[PPT] - Automorphism groups of some orbifold models of lattice VOAs Ching PowerPoint Presentation

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Automorphism groups of some orbifold models of lattice VOAs

Ching Hung Lam

Academia Sinica Based on joint works with Hiroki Shimakura and Koichi Betsumiya

June 25, 2019

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 1 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 2 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra. The main idea is to view V as a simple current extension of VLg ⊗ V ˆ

g Λg (proposed by H¨

hn);

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 2 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra. The main idea is to view V as a simple current extension of VLg ⊗ V ˆ

g Λg (proposed by H¨

hn);

VLg is a lattice VOA and Λg is a coinvariant lattice of the Leech lattice Λ .

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 2 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra. The main idea is to view V as a simple current extension of VLg ⊗ V ˆ

g Λg (proposed by H¨

hn);

VLg is a lattice VOA and Λg is a coinvariant lattice of the Leech lattice Λ . The key step is to compute the stabilizer StabAut (V )(VLg ⊗ V ˆ

g Λg )

using the theory of simple current extensions [Shimakura 2007].

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 2 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra. The main idea is to view V as a simple current extension of VLg ⊗ V ˆ

g Λg (proposed by H¨

hn);

VLg is a lattice VOA and Λg is a coinvariant lattice of the Leech lattice Λ . The key step is to compute the stabilizer StabAut (V )(VLg ⊗ V ˆ

g Λg )

using the theory of simple current extensions [Shimakura 2007]. It turns out Aut (V ) = Inn (V )StabAut (V )(VLg ⊗ V ˆ

g Λg ),

where Inn (V ) = exp(a(0)) | a ∈ V1}.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 2 / 26

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Main question

Try to compute the full automorphism group of a holomorphic vertex

perator algebra V of central charge 24.

Let V1 ∼ = g as a Lie algebra. The main idea is to view V as a simple current extension of VLg ⊗ V ˆ

g Λg (proposed by H¨

hn);

VLg is a lattice VOA and Λg is a coinvariant lattice of the Leech lattice Λ . The key step is to compute the stabilizer StabAut (V )(VLg ⊗ V ˆ

g Λg )

using the theory of simple current extensions [Shimakura 2007]. It turns out Aut (V ) = Inn (V )StabAut (V )(VLg ⊗ V ˆ

g Λg ),

where Inn (V ) = exp(a(0)) | a ∈ V1}. We need to know the groups Aut (VLg) and Aut (V ˆ

g Λg ).

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StabAut (V )(VLg ⊗ V ˆ

g Λg)

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StabAut (V )(VLg ⊗ V ˆ

g Λg)

Set V 1 = VLg and V 2 = V ˆ

g Λg .

Let f : (Irr(V 1), q1) → (Irr(V 2), −q2) be an isometry such that V =

M∈Irr(V 1)

M ⊗ f (M)

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StabAut (V )(VLg ⊗ V ˆ

g Λg)

Set V 1 = VLg and V 2 = V ˆ

g Λg .

Let f : (Irr(V 1), q1) → (Irr(V 2), −q2) be an isometry such that V =

M∈Irr(V 1)

M ⊗ f (M) Then S = {(M, f (M)) | M ∈ Irr(V 1)} is a maximal totally singular subspace of (Irr(V 1) ⊕ Irr(V 2), q1 + q2).

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StabAut (V )(VLg ⊗ V ˆ

g Λg)

Set V 1 = VLg and V 2 = V ˆ

g Λg .

Let f : (Irr(V 1), q1) → (Irr(V 2), −q2) be an isometry such that V =

M∈Irr(V 1)

M ⊗ f (M) Then S = {(M, f (M)) | M ∈ Irr(V 1)} is a maximal totally singular subspace of (Irr(V 1) ⊕ Irr(V 2), q1 + q2). By [Shimakura 2007], there is an exact sequence 1 → S∗ → NAut (V )(S∗) → StabAut (V 1⊗V 2)(S) → 1, where StabAut (V 1⊗V 2)(S) = {g ∈ Aut (V 1 ⊗ V 2) | S ◦ g = S} and S∗ = dual group of S.

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StabAut (V )(VLg ⊗ V ˆ

g Λg)

Set V 1 = VLg and V 2 = V ˆ

g Λg .

Let f : (Irr(V 1), q1) → (Irr(V 2), −q2) be an isometry such that V =

M∈Irr(V 1)

M ⊗ f (M) Then S = {(M, f (M)) | M ∈ Irr(V 1)} is a maximal totally singular subspace of (Irr(V 1) ⊕ Irr(V 2), q1 + q2). By [Shimakura 2007], there is an exact sequence 1 → S∗ → NAut (V )(S∗) → StabAut (V 1⊗V 2)(S) → 1, where StabAut (V 1⊗V 2)(S) = {g ∈ Aut (V 1 ⊗ V 2) | S ◦ g = S} and S∗ = dual group of S. Note: Aut (V 1 ⊗ V 2) = Aut (V 1) × Aut (V 2) since V 1 ≇ V 2.

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Let µi : Aut (V i) → O(Irr(V i), qi), i = 1, 2, be the group homomorphism induced from the g-conjugate action of Aut (V i) on Irr(V i),

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Let µi : Aut (V i) → O(Irr(V i), qi), i = 1, 2, be the group homomorphism induced from the g-conjugate action of Aut (V i) on Irr(V i), where O(Irr(V i), qi) denotes the isometry group of (Irr(V i), qi).

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Let µi : Aut (V i) → O(Irr(V i), qi), i = 1, 2, be the group homomorphism induced from the g-conjugate action of Aut (V i) on Irr(V i), where O(Irr(V i), qi) denotes the isometry group of (Irr(V i), qi).

Lemma

StabAut (V 1⊗V 2)(S)/(ker µ1 × ker µ2) ∼ = (Im µ1) ∩ f −1(Im µ2)f .

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Let K(V ) = {g ∈ Aut (V ) | g|V1 = idV1} and define Out(V ) = Aut (V )/K(V )Inn (V ).

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Let K(V ) = {g ∈ Aut (V ) | g|V1 = idV1} and define Out(V ) = Aut (V )/K(V )Inn (V ).

Proposition

Assume ker µ2 = id. Then we have Out(V ) ∼ = µ−1

L ((Im µ1) ∩ f −1(Im µ2)f )/W (V1),

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Let K(V ) = {g ∈ Aut (V ) | g|V1 = idV1} and define Out(V ) = Aut (V )/K(V )Inn (V ).

Proposition

Assume ker µ2 = id. Then we have Out(V ) ∼ = µ−1

L ((Im µ1) ∩ f −1(Im µ2)f )/W (V1),

where µL : O(Lg) → O(D(Lg), qLg) is the canonical group homomorphism and W (V1) the Weyl group of V1.

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Let K(V ) = {g ∈ Aut (V ) | g|V1 = idV1} and define Out(V ) = Aut (V )/K(V )Inn (V ).

Proposition

Assume ker µ2 = id. Then we have Out(V ) ∼ = µ−1

L ((Im µ1) ∩ f −1(Im µ2)f )/W (V1),

where µL : O(Lg) → O(D(Lg), qLg) is the canonical group homomorphism and W (V1) the Weyl group of V1.

Lemma

We have K(V ) < Inn (V ) and K(V ) = {exp(−2π √ −1x(0)) | x ∈ ˜ Q∗/Lg}, where ˜ Q = s

i=1 1 √ki Qi, Qi is the root lattice of gi and

V1 ∼ = g ∼ = g1 ⊕ · · · ⊕ gs.

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For ker µ1, let X(L) = {h ∈ O(L) | h = id on D(L) = L∗/L} and X(ˆ L) = {g ∈ O(ˆ L) | ¯ g ∈ X(L)}. Then we have

Lemma

ker µ1 = Inn (VLg)X(ˆ Lg) and Im µ1 ∼ = O(Lg)/X(Lg).

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Aut (V ˆ

g Λg)

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Aut (V ˆ

g Λg)

Recall that Aut (VL) = N(VL) O(ˆ L), where N(VL) =

exp(a(0)) | a ∈ (VL)1
= Inn (VL).

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Aut (V ˆ

g Λg)

Recall that Aut (VL) = N(VL) O(ˆ L), where N(VL) =

exp(a(0)) | a ∈ (VL)1
= Inn (VL).

Moreover, there is an exact sequence of [FLM88, Proposition 5.4.1] 1 → Hom(L, Z/2Z) → O(ˆ L)

ϕ

→ O(L) → 1.

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Aut (V ˆ

g Λg)

Recall that Aut (VL) = N(VL) O(ˆ L), where N(VL) =

exp(a(0)) | a ∈ (VL)1
= Inn (VL).

Moreover, there is an exact sequence of [FLM88, Proposition 5.4.1] 1 → Hom(L, Z/2Z) → O(ˆ L)

ϕ

→ O(L) → 1. When L(2) = {x ∈ L | x, x = 2} = ∅, the normal subgroup N(VL) = {exp(λα(0)) | α ∈ L, λ ∈ C} is abelian and we have N(VL) ∩ O(ˆ L) = Hom(L, Z/2Z) and Aut (VL)/N(VL) ∼ = O(L).

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Aut (V ˆ

g Λg)

Recall that Aut (VL) = N(VL) O(ˆ L), where N(VL) =

exp(a(0)) | a ∈ (VL)1
= Inn (VL).

Moreover, there is an exact sequence of [FLM88, Proposition 5.4.1] 1 → Hom(L, Z/2Z) → O(ˆ L)

ϕ

→ O(L) → 1. When L(2) = {x ∈ L | x, x = 2} = ∅, the normal subgroup N(VL) = {exp(λα(0)) | α ∈ L, λ ∈ C} is abelian and we have N(VL) ∩ O(ˆ L) = Hom(L, Z/2Z) and Aut (VL)/N(VL) ∼ = O(L). In particular, we have an exact sequence 1 → N(VL) → Aut (VL)

ϕ

→ O(L) → 1.

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Theorem

Let L be an even positive definite lattice with L(2) = ∅. Let g be a fixed point free isometry of L and ˆ g a lift of g in O(ˆ L). Then we have the following exact sequences. 1 − → Hom(L/(1 − g)L, C∗) − → NAut (VL)(ˆ g)

ϕ

− → NO(L)(g) − → 1; 1 − → Hom(L/(1 − g)L, C∗) − → CAut (VL)(ˆ g)

ϕ

− → CO(L)(g) − → 1.

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Theorem

Let L be an even positive definite lattice with L(2) = ∅. Let g be a fixed point free isometry of L and ˆ g a lift of g in O(ˆ L). Then we have the following exact sequences. 1 − → Hom(L/(1 − g)L, C∗) − → NAut (VL)(ˆ g)

ϕ

− → NO(L)(g) − → 1; 1 − → Hom(L/(1 − g)L, C∗) − → CAut (VL)(ˆ g)

ϕ

− → CO(L)(g) − → 1. It is clear that NAut (VL)(ˆ g) acts on V ˆ

g L and there is a group

homomorphism f : NAut (VL)(ˆ g)/ˆ g − → Aut (V ˆ

g L ).

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Theorem

Let L be an even positive definite lattice with L(2) = ∅. Let g be a fixed point free isometry of L and ˆ g a lift of g in O(ˆ L). Then we have the following exact sequences. 1 − → Hom(L/(1 − g)L, C∗) − → NAut (VL)(ˆ g)

ϕ

− → NO(L)(g) − → 1; 1 − → Hom(L/(1 − g)L, C∗) − → CAut (VL)(ˆ g)

ϕ

− → CO(L)(g) − → 1. It is clear that NAut (VL)(ˆ g) acts on V ˆ

g L and there is a group

homomorphism f : NAut (VL)(ˆ g)/ˆ g − → Aut (V ˆ

g L ).

The key question is to determine if f is injective and/or surjective.

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Theorem

Let L be an even positive definite lattice with L(2) = ∅. Let g be a fixed point free isometry of L and ˆ g a lift of g in O(ˆ L). Then we have the following exact sequences. 1 − → Hom(L/(1 − g)L, C∗) − → NAut (VL)(ˆ g)

ϕ

− → NO(L)(g) − → 1; 1 − → Hom(L/(1 − g)L, C∗) − → CAut (VL)(ˆ g)

ϕ

− → CO(L)(g) − → 1. It is clear that NAut (VL)(ˆ g) acts on V ˆ

g L and there is a group

homomorphism f : NAut (VL)(ˆ g)/ˆ g − → Aut (V ˆ

g L ).

The key question is to determine if f is injective and/or surjective.

Definition

An automorphism h ∈ Aut (V ˆ

g L ) is said to be an extra automorphism if it

is not in the image of f .

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Extra automorphisms

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Extra automorphisms

Let An be a root lattice of type An. Let hAn be an (n + 1)-cycle in Weyl(An) ∼ = Symn+1.

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Extra automorphisms

Let An be a root lattice of type An. Let hAn be an (n + 1)-cycle in Weyl(An) ∼ = Symn+1. Then the action of hAn on sln+1(C) is given by the conjugation of P, that is, hAn : A → P−1AP for A ∈ sl(n + 1, C),

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Extra automorphisms

Let An be a root lattice of type An. Let hAn be an (n + 1)-cycle in Weyl(An) ∼ = Symn+1. Then the action of hAn on sln+1(C) is given by the conjugation of P, that is, hAn : A → P−1AP for A ∈ sl(n + 1, C), and B−1PB = diag(ω, ω2, ..., 1) where P =   

1 · · · . . . ... ... . . . ... 1 1 · · ·

  

and B =

1 √n + 1    

ω ω2 · · · ωn 1 ω2 ω4 · · · ω2n 1 . . . ... ... . . . . . . ωn ω2n ... ωn2 1 1 1 · · · 1 1

    .

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Extra automorphisms

Let An be a root lattice of type An. Let hAn be an (n + 1)-cycle in Weyl(An) ∼ = Symn+1. Then the action of hAn on sln+1(C) is given by the conjugation of P, that is, hAn : A → P−1AP for A ∈ sl(n + 1, C), and B−1PB = diag(ω, ω2, ..., 1) where P =   

1 · · · . . . ... ... . . . ... 1 1 · · ·

  

and B =

1 √n + 1    

ω ω2 · · · ωn 1 ω2 ω4 · · · ω2n 1 . . . ... ... . . . . . . ωn ω2n ... ωn2 1 1 1 · · · 1 1

    . Define a map σAn : sl(n + 1, C) → sl(n + 1, C) by σAn(A) = B−1AB.

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Extra automorphisms

Let An be a root lattice of type An. Let hAn be an (n + 1)-cycle in Weyl(An) ∼ = Symn+1. Then the action of hAn on sln+1(C) is given by the conjugation of P, that is, hAn : A → P−1AP for A ∈ sl(n + 1, C), and B−1PB = diag(ω, ω2, ..., 1) where P =   

1 · · · . . . ... ... . . . ... 1 1 · · ·

  

and B =

1 √n + 1    

ω ω2 · · · ωn 1 ω2 ω4 · · · ω2n 1 . . . ... ... . . . . . . ωn ω2n ... ωn2 1 1 1 · · · 1 1

    . Define a map σAn : sl(n + 1, C) → sl(n + 1, C) by σAn(A) = B−1AB. By a direct calculation, it follows that σAnhAnσ−1

An (Eij) = B−1P−1BEstB−1PB = ωj−iEij.

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Let ρAn = 1

2(n − 1, n − 2, . . . , −(n − 2), −(n − 1)) be the Weyl vector.

Define ηAn = exp(

1 n+1(2πiρAn(0)).

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Let ρAn = 1

2(n − 1, n − 2, . . . , −(n − 2), −(n − 1)) be the Weyl vector.

Define ηAn = exp(

1 n+1(2πiρAn(0)).

Then the action of ηAn on sln+1(C) is given by ηAn : A → DAD−1.

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Let ρAn = 1

2(n − 1, n − 2, . . . , −(n − 2), −(n − 1)) be the Weyl vector.

Define ηAn = exp(

1 n+1(2πiρAn(0)).

Then the action of ηAn on sln+1(C) is given by ηAn : A → DAD−1.

Lemma

We have σAnhAnσ−1

An = ηAn and σAnηAnσ−1 An = h−1 An on sln+1(C).

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A.

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

Set X = L(ˆ ρ) = {α ∈ L | α, ˆ ρ ∈ Z}.

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

Set X = L(ˆ ρ) = {α ∈ L | α, ˆ ρ ∈ Z}. Then L = SpanZX ∪ R.

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

Set X = L(ˆ ρ) = {α ∈ L | α, ˆ ρ ∈ Z}. Then L = SpanZX ∪ R. Set h = hAk1 ⊗ · · · ⊗ hAkj , η = ηAk1 ⊗ · · · ⊗ ηAkj , σ = σAk1 ⊗ · · · ⊗ σAkj .

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

Set X = L(ˆ ρ) = {α ∈ L | α, ˆ ρ ∈ Z}. Then L = SpanZX ∪ R. Set h = hAk1 ⊗ · · · ⊗ hAkj , η = ηAk1 ⊗ · · · ⊗ ηAkj , σ = σAk1 ⊗ · · · ⊗ σAkj . Since they are inner automorphisms, we can extend them to VL by using the same exponential expressions.

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Let R = Ak1 ⊕ · · · ⊕ Akj be an orthogonal sum of simple root lattices of type A. Let L be an even overlattice of R and ˆ ρ = j

i=1 1 (ki+1)ρAki .

Set X = L(ˆ ρ) = {α ∈ L | α, ˆ ρ ∈ Z}. Then L = SpanZX ∪ R. Set h = hAk1 ⊗ · · · ⊗ hAkj , η = ηAk1 ⊗ · · · ⊗ ηAkj , σ = σAk1 ⊗ · · · ⊗ σAkj . Since they are inner automorphisms, we can extend them to VL by using the same exponential expressions.

Theorem

We have σ(V h

X) = V h X and σ induces an automorphism of V h X.

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Next, we discuss several explicit examples (10 cases mentioned by H¨

hn).

Table: Standard lift of g ∈ O(Λ)

Class Type rank(Λg ) |φg | ρg O(Λg ) R(V ˆ

g Λg )

2A 1828 16 2 1/2 2.O+

8 (2)

210 2C 212 12 4 3/4 211.Sym12 21042 3B 1636 12 3 2/3 6.PSU4(3).22 38 4C 142244 10 4 3/4 [213].Sym6 2246 5B 1454 8 5 4/5 (Frob20 × O+

4 (5))/2

56 6E 12223262 8 6 5/6 D12.(O+

4 (2) × O+ 4 (3))

2636 6G 2363 6 12 11/12 [211.34] 24.42.35 7B 1373 6 7 6/7 7.3.2.L2(7).2 75 8E 12214182 6 8 7/8 [212.3] 2.4.84 10F 22102 4 20 19/20 5.2.[28] 22.42.54

φg denotes the standard lift of g in Aut (VΛ).

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Holy construction for the Leech lattice

Let N be a Niemeier lattice with the root lattice R = R1 ⊕ · · · ⊕ Rj, where Ri’s are A, D or E type .

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Holy construction for the Leech lattice

Let N be a Niemeier lattice with the root lattice R = R1 ⊕ · · · ⊕ Rj, where Ri’s are A, D or E type . Let ρi be a Weyl vector of Ri and set ρ = 1

h

j

i=1 ρi,

where h is the Coxeter number of Ri.

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Holy construction for the Leech lattice

Let N be a Niemeier lattice with the root lattice R = R1 ⊕ · · · ⊕ Rj, where Ri’s are A, D or E type . Let ρi be a Weyl vector of Ri and set ρ = 1

h

j

i=1 ρi,

where h is the Coxeter number of Ri. Define N(ρ) = {x ∈ N | x, ρ ∈ Z}, and let α ∈ ρ + N such that α, α ∈ 2Z.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 13 / 26

SLIDE 51

Holy construction for the Leech lattice

Let N be a Niemeier lattice with the root lattice R = R1 ⊕ · · · ⊕ Rj, where Ri’s are A, D or E type . Let ρi be a Weyl vector of Ri and set ρ = 1

h

j

i=1 ρi,

where h is the Coxeter number of Ri. Define N(ρ) = {x ∈ N | x, ρ ∈ Z}, and let α ∈ ρ + N such that α, α ∈ 2Z. Then the lattice ˜ Nρ = SpanZN(ρ) ∪ {α} is isomorphic to the Leech lattice [Conway-Sloane, Chapter 24]. In particular, the Leech lattice contains a sublattice isometric to R(ρ) = {x ∈ R | x, ρ ∈ Z}.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 13 / 26

SLIDE 52

We verify that all coinvariant lattices mentioned in Table above can be realized as a lattice of the form L(ˆ ρ). The result is summarized in Table 2.

Table: Coinvariant lattices as L(ˆ ρ)

Class Type rank(Λg ) Niemeier R Glue 2A 1828 8 A24

1

A8

1

(18) 2C 212 12 A24

1

A12

1

(112) 3B 1636 12 A12

2

A6

2

(13, −13) 4C 142244 14 A8

3

A4

3A2 1

(111 − 1|11) 5B 1454 16 A6

4

A4

4

(1243) 6E 12223262 16 A4

5D4

A2

5A2 2A2 1

(11|11|11) 6G 2363 18 A4

5D4

A3

5A3 1

(551|111) 7B 1373 18 A4

6

A3

6

(124) 8E 12214182 18 A2

7D2 5

A2

7A3A1

(13|1|1) 10F 22102 20 A2

9D6

A2

9A2 1

(32|11)

Note that A5 > A2

2, D4 > A4 1, D5 > A3A2 1 and D6 > A6 1 as sublattices.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 14 / 26

SLIDE 53

Theorem

Let g ∈ O(Λ). Suppose CO(Λg)(g)/g acts faithfully on Λ∗

g/Λg. Then

the natural homomorphism µ2 : Aut (V ˆ

g Λg ) → O(R(V ˆ g Λg ), q)

is injective, i.e., ker µ2 = id.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 15 / 26

SLIDE 54

Automorphism groups

Class Type rank(Λg ) |φg | ρg O(Λg ) R(V ˆ

g Λg )

Aut (V ˆ

g Λg )

2A 1828 16 2 1/2 2.O+

8 (2)

210 O+

10(2)

2C 212 12 4 3/4 211.Sym12 21042 212.210.Sym12.Sym3 3B 1636 12 3 2/3 6.PSU4(3).22 38 Ω−

8 (3).2

4C 142244 10 4 3/4 [213].Sym6 2246 index 2 5B 1454 8 5 4/5 (Frob20 × O+

4 (5))/2

56 Ω+

6 (5).2

6E 12223262 8 6 5/6 D12.(O+

4 (2) × O+ 4 (3))

2636 index 2 6G 2363 6 12 11/12 [211.34] 24.42.35 7B 1373 6 7 6/7 7.3.2.L2(7).2 75 Ω5(7).2 8E 12214182 6 8 7/8 [212.3] 2.4.84 index 2 10F 22102 4 20 19/20 5.2.[28] 22.42.54 C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 16 / 26

SLIDE 55

2A element in O(Λ)

Assume that g belongs to the conjugacy class 2A of O(Λ);

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 17 / 26

SLIDE 56

2A element in O(Λ)

Assume that g belongs to the conjugacy class 2A of O(Λ); the coinvariant lattice Λg is isometric to √ 2E8 and Aut (V +

√ 2E8) ∼

= O+(10, 2).

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 17 / 26

SLIDE 57

2A element in O(Λ)

Assume that g belongs to the conjugacy class 2A of O(Λ); the coinvariant lattice Λg is isometric to √ 2E8 and Aut (V +

√ 2E8) ∼

= O+(10, 2). Let L be an even lattice of rank 16 such that D(L) ∼ = Irr(VL) ∼ = Z10

2 and

α|α ∈ Z for α ∈ L∗.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 17 / 26

SLIDE 58

2A element in O(Λ)

Assume that g belongs to the conjugacy class 2A of O(Λ); the coinvariant lattice Λg is isometric to √ 2E8 and Aut (V +

√ 2E8) ∼

= O+(10, 2). Let L be an even lattice of rank 16 such that D(L) ∼ = Irr(VL) ∼ = Z10

2 and

α|α ∈ Z for α ∈ L∗. Set N = √ 2L∗. Then D(N) ∼ = Z6

2. and N is a level 2 lattice. Such lattices

has been classified.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 17 / 26

SLIDE 59

2A element in O(Λ)

Assume that g belongs to the conjugacy class 2A of O(Λ); the coinvariant lattice Λg is isometric to √ 2E8 and Aut (V +

√ 2E8) ∼

= O+(10, 2). Let L be an even lattice of rank 16 such that D(L) ∼ = Irr(VL) ∼ = Z10

2 and

α|α ∈ Z for α ∈ L∗. Set N = √ 2L∗. Then D(N) ∼ = Z6

2. and N is a level 2 lattice. Such lattices

has been classified.

Proposition ([SV01, Theorem 2])

Up to isometry, there exist exactly 17 level 2 lattices of rank 16 with determinant 26. Moreover, they are uniquely determined by their root systems. The root systems and isometry groups of the lattices in the proposition above are summarized in Table below.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 17 / 26

SLIDE 60

Level 2 lattices N of rank 16 with D(N) ∼ = Z6

2

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A16

1

25 AGL4(2) W (A1) ≀ AGL4(2) A4

3(

√ 2A1)4 2341 W (D4) (W (A3)4 × W (A1)4).W (D4) D2

4 C4 2

23 2 × Sym4 (W (D4)2 × W (C2)4).(2 × Sym4) A2

5(

√ 2A2)2C2 3161 Dih8 (W (A5)2 × W (A2)2 × W (C2)).Dih8 A7( √ 2A3)C2

3

2141 Z2

2

(W (A7) × W (A3) × W (C3)2).Z2

2

D2

5 (

√ 2A3)2 42 Dih8 (W (D5)2 × W (A3)2).Dih8 C4

4

21 Sym4 W (C4) ≀ Sym4 D6C4( √ 2B3)2 22 Z2 (W (D6) × W (C4) × W (B3)2).Z2 A9( √ 2A4)( √ 2B3) 101 Z2 (W (A9) × W (A4) × W (B3)).Z2 E6( √ 2A5)C5 61 Z2 (W (E6) × W (A5) × W (C5)).Z2 C2

6 (

√ 2B4) 21 Z2 W (C6) ≀ 2 × W (B4) D8( √ 2B4)2 22 Z2 W (D8) × W (B4) ≀ Z2 D9( √ 2A7) 81 Z2 (W (D9) × W (A7)).Z2 C8F 2

4

1 Z2 W (C8) × W (F4) ≀ Z2 E7( √ 2B5)F4 21 1 W (E7) × W (B5) × W (F4) C10( √ 2B6) 21 1 W (C10) × W (B6) E8( √ 2B8) 21 1 W (B8) × W (E8) C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 18 / 26

SLIDE 61

Level 2 lattices N of rank 16 with D(N) ∼ = Z6

2

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A16

1

25 AGL4(2) W (A1) ≀ AGL4(2) A4

3(

√ 2A1)4 2341 W (D4) (W (A3)4 × W (A1)4).W (D4) D2

4 C4 2

23 2 × Sym4 (W (D4)2 × W (C2)4).(2 × Sym4) A2

5(

√ 2A2)2C2 3161 Dih8 (W (A5)2 × W (A2)2 × W (C2)).Dih8 A7( √ 2A3)C2

3

2141 Z2

2

(W (A7) × W (A3) × W (C3)2).Z2

2

D2

5 (

√ 2A3)2 42 Dih8 (W (D5)2 × W (A3)2).Dih8 C4

4

21 Sym4 W (C4) ≀ Sym4 D6C4( √ 2B3)2 22 Z2 (W (D6) × W (C4) × W (B3)2).Z2 A9( √ 2A4)( √ 2B3) 101 Z2 (W (A9) × W (A4) × W (B3)).Z2 E6( √ 2A5)C5 61 Z2 (W (E6) × W (A5) × W (C5)).Z2 C2

6 (

√ 2B4) 21 Z2 W (C6) ≀ 2 × W (B4) D8( √ 2B4)2 22 Z2 W (D8) × W (B4) ≀ Z2 D9( √ 2A7) 81 Z2 (W (D9) × W (A7)).Z2 C8F 2

4

1 Z2 W (C8) × W (F4) ≀ Z2 E7( √ 2B5)F4 21 1 W (E7) × W (B5) × W (F4) C10( √ 2B6) 21 1 W (C10) × W (B6) E8( √ 2B8) 21 1 W (B8) × W (E8)

Note: The group AGL4(2) can be regarded as a subgroup of Sym16 via the action on the first order Reed-Muller code RM(1, 4) of length 16, which is the glue code of the lattice with respect to A16

1 .

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 18 / 26

SLIDE 62

K(V ) and Out(V ) for the case g ∈ 2A

No. in [Sc93]

R( √ 2L∗) V1 dim V1 K(V ) Out(V ) 5 A16

1

A16

1,2

48 Z5

2

AGL4(2) 16 A4

3(

√ 2A1)4 A4

3,2A4 1,1

72 Z3

2 × Z4

W (D4) 22 A2

5(

√ 2A2)2C2 A2

5,2C2,1A2 2,1

96 Z3 × Z6 Dih8 25 D2

4 C2 2

D2

4,2C4 2,1

96 Z3

2

Z2 × Sym4 31 D2

5 (

√ 2A3)2 D2

5,2A2 3,1

120 Z2

4

Dih8 33 A7( √ 2A3)C2

3

A7,2C2

3,1A3,1

120 Z2 × Z4 Z2

2

38 C4

4

C4

4,1

144 Z2 Sym4 39 D6C4( √ 2B3)2 D6,2C4,1B2

3,1

144 Z2

2

Z2 40 A9( √ 2A4)( √ 2B3) A9,2A4,1B3,1 144 Z10 Z2 44 E6A( √ 2A5)C5 E6,2C5,1A5,1 168 Z6 Z2 47 D8( √ 2B4)2 D8,2B2

4,1

192 Z2

2

Z2 48 C2

6 (

√ 2B4) C2

6,1B4,1

192 Z2 Z2 50 D9( √ 2A7) D9,2A7,1 216 Z8 Z2 52 C8F 2

4

C8,1F 2

4,1

240 1 Z2 53 D7( √ 2B5)F4 E7,2B5,1F4,1 240 Z2 1 56 C10( √ 2B6) C10,1B6,1 288 Z2 1 62 ( √ 2B8)E8 B8,1E8,2 384 Z2 1 C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 19 / 26

SLIDE 63

3B element in O(Λ)

Assume that g belongs to the conjugacy class 3B of O(Λ); its cycle shape is 1636.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 20 / 26

SLIDE 64

3B element in O(Λ)

Assume that g belongs to the conjugacy class 3B of O(Λ); its cycle shape is 1636. The coinvariant lattice Λg is isometric to the Coxeter-Todd lattice K12 of rank 12.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 20 / 26

SLIDE 65

3B element in O(Λ)

Assume that g belongs to the conjugacy class 3B of O(Λ); its cycle shape is 1636. The coinvariant lattice Λg is isometric to the Coxeter-Todd lattice K12 of rank 12. Then Irr(V 2) ∼ = Z8

3 and Aut (V 2) ∼

= Ω−(8, 3):2, which is an index 2 subgroup of the full orthogonal group O(Irr(V 2), q2) = GO−(8, 3) ∼ = 2 × Ω−(8, 3):2.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 20 / 26

SLIDE 66

3B element in O(Λ)

Assume that g belongs to the conjugacy class 3B of O(Λ); its cycle shape is 1636. The coinvariant lattice Λg is isometric to the Coxeter-Todd lattice K12 of rank 12. Then Irr(V 2) ∼ = Z8

3 and Aut (V 2) ∼

= Ω−(8, 3):2, which is an index 2 subgroup of the full orthogonal group O(Irr(V 2), q2) = GO−(8, 3) ∼ = 2 × Ω−(8, 3):2. Hence µ2 : Aut (V 2) → O(Irr(V 2), q2) is injective.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 20 / 26

SLIDE 67

Let L be an even lattice of rank 12 such that D(L) ∼ = Irr(VL) ∼ = Z8

3

and α|α ∈ (2/3)Z for all α ∈ L∗.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 21 / 26

SLIDE 68

Let L be an even lattice of rank 12 such that D(L) ∼ = Irr(VL) ∼ = Z8

3

and α|α ∈ (2/3)Z for all α ∈ L∗. Set N = √ 3L∗. Then D(N) ∼ = Z4

3 and N is a level 3 lattice.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 21 / 26

SLIDE 69

Let L be an even lattice of rank 12 such that D(L) ∼ = Irr(VL) ∼ = Z8

3

and α|α ∈ (2/3)Z for all α ∈ L∗. Set N = √ 3L∗. Then D(N) ∼ = Z4

3 and N is a level 3 lattice.

Such lattices are also classified.

Proposition ([SV01, Theorem 3])

Up to isometry, there exist exactly 6 level 3 lattices of rank 12 with determinant 34. Moreover, they are uniquely determined by their root systems.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 21 / 26

SLIDE 70

Table: Level 3 lattices N of rank 12 with D(N) ∼ = Z4

3

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A6

2

31 Z2 × Sym6 (W (A2) ≀ Sym6).Z2 A5D4( √ 3A1)3 23 Dih12 (W (A5) × W (D4) × W (A1)3).Dih12 A8( √ 3A2)2 32 Z2

2

(W (A8) × W (A2)2).Z2

2

D7( √ 3A3)G2 41 Z2 (W (D7) × W (A3) × W (G2)).Z2 E6G3

2

1 Z2 × Sym3 (W (E6) × W (G2) ≀ Sym3).Z2 E7( √ 3A5) 61 Z2 (W (E7) × W (A5)).Z2

Table: K(V ) and Out(V ) for the case g ∈ 3B

No. in [Sc93]

R(N) V1 dim V1 K(V ) Out(V ) 6 A6

2

A6

2,3

48 Z3 Sym6 17 A5D4( √ 3A1)3 A5,3D4,3A3

1,1

72 Z3

2

Sym3 27 A8( √ 3A2)2 A8,3A2

2,1

96 Z2

3

Z2 32 E6G3

2

E6,3G2,13 120 1 Sym3 34 D7( √ 3A3)G2 D7,3A3,1G2,1 120 Z4 1 45 E7( √ 3A5) E7,3A5,1 168 Z6 1 C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 22 / 26

SLIDE 71

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 72

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454. The coinvariant sublattice Λg has rank 16 and the discriminant group Z4

5.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 73

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454. The coinvariant sublattice Λg has rank 16 and the discriminant group Z4

5.

In this case, Irr(V 2) ∼ = Z6

5 and Aut (V 2) is an index 2 subgroup of

GO+

6 (5).

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 74

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454. The coinvariant sublattice Λg has rank 16 and the discriminant group Z4

5.

In this case, Irr(V 2) ∼ = Z6

5 and Aut (V 2) is an index 2 subgroup of

GO+

6 (5).

Let L be an even lattice of rank 8 such that D(L) ∼ = Irr(VL) ∼ = Z6

5 and

α|α ∈ (2/5)Z for all α ∈ L∗.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 75

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454. The coinvariant sublattice Λg has rank 16 and the discriminant group Z4

5.

In this case, Irr(V 2) ∼ = Z6

5 and Aut (V 2) is an index 2 subgroup of

GO+

6 (5).

Let L be an even lattice of rank 8 such that D(L) ∼ = Irr(VL) ∼ = Z6

5 and

α|α ∈ (2/5)Z for all α ∈ L∗. Then N = √ 5L∗ is even and D(N) ∼ = Z2

5.

There are two such lattices and their root system are A2

4 and D6(

√ 5A2

1).

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 76

5B element in O(Λ)

Assume that g belongs to the conjugacy class 5B of O(Λ); note that its cycle shape is 1454. The coinvariant sublattice Λg has rank 16 and the discriminant group Z4

5.

In this case, Irr(V 2) ∼ = Z6

5 and Aut (V 2) is an index 2 subgroup of

GO+

6 (5).

Let L be an even lattice of rank 8 such that D(L) ∼ = Irr(VL) ∼ = Z6

5 and

α|α ∈ (2/5)Z for all α ∈ L∗. Then N = √ 5L∗ is even and D(N) ∼ = Z2

5.

There are two such lattices and their root system are A2

4 and D6(

√ 5A2

1).

Table: Level 5 lattices N of rank 8 with D(N) ∼ = Z2

5

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A2

4

1 Dih8 (2 × W (A4)) ≀ Sym2 D6( √ 5A2

1)

22 Sym2 (W (D6) × W (A1)2).2

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 23 / 26

SLIDE 77

Table: K(V ) and Out(V ) for the case g ∈ 5B

No. in [Sc93]

R(N) V1 dim V1 K(V ) Out(V ) 9 A5

4

A2

4,5

48 1 Z2

2

20 D6( √ 5A2

1)

D6,5A2

1,1

72 Z2

2

1

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 24 / 26

SLIDE 78

7B element in O(Λ)

Assume that g belongs to the conjugacy class 7B of O(Λ), which has the cycle shape 1373.

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 25 / 26

SLIDE 79

7B element in O(Λ)

Assume that g belongs to the conjugacy class 7B of O(Λ), which has the cycle shape 1373. The coinvariant lattice Λg has rank 18 and Irr(V 2) ∼ = Z5

7.

Moreover, Aut (V 2) is isomorphic to an index 2 subgroup of O(Irr(V 2), q2) ∼ = GO5(7).

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 25 / 26

SLIDE 80

7B element in O(Λ)

Assume that g belongs to the conjugacy class 7B of O(Λ), which has the cycle shape 1373. The coinvariant lattice Λg has rank 18 and Irr(V 2) ∼ = Z5

7.

Moreover, Aut (V 2) is isomorphic to an index 2 subgroup of O(Irr(V 2), q2) ∼ = GO5(7).

Table: Level 7 lattices N of rank 6 with D(N) ∼ = Z7

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A6 1 Z2 Z2 × W (A6)

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 25 / 26

SLIDE 81

7B element in O(Λ)

Assume that g belongs to the conjugacy class 7B of O(Λ), which has the cycle shape 1373. The coinvariant lattice Λg has rank 18 and Irr(V 2) ∼ = Z5

7.

Moreover, Aut (V 2) is isomorphic to an index 2 subgroup of O(Irr(V 2), q2) ∼ = GO5(7).

Table: Level 7 lattices N of rank 6 with D(N) ∼ = Z7

Root system R(N) N/Q O(N)/W (R(N)) Isometry group O(N) A6 1 Z2 Z2 × W (A6) The group structures of K(V ) and Out(V ) are as follows.

Table: K(V ) and Out(V ) for the case g ∈ 7B

No. in [Sc93]

R(N) V1 dim V1 K(V ) Out(V ) 11 A6 A6,7 48 1 1

C.H. Lam (A.S.) Orbifold VOAs June 25, 2019 25 / 26

SLIDE 82

Thank You

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