Lie algebras MOC

Associated Lie algebra of a positive definite even lattice

Let 𝐿 be a positive definite even rational lattice. Let 𝔥 =𝐿𝕂 for [[Characteristic|char⁡𝕂 =0]], and set

𝑐0:𝐿×𝐿→ℤ2(𝛼,𝛽)↦⟨𝛼,𝛽⟩+2ℤ

which is alternating ℤ-bilinear. Then 𝑐0 determines a unique 2 central extension of a free abelian group 𝐿

1→ℤ2𝜅↪ˆ𝐿𝜋↠𝐿→1

such that [𝑎,𝑏] =𝜅𝑐0(――𝑎,――𝑏)¹. Let Δ =𝐿2, ˆΔ =𝜋−1Δ, and define

𝔤=𝔥⊕𝕂⟨𝑥𝑎⟩𝑎∈ˆΔ⟨𝑥𝑎+𝑥𝑎𝜅:𝑎∈ˆΔ⟩

where 𝕂⟨𝑥𝑎⟩𝑎∈ˆΔ is a free module, whence follows dim⁡𝔤 =rank⁡𝐿 +|Δ|. Then 𝔤 is a quadratic Lie algebra under the bilinear bracket defined by

[𝔥,𝔥]=0[ℎ,𝑥𝑎]=−[𝑥𝑎,ℎ]=⟨ℎ,――𝑎⟩𝑥𝑎[𝑥𝑎,𝑥𝑏]=⎧{ {⎨{ {⎩――𝑎𝑎𝑏=1𝑥𝑎𝑏𝑎𝑏∈ˆΔ0𝑎𝑏∉ˆΔ∪{1,𝜅}

and the nonsingular bilinear form extending that of 𝔥 by

⟨𝔥,𝑥𝑎⟩=0⟨𝑥𝑎,𝑥𝑏⟩={1𝑎𝑏=10𝑎𝑏∉{1,𝜅}

for 𝑎,𝑏 ∈ˆΔ and ℎ ∈𝔥.² lie

With a nice choice of section

Let 𝜀0 :𝐿 ×𝐿 →ℤ2 be a bilinear map such that (cf. associated elementary 2-group of an even lattice)

𝜀0(𝛼,𝛼)=12⟨𝛼,𝛼⟩+2ℤ

then

𝜀0(𝛼,𝛽)−𝜀0(𝛽,𝛼)=⟨𝛼,𝛽⟩+2ℤ𝜀0(Δ,Δ)=1+2ℤ

and by the Correspondence between 2-cocycles and central extensions we have a central extension of the above form with a \Set-section 𝑠(−) :𝐿 ↪ˆ𝐿 such that

𝑠𝛼𝑠𝛽=𝑠𝛼+𝛽𝜅𝜀0(𝛼,𝛽)

and in particular 𝑠0 =1. Denote

𝜀(𝛼,𝛽)=(−1)𝜀0(𝛼,𝛽)

We then have

ˆΔ={𝑠𝛼,𝑠𝛼𝜅:𝛼∈Δ}

and are free to define 𝑥𝛼 =𝑥𝑠𝛼 for 𝛼 ∈Δ, and the commutation relations become

[𝔥,𝔥]=0[ℎ,𝑥𝛼]=−[𝑥𝛼,ℎ]=⟨ℎ,𝛼⟩𝑥𝛼[𝑥𝛼,𝑥𝛽]=⎧{ {⎨{ {⎩𝜀(𝛼,−𝛼)𝛼⟨𝛼,𝛽⟩=−2𝜀(𝛼,𝛽)𝑥𝛼+𝛽⟨𝛼,𝛽⟩=−10⟨𝛼,𝛽⟩≥0

for 𝑎,𝛽 ∈Δ and the nonsingular bilinear form is given by

⟨𝔥,𝑥𝑎⟩=⟨𝑥𝑎,𝔥⟩=0⟨𝑥𝛼,𝑥𝛽⟩={𝜀(𝛼,−𝛼)𝛼+𝛽=00𝛼+𝛽≠0

Proof of quadratic Lie algebra

It is clear that the bracket is alternating on 𝔥. For some 𝑎 ∈ˆΔ, we have ―――𝑎𝑎 ∈𝐿4 so [𝑥𝑎,𝑥𝑎] =0. Thus the bracket is alternating.

To prove that 𝔤 is a Lie algebra, it is sufficient to prove the ^Jacobi

𝐽=[𝑦1,[𝑦2,𝑦3]]+[𝑦2,[𝑦3,𝑦1]]+[𝑦3,[𝑦1,𝑦2]]=0

for 𝑦1,𝑦2,𝑦3 ∈𝔥 ∪{𝑥𝑎 :𝑎 ∈ˆΔ}. Clearly if 𝑦1,𝑦2,𝑦3 ∈𝔥 the identity holds. If 𝑦1,𝑦2 ∈𝔥 and 𝑦3 =𝑥𝑎

𝐽==[𝑦1,⟨𝑦2,――𝑎⟩𝑥𝑎]+[𝑦2,−⟨𝑦1,――𝑎⟩𝑥𝑎]=⟨𝑦2,――𝑎⟩⟨𝑦1,――𝑎⟩𝑥𝑎−⟨𝑦1,――𝑎⟩⟨𝑦2,――𝑎⟩𝑥𝑎=0

If 𝑦1 ∈𝔥, 𝑦2 =𝑥𝑎, and 𝑦3 =𝑥𝑏

𝐽=[𝑦1,[𝑥𝑎,𝑥𝑏]]−⟨𝑦1,――𝑏⟩[𝑥𝑎,𝑥𝑏]−⟨𝑦1,――𝑎⟩[𝑥𝑎,𝑥𝑏]

where in case 𝑎𝑏 ∈ˆΔ,

𝐽=(⟨𝑦1,――𝑎𝑏⟩−⟨𝑦1,――𝑏⟩−⟨𝑦1,――𝑎⟩)[𝑥𝑎,𝑥𝑏]=0

in case 𝑎𝑏 =1, ――𝑎 +――𝑏 =0 and thus

𝐽=[𝑦1,――𝑎]−⟨𝑦1,――𝑏⟩――𝑎−⟨𝑦1,――𝑎⟩――𝑎=0

anf case 𝑎𝑏 ∉ˆΔ ∪{1,𝜅}, we have [𝑥𝑎,𝑥𝑏] =0 and thus 𝐽 =0. Finally consider the case

𝑦1=𝑥𝑎𝑦2=𝑥𝑏𝑦3=𝑥𝑐

where in case ―――𝑎𝑏𝑐 ∉Δ ∪{0},

[𝑥𝑎,[𝑥𝑏,𝑥𝑐]]

---

develop | en | SemBr

Footnotes

1. where we denote 𝜋𝑥 =――𝑥. ↩

2. 1988. Vertex operator algebras and the Monster, §6.2, 126 ↩