The rank two lattice type vertex operator algebras V_{L}^{+} and their automorphism

groups

Chongying Dong and Robert L. Griess Jr.

21 September, 2004

Abstract. Let L be a positive definite even lattice and V_{L}^{+} be the fixed points of the lattice VOA V_{L} associated to L under an automorphism of V_{L} lifting the −1 isometry of L. For any positive rank, the full automorphism group of V_{L}^{+} is determined if L does not have vectors of norms 2 or 4. For any L of rank 2, a set of generators and the full automorphism group of V_{L}^{+} are determined.

01991 Mathematics Subject Classification. Primary 17B69.

0The first author is supported by NSF grants, a NSF grant of China and a research grant from UC Santa Cruz.

0The second author is supported by NSA grant USDOD-MDA904-03-1-0098.

Contents

1 Introduction 3

2 Background Definitions and Notations 4

3 Automorphism group of V_{L}^{+} with L_{1} = L_{2} = ∅ 5 3.1 Aut(M(1)^{+}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.2 Aut(V_{L}^{+}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Rank 2 lattices 9

5 About idempotents in small dimensional algebras 10 5.1 Idempotents of type 0 . . . . . . . . . . . . . . . . . . . . . . . . . . 11 5.2 Idempotents of type 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 12 5.3 Idempotents of type 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 13 5.4 Sums of idempotents . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

6 Automorphism group of V_{L}^{+} with rank(L) = 2 16 6.1 L_{1} = ∅ and rank(L_{2}) = 2; b = 0. . . . . . . . . . . . . . . . . . . . . 16 6.2 L_{1} = ∅ and rank(L_{2}) = 2; b = 2. . . . . . . . . . . . . . . . . . . . . 18 6.3 Alternate proof for b = 2 . . . . . . . . . . . . . . . . . . . . . . . . . 20 6.4 L_{1} = ∅ and rank(L_{2}) = 2; b = 1. . . . . . . . . . . . . . . . . . . . . 21

7 Aut(V_{L}^{+}), for L_{1} = ∅, rank(L_{2}) = 1 22

8 Aut(V_{L}^{+}), for L_{1} ≠ ∅ 23

8.1 rank(L_{1}) = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 8.1.1 L has type A_{1}^{2} . . . . . . . . . . . . . . . . . . . . . . . . . . 23 8.1.2 L has type A_{2} . . . . . . . . . . . . . . . . . . . . . . . . . . 23

8.2 rank(L_{1}) = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 8.2.1 L rectangular. . . . . . . . . . . . . . . . . . . . . . . . . . . 25 8.2.2 L not rectangular . . . . . . . . . . . . . . . . . . . . . . . . 26

9 Appendix: Algebraic rules 28

1 Introduction

This article continues a program to study automorphism groups of vertex operator algebras. See references in the survey [G2] and the more recent articles [G1], [DG1], [DG2], [DGR] and [DN1].

Here we investigate the fixed point subVOA of a lattice type VOA with respect to a group of order 2 lifting the −1 map on a positive definite lattice. We can obtain a definitive answer for the automorphism group of this subVOA in two extreme cases. The first is where the lattice has no vectors of norms 2 or 4, and the second is where the lattice has rank 2.

We use the standard notation V_{L} for a lattice VOA, based on the positive definite even integral lattice, L. For a subgroup G of Aut(L), V_{L}^{G} denotes the subVOA of points fixed by G. When G is a group of order 2 lifting −1_{L}, it is customary to write V_{L}^{+} for the fixed points (though, strictly speaking, G is defined only up to conjugacy; see the discussion in [DGH] or [GH]).

The rank 2 case is a natural extension of work on the rank 1 case, where Aut(V_{L}^{G}) was determined for all rank 1 lattices L and all choices of finite group G ≤ Aut(V_{L}). The styles of proofs are different. In the rank 1 case, there was heavy analysis of the representation theory of the principal Virasoro subVOA on the ambient VOA. In the rank 2 case, there is a lot of work on idempotents, solving nonlinear equations as well as work with several subVOAs associated to Virasoro elements. For rank 2, the case of nontrivial degree 1 part is harder to settle than in rank 1.

Our strategy follows this model. Let V be one of our V_{L}^{+}. We get information about G := Aut(V) by its action on the finite dimensional algebra A := (V_{2}, 1^{st}). We take a subset S of A which is G-invariant and understand S well enough to limit the possibilities for G (usually, there are no automorphisms besides the ones naturally inherited from V_{L}). A natural choice for S is the set of idempotents or conformal vectors. Usually, S spans A, or at least generates A. In the main case of rank 2 lattice, we prove that Aut(V) fixes a subalgebra of A which is the natural M(1)_{2}^{+}. The structure of V is controlled by M(1)^{+}, which is generated by M(1)_{2}^{+} and its eigenspaces, so we eventually determine G.

For several results, we give more than one proof.

For the case of a lattice L without roots, the automorphism group of V_{L}^{+} was studied in the recent article [S].

We thank Harm Derksen for help with computer algebra.

2 Background Definitions and Notations

Notation 2.1. Let L be an even integral lattice. For an integer m, define L_{m} := {x ∈ L|(x, x) = 2m}. Let H := ℂ ⊗ L, the ambient complex vector space. For a subset S of L, define rank(S) to be the rank of the sublattice spanned by S.

Definition 2.2. For a lattice, L, the group of automorphisms of the free abelian group L which preserves the bilinear form is called the group of automorphisms, the isometry group, the group of units or the orthogonal group of L. This group is denoted Aut(L) or O(L). We will use the notation O(L) in this article, as well as the associated SO(L) for the elements of determinant 1, P O(L) for O(L)/{±1} and P SO(L) for SO(L)/SO(L) ∩ {±1}.

Definition 2.3. For an even integral lattice, L, we let L̂ be the 2-fold cover of L described in [FLM], [DGH], [GH]. We may write bars for the map L̂ → L. The the group of automorphisms, the isometry group, the group of units or orthogonal group is the set of group automorphisms of L̂ which preserve the bilinear form on the quotient of L̂ by the normal subgroup of order 2. It is denoted Aut(L̂) or O(L̂) and has shape 2^{rank(L)}.O(L). We use bars to denote the natural map O(L̂) → O(L).

We list some notations for work with lattice type VOAs. List of Notations

D(L)  the discriminant group of the integral lattice L is D(L) := L^{∗}/L.
e^{α}  standard basis element for ℂ[L]
FVOA  framed vertex operator algebra [DGH]
LVOA  lattice vertex operator algebra [FLM]
LVOA type  the fixed points of a lattice vertex operator algebra under a

finite group of automorphisms [DG1, DGR]

LVOA+  V_{L}^{+} for an even lattice L
LVOAG(L)  the subgroup of Aut(V_{L}), for an even integral lattice L,

as described in [DN1]; it is denoted ℕ(L̂) and is an extension of the form T.Aut(L) (possibly nonsplit), where T is a natural copy of the torus ℂ ⊗ L/L^{∗} obtained by exponentiating the maps 2πx_{0}, for x ∈ V_{1}; the quotient of this group by the normal subgroup T is naturally isomorphic to Aut(L). Also, ℕ(L̂) is the product of subgroups T S, where S ≅ O(L̂) and S ∩ T = {x ∈ T |x^{2} = 1} ≅ ℤ_{2}^{rank(L)}. We may take S to be the centralizer in LVOAG(L) of a lift of −1; it has the form 2^{rank(L)}.Aut(L) and in fact any such S has this form. Denote the groups S, T by 𝕆(L̂) and 𝕋(L̂), respectively.

LVOA group for L this means LVOAG(L).
LVOAG  this means LVOAG(L), for some L
LVOAG^{+}(L)  this is the centralizer in LVOAG(L) of a lift of −1 modulo the

group of order 2 generated by the lift;

it has the form 2^{rank(L)}.[Aut(L)/⟨−1⟩]; it is the inherited group

LVOAG^{+}  this means LVOAG^{+}(L), for some L.
LVOA^{+}-group  same as LVOAG^{+}
M(1), M(1)^{+}  See Section 3.
ℕ(L̂)  See LVOAG(L)
o  linear map from V to End(V)
𝕆(L̂)  See LVOAG(L)
𝕋(L̂)  See LVOAG(L)
v_{α}  e^{α} + e^{−α}
𝕏 or 𝕏(L):  given an even integral lattice, L, this is a group of shape 2^{1+rank(L)}

for which commutation corresponds to inner products modulo 2; see an appendix of [GH].

𝕏𝕆 or 𝕏𝕆(L̂)  an extension of 𝕏 upwards by O(L).
𝕏ℙ𝕆 or 𝕏ℙ𝕆(L̂)  a quotient of 𝕏𝕆 by a central involution which corresponds to −1_{L}

under the natural epimorphism to O(L).

Remark 2.4. If (L, L) ⊂ 2ℤ, L̂ ≅ L × ⟨±1⟩. Thus O(L̂) contains a copy of O(L) which complements the normal subgroup of order 2^{rank(L)} consisting of automorphisms which are trivial on the quotient group L of L̂. This splitting passes to the groups P O(L̂) and 𝕏ℙ𝕆(L).

3 Automorphism group of V_{L}^{+} with L_{1} = L_{2} = ∅

In this section, we determine the automorphism group of V_{L}^{+} with L_{1} = L_{2} = ∅ and assume only that rank(L) > 1. The automorphism group of V_{L}^{+} in the case rank(L) = 1 is determined in [DG1] without any restriction on L. The assumption that L_{1} = L_{2} = ∅ ensures that any automorphism of V_{L}^{+} preserves the subspace M(1)_{2}^{+}, which can be identified with the Jordan algebra S^{2}H.

Since M(1)^{+} is generated by M(1)_{2}^{+} if dim H > 1 and V_{L}^{+} is a direct sum of eigenspaces for M(1)_{2}^{+} (cf. [AD]), the structure of Aut(V_{L}^{+}) can be determined easily. We shall use a classic result.

Proposition 3.1. The automorphism group of the Jordan algebra of symmetric n×n matrices is P O(n, ℂ), acting by conjugation.

Proof. [J]. □

3.1 Aut(M(1)^{+})

We first recall the construction of M(1)^{+}. Let H be a n-dimensional complex vector space with a nondegenerate symmetric bilinear form (·, ·) and Ĥ = H ⊗ℂ[t, t^{−1}]⊕ℂ c the corresponding affine Lie algebra. Consider the induced Ĥ-module

M(1) = U(Ĥ) ⊗_{U(H⊗ℂ[t]⊕ℂc)} ℂ ≃ S(H ⊗ t^{−1}ℂ[t^{−1}]) (linearly)

where H ⊗ ℂ[t] acts trivially on ℂ, and c acts as 1. For α ∈ H and n ∈ ℤ we set α(n) := α ⊗ t^{n}. Let τ be the automorphism of M(1) such that

τ(α_{1}(−n_{1}) ⋯ α_{k}(−n_{k})) = (−1)^{k}α_{1}(−n_{1}) ⋯ α_{k}(−n_{k})

for α_{i} ∈ H and n_{1} ≥ ⋯ ≥ n_{k} ≥ 1. Then M(1)^{+} is the fixed point subspace of τ.

Proposition 3.2. The automorphism group of M(1)^{+} is P O(n, ℂ).

Proof. We first deal with the case that dim H > 1. Then M(1)^{+} is generated by M(1)_{2}^{+} (cf. [DN2]), which is a Jordan algebra under u · v = u_{1}v for u, v ∈ M(1)_{2}^{+}. So any automorphism of M(1)^{+} restricts to an automorphism of the Jordan algebra M(1)_{2}^{+}. On the other hand, the automorphism group of M(1) is O(n, ℂ) [DM2], which preserves M(1)^{+}. Clearly, the kernel of the action of O(n, ℂ) on M(1)^{+} is {±1}. As a result P O(n, ℂ) is a subgroup of the automorphism group of M(1)^{+}. By Proposition 3.1, any automorphism of M(1)_{2}^{+} extends to an automorphism of M(1)^{+}.

We now assume that dim H = 1. Then M(1)^{+} is not generated by M(1)_{2}^{+}. By Lemma 2.6 and Theorem 2.7 of [DG1] for any nonnegative even integer n there is a unique lowest weight vector u^{n} (up to scalar multiple) of weight n^{2} and M(1)^{+} is generated by the Virasoro vector and u^{n}. Using the fusion rule given in Lemma 2.6 of [DG1] we immediately see that the automorphism group of M(1)^{+} in this case is trivial. Clearly, P O(1, ℂ) = 1. This finishes the proof. □

3.2 Aut(V_{L}^{+})

First we review from [B] and [FLM] the construction of lattice vertex operator algebra V_{L} for any positive definite even lattice L. Let H = ℂ ⊗_{ℤ} L. Recall that L̂ is the canonical central extension of L by the cyclic group ⟨±1⟩ such that the commutator map is given by c(α, β) = (−1)^{(α,β)}. We fix a bimultiplicative 2-cocycle ϵ : L × L → ⟨±1⟩ such that ϵ(α, β)ϵ(β, α) = c(α, β) for α, β ∈ L. Form the induced L̂-module

ℂ{L} = ℂ[L̂] ⊗_{ℂ[⟨±1⟩]} ℂ ≃ ℂ[L] (linearly),

where ℂ[·] denotes the group algebra and −1 acts on ℂ as multiplication by −1. For a ∈ L̂, write ι(a) for a ⊗ 1 in ℂ{L}. Then the action of L̂ on ℂ{L} is given by: a · ι(b) = ι(ab) for a, b ∈ L̂. If (L, L) ⊂ 2ℤ then ℂ{L} and ℂ[L] are isomorphic algebras. The lattice vertex operator algebra V_{L} is defined to be M(1) ⊗ ℂ{L}, as a vector space.

Then O(L̂) is a naturally defined subgroup of Aut(L̂) and Hom(L, ℤ/2ℤ) may be identified with a subgroup of O(L̂) (see [FLM] , [DN1], [GH]) and there is an exact sequence

1 → Hom(L, ℤ/2ℤ) → O(L̂) →^{−} O(L) → 1.

It is proved in [DN1] that Aut(V_{L}) has shape N ·O(L̂) where N is the normal subgroup of Aut(V_{L}) generated by e^{u_{0}} for u ∈ (V_{L})_{1}. Note that Hom(L, ℤ/2ℤ) can furthermore be identified with the intersection of N and O(L̂). See the List of Notations.

Let e : L → L̂ be a section associated to the 2-cocycle ϵ, written α ↦ e_{α}. Let θ be the automorphism θ of L̂ of order 2 such that θe_{α} = e_{−α} for α ∈ L. Then θ extends to an automorphism of V_{L}, still denoted by θ, such that θ|_{M(1)} is identified with τ and θι(a) = ι(θa) for all a ∈ L̂. Set e^{α} = ι(e_{α}). Then θe^{α} = e^{−α}.

Let V_{L}^{+} be the fixed points of θ. In order to determine the automorphism group of V_{L}^{+}, it is important to understand which automorphism of V_{L} restricts to an automorphism of V_{L}^{+}. Clearly, the centralizer of θ in Aut(V_{L}) acts on V_{L}^{+}. So we get an action of O(L̂)/⟨±1⟩ on V_{L}^{+}. Let h ∈ H. Then e^{2πih(0)} preserves V_{L}^{+} if and only if (h, α) ≡ (h, −α) modulo ℤ for any α ∈ L. That is, h ∈ \frac{1}{2}L^{∗} where L^{∗} is the dual lattice of L.

Lemma 3.3. The subgroup of Aut(V_{L}^{+}) which preserves M(1)_{2}^{+} is just the LVOA^{+}- group.

Proof. Let n := dim(H). Let σ ∈ Aut(V_{L}^{+}) such that σM(1)_{2}^{+} ⊂ M(1)_{2}^{+}. Then σ|_{M(1)_{2}^{+}} ∈ P O(n, ℂ) as in 3.2. Note that M(1)^{+} is generated by M(1)_{2}^{+} as rank(L) > 1 (see the proof of Proposition 3.2). So, σ preserves M(1)^{+}.

For any α ∈ L, let V_{L}^{+}(α) be the M(1)^{+}-submodule generated by v_{α} := e^{α} + e^{−α}. Then V_{L}^{+}(α) is an irreducible M(1)^{+}-module, V_{L}^{+}(α) and V_{L}^{+}(β) are isomorphic M(1)^{+}-modules if and only if α = ±β (cf. [AD]). Moreover, if α ≠ 0 then V_{L}^{+}(α) is isomorphic to M(1) ⊗ e^{α} (cf. [AD]).

Note that V_{L}^{+} = ∑_{α∈L} V_{L}^{+}(α). Let S be a subset of L such that |S∩{±α}| = 1 for any α ∈ L. Then for any two different α, β ∈ S, V_{L}^{+}(α) and V_{L}^{+}(β) are nonisomorphic M(1)^{+}-modules and

V_{L}^{+} = ⊕_{α∈S}V_{L}^{+}(α)

is a direct sum of nonisomorphic irreducible M(1)^{+}-modules.

Let α ∈ L. Since σ preserves M(1)^{+}, it sends V_{L}^{+}(α) to V_{L}^{+}(β) for some β ∈ L. The vector v_{α} is the unique lowest weight vector (up to a scalar) of V_{L}^{+}(α). This implies that σ(v_{α}) = λv_{β} for some nonzero scalar λ ∈ ℂ (depending on α and β).

For a vertex operator algebra V and a homogeneous v ∈ V we set o(v) = v_{wtv−1} and extend to all of V linearly. Note that v_{α} is an eigenvector for o(v) for v ∈ M(1)_{2}^{+}. In fact, o(h_{1}(−1)h_{2}(−1))v_{α} = (h_{1}, α)(h_{2}, α)v_{α} for h_{i} ∈ H. Recall the proof of Proposition 3.2. We can regard the restriction of σ to (V_{L}^{+})_{2} ≅ M(1)_{2}^{+} as an element of O(n, ℂ), well-defined modulo ±1. Then σ(h_{1}(−1)h_{2}(−1)) = (σh_{1})(−1)(σh_{2})(−1). Note that σ^{−1} is the adjoint of σ. Then,

(h_{1}, α)(h_{2}, α)λv_{β} = σ((h_{1}, α)(h_{2}, α)v_{α}) = σ(o(h_{1}(−1)h_{2}(−1))v_{α})
= o(σ(h_{1}(−1)h_{2}(−1)))λv_{β} = (σh_{1}, β)(σh_{2}, β)λv_{β}.

Since the h_{i} are arbitrary, σα = ±β. Thus σ maps L onto L so induces an isometry of L which is well defined modulo ⟨±1⟩.

Multiplying σ by an element from LVOAG^{+}(L) (which comes from ℕ(L̂)), we can assume that σ|_{M(1)^{+}} = id_{M(1)^{+}} Then σv_{α} = λ_{α}v_{α} for some nonzero λ_{α} ∈ ℂ. Since V_{L}^{+}(α) is an irreducible M(1)^{+}-module we see that σ acts as the scalar λ_{α} on V_{L}^{+}(α). Clearly, λ_{α} = λ_{−α}. Note that

Y (v_{α}, z)v_{β} = E^{−}(−α, z)ϵ(α, β)e^{α+β}z^{(α,β)} + E^{−}(−α, z)ϵ(α, −β)e^{α−β}z^{−(α,β)}
+E^{−}(α, z)ϵ(−α, β)e^{−α+β}z^{−(α,β)} + E^{−}(α, z)ϵ(α, β)e^{−α−β}z^{(α,β)}

where

E^{−}(α, z) = exp(∑_{n<0} \frac{α(n)z^{−n}}{n}).

Thus, if n is sufficiently negative, (v_{α})_{n}(v_{β}) = u + v for some nonzero u ∈ V_{L}^{+}(α + β) and v ∈ V_{L}^{+}(−α + β). This gives λ_{α}λ_{β} = λ_{α+β} = λ_{α−β} by applying σ to (v_{α})_{n}(v_{β}) = u + v. So α ↦ λ_{α} defines a character of abelian group L/2L of order 2^{n}. Clearly, any character λ : L/2L → ⟨±1⟩ defines an automorphism σ which acts on V_{L}^{+}(α) as λ_{α}. As a result, the subgroup of Aut(V_{L}^{+}) which acts trivially on M(1)^{+} is isomorphic the dual group of L/2L and is exactly the subgroup of O(L̂)/⟨±1⟩ which we identified as Hom(L, ℤ/2ℤ). As a result, the subgroup of Aut(V_{L}^{+}) which preserves M(1)_{2}^{+} is exactly the group O(L̂)/⟨±1⟩, as desired. □

Proposition 3.4. Let L be a positive definite even lattice such that L_{1} = L_{2} = ∅. Then Aut(V_{L}^{+}) is the inherited group, i.e., the LVOA^{+}-group.

Proof. In this case we have (V_{L}^{+})_{2} = M(1)_{2}^{+}. Thus, any automorphism of V_{L}^{+} preserves M(1)_{2}^{+}. By Lemma 3.3, Aut(V_{L}^{+}) is the LVOA^{+}-group. □

4 Rank 2 lattices

All lattices in this article are positive definite. Throughout this article, L denotes an even integral lattice. We recall a general result.

Lemma 4.1. Let L be a lattice and M a sublattice.

(i) If |L : M| is finite, det(M) = det(L)|L : M|^{2}.

(ii) If M is a direct summand of L, L/[M + ann_{L}(M)] embeds in D(M).

Proof. These are standard results. For example, see [G3]. □

We need to sort out rank 2 lattices by whether they contain roots or elements of order 4, due to their contributions to low degree terms of the lattice VOA. We shall use the notations 2.1.

Lemma 4.2. Suppose that rank(L_{1}) = 2. Then L_{1} spans L and L is one of L_{A_{1}^{2}} or L_{A_{2}}.

Proof. The span of L_{1} is isometric to L_{A_{1}^{2}} or L_{A_{2}}. Each of these is a maximal even integral lattice under containment. □

Lemma 4.3. Suppose that rank(L_{1}) = 1. Let r ∈ L_{1} and let s generate ann_{L}(r). Then (s, s) ≥ 4 and if L > span{r, s}, then 14 ≤ (s, s) ∈ 6 + 8ℤ.

Proof. Note that ℤr is a direct summand of L. We have (s, s) ≥ 4. In case L > N := span{r, s}, L/N has order 2, by 4.1. If x represents the nontrivial coset, (x, x) ≥ 4 then (2x, 2x) ≥ 16. Also, (2x, 2x) ∈ 8ℤ. Since (x, r) is odd, if we write 2x = pr + qs, for integers p, q ∈ ℤ, then p is odd and so q^{2}(s, s) ∈ 6 + 8ℤ. It follows that q is odd and (s, s) ∈ 6 + 8ℤ. □

Lemma 4.4. Suppose that L_{1} = ∅ and rank(L_{2}) = 2. If r, s are linearly independent

norm 4 elements, then they span L and have Gram matrix G = ( \begin{matrix}4 & b \\ b & 4 \end{matrix} ), for some

b ∈ {0, ±1, ±2}.

Proof. If L ≠ N := span{r, s}, then det(N) = 16 − b^{2} is divisible by a perfect square, whence b = 0 or b = ±2 and the index is 2. Actually, b = 0 does not occur here since \frac{1}{2}r, \frac{1}{2}s ∉ L implies that \frac{1}{2}(r+s) ∈ L_{1}, a contradiction. So, b = ±2. Clearly, span\{r, s\} ≅ \sqrt{2}L_{A_{2}}. However, any integral lattice containing the latter with index 2 is odd, a contradiction. Therefore, L = N and the Gram matrix is as above. Positive definiteness implies that |b| < 4 and rootlessness implies that b ≠ ±3. □

Lemma 4.5. Suppose that L_{1} = ∅ and rank(L_{2}) = 1. Let r ∈ L_{2} and let s generate ann_{L}(x). Then (s, s) ≥ 6 and L/span{r, s} is a subgroup of ℤ_{4}.

If the order of L/span{r, s} is 2, 8 ≤ (s, s) ∈ 4 + 8ℤ.

If the order of L/span{r, s} is 4, 28 ≤ (s, s) ∈ 28 + 32ℤ.

Proof. Let x be in a nontrivial coset of N := span{r, s} in L.

If (x, r) ∈ 2 + 4ℤ, 2x = pr + qs, where p is odd. We have (x, x) ≥ 6, p odd and q ≠ 0. Therefore, (2x, 2x) ∈ 8ℤ, 24 ≤ 4p^{2} + (s, s)q^{2}, whence q is odd and (s, s) ∈ 4 + 8ℤ.

If (x, r) ∈ 1 + 2ℤ, (4x, 4x) ∈ 32ℤ. We have (x, x) ≥ 6, whence (4x, 4x) ≥ 96. If we write 4x = pr + qs, we have 4p^{2} + q^{2}(s, s) ∈ 32ℤ. Since p is odd, p^{2} ∈ 1 + 8ℤ and 4p^{2} ∈ 4 + 32ℤ. Since (s, s) is even, q is odd, q^{2} ∈ 1 + 8ℤ and (s, s) ∈ 4 + 8ℤ. It follows that \frac{1}{4}q^{2}(s, s) ∈ 7 + 8ℤ whence \frac{1}{4}(s, s) ∈ 7 + 8ℤ. □

5 About idempotents in small dimensional algebras

We can derive a lot of information about the automorphism group of a vertex operator algebra by restricting to low degree homogeneous pieces. For the V_{L}^{+} problem, the degree 2 piece and its product x, y ↦ x_{1}y give an algebra which is useful to study. Here, for rank(L) = 2, we concentrate on some commutative algebras of dimension around 5. Commutativity of (V_{L}^{+}, 1^{st}) is implied if L_{1} = ∅, which is so for b ∈ {0, ±1, ±2} as in 4.4.

It does not seem advantageous to give particular values to b most of the time, so we keep it as an unspecified constant in case these arguments might be a model for future work. In the present work, we shall note limits on b, as needed.

Notation 5.1. Let S be the Jordan algebra of degree 2 symmetric matrices and suppose that A is a commutative 5 dimensional algebra of the form A = S⊕ℂv_{r}⊕ℂv_{s}. Suppose that v_{r} × v_{s} = 0 and that the notations of Appendix: Algebraic rules apply here, with the usual inner products and algebra product. Let w = p + c_{r}v_{r} + c_{s}v_{s} be an idempotent. Suppose also that t is a norm 4 vector orthogonal to r. Let a_{1}, a_{2}, a_{3} be scalars so that p = a_{1}r^{2} + a_{2}rt + a_{3}t^{2}.

Remark 5.2. We note that the basis r, s of H has dual basis r^{∗}, s^{∗}, where r^{∗} = \frac{1}{16−b^{2}} (4r − bs) and s^{∗} = \frac{1}{16−b^{2}} (4s − br). The identity of A is \frac{1}{4} \frac{1}{16−b^{2}} (rr^{∗} + ss^{∗}) = \frac{1}{4} \frac{1}{16−b^{2}} (4r^{2} + 4s^{2} − 2brs).

Notation 5.3. If w is an element of A, write w = p+q for p ∈ S^{2}H and q ∈ ℂv_{r}⊕ℂv_{s}. Call the element w̄ := p − q the conjugate element. The components p, q are called the P -part and the Q-part of w. Extend this notation to subscripted elements: w_{i} = p_{i} + q_{i}, w̄_{i} = p_{i} − q_{i}, for indices i.

Remark 5.4. In 5.3, q^{2} ∈ S^{2}H since v_{r} × v_{s} = 0. Also w = p + q is an idempotent if and only if p = p^{2} + q^{2} and q = 2p × q. Therefore, w = p + q is an idempotent if and only if the conjugate p − q is an idempotent.

Lemma 5.5. Suppose that w_{1} and w_{2} are idempotents and their sum is an idempotent. Then w_{1} × w_{2} = 0 and (w_{1}, w_{2}) = 0.

Proof. We have (w_{1} + w_{2})^{2} = w_{1}^{2} + 2w_{1} × w_{2} + w_{2}^{2}, whence w_{1} × w_{2} = 0. Also, (w_{1}, w_{2}) = (w_{1}^{2}, w_{2}) = (w_{1}, w_{1} × w_{2}) = 0. □

Definition 5.6. Throughout this article, an idempotent is not zero or the identity, unless the context clearly allows the possibility. We call an idempotent w of type 0, 1, 2, respectively, if it has Q-part which is 0, is a multiple of v_{r} or v_{s}, or is not a multiple of either v_{r} or v_{s}.

Lemma 5.7. Then (i) r^{2} × s^{2} = 4brs, r^{2} × r^{2} = 16r^{2}, s^{2} × s^{2} = 16s^{2}, rs × rs = 4r^{2} + 4s^{2}+2brs; x^{2}×v_{r} = (x, r)^{2}v_{r} = \frac{1}{2}(x^{2}, r^{2})v_{r}; r^{2}×rs = 8rs+2b^{2}r^{2}, s^{2}×rs = 8rs+2b^{2}s^{2}; also v_{r} × v_{r} = r^{2}, v_{r} × v_{s} = 0, v_{s} × v_{s} = s^{2}.

(ii) (r^{2}, r^{2}) = 32 = (s^{2}, s^{2}), (r^{2}, s^{2}) = 2b^{2}, (rs, rs) = 16 + b^{2}, (rs, r^{2}) = 8b = (rs, s^{2}) and (v_{r}, v_{r}) = 2 = (v_{s}, v_{s}) and (v_{r}, v_{s}) = 0.

Proof. See the Appendix (and take a = d = 4, b ≠ 0, ±2). □

5.1 Idempotents of type 0

Lemma 5.8. These are just idempotents in the Jordan algebra of symmetric matrices. They are ordinary idempotent matrices which are symmetric. Up to conjugacy by orthogonal transformation, they are diagonal matrices with diagonal entries only 1 and 0.

5.2 Idempotents of type 1

Notation 5.9. The next few results apply to the case of an idempotent of type 1, i.e., the form w = p + c_{r}v_{r}, where c_{r} ≠ 0. In such a case, w = w^{2} = p^{2} + c_{r}^{2}r^{2} + c_{r}(p, r^{2})v_{r} (see the Appendix : Algebraic Rules) From c_{r} ≠ 0, we get (p, r^{2}) = 1. We continue to use the notation of 5.1.

Lemma 5.10. Suppose that c_{r} ≠ 0 and c_{s} = 0. We have a_{1} = 16a_{1}^{2} + 4a_{2}^{2} + c_{r}^{2}; a_{2} = 16a_{2}(a_{1} + a_{3}); a_{3} = 16a_{3}^{2} + 4a_{2}^{2} and (p, r^{2}) = 1.

Proof. Compute p+c_{r}v_{r} = w = w^{2} = p^{2} +c_{r}^{2}r^{2} +c_{r}(p, r^{2})v_{r} (see 5.7) and expand in the basis r^{2}, rt, t^{2}, v_{r}. □

Corollary 5.11. a_{1} = \frac{1}{32}.

Proof. We have 1 = (p, r^{2}) = a_{1}(r^{2}, r^{2}) = 32a_{1}, whence a_{1} = \frac{1}{32}. □

Lemma 5.12. Suppose that c_{r} ≠ 0 and c_{s} = 0. Then (A1) a_{1} = 16a_{1}^{2} + 4a_{2}^{2} + c_{r}^{2}; (A2) a_{2} = 16a_{2}(a_{1} + a_{3}); and (A3) a_{3} = 16a_{3}^{2} + 4a_{2}^{2}.

Proof. Compute p^{2} = (16a_{1}^{2} + 4a_{2}^{2})r^{2} + 16(a_{1}a_{2} + a_{3}a_{2})rt + (16a_{3}^{2} + 4a_{2}^{2})t^{2} and use w = w^{2} = p^{2} + c_{r}^{2}r^{2} + c_{r}v_{r}. □

Lemma 5.13. Suppose that c_{r} ≠ 0 and c_{s} = 0. If a_{2} = 0, then a_{3} ∈ \{0, \frac{1}{16}\} and c_{r} = ±\frac{1}{8}.

Proof. We deduce from (A3) that a_{3} = 16a_{3}^{2}, then c_{r} = ±\frac{1}{8}. □

Lemma 5.14. Suppose that c_{r} ≠ 0 and c_{s} = 0. Then a_{2} = 0.

Proof. If a_{2} ≠ 0, then from (A2), 1 = 16(a_{1} + a_{3}) and we get a_{3} = \frac{1}{32}. Next, use (A3) to get \frac{1}{64} = 4a_{2}^{2}. Finally use (A1) to get c_{r} = 0. □

Theorem 5.15. Assume that c_{r} ≠ 0 and c_{s} = 0. Then (i) a_{1} = \frac{1}{32}, a_{2} = 0, c_{r} = ±\frac{1}{8}; and (ii) either (w, w) = \frac{1}{16} and a_{3} = 0; or (w, w) = \frac{3}{16} and a_{3} = \frac{1}{16}. All of the above cases occur. If an idempotent occurs, so does its complementary

idempotent.

Proof. This is a summary of preceding results. □

Lemma 5.16. If w is an idempotent of type 1, then

(i) if (w, w) = \frac{1}{16}, the eigenvalues for ad(w) are 1, 0, 0, \frac{1}{4}, \frac{b^{2}}{32}. Eigenvectors for these respective eigenspaces are w, 1 − w, t^{2}, rt, v_{s};

(ii) if (w, w) = \frac{3}{16}, the eigenvalues for ad(w) are 0, 1, 1, \frac{3}{4}, 1 − \frac{b^{2}}{32}. Eigenvectors for these respective eigenspaces are w, 1 − w, t^{2}, rt, v_{s}.

If \frac{b^{2}}{32} ≠ 0, 1, \frac{1}{4}, \frac{3}{4}, the multiplicities of 0 and 1 are 2 and 1 in case (i) and 1 and 2 in case (ii).

Proof. Straightforward calculation. Note that \frac{b^{2}}{32} ≠ 0, 1, \frac{1}{4}, \frac{3}{4} follows if b is rational □

Corollary 5.17. If w is a type 1 idempotent and is the sum of two nonzero idempotents, w_{1}, w_{2}, then w has the form \frac{1}{32}r^{2} + \frac{1}{16}t^{2} ± \frac{1}{8}v_{r} and w_{1}, w_{2} are, up to order, \frac{1}{32}r^{2} ± \frac{1}{8}v_{r} and \frac{1}{16}t^{2}.

Proof. If w is such a sum, each w_{i} is in the 1-eigenspace of ad(w), which must be more than 1-dimensional. This means that w has norm \frac{3}{16} and one of the w_{i}, say for i = 1, has type 1 and Q-part ±\frac{1}{8}v_{r}. Therefore, w_{2} has type 0, whence norm \frac{1}{8}. This means that w_{1} has norm \frac{1}{16} and so we know that w_{1} has shape \frac{1}{32}r^{2} ± \frac{1}{8}v_{r} and w = \frac{1}{16}t^{2}. □

5.3 Idempotents of type 2

Hypothesis 5.18. We assume in this subsection that the parameter b ≠ 0, ±2, ±3 (which means b = ±1). Then the algebra (V_{2}, 1^{st}) is commutative since V_{1} = 0.

Notation 5.19. p = c(r^{2} + s^{2}) + drs, v = c_{r}v_{r} + c_{s}v_{s}.

Lemma 5.20. If c_{r} and c_{s} are nonzero, then there are at most 8 possibilities for w. In more detail, there are at most two values of c (and, correspondingly, of d). We have c_{r}^{2} = c_{s}^{2} and this common value depends on c (or on d).

Proof. Compute p + c_{r}v_{r} + c_{s}v_{s} = w = w^{2} = p^{2} + c_{r}^{2}r^{2} + c_{s}^{2}s^{2} + c_{r}(p, r^{2})v_{r} + c_{s}(p, s^{2})v_{s}. Since c_{r} and c_{s} are nonzero, (p, r^{2}) = 1 = (p, s^{2}).

Since (r^{2}, r^{2}) = 32 = (s^{2}, s^{2}), (rs, r^{2}) = 8b = (rs, s^{2}) and (r^{2}, s^{2}) = 16 + b^{2}, we have p = c(r^{2} +s^{2})+drs. for some scalars, c, d. The previous paragraph then implies that 1 = (32 + 2b^{2})c + 8bd. Since b ≠ 0,

(e1)  d = \frac{1}{8b}(2c(32 + 2b^{2}) − 1)

is a linear expression in c.

Now, p^{2} = (16c^{2} +4d^{2})(r^{2} +s^{2})+(2bc^{2} +2bd^{2})rs and so w^{2} = (16c^{2} +4d^{2} +c_{r}^{2})r^{2} + (16c^{2} + 4d^{2} + c_{s}^{2})s^{2} + (2bc^{2} + 2bd^{2})rs + c_{r}v_{r} + c_{s}v_{s}. It follows that c_{r}^{2} = c_{s}^{2}.

We compare coefficients of r^{2} and get

(e2)  c = 16c^{2} + 4d^{2} + c_{r}^{2}.

We compare coefficients of rs and get

(e3)  d = 2bc^{2} + 2bd^{2}.

Since d is a linear expression in c, c satisfies a quadratic equation, depending on b but not c_{r}^{2} = c_{s}^{2}. The degree of this equation really is 2 since b ≠ 0 real implies that the top coefficient is nonzero.

It follows that the ordered pair d, c has at most two possible values. For each, there is a unique value for c_{r}^{2}, hence at most two possible values for c_{r} (and the same two for c_{s}). Therefore there are at most eight idempotents of type 2. □

Lemma 5.21. c ≠ 0 and d ≠ 0.

Proof. Suppose that c = 0. We then have p = drs, d = \frac{−1}{8b}. On the other hand, since w = drs + v is an idempotent, the coefficient for w^{2} at rs is d = 8bc^{2} + 2bd^{2} = 2bd^{2} = 2bd^{2}, which implies that 1 = 2bd. This is incompatible with d = \frac{−1}{8b}.

If d = 0, equation (e3) implies that c = 0, which is false. □

Lemma 5.22. If w is a type 2 idempotent, w = c(r^{2} + s^{2}) + drs + c_{r}v_{r} + c_{s}v_{s} and 1 − w is the complementary idempotent, expanded similarly as 1 − w = c′(r^{2} + s^{2}) + d′rs + c′_{r}v_{r} + c′_{s}v_{s}, then c′_{r} = −c_{r} ≠ c_{r}, c′_{s} = −c_{s}, c ≠ c′ and d ≠ d′. In particular, in the notation of 5.20, the function c ↦ c_{r}^{2} is 2-to-1 and so only one value of c_{r}^{2} occurs for type 2 idempotents.

Proof. If it were true that c = c′, then w = \frac{1}{2}𝕀 + v and 1 − w = \frac{1}{2}𝕀 − v. Since these are idempotents, v^{2} = \frac{1}{2}𝕀. However, this is impossible as b ≠ ±2 implies that v^{2} is a multiple of r^{2} + s^{2} and 𝕀 is not a linear combination of r^{2}, s^{2} for b ≠ 0 (see 5.2). □

5.4 Sums of idempotents

Hypothesis 5.23. We continue to take b = ±1. Results of the previous subsection apply.

In the arguments in this section, we allow the symbol b to be any odd integer, though the lattice is positive definite only for b = ±1.

Lemma 5.24. Suppose that w_{1}, w_{2} are two idempotents of type 1. If w_{1} + w_{2} is an idempotent, then w_{1} + w_{2} does not have type 1 or type 2.

Proof. We eliminate the sum having type 1 with Lemmas 5.17. To eliminate a sum having type 2, we note that for type 1 idempotents, we have a_{2} = 0 by 5.15, whereas d ≠ 0 for type 2, by 5.21. □

Lemma 5.25. If w_{1}, w_{2} are idempotents of type 2 and not complementary, their sum is not an idempotent.

Proof. Assume that the sum w is an idempotent. From 5.24, the sum has type 0, so has the form \frac{1}{16}u^{2}, for some vector u ∈ H of norm 4. The eigenvalues of ad(u) are 1, 0, \frac{1}{2} and \frac{1}{16}(u, r)^{2}, \frac{1}{16}(u, s)^{2}, with respective eigenvectors u^{2}, 1 − u^{2}, \frac{1}{2}uu′, v_{r}, v_{s}, where u′ spans the orthogonal of u in H.

Now, w_{1}, w_{2} are linearly independent (or else they are equal, which is impossible). This means that the eigenvalue 1 has multiplicity at least 2. So, at least one of (u, r)^{2}, (u, s)^{2} is 16. Since w_{1}, w_{2} lie in the 1-eigenspace of ad(u) and both w_{i} have type 2, both these square norms must be 16, i.e., m := (u, r) = ±4 and n := (u, s) = ±4. Since r, s form a basis and the form is nonsingular, this forces u = mr^{∗} + ns^{∗}, where r^{∗}, s^{∗} is the dual basis. We have 4 = (u, u) = 16(r^{∗}, r^{∗}) + 2mn(r^{∗}, s^{∗}) + 16(s^{∗}, s^{∗}). The right side is \frac{1}{16−b^{2}} [16(4r−bs, 4r−bs)+2mn(4r−bs, 4s−br)+16(4s−br, 4s−br)].

Since b is an odd integer, the above rational number in reduced form clearly has numerator divisible by 16, so does not equal 4, a contradiction. □

Lemma 5.26. The sum of a type 1 and type 2 idempotent is not an idempotent.

Proof. Assume that w := w_{1} + w_{2} is an idempotent. Obviously it does not have type 0. By 5.17, it does not have type 1.

We conclude that w has type 2. However, the coefficients of w at r^{2} and s^{2} must be equal for type 2, a contradiction since this forces the P -part if the type 1 idempotent to be 0. □

Corollary 5.27. The only idempotents which are a proper summand of some nontrivial idempotent are the ones of type 1 and norm \frac{1}{16}. There are 4 such and they come in orthogonal pairs, which are just pair of idempotents and their conjugates.

Corollary 5.28. Aut(A) is a dihedral group of order 8.

Proof. The automorphism group preserves and acts faithfully on the set J of type 1 idempotents of norm \frac{1}{16}, the complete set of idempotents which are proper summands of proper idempotents, and furthermore preserves the partition defined by orthogonality. The orthogonal in A of the nonsingular subspace span(J) is spanned by v := r^{2}+s^{2}− \frac{16+b^{2}}{4b} rs. We claim that if an automorphism acts trivially on span(J), it acts trivially on A. This is so because rs ∈ span{r^{2} ×s^{2}, r^{2}, s^{2}} and {rs}∪J spans A.

This proves that the automorphism group of A embeds in a dihedral group of order 8. This embedding is an isomorphism onto since the LVOA^{+}- group embeds in Aut(A). □

Proposition 5.29. Aut(V_{L}^{+}) is just the LVOA^{+}-group, isomorphic to Dih_{8}.

Proof. In this case we have (V_{L}^{+})_{2} = M(1)_{2}^{+}. Thus any automorphism of V_{L}^{+} preserves M(1)_{2}^{+}. Now use 3.3. □

6 Automorphism group of V_{L}^{+} with rank(L) = 2

In this section, we assume that the rank of L is equal to 2. If L_{1} = L_{2} = ∅, the automorphism group of V_{L}^{+} was determined in Proposition 3.4. So in this section we assume that L_{1} or L_{2} is not empty.

6.1 L_{1} = ∅ and rank(L_{2}) = 2; b = 0.

Note that L is generated by L_{2}. We will discuss the automorphism group according to the value b in the Gram matrix G (see 4.4).

First we assume that b = 0 in the Gram matrix G. Then L ≅ \sqrt{2}L_{A_{1}} ⊥ \sqrt{2}L_{A_{1}}, where L_{A_{1}} is the root lattice of type A_{1}. Let L = ℤα_{1} ⊕ ℤα_{2} with (α_{i}, α_{j}) = 4δ_{ij} for i, j = 1, 2. Set

ω_{1} = \frac{1}{16}α_{1}(−1)^{2} + \frac{1}{4}(e^{α_{1}} + e^{−α_{1}}),

ω_{2} = \frac{1}{16}α_{2}(−1)^{2} − \frac{1}{4}(e^{α_{2}} + e^{−α_{2}}).

We also use α_{2} to define ω_{3} and ω_{4} in the same fashion. Then ω_{i} for i = 1, 2, 3, 4 are commutative Virasoro vectors of central charge \frac{1}{2} (see [DMZ] and [DGH]). It is well-known that (V_{L}^{+})_{2} is a commutative (nonassociative) algebra under u × v = u_{1}v since the degree 1 part is 0 (cf. [FLM]). Let X be the span of ω_{i} for all i.

Lemma 6.1. If u ∈ (V_{L})_{2} is a Virasoro vector of central charge 1/2 then u = ω_{i} for some i.

Proof. The space (V_{L})_{2} is 5-dimensional with a basis

{ω_{1}, ω_{2}, ω_{3}, ω_{4}, α_{1}(−1)α_{2}(−1)}.

Let u = ∑_{i=1}^{4} c_{i}ω_{i} + xα_{1}(−1)α_{2}(−1) ∈ (V_{L})_{2} be a Virasoro vector of central charge 1/2. Then u × u = 2u. Note that ω_{i} × ω_{j} = δ_{i,j}2ω_{i} for i, j = 1, 2, 3, 4, ω_{i} × α_{1}(−1)α_{2}(−1) = \frac{1}{2}α_{1}(−1)α_{2}(−1) and α_{1}(−1)α_{2}(−1) × α_{1}(−1)α_{2}(−1) = 4α_{1}(−1)^{2} + 4α_{2}(−1)^{2}. So we have a nonlinear system

2c_{i} = 2c_{i}^{2} + 32x^{2}, i = 1, 2, 3, 4

2x = ∑_{i=1}^{4} xc_{i}.

If x ≠ 0 then ∑_{i=1}^{4} c_{i} = 2 and 2 = ∑_{i=1}^{4} c_{i}^{2} + 64x^{2}.

Since the central charge of u is 1/2 we have

\frac{1}{4} = u_{3}u = ∑_{i=1}^{4} \frac{c_{i}^{2}}{4} + 16x^{2}

and 1 = ∑_{i=1}^{4} c_{i}^{2} + 64x^{2}. This is a contradiction. So x = 0. This implies that c_{i} = 0, 1 and u = ω_{i} for some i. □

By Lemma 6.1, any automorphism σ of V_{L}^{+} induces a permutation of the four ω_{i}. It is known from [FLM] that (V_{L}^{+})_{2} has a nondegenerate symmetric bilinear form

(·, ·) given by (u, v) = u_{3}v for u, v ∈ (V_{L}^{+})_{2}. The orthogonal complement of X in (V_{L}^{+})_{2} with respect to the form is spanned by α_{1}(−1)α_{2}(−1). Thus σα_{1}(−1)α_{2}(−1) = λα_{1}(−1)α_{2}(−1) for some nonzero constant λ. Since α_{1}(−1)α_{2}(−1)×α_{1}(−1)α_{2}(−1) = α_{1}(−1)^{2} + α_{2}(−1)^{2} which is a multiple of the Virasoro element ω. This shows that λ = ±1.

On the other hand,

V_{L}^{+} ≅ V_{\sqrt{2}L_{A_{1}}}^{+} ⊗ V_{\sqrt{2}L_{A_{1}}}^{+} ⊕ V_{\sqrt{2}L_{A_{1}}}^{−} ⊗ V_{\sqrt{2}L_{A_{1}}}^{−}.

By Corollary 3.3 of [DGH],

V_{L}^{+} ≅ L(\frac{1}{2}, 0)^{⊗4} ⊕ L(\frac{1}{2}, \frac{1}{2})^{⊗4}.

So if the restriction of σ to X is identity, then the action of σ on V_{\sqrt{2}L_{A_{1}}}^{+} ⊗ V_{\sqrt{2}L_{A_{1}}}^{+} is trivial and on V_{\sqrt{2}L_{A_{1}}}^{−} ⊗V_{\sqrt{2}L_{A_{1}}}^{−} is ±1. Indeed, there is automorphism τ of V_{L}^{+} such that τ acts trivially on V_{\sqrt{2}L_{A_{1}}}^{+} ⊗ V_{\sqrt{2}L_{A_{1}}}^{+} and acts as −1 on V_{\sqrt{2}L_{A_{1}}}^{−} ⊗ V_{\sqrt{2}L_{A_{1}}}^{−} by the fusion role for V_{\sqrt{2}L_{A_{1}}}^{+} (see [ADL]). As V_{\sqrt{2}L_{A_{1}}}^{+} ⊗V_{\sqrt{2}L_{A_{1}}}^{+} is generated by ω_{i} for i = 1, 2, 3, 4, any automorphism preserves V_{\sqrt{2}L_{A_{1}}}^{+} ⊗ V_{\sqrt{2}L_{A_{1}}}^{+} and its irreducible module V_{\sqrt{2}L_{A_{1}}}^{−} ⊗ V_{\sqrt{2}L_{A_{1}}}^{−} (cf. [DM1]). As a result, ⟨τ⟩ is a normal subgroup of Aut(V_{L}^{+}) isomorphic to ℤ_{2}.

Next we show how Sym_{4} can be realized as a subgroup of Aut(V_{L}^{+}) by showing that any permutation σ ∈ Sym_{4} gives rise to an automorphism of V_{L}^{+}. But it is clear that Sym_{4} acts on V_{L}^{+} by permuting the tensor factors. In order to see that Sym_{4} acts on V_{L}^{+} as automorphisms, it is enough to show that σ(Y (u, z)v) = Y (σu, z)σv for σ ∈ Sym_{4} and u, v ∈ V_{L}^{+}. There are 4 different ways to choose u, v. We only discuss the case that u, v ∈ L(\frac{1}{2}, \frac{1}{2})^{⊗4} since the other cases can be dealt with in a similar fashion. Let u = u^{1} ⊗ u^{2} ⊗ u^{3} ⊗ u^{4} and v = v^{1} ⊗ v^{2} ⊗ v^{3} ⊗ v^{4} where u_{i}, v_{i} are tensor factors in the i-th L(\frac{1}{2}, \frac{1}{2}). Let Y be a nonzero intertwining operator of type \binom{L( \frac{1}{2} ,0)}{L( \frac{1}{2} , \frac{1}{2} ) L( \frac{1}{2} , \frac{1}{2} )}. Then, up to a constant,

Y (u, z)v = Y(u_{1}, z)v_{1} ⊗ Y(u_{2}, z)v_{2} ⊗ Y(u_{3}, z)v_{3} ⊗ Y(u_{4}, z)v_{4}

(see [DMZ]). Since σ is a permuation, it is trivial to verify that σ(Y (u, z)v) = Y (σu, z)σv.

So we have proved the following:

Proposition 6.2. If b = 0 in the Gram matrix G then L ≅ \sqrt{2}L_{A_{1}} × \sqrt{2}L_{A_{1}} and Aut(V_{L}^{+}) ≅ Sym_{4} × ℤ_{2}.

Remark 6.3. Here is a different proof that Aut(V_{L}^{+}) contains a copy of Sym_{4} × ℤ_{2}, using the theory of finite subgroups of Lie groups. Our lattice L lies in M ≅ L_{A_{1}^{2}}. Take V_{M}, which is a lattice VOA. By [DN1] , V_{M} has automorphism group isomorphic to P SL(2, ℂ) ≀ 2. In P SL(2, ℂ), there is up to conjugacy a unique four group and its normalizer is isomorphic to Sym_{4}. Correspondingly, in P SL(2, ℂ) ≀ 2 there is a subgroup isomorphic to Sym_{4}≀2. In this, take a subgroup H of the form 2^{4}:[Sym_{3}×2]. Let t be an involution of H which maps to the central involution of H/O_{2}(H) ≅ Sym_{3} × 2 and take R := C_{O_{2}(H)}(t) ≅ 2^{2}. Take the fixed points V_{M}^{R}. We have that V_{M}^{R} is isomorphic to our V_{L}^{+}. So, V_{L}^{+} gets an action of H/R ≅ 2^{2}:[Sym_{3} × 2] ≅ Sym_{4} × 2.

6.2 L_{1} = ∅ and rank(L_{2}) = 2; b = 2.

Next we assume that b in the Gram matrix is 2. Then L ≅ \sqrt{2}L_{A_{2}}. Then L = ℤα_{1} + ℤα_{2} with (α_{i}, α_{i}) = 4 and (α_{1}, α_{2}) = 2. As before we define ω_{1}, ω_{2} by using α_{1}, ω_{3}, ω_{4} by using α_{2} and ω_{5}, ω_{6} by using α_{1} + α_{2}. Then ω_{i}, for i = 1, ..., 6 form a basis of (V_{L}^{+})_{2}.

Lemma 6.4. If u ∈ (V_{L}^{+})_{2} is a Virasoro vector of central charge 1/2, then u = ω_{i} for some i.

Proof. First proof. (There will be a second proof in the next section.)

Let u = ∑_{i=1}^{6} c_{i}ω_{i} for some c_{i} ∈ ℂ. Then u is a Virasoro vector of central charge 1/2 if and only if (u, u) = 1/4 and u × u = 2u. Note that

(ω_{i}, ω_{i}) = 1/4, (ω_{2j−1}, ω_{2j}) = 0, 1 ≤ i ≤ 6, j = 1, 2, 3

(ω_{1}, ω_{k}) = (ω_{2}, ω_{k}) = \frac{1}{32}, k = 3, 4, 5, 6.

So we have

(u, u) = \frac{1}{4} ∑_{i=1}^{6} c_{i}^{2} + \frac{1}{16} ∑_{j=1,2} ∑_{j<k≤3}(c_{2j−1}c_{2k−1} + c_{2j−1}c_{2k} + c_{2j}c_{2k−1} + c_{2j}c_{2k}) = \frac{1}{4}.

In order to compute u × u we need the following multiplication table in (V_{L}^{+})_{2} :

ω_{2i−1} × ω_{2i} = 0, i = 1, 2, 3

ω_{1} × ω_{3} = \frac{1}{4}(ω_{1} + ω_{3} − ω_{6}), ω_{2} × ω_{3} = \frac{1}{4}(ω_{2} + ω_{3} − ω_{5})

ω_{1} × ω_{4} = \frac{1}{4}(ω_{1} + ω_{4} − ω_{5}), ω_{2} × ω_{4} = \frac{1}{4}(ω_{2} + ω_{4} − ω_{6})

ω_{1} × ω_{5} = \frac{1}{4}(ω_{1} + ω_{5} − ω_{4}), ω_{2} × ω_{5} = \frac{1}{4}(ω_{2} + ω_{5} − ω_{3})

ω_{1} × ω_{6} = \frac{1}{4}(ω_{1} + ω_{6} − ω_{3}), ω_{2} × ω_{6} = \frac{1}{4}(ω_{2} + ω_{6} − ω_{4})

ω_{3} × ω_{5} = \frac{1}{4}(ω_{3} + ω_{5} − ω_{2}), ω_{4} × ω_{5} = \frac{1}{4}(ω_{4} + ω_{5} − ω_{1})

ω_{3} × ω_{6} = \frac{1}{4}(ω_{3} + ω_{6} − ω_{1}), ω_{2} × ω_{6} = \frac{1}{4}(ω_{2} + ω_{6} − ω_{2}).

Then u × u = 2u if and only if

c_{1}^{2} + \frac{1}{4}(c_{1}c_{3} + c_{1}c_{4} + c_{1}c_{5} + c_{1}c_{6} − c_{3}c_{6} − c_{4}c_{5}) = c_{1}

c_{2}^{2} + \frac{1}{4}(c_{2}c_{3} + c_{2}c_{4} + c_{2}c_{5} + c_{2}c_{6} − c_{3}c_{5} − c_{4}c_{6}) = c_{2}

c_{3}^{2} + \frac{1}{4}(c_{1}c_{3} + c_{2}c_{3} + c_{3}c_{5} + c_{3}c_{6} − c_{1}c_{6} − c_{2}c_{5}) = c_{3}

c_{4}^{2} + \frac{1}{4}(c_{1}c_{4} + c_{2}c_{4} + c_{4}c_{5} + c_{4}c_{6} − c_{1}c_{5} − c_{2}c_{6}) = c_{4}

c_{5}^{2} + \frac{1}{4}(c_{1}c_{5} + c_{2}c_{5} + c_{3}c_{5} + c_{4}c_{6} − c_{1}c_{4} − c_{2}c_{3}) = c_{5}

c_{6}^{2} + \frac{1}{4}(c_{1}c_{6} + c_{2}c_{6} + c_{3}c_{6} + c_{4}c_{6} − c_{1}c_{3} − c_{2}c_{4}) = c_{6}.

There are exactly 6 solutions to this linear system: c_{i} = 1 and c_{j} = 0 if j ≠ i where i = 1, ..., 6. We thank Harm Derksen for obtaining this result with the MacCauley software package. This finishes the proof of the lemma. □

Proposition 6.5. If b = 2 in the Gram matrix, then L ≅ \sqrt{2}L_{A_{2}} and Aut(V_{L}^{+}) is the LVOA^{+}-group.

Proof. First note that the Weyl group acts on L, preserving and acting as Sym_{3} on the set

{{±α_{1}}, {±α_{2}}, {±(α_{1} + α_{2})}}.

Now let σ ∈ Aut(V_{L}^{+}). Set X_{i} = {ω_{2i−1}, ω_{2i}} for i = 1, 2, 3. Then X_{i} are the only orthogonal pairs in X = X_{1} ∪ X_{2} ∪ X_{3}. Since σX = X we see that σ induces a permutation on the set {X_{1}, X_{2}, X_{3}}.

The above shows that 𝕆(L̂) induces Sym_{3} on this 3-set. We may therefore assume that σ preserves each X_{i}. In this case σ acts trivially on α_{1}(−1)^{2}, α_{2}(−1)^{2}, (α_{1} + α_{2})(−1)^{2}. That is, σ acts trivially on the subVOA they generate, which is isomorphic to M(1)^{+}. As a result, σ is in the LVOA^{+}-group. □

6.3 Alternate proof for b = 2

The system of equations in the variables c_{i} which occurred in the proof of 6.4 can be replaced by an equivalent system 6.7 which looks more symmetric. The old system was solved with software package MacCauley but not with Maple. The new system was solved with Maple and gives the same result as before.

Notation 6.6. Let r and s be independent norm 4 elements so that t := −r − s has norm 4. Let w be an idempotent w = p + q, where p = ar^{2} + bs^{2} + ct^{2} and q = dv_{r} + ev_{s} + ev_{t} which satisfies (w, w) = \frac{1}{16}. Since (L, L) ≤ 2ℤ, we may and do assume that the epsilon-function is identically 1. It follows that v_{r} × v_{s} = v_{t} and similarly for all permutations of {r, s, t}.

Lemma 6.7. From w^{2} = w, we have equations

(e1)  a = 16a^{2} + 4ab + 4ac − 4bc + d^{2}

(e2)  b = 16b^{2} + 4bc + 4ba − 4ac + e^{2}
(e3)  c = 16c^{2} + 4cq + 4cb − 4ab + f^{2}
(e4)  d = 2d(16a + 4b + 4c) + 2ef
(e5)  e = 2e(4a + 16b + 4c) + 2df
(e6)  f = 2f(4a + 4b + 16c) + 2de

and from (w, w) = \frac{1}{16}, we get the equation

(e7)  \frac{1}{16} = 32(a^{2} + b^{2} + c^{2}) + 16(ab + ac + bc) + 2(d^{2} + e^{2} + f^{2}).

Proof. Straightforward from Appendix: Algebraic rules. □

Proposition 6.8. There are just 6 solutions (a, b, c, d, e, f) ∈ ℂ^{6} to the equations (e1), …, (e7). They are (\frac{1}{32}, 0, 0, \frac{1}{8}, 0, 00), (\frac{1}{32}, 0, 0, −\frac{1}{8}, 0, 00) and ones obtained from these by powers of the permutation (abc)(def).

Proof. This follows from use of the solve command in the software package Maple. □

Remark 6.9. If we omit (e7), there are infinitely many solutions with d = e = f = 0. The reason is that the Jordan algebra of symmetric degree 2 matrices has infinitely many idempotents. It seems possible that the system in Lemma 6.7 could be solved by hand.

6.4 L_{1} = ∅ and rank(L_{2}) = 2; b = 1.

We now deal with the cases b = 1 in the Gram matrix.

Proposition 6.10. If b = 1 in the Gram matrix, then Aut(V_{L}^{+}) is the LVOA^{+} group.

Proof. By Corollary 5.27, any automorphism of V_{L}^{+} preserves M(1)_{2}^{+}, the result follows from Lemma 3.3 □

7 Aut(V_{L}^{+}), for L_{1} = ∅, rank(L_{2}) = 1

In this case we can assume that L_{2} = {2α_{1}, −2α_{1}}. Let α_{2} ∈ H such that (α_{i}, α_{j}) = δ_{i,j}. Then (V_{L}^{+})_{2} is 4-dimensional with basis v_{2α_{1}}, \frac{1}{2}α_{1}(−1)^{2}, \frac{1}{2}α_{2}(−1)^{2}, α_{1}(−1)α_{2}(−1).

Lemma 7.1. Any automorphism of V_{L}^{+} preserves the subspace S^{2}H of (V_{L}^{+})_{2} spanned by \frac{1}{2}α_{1}(−1)^{2}, \frac{1}{2}α_{2}(−1)^{2}, α_{1}(−1)α_{2}(−1).

Proof. Since Virasoro vectors of central charge 1 in S^{2}H span S^{2}H, it is enough to show that any Virasoro vector of central charge 1 lies in S^{2}H.

Let t = d_{1} \frac{α_{1}^{2}}{2} + d_{2} \frac{α_{2}^{2}}{2} + d_{3}v_{2α_{1}} + d_{4}α_{1}α_{2} be a Virasoro vector of central charge 1 with d_{3} ≠ 0. Then we must have t × t = 2t and (t, t) = 1/2. A straightforward computation shows that

t × t = d_{1}^{2}α_{1}^{2} + d_{2}^{2}α_{2}^{2} + d_{3}^{2}(2α_{1})^{2} + d_{4}^{2}(α_{1}^{2} + α_{2}^{2}) + 4d_{1}d_{3}v_{2α_{1}} + 2d_{1}d_{4}α_{1}α_{2} + 2d_{2}d_{4}α_{1}α_{2}.

This gives four equations

d_{1} = d_{1}^{2} + 4d_{3}^{2} + d_{4}^{2}
d_{2} = d_{2}^{2} + d_{4}^{2}
d_{3} = 2d_{1}d_{3}
d_{4} = d_{1}d_{4} + d_{2}d_{4}.

The relation (t, t) = 1/2 gives one more equation:

\frac{1}{2} = \frac{1}{2}d_{1}^{2} + \frac{1}{2}d_{2}^{2} + d_{4}^{2} + 2d_{3}^{2}.

Thus

1 = d_{1} + d_{2}.

Since d_{3} ≠ 0, d_{1} = 1/2 and d_{2} = 1/2. So we have

\frac{1}{4} = 4d_{3}^{2} + d_{4}^{2}, \frac{1}{4} = 2d_{3}^{2} + d_{4}^{2}.

This forces d_{3} = 0, a contradiction. □

Proposition 7.2. In this case, Aut(V_{L}^{+}) is the LVOA^{+}-group.

Proof. By Corollary 5.27, any automorphism of V_{L}^{+} preserves M(1)_{2}^{+}, the result follows from Lemma 3.3 □

8 Aut(V_{L}^{+}), for L_{1} ≠ ∅

Finally we deal with the case that L_{1} ≠ ∅. There are two cases: rank(L_{1}) = 2 or rank(L_{1}) = 1.

8.1 rank(L_{1}) = 2

In this case L = L_{A_{1}^{2}} or L = L_{A_{2}} because these are the only rank 2 root lattices possible and each is a maximal even integral lattice in its rational span.

8.1.1 L has type A_{1}^{2}

If L ≅ L_{A_{1}^{2}} then Aut(V_{L}) ≅ P SL(2, ℂ) ≀ 2 and

V_{L}^{+} ≅ V_{L_{A_{1}}}^{+} ⊗ V_{L_{A_{1}}}^{+} ⊕ V_{L_{A_{1}}}^{−} ⊗ V_{L_{A_{1}}}^{−}.

Since the connected component of the identity in Aut(V_{L}) contains a lift of −1_{L}, we may assume that such a lift is in a given maximal torus, so is equal to the automorphism e^{πiβ(0)/2}, where β is a sum of orthogonal roots.

It follows that V_{L}^{+} ≅ V_{K}, where K = 2L + ℤβ. The result [DN1] implies that Aut(V_{L}^{+}) ≅ Aut(V_{K}), which is the LVOA group 𝕋_{2}.Dih_{8}.

8.1.2 L has type A_{2}

Here, (V_{L}^{+})_{1} is a 3-dimensional Lie algebra isomorphic to sl(2, ℂ). The difficult part in this case is to determine the vertex operator subalgebra generated by (V_{L}^{+})_{1}. Let L_{A_{2}} = ℤα_{1} + ℤα_{2} such that (α_{i}, α_{i}) = 2 and (α_{1}, α_{2}) = −1. The set of roots in L is L_{1} = {±α_{i}|i = 1, 2, 3} where α_{3} = α_{1}+α_{2}. The positive roots are {α_{i}|i = 1, 2, 3}. The space (V_{L}^{+})_{1} is 3-dimensional with a basis v_{α_{i}} for i = 1, 2, 3 and (V_{L}^{−})_{1} is 5-dimensional with a basis α_{1}(−1), α_{2}(−1), e^{α_{i}} − e^{−α_{i}} for i = 1, 2, 3. It is a straightforward to verify that (v_{α_{i}})_{−1}v_{α_{i}} for i = 1, 2, 3 and α_{i}(−1)^{2} for i = 1, 2, 3 span the same space. Thus ω = \frac{1}{4}α_{1}(−1)^{2} + \frac{1}{12}(α_{1}(−1) + 2α_{2}(−1))^{2} lies in the vertex operator algebra generated by (V_{L}^{+})_{1}.

In order to determine the vertex operator algebra generated by (V_{L}^{+})_{1} we need to recall the standard modules for affine algebra

A_{1}^{(1)} = ŝl(2, ℂ) = sl(2, ℂ) ⊗ ℂ[t, t^{−1}] ⊕ ℂK

(cf. [DL]). We use the standard basis {α, x_{α}, x_{−α}} for sl(2, ℂ) such that

[α, x_{±α}] = ±2x_{±α}, [x_{α}, x_{−α}] = α.

We fix an invariant symmetric nondegenerate bilinear form on sl(2, ℂ) such that (α, α) = 2. The level k standard A_{1}^{(1)}-modules are parametrized by dominant integral linear weights \frac{i}{2}α for i = 0, ..., l such that the highest weight of the A_{1}^{(1)}-module, viewed as a linear form on ℂα⊕ℂK ⊂ ŝl(2, ℂ), is given by \frac{i}{2}α and the correspondence K ↦ k. Let us denote the corresponding standard A_{1}^{(1)}-module by L(k, \frac{i}{2}α). It is well known that L(k, 0) is a simple rational vertex operator algebra and L(k, \frac{i}{2}α) for i = 0, ..., k is a complete list of irreducible L(k, 0)-modules (cf. [DL], [FZ] and [L2]). Note that

L(k, \frac{i}{2}α) = ⊕_{n=0}^{∞}L(k, \frac{i}{2}α)_{λ_{i}+n}

where λ_{i} = \frac{i(i+2)}{4(k+2)} and L(k, \frac{i}{2}α)_{λ_{i}+n} is the eigenspace of L(0) with eigenvalue λ_{i} + n (cf. [DL]). In fact, the lowest weight space L(k, \frac{i}{2}α)_{λ_{i}} of L(k, \frac{i}{2}α) is an irreducible sl(2, ℂ)-module of dimension i + 1.

Since V_{L_{A_{2}}} is a unitary module for affine algebra A_{2}^{(1)} (cf. [FK]), the vertex operator algebra V generated by (V_{L}^{+})_{1} is isomorphic to the standard level k A_{1}^{(1)}-module L(k, 0) for some nonnegative integer k. Let {v_{1}, v_{2}, v_{3}} be an orthonormal basis of sl(2, ℂ) with respect to the standard bilinear form. Then ω′ = \frac{1}{2(k+2)} ∑_{i=1}^{3} v_{i}(−1)^{2}1 ∈ V is the Segal-Sugawara Virasoro vector. Let

Y (ω′, z) = ∑_{n∈ℤ}L(n)′z^{−n−2}.

Then

[L(n) − L(n)′, u_{m}] = 0

for m, n ∈ ℤ and u ∈ V. So L(−2) − L(−2)′ acts as a constant on V as V is a simple vertex operator algebra. As a result, L(−2) − L(−2)′ = 0 since the left side is both a constant and an operator which shifts degree by 2. The creation axiom for VOAs implies that, ω′ = ω. Since the central charge of ω is 2, the central charge \frac{3k}{2(k+2)} of ω′ is also 2. This implies that k = 4 and V ≅ L(4, 0). Now V_{L}^{+} is a L(4, 0)-module and the quotient module V_{L}^{+}/V has minimal weight (as inherited from V_{L}^{+}) greater than 1. On the other hand, the minimal weight of the irreducible L(4, 0)-module L(4, \frac{i}{2}α) is \frac{i(i+2)}{4(4+2)} which is less than 2 for 0 ≤ i ≤ 4. Since every irreducible is one of these, we conclude V_{L}^{+} = V = L(4, 0). Since V_{L}^{−} is an irreducible V_{L}^{+}-module with minimal weight 1, we immediately see that V_{L}^{−} = L(4, 2α).

So we have proved the following: Proposition 8.1. If rank(L_{1}) = 2 there are two cases.

(1) If L = L_{A_{1}^{2}} then V_{L}^{+} is again a lattice vertex operator algebra V_{K} where K is generated by β_{1}, β_{2} with (β_{i}, β_{i}) = 4 and (β_{1}, β_{2}) = 0. The automorphism group of V_{L}^{+} is the LVOA^{+} group which is isomorphic to the LVOA-group for lattice K.

(2) If L = L_{A_{2}}, then V_{L}^{+} is isomorphic to the vertex operator algebra L(4, 0) and Aut(V_{L}^{+}) is isomorphic to P SL(2, ℂ) which is the automorphism group of sl(2, ℂ).

8.2 rank(L_{1}) = 1

8.2.1 L rectangular.

We first assume that L = ℤr + ℤs such that (r, r) = 2, (s, s) ∈ 6 + 8ℤ and (r, s) = 0. Then V_{L} = V_{L_{A_{1}}} ⊗ V_{ℤs} and

V_{L}^{+} = V_{L_{A_{1}}}^{+} ⊗ V_{ℤs}^{+} ⊕ V_{L_{A_{1}}}^{−} ⊗ V_{ℤs}^{−}.

Lemma 8.2. A group of shape (ℂβ/ℤ\frac{1}{4}β · ℤ_{2}) × ℤ_{2} acts on V_{L}^{+} as automorphisms.

Proof. We have already mentioned that V_{L_{A_{1}}}^{+} is isomorphic to V_{ℤβ} for (β, β) = 8 and V_{L_{A_{1}}}^{−} is isomorphic to V_{ℤβ+ \frac{1}{2} β} as V_{L_{A_{1}}}^{+}-modules. We also know from [DN1] that Aut(V_{ℤβ}) is isomorphic to ℂβ/(ℤ\frac{1}{8}β) · ℤ_{2} where the generator of ℤ_{2} is induced from the −1 isometry of the lattice ℤβ. The action of λβ ∈ ℂβ is given by the operator e^{2πiλβ(0)}. Note that ℂβ acts on V_{ℤβ+ \frac{1}{2} β} in the same way. But the kernel of the action of ℂβ on V_{ℤβ+ \frac{1}{2} β} is ℤ\frac{1}{4}β instead of ℤ\frac{1}{8}β. As a result, the torus ℂβ/ℤ\frac{1}{4}β acts on both V_{ℤβ} and V_{ℤβ+ \frac{1}{2} β}. By [DG], Aut(V_{ℤs}^{+}) is isomorphic to \frac{1}{2}ℤs/ℤs ≅ ℤ_{2} which also acts on V_{ℤs}^{−}. So the group (ℂβ/ℤ\frac{1}{4}β · ℤ_{2}) × ℤ_{2} acts on V_{L}^{+} as automorphisms. □

In order to determine Aut(V_{L}^{+}) in this case we need to recall the notion of commutant from [FZ].

Definition 8.3. Let V = (V, Y, 1, ω) be a vertex operator algebra and U = (U, Y, 1, ω′) be vertex operator subalgebra with a different Virasoro vector ω′. The commutant U^{c} of U in V is defined by

U^{c} := {v ∈ V |u_{n}v = 0, u ∈ U, n ≥ 0}.

Remark 8.4. The above space U^{c} is the space of vacuum-like vectors for U (see [L1]).

Lemma 8.5. Let V be a vertex operator algebra and U^{i} = (U^{i}, Y, 1, ω^{i}) are simple vertex operator subalgebras of V with Virasoro vector ω^{i} for i = 1, 2 such that ω = ω^{1} + ω^{2}. We assume that V has a decomposition

V ≅ ⊕_{i=0}^{p}P^{i} ⊗ Q^{i}

as U^{1} ⊗ U^{2}-module such that P^{0} ≅ U^{1}, Q^{0} ≅ U^{2}, the P^{i} are inequivalent U^{1}-modules and the Q^{i} are inequivalent U^{2}-modules. Then (U^{1})^{c} = U^{2} and (U^{2})^{c} = U^{1}.

Proof. It is enough to prove that (U^{2})^{c} ⊂ U^{1}. Let v ∈ (U^{2})^{c}. Then v is a vacuum-like vector for U^{2}. Then the U^{2}-submodule generated by v is isomorphic to U^{2} (see [L1]). Since V is a completely reducible U^{2}-module and any U^{2}-submodule isomorphic to U^{2} is contained in U^{1} ⊗ U^{2}. In particular, v ∈ U^{1} ⊗ U^{2}. This forces v ∈ U^{1}. □

Proposition 8.6. The group Aut(V_{L}^{+}) is isomorphic to ((ℂβ/ℤ\frac{1}{4}β) · ℤ_{2}) × ℤ_{2}. This can be interpreted as an action of ℕ(ℤ̂β) × ℤ_{2}, where (β, β) = 8.

Proof. We have already shown 8.2 that the group ((ℂβ/ℤ\frac{1}{4}β) · ℤ_{2}) × ℤ_{2} acts on V_{L}^{+} as automorphisms.

Let σ be an automorphism of V_{L}^{+}. Then σβ(−1) = λβ(−1) for some nonzero λ ∈ ℂ as (V_{L}^{+})_{1} is spanned by β(−1). This implies that σβ(n)σ^{−1} = λβ(n) for n ∈ ℤ. Since V_{ℤs}^{+} is precisely the subspace of V_{L}^{+} consisting of vectors killed by β(n) for n ≥ 0, we see that σV_{ℤs}^{+} ⊂ V_{ℤs}^{+}. Thus σ|_{V_{ℤs}^{+}} is an automorphism of V_{ℤs}^{+}. On the other hand, V_{L_{A_{1}}}^{+} is the commutant of V_{ℤs}^{+} in V_{L}^{+}by Lemma 8.5.

The above show that σ induces an automorphism of the tensor factor V_{L_{A_{1}}}^{+}. The restriction of σ to V_{L_{A_{1}}}^{+} ⊗ V_{ℤs}^{+} is a product σ_{1} ⊗ σ_{2} for some σ_{1} ∈ Aut(V_{L_{A_{1}}}^{+}) and σ_{2} ∈ Aut(V_{ℤs}^{+}). Multiplying σ by σ_{2} we can assume that σ = 1 on V_{ℤs}^{+}. As we have already mentioned, Aut(V_{L_{A_{1}}}^{+}) is isomorphic to (ℂβ/ℤ\frac{1}{8}β)·ℤ_{2}. Since (ℂβ/ℤ\frac{1}{8}β) acts trivially on β(−1) and the outer factor ℤ_{2} is represented in Aut(V_{L}^{+}) by action of ±1 on β(−1). As a result σβ(−1) = ±β(−1). Now multiplying σ by an outer element of (ℂβ/ℤ\frac{1}{4}β) · ℤ_{2}, we can assume that σβ(−1) = β(−1).

Set W_{nβ} = M(1) ⊗ e^{nβ} ⊗ V_{ℤs}^{+} and W_{nβ+β/2} = M(1) ⊗ e^{nβ+β/2} ⊗ V_{ℤs}^{−} for n ∈ ℤ where M(1) = ℂ[β(−n)|n > 0]. Then V_{L}^{+} = ⊕_{n∈ \frac{1}{2} ℤ}W_{nβ} and u_{m}v ∈ W_{μ+ν} for u ∈ W_{μ} and v ∈ W_{ν}, and n ∈ ℤ. Note that W_{μ} is the eigenspace of β(0) with eigenvalue (β, μ). Since σβ(−1) = β(−1) we see that σ acts on each W_{μ} as a constant λ_{μ} and λ_{μ}λ_{ν} = λ_{μ+ν}. As a result, σ = e^{2πiγ(0)} for some γ ∈ ℂβ. That is, σ lies in ℂβ/ℤ\frac{1}{4}β. This completes the proof. □

8.2.2 L not rectangular

Next we assume that L ≠ ℤr ⊥ ℤs. Then L = ℤr ⊕ ℤ\frac{1}{2}(s + t) where (s, s) ∈ 6 + 8ℤ and (s, s) ≥ 14 (see 4.3). Let K = ℤr ⊕ ℤs. Then L = K ∪ (K + \frac{1}{2}(r + s)) and V_{L} = V_{ℤr} ⊗ V_{ℤs} ⊕ V_{(ℤ+ \frac{1}{2} )r} ⊗ V_{(ℤ+ \frac{1}{2} )s}. Thus

V_{L}^{+} = V_{ℤr}^{+} ⊗ V_{ℤs}^{+} ⊕ V_{ℤr}^{−} ⊗ V_{ℤs}^{−} ⊕ V_{(ℤ+ \frac{1}{2} )r}^{+} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{+} ⊕ V_{(ℤ+ \frac{1}{2} )r}^{−} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{−}

and

V_{L}^{+} = V_{K}^{+} ⊕ V_{K+ \frac{1}{2} (s+t)}^{+}.

As before, we note that V_{ℤr}^{+} is isomorphic to V_{ℤβ} with (β, β) = 8.

Proposition 8.7. Assume that rank(L_{1}) = 1, L ≠ ℤr + ℤs, r, s as above. Then Aut(V_{L}^{+}) ≅ (ℂβ/\frac{1}{2}ℤβ) · ℤ_{2}, where (β, β) = 8. The action is trivial on the subVOA V_{ℤs}^{+} and leaves V_{ℤr}^{+} invariant. A generator of the quotient ℤ_{2} comes from the −1 isometry of \frac{1}{4}ℤβ and α ∈ ℂβ acts as e^{2πiα(0)}.

Proof. Note that V_{K}^{+} is a subalgebra of V_{L}^{+} and V_{K+ \frac{1}{2} (r+s)}^{+} is an irreducible V_{K}^{+}-module. By Proposition 8.6,

Aut(V_{K}^{+}) = ((ℂβ/\frac{1}{4}ℤβ)·ℤ_{2}) × ℤ_{2}.

As we have already mentioned that V_{ℤr}^{+} is isomorphic to V_{ℤβ} with (β, β) = 8 and V_{ℤr}^{−} is isomorphic to V_{ℤβ+ \frac{1}{2} β} as V_{ℤβ}-module. It is easy to see that V_{(ℤ+ \frac{1}{2} )β}^{±} is isomorphic to V_{(ℤ± \frac{1}{4} )β} as V_{ℤβ}-module. So the action of ℂβ/ℤ\frac{1}{4}β on V_{K}^{+} cannot be extended to

an action of V_{L}^{+}. But the torus ℂβ/\frac{1}{2}ℤβ does acts on V_{L}^{+}. As a result, ℕ(ℤ̂\frac{1}{2}β) ≅ (ℂβ/\frac{1}{2}ℤβ) · ℤ_{2} is a subgroup of Aut(V_{L}^{+}).

The same argument used in the proof of Proposition 8.6 shows that that any automorphism σ of V_{L}^{+} preserves V_{ℤr}^{+} ⊗ V_{ℤs}^{+}. Since V_{ℤr}^{+} ⊗ V_{ℤs}^{+}, V_{ℤr}^{−} ⊗ V_{ℤs}^{−}, V_{(ℤ+ \frac{1}{2} )r}^{+} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{+}, V_{(ℤ+ \frac{1}{2} )r}^{−} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{−} are inequivalent irreducible V_{ℤr}^{+} ⊗ V_{ℤs}^{+}-modules (see [DM1] and [DLM]), we see that σ preserves

V_{K}^{+} = V_{ℤr}^{+} ⊗ V_{ℤs}^{+} ⊕ V_{ℤr}^{−} ⊗ V_{ℤs}^{−}.

Since ℂβ/\frac{1}{4}ℤβ · ℤ_{2} is a quotient group of ℂβ/\frac{1}{2}ℤβ · ℤ_{2}, we can multiply σ by an element of ℂβ/\frac{1}{2}ℤβ · ℤ_{2} and assume that σ acts trivially on the first tensor factor of V_{K}^{+}. If σ is the identity on V_{K}^{+}, then σ is either 1 or −1 on V_{K+ \frac{1}{2} (r+s)}^{+}. If σ is −1 on V_{K+ \frac{1}{2} (r+s)}^{+} then σ = e^{πi \frac{1}{2} β(0)} is an element of ℂβ/\frac{1}{2}ℤβ · ℤ_{2}.

If σ is not identity on V_{K}^{+} then we must have σ = e^{πi \frac{1}{(s,s)} s(0)} on V_{K}^{+}. We will get a contradiction in this case. Notice that the lowest weight space of V_{(ℤ+ \frac{1}{2} )r}^{+} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{+} is 1-dimensional and spanned by u = (e^{r/2} + e^{−r/2}) ⊗ (e^{s/2} + e^{−s/2}). Since σ preserves V_{(ℤ+ \frac{1}{2} )r}^{+} ⊗ V_{(ℤ+ \frac{1}{2} )s}^{+}, it must map u to λu for some nonzero constant λ. Note that u_{\frac{1}{4} (r+s,r+s)−1}u = 4. This forces λ = ±1. On the other hand,

u_{− \frac{1}{4} (r+s,r+s)−1}u = (e^{r} + e^{−r}) ⊗ (e^{s} + e^{−s}) + ⋯

has nontrivial projection to the −1 eigenspace of σ in V_{K}^{+}. This forces λ = ±i, a contradiction. □

9 Appendix: Algebraic rules

For the symmetric matrices of degree n, there is a widely used basis, Jordan product and inner product, which we review here. (This section is taken almost verbatim from [G10]).

Proposition 9.1. H is a vector space of finite dimension n with nondegenerate symmetric bilinear form (·, ·).

r, s, … stand for elements of H and rs stands for the symmetric tensor r⊗s+s⊗r. rs × pq = (r, p)sq + (r, q)sp + (s, p)rq + (s, q)rp. (rs, pq) = (r, p)(s, q) + (r, q)(s, p) rs × v_{t} = (r, t)(s, t)v_{t}.

Definition 9.2. The Symmetric Bilinear Form. Source: [FLM], p.217. This form is associative with respect to the product (Section 3). We write H for H_{1}. The set of all g^{2} and x_{α}^{+} spans V_{2}.

(2.2.1)  ⟨g^{2}, h^{2}⟩ = 2⟨g, h⟩^{2},

whence

(2.2.2)  ⟨pq, rs⟩ = ⟨p, r⟩⟨q, s⟩ + ⟨p, s⟩⟨q, r⟩, for p, q, r, s ∈ H.

(2.2.3)  ⟨x_{α}^{+}, x_{β}^{+}⟩ = \begin{cases}2 & α = ±β \\ 0 & else \end{cases}

(2.2.4)  ⟨g^{2}, x_{β}^{+}⟩ = 0.

Definition 9.3. In addition, we have the distinguished Virasoro element ω and identity 𝕀 := \frac{1}{2}ω on V_{2} (see Section 3). If h_{i} is a basis for H and h_{i}^{∗} the dual basis, then ω = \frac{1}{2} ∑_{i} h_{i}h_{i}^{∗}.

Remark 9.4.

(2.4.1)  ⟨g^{2}, ω⟩ = ⟨g, g⟩

(2.4.2)  ⟨g^{2}, 𝕀⟩ = \frac{1}{2}⟨g, g⟩

(2.4.3)  ⟨𝕀, 𝕀⟩ = dim(H)/8

(2.4.4)  ⟨ω, ω⟩ = dim(H)/2

If {x_{i} | i = 1, … ℓ} is an ON basis,

(2.4.5)  𝕀 = \frac{1}{4} ∑_{i=0}^{ℓ} x_{i}^{2}

(2.4.6)  ω = \frac{1}{2} ∑_{i=0}^{ℓ} x_{i}^{2}.

Definition 9.5. The product on V_{2}^{F} comes from the vertex operations. We give it on standard basis vectors, namely xy ∈ S^{2}H_{1}, for x, y ∈ H_{1} and v_{λ} := e^{λ} + e^{−λ}, for λ ∈ L_{2}. (This is the same as x_{λ}^{+}, used in [FLM].) Note that (3.1.1) give the Jordan algebra structure on S^{2}H_{1}, identified with the space of symmetric 8 × 8 matrices, and with ⟨x, y⟩ = \frac{1}{8}tr(xy). The function ε below is a standard part of notation for lattice VOAs.

(3.1.1)  x^{2} × y^{2} = 4⟨x, y⟩xy,  pq × y^{2} = 2⟨p, y⟩qy + 2⟨q, y⟩py,
pq × rs = ⟨p, r⟩qs + ⟨p, s⟩qr + ⟨q, r⟩ps + ⟨q, s⟩pr;

(3.1.2)  x^{2} × v_{λ} = ⟨x, λ⟩^{2}v_{λ},  xy × v_{λ} = ⟨x, λ⟩⟨y, λ⟩v_{λ}

(3.1.3)  v_{λ} × v_{μ} = \begin{cases}0 & ⟨λ, μ⟩ ∈ \{0, ±1, ±3\}; \\ ε⟨λ, μ⟩v_{λ+μ} & ⟨λ, μ⟩ = −2; \\ λ^{2} & λ = μ. \end{cases}

Some consequences are these:

Corollary 9.6. If x_{1}, … is a basis and y_{1}, … is the dual basis, then 𝕀 := \frac{1}{4} ∑_{i=1}^{n} x_{i}y_{i} is the identity of the algebra S^{2}H.

(𝕀, 𝕀) = \frac{n}{8}.

References

[AD] T. Abe and C. Dong, Classification of irreducible modules for vertex operator algebra V_{L}^{+} : general case, J. Algebra 273 (2004), 657–685.

[ADL] T. Abe, C. Dong and H. Li, Fusion rules for the vertex operator algebras M(1)^{+} and V_{L}^{+}, Comm. Math. Phys., to appear.

[B] R. E. Borcherds, Vertex algebras, Kac-Moody algebras, and the Monster, Proc. Natl. Acad. Sci. USA 83 (1986), 3068-3071.

[D] C. Dong, Vertex algebras associated with even lattices, J. Algebra 161 (1993), 245-265.

[DG1] C. Dong and R.L. Griess Jr., Rank one lattice type vertex operator algebras and their automorphism groups, J. Algebra 208 (1998), 262-275.

[DG2] C. Dong and R.L. Griess Jr., Automorphism groups of finitely generated vertex operator algebras, Michigan Math. J. 50 (2002), 227–239.

[DGR] C. Dong and R.L. Griess Jr. and A. J. E. Ryba, Rank one lattice type vertex operator algebras and their automorphism groups, J. Algebra 217 (1999), 701–710.

[DGH] C. Dong, R. Griess Jr. and G. Hoehn, Framed vertex operator algebras, codes and the moonshine module, Comm. Math. Phys. 193 (1998), 407–448.

[DL] C. Dong and J. Lepowsky, Generalized vertex algebras and relative vertex operators, Progress in Math., Vol.112, Birkhäuser, Boston, 1993.

[DLM] C. Dong, H. Li and G. Mason, Compact automorphism groups of vertex operator algebras, International Math. Research Notices 18 (1996), 913–921.

[DM1] C. Dong and G. Mason, On quantum Galois theory, Duke Math. J. 86 (1997), 305–321.

[DM2] C. Dong and G. Mason, Vertex operator algebras and their automorphism groups, In: Proceedings of International Conference on Representation Theory (Shanghai, 1998), China Higher Education Press and Springer-Verlag, Beijing, 2000, 145-166.

[DN1] C. Dong and K. Nagatomo, Automorphism groups of lattice vertex algebras, Contemp. Math. 248 (1999), 117-133.

[DN2] C. Dong and K. Nagatomo, Classification of irreducible modules for the vertex operator algebra M(1)^{+} II. Higher Rank, J. Algebra, 240 (2001), 389–325.

[FLM] I. Frenkel, J. Lepowsky and A. Meurman, Vertex Operator Algebras and the Monster, Pure and Appl. Math., Vol. 134, Academic Press, Boston, 1988.

[FZ] I. Frenkel and Y. Zhu, Vertex operator algebras associated to representations of affine and Virasoro algebras, Duke Math. J. 66 (1992), 123–168.

[G1] R. Griess, Jr., GNAVOA, I. Studies in groups, nonassociative algebras and vertex operator algebras. Vertex operator algebras in mathematics and physics (Toronto, ON, 2000), 71–88, Fields Inst. Commun., 39, AMS, 2003.

[G2] Robert L. Griess, Jr., Automorphisms of vertex operator algebras, a survey, Proceedings of the Raleigh Conference on affine algebras, quantum affine algebras and related topics, 21-24 May, 1998. Contemporary Mathematics, volume 248, American Mathematical Society, 1999.

[G3] Robert L. Griess, Jr., Positive definite lattices of rank at most 8, Journal of Number Theory, 103 (2003), 77-84.

[Gr10] R. Griess, A vertex operator algebra related to E_{8} wth automorphism group O^{+}(10, 2), article inThe Monster and Lie Algebras, ed. J. Ferrar and K. Harada, de Gruyter, 1998.

[GH] R. Griess Jr. and G. Hoehn, Frame stabilizers for the lattice vertex operator algebra of type E_{8}, with Gerald Höhn, J. Reine. Angew. Math. 561 (2003), 1-37.

[J] Nathan Jacobson, ”Isomorphism of Jordan Rings”, Amer. J. Math. 70, (1948), 317-326.

[L1] H. Li, Symmetric invariant bilinear forms on vertex operator algebras, J. Pure Appl. Algebra 96 (1994), 279–297.

[L2] H. Li, Local systems of vertex operators, vertex superalgebras and modules, J. Pure Appl. Alg, 109 (1996), 143–195.

[S] H. Shimakura, The automorphism group of the vertex operator algebra V_{L}^{+} for an even lattice L without roots, J. Algebra 280 (2004), 29-57.