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A simple ordinary module admits a nonnegative grading

Statement

Let $V$ be a conformal vertex algebra of CFT type and let $S$ be a nonzero simple ordinary $V$ -module. Then $S$ admits a grading $S = ⨁_{n \geq 0} S (n)$ satisfying the admissible mode-shift rule; hence $S$ is an admissible module.

Statement references

Admissible module for a vertex operator algebra

Ordinary module

Facts & Assumptions

Given: $V$ is a conformal vertex algebra of CFT type and $S$ is a nonzero simple ordinary $V$ -module.

[F1]

Conformal vertex algebra of CFT type: $V = ⨁_{r \geq 0} V_{r}$ and $L (0) a = r a$ for $a \in V_{r}$ .

[F2]

Module for an ordinary vertex algebra: the module Jacobi identity supplies the conformal mode commutator, and a simple module is nonzero with no proper nonzero submodule; hence any nonzero vector generates $S$ .

[F3]

Ordinary module: $S$ is a direct sum of finite-dimensional $L (0)$ -eigenspaces whose occurring weights have real parts bounded below.

[F4]

Admissible module for a vertex operator algebra: an admissible grading must satisfy $a_{(m)}^{M} S (n) \subseteq S (n + r - m - 1)$ for $a \in V_{r}$ , with negative grades interpreted as zero.

Proof

1.1givenF1F2algebra

For homogeneous $a \in V_{r}$ , the module Jacobi identity gives $[L_{S} (0), a_{(m)}^{S}] = (r - m - 1) a_{(m)}^{S}$ . Indeed, the commutator formula at the conformal mode $ω_{(1)} = L (0)$ has only the terms $(ω_{(0)} a)_{(m + 1)} + (ω_{(1)} a)_{(m)}$ ; these equal $- (m + 1) a_{(m)}^{S} + r a_{(m)}^{S}$ because $ω_{(0)} a = L (- 1) a = T a$ . Thus each mode shifts an $L (0)$ -eigenvalue by the integer $r - m - 1$ .

2.1givenF2F3step 1.1choose

Choose a nonzero eigenvector $s \in S_{[h]}$ . By [F2], the submodule it generates is all of $S$ ; each homogeneous mode shifts the eigenvalue by an integer by step 1.1. Thus every occurring eigenvalue lies in $h + Z$ . By [F3], the integer indices of nonzero eigenspaces are bounded below; choose their least value $k_{0}$ and set $S (n) = S_{h + k_{0} + n}$ for $n \geq 0$ . The ordinary eigenspace decomposition makes $S = ⨁_{n \geq 0} S (n)$ .

3.1givenF1F4step 2.1algebra∎

For $a \in V_{r}$ , step 1.1 gives $a_{(m)}^{S} S (n) \subseteq S (n + r - m - 1)$ ; if the index is negative, minimality of $k_{0}$ makes that eigenspace zero. This is exactly [F4], so the constructed grading is admissible.

Proof status

This is a local mode-shift proof. It establishes that each simple ordinary summand is admissible; it does not show that a weak-module decomposition supplied by regularity respects a separately specified admissible grading.

A simple ordinary module admits a nonnegative grading

Statement

Statement references

Facts & Assumptions

Proof

Proof status

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