Natural Heisenerg algebras

Let $𝔥$ be a nondegenerate finite-dimensional quadratic space over $𝕂$ ¹ endowed with an abelian bracket, and $𝔤 ℤ = ˜ 𝔥$ and $𝔤 ℤ + 12 = ˜ 𝔥 [- 1]$ denote untwisted and twisted affine Lie algebras respectively, which are each triangular. Then the commutator ideals²

ˆ 𝔥 ℤ = 𝔤' ℤ = ˜ 𝔥' = 𝖵 𝖾 𝖼 𝗍 𝕂 𝕂 𝑐 \oplus ⨁ 𝑛 \in ℤ ∖ {0} 𝔥 \otimes 𝑡 𝑛 ˆ 𝔥 ℤ + 12 = 𝔤' ℤ + 12 = ˜ 𝔥 [- 1]' = 𝖵 𝖾 𝖼 𝗍 𝕂 𝕂 𝑐 \oplus ⨁ 𝑛 \in ℤ + 12 𝔥 \otimes 𝑡 𝑛 = ˆ 𝔥 [- 1]

define Heisenberg algebras, called the $ℤ$ - and $(ℤ + 12)$ -natural Heisenberg algebras³ induced by $𝔥$ . #m/def/lie For $𝑍 = ℤ$ or $𝑍 = ℤ + 12$ , we have the commutation relations

[𝑐, ˆ 𝔥 𝑍] = 0 [𝑥 \otimes 𝑡 𝑚, 𝑦 \otimes 𝑡 𝑛] = ⟨ 𝑥, 𝑦 ⟩ 𝑚 𝛿 𝑚 + 𝑛 𝑐

for $𝑥, 𝑦 \in 𝔥$ and $𝑚, 𝑛 \in 𝑍 ∖ {0}$ .⁴

Modules

Heisenberg modules

Let $𝑍 = ℤ$ or $𝑍 = ℤ + 12$ and $𝑘 \in 𝕂$ . The $ˆ 𝔥 𝑍$ -Heisenberg module is then isomorphic as a $𝕂$ -graded vector space

𝑀 (𝑘) ≅ 𝖦 𝗋 𝕂 𝖵 𝖾 𝖼 𝗍 𝕂 𝑆 ∙ (ˆ 𝔥 𝑍)

setting $𝑘 = 1$ and denoting $(ℎ \otimes 𝑡 𝑛) ⊙ 𝑣 = ℎ (𝑛) 𝑣$ we have

𝑐 ⊙ 𝑓 = 𝑘 𝑓 ℎ (- 𝑛) ⊙ 𝑓 = ℎ (- 𝑛) 𝑓 ℎ (𝑛) ⊙ 𝑓 = 𝑘 𝑛 𝜕 𝑓 𝜕 ℎ (- 𝑛)

for $ℎ \in 𝔥$ and $𝑛 \in ℕ$ . In addition $𝑆 ∙ (ˆ 𝔥 𝑍)$ is a $𝕂$ -graded irreducible $𝔤 𝑍$ module where $𝑑$ acts as a degree operator, and if $𝑍 = ℤ$ , $𝔥 = 𝔥 \otimes 𝑡 0$ act trivially.⁵

Triangular modules

Defining the linear form

𝜆 : 𝔤 0 𝑍 \to 𝕂 𝑐 \mapsto 1 𝑑 \mapsto 0 𝔥 \mapsto 0 (𝑍 = ℤ)

Then then the triangular module and Heisenberg module $𝑀 (𝜆) = 𝑀 (1)$ as $𝔤 𝑍$ -modules. Thus given the involutive antiautomorphism

𝜔 : 𝑐 \mapsto 𝑐 𝑑 \mapsto 𝑑 ℎ \otimes 𝑡 𝑛 \mapsto ℎ \otimes 𝑡 - 𝑛 (ℎ \in 𝔥)

there exists a unique contravariant form

𝑏 : 𝑆 ∙ (ˆ 𝔥 - 𝑍) \times 𝑆 ∙ (ˆ 𝔥 - 𝑍) \to 𝕂

with the properties

(𝑑 ⊙ 𝑣, 𝑤) = (𝑣, 𝑑 ⊙ 𝑤) ((ℎ \otimes 𝑡 𝑛) ⊙ 𝑣, 𝑤) = (𝑣, (ℎ \otimes 𝑡 - 𝑛) ⊙ 𝑤) (1, 1) = 1

for $ℎ \in 𝔥$ , $𝑛 \in 𝑍$ , and $𝑣, 𝑤 \in 𝑆 ∙ (ˆ 𝔥 - 𝑍)$ . The first conditions implies that $𝑏 (𝑣, 𝑤) = 0$ if $𝑣, 𝑤$ are homogenous of different degrees.⁶

Natural Heisenberg module $𝑀 𝛼$

Let $𝑍 = ℤ$ or $𝑍 = ℤ + 12$ and let $𝑆 ∙ (ˆ 𝔥 - 𝑍)$ denote the $𝔤 𝑍$ - and $ˆ 𝔥 𝑍$ -module $𝑀 (1)$ . Let $𝛼 \in 𝔥$ , $𝜇 \in 𝕂$ , and $𝕂 𝑣 𝛼$ be a 1-dimensional $𝔤 𝑍$ -module defined by

ℎ ⊙ 𝑣 𝛼 = ⟨ ℎ, 𝛼 ⟩ 𝑣 𝛼 ℎ \in 𝔥 ℎ ⊙ 𝑣 𝛼 = 0 ℎ \in ˆ 𝔤 𝑍 𝑑 ⊙ 𝑣 𝛼 = 𝜇 𝑣 𝛼

We define the natural $𝔤 𝑍$ -module $𝑀 𝛼$ to be the tensor product of graded vector spaces⁷

𝑀 = 𝑆 ∙ (ˆ 𝔥 - 𝑍) \otimes 𝕂 𝕂 𝑣 𝛼

with the $𝔤 𝑍$ -action given by the Tensor product of Lie algebra representations

ℎ ⊙ (𝑓 \otimes 𝑣 𝛼) = (ℎ ⊙ 𝑓) \otimes 𝑣 𝛼 + 𝑓 \otimes (ℎ ⊙ 𝑣 𝛼)

In the case $𝛼 = 0$ or $𝑍 = ℤ + 12$ ⁸ this amounts to a shifted graded module by $𝜇$ . If $𝑍 = ℤ$

𝑀 𝛼 = I n d 𝔤 𝑍 𝔤 0 𝑍 \oplus 𝔤 + 𝑍 𝕂 𝑣 𝛼

Conventionally we will take⁹

𝜇 = ⎧ { {⎨ { {⎩ - 12 ⟨ 𝛼, 𝛼 ⟩ + 124 d i m 𝔥 𝑍 = ℤ - 148 d i m 𝔥 𝑍 = ℤ + 12

Virasoro representation

Letting ${ℎ 𝑖} d i m 𝔥 𝑖 = 1$ be an orthonormal basis of $𝔥$ , extending the ground field if necessary, we define the following (basis-independent) operators on $𝑀 𝛼$

𝐿 (𝑛) = 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 ℎ 𝑖 (𝑛 - 𝑘) ℎ 𝑖 (𝑘) 𝑛 \in ℤ 𝐿 (0) = 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 ℎ 𝑖 (- | 𝑘 |) ℎ 𝑖 (| 𝑘 |) + 𝛽 0 d i m 𝔥

where

𝛽 0 = {0 𝑍 = ℤ 116 𝑍 = ℤ + 12

Then

𝜑 : 𝔳 \to E n d 𝕂 𝑀 𝐿 𝑛 \mapsto 𝐿 (𝑛) 𝑐 \mapsto d i m 𝔥

is a graded representation of the Virasoro algebra $𝔳$ .¹⁰

Proof

For $ℎ \in 𝔥$ , $𝑚 \in ℤ$ , and $𝑘 \in 𝑍$ , it follows from the commutation relations on $𝑀$ that

[𝐿 (𝑚), ℎ (𝑘)] = - 12 d i m 𝔥 \sum 𝑖 = 1 \sum ℓ \in 𝑍 [ℎ (𝑘), ℎ 𝑖 (𝑚 - ℓ) ℎ 𝑖 (ℓ)] = - 12 d i m 𝔥 \sum 𝑖 = 1 \sum ℓ \in 𝑍 [ℎ (𝑘), ℎ 𝑖 (𝑚 - ℓ)] ℎ 𝑖 (ℓ) - 12 d i m 𝔥 \sum 𝑖 = 1 \sum ℓ \in 𝑍 ℎ 𝑖 (𝑚 - ℓ) [ℎ (𝑘), ℎ 𝑖 (ℓ)] = - 12 d i m 𝔥 \sum 𝑖 = 1 \sum ℓ \in 𝑍 ⟨ ℎ, ℎ 𝑖 ⟩ 𝑘 𝛿 𝑘 + 𝑚 - ℓ ℎ 𝑖 (ℓ) - 12 d i m 𝔥 \sum 𝑖 = 1 \sum ℓ \in 𝑍 ℎ 𝑖 (𝑚 - ℓ) ⟨ ℎ, ℎ 𝑖 ⟩ 𝑘 𝛿 𝑘 + ℓ = - 12 d i m 𝔥 \sum 𝑖 = 1 𝑘 ⟨ ℎ, ℎ 𝑖 ⟩ ℎ 𝑖 (𝑘 + 𝑚) - 12 d i m 𝔥 \sum 𝑖 = 1 𝑘 ⟨ ℎ, ℎ 𝑖 ⟩ ℎ 𝑖 (𝑘 + 𝑚) = - 𝑘 ℎ (𝑘 + 𝑚)

Hence for $𝑚, 𝑛 \in ℤ$ with $𝑛 \neq 0$ and $𝑚 + 𝑛 \neq 0$ ,

[𝐿 (𝑚), 𝐿 (𝑛)] = 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 [𝐿 (𝑚), ℎ 𝑖 (𝑛 - 𝑘) ℎ 𝑖 (𝑘)] = 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 ℎ 𝑖 (𝑛 - 𝑘) [𝐿 (𝑚), ℎ 𝑖 (𝑘)] + 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 [𝐿 (𝑚), ℎ 𝑖 (𝑛 - 𝑘)] ℎ 𝑖 (𝑘) = - 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 (𝑘 ℎ 𝑖 (𝑛 - 𝑘) ℎ 𝑖 (𝑘 + 𝑚) + (𝑛 - 𝑘) ℎ 𝑖 (𝑚 + 𝑛 - 𝑘) ℎ 𝑖 (𝑘)) = - 12 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 ((𝑘 - 𝑚) ℎ 𝑖 (𝑛 - 𝑘 + 𝑚) ℎ 𝑖 (𝑘) + (𝑛 - 𝑘) ℎ 𝑖 (𝑚 + 𝑛 - 𝑘) ℎ 𝑖 (𝑘)) = (𝑚 - 𝑛) 𝐿 (𝑚 + 𝑛)

The only remaining case is essentially that $𝑛 \neq 0$ and $𝑚 + 𝑛 = 0$ , since the $𝑛 = 0$ case may be reduced to either zero or another case by the alternating property. In this case, from the expression

𝐿 (- 𝑚) = 12 d i m 𝔥 \sum 𝑖 = 1 (\sum 𝑘 \in 𝑍 : 𝑘 \leq 𝑚 ℎ 𝑖 (𝑘 - 𝑚) ℎ 𝑖 (- 𝑘) + \sum 𝑘 \in 𝑍 : 𝑘 > 𝑚 ℎ 𝑖 (- 𝑘) ℎ 𝑖 (𝑘 - 𝑚))

it follows

[𝐿 (𝑚), 𝐿 (- 𝑚)] = 12 d i m 𝔥 \sum 𝑖 = 1 (\sum 𝑘 \in 𝑍 : 𝑘 \leq 𝑚 [𝐿 (𝑚), ℎ 𝑖 (𝑘 - 𝑚) ℎ 𝑖 (- 𝑘)] + \sum 𝑧 \in 𝑍 : 𝑘 > 𝑚 [𝐿 (𝑚), ℎ 𝑖 (- 𝑘) ℎ 𝑖 (𝑘 - 𝑚)]) = 12 d i m 𝔥 \sum 𝑖 = 1 (\sum 𝑘 \in 𝑍 : 𝑘 \leq 𝑚 ([𝐿 (𝑚), ℎ 𝑖 (𝑘 - 𝑚)] ℎ 𝑖 (- 𝑘) + ℎ 𝑖 (𝑘 - 𝑚) [𝐿 (𝑚), ℎ 𝑖 (- 𝑘)]) + \sum 𝑧 \in 𝑍 : 𝑘 > 𝑚 ([𝐿 (𝑚), ℎ 𝑖 (- 𝑘)] ℎ 𝑖 (𝑘 - 𝑚) + ℎ 𝑖 (- 𝑘) [𝐿 (𝑚), ℎ 𝑖 (𝑘 - 𝑚)])) = 12 d i m 𝔥 \sum 𝑖 = 1 (\sum 𝑘 \in 𝑍 : 𝑘 \leq 𝑚 ((𝑚 - 𝑘) ℎ 𝑖 (𝑘) ℎ 𝑖 (- 𝑘) + 𝑘 ℎ 𝑖 (𝑘 - 𝑚) ℎ 𝑖 (𝑚 - 𝑘)) + \sum 𝑧 \in 𝑍 : 𝑘 > 𝑚 (𝑘 ℎ 𝑖 (𝑚 - 𝑘) ℎ 𝑖 (𝑘 - 𝑚) + (𝑚 - 𝑘) ℎ 𝑖 (- 𝑘) ℎ 𝑖 (𝑘))) = 12 d i m 𝔥 \sum 𝑖 = 1 (\sum 𝑘 \in 𝑍 : 𝑘 \leq 𝑚 (𝑚 - 𝑘) ℎ 𝑖 (𝑘) ℎ 𝑖 (- 𝑘) + \sum 𝑘 \in 𝑍 : 𝑘 \leq 0 (𝑚 + 𝑘) ℎ 𝑖 (𝑘) ℎ 𝑖 (- 𝑘) + \sum 𝑧 \in 𝑍 : 𝑘 > 0 (𝑚 + 𝑘) ℎ 𝑖 (- 𝑘) ℎ 𝑖 (𝑘) + \sum 𝑧 \in 𝑍 : 𝑘 > 𝑚 (𝑚 - 𝑘) ℎ 𝑖 (- 𝑘) ℎ 𝑖 (𝑘)) = 2 𝑚 𝐿 (0) + 𝛾 𝑚, - 𝑚

where $𝛾 𝑚, - 𝑚$ is some constant which we get from the $𝛽 0$ term and reversing the order of some $ℎ 𝑖 (- 𝑘) ℎ (𝑘)$ where necessary.¹¹

We will compute $𝛾 𝑚, - 𝑚$ using the application of of $[𝐿 (𝑚), 𝐿 (- 𝑚)]$ to a vacuum vector of $𝑀$ , e.g. $𝑣 𝛼$ . We note the following facts: From above there is a unique contravariant form such that $(𝑣 𝛼, 𝑣 𝛼) = 1$ and

(ℎ (𝑛) 𝑣, 𝑤) = (𝑣, ℎ (- 𝑛) 𝑤) (𝑑 𝑣, 𝑤) = (𝑣, 𝑑 𝑤)

for $ℎ \in 𝔥$ , $𝑛 \in 𝑍$ , and $𝑣, 𝑤 \in 𝑀$ . It follows by the definition of $𝐿 (𝑛)$ that

(𝐿 (𝑛) 𝑣, 𝑤) = (𝑣, 𝐿 (- 𝑛) 𝑤)

for $𝑛 \in ℤ$ and $𝑣, 𝑤 \in 𝑀$ . We also have

d i m 𝔥 \sum 𝑖 = 1 ℎ 𝑖 (0) 2 𝑣 𝛼 = ⟨ 𝛼, ℎ 𝑖 ⟩ ⟨ 𝛼, ℎ 𝑖 ⟩ 𝑣 𝛼 = ⟨ 𝛼, 𝛼 ⟩ 𝑣 𝛼

Now consider the case $𝑚 > 0$ . Then

𝛾 𝑚, - 𝑚 = (𝑣 𝛼, 𝛾 𝑚, - 𝑚 𝑣 𝛼) = (𝑣 𝛼, ([𝐿 (𝑚), 𝐿 (- 𝑚)] - 2 𝑚 𝐿 (0)) 𝑣 𝛼) = (𝑣 𝛼, (𝐿 (𝑚) 𝐿 (- 𝑚) - 2 𝑚 𝐿 (0)) 𝑣 𝛼) = (𝐿 (- 𝑚) 𝑣 𝛼, 𝐿 (- 𝑚) 𝑣 𝛼) - 2 𝑚 (𝑣 𝛼, 𝐿 (0) 𝑣 𝛼) = 14 d i m 𝔥 \sum 𝑖 = 1 d i m 𝔥 \sum 𝑗 = 1 (\sum 𝑘 \in 𝑍 : 0 \leq 𝑘 \leq 𝑚 ℎ 𝑖 (𝑘 - 𝑚) ℎ 𝑖 (- 𝑘) 𝑣 𝛼, \sum ℓ \in 𝑍 : 0 \leq ℓ \leq 𝑚 ℎ 𝑗 (ℓ - 𝑚) ℎ 𝑗 (- ℓ) 𝑣 𝛼) = - 𝑚 (𝑣 𝛼, d i m 𝔥 \sum 𝑖 = 1 ℎ 𝑖 (0) 2 𝑣 𝛼) [0 \in 𝑍] - 2 𝑚 𝛽 0 d i m 𝔥 = 14 d i m 𝔥 \sum 𝑖 = 1 (𝑣 𝛼, (\sum 𝑘 \in 𝑍 : 0 \leq 𝑘 \leq 𝑚 ℎ 𝑖 (𝑘) ℎ 𝑖 (𝑚 - 𝑘)) (\sum ℓ \in 𝑍 : 0 \leq ℓ \leq 𝑚 ℎ 𝑖 (ℓ - 𝑚) ℎ 𝑖 (- ℓ)) 𝑣 𝛼) = - 𝑚 ⟨ 𝛼, 𝛼 ⟩ [0 \in 𝑍] - 2 𝑚 𝛽 0 d i m 𝔥

where we have used an Iverson bracket and the fact that for $𝑖 \neq 𝑗$ we can commute the positively graded operators to annihilate $𝑣 𝛼$ . Now consider each of the terms

𝜂 𝑚, 𝑖, 𝑘, ℓ = (𝑣 𝛼, ℎ 𝑖 (𝑘) ℎ 𝑖 (𝑚 - 𝑘) ℎ 𝑖 (ℓ - 𝑚) ℎ 𝑖 (- ℓ) 𝑣 𝛼)

where $0 \leq 𝑘, ℓ \leq 𝑚$ , so that

𝛾 𝑚, - 𝑚 = 14 d i m 𝔥 \sum 𝑖 = 1 \sum 𝑘 \in 𝑍 : 0 \leq 𝑘 \leq 𝑚 \sum ℓ \in 𝑍 : 0 \leq ℓ \leq 𝑚 𝜂 𝑚, 𝑖, 𝑘, ℓ - 𝑚 ⟨ 𝛼, 𝛼 ⟩ [0 \in 𝑍] - 2 𝑚 𝛽 0 d i m 𝔥

We have the following cases

$𝜂 𝑚, 𝑖, 𝑘, ℓ = 0$ for $ℓ \notin {𝑘, 𝑚 - 𝑘}$ , since we may again commute and annihilate;
$𝜂 𝑚, 𝑖, 𝑘, ℓ = 𝑚 (𝑣 𝛼, ℎ 𝑖 (0) 2 𝑣 𝛼)$ for $𝑘 \in {0, 𝑚}$ whence $ℓ \in {0, 𝑚}$ ;¹²
$𝜂 𝑚, 𝑖, 𝑘, ℓ = 1$ otherwise

Thus

𝛾 𝑚, - 𝑚 = 14 d i m 𝔥 \sum 𝑖 = 1 ⎛ ⎜ ⎜ ⎝ \sum 𝑘 \in {0, 𝑚} \cap 𝑍 \sum ℓ \in {0, 𝑚} \cap 𝑍 𝑚 (𝑣 𝛼, ℎ 𝑖 (0) 2 𝑣 𝛼) + \sum 𝑘 \in 𝑍 : 0 < 𝑘 < 𝑚 \sum ℓ \in {𝑘, 𝑚 - 𝑘} 1 ⎞ ⎟ ⎟ ⎠ = - 𝑚 ⟨ 𝛼, 𝛼 ⟩ [0 \in 𝑍] - 2 𝑚 𝛽 0 d i m 𝔥 = 𝑚 ⟨ 𝛼, 𝛼 ⟩ [0 \in 𝑍] + 14 (d i m 𝔥) ∣ {(𝑘, ℓ) \in 𝑍 2 : 0 < 𝑘 < 𝑚, ℓ \in {𝑘, 𝑚 - 𝑘}} ∣ = - 𝑚 ⟨ 𝛼, 𝛼 ⟩ [0 \in 𝑍] - 2 𝑚 𝛽 0 d i m 𝔥 = 14 (d i m 𝔥) ∣ {(𝑘, ℓ) \in 𝑍 2 : 0 < 𝑘 < 𝑚, ℓ \in {𝑘, 𝑚 - 𝑘}} ∣ - 2 𝑚 𝛽 0 d i m 𝔥 = 12 (d i m 𝔥) (\sum 𝑘 \in 𝑍 : 0 < 𝑘 < 𝑚 𝑘 (𝑚 - 𝑘) - 4 𝑚 𝛽 0) = d i m 𝔥 12 (𝑚 3 - 𝑚)

as required.

In fact, the choice $𝜑 (𝑐) = d i m 𝔥$ and $𝔭 = 𝕂 𝑑 - 1 + 𝕂 𝑑 0 + 𝕂 𝑑 1$ being trivial uniquely determine the central term in the commutation relations for the Virasoro algebra. Letting $𝐿' 𝑛 = 𝐿 𝑛 - 124 𝛿 𝑛 𝑐$ , the above choice of $𝜇$ gives

𝐿' 0 ⊙ 𝑣 = - 𝑑 ⊙ 𝑣

for $𝑣 \in 𝑀 𝛼$ . The $𝐿 (0)$ -eigenvalue of a homogenous element $𝑣 \in 𝑀 𝛼$ is termed the weight and denoted $w t 𝑣$ so that

d e g 𝑣 = - w t 𝑣 + 124 d i m 𝔥

and

w t 𝑣 𝛼 = {112 ⟨ 𝛼, 𝛼 ⟩ 𝑍 = ℤ 116 d i m 𝔥 𝑍 = ℤ + 12

Natural Heisenerg algebras

Modules

Heisenberg modules

Triangular modules

Natural Heisenberg module 𝑀𝛼

Virasoro representation

See also

Footnotes

Natural Heisenberg module $𝑀 𝛼$