LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Strong generators generate the lowest space of a cyclic module

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Let $V = ⨁_{d \geq 0} V_{d}$ be a conformal vertex algebra with $V_{0} = C 1$ , strongly generated by homogeneous positive-weight states $S$ , meaning $V = span {s_{(- k_{1})}^{1} \dots s_{(- k_{ℓ})}^{ℓ} 1 : ℓ \geq 0, s^{i} \in S, k_{i} \geq 1},$ where the $ℓ = 0$ word is $1$ . Let $M = ⨁_{r \geq 0} M (r)$ be an admissible $V$ -module, so a homogeneous $a \in V_{d}$ satisfies $a_{(m)} M (r) \subset M (r + d - m - 1)$ . If $M$ is cyclic on $u \in M (0)$ , meaning $M = span {a_{(m_{1})}^{1} \dots a_{(m_{k})}^{k} u : k \geq 0, a^{i} \in V, m_{i} \in Z},$ where the $k = 0$ word is $u$ , then $M (0) = span {o (s_{1}) \dots o (s_{k}) u : k \geq 0, s_{i} \in S}, o (a) = a_{(wt a - 1)} .$ Moreover, for each homogeneous state $a$ , a noncommutative polynomial in the generator zero modes, depending only on $a$ and a fixed strong spanning expression for it, equals $o (a)$ on the lowest space of every admissible module. Thus the same generator zero-mode matrices determine the same full lowest-space zero-mode action. The statement does not require semisimple $L (0)$ , irreducibility, $C_{2}$ -cofiniteness, the existence of a universal induced module, or the general Zhu correspondence.

Statement references

Conformal vertex algebra of CFT type

Module for an ordinary vertex algebra

Admissible module for a vertex operator algebra

Facts & Assumptions

Given: The objects, hypotheses and conventions in the Statement.

[A1]

The CFT-type grading, nonnegative admissible module grading, strong generating set $S$ , and cyclic vector $u$ are precisely the hypotheses of the Statement.

[F1]

Conformal vertex algebra of CFT type gives the nonnegative integer state weights and vacuum line.

[F2]

Module for an ordinary vertex algebra supplies the full module Jacobi identity; taking its formal coefficients gives the commutator and iterate identities displayed in the proof.

[F3]

Admissible module for a vertex operator algebra specifies the degree shift and the vanishing of negative module degrees. Strong generation here means that $V$ is spanned by finite words of strictly negative modes of states in $S$ applied to $1$ .

[A2]

Jacobi and residue calculus from the state-field locality axioms proves the homogeneous state-product weight rule and the module commutator and iterate coefficient identities used below. Its complete current proof supplies exactly these formal identities.

Proof

1.1givenA2F1algebra

Give a mode $a_{(m)}$ of homogeneous $a$ degree $wt (a) - m - 1$ . The commutator identity in [A2] and its state-product weight rule show every term in $[a_{(m)}, b_{(n)}]$ has degree $wt (a) - m - 1 + wt (b) - n - 1$ . The sum is finite by state truncation. Thus mode commutators preserve total degree.

1.2A2algebra

Let $a, b$ be homogeneous of weights $d, e$ , with $d \geq 1$ , and define the inhomogeneous states $C_{k} (a, b) = \sum_{i = 0}^{d} (\binom{d}{i}) a_{(i - k)} b$ for $k \geq 1$ . Extend $o$ linearly over homogeneous components. For the summand indexed by $i$ , put $n = i - k$ and $m = d + e - n - 2$ , the zero-mode index of $a_{(n)} b$ . The iterate identity in [A2] gives $(a_{(n)} b)_{(m)} = \sum_{j \geq 0} (- 1)^{j} (\binom{n}{j}) (a_{(n - j)} b_{(m + j)} - (- 1)^{n} b_{(m + n - j)} a_{(j)}) .$ The first term has rightmost mode $b_{(m + j)}$ of degree $n - d + 1 - j = i - k - d + 1 - j$ .

2.1givenA1F3step 1.1algebra

Reorder a finite word of modes so positive-degree modes lie left, zero-degree modes middle, and negative-degree modes right. Swap any adjacent factors in the wrong order using $A B = B A + [A, B]$ : the swapped word has one fewer inversion, and every commutator term has shorter length. Induction on length and then inversion count terminates. On applying the reordered word to $u \in M (0)$ , a rightmost negative-degree mode kills $u$ by admissibility. A word containing only nonnegative modes and at least one positive-degree mode has positive total degree. Since $M$ is cyclic on $u$ , its degree-zero space is therefore spanned by words in $o (a) = a_{(wt (a) - 1)}$ , $a \in V$ , applied to $u$ .

2.2givenF2F3A2step 1.2algebra

Because $i \leq d$ , this first term kills $M (0)$ for every $k \geq 2$ . For $k = 1$ it can survive only at $i = d, j = 0$ , when it is exactly $o (a) o (b)$ . In the second term $a_{(j)}$ kills $M (0)$ for $j \geq d$ . For each fixed $0 \leq j < d$ , the mode $b_{(m + n - j)} = b_{(d + e - 2 - j)}$ is independent of $i$ . Its coefficient after summing over $i$ is, up to the fixed sign $(- 1)^{j - k + 1}$ , $\sum_{i = 0}^{d} (- 1)^{i} (\binom{d}{i}) (\binom{i - k}{j}) = 0.$ Indeed $i \mapsto (\binom{i - k}{j})$ is a polynomial of degree $j < d$ , so its $d$ th finite difference vanishes; this remains true for negative upper entries by the polynomial definition of the binomial coefficient. Module truncation makes the original iterate sums finite on each vector. Therefore $o (C_{1} (a, b)) = o (a) o (b)$ and $o (C_{k} (a, b)) = 0$ for $k \geq 2$ on $M (0)$ .

3.1givenA1F1F3A2step 2.1step 2.2algebra∎

Induct on state weight to express every $o (v)$ on $M (0)$ as a noncommutative polynomial in $o (S)$ . Weight zero gives the identity. A strong spanning monomial of positive weight is $s_{(- k)} b$ with $s \in S$ , $k \geq 1$ , and $b$ a shorter strong monomial. For $k = 1$ , expand $C_{1} (s, b)$ : its first term is $s_{(- 1)} b$ , while every other term $s_{(i - 1)} b$ has weight $wt (s) + wt (b) - i$ , strictly below the target weight; also $b$ has lower weight because $wt (s) \geq 1$ . The identity in step 2.2 expresses its zero mode using $o (s) o (b)$ and those lower-weight zero modes. For $k \geq 2$ , expand $C_{k} (s, b)$ ; its first term is $s_{(- k)} b$ , and each remaining term $s_{(i - k)} b$ has weight $wt (s) + wt (b) + k - 1 - i$ , strictly lower than the target. The identity $o (C_{k}) = 0$ gives the reduction. Negative-weight terms vanish because $V_{d} = 0$ for $d < 0$ . Fix a strong spanning expression for each state: the induction decreases weight and each $C_{k}$ is finite, so it terminates and yields a polynomial independent of $M$ . Combining with step 2.1 proves both assertions.

Depends on

Dependency tree · 2 levels

Sources

Adamović and Milas, On the triplet vertex algebra W(p)