Associated Lie algebra of a positive definite even lattice

Let $𝐿$ be a positive definite even rational lattice. Let $𝔥 = 𝐿 𝕂$ for $c h a r 𝕂 = 0$ , and set

𝑐 0 : 𝐿 \times 𝐿 \to ℤ 2 (𝛼, 𝛽) \mapsto ⟨ 𝛼, 𝛽 ⟩ + 2 ℤ

which is alternating $ℤ$ -bilinear. Then $𝑐 0$ determines a unique 2-central extension of a free abelian group $𝐿$

1 \to ℤ 2 𝜅 ↪ ˆ 𝐿 𝜋 ↠ 𝐿 \to 1

such that $[𝑎, 𝑏] = 𝜅 𝑐 0 (―― 𝑎, ―― 𝑏)$ ¹. Let $Δ = 𝐿 2$ , $ˆ Δ = 𝜋 - 1 Δ$ , and define

𝔤 = 𝔥 \oplus 𝕂 ⟨ 𝑥 𝑎 ⟩ 𝑎 \in ˆ Δ ⟨ 𝑥 𝑎 + 𝑥 𝑎 𝜅 : 𝑎 \in ˆ Δ ⟩

where $𝕂 ⟨ 𝑥 𝑎 ⟩ 𝑎 \in ˆ Δ$ is a free module, whence follows $d i m 𝔤 = r a n k 𝐿 + | Δ |$ . Then $𝔤$ is a quadratic Lie algebra under the bilinear bracket defined by

[𝔥, 𝔥] = 0 [ℎ, 𝑥 𝑎] = - [𝑥 𝑎, ℎ] = ⟨ ℎ, ―― 𝑎 ⟩ 𝑥 𝑎 [𝑥 𝑎, 𝑥 𝑏] = ⎧ { {⎨ { {⎩ ―― 𝑎 𝑎 𝑏 = 1 𝑥 𝑎 𝑏 𝑎 𝑏 \in ˆ Δ 0 𝑎 𝑏 \notin ˆ Δ \cup {1, 𝜅}

and the nonsingular bilinear form extending that of $𝔥$ by

⟨ 𝔥, 𝑥 𝑎 ⟩ = 0 ⟨ 𝑥 𝑎, 𝑥 𝑏 ⟩ = {1 𝑎 𝑏 = 1 0 𝑎 𝑏 \notin {1, 𝜅}

for $𝑎, 𝑏 \in ˆ Δ$ and $ℎ \in 𝔥$ .² #m/def/lie

With a nice choice of section

Let $𝜀 0 : 𝐿 \times 𝐿 \to ℤ 2$ be a bilinear map such that (cf. associated elementary 2-group of an even lattice)

𝜀 0 (𝛼, 𝛼) = 12 ⟨ 𝛼, 𝛼 ⟩ + 2 ℤ

then

𝜀 0 (𝛼, 𝛽) - 𝜀 0 (𝛽, 𝛼) = ⟨ 𝛼, 𝛽 ⟩ + 2 ℤ 𝜀 0 (Δ, Δ) = 1 + 2 ℤ

and by the Correspondence between 2-cocycles and central extensions we have a central extension of the above form with a $𝖲 𝖾 𝗍$ -section $𝑠 (-) : 𝐿 ↪ ˆ 𝐿$ such that

𝑠 𝛼 𝑠 𝛽 = 𝑠 𝛼 + 𝛽 𝜅 𝜀 0 (𝛼, 𝛽)

and in particular $𝑠 0 = 1$ . Denote

𝜀 (𝛼, 𝛽) = (- 1) 𝜀 0 (𝛼, 𝛽)

We then have

ˆ Δ = {𝑠 𝛼, 𝑠 𝛼 𝜅 : 𝛼 \in Δ}

and are free to define $𝑥 𝛼 = 𝑥 𝑠 𝛼$ for $𝛼 \in Δ$ , and the commutation relations become

[𝔥, 𝔥] = 0 [ℎ, 𝑥 𝛼] = - [𝑥 𝛼, ℎ] = ⟨ ℎ, 𝛼 ⟩ 𝑥 𝛼 [𝑥 𝛼, 𝑥 𝛽] = ⎧ { {⎨ { {⎩ 𝜀 (𝛼, - 𝛼) 𝛼 ⟨ 𝛼, 𝛽 ⟩ = - 2 𝜀 (𝛼, 𝛽) 𝑥 𝛼 + 𝛽 ⟨ 𝛼, 𝛽 ⟩ = - 1 0 ⟨ 𝛼, 𝛽 ⟩ \geq 0

for $𝑎, 𝛽 \in Δ$ and the nonsingular bilinear form is given by

⟨ 𝔥, 𝑥 𝑎 ⟩ = ⟨ 𝑥 𝑎, 𝔥 ⟩ = 0 ⟨ 𝑥 𝛼, 𝑥 𝛽 ⟩ = {𝜀 (𝛼, - 𝛼) 𝛼 + 𝛽 = 0 0 𝛼 + 𝛽 \neq 0

Proof of quadratic Lie algebra

It is clear that the bracket is alternating on $𝔥$ . For some $𝑎 \in ˆ Δ$ , we have $――― 𝑎 𝑎 \in 𝐿 4$ so $[𝑥 𝑎, 𝑥 𝑎] = 0$ . Thus the bracket is alternating.

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

𝐽 = [𝑦 1, [𝑦 2, 𝑦 3]] + [𝑦 2, [𝑦 3, 𝑦 1]] + [𝑦 3, [𝑦 1, 𝑦 2]] = 0

for $𝑦 1, 𝑦 2, 𝑦 3 \in 𝔥 \cup {𝑥 𝑎 : 𝑎 \in ˆ Δ}$ . Clearly if $𝑦 1, 𝑦 2, 𝑦 3 \in 𝔥$ the identity holds. If $𝑦 1, 𝑦 2 \in 𝔥$ and $𝑦 3 = 𝑥 𝑎$

𝐽 = = [𝑦 1, ⟨ 𝑦 2, ―― 𝑎 ⟩ 𝑥 𝑎] + [𝑦 2, - ⟨ 𝑦 1, ―― 𝑎 ⟩ 𝑥 𝑎] = ⟨ 𝑦 2, ―― 𝑎 ⟩ ⟨ 𝑦 1, ―― 𝑎 ⟩ 𝑥 𝑎 - ⟨ 𝑦 1, ―― 𝑎 ⟩ ⟨ 𝑦 2, ―― 𝑎 ⟩ 𝑥 𝑎 = 0

If $𝑦 1 \in 𝔥$ , $𝑦 2 = 𝑥 𝑎$ , and $𝑦 3 = 𝑥 𝑏$

𝐽 = [𝑦 1, [𝑥 𝑎, 𝑥 𝑏]] - ⟨ 𝑦 1, ―― 𝑏 ⟩ [𝑥 𝑎, 𝑥 𝑏] - ⟨ 𝑦 1, ―― 𝑎 ⟩ [𝑥 𝑎, 𝑥 𝑏]

where in case $𝑎 𝑏 \in ˆ Δ$ ,

𝐽 = (⟨ 𝑦 1, ―― 𝑎 𝑏 ⟩ - ⟨ 𝑦 1, ―― 𝑏 ⟩ - ⟨ 𝑦 1, ―― 𝑎 ⟩) [𝑥 𝑎, 𝑥 𝑏] = 0

in case $𝑎 𝑏 = 1$ , $―― 𝑎 + ―― 𝑏 = 0$ and thus

𝐽 = [𝑦 1, ―― 𝑎] - ⟨ 𝑦 1, ―― 𝑏 ⟩ ―― 𝑎 - ⟨ 𝑦 1, ―― 𝑎 ⟩ ―― 𝑎 = 0

anf case $𝑎 𝑏 \notin ˆ Δ \cup {1, 𝜅}$ , we have $[𝑥 𝑎, 𝑥 𝑏] = 0$ and thus $𝐽 = 0$ . Finally consider the case

𝑦 1 = 𝑥 𝑎 𝑦 2 = 𝑥 𝑏 𝑦 3 = 𝑥 𝑐

where in case $――― 𝑎 𝑏 𝑐 \notin Δ \cup {0}$ ,

[𝑥 𝑎, [𝑥 𝑏, 𝑥 𝑐]]

#state/develop | #lang/en | #SemBr

where we denote $𝜋 𝑥 = ―― 𝑥$ . ↩
1988. Vertex operator algebras and the Monster, §6.2, 126 ↩

Associated Lie algebra of a positive definite even lattice

Footnotes