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 $𝔥$ . 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

Table of Contents

Natural Heisenerg algebras

Modules

Heisenberg modules

Triangular modules

Natural Heisenberg module $𝑀 𝛼$

Virasoro representation

See also

Graph View

Backlinks

Table of Contents

Natural Heisenerg algebras

Modules

Heisenberg modules

Triangular modules

Natural Heisenberg module 𝑀𝛼

Virasoro representation

See also

Footnotes

Graph View

Backlinks

Natural Heisenberg module $𝑀 𝛼$