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The completed local mode algebra acts on conformal modules

Statement

Let $V$ be a conformal vertex algebra of CFT type and $M$ a lower-truncated $V$ -module. The bracket $[a \otimes f d t, b \otimes g d t] = \sum_{j \geq 0} (a_{(j)} b) \otimes f^{(j)} g d t / j!$ on the quotient ${\hat{L}}_{t} (V) = L_{t} (V) d t / D_{t} (L_{t} (V) d t)$ makes it a Lie algebra, and the residue formula defines a representation on $M$ . For each completed element and $m \in M$ , its action is a finite sum; no topology on $M$ or completion of $M$ is used.

Statement references

Completed local mode Lie algebra of a conformal vertex algebra

Facts & Assumptions

Given: The local mode algebra and residue action in Completed local mode Lie algebra of a conformal vertex algebra, the module Jacobi identity in Module for an ordinary vertex algebra, and formal residues in Formal distributions and fields on a vector space.

[F1]

Completed local mode Lie algebra of a conformal vertex algebra: the bracket of $a \otimes f d t$ and $b \otimes g d t$ is $\sum_{j \geq 0} (a_{(j)} b) \otimes f^{(j)} g d t / j!$ , and exact representatives are $T a \otimes f d t + a \otimes f^{'} d t$ .

[F2]

Module for an ordinary vertex algebra: the module Jacobi identity implies the mode commutator identity, with $Y_{M} (a, z) = \sum_{m \in Z} a_{(m)}^{M} z^{- m - 1}$ and lower truncation on each fixed module vector.

[F3]

Formal distributions and fields on a vector space: ${Res}_{z} z^{m} Y_{M} (a, z) = a_{(m)}^{M}$ and formal residue is coefficient extraction.

[F4]

Conformal vertex algebra of CFT type: every conformal-weight space $V_{d}$ is finite dimensional, so $V_{\leq N}$ is finite dimensional for each $N$ .

[F5]

A formal chiral operation glues over a pointed curve: the local residue bracket is skew-symmetric and satisfies the chiral Jacobi identity.

[F6]

Mutually local creative generators determine a state–field map: the creative state determines a translation-covariant local field uniquely; the $n$ th-product field has creative state $A_{(n)} b$ .

Proof

1.1F1F2F3algebra

For monomials $f = t^{m}$ and $g = t^{n}$ , the bracket in [F1] is $\sum_{j \geq 0} (\binom{m}{j}) (a_{(j)} b) \otimes t^{m + n - j} d t$ . Applying the module Jacobi identity and taking the two mode residues gives $[a_{(m)}^{M}, b_{(n)}^{M}] = \sum_{j \geq 0} (\binom{m}{j}) (a_{(j)} b)_{(m + n - j)}^{M}$ , exactly the image of this bracket under the residue action. The sum is finite because $a_{(j)} b = 0$ for all sufficiently large $j$ .

1.2F1F2F3F4F5F6algebra

The exact representative in [F1] acts by ${Res}_{t} (f Y_{M} (T a, t) + f^{'} Y_{M} (a, t)) d t = {Res}_{t} \partial_{t} (f Y_{M} (a, t)) d t = 0$ , since a formal derivative has zero residue; the module derivative identity follows by taking the $L_{M} (- 1)$ residue in its Jacobi axiom. For the bracket, translation covariance and the uniqueness of creative state in [F6] give $(T a)_{(0)} b = 0$ and $(T a)_{(j)} b = - j a_{(j - 1)} b$ for $j \geq 1$ . Hence $[D_{t} (a \otimes f d t), b \otimes g d t]$ is the sum $- \sum_{j \geq 1} (a_{(j - 1)} b) \otimes f^{(j)} g d t / (j - 1)! + \sum_{j \geq 0} (a_{(j)} b) \otimes f^{(j + 1)} g d t / j! = 0$ , by reindexing. Thus exact representatives form a left bracket ideal; skew-symmetry in [F5] gives the right ideal property. For a completed element, its coefficient states lie in one finite-dimensional $V_{\leq N}$ ; choose a basis of that space. Lower truncation kills all sufficiently high modes of each basis vector on the fixed input $m$ , and the maximum of the finitely many bounds shows the completed action is finite.

2.1F1F3F5step 1.2algebra

The chiral Jacobi identity in [F5], after taking residues in the three formal variables, gives $[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0$ for monomial representatives. Every nested bracket is finite because each pair of states has finite truncation order. For Laurent-series representatives each fixed coefficient is a finite sum, since their exponents are bounded below and their state coefficients lie in finite-dimensional weight truncations. Thus the bracket is a Lie bracket on the completed derivative quotient.

3.1F1F2F3F4step 1.1step 1.2∎

The representation identity on monomials from step 1.1 is exactly the Lie bracket relation. The pointwise finiteness in step 1.2 extends both sides to completed elements; equality of their finite coefficient sums proves that the completed action is a representation.

The completed local mode algebra acts on conformal modules

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources