ℤ, respectively \frac{8}{3} + 2ℤ are 4 and 4, respectively.

Proof. (i) By (6.31), u has norm 4 or 12. The listed generators span F = F_{M} +F_{N} since F_{M} = span\{u, v, \frac{1}{2}(u + v)\} and F_{N} = span\{u, w, \frac{1}{2}(u + w)\}. If (u, u) = 12, then v and w have norm 4 and \frac{1}{2}(v + w) is a root in F, whereas L is rootless. So, (u, u) = 4 and (v, v) = 12 = (w, w).

Now we prove (ii). Since we have already discussed the case of rank(F) equal to 2 and 4, we assume rank(F) = 3, for which we may use the earlier results. In the above notation, we may assume that F = span\{u, v, w, \frac{1}{2}(u+v), \frac{1}{2}(u+w), \frac{1}{2}(v +w)\} and (u, u) = 4 and (v, v) = 12 = (w, w). Then O_{3}(D(F)) ≅ 3^{2} and the pair of elements \frac{1}{3}v, \frac{1}{3}w map modulo F to generators of O_{3}(D(F)). Since their norms are \frac{4}{3}, \frac{4}{3} and they are orthogonal, the rest of (ii) follows. □

6.2.3 DIH_{12}: A comparison of eigenlattices

Notation 6.34. Let ν be the usual additive 3-adic valuation on ℚ, with ν(3^{k}) = k. Set P := Mg ∩ J, K := P + P g = ℤ[⟨g⟩]P, R := ann_{K}(P). Note that P is the (−1)-eigenlattice of t^{g} in both J and K while R is the (+1)-eigenlattice of t^{g} in K.

We study the actions of u on J, J_{M}, P, K and R.

Lemma 6.35. J = J_{M} = J_{N} and D(J) ≅ 3^{6}.

Proof. Since J contains J_{M} with finite index, we may use (6.20). □

Notation 6.36. Define integers r := rank(P^{+}(u)), s := rank(P^{−}(u)). We have r = rank(R^{−}(u)), s = rank(R^{+}(u)) and r + s = 6.

Corollary 6.37. (r, s) ∈ {(2, 4), (4, 2), (6, 0)}.

Proof. Since P^{−} + R^{−} has finite index 3^{p} in J^{−}(u) = N ∩ J ≅ EE_{6} and det(P^{−} + R^{−}) = 2^{m}3^{1+r}, for some m ≥ 6, the determinant index formula implies that r is even. Similarly, we get s is even. If r = 0, then s = 6 and J ∩ Mg = J ∩ N, and so z := t^{g}u is the identity on J. Since t^{g} ≠ u, we have a contradiction to the DIH_{4} theory since the common negated space for t^{g} and u is at least 6-dimensional. So, r ≠ 0. □

6.3 s = 0

Lemma 6.38. If s = 0, then the pair Mg, N is in case DIH_{4}(15) or DIH_{4}(16).

Proof. In this case, Mg∩N is RSSD in F_{N} so is isometric to AA_{2} or AA_{1} or \sqrt{6}A_{1}. Then DIH_{4} theory implies that Mg ∩ N is isometric to AA_{1} or 0. □

Lemma 6.39. s ≠ 0.

Proof. Suppose that s = 0. Then u acts as 1 on Mg ∩ J. The sublattice (Mg ∩ J) ⊥ (N ∩ J) has determinant 2^{12}3^{2} is contained in T el(J, u), which has determinant 2^{12}det(J) = 2^{12}3^{6}. Since Mg ∩ J = J^{−}(t^{g}), it follows from determinant considerations that N ∩ J is contained with index 3^{2} in J^{+}(t^{g}). Since s = 0, J^{+}(t^{g}) ≤ J^{−}(u) = N ∩ J and we have a contradiction. □

6.4 s ∈ {2, 4}

Lemma 6.40. If s > 0 and det(P^{−}(u)) is not a power of 2, then s = 2 and P^{−}(u) ≅ 2A_{2} and the pair Mg, N is in case DIH_{4}(12).

Proof. Since P^{−}(u) is RSSD in P and det(P^{−}(u)) is not a power of 2, (D.16) implies that P^{−}(u) ≅ AA_{5} or 2A_{2} or (2A_{2})(AA_{1})^{m}. Since s = rank(P^{−}(u)) is 2 or 4, we have P^{−}(u) ≅ 2A_{2} or (2A_{2})(AA_{1})^{2}. By DIH_{4} theory, Mg ∩ N ≅ DD_{4}. Since P^{−}(u) is contained in Mg ∩ N, P^{−}(u) ≅ 2A_{2}. □

Lemma 6.41. (i) If s > 0 and det(P^{−}(u)) is a power of 2, then s = 2 and P^{−}(u) ≅ AA_{1} ⊥ AA_{1} or s = 4 and P^{−}(u) ≅ DD_{4}.

(ii) If P^{−}(u) ≅ DD_{4}, the pair Mg, N is in case DIH_{4}(12).

(iii) If P^{−}(u) ≅ AA_{1} ⊥ AA_{1}, the pair Mg, N is in case DIH_{4}(14). In particular, F_{M} ∩ F_{N} = 0 and so F = F_{M} ⊥ F_{N}.

Proof. (i) Use (D.16) and evenness of s.

(ii) This follows from DIH_{4} theory since Mg ∩ N contains a copy of DD_{4} and Mg ≠ N.

(iii) Since dim(P^{−}(u)) = 2, it suffices by DIH_{4} theory to prove that dim(Mg ∩ N) ≠ 4. Assume by way of contradiction that dim(Mg ∩ N) = 4. Then Mg ∩ N ≅ DD_{4} and rank(Mg ∩ N ∩ F) = 2. This means that F = F_{M} = F_{N} ≅ AA2. Therefore, Mg ∩ N ≅ DD_{4} contains the sublattice P^{−}(u) ⊥ F ≅ AA_{1} ⊥ AA_{1} ⊥ AA_{2}, which is impossible. □

Lemma 6.42. Suppose that P^{−}(u) is isometric to 2A_{2}. Then rank(L) = 14 and L has roots.

Proof. We have det(P^{−}(u) ⊥ R^{−}(u)) = 2^{2}3 · 2^{6}3^{4} = 2^{8}3^{5}. Since N ∩ J covers J/K_{M}, the determinant formula implies that |J : K_{M}| = 3^{2} and so det(J) = 3^{4}. Now use (6.20). □

Lemma 6.43. P^{−}(u) is not isometric to AA_{1} ⊥ AA_{1}.

Proof. Suppose P^{−}(u) ≅ AA_{1} ⊥ AA_{1}. Then, P^{+}(u) ≅ \sqrt{2}Q, where Q is the rank 4 lattice which is described in (D.19). Also, R^{+} ≅ \sqrt{6}A_{1}^{2} and R^{−} ≅ \sqrt{6}Q. Then P^{−}(u) ⊥ R^{−}(u) embeds in EE_{6}. Since sublattices of E_{6} which are isometric to A_{1}^{2} are in a single orbit under O(E_{6}), it follows that \sqrt{3}Q embeds in Q. However, this is in contradiction with (D.24). □

To summarizes our conclusion, we have the proposition.

Proposition 6.44. P^{−}(u) = Mg ∩ N ≅ DD4 and the pair is in case DIH_{4}(12).

6.5 Uniqueness of the case DIH_{12}(16)

As in other sections, we aim to use (4.1) for the case (6.41)(ii).

The input M, N determines the dihedral group ⟨t, u⟩ and therefore Mg and Mg + N. By DIH_{4} theory, the isometry type of Mg + N is determined up to isometry. Since Mg + N has finite index in M + N, M + N is determined by (4.1). Thus, Theorem (6.1) is proved.

7 DIH_{10} theory

Notation 7.1. Define t := t_{M}, h := t_{M}t_{N}. We suppose h has order 5. Let g := h^{3}. Then g also has order 5 and D :=< t_{M}, t_{M} >=< t, g >. In addition, we have N = Mg. Define F := M ∩ N, J := ann_{L}(F). Note that F is the common negated lattice for t_{M} and t_{N} in L, so is the fixed point sublattice for g and is a direct summand of L (A.10).

Definition 7.2. Define the integer s by 5^{s} := |L/(J + F)|.

Lemma 7.3. Equivalent are (i) L = J + F; (ii) s = 0; (iii) F = 0; (iv) J = L.

Proof. Trivially, L = J + F and s = 0 are equivalent. These conditions follow if F = 0 or if J = 0 (but the latter does not happen since g has order 5). If L = J +F holds, then M = (J ∩ M) ⊥ F which implies that F = 0 and J = L since M ≅ EE_{8} is orthogonally decomposable. □

Lemma 7.4. (i) g acts trivially on both F and L/J.

(ii) g − 1 induces an embedding L/F → J.

(iii) g − 1 induces an embedding L/(J + F) → J/J(g − 1), whose rank is at most \frac{1}{4}rank(J) since (g − 1)^{4} induces the map 5w on J, where w = −g^{2} + g − 1 induces an invertible linear map on J.

(iv) s ≤ \frac{1}{4}rank(J), so that s = 0 and F = 0 or s ∈ \{1, 2, 3, 4\} and F ≠ 0.

(v) The inclusion M ≤ L induces an isomorphism M/((M ∩ J) + F) ≅ L/(J + F) ≅ 5^{s}, an elementary abelian group.

Proof. (i) and (ii) are trivial.

(iii) This is equivalent to some known behavior in the ring of integers ℤ[e^{2πi/5}], but we give a self-contained proof here. We calculate (g − 1)^{4} = g^{4} − 4g^{3} + 6g^{2} − 4g + 1 = (g^{4} + g^{3} + g^{2} + g + 1) + 5w, which in End(J) is congruent to 5w. Note that the images of g + 1 and g^{3} + 1 are non zero-divisors (e.g., because (g + 1)(g^{4} − g^{3} + g^{2} − g + 1) = g^{5} + 1 = 2, and 2 is a non zero-divisor) and are associates in End(J) so that their ratio w is a unit. For background, we mention [GHig].

(iv) The Jordan canonical form for the action of g − 1 on J/5J is a direct sum of degree 4 indecomposable blocks, by (iii), since (g − 1)^{4} has determinant 5^{rank(J)}. Since the action of g on L/J is trivial, s ≤ \frac{1}{4}rank(J). Since rank(J) ≤ 16, s ≤ 4. For the case s = 0, see (7.3).

(v) Since N = Mg, N and M are congruent modulo J. Therefore L = M +N = M + J and so 5^{s} ≅ L/(J + F) = (M + J)/(J + F) = (M + (J + F))/(J + F) ≅ M/(M ∩(J +F)) (by a basic isomorphism theorem) and this equals M/((M ∩J)+F) (by the Dedekind law).

Since L(g−1) ≤ J, (g−1) annihilates L/(J +F). Since (g−1)^{4} takes (L/(J +F)) to 5(L/(J + F)), it follows that 5L ≤ J + F. That is, L/(J + F) is an elementary abelian 5-group. □

Lemma 7.5. s = 0, 1, 2 or 3 and F = M ∩ N ≅ 0, AA_{4}, \sqrt{2}M(4, 25) or \sqrt{2}A_{4}(1).

Proof. We have that \frac{1}{\sqrt{2}}M ≅ E_{8}. The natural map of \frac{1}{\sqrt{2}}M to D(\frac{1}{\sqrt{2}}F) is onto and has kernel \frac{1}{\sqrt{2}} ((M ∩ J) ⊥ F)). Therefore, D(\frac{1}{\sqrt{2}}F) ≅ 5^{s} is elementary abelian. Now apply (D.9) to get the possibilities for \frac{1}{\sqrt{2}}F and hence for F. Note that M = N is impossible here, since t_{M} ≠ t_{N}. □

7.1 DIH_{10}: Which ones are rootless?

From Lemma 7.5, s = 0, 1, 2 or 3. We shall eliminate the case s = 1, s = 2 and s = 3, proving that s = 0 and F = 0.

Lemma 7.6. If L = M + N is integral and rootless, then F = M ∩ N = 0.

Proof. By Lemma 7.5, we know that M ∩ N ≅ 0, AA_{4} \sqrt{2}M(4, 25) or \sqrt{2}A_{4}(1) since M ≠ N. We shall eliminate the cases M ∩ N ≅ AA_{4}, \sqrt{2}M(4, 25) and \sqrt{2}A_{4}(1).

Case: F = M ∩ N ≅ AA_{4}. In this case, M ∩ J ≅ N ∩ J ≅ AA_{4}. Therefore, there exist α ∈ F^{∗} and β ∈ (M ∩ J)^{∗} such that M = span_{ℤ}{F + (M ∩ J), α + β}. Without loss, we may assume (α, α) = 12/5 and (β, β) = 8/5. Let γ = βg. Then, (α + β)g = α + γ ∈ N and we have N = span_{ℤ}{F + (N ∩ J), α + γ}. Since L is integral and rootless and since α + β ∈ L has norm 4, by (D.11),

0 ≥ (α + β, (α + β)g) = (α + β, α + γ) = (α, α) + (β, γ) = \frac{12}{5} + (β, γ).

Thus, we have (β, γ) ≤ −\frac{12}{5}. However, by the Schwartz inequality,

|(β, γ)| ≤ \sqrt{(β, β)(γ, γ)} = \frac{8}{5},

which is a contradiction.

Case: F = M ∩ N ≅ \sqrt{2}M(4, 25). In this case, M ∩ J ≅ N ∩ J ≅ \sqrt{2}M(4, 25), also. Let \sqrt{2}u, \sqrt{2}v, \sqrt{2}w, \sqrt{2}x be a set of orthogonal elements in F ≅ \sqrt{2}M(4, 25) such that their norms are 4, 8, 20, 40, respectively (cf. (B.7)). Let \sqrt{2}u′, \sqrt{2}v′, \sqrt{2}w′, \sqrt{2}x′ be a sequence of pairwise orthogonal elements in M ∩ J such that their norms are 4, 8, 20, 40, respectively. By the construction in (B.8) and the uniqueness assertion, we may assume that the element γ = \frac{\sqrt{2}}{5}(w + x + x′) is in M. Since γ has norm 4, by (D.11),

0 ≥ (γ, γg) = \frac{2}{25}(w + x + x′, w + x + x′g) = \frac{60}{25} + \frac{2}{25}(x′, x′g).

Thus, we have (x′, x′g) ≤ −30. By the Schwartz inequality,

|(x′, x′g)| ≤ \sqrt{(x′, x′)(x′g, x′g)} = 20,

which is again a contradiction.

Case: F = M ∩ N ≅ \sqrt{2}A_{4}(1). Since F is a direct summand of M and N, we have M ∩ J ≅ N ∩ J ≅ \sqrt{2}A_{4}(1) by (D.13). Recall that (A_{4}(1))^{∗} ≅ \frac{1}{\sqrt{5}}A_{4}.

By the construction in (D.14), there exists α ∈ 2F^{∗} with (α, α) = 2×8/5 = 16/5 and α_{M} ∈ 2(M ∩ J)^{∗} with (α_{M}, α_{M}) = 2 × 2/5 = 4/5 such that y = α + α_{M} ∈ M.

Since (y, y) = 4, by (D.11),

0 ≥ (y, yg) = (α + α_{M}, α + α_{M}g) = (α, α) + (α_{M}, α_{M}g)

and we have (α_{M}, α_{M}g) ≤ −(α, α) = −16/5. However, by the Schwartz inequality,

|(α_{M}, α_{M}g)| ≤ \sqrt{(α_{M}, α_{M})(α_{M}g, α_{M}g)} = 4/5,

which is a contradiction. □

7.2 DIH_{10}: An orthogonal direct sum

For background, we refer to (B.3), (B.10), (D.10) – (D.12). Our goal here is to build up an orthogonal direct sum of four copies of AA_{4} inside L. We do so one summand at a time. This direct sum shall determine L (see the following subsection).

Notation 7.7. Define Z(i) := {x ∈ M|(x, x) = 4, (x, xg) = i}. Note that (x, xg) = −3, −2, −1, 0, or 1 by Lemma D.11.

Lemma 7.8. For u, v ∈ M, (u, vg) = (u, vg^{−1}) = (ug, v).

Proof. Since t preserves the form, (u, vg) = (ut, vgt) = (−u, vtg^{−1}) = (−u, −vg^{−1}) = (u, vg^{−1}). This equals (ug, v) since g preserves the form. □

Lemma 7.9. If u, v, w is any set of norm 4 vectors so that u + v + w = 0, then one or three of u, v, w lies in Z(−2) ∪ Z(0). In particular, Z(−2) ∪ Z(0) ≠ ∅.

Proof. Suppose that we have norm 4 vectors u, v, w so that u + v + w = 0. Then 0 = (u + v + w, ug + vg + wg) = (u, ug) + (v, vg) + (w, wg) + (u, vg) + (ug, v) + (u, wg) + (ug, w) + (v, wg) + (vg, w) ≡ (u, ug) + (v, vg) + (w, wg)(mod 2), by (7.8), whence evenly many of (u, ug), (v, vg), (w, wg) are odd. □

Now we look at D-submodules of L and decompositions.

Definition 7.10. Let M_{4} = {α ∈ M| (α, α) = 4}. Define a partition of M_{4} into sets M_{4}^{1} := {α ∈ M_{4}| αℤ[D] ≅ A_{4}(1)} and M_{4}^{2} := {α ∈ M_{4}| αℤ[D] ≅ AA_{4}} (cf. (D.11)). For α, β ∈ M_{4}^{2}, say that α and β are equivalent if and only if αℤ[D] = βℤ[D]. Define the partition N_{4} = N_{4}^{1} ∪ N_{4}^{2} and equivalence relation on N_{4}^{2} similarly.

Remark 7.11. The linear maps g^{i} + g^{−i} take M into M since they commute with t. Also, g^{2} +g^{3} and g +g^{4} are linear isomorphisms of M onto M since their product is −1. Note that they may not preserve inner products.

Lemma 7.12. M_{4} = M_{4}^{2} and M_{4}^{1} = ∅.

Proof. Supposing the lemma to be false, we take α ∈ M_{4}^{1}. Then the norm of α(g^{2} + g^{3}) is 4 + 2(αg^{2}, αg^{3}) + 4 = 8 − 2 = 6 (cf. (D.11)), which is impossible since M ≅ EE_{8} is doubly even. □

Lemma 7.13. Let α ∈ M_{4}. Then M ∩ αℤ[D] ≅ AA_{1}^{2}.

Proof. Let α ∈ M_{4}. Then by (7.12), αℤ[D] ≅ AA_{4}. In this case, we have either (1) (α, αg) = −2 and (α, αg^{2}) = 0 or (2) (α, αg) = 0 and (α, αg^{2}) = −2.

In case (1), we have α(g^{2}+g^{3}) ∈ M_{4} and α and α(g^{2}+g^{3}) generate a sublattice of type AA_{1}^{2} in M ∩αℤ[D]. Similarly, αg, α(g^{2}+g^{3})g generate a sublattice of type AA_{1}^{2} in N ∩αℤ[D]. Since M ∩N = 0, we have rank(M ∩αℤ[D]) = rank(N ∩αℤ[D]) = 2. Moreover, {α, α(g^{2} + g^{3}), αg, α(g^{3} + g^{4})} forms a ℤ-basis of an AA_{4}-sublattice of αℤ[D] ≅ AA_{4}. Thus, {α, α(g^{2} + g^{3}), αg, α(g^{3} + g^{4})} is also a basis of αℤ[D] and span_{ℤ}{α, α(g^{2} + g^{3})} is summand of αℤ[D]. Hence, M ∩ αℤ[D] = span_{ℤ}{α, α(g^{2} + g^{3})} ≅ AA_{1}^{2} as desired.

In case (2), we have α(g+g^{4}) ∈ M_{4} and thus M ∩αℤ[D] = span_{ℤ}{α, α(g+g^{4})} ≅ AA_{1}^{2} by an argument as in case (1). □

Lemma 7.14. Suppose that α ∈ M_{4}^{2}, β ∈ N_{4}^{2} and αℤ[D] = βℤ[D]. Let the equivalence class of α be {±α, ±α′} and let the equivalence class of β be {±β, ±β′}. After interchanging β and one of ±β′ if necessary, the Gram matrix of α, α′, β, β′ is

2 (\begin{matrix}2 & 0 & 0 & 1 \\ 0 & 2 & 1 & −1 \\ 0 & 1 & 2 & 0 \\ 1 & −1 & 0 & 2 \end{matrix}) = (\begin{matrix}4 & 0 & 0 & 2 \\ 0 & 4 & 2 & −2 \\ 0 & 2 & 4 & 0 \\ 2 & −2 & 0 & 4 \end{matrix}) .

Proof. We think of A_{4} as the 5-tuples in ℤ^{5} with zero coordinate sum. Index coordinates with integers mod 5: 0, 1, 2, 3, 4. Consider g as addition by 1 mod 5 and t as negating indices modulo 5. We may take α := \sqrt{2}(0, 1, 0, 0, −1), α′ := \sqrt{2}(0, 0, 1, −1, 0). We define β := αg, β′ := α′g. The computation of the Gram matrix is straightforward. □

Lemma 7.15. Let m ≥ 1. Suppose that U is a rank 4m ℤ[D]-invariant sublattice of L which is generated as a ℤ[D]-module by S, a sublattice of U ∩ M which is isometric to AA_{1}^{2m}. Then ann_{M}(U) contains a sublattice of type AA_{1}^{8−2m}.

Proof. We may assume that m ≤ 3. Since t inverts g and g is fixed point free on L, U^{−}(t) = U ∩ M has rank 2m. Let S be a sublattice of U ∩ M of type AA_{1}^{2m}. Then S has finite index in U ∩ M. Let W := ann_{L}(U), a direct summand of L of rank 16 − 4m. The action of g on W is fixed point free and t inverts g under conjugation, so W ∩ M = ann_{M}(U) is a direct summand of M of rank 8 − 2m. It is contained in hence is equal to the annihilator in M of S, by rank considerations, so is isometric to DD_{6}, DD_{4}, AA_{1}^{4} or AA_{1}^{2}. Each of these lattices contains a sublattice of type AA_{1}^{8−2m}. □

Corollary 7.16. L contains an orthogonal direct sum of four D-invariant lattices, each isometric to AA_{4}.

Proof. We prove by induction that for k = 0, 1, 2, 3, 4, L contains an orthogonal direct sum of k D-invariant lattices, each isometric to AA_{4}. This is trivial for k = 0. If 0 ≤ k ≤ 3, let U be such an orthogonal direct sum of k copies of AA_{4}. Then M ∩ U ≅ AA_{1}^{2k} and thus ann_{M}(U) contains a norm 4 vector, say α. By (7.12), αℤ[D] ≅ AA_{4}. So, U ⊥ αℤ[D] is an orthogonal direct sum of (k + 1) D-invariant lattices, each isometric to AA_{4}. □

Corollary 7.17. L = M + N is unique up to isometry.

Proof. Uniqueness follows from the isometry type of U (finite index in L) and (4.1). We take the finite index sublattices M_{1} := M ∩ U and N_{1} := N ∩ U and use (7.14). An alternate proof is given by the glueing in (7.18) □

7.3 DIH_{10}: From AA_{4}^{4} to L

We discuss the glueing from a sublattice U = U_{1} ⊥ U_{2} ⊥ U_{3} ⊥ U_{4}, as in (7.16) to L. We assume that each U_{i} is invariant under D.

By construction, M/(M ∩ U) ≅ 2^{4}, M ∩ U ≅ AA_{1}^{8}. A similar statement is true with N in place of M. Since L = M + N, it follows that L/U is a 2-group. Since g acts fixed point freely on L/U, L/U is elementary abelian of order 2^{4} or 2^{8}. Also, L/U is the direct sum of C_{L/U}(t) and C_{L/U}(u), and each of the latter groups is elementary abelian of order |L : U|^{\frac{1}{2}} . So |L : U|^{\frac{1}{2}} = |M : M ∩ U|^{2} = (2^{4})^{2} = 2^{8}. Therefore, det(L) = 5^{4} and the Smith invariant sequence of L is 1^{12}5^{4}.

Proposition 7.18. The glueing from U to L may be identified with the direct sums of these two glueings from U ∩ M to M and U ∩ N to N. Each glueing is based on the extended Hamming code with parameters [8,4,4] with respect to the orthogonal frame.

7.4 DIH_{10}: Explicit glueing and tensor products

In this section, we shall give the glue vectors from U = U_{1} ⊥ U_{2} ⊥ U_{3} ⊥ U_{4} to L explicitly in Proposition 7.23 (cf. Proposition 7.18). We also show that L contains a sublattice isomorphic to a tensor product A_{4} ⊗ A_{4}.

Notation 7.19. Recall that M ∩ U_{i} ≅ AA_{1} ⊥ AA_{1} for i = 1, 2, 3, 4. Let α_{i} ∈ M_{4} ∩ U_{i}, i = 1, 2, 3, 4, such that (α_{i}, α_{i}g) = −2. Note that such α_{i} exists because if (α_{i}, α_{i}g) ≠ −2, then (α_{i}, α_{i}g) = 0 and (α_{i}, α_{i}g^{2}) = −2. In this case, α̃_{i} = α_{i}(g + g^{4}) ∈ M_{4} ∩ U_{i} and (α̃_{i}, α̃_{i}g) = −2 (7.11).

Set α′_{i} := α_{i}(g^{2} + g^{3}) for i = 1, 2, 3, 4. Then α′_{i} ∈ M_{4} ∩ U_{i} and M ∩ U_{i} = span_{ℤ}{α_{i}, α′_{i}} (7.11).

Lemma 7.20. Use the same notation as in (7.19). Then for all i = 1, 2, 3, 4, we have (α_{i}, α_{i}g) = −2, (α_{i}, α_{i}g^{2}) = 0, (α_{i}, α′_{i}) = 0, (α′_{i}, α′_{i}g) = 0 and (α_{i}, α′_{i}g) = −2.

Proof. By definition, (α_{i}, α_{i}g) = −2 and (α_{i}, α_{i}g^{2}) = 0. Thus, we have (α′_{i}, α_{i}) = (α_{i}(g^{2} + g^{3}), α_{i}) = 0. Also,

(α′_{i}, α′_{i}g) = (α_{i}(g^{2} + g^{3}), α_{i}(g^{2} + g^{3})g)
= (α_{i}g^{2}, α_{i}g^{3}) + (α_{i}g^{2}, α_{i}g^{4}) + (α_{i}g^{3}, α_{i}g^{3}) + (α_{i}g^{3}, α_{i}g^{4})
= −2 + 0 + 4 − 2 = 0

and

(α_{i}, α′_{i}g) = (α_{i}, α_{i}(g^{2} + g^{3})g) = (α_{i}, α_{i}g^{3}) + (α_{i}, α_{i}g^{4}) = 0 − 2 = −2.

□

Remark 7.21. Since M and U are doubly even and since \frac{1}{\sqrt{2}}(U ∩ M) ≅ (A_{1})^{8} and (A_{1})^{∗} = \frac{1}{2}A_{1}, for any β ∈ M \ (U ∩ M),

β = ∑_{i=1}^{4}(\frac{b_{i}}{2} α_{i} + \frac{b′_{i}}{2} α′_{i}) where b_{i}, b′_{i} ∈ ℤ with some b_{i}, b′_{i} odd.

Lemma 7.22. Let β ∈ M \ (U ∩ M) with (β, β) = 4. Then, one of the following three cases holds.

(i) |b_{i}| = 1 and b′_{i} = 0 for all i = 1, 2, 3, 4;

(ii) |b′_{i}| = 1 and b_{i} = 0 for all i = 1, 2, 3, 4;

(iii) There exists a 3-set {i, j, k} ⊂ {1, 2, 3, 4} such that b_{i}^{2} = b_{j}^{2} = 1 and b_{i}^{′2} = b_{k}^{′2} = 1.

Proof. Let β = ∑_{i=1}^{4}(\frac{b_{i}}{2} α_{i} + \frac{b′_{i}}{2} α′_{i}) ∈ M \ (U ∩ M) with (β, β) = 4. Then we have ∑_{i=1}^{4}(b_{i}^{2} + b_{i}^{′2}) = 4. Since no |b_{i}| or |b′_{i}| is greater than 1 (or else no b_{i} or b′_{i} is odd), b_{i}, b′_{i} ∈ {−1, 0, 1}. Moreover, (β, βg) = 0 or −2 since β ∈ M_{4}. By (7.20),

\begin{matrix}(β, βg) = & (∑_{i=1}^{4}(\frac{b_{i}}{2} α_{i} + \frac{b′_{i}}{2} α′_{i}), ∑_{i=1}^{4}(\frac{b_{i}}{2} α_{i}g + \frac{b′_{i}}{2} α′_{i}g)) \\ = & \frac{1}{4} ∑_{i=1}^{4} (b_{i}^{2}(−2) + 2b_{i}b′_{i}(−2)) = −\frac{1}{2} ∑_{i=1}^{4} (b_{i}^{2} + 2b_{i}b′_{i}) . \end{matrix}

If (β, βg) = −2, then

∑_{i=1}^{4} (b_{i}^{2} + 2b_{i}b′_{i}) = 4 = ∑_{i=1}^{4}(b_{i}^{2} + b_{i}^{′2}).

and hence we have (*) ∑_{i=1}^{4} b′_{i}(b′_{i} − 2b_{i}) = 0.

Set k_{i} := b′_{i}(b′_{i} − 2b_{i}). The values of k_{i}, for all b_{i}, b′_{i} ∈ {−1, 0, 1}, are listed in Table 9.

Table 9: Values of k_{i}

\begin{matrix}b′_{i} & 0 & −1 & −1 & −1 & 1 & 1 & 1 \\ b_{i} & −1, 0, 1 & −1 & 0 & 1 & −1 & 0 & 1 \\ k_{i} = b′_{i}(b′_{i} − 2b_{i}) & 0 & −1 & 1 & 3 & 3 & 1 & −1 \end{matrix}

Note that k_{i} = 0, ±1 or 3 for all i = 1, 2, 3, 4. Therefore, up to the order of the indices, the values for (k_{1}, k_{2}, k_{3}, k_{4}) are (3, −1, −1, −1), (1, −1, 1, −1), (1, −1, 0, 0) or (0, 0, 0, 0).

However, for (k_{1}, k_{2}, k_{3}, k_{4}) = (3, −1, −1, −1) or (1, −1, 1, −1), b_{i}^{′2} = b_{i}^{2} = 1 for all i = 1, 2, 3, 4 and then ∑_{i=1}^{4}(b_{i}^{2} + b_{i}^{′2}) = 8 > 4. Therefore, (k_{1}, k_{2}, k_{3}, k_{4}) = (1, −1, 0, 0) or (0, 0, 0, 0).

If (k_{1}, k_{2}, k_{3}, k_{4}) = (1, −1, 0, 0), then we have, up to order, (b′_{1})^{2} = 1 (whence k_{1} = 1), b_{1} = 0, b′_{2} = b_{2} = ±1 (whence k_{2} = −1) and b′_{3} = b′_{4} = 0. Since ∑_{i=1}^{4}(b_{i}^{2} + b_{i}^{′2}) = 4, b_{3}^{2} + b_{4}^{2} = 1 and hence we have (iii).

If k_{i} = b′_{i}(b′_{i} − 2b_{i}) = 0 for all i = 1, 2, 3, 4, then b′_{i} = 0 for all i and we have (i). Note that if b′_{i} ≠ 0, b′_{i} − 2b_{i} ≠ 0.

Now assume (β, βg) = 0. Then ∑_{i=1}^{4} (b_{i}^{2} + 2b_{i}b′_{i}) = 0. Note that this equation is the same as the above equation (*) in the case for (β, βg) = −2 if we replace b_{i} by b′_{i} and b′_{i} by −b_{i} for i = 1, 2, 3, 4. Thus, by the same argument as in the case for (β, βg) = −2, we have either b_{i}^{2} + 2b_{i}b′_{i} = 0 for all i = 1, 2, 3, 4 or b_{1}^{2} + 2b_{1}b′_{1} = 1, b_{2}^{2} + 2b_{2}b′_{2} = −1, and b_{3} = b_{4} = 0. In the first case, we have b_{i} = 0 and b_{i}^{′2} = 1 for all i = 1, 2, 3, 4, that means (ii) holds. For the later cases, we have b_{1} = 1, b′_{1} = 0, b_{2} = 1, b′_{2} = −1, b_{3} = b_{4} = 0, and b_{3}^{′2} + b_{4}^{′2} = 1 and thus (iii) holds. □

Proposition 7.23. By rearranging the indices if necessary, we have

M = span_{ℤ} \left\{{M ∩ U, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}), \frac{1}{2}(α′_{1} + α′_{2} + α′_{3} + α′_{4}), \atop \frac{1}{2}(α_{1} + α_{2} + α′_{2} + α′_{4}), \frac{1}{2}(α_{1} + α_{3} + α′_{2} + α′_{3})}\right\}

and

N = Mg = span_{ℤ} \left\{{N ∩ U, \frac{1}{2}(β_{1} + β_{2} + β_{3} + β_{4}), \frac{1}{2}(β′_{1} + β′_{2} + β′_{3} + β′_{4}), \atop \frac{1}{2}(β_{1} + β_{2} + β′_{2} + β′_{4}), \frac{1}{2}(β_{1} + β_{3} + β′_{2} + β′_{3})}\right\} ,

where β_{i} = α_{i}g and β′_{i} = α′_{i}g for all i = 1, 2, 3, 4.

Proof. By (7.22), the norm 4 vectors in M \ (U ∩ M) are of the form

\frac{1}{2}(±α_{1} ± α_{2} ± α_{3} ± α_{4}), \frac{1}{2}(±α′_{1} ± α′_{2} ± α′_{3} ± α′_{4}), or \frac{1}{2}(±α_{i} ± α_{j} ± α′_{j} ± α′_{k}),

where i, j, k are distinct elements in {1, 2, 3, 4}.

Since M ≅ EE_{8} and U ∩M ≅ (AA_{1})^{8}, the cosets of M/(U ∩M) can be identified with the codewords of the Hamming [8,4,4] code H_{8}.

Let φ : M/(U ∩ M) → H_{8} be an isomorphism of binary codes. For any β ∈ M, we denote the coset β + U ∈ M/(U ∩ M) by β̄. We shall also arrange the index set such that the first 4 coordinates correspond to the coefficient of \frac{1}{2}α_{1}, \frac{1}{2}α_{2}, \frac{1}{2}α_{3} and \frac{1}{2}α_{4} and the last 4 coordinates correspond to the coefficient of \frac{1}{2}α′_{1}, \frac{1}{2}α′_{2}, \frac{1}{2}α′_{3} and \frac{1}{2}α′_{4}.

Since (1, …, 1) ∈ H_{8}, we have

\frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4} + α′_{1} + α′_{2} + α′_{3} + α′_{4}) ∈ M.

We shall also show that \frac{1}{2}(α_{1}+α_{2}+α_{3}+α_{4}) ∈ M and hence \frac{1}{2}(α′_{1}+α′_{2}+α′_{3}+α′_{4}) ∈ M.

Since M/(M ∩ U) ≅ H_{8}, there exist β_{1}, β_{2}, β_{3}, β_{4} ∈ M \ (U ∩ M) such that φ(β̄_{1}), φ(β̄_{2}), φ(β̄_{3}), φ(β̄_{4}) generates the Hamming code H_{8}. By (7.22), their projections to the last 4 coordinates are all even and thus spans an even subcode of ℤ_{2}^{4}, which has dimension ≤ 3. Therefore, there exists a_{1}, a_{2}, a_{3}, a_{4} ∈ {0, 1}, not all zero such that φ(a_{1}β̄_{1} + a_{2}β̄_{2} + a_{3}β̄_{3} + a_{4}β̄_{4}) projects to zero and so must equal (11110000). Therefore, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}) ∈ M.

Since |M/(U ∩ M)| = 2^{4}, there exists β′ = \frac{1}{2}(α_{i} + α_{j} + α′_{j} + α′_{k}) and β′′ = \frac{1}{2}(α_{m} + α_{n} + α′_{n} + α′_{ℓ}) such that

M = span_{ℤ} \{M ∩ U, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}), \frac{1}{2}(α′_{1} + α′_{2} + α′_{3} + α′_{4}), β′, β′′\} .

Note that

β′ + β′′ = \frac{1}{2}((α_{i} + α_{j} + α_{m} + α_{n}) + (α′_{j} + α′_{k} + α′_{n} + α′_{ℓ})).

Let A := ({i, j} ∪ {m, n}) − ({i, j} ∩ {m, n}) and A′ := ({j, k} ∪ {n, ℓ}) − ({j, k} ∩ {n, ℓ}). We shall show that |{i, j} ∩ {m, n}| = |{j, k} ∩ {n, ℓ}| = 1 and |A ∩ A′| = 1.

Since φ(β̄′ + β̄′′) ∈ H_{8} but φ(β̄′ + β̄′′) ∉ span_{ℤ_{2}}{(11110000), (00001111)}, by (7.22), we have

β′ + β′′ ∈ \frac{1}{2}(α_{p} + α_{q} + α′_{p} + α′_{r}) + M ∩ U,

for some p, q ∈ {i, j, m, n}, p, r ∈ {j, k, n, ℓ} such that p, q, r are distinct.

That means \frac{1}{2}(α_{i}+α_{j} +α_{m}+α_{n}) ∈ \frac{1}{2}(α_{p}+α_{q})+M ∩U and \frac{1}{2}(α′_{j} +α′_{k}+α′_{n}+α′_{ℓ}) ∈ \frac{1}{2}(α′_{p} +α′_{r})+M ∩U. It implies that A = (\{i, j\}∪\{m, n\})−(\{i, j\}∩\{m, n\}) = \{p, q\} and A′ = ({j, k} ∪ {n, ℓ}) − ({j, k} ∩ {n, ℓ}) = {p, r}. Hence, |{i, j} ∩ {m, n}| = |{j, k} ∩ {n, ℓ}| = 1 and |A ∩ A′| = 1.

By rearranging the indices if necessary, we may assume β′ = \frac{1}{2}(α_{1}+α_{2}+α′_{2}+α′_{4}), β′′ = \frac{1}{2}(α_{1} + α_{3} + α′_{2} + α′_{3}) and hence

M = span_{ℤ} \left\{{M ∩ U, \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}), \frac{1}{2}(α′_{1} + α′_{2} + α′_{3} + α′_{4}), \atop \frac{1}{2}(α_{1} + α_{2} + α′_{2} + α′_{4}), \frac{1}{2}(α_{1} + α_{3} + α′_{2} + α′_{3})}\right\}

Now let β_{i} = α_{i}g and β′_{i} = α′_{i}g for all i = 1, 2, 3, 4. Then

N = Mg = span_{ℤ} \left\{{N ∩ U, \frac{1}{2}(β_{1} + β_{2} + β_{3} + β_{4}), \frac{1}{2}(β′_{1} + β′_{2} + β′_{3} + β′_{4}), \atop \frac{1}{2}(β_{1} + β_{2} + β′_{2} + β′_{4}), \frac{1}{2}(β_{1} + β_{3} + β′_{2} + β′_{3})}\right\} ,

as desired. □

Next we shall show that L contains a sublattice isomorphic to a tensor product A_{4} ⊗ A_{4}.

Notation 7.24. Take

γ_{0} := \frac{1}{2}(−α_{1} + α_{2} + α′_{2} − α′_{4}),
γ_{1} := α_{1},
γ_{2} := − \frac{1}{2}(α_{1} + α_{2} + α_{3} + α_{4}),
γ_{3} := α_{3},
γ_{4} := − \frac{1}{2}(α_{3} − α_{4} + α′_{2} − α′_{4})

in M (cf. (7.23)) and set R := span_{ℤ}{γ_{1}, γ_{2}, γ_{3}, γ_{4}}. Then R ≅ AA_{4}. Note that γ_{0} = −(γ_{1} + γ_{2} + γ_{3} + γ_{4}).

Lemma 7.25. For any i = 0, 1, 2, 3, 4 and j = 1, 2, 3, 4, we have

(i) (γ_{i}, γ_{i}g) = (γ_{i}, γ_{i}g^{4}) = −2 and (γ_{i}, γ_{i}g^{2}) = (γ_{i}, γ_{i}g^{3}) = 0; (ii) (γ_{j−1}, γ_{j}g) = (γ_{j−1}, γ_{j}g^{4}) = 1 and

(γ_{j−1}, γ_{j}g^{2}) = (γ_{j−1}, γ_{j}g^{3}) = 0; (iii) (γ_{i}, γ_{j}g^{k}) = 0 for any k if |i − j| > 1 (iv) γ_{i}ℤ[D] ≅ AA_{4}.

Proof. Straightforward. □

Proposition 7.26. Let T = Rℤ[D]. Then T ≅ A_{4} ⊗ A_{4}.

Proof. By (iv) of (7.25), γ_{i}ℤ[D] = span_{ℤ}{γ_{i}g^{j} | j = 0, 1, 2, 3, 4} ≅ AA_{4}.

Let {e_{1}, e_{2}, e_{3}, e_{4}} be a fundamental basis of A_{4} and denote e_{0} = −(e_{1} + e_{2} + e_{3} + e_{4}). Now define a linear map φ : T → A_{4} ⊗ A_{4} by φ(γ_{i}g^{j}) = e_{i} ⊗ e_{j}, for i, j = 1, 2, 3, 4. By the inner product formulas in (7.25),

(γ_{i}g^{j}, γ_{k}g^{ℓ}) = (γ_{i}, γ_{k}g^{ℓ−j}) = \begin{cases}4 & if i = k, j = ℓ, \\ −2 & if i = k, |j − ℓ| = 1, \\ 1 & if |i − k| = 1, |j − ℓ| = 1, \\ 0 & if |i − k| > 1, |j − ℓ| > 1. \end{cases}

Hence, (γ_{i}g^{j}, γ_{k}g^{ℓ}) = (e_{i} ⊗ e_{j}, e_{k} ⊗ e_{ℓ}) for all i, j, k, ℓ and φ is an isometry. □

A General results about lattices

Lemma A.1. Let p be a prime number, f(x) := 1 + x + x^{2} + ⋯ + x^{p−1}. Let L be a ℤ[x]-module. For v ∈ L, pv ∈ L(x − 1) + Lf(x).

Proof. We may write f(x) = ∑_{i=0}^{p−1} x^{i} = ∑_{i=0}^{p−1}((x − 1) + 1)^{i} = (x − 1)h(x) + p, for some h(x) ∈ ℤ[x]. Then if v ∈ L, pv = v(f(x) − (x − 1)h(x)). □

Lemma A.2. Suppose that the four group D acts on the abelian group A. If the fixed point subgroup of D on A is 0, then A/T el(A, D) is an elementary abelian 2-group.

Proof. Let a ∈ A and let r ∈ D. We claim that a(r + 1) is an eigenvector for D. It is clearly an eigenvector for r. Take s ∈ D so that D = ⟨r, s⟩. Then a(r + 1)s = a(r + 1)(s + 1) − a(r + 1) = −a(r + 1) since a(r + 1)(s + 1) is a fixed point. So, a(r + 1) is an eigenvector for D.

To prove the lemma, we just calculate that a(1 + r) + a(1 + s) + a(1 + rs) = 2a + a(1 + sr + s + rs) = 2a since a(1 + sr + s + rs) is a fixed point. □

Lemma A.3. Suppose that X is a lattice of rank n and Y is a sublattice of rank m. Let p be a prime number. Suppose that the p-rank of D(X) is r. Then, (Y ∩ pX^{∗})/(Y ∩ pX) has p-rank at least r + m − n. In particular, the p-rank of D(Y) is at least r + m − n; and if r + m > n, then p divides det(Y).

Proof. We may assume that Y is a direct summand of X. The quadratic space X/pX has dimension n over 𝔽_{p} and its radical pX^{∗}/pX has dimension r. The image of Y in X/pX is Y +pX/pX, and it has dimension m since Y is a direct summand of X. Let q be the quotient map X/pX to (X/pX)/(pX^{∗}/pX) ≅ X/pX^{∗} ≅ p^{r}. Then dim(q(Y + pX/pX)) ≤ n − r, so that dim(Ker(q) ∩ (Y + pX/pX)) ≥ m − (n − r) = r + m − n.

We note that Ker(q) = pX^{∗}/pX, so the above proves r + m − n ≤ rank((Y ∩ pX^{∗})+pX/pX) = rank(Y ∩pX^{∗})/(Y ∩pX^{∗} ∩pX)) = rank((Y ∩pX^{∗})/(Y ∩pX)) = rank((Y ∩ pX^{∗})/pY). Note that (Y ∩ pX^{∗})/pY ≅ (\frac{1}{p}Y ∩ X^{∗})/Y ≤ Y^{∗}/Y, which implies the inequality of the lemma. □

Lemma A.4. Suppose that Y is an integral lattice such that there exists an integer r > 0 so that D(Y) contains a direct product of rank(Y) cyclic groups of order r. Then \frac{1}{\sqrt{r}} Y is an integral lattice.

Proof. Let Y < X < Y^{∗} be a sublattice such that X/Y ≅ (ℤ_{r})^{rank Y}. Then x ∈ X if and only if rx ∈ Y. Let y, y′ ∈ Y. Then (\frac{1}{\sqrt{r}} y, \frac{1}{\sqrt{r}} y′) = (\frac{1}{r}y, y′) ∈ (X, Y) ≤ (Y^{∗}, Y) = ℤ. □

Lemma A.5. Suppose that X is an integral lattice and that there is an integer s ≥ 1 so that \frac{1}{\sqrt{s}}X is an integral lattice. Then the subgroup sD(X) is isomorphic to D(\frac{1}{\sqrt{s}}X) and D(X)/sD(X) is isomorphic to s^{rank(X)}.

Proof. Study the diagram below, in which the horizontal arrows are multiplication by \frac{1}{\sqrt{s}}. The hypothesis implies that the finite abelian group D(X) is a direct sum of cyclic groups, each of which has order divisible by s.

\begin{matrix}X^{∗} & ⟶ & \frac{1}{\sqrt{s}}X^{∗} \\ | & & | \\ sX^{∗} & ⟶ & \sqrt{s}X^{∗} \\ | & & | \\ X & ⟶ & \frac{1}{\sqrt{s}}X \end{matrix}.

We take a vector a ∈ ℝ⊗X and note that a ∈ (\frac{1}{\sqrt{s}}X)^{∗} if and only if (a, \frac{1}{\sqrt{s}}X) ≤ ℤ if and only if (\frac{1}{\sqrt{s}}a, X) ≤ ℤ if and only if \frac{1}{\sqrt{s}}a ∈ X^{∗} if and only if a ∈ \sqrt{s}X^{∗}. This proves the first statement. The second statement follows because D(X) is a direct sum of rank(X) cyclic groups, each of which has order divisible by s. □

Lemma A.6. Let y be an order 2 isometry of a lattice X. Then X/T el(X, y) is an elementary abelian 2-group. Suppose that X/T el(X, y) ≅ 2^{c}. Then we have det(X^{+}(y))det(X^{−}(y)) = 2^{2c}det(X) and for ε = ±, the image of X in D(X^{ε}(y)) is 2^{c}. In particular, for ε = ±, det(X^{ε}(y)) divides 2^{c}det(X) and is divisible by 2^{c}. Finally, c ≤ rank(X^{ε}(y)), for ε = ±, so that c ≤ \frac{1}{2}rank(X).

Proof. See [GrE8]. □

Lemma A.7. Suppose that t is an involution acting on the abelian group X. Suppose that Y is a t-invariant subgroup of odd index so that t acts on X/Y as a scalar c ∈ {±1}. Then for every coset x + Y of Y in X, there exists u ∈ x + X so that ut = cu.

Proof. First, assume that c = 1. Define n := \frac{1}{2}(|X/Y |+1), then take u := nx(t+1). This is fixed by t and u ≡ 2nx ≡ x(mod Y).

If c = −1, apply the previous argument to the involution −t. □

Lemma A.8. If X and Y are abelian groups with |X : Y | odd, an involution r acts on X, and Y is r-invariant, then X/T el(X, r) ≅ Y/T el(Y, r).

Proof. Since X/Y has odd order, it is the direct sum of its two eigenspaces for the action of r. Use (A.7) to show that Y +T el(X, r) = X and Y ∩T el(X, r) = T el(Y, r). □

Lemma A.9. Suppose that X is an integral lattice which has rank m ≥ 1 and there exists a lattice W, so that X ≤ W ≤ X^{∗} and W/X ≅ 2^{r}, for some integer r ≥ 1. Suppose further that every nontrivial coset of X in W contains a vector with noninteger norm. Then r = 1.

Proof. Note that if u + X is a nontrivial coset of X in W, then (u, u) ∈ \frac{1}{2} + ℤ.

Let ϕ : X → Y be an isometry of lattices, extended linearly to a map between duals. Let Z be the lattice between X ⊥ Y and W ⊥ ϕ(W) which is diagonal with respect to ϕ, i.e., is generated by X ⊥ Y and all vectors of the form (x, xϕ), for x ∈ W.

Then Z is an integral lattice. In any integral lattice, the even sublattice has index 1 or 2. Therefore, r = 1 since the nontrivial cosets of X ⊥ Y in Z are odd. □

Lemma A.10. Suppose that the integral lattice L has no vectors of norm 2 and that L = M + N, where M ≅ N ≅ EE_{8}. The sublattices M, N, F = M ∩ N are direct summands of L = M + N.

Proof. Note that L is the sum of even lattices, so is even. Therefore, it has no vectors of norm 1 or 2. Since M by definition defines the summand S of negated vectors by t_{M}, we get S = M because M ≤ S ≤ \frac{1}{2}M and the minimum norm of L is 4. A similar statement holds for N. The sublattice F is therefore the sublattice of vectors fixed by both t_{M} and t_{N}, so it is clearly a direct summand of L. □

Lemma A.11. D ≅ Dih_{6}, ⟨g⟩ = O_{3}(D) and t is an involution in D. Suppose that D acts on the abelian group A, 3A = 0 and A(g − 1)^{2} = 0. Let ε = ±1. If v ∈ A and vt = εv, then v(g − 1)t = −εv(g − 1).

Proof. Calculate v(g − 1)t = vt(g^{−1} − 1) = εv(g^{−1} − 1). Since (g − 1)^{2} = 0, g acts as 1 on the image of g − 1, which is the image of (g^{−1} − 1). Therefore, εv(g^{−1} − 1) = εv(g^{−1} − 1)g = εv(1 − g) = −εv(g − 1). □

Lemma A.12. Suppose that X is an integral lattice and Y has finite index, m, in X. Then D(X) is a subquotient of D(Y) and |D(X)|m^{2} = |D(Y)|. The groups have isomorphic Sylow p-subgroups if p is a prime which does not divide m.

Proof. Straightforward. □

Lemma A.13. Suppose that X is a lattice, that Y is a direct summand and Z := ann_{X}(Y). Let n := |X : Y ⊥ Z|.

(i) The image of X in D(Y) has index dividing (det(Y), det(X)). In particular, if (det(Y), det(X)) = 1, X maps onto D(Y)

(ii) Let A := ann_{X^{∗}}(Y). Then X^{∗}/(Y ⊥ A) ≅ D(Y).

(iii) There are epimorphisms of groups φ_{1} : D(X) → X^{∗}/(X + A) and φ_{2} : D(Y) → X^{∗}/(X + A).

(iv) We have isomorphisms Ker(φ_{1}) ≅ (X + A)/X and Ker(φ_{2}) ≅ ψ(X)/Y ≅ X/(Y ⊥ Z). The latter is a group of order n.

In particular, Im(φ_{1}) ≅ Im(φ_{2}) has order \frac{1}{n}det(Y).

(v) If p is a prime which does not divide n, then O_{p}(D(Y)) injects into O_{p}(D(X)). This injection is an isomorphism onto if (p, det(Z)) = 1.

Proof. (i) This is clear since the natural map ψ : X^{∗} → Y^{∗} is onto and X has index det(X) in X^{∗}.

(ii) The natural map X^{∗} → Y^{∗} followed by the quotient map ζ : Y^{∗} → Y^{∗}/Y has kernel Y ⊥ A.

(iii) Since D(X) = X^{∗}/X, we have the first epimorphism. Since X +A ≥ Y +A, existence of the second epimorphism follows from (ii).

(iv) First, Ker(ζψ) = Y ⊥ A follows from (ii) and the definitions of ψ and ζ. So, Ker(φ_{1}) ≅ (X +A)/(Y +A). Note that (X +A)/(Y +A) = (X +(Y +A))/(Y + A) ≅ X/((Y + A) ∩ X) = X/(Y + Z)). The latter quotient has order n. For the order statement, we use the formula |Im(ψ)||Ker(ψ)| = |D(Y)|. For the second isomorphism, use Y^{∗} ≅ X^{∗}/A and D(Y) = Y^{∗}/Y ≅ (X^{∗}/A)/((Y + A)/A) and note that in here the image of X is ((X + A)/A)/((Y + A)/A) ≅ (X + A)/(Y + A).

(v) Let P be a Sylow p-subgroup of D(Y). Then P ∩Ker(ψ) = 0 since (p, n) = 1. Therefore P injects into Im(ψ). The epimorphism φ has kernel A/Z = D(Z). So, P is isomorphic to a Sylow p-subgroup of D(X) if (p, det(Z)) = 1. □

Lemma A.14. Suppose that X is an integral lattice and E is an elementary abelian 2-group acting in X. If H is an orthogonal direct summand of T el(X, E) and H is a direct summand of X, then the odd order Sylow groups of D(H) embed in D(X). In other notation, O_{2′}(D(H)) embeds in O_{2′}(D(X))

Proof. Apply (A.13) to Y = H, n a divisor of |X : T el(X, E)|, which is a power of 2. □

Lemma A.15. Suppose that X is a lattice and Y is a direct summand and Z := ann_{X}(Y). Let n := |X : Y ⊥ Z|.

Suppose that v ∈ X^{∗} and that v has order m modulo X. If (m, det(Z)) = 1, there exists w ∈ Z so that v − w ∈ Y^{∗} ∩ X^{∗}. Therefore, the coset v + X contains a representative in Y^{∗}. Furthermore, any element of Y^{∗} ∩ (v + X) has order m modulo Y.

Proof. The image of v in D(Z) is zero, so the restriction of the function v to Z is the same as taking a dot product with an element of Z. In other words, the projection of v to Z^{∗} is already in Z. Thus, there exists w ∈ Z so that v−w ∈ ann_{X^{∗}}(Z) = Y^{∗}.

Define u := v − w. Then mu = mv − mw ∈ X. Since v − w ∈ Y^{∗}, mu ∈ X ∩ Y^{∗}, which is Y since Y is a direct summand of X.

Now consider an arbitrary u ∈ (v + X) ∩ Y^{∗}. We claim that its order modulo Y is m. There is x ∈ X so that u = x + v. Suppose k > 0. Then ku = kx + kv is in X if and only if kv ∈ X, i.e. if and only if m divides k. □

Lemma A.16. Let D be a dihedral group of order 2n, n > 2 odd, and Y a finitely generated free abelian group which is a ℤ[D]-module, so that an element 1 ≠ g ∈ D of odd order acts with zero fixed point subgroup on Y. Let r be an involution of D outside Z(D). Then Y/T el(Y, r) is elementary abelian of order 2^{\frac{1}{2} rank(Y)}. Consequently, det(T el(Y, r)) = 2^{rank(Y)}det(Y).

Proof. The first statement follows since the odd order g is inverted by r and acts without fixed points on Y. The second statement follows from the index formula for determinants. □

Lemma A.17. Suppose that X ≅ E_{8} and that Y is a sublattice such that X/Y ≅ 3^{4} and Y ≅ \sqrt{3}E_{8}. Then there exists an element g of order 3 in O(X) so that X(g − 1) = Y. In particular, Y defines a partition on the set of 240 roots in X where two roots are equivalent if and only if their difference lies in Y.

Proof. Note that Y has 80 nontrivial cosets in X and the set Φ of roots has cardinality 240. Let x, y ∈ Φ such that x − y ∈ Y and x, y are linearly independent. Then 6 ≤ (x − y, x − y) = 4 − 2(x, y), whence |(x, y)| ≤ 2. Therefore (x, y) = −1, since x, y are linearly independent roots. Let z be a third root which is congruent to x and y modulo Y. Then (x, z) = −1 = (y, z) by the preceding discussion.

Therefore the projection of z to the span of x, y must be −x − y, which is a root. Therefore, x + y + z = 0.

It follows that a nontrivial coset of Y in X contains at most three roots. By counting, a nontrivial coset of Y in X contains exactly three roots.

Let P := span{x, y} ≅ A_{2}, Q := ann_{X}(P). We claim that Y ≤ P ⊥ Q, which has index 3 in X. Suppose the claim is not true. Then the structure of P^{∗} means that there exists r ∈ Y so that (r, x − y) is not divisible by 3. Since x − y ∈ Y, we have a contradiction to Y ≅ \sqrt{3}E_{8}. The claim implies that g_{P}, an automorphism of order 3 on P, extended to X by trivial action on Q, leaves Y invariant (since it leaves invariant any sublattice between P ⊥ Q and P^{∗} ⊥ Q^{∗}). Moreover, X(g_{P} − 1) = ℤ(x − y).

Now take a root x′ which is in Q. Let y′, z′ be the other members of its equivalence class of x′. We claim that these are also in Q. We know that x′ − y′ ∈ Y ≤ P ⊥ Q, so P′ := span{x′, y′, z′} ≤ P ⊥ Q. Now, we claim that the projection of P′ to P is 0. Suppose otherwise. Then the projection of some root u ∈ P′ to P is nonzero. Therefore the projection is a root, i.e. u ∈ P. But then u is in the equivalence class of x or −x and so P′ = P, a contradiction to (x′, P) = 0.

We now have that the class of x′ spans a copy of A_{2} in Q. We may continue this procedure to get a sublattice U = U_{1} ⊥ U_{2} ⊥ U_{3} ⊥ U_{4} of X such that U_{i} ≅ A_{2} for of X with the property that if g_{i} is an automorphism of order 3 on U_{i} extended to X by trivial action on ann_{X}(U_{i}), then each U(g_{i} − 1) ≤ Y and, by determinants, g := g_{1}g_{2}g_{3}g_{4} satisfies U(g − 1) = Y. □

Proposition A.18. Suppose that T ≅ A_{2} ⊗ E_{8}.

(i) Then D(T) ≅ 3^{8} and the natural \frac{1}{3}ℤ/ℤ-valued quadratic form has maximal Witt index; in fact, there is a natural identification of quadratic spaces D(T) with E_{8}/3E_{8}, up to scaling.

(ii) Define O_{k} := {X|T ≤ X ≤ T^{∗}, X is an integral lattice, dim(X/T) = k} (dimension here means over 𝔽_{3}).

(a) O_{k} is nonempty if and only if 0 ≤ k ≤ 4;

(b) O_{k} consists of even lattices for each k, 0 ≤ k ≤ 4;

(c) On T^{∗}/T, the action of g, the isometry of order 3 on T corresponding to an order 3 symmetry of the A_{2} tensor factor, is trivial. Therefore any lattice between T and T^{∗} is g-invariant.

(iii) the lattices in O_{k} embed in E_{8} ⊥ E_{8}. For a fixed k, the embeddings are unique up to the action of O(E_{8} ⊥ E_{8}).

(iv) the lattices in O_{k} are rootless if and only if k = 0.

Proof. (i) This follows since the quotient T^{∗}/T is covered by \frac{1}{3}P, where P ≅ \sqrt{6}E_{8} is ann_{T} (E), where E is one of the three EE_{8}-sublattices of T.

(ii) Observe that X ∈ O_{k} if and only if X/T is a totally singular subspace of D(T). This implies (a). An integral lattice is even if it contains an even sublattice of odd index. This implies (b). For (c), note that T^{∗}/T is covered by \frac{1}{3}P and P (g − 1) ≤ T (g − 1)^{2} = 3T. Hence g acts trivially on D(T) = T^{∗}/T. This means any lattice Y such that T ≤ Y ≤ T^{∗} is g-invariant.

(iii) and (iv) First, we take X ∈ O_{4} and prove X ≅ E_{8} ⊥ E_{8}. Such an X is even, unimodular and has rank 16, so is isometric to HS_{16} or E_{8} ⊥ E_{8}. Since X has a fixed point free automorphism of order 3, X ≅ E_{8} ⊥ E_{8}. Such an automorphism fixes both direct summands. Call these summands X_{1} and X_{2}. Define Y_{i} := X_{i}(g − 1), for i = 1, 2. Thus, Y_{i} ≅ \sqrt{3}E_{8}.

The action of g on X_{i} ≅ E_{8} is unique up to conjugacy, namely as a diagonally embedded cyclic group of order 3 in a natural O(A_{2})^{4} subgroup of O(X_{i}) ≅ W eyl(E_{8}) (this follows from the corresponding conjugacy result for O^{+}(8, 2) ≅ W eyl(E_{8})/Z(W eyl(E_{8})).

We consider how T embeds in X. Since |X : T | = 3^{4}, |X_{i} : T ∩ X_{i}| divides 3^{4}. Since T ∩ X_{i} ≥ X_{i}(g − 1) and T has no roots, (A.17) implies that T ∩ X_{i} = Y_{i}, for i = 1, 2.

If U ∈ O_{k}, T ≤ U ≤ X, then rootlessness of U implies that U ∩ X_{i} = Y_{i} for i = 1, 2. Therefore, U = T, i.e. k = 0. □

Lemma A.19. Let q be an odd prime power and let (V, Q) be a finite dimensional quadratic space over 𝔽_{q}. Let c be a generator of 𝔽_{q}^{×}.

If dim(V) is odd, there exists g ∈ GL(V) so that gQ = c^{2}Q.

If dim(V) is even, there exists g ∈ GL(V) so that gQ = cQ.

Proof. The scalar transformation c takes Q to c^{2}Q. This proves the result in case dim(Q) is odd. Now suppose that dim(V) is even.

Suppose that V has maximal Witt index. Let V = U ⊕ U′, where U, U′ are each totally singular. We take g to be c on U and 1 on U′.

Suppose that V has nonmaximal Witt index. The previous paragraph allows us to reduce the proof to the case dim(V) = 2 with V anisotropic. (One could also observe that if we write V as the orthogonal direct sum of nonsingular 2-spaces, the result follows from the case dim(V) = 2.) Then V may be identified with 𝔽_{q^{2}} and Q with a scalar multiple of the norm map. We then take g to be multiplication by a scalar b ∈ 𝔽_{q^{2}} such that b^{q+1} = c. □

B Characterizations of lattices of small rank

Some results in this section are in the literature. We collect them here for convenience.

Lemma B.1. Let J be a rank 2 integral lattice. If det(J) ∈ {1, 2, 3, 4, 5, 6}, then J contains a vector of norm 1 or 2. If det(J) ∈ {1, 2}, J is rectangular. If J is even, J ≅ A_{1} ⊥ A_{1} or A_{2}.

Proof. The first two statements follows from values of the Hermite function. Suppose that J is even. Then J has a root, say u. Then ann_{J}(u) has determinant \frac{1}{2}det(J) or 2 det(J). If ann_{J}(u) = \frac{1}{2}det(J), then J is an orthogonal direct sum ℤu ⊥ ℤv, for some vector v ∈ J. For det(J) to be at most 6 and J to be even, (v, v) = 2 and J is the lattice A_{1} ⊥ A_{1}. Now assume that ann_{J}(u) has determinant 2 det(J), an integer at most 12. Let v be a basis for ann_{J}(u). Then \frac{1}{2}(u + v) ∈ J and so \frac{1}{4}(2 + (v, v)) ∈ 2ℤ since J is assumed to be even. Therefore, 2 + (v, v) ∈ 8ℤ. Since (v, v) ≤ 12, (v, v) = 6. Therefore, \frac{1}{2}(u + v) is a root and we get J ≅ A_{2}. □

Lemma B.2. Let J be a rank 3 integral lattice. If det(J) ∈ {1, 2, 3}, then J is rectangular or J is isometric to ℤ ⊥ A_{2}. If det(J) = 4, J is rectangular or is isometric to A_{3}.

Proof. If J contains a unit vector, J is orthogonally decomposable and we are done by (B.1). Now use the Hermite function: H(3, 2) = 1.67989473 …, H(3, 3) = 1.92299942 … and H(3, 4) = 2.11653473 . . .. We therefore get an orthogonal decomposition unless possibly det(J) = 4 and J contains no unit vector. Assume that this is so.

If D(J) is cyclic, the lattice K = J + 2J^{∗} which is strictly between J and J^{∗} is integral and unimodular, so is isomorphic to ℤ^{3}. So, J has index 2 in ℤ^{3}, and the result is easy to check. If D(J) ≅ 2 × 2, we are done by a similar argument provided a nontrivial coset of J in J^{∗} contains a vector of integral norm. If this fails to happen, we quote (A.9) to get a contradiction. □

Lemma B.3. Suppose that X is an integral lattice which has rank 4 and determinant 4. Then X embeds with index 2 in ℤ^{4}. If X is odd, X is isometric to one of 2ℤ ⊥ ℤ^{3}, A_{1} ⊥ A_{1} ⊥ ℤ^{2}, A_{3} ⊥ ℤ. If X is even, X ≅ D_{4}.

Proof. Clearly, if X embeds with index 2 in ℤ^{4}, X may be thought of as the annihilator mod 2 of a vector w ∈ ℤ of the form (1, …, 1, 0, …, 0). The isometry types for X correspond to the cases where the weight of w is 1, 2, 3 and 4. It therefore suffices to demonstrate such an embedding.

First, assume that D(X) is cyclic. Then X + 2X^{∗} is an integral lattice (since (2x, 2y) = (4x, y), for x, y ∈ X^{∗}) and is unimodular, since it contains X with index 2. Then the classification of unimodular integral lattices of small rank implies X + 2X^{∗} ≅ ℤ^{4}, and the conclusion is clear.

Now, assume that D(X) is elementary abelian. By (A.9), there is a nontrivial coset u + X of X in X^{∗} for which (u, u) is an integer. Therefore, the lattice X′ := X + ℤu is integral and unimodular. By the classification of unimodular integral lattices, X′ ≅ ℤ^{4}. □

Theorem B.4. Let L be a unimodular integral lattice of rank at most 8. Then L ≅ ℤ^{n} or L ≅ E_{8}.

Proof. This is a well-known classification. The article [GrE8] has an elementary proof and discusses the history. □

The next result is well known. The proof may be new.

Proposition B.5. Let X be an integral lattice of determinant 3 and rank at most 6. Then X is rectangular; or X ≅ A_{2} ⊥ ℤ^{m}, for some m ≤ 4; or X ≅ E_{6}.

Proof. Let u ∈ X^{∗} \ X. Since 3u ∈ X, (u, u) ∈ \frac{1}{3}ℤ. Since det(X^{∗}) = \frac{1}{3}, (u, u) ∈ \frac{1}{3} + ℤ or \frac{2}{3} + ℤ.

Suppose (u, u) ∈ \frac{1}{3} + ℤ. Let T ≅ A_{2}. Then we quote (D.6) to see that there is a unimodular lattice, U, which contains X ⊥ T with index 3.

Suppose U is even. By the classification (B.4), U ≅ E_{8}. A well-known property of E_{8} is that all A_{2}-sublattices are in one orbit under the Weyl group. Therefore, X ≅ E_{6}.

If U is not even, U ≅ ℤ^{n}, for some n ≤ 8. Any root in ℤ^{n} has the form (±1, ±1, 0, 0, …, 0, 0). It follows that every A_{2} sublattice of ℤ^{n} is in one orbit under the isometry group 2 ≀ Sym_{n}. Therefore X = ann_{U}(T) is rectangular.

Suppose (u, u) ∈ \frac{2}{3} +ℤ. Then we consider a unimodular lattice W which contains X ⊥ ℤv with index 3, where (v, v) = 3. By the classification, W ≅ ℤ^{7}. Any norm 3 vector in ℤ^{n} has the form (±1, ±1, ±1, 0, 0, …, 0) (up to coordinate permutation). Therefore, ann_{W} (v) must be isometric to ℤ^{4} ⊥ A_{2}. □

Lemma B.6. If M is an even integral lattice of determinant 5 and rank 4, then M ≅ A_{4}.

Proof. Let u ∈ M^{∗} so that u + M generates D(M). Then (u, u) = \frac{k}{5}, where k is an integer. Since 5u ∈ M, k is an even integer. Since H(4, \frac{1}{5}) = 1.029593054 …, a minimum norm vector in M^{∗} does not lie in M, since M is an even lattice. We may assume that u achieves this minimum norm. Thus, k ∈ {2, 4}.

Suppose that k = 4. Then we may form M ⊥ ℤ5v, where (v, v) = \frac{1}{5}. Define w := u+v. Thus, P := M+ℤw is a unimodular integral lattice. By the classification, P ≅ ℤ^{5}, so we identify P with ℤ^{5}. Then M = ann_{P} (y) for some norm 5 vector y. The only possibilities for such y ∈ P are (2, 1, 0, 0, 0), (1, 1, 1, 1, 1), up to monomial transformations. Since M is even, the latter possibility must hold and we get M ≅ A_{4}.

Suppose that k = 2. We let Q be the rank 2 lattice with Gram matrix (\begin{matrix}3 & 2 \\ 2 & 3 \end{matrix}).

So, det(Q) = 5 and there is a generator v ∈ Q^{∗} for Q^{∗} modulo Q which has norm \frac{3}{5}. We then form M ⊥ Q and define w := u + w. Then P := M + Q + ℤw is an integral lattice of rank 6 and determinant 1. By the classification, P ≅ ℤ^{6}. In P, M is the annihilator of a pair of norm 3 vector, say y and z. Each corresponds in ℤ^{6} to some vector of shape (1, 1, 1, 0, 0, 0), up to monomial transformation. Since M is even, the 6-tuples representing y and z must have supports which are disjoint 3-sets. However, since (y, z) = 2, by the Gram matrix, we have a contradiction. □

Notation B.7. We denote by M(4, 25) an even integral lattice of rank 4 and determinant 25. By (B.8), it is unique.

Lemma B.8. (i) There exists a unique, up to isometry, rank 4 even integral lattice whose discriminant group has order 25.

(ii) It is isometric to a glueing of the orthogonal direct sum A_{2} ⊥ \sqrt{5}A_{2} by a glue vector of the shape u + v, where u is in the dual of the first summand and (u, u) = \frac{2}{3}, and where v is in the dual of the second summand and has norm \frac{10}{3}.

(iii) The set of roots forms a system of type A_{2}; in particular, the lattice does not contain a pair of orthogonal roots.

(iv) The isometry group is isomorphic to Sym_{3}×Sym_{3}×2, where the first factor acts as the Weyl group on the first summand in (iii) and trivially on the second, the second factor acts as the Weyl group on the second summand of (iii) and trivially on the first, and where the third direct factor acts as −1 on the lattice.

(v) The isometry group acts transitively on (a) the six roots; (b) the 18 norm 4 vectors; (c) ordered pairs of norm 2 and norm 4 vectors which are orthogonal; (d) length 4 sequences of orthogonal vectors whose norms are 2, 4, 10, 20.

(vi) An orthogonal direct sum of two such embeds as a sublattice of index 5^{2} in E_{8}.

Proof. The construction of (ii) shows that such a lattice exists and it is easy to deduce (iii), (iv), and (v).

We now prove (i). Suppose that L is such a lattice. We observe that if the discriminant group were cyclic of order 25, the unique lattice strictly between L and its dual would be even and unimodular. Since L has rank 4, this is impossible. Therefore, the discriminant group has shape 5^{2}.

Since H(4, 25) = 3.44265186 …, L contains a root, say u. Define N := ann_{L}(u). Since H(3, 50) = 4.91204199 …, N contains a norm 2 or 4 element, say v.

Define R := ℤu ⊥ ℤv and P := ann_{L}(R), a sublattice of rank 2 and determinant 2(v, v)·25. Also, the Sylow 5-group of D(P) has exponent 5. Then P ≅ \sqrt{5}J, where J is an even, integral lattice of rank 2 and det(J) = 2(v, v). Since det(J) is even and the rank of the natural bilinear form on J/2J is even, it follows that J = \sqrt{2}K, for an integral, positive definite lattice K. We have det(K) = \frac{1}{2}(v, v) ∈ \{1, 2\} and so K is rectangular. Also, P ≅ \sqrt{10}K. So, P has rectangular basis w, x whose norm sequence is 10, 5(v, v)

Suppose that (v, v) = 2. Also, L/(R ⊥ P) ≅ 2 × 2. A nontrivial coset of R ⊥ P contains an element of the form \frac{1}{2}y + \frac{1}{2}z, where y ∈ span\{u, v\} and z ∈ span\{w, x\}. We may furthermore arrange for y = au + bv, z = cw + dx, where a, b, c, d ∈ {0, 1}. For the norm of \frac{1}{2}y + \frac{1}{2}z to be an even integer, we need a = b = c = d = 0 or a = b = c = d = 1. This is incompatible with L/(R ⊥ P) ≅ 2 × 2. Therefore, (v, v) = 4.

We have L/(R ⊥ P) ≅ 5^{2}. Therefore, \frac{1}{5}w and \frac{1}{5}x are in L^{∗} but are not in L Form the orthogonal sum L ⊥ ℤy, where (y, y) = 5. Define w := \frac{1}{5}x + \frac{1}{5}y. Then

(w, w) = 1. Also, Q := L + ℤw has rank 5, is integral and contains L ⊥ ℤy with index 5, so has determinant 5. Since w is a unit vector, S := ann_{Q}(w) has rank 4 and determinant 5, so S ≅ A_{4}. Therefore, Q = ℤw ⊥ S and L = ann_{Q}(y) for some y ∈ Q of norm 5, where y = e + f, e ∈ ℤw, f ∈ S. Since S has no vectors of odd norm, e ≠ 0 has odd norm. Since (y, y) = 5 and since (e, e) is a perfect square, (e, e) = 1 and (f, f) = 4. Since O(A_{4}) acts transitively on norm 4 vectors of A_{4}, f is uniquely determined up to the action of O(S). Therefore, the isometry type of L is uniquely determined.

It remains to prove (vii). For one proof, use (B.9). Here is a second proof. We may form an orthogonal direct sum of two such lattices and extend upwards by certain glue vectors.

Let M_{1} and M_{2} be two mutually orthogonal copies of L. Let u, v, w, x be the orthogonal elements of M_{1} of norm 2, 4, 10, 20 as defined in the proof of (i). Let u′, v′, w′, x′ be the corresponding elements in M_{2}. Set

γ = \frac{1}{5}(w + x + x′) and γ′ = \frac{1}{5}(x + w′ + x′).

Their norm are both 2. By computing the Gram matrix, it is easy to show that E = span_{ℤ}{M_{1}, M_{2}, γ, γ′} is integral and has determinant 1. Thus, E is even and so E ≅ E_{8}. □

Lemma B.9. Let p be a prime which is 1(mod 4). Suppose that M, M′ are lattices such that D(M) and D(M′) are elementary abelian p-groups which are isometric as quadratic spaces over 𝔽_{p}. Let ψ be such an isometry and let c ∈ 𝔽_{p} be a square root of −1. Then the overlattice N of M ⊥ M′ spanned by the “diagonal cosets” {α + cαψ|α ∈ D(M)} is unimodular. Also, N is even if M and M′ are even.

Proof. The hypotheses imply that N contains M + M′ with index |det(M)|, so is unimodular. It is integral since the space of diagonal cosets so indicated forms a maximal totally singular subspace of the quadratic space D(M) ⊥ D(M′). The last sentence follows since |N : M ⊥ M′| is odd. □

Lemma B.10. An even rank 4 lattice with discriminant group which is elementary abelian of order 125 is isometric to \sqrt{5}A_{4}^{∗}.

Proof. Suppose that L is such a lattice. Then det(\sqrt{5}L^{∗}) = 5. We may apply the result (B.6) to get \sqrt{5}L^{∗} ≅ A_{4}. □

Lemma B.11. An even integral lattice of rank 4 and determinant 9 is isometric to A_{2}^{2}.

Proof. Let M be such a lattice. Since H(4, 9) = 2.66666666 … and H(3, 18) = 3.494321858 …, M contains an orthogonal pair of roots, u, v. Define P := ℤu ⊥ ℤv. The natural map M → D(P) is onto since (detM, detP) = 1. Therefore, Q := ann_{M}(P) has determinant 36 and the image of M in D(Q) is 2×2. Therefore, Q has a rectangular basis w, x, each of norm 6 or with respective norms 2, 18.

We prove that 2, 18 does not occur. Suppose that it does. Then there is a sublattice N isometric to A_{1}^{3}. Since there are no even integer norm vectors in N^{∗} \ N, N is a direct summand of M. By coprimeness, the natural map of M to D(N) ≅ 2^{3} is onto. Then the natural map of M to D(ℤx) has image isomorphic to 2^{3}. Since D(ℤx) is cyclic, we have a contradiction.

Since M/(P ⊥ Q) ≅ 2 × 2 and M is even, it is easy to see that M is one of M_{1} := span\{P, Q, \frac{1}{2}(u + w), \frac{1}{2}(v + x)\} or M_{2} := span\{P, Q, \frac{1}{2}(u + x), \frac{1}{2}(v + w)\}. These two overlattices are isometric by the isometry defined by u ↦ u, v ↦ v, w ↦ x, x ↦ w. It is easy to see directly that they are isometric to A_{2}^{2}. For example, M_{1} = span\{u, x, \frac{1}{2}(u + x)\} ⊥ span\{v, w, \frac{1}{2}(v + w)\}. □

C Nonexistence of particular lattices

Lemma C.1. Let X ≅ ℤ^{2}. There is no sublattice of X whose discriminant group is 3 × 3.

Proof. Let Y be such a sublattice. Its index is 3. Let e, f be an orthonormal basis of X. Then Y contains W := span{3e, 3f} with index 3. Let v ∈ Y \ W, so that Y = W + ℤv. If v is e or f, clearly D(Y) is cyclic of order 9. We may therefore assume that v = e + f or e − f. Then Y is spanned by 3e and e ± f, and so its Smith invariant sequence 1, 9. This final contradiction completes the proof. □

Lemma C.2. There does not exist an even rank 4 lattice of determinant 3.

Proof. Let L be such a lattice and let u ∈ L^{∗} so that u generates L^{∗} modulo L. Then (u, u) = \frac{k}{3} for some integer k > 0. Since L is even, k is even. Since H(4, \frac{1}{3}) = 1.169843567 ⋯ < 4/3, we may assume that k = 2.

We now form L ⊥ ℤ(3v), where (v, v) = \frac{1}{3}. Define w := u + v. The lattice P := L + ℤw is unimodular, so is isometric to ℤ^{5}. Since det(M) = 3, M = ann_{P} (y) for some vector y of norm 3. This forces M to be isometric to A_{2} ⊥ ℤ^{2}, a contradiction to evenness. □

Corollary C.3. There does not exist an even rank 4 lattice whose discriminant group is elementary abelian of order 3^{3}.

Proof. If M is such a lattice, then 3M^{∗} has rank 4 and determinant 3. Now use (C.2). □

D Properties of particular lattices

Lemma D.1. Suppose that M ≠ 0 is an SSD sublattice of E_{8} and that rank(M) ≤ 4. Then M contains a root, r, and ann_{M}(r) is an SSD sublattice of E_{8} of rank (rank(M) − 1). Also M ≅ A_{1}^{rank(M)} or D_{4}.

Proof. Let L := E_{8}. Let d := det(M), a power of 2 and k := rank(M). Note that D(M) is elementary abelian of rank at most k. If d = 2^{k}, then \frac{1}{\sqrt{2}}M is unimodular, hence is isometric to ℤ^{k}, and the conclusion holds. So, we assume that d < 2^{k}. For n ≤ 4 and d|8, it is straightforward to check that the Hermite function H satisfies H(n, d) < 4. Therefore, M contains a root, say r.

Suppose that M is a direct summand of L. By (2.8), N := ann_{M}(r) is RSSD in L, hence is SSD in L by (2.7) and we apply induction to conclude that N is an orthogonal sum of A_{1}s. So M contains M′ an orthogonal sum of A_{1}s with index 1 or 2. Furthermore, det(M′) = 2^{k}. If the index were 1, we would be done, so we assume the index is 2. Since d > 1, d = 2, 4 or 8. By the index formula for determinants, 2^{2} is a divisor of d. Therefore, d = 4 or 8. However, if d = 8, then det(M′) = 32, which is impossible since rank(M′) ≤ 4. Therefore, d = 4 and rank(M′) = 4. It is trivial to deduce that M ≅ D_{4}.

We now suppose that M is not a direct summand of L. Let S be the direct summand of L determined by M. Then S is SSD and the above analysis says S is isometric to some A_{1}^{m} or D_{4}. The only SSD sublattices of A_{1}^{m} are the orthogonal direct summands. The only SSD sublattices of D_{4} which are proper have determinant 2^{4} and so equal twice their duals and therefore are isometric to A_{1}^{4}. □

Lemma D.2. Suppose that M is an SSD sublattice of E_{8}. Then M is one of the sublattices in Table 10 and Table 11.

Proof. We may assume that 1 ≤ rank(M) ≤ 7. First we show that M contains a root.

If rank(M) ≤ 4, this follows from (D.1). If rank(M) ≥ 4, then N := ann_{L}(M) has rank at most 4, so is isometric to one of A_{1}^{k} or D_{4}.

Table 10: SSD sublattices of E_{8} which span direct summands

Rank  Type
0  0
1  A_{1}
2  A_{1} ⊥ A_{1}
3  A_{1} ⊥ A_{1} ⊥ A_{1}
4 A_{1} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, D_{4}
5  D_{4} ⊥ A_{1}
6  D_{6}
7  E_{7}
8  E_{8}

Table 11: SSD sublattices of E_{8} which do not span direct summands

Rank  Type  contained in the summand
4  A_{1} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}  D_{4}
5  A_{1}^{⊥5}  D_{4} ⊥ A_{1}
6  A_{1}^{⊥6}, D_{4} ⊥ A_{1} ⊥ A_{1}  D_{6}
7 A_{1}^{⊥7}, D_{4} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, D_{6} ⊥ A_{1}  E_{7}
8 A_{1}^{⊥8}, D_{4} ⊥ A_{1}^{⊥4}, D_{4} ⊥ D_{4}, E_{8}
D_{6} ⊥ A_{1} ⊥ A_{1}, E_{7} ⊥ A_{1}

Suppose that rank(N) = 4. If N ≅ A_{1}^{4} and so ann_{L}(N) ≅ A_{1}^{4}, which contains M and whose only SSD sublattices are orthogonal direct summands, so M = ann_{L}(N) and the result follows in this case. If N ≅ D_{4}, then M ≅ D_{4} or A_{1}^{4} by an argument in the proof of (D.1).

We may therefore assume that rank(N) ≤ 3, whence N ≅ A_{1}^{rank(N)} and rank(M) ≥ 5. Furthermore, we may assume that rank(M) > rank(D(M)), or else we deduce that M ≅ A_{1}^{rank(M)}. It follows that det(M) is a proper divisor of 128.

Note that D(M) has rank which is congruent to rank(M) mod 2 (this follows from the index determinant formula plus the fact that D(M) is an elementary abelian 2-group). Therefore, since rank(M) ≤ 7, det(M) is a proper divisor of 64, i.e. is a divisor of 32.

For any d ≥ 2, H(n, d) is an increasing function of n for n ∈ [5, ∞). For fixed n, H(n, d) is increasing as a function of d. Since H(7, 32) = 3.888997243 …, we conclude that M contains a root, say r.

Since L/(M ⊥ N) is an elementary abelian 2-group by (A.6), M ⊥ N ≥ 2L. Also, r + 2L contains a frame, F, a subset of 16 roots which span an A_{1}^{8}-sublattice of L. Since roots are orthogonally indecomposable in L, F = (F ∩ M) ∪ (F ∩ N). It follows that M contains a sublattice M′ spanned by F ∩ M, M′ ≅ A_{1}^{rank(M)}, and so M is generated by M′ and glue vectors of the form \frac{1}{2}(a + b + c + d), where a, b, c, d are linearly independent elements of F ∩ M. It is now straightforward to obtain the list in the conclusion by considering the cases of rank 5, 6 and 7 and subspaces of the binary length 8 Hamming code. □

Lemma D.3. Let X ≅ D_{6} and let 𝔖 = {Y ⊂ X| Y ≅ DD_{6}}. Then O(X) acts transitively on 𝔖.

Proof. Let X ≅ D_{6} and R = 2X^{∗}. Since D_{6}^{∗}/D_{6} ≅ ℤ_{2} × ℤ_{2}, we have X ≥ R ≥ 2X and R/2X ≅ D_{6}^{∗}/D_{6} ≅ ℤ_{2} × ℤ_{2}. Thus, the index of R in X is 2^{6}/2^{2} = 2^{4}.

Let ¯ : X → X/2X be the natural projection. Then for any Y ∈ 𝔖, Ȳ is a totally isotropic subspace of X̄. Note that R/2X is the radical of X̄ and thus X/R ≅ 2^{4} is nonsingular. Therefore, dim(Y + R)/R ≤ 2 and Y/(Y ∩ R) ≅ (Y + R)/R also has dimension ≤ 2.

First we shall show that Y ≥ 2X and dim(Y + R)/R = 2. Consider the tower

Y ≥ Y ∩ R ≥ Y ∩ 2X.

Since Y ∩R+2X is doubly even but R is not, R ≠ Y ∩R+2X and (Y ∩R+2X)/2X ≨ R/2X. Thus (Y ∩ R + 2X)/2X ≅ Y ∩ R/Y ∩ 2X has dimension ≤ 1 and hence

|Y : Y ∩ 2X| = |Y : Y ∩ R| · |Y ∩ R : Y ∩ 2X| ≤ 2^{3}.

However, det(Y) = 2^{8} and det(2X) = 2^{12}4 = 2^{14}. Therefore, |Y : Y ∩ 2X| ≥ 2^{3} and hence |Y : Y ∩ 2X| = 2^{3}. This implies Y ∩ 2X = 2X, i.e., Y ≥ 2X. It also implies that |Y + R : R| = |Y : Y ∩ R| = 2^{2} and hence (Y + R)/R is a maximal isotropic subspace of X/R ≅ ℤ_{2}^{4}.

Let 2X ≤ M ≤ X be such that M/2X is maximal totally isotropic subspace. Then M ≥ R and \frac{1}{\sqrt{2}}M is an integral lattice. Set M_{even} = \{α ∈ M| \frac{1}{2}(α, α) ∈ 2ℤ\}. Then M_{even} is a sublattice of M of index 1 or 2. If Y is contained in such M, then Y = M_{even}. That means Y is uniquely determined by M.

Finally, we shall note that the Weyl group acts on X/R as the symmetric group Sym_{6}. Moreover, Sym_{6} acts faithfully on X/R ≅ ℤ_{2}^{4} and fixes the form ( , ), so it acts as Sp(4, 2). Thus it acts transitively on maximal totally isotropic subspace and we have the desired conclusion. □

Lemma D.4. Let X ≅ D_{6} and let Y ≅ DD_{6} be a sublattice of X. Then there exists a subset {η_{1}, …, η_{6}} ⊂ X with (η_{i}, η_{j}) = 2δ_{i,j} such that

Y = span_{ℤ}{η_{i} ± η_{j}| i, j = 1, …, 6}

and

X = span_{ℤ} \{η_{1}, η_{2}, η_{4}, η_{6}, \frac{1}{2}(−η_{1} + η_{2} − η_{3} + η_{4}), \frac{1}{2}(−η_{3} + η_{4} − η_{5} + η_{6})\} .

Proof. We shall use the standard model for D_{6}, i.e.,

D_{6} = {(x_{1}, x_{2}, …, x_{6}) ∈ ℤ^{6}| x_{1} + ⋯ + x_{6} ≡ 0 mod 2}.

Let β_{1} = (1, 1, 0, 0, 0, 0), β_{2} = (−1, 1, 0, 0, 0, 0), β_{3} = (0, 0, 1, 1, 0, 0), β_{4} = (0, 0, −1, 1, 0, 0), β_{5} = (0, 0, 0, 0, 1, 1), and β_{6} = (0, 0, 0, 0, −1, 1). Then, (β_{i}, β_{j}) = 2δ_{i,j} and

W = span_{ℤ}{β_{i} ± β_{j}| i, j = 1, 2, 3, 4, 5, 6} ≅ DD_{6}.

Note also that {(1, 1, 0, 0, 0, 0), (−1, 1, 0, 0, 0, 0), (0, 0, −1, 1, 0, 0), (0, 0, 0, 0, −1, 1), (−1, 0, −1, 0, 0, 0), (0, 0, −1, 0, −1, 0)} forms a basis for X (since their Gram matrix has determinant 4). By expressing them in β_{1}, …, β_{6}, we have

X = span_{ℤ} \{β_{1}, β_{2}, β_{4}, β_{6}, \frac{1}{2}(−β_{1} + β_{2} − β_{3} + β_{4}), \frac{1}{2}(−β_{3} + β_{4} − β_{5} + β_{6})\} .

Let Y ≅ DD_{6} be a sublattice of X. Then by Lemma D.3, there exists g ∈ O(X) such that Y = W g. Now set η_{i} = β_{i}g and we have the desired result. □

Lemma D.5. Let X ≅ D_{4} and let Y ≅ DD_{4} be a sublattice of X. Then Y = 2X^{∗} and hence X ≤ \frac{1}{2}Y.

Proof. The radical of the form on X/2X is 2X^{∗}/2X. If W is any A_{2}-sublattice of X, its image in X/2X complements 2X^{∗}/2X. Therefore, every element of X \ 2X^{∗} has norm 2(mod 4). It follows that Y ≤ 2X^{∗}. By determinants, Y = 2X^{∗}. □

Lemma D.6. Let L be the A_{2}-lattice, with basis of roots r, s. Let g ∈ O(L) and |g| = 3. (i) Then L^{∗} ≅ \frac{1}{\sqrt{3}}L and every nontrivial coset of L in L^{∗} has minimum norm \frac{2}{3}. All norms in such a coset lie in \frac{2}{3} + 2ℤ. (ii) If x is any root, ann_{L}(x) = ℤ(xg − xg^{2}) and xg − xg^{2} has norm 6.

Proof. (i) The transformation g : r ↦ s, s ↦ −r − s is an isometry of order 3 and h := g − g^{−1} satisfies h^{2} = −3 and (xh, yh) = 3(x, y) for all x, y ∈ ℚ ⊗ L. Furthermore, g acts indecomposably on L/3L ≅ 3^{2}. We have \frac{1}{3}L = Lh^{−2} ≥ Lh^{−1} ≥ L, with each containment having index 3 (since h^{2} = −3). Since L^{∗} lies strictly between L and \frac{1}{3}L and is g-invariant, L^{∗} = Lh^{−1}.

Since L^{∗} ≅ \frac{1}{\sqrt{3}}L, the minimum norm in L^{∗} is \frac{2}{3} by (D.6). The final statement follows since the six roots ±r, ±s, ±(r + s) fall in two orbits under the action of ⟨g⟩, the differences rg^{i} − sg^{i} lie in 3L^{∗} and r and −r are not congruent modulo 3L^{∗}.

(ii) The element xg − xg^{2} has norm 6 and is clearly in ann_{L}(x). The sublattice ℤx ⊥ ℤ(xg − xg^{2}) has norm 2 · 6 = 12, so has index 3 in L, which has determinant

3. Since L is indecomposable, ann_{L}(x) is not properly larger than ℤ(xg − xg^{2}). □

Lemma D.7. Let X ≅ A_{2} and let Y ≤ X, |X : Y | = 3. Then either Y = 3X^{∗}

and its Smith invariant sequence 3, 9; or Y has Gram matrix (\begin{matrix}2 & −3 \\ −3 & 18 \end{matrix}), which has

Smith invariant sequence 1, 27. In particular, such Y has D(Y) of rank 2 if and only if Y = 3X^{∗}.

Proof. Let r, s, t be roots in X such that r+s+t = 0. Any two of them form a basis for X. The sublattices span{r, 3s}, span{s, 3t}, span{t, 3r} of index 3 are distinct (since their sets of roots partition the six roots of X) and the index 3 sublattice 3X^{∗} contains no roots. Since there are just four sublattices of index 3 in X, we have listed all four. It is straightforward to check the assertions about the Gram matrices. Note that 3X^{∗} ≅ \sqrt{3}X. □

Proposition D.8. Suppose that M is a sublattice of L ≅ E_{8}, that M is a direct summand of M, that M has discriminant group which is elementary abelian of order 3^{s}, for some s. Then M is 0, L, or is a natural A_{2}, A_{2} ⊥ A_{2}, or E_{6} sublattice. The respective values of s are 0, 0, 1, 2 and 1. In case M is not a direct summand, the list of possibilities expands to include A_{2} ⊥ A_{2} ⊥ A_{2}, A_{2} ⊥ A_{2} ⊥ A_{2} ⊥ A_{2} and A_{2} ⊥ E_{6} sublattices.

Proof. One of M and ann_{L}(M) has rank at most 4 and the images of L in their discriminant group are isomorphic. Therefore, s ≤ 4. If s were equal to 4, then both M and ann_{L}(M) would have rank 4, and each would be isometric to \sqrt{3} times some rank 4 integral unimodular lattice. By (B.4), each would be isomorphic to \sqrt{3}ℤ^{4}, which would contradict their evenness. Therefore, s ≤ 3.

The second statement is easy to derive from the first, which we now prove. We may replace M by its annihilator in L if necessary to assume that r :=

rank(M) ≤ 4. Since M is even and det(M) is a power of 3, r is even. We may assume that r ≥ 2 and that s ≥ 1. If r = 2, M ≅ A_{2} (B.1). We therefore may and do assume that r = 4.

For s ∈ {1, 3}, we quote (C.2) and (C.3) to see that there is no such M. If s = 2 we quote (B.11) to identify M. □

Proposition D.9. Suppose that M is a sublattice of L ≅ E_{8}, that M is a direct summand, that M has discriminant group which is elementary abelian of order 5^{s}, for some s ≤ 4. Then M is 0, a natural A_{4} sublattice, the rank 4 lattice M(4, 25) (cf. (B.7)), the rank 4 lattice \sqrt{5}A_{4}^{∗} ≅ A_{4}(1) (cf. (D.10)) or L. The respective values of s are 0, 1, 2, 3 and 0.

Proof. We may replace M by its annihilator in L if necessary to assume that r := rank(M) ≤ 4. Since M is even and det(M) is a power of 5, r is even. We may assume that r ≥ 2 and that s ≥ 1. If r = 2, det(M) ≡ 3(mod 4) (consider the

form of a Gram matrix (\begin{matrix}a & b \\ b & d \end{matrix}), which has even entries on the diagonal and odd

determinant, whence b is odd). This is not possible since det(M) is a power of 5. We therefore may and do assume that r = 4. Suppose that s = 4, i.e., that D(M) ≅ 5^{4}. Then M ≅ \sqrt{5}J, where J is an

integral lattice of determinant 1. Then J ≅ ℤ^{rank(M)}, which is not an even lattice. This is a contradiction since M is even. We conclude s = rank(D(M)) ≤ 3. If s = 0, M is a rank 4 unimodular integral lattice, hence is odd by (B.4), a contradiction. Therefore, 1 ≤ s ≤ 3. The results (B.6), (B.8) and (B.10) identify M. □

Notation D.10. The lattice A_{4}(1) is defined by the Gram matrix

(\begin{matrix}4 & −1 & −1 & −1 \\ −1 & 4 & −1 & −1 \\ −1 & −1 & 4 & −1 \\ −1 & −1 & −1 & 4 \end{matrix}) .

It is spanned by vectors v_{1}, ⋯, v_{5} which satisfy v_{1} + v_{2} + v_{3} + v_{4} + v_{5} = 0 and (v_{i}, v_{j}) = −1 + 5δ_{ij}. Its isometry group contains Sym_{5} × ⟨−1⟩.

Lemma D.11. Suppose that u is a norm 4 vector in an integral lattice U where D acts as isometries so that 1+g+g^{2}+g^{3}+g^{4} acts as 0. Then one of three possibilities occurs.

(i) The unordered pair of scalars (u, ug) = (u, g^{4}u) and (u, ug^{2}) = (u, ug^{3}) equals the unordered set {0, −2}; or

(ii) The unordered pair of scalars (u, ug) = (u, g^{4}u) and (u, ug^{2}) = (u, ug^{3}) equals the unordered set {−3, 1}; or

(iii) (u, ug) = (u, ug^{2}) = (u, ug^{3}) = (u, g^{4}u) = −1.

The isometry types of the lattice span{u, ug, ug^{2}, ug^{3}, ug^{4}} in these respective cases are AA_{4}, A_{4}, A_{4}(1).

Proof. Straightforward. □

Lemma D.12. Let X ≅ A_{4}(1) (D.10). Then (i) X is rootless and contains exactly 10 elements of norm 4; (ii) Suppose u ∈ X has norm 4. Then ann_{X}(u) ≅ \sqrt{5}A_{3}. (iii) O(X) ≅ 2 × Sym_{5}; furthermore, if X_{4} is the set of norm 4 vectors and O is

an orbit of a subgroup of order 5 in O(X) on X_{4}, then the subgroup of O(X) which preserves O is a subgroup isomorphic to Sym_{5}, and is the subgroup generated by all reflections.

(iv) Suppose that Y ≅ A_{4} and that g ∈ O(Y) has order 5, then Y (g − 1) ≅ X. Also, O(Y) ∩ O(Y (g − 1)) = N_{O(Y)}(⟨g⟩) ≅ 2 × 5:4, where the right direct factor is a Frobenius group of order 20.

(v) D(X) is elementary abelian.

Proof. (i) If the set of roots R in X were nonempty, then R would have an isometry of order 5. Since the rank of R is at most 4, R would be an A_{4}-system and so the sublattice of X which R generates would be A_{4}, which has determinant 5. Since det(X) = 5^{3}, this is a contradiction.

By construction, X has an cyclic group Z of order 10 in O(X) which has an orbit of 10 norm 4 vectors, which are denoted ±v_{i} in (D.10). Suppose that w is a norm 4 vector outside the previous orbit. Let g ∈ Z have order 5. Then w + wg + wg^{2} + wg^{3} + wg^{4} = 0. Therefore 0 = (v, w + wg + wg^{2} + wg^{3} + wg^{4}), which means that there exists an index i so that (w, vg^{i}) is even. Since vg^{i} and w are linearly independent, (vg^{i}, w) is not ±4 and the sublattice X′ which vg^{i} and w span has rank 2. Since (vg^{i}, w) ∈ {−2, 0, 2}, X′ ≅ AA_{1} ⊥ AA_{1} or AA_{2}. This contradicts (A.3) (in that notation, n = 4, m = 2, p = 5, r = 3).

(ii) Let K := ann_{X}(u). Since (u, u) = 4 is relatively prime to det(X), the natural map X → D(ℤu) is onto. Therefore ℤu ⊥ K has index 4 in X. Hence det(ℤu ⊥ K) = 4^{2}·5^{3} and detK = 4·5^{3}. Since D(X) ≤ D(ℤu ⊥ K) = D(ℤu)×D(K) and the Smith invariant sequence of X is 1,5,5,5, D(K) contains an elementary group 5^{3}. Moreover, the image of X in D(K) isomorphic to the image of X in D(ℤu), which is isomorphic to ℤ_{4}. Therefore, D(K) = ℤ_{4} ×ℤ_{5}^{3} by determinants. Hence, the Smith invariant sequence for K is 5, 5, 20 and so K ≅ \sqrt{5}W, for an integral lattice W such that D(W) = 4. Since X is even, W is even. We identify W with A_{3} by (B.2).

(iii) We use the notation in the proof of (i). By (D.11), for any two distinct vectors of the form vg^{i}, the inner product is −1, so the symmetric group on the set of all vg^{i} acts as isometries on the ℤ-free module spanned by them, and on the quotient of this module by the ℤ-span of v + vg + vg^{2} + vg^{3} + vg^{4}, which is isometric to X.

Pairs of elements of norm 4 fall into classes according to their inner products: ±4, ±1. An orbit of an element of order 5 on X_{4} gives pairs only with inner products 4, −1 (since the sum of these five values is 0). There are two such orbits and an inner product between norm 4 vectors from different orbits is one of −4, 1. The map −1 interchanges these two orbits. Therefore, the stabilizer of O has index 2 in O(X). It contains the map which interchanges distinct vg^{i} and vg^{j} and fixes other vg^{k} in the orbit. Such a map is a reflection on the ambient vector space. Since Sym_{5} has just two classes of involutions, it is clear that every reflection in O(X) = ⟨−1⟩ × Stab_{O(X)}(O) is contained in Stab_{O(X)}(O).

(iv) We have Y (g − 1) = span_{ℤ}{vg^{i} − vg^{j}|i, j ∈ ℤ}. By checking a Gram matrix, one sees that it is isometric to X. We consider O(Y) ∩ O(Y (g − 1)), which clearly contains N_{O(Y)}(⟨g⟩). We show that this containment is equality. We take for Y the standard model, the set of coordinate sum 0 vectors in ℤ^{5}. Take v ∈ Y (g − 1), a norm 4 vector. It has shape (1, 1, −1, −1, 0) (up to reindexing). The coordinate permutation t which transposes the last two coordinates is not in O(Y (g −1)) (since v(t − 1) has norm 2). Therefore O(Y) does not stabilize Y (g − 1). Since N_{O(Y)}(⟨g⟩) is a maximal subgroup of O(Y), it equals O(Y) ∩ O(Y (g − 1)).

(v) Since A_{4}(1) ≅ \sqrt{5}A_{4}^{∗} by (B.10), (A_{4}(1))^{∗} ≅ \frac{1}{\sqrt{5}}A_{4}. Thus, 5A_{4}(1)^{∗} < A_{4}(1) and D(A_{4}(1)) is elementary abelian. □

Lemma D.13. Let X ≅ A_{4}(1) be a sublattice of E_{8}. If X is a direct summand, then ann_{E_{8}}(X) ≅ A_{4}(1).

Proof. . Let Y ≅ E_{8} and let X ≅ A_{4}(1) be a sublattice of Y. Since X is a direct summand, the natural map Y → D(X) is onto. Similarly, the natural map from Y → D(ann_{Y} (X)) is also onto and these two images are isomorphic. Thus, D(ann_{Y} (X)) ≅ D(X) ≅ 5^{3}. Hence, ann_{Y} (X) is isomorphic to A_{4}(1) by (B.10).□

Remark D.14. Note that A_{4}(1) can be embedded into E_{8} as a direct summand. Recall that

\begin{matrix}A_{4}(1) ≅ \sqrt{5}A_{4}^{∗} \\ = span_{ℤ}\{\frac{1}{\sqrt{5}}(1, 1, 1, 1, −4), \frac{1}{\sqrt{5}}(1, 1, 1, −4, 1), \frac{1}{\sqrt{5}}(1, 1, −4, 1, 1), \frac{1}{\sqrt{5}}(1, −4, 1, 1, 1)\}. \end{matrix}

Then,

(A_{4}(1))^{∗} ≅ \frac{1}{\sqrt{5}}A_{4} = \{\frac{1}{\sqrt{5}}(x_{1}, …, x_{5}) | ∑_{i=1}^{5} x_{i} = 0 and x_{i} ∈ ℤ, i = 1, …, 5\} .

Let

Y = span_{ℤ} \{ {A_{4}(1) ⊥ A_{4}(1), \frac{1}{\sqrt{5}}(1, −1, 0, 0, 0 | 2, −2, 0, 0, 0) \atop \frac{1}{\sqrt{5}}(0, 1, −1, 0, 0 | 0, 2, −2, 0, 0), \frac{1}{\sqrt{5}}(0, 0, 1, −1, 0 | 0, 0, 2, −2, 0,)}\} .

Then Y is a rank 8 even lattice and |Y : A_{4}(1) ⊥ A_{4}(1)| = 5^{3}. Thus det(Y) = 1 and Y ≅ E_{8}. Clearly, A_{4}(1) is a direct summand by the construction.

Remark D.15. We have O(E_{6}) ≅ W eyl(E_{6}) × ⟨−1⟩. Thus, outer involutions are negatives of inner involutions. The next result does not treat inner and outer cases differently.

Lemma D.16. Let t ∈ O(E_{6}) be an involution. The negated sublattice for t is either SSD (so occurs in the list for E_{8} (D.2)) or is RSSD but not SSD and is isometric to one of AA_{2}, AA_{2} ⊥ A_{1}, AA_{2} ⊥ A_{1} ⊥ A_{1}, AA_{2} ⊥ A_{1} ⊥ A_{1} ⊥ A_{1}, A_{5}, A_{5} ⊥ A_{1}, E_{6}. Moreover, the isometry types of the RSSD sublattices determine them uniquely up to the action of O(E_{6}).

Proof. Let S be the negated sublattice and assume that it is not SSD. Then the image of E_{6} in D(S) has index 3 and is an elementary abelian 2-group, so that det(S) = 2^{a}3, where a ≤ rank(S). Note that rank(S) ≥ 2. Now, let T := ann_{E_{6}}(S), a sublattice of rank at most 4. Since det(S ⊥ T) = 2^{2a}3, det(T) = 2^{a} and the image of E_{6} in D(T) is all of D(T). Therefore, T is SSD and we may find the isometry type of T among the SSD sublattices of E_{8}. As we search through SSD sublattices of rank at most 4 (all have the form A_{1}^{m} or D_{4}), it is routine to determine the annihilators of their embeddings in E_{6}. □

Lemma D.17. Suppose that R ⊥ Q is an orthogonal direct sum with Q ≅ AA_{2} and R ≅ D_{4}. Let ϕ : D(R) → D(Q) be any monomorphism (recall that D(R) ≅ 2 × 2 and D(Q) ≅ 2 × 2 × 3). Then the lattice X which is between R ⊥ Q and R^{∗} ⊥ Q^{∗} and which is the diagonal with respect to ϕ is isometric to E_{6}. Furthermore, if X is a lattice isometric to E_{6} which contains R ⊥ Q, then X is realized this way.

Proof. Such X have determinant 3. The cosets of order 2 for D_{4} in its dual have odd integer norms (the minimum is 1). The cosets of order 2 for AA_{2} in its dual have odd integer norms (the minimum is 1). It follows that such X above are even lattices. By a well-known characterization, X ≅ E_{6} (cf. (B.5)).

Conversely, suppose that X is a lattice containing R ⊥ Q, X ≅ E_{6}. Since det(X) is odd, the image of the natural map X → D(R) is onto. Therefore, |X : R ⊥ Q| = 4. The image of X in D(Q) is isomorphic to the image of X in D(R). The last statement follows. □

Corollary D.18. (i) Let Y be a sublattice of X ≅ E_{6} so that Y ≅ D_{4}. Then ann_{X}(Y) ≅ AA_{2}.

(ii) Let U be a sublattice of X ≅ E_{6} so that U ≅ AA_{2} and X/(U ⊥ ann_{X}(U)) is an elementary abelian 2-group. Then X/(U ⊥ ann_{X}(U)) ≅ 2^{2} and ann_{X}(U) ≅ D_{4}.

Proof. (i) Let Z := ann_{X}(Y). Since (det(X), det(Y)) = 1, the natural map of X to D(Y) ≅ 2 × 2 is onto, so the natural map of X to D(Z) ≅ 2 × 2 × 3 has image 2 × 2. Since rank(Z) = 2, this means \frac{1}{\sqrt{2}}Z is an integral lattice of determinant 2. It is not rectangular, or else there exists a root of X whose annihilator contains Y, whereas a root of E_{6} has annihilator which is an A_{5}-sublattice, which does not contain a D_{4}-sublattice (since an A_{5} lattice does not contain an A_{1}^{4}-sublattice). Therefore, by (B.1), \frac{1}{\sqrt{2}}Z ≅ A_{2}.

(ii) Use (B.3). □

Notation D.19. We define two rank 4 lattices X, Q. First, X ≅ A_{1}^{2}A_{2}, D(X) ≅ 2^{2} × 3. Let X have the decomposition into indecomposable summands X = X_{1} ⊥ X_{2} ⊥ X_{3}, where X_{1} ≅ X_{2} ≅ A_{1} and X_{3} ≅ A_{2}. Let α_{1} ∈ X_{1}, α_{2} ∈ X_{2}, α_{3}, α_{4} ∈ X_{3} be roots with (α_{3}, α_{4}) = −1.

We define Q ≅ ann_{E_{6}}(P), where P is a sublattice of E_{6} isometric to A_{1}^{2}. Then D(Q) ≅ 2^{2} × 3 and rank(Q) = 4. Then Q is not a root lattice (because in E_{6}, the annihilator of an A_{1}-sublattice is an A_{5}-sublattice; in an A_{5}-lattice, the annihilator of an A_{1}-sublattice is not a root lattice).

We use the standard model for E_{6}, the annihilator in the standard model of E_{8} of J := span{(1, −1, 0, 0, 0, 0, 0, 0), (0, 1, −1, 0, 0, 0, 0, 0)}. So, E_{6} is the set of E_{8} vectors with equal first three coordinates.

We may take P to be the span of (0, 0, 0, 1, 1, 0, 0, 0) and (0, 0, 0, 1, −1, 0, 0, 0). Therefore, Q = span{u, Q_{1}, w}, where u = (2, 2, 2, 0, 0, 0, 0, 0), Q_{1} is the D_{3}- sublattice supported on the last three coordinates, and w := (1, 1, 1, 0, 0, 1, 1, 1).

Lemma D.20. The action of O(Q_{1}) ≅ 2 × Sym_{3} extends to an action on Q. This action is faithful on Q/3Q^{∗}.

Proof. The action of O(Q_{1}) ≅ 2 × Sym_{3} extends to an action on Q by letting reflections in roots of Q_{1} act trivially on u and by making the central involution of O(Q_{1}) act as −1 on Q. The induced action on Q/3Q^{∗} is faithful since Q_{1} maps onto Q/3Q^{∗} (because (3, det(Q_{1})) = 1) and the action on Q_{1}/3Q_{1} is faithful. In more detail, the action of O_{2}(O(Q_{1})) ≅ 2^{3} is by diagonal matrices and any normal subgroup of O(Q_{1}) meets O_{2}(O(Q_{1})) nontrivially. □

Lemma D.21. We use notation (D.19). Then X contains a sublattice Y ≅ \sqrt{3}Q and X > Y > 3X.

Proof. We define β_{1} := α_{1} + α_{2} + α_{3}, β_{2} := −2α_{3} − α_{4}, β_{3} := α_{3} + 2α_{4}. Then Y_{1} := span\{β_{1}, β_{2}, β_{3}\} ≅ \sqrt{3}D_{3}. The vector β_{4} := 3α_{1} − 3α_{2} is orthogonal to Y_{1} and has norm 36. Finally, define γ := \frac{1}{2}β_{4} + \frac{1}{2}(β_{1} + 2β_{2} + 3β_{3}) = 2α_{1} − α_{2} + 2α_{4}. Then Y := span{Y_{1}, β_{4}, γ} is the unique lattice containing Y_{1} ⊥ ℤβ_{4} with index 2 whose intersection with \frac{1}{2}Y_{1} is Y_{1} and whose intersection with \frac{1}{2}ℤβ_{4} is ℤβ_{4}. There is an analogous characterization for Q and \sqrt{3}Q. We conclude that Y ≅ \sqrt{3}Q.

Moreover, by direct calculation, it is easy to show that

\begin{aligned}3α_{1} & = γ + β_{1} − β_{3}, 3α_{2} = γ + β_{1} − β_{3} − β_{4}, \\ 3α_{3} & = β_{1} + 2β_{3} + β_{4} − 2γ, 3α_{4} = 2γ − (β_{1} + β_{2} + β_{3} + β_{4}). \end{aligned}

Hence, Y also contains 3X. □

Lemma D.22. Let M be the set of rank n integral lattices. For q ∈ ℤ, let M(q) be the set of X ∈ M such that X ≤ qX^{∗}. Suppose that q is a prime, X, Y ∈ M(q), Y ≥ X and q divides |Y : X|. Then q^{n+2} divides det(X). In particular, if q^{n+2} does not divide det(X), then X is not properly contained in a member of M(q).

Proof. Use the index formula for determinants of lattices. □

Proposition D.23. For an integral lattice K, define K̃ := K + 2K^{∗}. Let

A := \{(R, S)|R ≤ S ≤ ℝ^{4}, R ≅ \sqrt{3}Q, S ≅ X\}
B := \{(T, U)|T ≤ U ≤ ℝ^{4}, T ≅ \sqrt{3}X, U ≅ Q\}
A′ := {(R, S)|R ≤ S ≤ ℝ^{4}, R ≤ 3R^{∗}, det(R) = 2^{2}3^{5}, S ≅ X}
B′ := {(T, U)|T ≤ U ≤ ℝ^{4}, T ≤ 3T^{∗}, det(T) = 2^{2}3^{5}, U ≅ Q}.

Then (i) A = A′ ≠ ∅ and B = B′ ≠ ∅;

(ii) the map (T, U) ↦ (\sqrt{3}U, \frac{1}{\sqrt{3}}T) gives a bijection from B onto A; furthermore if (T, U) ∈ B, then T ≥ 3Ũ and if (R, S) ∈ A, then R ≥ 3S̃;

(iii) O(ℝ^{4}) has one orbit on A and on B.

Proof. Clearly, A ⊆ A′ and B ⊆ B′. From (D.21), A ≠ ∅ and B ≠ ∅. Moreover, the formula in (ii) gives a bijection between A and B.

Now, let (E, F) be in A′ ∪ B′.

We claim that 3F̃ = F ∩ 3F^{∗}. We prove this with the theory of modules over a PID. Since D(F) ≅ 2^{2} × 3, there exists a basis a, b, c, d of F^{∗} so that a, b, 2c, 6d is a basis of F. Then a, b, 2c, 2d is a basis of F̃. Since 3F^{∗} has basis 3a, 3b, 3c, 3d, F ∩ 3F^{∗} has basis 3a, 3b, 6c, 6d. The claim follows.

Note that F/3F̃ is an elementary abelian 3-group of rank 3 and the claim implies that it is a nonsingular quadratic space. Therefore, its totally singular subspaces have dimension at most 1.

We now study E′ := E + 3F̃, which maps onto a totally singular subspace of F/3F̃. Since totally singular subspaces have dimension at most 1, |F : E′| is divisible by 3^{2} and so its determinant is |F : E′|^{2}det(F) = |F : E′|^{2}2^{2}3. However, E′ contains E, which has determinant 2^{2}3^{5}. We conclude that E = E′ has index 9 in F. Therefore, E = E′ ≥ 3F̃. The remaining parts of (ii) follow.

Let (T, U) ∈ B. The action of O(U) ≅ Sym_{4} × 2 on U/3Ũ is that of a monomial group with respect to a basis of equal norm nonsingular vectors (D.20). It follows that the action is transitive on maximal totally singular subspaces, of which T/3Ũ is one. This proves transitivity for B. Therefore B = B′ and, using the O(U)- equivariant bijection (ii), A = A′. □

Corollary D.24. \sqrt{3}Q does not embed in Q and \sqrt{3}X does not embed in X.

Proof. Use (D.23), (D.21) and the fact that X is not isometric to Q (X is a root lattice and Q is not). □

Lemma D.25. Suppose that S ⊥ T is an orthogonal direct sum with S ≅ A_{2}, T ≅ E_{6}. The set of E_{8} lattices which contain S ⊥ T is in bijection with {X| S ⊥ T ≤ X ≤ S^{∗} ⊥ T^{∗}, |X : S ⊥ T | = 3, S^{∗} ∩ X = S, T^{∗} ∩ X = T }.

Proof. This is clear since any E_{8} lattice containing S ⊥ T lies in S^{∗} ⊥ T^{∗} and since the nontrivial cosets of S in S^{∗} have norms in \frac{2}{3} + 2ℤ and the nontrivial cosets of T in T^{∗} have norms in \frac{4}{3} + 2ℤ. □

Lemma D.26. Let X ≅ D_{4} and let H ≅ AA_{1} be a sublattice of X. Then the image of the natural map X^{∗} to H^{∗} is H^{∗} = \frac{1}{4}H.

Proof. A generator of H has norm 4, so H is a direct summand of L. In general, if W is a lattice and Y is a direct summand of W, the natural map W^{∗} → Y^{∗} is onto. The lemma follows. □

Lemma D.27. (i) Up to the action of the root reflection group of D_{4}, there is a unique embedding of AA_{2} sublattices.

(ii) We have transitivity of O(D_{4}) on the set of A_{2} sublattices and on the set of AA_{2}-sublattices. In D_{4}, the annihilator of an AA_{2} sublattice is an A_{2} sublattice, and the annihilator of an A_{2} sublattice is an AA_{2}-sublattice.

Proof. (i) Let X ≅ D_{4} and Y ≅ AA_{2}. Since every element of Y has norm divisible by 4, Y ≤ 2X^{∗}. Now let s := f − 1, where f ∈ O_{2}(W eyl(X)), f^{2} = −1. Then s^{−1} takes 2X^{∗} to X and takes Y to an A_{2} sublattice of X. Now use the well-known results that A_{2} sublattices form one orbit under W eyl(X) and O(A_{2}) ≅ Dih_{12} is induced on an A_{2} sublattice of D_{4} by its stabilizer in W eyl(D_{4}).

(ii) We may take Y := span{(−2, 0, 0, 0), (1, 1, 1, 1)} ≅ AA_{2}. Its annihilator is Z := span{(0, 1, −1, 0), (0, 0, 1, −1)} ≅ A_{2}. Trivially, ann_{X} = (Z) = Y. □

Lemma D.28. Let X ≅ E_{6} and Y, Z sublattices such that Z ≅ D_{4} and Y := ann_{X}(Z) ≅ AA_{2}. Define W := 2Y^{∗} (alternatively, W may be characterized by the property that Y ≤ W ≤ Y^{∗}, W/Y ≅ 3). Then W ≤ X^{∗}.

Proof. By coprimeness, the natural map X → Z^{∗} is onto, and the image of X in D(Z) has order 4. Therefore, the image of the natural map X → D(Y) has order 4 and so the image of the natural map X → Y^{∗} is \frac{1}{2}Y. The dual of \frac{1}{2}Y is 2Y^{∗}, which contains Y with index 3 and satisfies (X, 2Y^{∗}) ≤ ℤ. □

Lemma D.29. We have (E_{6}^{∗}, E_{6}^{∗}) = \frac{1}{3}ℤ and the norms of vectors in E_{6}^{∗} \ E_{6} are in \frac{4}{3} + ℤ.

Proof. This follows from the fact that E_{6} has a sublattice of index 3 which is isometric to A_{2}^{3} and the facts that (A_{2}^{∗}, A_{2}^{∗}) = \frac{2}{3}ℤ and that a glue vector for A_{2}^{3} in E_{6} has nontrivial projection to the spaces spanned by each of the three summands. □

Hypothesis D.30. L is a rank 12 even integral lattice, D(L) ≅ 3^{k}, for some integer k, L is rootless and L^{∗} contains no vector of norm \frac{2}{3}.

Lemma D.31. The quadratic space D(L) in (D.30) has nonmaximal Witt index if k is even.

Proof. If the Witt index were maximal for k is even, there would exist a lattice M which satisfies 3L ≤ 3M ≤ L and 3M/3L is a totally singular space of dimension \frac{k}{2}. Such an M is even and unimodular. A well-known theorem says that rank(M) ∈ 8ℤ, a contradiction. □

Proposition D.32. Let L, L′ be two lattices which satisfy hypothesis (D.30) for k even, and which have the same determinant. There exists an embedding of L ⊥ L′ into the Leech lattice.

Proof. We form L ⊥ L′. The quadratic spaces D(L), D(L)′ have nonmaximal Witt index.

Let g be a linear isomorphism from D(L) to D(L′) which takes the quadratic form on D(L) to the negative of the quadratic form on D(L′) (A.19).

Now, form the overlattice J by gluing from D(L) to D(L′) with g. Clearly, J has rank 24, is even and unimodular. The famous characterization of the Leech lattices reduces the proof to showing that J is rootless.

Suppose that J has a root, s. Write s = r + r′ as a sum of its projections to the rational spaces spanned by L, L′ respectively. The norm of any element x ∈ L^{∗} has the form a/3, where a is an even integer at least 4. The norm of any element x ∈ L^{′∗} has the form b/3, where b is an even integer at least 4. Therefore, we may assume that r, r′ have respective norms at least \frac{4}{3}. Then (s, s) ≥ \frac{8}{3} > 2, a contradiction. □

Lemma D.33. Let L be an even integral rootless lattice with D(L) ≅ 3^{k}, for an integer k, and an automorphism g of order 3 without eigenvalue 1 such that L^{∗}(g − 1) ≤ L. Then L satisfies hypothesis (D.30).

Proof. We need to show that if v ∈ L^{∗}, then (v, v) ≥ \frac{4}{3}. This follows since v(g − 1) ∈ L, (v(g − 1), v(g − 1)) = 3(v, v) and L is rootless. □

Corollary D.34. If L, L′ satisfy hypotheses of (D.33) and each of L, L′ is not properly contained in a rank 12 integral rootless lattice (such an overlattice satisfies (D.33)), then L ≅ L′ and k = 6.

Proof. Let Λ be the Leech lattice. We use results from [Gr12] which analyze the elements of order 3 in Λ.

Take two copies L_{1}, L_{2} of L. We have by (D.32), an embedding of L_{1} ⊥ L_{2} in Λ. Identify L_{1} ⊥ L_{2} with a sublattice of Λ.

Since L_{1}, L_{2} are not properly contained in another lattice which satisfies (D.33) and since Λ is rootless, L_{1} and L_{2} are direct summands of Λ. Since they are direct summands, L_{2} = ann_{Λ}(L_{1}), L_{1} = ann_{Λ}(L_{2}) and the natural maps of Λ to D(L_{1}) and D(L_{2}) are onto. The gluing construction shows that the automorphism g of order 3 in L as in (D.33) extends to an automorphism of Λ by given action on L_{2} and trivial action on L_{1}. Denote the extension by g.

We now do the same for L′, g′ in place of L, g.

From Theorem 10.35 of [Gr12], g and g′ are conjugate in O(Λ) and det(L) = det(L′) = 3^{6}. A conjugating element takes the fixed point sublattice L_{1} of g to the fixed point sublattice L′_{1} of g′. Therefore, L and L′ are isometric. □

Corollary D.35. The Coxeter-Todd lattice is not properly contained in an integral, rootless lattice.

Proof. Embed the Coxeter-Todd lattice P in a lattice Q satisfying the hypothesis of (D.34). Since det(P) = 3^{6} = det(Q), P = Q. □

Lemma D.36. Let X ≅ E_{8}, P ≤ X, P ≅ E_{6} and Q := ann_{X}(P).

(i) There exists a sublattice R ≅ A_{2} so that R ∩ (P ∪ Q) contains no roots. (ii) If r ∈ R is a root, then the orthogonal projection of r to P has norm \frac{4}{3} and

the projection to Q has norm \frac{2}{3}.

Proof. (i) We may pass to a sublattice Q_{1} ⊥ Q_{2} ⊥ Q_{3} ⊥ Q of type A_{2}^{4}, where P ≥ Q_{1} ⊥ Q_{2} ⊥ Q_{3}. Then X is described by a standard gluing with a tetracode, the subspace of 𝔽_{3}^{4} spanned by (0, 1, 1, 1), (1, 0, 1, 2), and elements v_{i} of the dual of Q_{i} (Q_{4} := Q) where v_{i} has norm \frac{2}{3}. Then for example take R to be the span of v_{2} + v_{3} + v_{4}, v_{1} + v_{3} − w, where w ∈ v_{4} + Q has norm \frac{2}{3} but (w, v_{4}) = −\frac{1}{3}. See (D.6).

(ii) This follows since the norms in any nontrivial coset of Q in Q^{∗} is \frac{2}{3} + 2ℤ. □

E Values of the Hermite function

Notation E.1. Let n and d be positive integers. Define the Hermite function

H(n, d) := (\frac{4}{3})^{\frac{n−1}{2}} d^{(1/n)}.

Theorem E.2 (Hermite: cf. proof in [Kn], p. 83). If a positive definite rank n lattice has determinant d, it contains a nonzero vector of norm ≤ H(n, d).

Table 12: Values of the Hermite function H(n, d); see [Kn], p.83.

\begin{matrix}n & d & H(n, d) \\ 2 & 2 & 1.632993162 \\ 2 & 3 & 2.000000000 \\ 2 & 4 & 2.309401077 \\ 2 & 5 & 2.581988897 \\ 2 & 6 & 2.828427125 \\ 2 & 7 & 3.055050464 \\ 2 & 8 & 3.265986324 \\ 2 & 9 & 3.464101616 \\ 2 & 12 & 4.000000000 \\ 2 & 20 & 5.163977796 \\ 2 & 24 & 5.656854249 \\ 5 & 6 & 2.543945033 \end{matrix} \begin{matrix}n & d & H(n, d) \\ 3 & 2 & 1.679894733 \\ 3 & 3 & 1.922999426 \\ 3 & 4 & 2.116534735 \\ 3 & 5 & 2.279967929 \\ 3 & 6 & 2.422827457 \\ 3 & 8 & 2.666666666 \\ 3 & 10 & 2.872579586 \\ 3 & 12 & 3.052571313 \\ 3 & 16 & 3.359789466 \\ 3 & 24 & 3.845998854 \\ 3 & 50 & 4.912041997 \\ 6 & 3 & 2.465284531 \end{matrix} \begin{matrix}n & d & H(n, d) \\ 4 & 2 & 1.830904128 \\ 4 & 3 & 2.026228495 \\ 4 & 4 & 2.177324216 \\ 4 & 5 & 2.302240057 \\ 4 & 6 & 2.409605343 \\ 4 & 7 & 2.504278443 \\ 4 & 8 & 2.589289450 \\ 4 & 9 & 2.666666668 \\ 4 & 12 & 2.865519818 \\ 4 & 25 & 3.442651865 \\ 4 & 125 & 5.147965271 \\ 7 & 32 & 3.888997243 \end{matrix}

F Embeddings of NREE8 pairs in the Leech lattice

If M, N is an NREE8 pair, then except for the case DIH_{4}(15), L = M + N can be embedded in the Leech lattice Λ. In this section, we shall describe such embeddings explicitly.

In the exceptional case DIH_{4}(15), M ∩ N ≅ AA_{1} and |t_{M}t_{N}| = 2. See (5.5).

F.1 Leech lattice and its isometry group

We shall recall some notations and review certain basic properties of the Leech lattice Λ and its isometry group O(Λ), which is also known as Co_{0}, a perfect group of order 2^{22} · 3^{9} · 5^{4} · 7^{2} · 11 · 13 · 23.

Let Ω = {1, 2, 3, …, 24} be a set of 24 element and let G be the extended Golay code of length 24 indexed by Ω. A subset S ⊂ Ω is called a G-set if S = supp α for some codeword α ∈ G. We shall identify a G-set with the corresponding codeword in G. A G-set O is called an octad if |O| = 8 and is called a dodecad if |O| = 12. A sextet is a partition of Ω into six 4-element sets of which the union of any two forms a octad. Each 4-element set in a sextet is called a tetrad.

For explicit calculations, we shall use the notion of hexacode balance to denote the codewords of the Golay code and the vectors in the Leech lattice. First we arrange the set Ω into a 4 × 6 array such that the six columns forms a sextet.

For each codeword in G, 0 and 1 are marked by a blanked and non-blanked space, respectively, at the corresponding positions in the array. For example, (1^{8}0^{16}) is denoted by the array

\begin{matrix}∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \end{matrix}

The following is a standard construction of the Leech lattice.

Definition F.1 (Standard Leech lattice [CS, Gr12]). Let e_{i} := \frac{1}{\sqrt{8}} (0, …, 4, …, 0) for i ∈ Ω. Then (e_{i}, e_{j}) = 2δ_{i,j}. Denote e_{X} := ∑_{i∈X} e_{i} for X ∈ G. The standard Leech lattice Λ is a lattice of rank 24 generated by the vectors:

\frac{1}{2}e_{X} , where X is a generator of the Golay code G;

\frac{1}{4}e_{Ω} − e_{1} ;

e_{i} ± e_{j} , i, j ∈ Ω.

Remark F.2. By arranging the set Ω into a 4 × 6 array, every vector in the Leech lattice Λ can be written as the form

X = \frac{1}{\sqrt{8}} [X_{1}X_{2}X_{3}X_{4}X_{5}X_{6}] , juxtaposition of column vectors.

For example,

\frac{1}{\sqrt{8}} 2 2 0 0 0 0 2 2 0 0 0 0 2 2 0 0 0 0 2 2 0 0 0 0

denotes the vector \frac{1}{2} e_{A}, where A is the codeword

\begin{matrix}∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \end{matrix} .

Definition F.3. A set of vectors {±β_{1}, …, ±β_{24}} ⊂ Λ is called a frame of Λ if (β_{i}, β_{j}) = 8δ_{i,j} for all i, j ∈ {1, …, 24}. For example, {±2e_{1}, …, ±2e_{24}} is a frame and we call it the standard frame.

Next, we shall recall some basic facts about the involutions in O(Λ).

Let F = {±β_{1}, …, ±β_{24}} be a frame. For any subset S ⊂ Ω, we can define an isometry ε_{S}^{F} : ℝ^{24} → ℝ^{24} by ε_{S}^{F}(β_{i}) = −β_{i} if i ∈ S and ε_{S}^{F}(β_{i}) = β_{i} if i ∉ S. The involutions in O(Λ) can be characterized as follows:

Theorem F.4 ([CS, Gr12]). There are exactly 4 conjugacy classes of involutions in O(Λ). They correspond to the involutions ε_{S}^{F}, where F is a frame and S ∈ G is an octad, the complement of an octad, a dodecad, or the set Ω. Moreover, the eigen-sublattice {v ∈ Λ | ε_{S}^{F}(v) = −v} is isomorphic to EE_{8}, BW_{16}, DD_{12}^{+} and Λ, respectively, where BW_{16} is the Barnes-Wall lattice of rank 16.

F.2 Standard EE_{8}s in the Leech lattice

We shall describe some standard EE_{8}s in the Leech lattice in this subsection.

F.2.1 EE_{8} corresponding to octads in different frames

Let F = {±β_{1}, …, ±β_{24}} ⊂ Λ be a frame and denote α_{i} := β_{i}/2. For any octad O, denote

E_{F}(O) = span \{α_{i} ± α_{j}, i, j ∈ O, \frac{1}{2} ∑_{i∈O} α_{i}\} .

Then E_{F}(O) is a sublattice of Λ isomorphic to EE_{8}. If {±2e_{1}, …, ±2e_{24}} is the standard frame, we shall simply denote E_{F}(O) by E(O).

Next we shall consider another frame. Let

A = \frac{1}{2} [\begin{matrix}−1 & 1 & 1 & 1 \\ 1 & −1 & 1 & 1 \\ 1 & 1 & −1 & 1 \\ 1 & 1 & 1 & −1 \end{matrix}] .

Notation F.5. Define a linear map ξ : Λ → Λ by Xξ = AXD, where

X = \frac{1}{\sqrt{8}} [X_{1}X_{2}X_{3}X_{4}X_{5}X_{6}]

is a vector in the Leech lattice Λ and D is the diagonal matrix

(\begin{matrix}−1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}) .

Recall that ξ defines an isometry of Λ (cf. [CS, p. 288] and [Gr12, p. 97]).

Let F := {±2e_{1}, …, ±2e_{24}} be the standard frame. Then F_{ξ} = {±2e_{1}, …, ±2e_{24}}ξ is also a frame. In this case, E(O)ξ = E_{F_{ξ}}(O) is also isomorphic to EE_{8} for any octad O. Note that if F is a frame and g ∈ O(Λ), then Fg is also frame.

F.2.2 EE_{8} associated to an even permutation in octad stabilizer

The subgroup of Sym_{Ω} which fixes G setwise is the Mathieu group M_{24}, which is a simple group of order 2^{10}·3^{3}·5·7·11·23. Recall that M_{24} is transitive on octads. The stabilizer of an octad is the group 2^{4}:Alt_{8} ≅ AGL(4, 2) and it acts as the alternating group Alt_{8} on the octad. If we fix a particular point outside the octad, then every even permutation on the octad can be extended to a unique element of M_{24} which fixes the point.

Let σ = (ij)(kℓ) ∈ Sym(O) be a product of 2 disjoint transpositions on the standard octad O. Then σ determines a sextet which contains {i, j, k, ℓ} as a tetrad and σ extends uniquely to an element σ̃ which fixes a particular point outside the octad. Note that σ̃ fixes 2 tetrads and keeps the other 4 tetrads invariant. Moreover, σ̃ has a rank 8 (−1)-eigenlattice which we call E, and that E is isometric to EE_{8}.

Take σ̃ to be the involution (UP6) listed in [Gr12, pp. 49–52].

(UP 6)

Then σ̃ stabilizes the octad

\begin{matrix}∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \end{matrix}

and determines as above the sublattice

E = span_{ℤ} \{±α_{i} ± α_{j}, \frac{1}{2} ∑_{i=1}^{8} ϵ_{i}α_{i}\} ,

where

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}2 & −2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −2 \\ 0 & 0 & 0 & 0 & 2 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & −2 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −2 \\ 0 & 0 & 0 & 0 & 2 & 0 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 0 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 & 0 \end{matrix} , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & −2 \\ 0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \end{matrix} ,

i, j = 1, …, 8 and ϵ_{i} = ±1 such that ∏_{i=1}^{8} ϵ_{i} = 1. Then E is a sublattice in Λ which is isomorphic to EE_{8}. By our construction, it is also clear that σ̃ acts as −1 on E and 1 on ann_{Λ}(E).

Recall that σ̃ is acting on Λ from the right according to our convention.

F.3 EE_{8} pairs in the Leech lattice

In this subsection, we shall describe certain NREE8 pairs M, N explicitly inside the Leech lattice. By using the uniqueness theorem (cf. Theorem 4.1), we know that our examples are actually isomorphic to the lattices in Table 1. It turns out that except for DIH_{4}(15), all lattices in Table 1 can be embedded into the Leech lattice.

We shall note that the lattice L = M +N is uniquely determined (up to isometry) by the rank of L and the order of the dihedral group D := ⟨t_{M}, t_{N}⟩ except for DIH_{8}(16, 0) and DIH_{8}(16, DD_{4}). Extra information about ann_{M}(N) is needed to distinguish them.

Let M and N be EE_{8} sublattices of the Leech lattice Λ. Let t := t_{M} and u := t_{N} be the involutions of Λ such that t and u act on M and N as −1 and act as 1 on M^{⊥} and N^{⊥}, respectively. Set g := ut and D := ⟨t, u⟩, the dihedral group generated by t and u.

Notation F.6. In this subsection, O, O′, O′′, etc denote some arbitrary octads while O_{1}, O_{2}, O_{3}, and O_{4} denote the fixed octads given as follows.

\begin{matrix}O_{1} = \begin{matrix}∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \end{matrix}  , & O_{2} = \begin{aligned}& ∗ & ∗ & ∗ & ∗ & ∗ \\ ∗ \\ ∗ \\ ∗ \end{aligned} , \\ O_{3} =  \begin{matrix}∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \\ ∗ & ∗ \end{matrix} , & O_{4} =  \begin{matrix}∗ & ∗ & ∗ & ∗ \\ ∗ & ∗ & ∗ & ∗ \end{matrix} . \end{matrix}

Remark F.7. All the Gram matrices in this subsection are computed by multiplying the matrix A by its transpose A^{t}, where A is the matrix whose rows form an ordered basis given in each case. The Smith invariants sequences are computed using the command ismith in Maple 8.

F.3.1 |g| = 2.

In this case, M ∩ N ≅ 0, AA_{1}, AA_{1} ⊥ AA_{1} or DD_{4}.

Case: DIH_{4}(15): This case does not embed into Λ.

If M ∩ N ≅ AA_{1}, then L = M + N ≅ DIH_{4}(15) contains a sublattice isometric to AA_{1} ⊥ EE_{7} ⊥ EE_{7}, which cannot be embedded in the Leech lattice Λ because the (−1)-eigenlattice of the involution g = t_{M}t_{N} has rank 14 but there is no such involution in O(Λ) (cf. Theorem F.4).

Notation F.8. Let O = {i_{1}, …, i_{8}} and O′ = {j_{1}, …, j_{8}} be 2 distinct octads and denote M := E(O) and N := E(O′). Since the Golay code G is a type II code (doubly even) and the minimal norm of G is 8, |O ∩ O′| is either 0, 2, or 4.

DIH_{4}(16)

When |O ∩ O′| = 0, clearly M ∩ N = 0 and M + N ≅ EE_{8} ⊥ EE_{8}.

DIH_{4}(14)

Suppose O ∩ O′ = {i_{1}, i_{2}} = {j_{1}, j_{2}}. Then |O ∩ O′| = 2 and F = M ∩ N = span_{ℤ}{e_{i_{1}} +e_{i_{2}}, e_{i_{1}} −e_{i_{2}}} ≅ AA_{1} ⊥ AA_{1}. In this case, ann_{M}(F) ≅ ann_{N}(F) ≅ DD_{6} and L contains a sublattice of type AA_{1} ⊥ AA_{1} ⊥ DD_{6} ⊥ DD_{6} which has index 2^{4} in L. Note that L is of rank 14. By computing the Gram matrices, it is easy to check that

\{\frac{1}{2}(e_{i_{1}} + ⋯ + e_{i_{8}})\} ∪ \{−e_{i_{k}} + e_{i_{k−1}}| 7 ≥ k ≥ 3\} ∪ \{−e_{i_{2}} + e_{i_{1}}, −e_{i_{1}} − e_{i_{2}}\}

is a basis of M and

\{−e_{i_{2}} + e_{i_{1}}, −e_{i_{1}} − e_{i_{2}}\} ∪ \{e_{j_{k−1}} − e_{j_{k}}| 3 ≤ k ≤ 7\} ∪ \{\frac{1}{2}(e_{j_{1}} + ⋯ + e_{j_{8}})\}

is a basis of N. Thus,

\begin{matrix}\{\frac{1}{2}(e_{i_{1}} + ⋯ + e_{i_{8}})\} ∪ \{−e_{i_{k}} + e_{i_{k−1}}| 8 ≥ k ≥ 3\} ∪ \{−e_{i_{2}} + e_{i_{1}}, −e_{i_{1}} − e_{i_{2}}\} \\ ∪ \{e_{j_{k−1}} − e_{j_{k}}| 3 ≤ k ≤ 8\} ∪ \{\frac{1}{2}(e_{j_{1}} + ⋯ + e_{j_{8}})\} \end{matrix}

is a basis of L and the Gram matrix of L is given by

[ \begin{matrix}4 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & −2 & 2 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & −2 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 0 & 0 & 0 & −2 & 0 & 4 & −2 & 0 & 0 & 0 & 0 & −2 \\ 1 & 0 & 0 & 0 & 0 & 2 & −2 & −2 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & −2 & 0 & 0 & 0 & 0 & 0 & 4 \end{matrix} ] .

The Smith invariant sequence is 11112222222244.

DIH_{4}(12)

Suppose O ∩ O′ = {i_{1}, i_{2}, i_{3}, i_{4}} = {j_{1}, j_{2}, j_{3}, j_{4}} (cf. Notation F.8). Then |O ∩ O′| = 4 and F = M ∩ N = span_{ℤ}{e_{i_{k}} ± e_{i_{l}}| 1 ≤ k < l ≤ 4} ≅ DD_{4}. Thus, ann_{M}(F) ≅ ann_{N}(F) ≅ DD_{4}. In this case, L is of rank 12 and it contains a sublattice of type DD_{4} ⊥ DD_{4} ⊥ DD_{4} which has index 2^{4} in L. Note that {e_{i_{1}} + e_{i_{2}}, e_{i_{1}} − e_{i_{2}}, e_{i_{2}} − e_{i_{3}}, e_{i_{3}} − e_{i_{4}}} is a basis of F = M ∩ N. A check of Gram matrices also shows that

\{e_{i_{1}} +e_{i_{2}}, e_{i_{1}} −e_{i_{2}}, e_{i_{2}} −e_{i_{3}}, e_{i_{3}} −e_{i_{4}}\}∪\{e_{i_{4}} −e_{i_{5}}, e_{i_{5}} −e_{i_{6}}, e_{i_{6}} −e_{i_{7}}, \frac{−1}{2}(e_{i_{1}} +⋯+e_{i_{8}})\}

is a basis of M and

\{e_{i_{1}} +e_{i_{2}}, e_{i_{1}} −e_{i_{2}}, e_{i_{2}} −e_{i_{3}}, e_{i_{3}} −e_{i_{4}}\}∪\{e_{j_{4}} −e_{j_{5}}, e_{j_{5}} −e_{j_{6}}, e_{j_{6}} −e_{j_{7}}, \frac{−1}{2}(e_{j_{1}} +⋯+e_{j_{8}})\}

is a basis of N. Therefore, L = M + N has a basis

{e_{i_{1}} + e_{i_{2}}, e_{i_{1}} − e_{i_{2}}, e_{i_{2}} − e_{i_{3}}, e_{i_{3}} − e_{i_{4}}}
∪\{e_{i_{4}} − e_{i_{5}}, e_{i_{5}} − e_{i_{6}}, e_{i_{6}} − e_{i_{7}}, \frac{−1}{2}(e_{i_{1}} + ⋯ + e_{i_{8}})\}
∪\{e_{j_{4}} − e_{j_{5}}, e_{j_{5}} − e_{j_{6}}, e_{j_{6}} − e_{j_{7}}, \frac{−1}{2}(e_{j_{1}} + ⋯ + e_{j_{8}})\}

and the Gram matrix of L is given by

[ \begin{matrix}4 & 0 & 2 & 0 & 0 & 0 & 0 & −2 & 0 & 0 & 0 & −2 \\ 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 2 & 0 & 0 & −1 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & −1 & 0 & 0 & 2 \\ 0 & 0 & 0 & −2 & 2 & 0 & 0 & −1 & 4 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ −2 & 0 & 0 & 0 & −1 & 0 & 0 & 2 & 0 & 0 & 0 & 4 \end{matrix} ]

whose Smith invariant sequence is 111122222244.

F.3.2 |g| = 3.

In this case, M ∩ N = 0 or AA_{2}. DIH_{6}(16) Notation F.9. Let M := E(O_{1}) ≅ EE_{8}, where O_{1} is the octad described in Notation (F.6). We choose a basis {β_{1}, …, β_{8}} of M, where

β_{1} = \frac{1}{\sqrt{8}} \begin{matrix}4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  β_{2} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 4 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,
β_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  β_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,
β_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  β_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,
β_{7} = \frac{1}{\sqrt{8}} \begin{matrix}−4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  β_{8} = \frac{1}{\sqrt{8}} 2 2 0 0 0 0 2 2 0 0 0 0 2 2 0 0 0 0 2 2 0 0 0 0 .

Let N be the lattice generated by the vectors

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}2 & −2 & 2 & −2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 2 & 0 & 2 & 0 & 2 \\ −2 & 0 & −2 & 0 & 0 & −2 \\ 0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & −2 \\ 2 & −2 & 2 & −2 & 2 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 0 & 2 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & 2 & −2 & 2 \\ −2 & 0 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −2 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & −2 \\ 0 & 0 & 0 & 0 & 2 & 0 \\ 2 & −2 & 2 & −2 & 2 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & −2 \\ 0 & 2 & 0 & 2 & −2 & 2 \\ −2 & 0 & −2 & 0 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & −2 & −2 & −2 & 0 & −2 \\ 0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 \end{matrix} , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}1 & 1 & 1 & 1 & −3 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \end{matrix} .

By checking the inner products, it is easy to shows that N ≅ EE_{8}. Note that α_{1}, …, α_{7} are supported on octads and thus N ≤ Λ by (F.1).

In this case, M ∩N = 0. Then {β_{1}, β_{2}, …, β_{8}, α_{1}, …, α_{8}} is a basis of L = M +N and the Gram matrix of L = M + N is given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 4 & −2 & 0 & 0 & 0 & −2 & 0 & −1 & 2 & −1 & 0 & 0 & 0 & −1 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 2 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 & 4 & −2 & 0 & −1 & 0 & 0 & 0 & 0 & 2 & −1 \\ 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 2 \\ 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 2 & −1 & 0 & 0 & 0 & −1 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & −2 & 0 \\ 0 & −1 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & −1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 \\ 0 & −1 & 0 & 0 & 0 & 0 & 2 & −1 & 0 & −2 & 0 & 0 & 0 & 0 & 4 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 & −1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 \end{matrix} ]

By looking at the Gram matrix, it is clear that L = M + N ≅ A_{2} ⊗ E_{8}. The Smith invariant sequence is 1111111133333333. DIH_{6}(12)

Let M := E(O_{2}) and N := Mξ, where O_{2} is the octad described in Notation (F.6) and ξ is the isometry defined in Notation (F.5).

Notation F.10. Set

γ_{1} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  γ_{2} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

γ_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & −4 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  γ_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 4 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

γ_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 4 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  γ_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 4 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

γ_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,  γ_{8} = \frac{1}{\sqrt{8}} 0 2 2 2 2 2 2 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 0 0 .

Then {γ_{1}, …, γ_{8}} is a basis of M and {γ_{1}ξ, …, γ_{8}ξ} is a basis of N.

By the definition of ξ, it is easy to show that γ_{1}ξ = −γ_{1}, γ_{2}ξ = −γ_{2}. Moreover, for any α ∈ M = E(O_{2}), αξ is supported on O_{2} if and only if α ∈ span_{ℤ}{γ_{1}, γ_{2}}. Hence, F = M ∩ N = span_{ℤ}{γ_{1}, γ_{2}} ≅ AA_{2}. Then ann_{M}(F) ≅ ann_{N}(F) ≅ EE_{6} and L = M + N is of rank 14.

Note that {γ_{1}, γ_{2}, γ_{3}, …, γ_{8}} is a basis of M and {γ_{1}, γ_{2}, γ_{3}ξ, …, γ_{8}ξ} is a basis of N. Therefore,

{γ_{1}, γ_{2}} ∪ {γ_{3}, …, γ_{8}} ∪ {γ_{3}ξ, …, γ_{8}ξ}

is a basis of L and the Gram matrix of L is given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 4 & −2 & 0 & 0 & 0 & −2 & 0 & −2 & 0 & 0 & 0 & 2 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & −2 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 1 & −2 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 1 & −2 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 1 & −2 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 & 4 & −2 & 2 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & −1 & −2 \\ 0 & −2 & 0 & 1 & 0 & 0 & 2 & 0 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & −2 & 1 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & −2 & 1 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & −2 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 \\ 0 & 2 & −2 & 0 & 0 & 0 & 0 & −1 & 0 & 0 & 0 & 0 & 4 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & −2 & 0 & 0 & 0 & 0 & 2 & 4 \end{matrix} ]

whose Smith invariant sequence is 1111 11111 33366.

Recall that ann_{M}(F) + ann_{N}(F) ≅ A_{2} ⊗ E_{6} (cf. (3.2)) and thus L contains a sublattice isometric to AA_{2} ⊥ (A_{2} ⊗ E_{6}).

F.3.3 |g| = 4.

In this case, M ∩ N = 0 or AA_{1}. There are 2 subcases for M ∩ N = 0.

DIH_{8}(16, 0)

Let M := E(O_{1}), where O_{1} is the octad as described in Notation (F.6).

Take {β_{1}, …, β_{8}} as defined in Notation (F.9). Then it is a basis of M = E(O_{1}). Let N be the EE_{8} sublattice generated by

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ −2 & −2 & −2 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & −2 & −2 & −2 & 0 & 0 \end{matrix} α_{2} = \frac{1}{\sqrt{8}} 0 0 0 0 0 0 4 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}−4 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}2 & 0 & 2 & 0 & 0 & 0 \\ −2 & 0 & −2 & 0 & 0 & 0 \\ 2 & 0 & 2 & 0 & 0 & 0 \\ 2 & 0 & 2 & 0 & 0 & 0 \end{matrix}

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 2 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 2 & 0 & 0 \\ 2 & 0 & 2 & 0 & 0 & 0 \\ −2 & 0 & −2 & 0 & 0 & 0 \end{matrix}

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & −4 & 0 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 2 & 0 & 2 & 0 & 0 \\ 0 & −2 & 0 & −2 & 0 & 0 \\ 0 & 2 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 2 & 0 & 0 \end{matrix}

In this case, M ∩ N = 0 and ann_{N}(M) = ann_{M}(N) = 0. Moreover, the set {β_{1}, …, β_{8}, α_{1}, … α_{8}} forms a basis of L = M + N and the Gram matrix of L is

[ \begin{matrix}4 & 0 & −2 & 0 & 0 & 0 & 0 & −2 & 1 & −2 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 1 & −2 & −2 & 2 & 0 & 0 & 0 & 0 \\ −2 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & −1 & 2 & 0 & −2 & 2 & −1 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & −2 & 2 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & −1 & 0 & 0 & 1 & 0 & −2 & 2 & −1 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & −1 & 0 & 0 & 0 & 0 & 1 & 0 & −2 \\ −2 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & −2 & 1 & −1 & 1 & −1 & 1 & −1 & 1 \\ 1 & 1 & −1 & 1 & −1 & 1 & −1 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & −2 & 2 & 0 & 0 & 0 & 0 & 1 & −2 & 4 & 0 & −2 & 0 & 0 & 0 & 0 \\ 2 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 0 & 0 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & 2 & −2 & 0 & 1 & 0 & 0 & 1 & 0 & −2 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 0 & 2 & −2 & 0 & 0 & 0 & −1 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 \\ 0 & 0 & −1 & 2 & −2 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 0 & 2 & −2 & 0 & −1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 \\ 0 & 0 & 0 & 0 & −1 & 2 & −2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 \end{matrix} ] .

The Smith invariant sequence is 1111111122222222.

It is clear that L ≤ ann_{Λ}(E(O_{3})) (see (F.6) for the definition of O_{3}). On the other hand, det(L) = 2^{8} = det(ann_{Λ}(E(O_{3}))). Hence, L = ann_{Λ}(E(O_{3})) is isomorphic to BW_{16} (cf. Section (5.2.2)).

DIH_{8}(16, DD_{4})

Define M := E(O_{2})ξ and N := E(O_{4}), where O_{2} and O_{4} are defined as in Notation (F.6). We shall use the set {γ_{1}ξ, …, γ_{8}ξ} defined in Notation (F.10) as a basis of M and the set {α_{1}, …, α_{8}} as a basis of N, where

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & −4 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −4 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & −4 \\ 0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} 0 0 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & −2 & −2 & −2 & −2 \\ 0 & 0 & −2 & −2 & −2 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} .

Recall that

O_{2} = \begin{aligned}& ∗ & ∗ & ∗ & ∗ & ∗ \\ ∗ \\ ∗ \\ ∗ \end{aligned} and O_{4} = \begin{matrix}∗ & ∗ & ∗ & ∗ \\ ∗ & ∗ & ∗ & ∗ \end{matrix}

and

\frac{1}{2} [\begin{matrix}−1 & 1 & 1 & 1 \\ 1 & −1 & 1 & 1 \\ 1 & 1 & −1 & 1 \\ 1 & 1 & 1 & −1 \end{matrix}] [\begin{matrix}±2 \\ 0 \\ 0 \\ 0 \end{matrix}] = ± [\begin{matrix}−1 \\ 1 \\ 1 \\ 1 \end{matrix}] .

Therefore, no vector in M = E(O_{2})ξ can be supported on O_{4} and hence M ∩N = 0. Moreover, we have

ann_{M}(N) = {γξ ∈ M| (γξ, α_{i}) = 0, for all i = 1, …, 8}
= {γξ ∈ M| supp γ ∩ O_{4} = ∅}
= span_{ℤ}{γ_{1}ξ, γ_{2}ξ, γ_{3}ξ, γ_{7}ξ} ≅ DD_{4}.

Set

\begin{matrix}δ_{1} = \begin{matrix}0 & 0 & −1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{matrix} , δ_{2} = \begin{matrix}0 & 0 & 0 & −1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \end{matrix} , \\ δ_{3} = \begin{matrix}0 & 0 & 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} , δ_{4} = \begin{matrix}0 & 0 & 0 & 0 & 0 & −1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} . \end{matrix}

Then

ann_{N}(M) ={α ∈ E(O_{4})|(α, γ_{i}ξ) = 0 for all i = 1, …, 8}
={α ∈ E(O_{4})|(α, δ_{i}) = 0 for all i = 1, 2, 3, 4}

=\frac{1}{\sqrt{8}}span_{ℤ} \{ {0 0 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , 0 0 0 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 , \atop 0 0 0 0 4 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 0 0 0 0 , \begin{matrix}0 & 0 & −2 & −2 & −2 & −2 \\ 0 & 0 & −2 & −2 & −2 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}} \}

≅DD_{4}

In this case, L = M + N is of rank 16 and {α_{1}, …, α_{8}, γ_{1}ξ, …, γ_{8}ξ} is a basis of L.

The Gram matrix of L is given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −2 & 0 & 0 & −1 \\ −2 & 4 & −2 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & −1 & 0 & 1 & 0 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −2 & 0 & 1 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 0 & 1 & −1 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −2 & 1 \\ 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 0 & −1 \\ 0 & 2 & 0 & 0 & 0 & 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 \\ 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ −2 & 0 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 2 \\ 0 & 1 & −2 & 0 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 1 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ −1 & 0 & 1 & −1 & 1 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 4 \end{matrix} ]

whose Smith invariant sequence is 1111111122224444.

A check of the Gram matrices also shows that

ann_{M}(ann_{M}(N)) = span_{ℤ}{γ_{5}ξ, γ_{6}ξ, γ′_{7}ξ, γ′_{8}ξ} ≅ DD_{4}

and

ann_{N}(ann_{N}(M)) = spanℤ{α_{1}, α_{3}, α_{5}, α′_{8}} ≅ DD_{4},

where

γ′_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 4 & −4 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , γ′_{8} = \frac{1}{\sqrt{8}} 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ,

α′_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & −2 & −2 & −2 & −2 \\ 0 & 0 & 2 & 2 & 2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} .

Let K = ann_{M}(ann_{M}(N)) + ann_{N}(ann_{N}(M)). Then K is generated by

γ_{5}ξ, γ_{6}ξ, γ′_{7}ξ, γ′_{8}ξ, α_{1}, α_{3}, α_{5}, α′_{8},

The determinant of K is 2^{8} and thus it is also isometric to EE_{8}.

DIH_{8}(15)

When M ∩ N ≅ AA_{1}, this is the only possible case.

Let σ_{1} and σ_{2} be the involutions given as follows.

σ_{1} = (UP 12) , σ_{2} = (UP 11)

Then,

σ_{1}σ_{2} = ( UP 12×UP 11 )  is of order 4.

Let M and N be the EE_{8} lattices corresponding to σ_{1} and σ_{2}, respectively. Then,

M ∩ N = span_{ℤ} \{\frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & −2 & 2 \\ 0 & 0 & 2 & −2 & −2 & 2 \end{matrix} \} ≅ AA_{1}.

Let

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −4 \\ 0 & 0 & 0 & 0 & 4 & 0 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & −2 & 2 \\ 0 & 0 & 2 & −2 & −2 & 2 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & −2 & −2 & 0 & 0 \\ 0 & 0 & 2 & 2 & 0 & 0 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 4 & −4 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 0 & 0 & −2 & 2 \\ −2 & 2 & 0 & 0 & 0 & 0 \\ 2 & −2 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 2 \\ 0 & −2 & −2 & 0 & 2 & 0 \\ 0 & 2 & 2 & 0 & 0 & −2 \end{matrix} ,

α′_{1} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −4 & 4 & 0 & 0 \end{matrix} ,

α′_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & −4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & 2 & −2 \\ 0 & 0 & −2 & −2 & 2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{6} = \frac{1}{\sqrt{8}} \begin{matrix}2 & 0 & 0 & 0 & 0 & 0 \\ 0 & −2 & −2 & −2 & 2 & 2 \\ 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & 0 \\ 0 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{8} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 0 & 2 \\ 0 & 0 & −2 & 0 & 0 & 2 \\ 0 & 2 & 0 & 0 & 2 & −2 \end{matrix} .

Note that M ∩ N = span_{ℤ}{α_{2}}. A check of Gram matrices also shows that {α_{1}, α_{2}, α_{3}, …, α_{8}} is a basis of M and {α′_{1}, α_{2}, α′_{3}, …, α′_{8}} is a basis of N. Thus, {α_{1}, α_{2}, α_{3}, …, α_{8}}∪{α′_{1}, α′_{3}, …, α′_{8}} is a basis of L = M +N and the Gram matrix of L is given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −1 & 0 & 0 & 0 & 0 \\ −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & −2 & −2 & 0 & 0 & 0 & 0 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & −2 & 2 & 2 & −1 & 0 & 0 & 0 & −1 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & −1 & 1 & 1 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 0 & 0 & 0 & 0 & 4 & −1 & −1 & 0 & 1 & −1 & 1 & 2 \\ 0 & −2 & 2 & 0 & 0 & 0 & 0 & −1 & 4 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & −2 & 2 & −1 & 0 & 0 & 0 & −1 & 0 & 4 & −2 & 0 & 0 & 0 & −2 \\ −1 & 0 & −1 & 2 & 0 & −1 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 0 & 0 & −1 & 0 & 1 & 0 & 1 & 0 & 0 & −2 & 4 & −2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & −1 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 0 & 0 & −1 & 0 & 1 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ 0 & 0 & −1 & 0 & 0 & 0 & 0 & 2 & 0 & −2 & 0 & 0 & 0 & 0 & 4 \end{matrix} ]

The Smith invariant sequence for L is 111111111144444.

F.3.4 |g| = 5.

In this case, M ∩ N = 0 and ann_{M}(N) = ann_{N}(M) = 0.

DIH_{10}(16) Let σ_{1} and σ_{2} be the involutions given as follows:

σ_{1} =  (UP 6)  , σ_{2} =  (UP 11)

Then,

σ_{1}σ_{2} = ( UP 6×UP 11 )  is of order 5.

Let M and N be the EE_{8} lattices corresponding to σ_{1} and σ_{2}, respectively. Then, M ∩ N = 0 and ann_{M}(N) = ann_{N}(M) = 0. The lattice L = M + N is of rank 16. By Gram matrices, it is easy to check

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 2 & 0 & −2 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & −2 \\ 0 & −2 & 0 & 0 & 2 & 0 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & −2 & −2 & 0 \\ 0 & 0 & 2 & 0 & 0 & −2 \\ 0 & 0 & 2 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 & −2 & 0 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & −2 \\ 0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 2 & 2 & 0 & 0 \\ 0 & 0 & 2 & −2 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −2 & 2 & 0 & 0 & −2 \\ 0 & 2 & −2 & 0 & 2 & 0 \end{matrix} ,

form a basis of M and

α′_{1} = \frac{1}{\sqrt{8}} \begin{matrix}4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −4 & 0 & 0 & 0 & 0 \end{matrix} , α′_{2} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & −2 & −2 & 2 & 2 \\ 0 & −2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & 2 & −2 \\ 0 & 0 & −2 & −2 & 2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & −4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & −2 & 2 \\ 0 & 0 & −2 & 2 & −2 & 2 \end{matrix} ,

α′_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 4 & −4 \end{matrix} , α′_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & −2 & 0 & −2 & 0 & 2 \\ 0 & 2 & −2 & 0 & 0 & 2 \\ 0 & 0 & 2 & −2 & 0 & 0 \end{matrix}

form a basis of N. In addition, {α_{1}, …, α_{8}, α′_{1}, …, α′_{8}} is a basis of L = M + N. The Gram matrix of L is then given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 1 & −1 & 1 & −1 & 1 & 0 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 1 & −2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & −1 & 1 & 0 & 0 & 1 & −1 & −1 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & −2 & 0 & 0 & 0 & 1 & −2 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 1 & −2 & 1 & −1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 1 & −2 & 2 & 0 & −1 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 0 & 0 & 4 & −1 & 1 & 0 & −1 & 2 & −1 & 1 & −2 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0 & −1 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & −1 & 1 & −1 & 0 & 0 & 1 & 1 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & −2 & 1 & 0 & 1 & −2 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & −1 & 1 & 0 & 1 & −2 & 2 & −1 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & −2 & 1 & 0 & 2 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & −2 \\ 0 & −1 & 0 & 1 & 1 & −1 & −1 & −1 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 1 & 0 & −1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ 0 & 0 & 0 & −1 & 1 & 1 & 0 & −2 & 0 & 0 & 0 & 0 & −2 & 0 & 0 & 4 \end{matrix} ]

The Smith invariant sequence is 1111111111115555.

F.3.5 |g| = 6.

In this case, M ∩ N = 0, and ann_{N}(M) ≅ ann_{M}(N) ≅ AA_{2}.

DIH_{12}(16)

Let σ_{1} and σ_{2} be the involutions given as follows:

σ_{1} = (UP 11×UP 6) , σ_{2} =  (UP 10)  ,

Then,

(UP 11×UP 6×UP 10) σ_{1}σ_{2} =  is of order 6.

Let M and N be the EE_{8} lattices corresponding to σ_{1} and σ_{2}, respectively. Then, M ∩ N = 0. Moreover, we have

\begin{aligned}ann_{M}(N) & = \frac{1}{\sqrt{8}}span_{ℤ} \{ \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 0 & 0 & 0 & 0 & 0 \\ −4 & 0 & 0 & 0 & 0 & 0 \end{matrix} , \begin{matrix}0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 2 & 0 & 2 & 0 \\ 2 & 0 & −2 & 0 & 0 & −2 \end{matrix} \} \\ & ≅ AA_{2} \end{aligned}

and

\begin{aligned}ann_{N}(M) & = \frac{1}{\sqrt{8}}span_{ℤ} \{ \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & 0 \\ 0 & −4 & 0 & 0 & 0 & 0 \end{matrix} , \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 & −2 \\ 0 & −2 & 0 & −2 & 0 & 2 \\ 0 & −2 & 0 & 2 & −2 & 0 \end{matrix} \} \\ & ≅ AA_{2}. \end{aligned}

In this case, L = M + N is of rank 16. By Gram matrices, it is easy to check

α_{1} = \frac{1}{\sqrt{8}} \begin{matrix}4 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{2} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 2 & 0 & −2 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & −2 & 0 \\ −2 & 0 & 0 & 0 & 0 & 2 \end{matrix} ,

α_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 4 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & −2 & 2 & 0 \\ 0 & 0 & 2 & 0 & 0 & 2 \\ 0 & 0 & 0 & −2 & 2 & 0 \\ 0 & 0 & −2 & 0 & 0 & −2 \end{matrix} ,

α_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & 0 & 0 \end{matrix} , α_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & −2 & 2 & 0 & 0 \\ 0 & 0 & −2 & −2 & 0 & 0 \end{matrix} ,

α_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 & 0 \end{matrix} , α_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 2 & 0 & −2 & 0 \\ 2 & 0 & −2 & 0 & 0 & 2 \end{matrix} ,

form a basis of M and

α′_{1} = \frac{1}{\sqrt{8}} \begin{matrix}4 & 0 & 0 & 0 & 0 & 0 \\ 0 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{2} = \frac{1}{\sqrt{8}} \begin{matrix}−2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & −2 & 2 & −2 & 2 \\ 0 & 2 & 0 & 0 & 0 & 0 \\ 0 & −2 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{3} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{4} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & 2 & −2 \\ 0 & 0 & −2 & −2 & 2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α′_{5} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & 0 & −4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α′_{6} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 2 & 2 & −2 \\ 0 & 0 & −2 & −2 & 2 & 2 \end{matrix} ,

α′_{7} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & −4 & 0 \end{matrix} , α′_{8} = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & −2 & 2 \\ 0 & −2 & −2 & 0 & 2 & 0 \\ 0 & 2 & 2 & 0 & 0 & −2 \end{matrix} .

form a basis for N. Note that {α_{1}, …, α_{8}, α′_{1}, ⋯, α′_{8}} is a basis of L and the Gram matrix of L is given by

[ \begin{matrix}4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 0 & 1 & −1 & 1 & 0 & 0 & −1 \\ 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 1 & −2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & −1 & 0 & 0 & 1 \\ 0 & 0 & 0 & −2 & 4 & −2 & 0 & −2 & 0 & 0 & 0 & 0 & 0 & −1 & 0 & 1 \\ 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & −1 & 0 & 1 & −1 & 2 & −1 & −1 \\ 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 & 0 & 1 & 0 & −1 & 0 & −1 & 2 & 1 \\ 0 & 0 & 0 & 0 & −2 & 0 & 0 & 4 & 0 & 0 & 0 & −1 & 2 & 0 & 0 & −2 \\ 2 & −1 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & −2 & 0 & 0 & 0 & 0 & 0 & 0 \\ −1 & 0 & 1 & 0 & 0 & −1 & 1 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & −2 & 1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 & 0 \\ 0 & −1 & 1 & 0 & 0 & 1 & −1 & −1 & 0 & 0 & −2 & 4 & −2 & 0 & 0 & 0 \\ 0 & 1 & 0 & −1 & 0 & −1 & 0 & 2 & 0 & 0 & 0 & −2 & 4 & −2 & 0 & −2 \\ 0 & 0 & 0 & 0 & −1 & 2 & −1 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & −2 & 0 \\ 0 & 0 & 0 & 0 & 0 & −1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & −2 & 4 & 0 \\ 0 & −1 & 0 & 1 & 1 & −1 & 1 & −2 & 0 & 0 & 0 & 0 & −2 & 0 & 0 & 4 \end{matrix} ]

The Smith invariant sequence is 1111111111116666.

Now let g = (σ_{1}σ_{2})^{2}. Then

g =

Note that Mg is also isometric to EE_{8} and it has a basis

α_{1}g = \frac{1}{\sqrt{8}} \begin{matrix}−4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{2}g = \frac{1}{\sqrt{8}} \begin{matrix}2 & 0 & 0 & 0 & −2 & 2 \\ 0 & −2 & −2 & 2 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 0 & 0 & 0 & 0 \end{matrix} ,

α_{3}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , α_{4}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 2 & −2 \\ 0 & 0 & −2 & 2 & 0 & 2 \\ 0 & 0 & 2 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & −2 & −2 \end{matrix} ,

α_{5}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & −4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 4 \end{matrix} , α_{6}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & −2 & −2 & −2 \\ 0 & 0 & 2 & 2 & 2 & −2 \end{matrix} ,

α_{7}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & −4 \\ 0 & 0 & 0 & 4 & 0 & 0 \end{matrix} , α_{8}g = \frac{1}{\sqrt{8}} \begin{matrix}0 & 0 & 0 & 0 & −2 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ −2 & 0 & 2 & 0 & 2 & 0 \\ 2 & 0 & −2 & 0 & 0 & −2 \end{matrix} .

Hence we have

\begin{aligned}Mg ∩ N & = \frac{1}{\sqrt{8}}span_{ℤ} \{ {\begin{matrix}4 & 0 & 0 & 0 & 0 & 0 \\ 0 & −4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 4 & −4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix} , \atop \begin{matrix}0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & −2 & −2 & 2 \\ 0 & 0 & 2 & −2 & 2 & −2 \end{matrix} , \begin{matrix}−2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & −2 & 2 & 0 & 0 \\ 0 & 0 & −2 & 0 & 2 & 0 \\ 0 & 0 & −2 & 0 & 0 & 2 \end{matrix}} \} \\ & ≅ DD_{4} \end{aligned}

Note that

t_{Mg} = g^{−1}t_{M}g =

and it commutes with t_{N}. In this case, t_{Mg} and t_{N} generates a dihedral group of order 4 and Mg + N is isometric to the lattice DIH_{4}(12).

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