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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Screening spanning criterion for strong generation

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Let $V = ⨁_{m \geq 0} V_{m}$ be a conformal vertex algebra of CFT type, with vacuum $1$ , conformal vector $ω$ , and translation $T = L (- 1)$ . Thus $L (0) ∣_{V_{m}} = m I$ and $V_{0} = C 1$ . A nonzero vector $u$ is a Virasoro highest-weight vector of weight $λ$ if $L (0) u = λ u$ and $L (r) u = 0$ for every $r > 0$ . A Virasoro descendant of $u$ means $L (- r_{1}) \dots L (- r_{s}) u$ for some $s \geq 0$ and integers $r_{i} \geq 1$ .

Let $Q : V \to V$ be complex-linear and weight-preserving, meaning $Q (V_{m}) \subseteq V_{m}$ for every $m \geq 0$ . Say that $Q$ is a derivation of the vertex products if, for every $a, b \in V$ and $r \in Z$ , $Q (a_{(r)} b) = (Q a)_{(r)} b + a_{(r)} (Q b) .$ Assume this derivation identity and $Q L (n) = L (n) Q$ for every $n \in Z$ . Suppose $Q ω = 0$ , $Q F = H$ , $Q H = E$ , and $Q E = 0$ , where $F, H, E$ are homogeneous positive-weight states.

Here a vertex subalgebra is a linear subspace containing $1$ , stable under $T$ , and closed under every mode product $a_{(r)} b$ of its elements. Let $K \subseteq V$ be a vertex subalgebra containing $ω, F, H, E$ . Assume every nonzero $Q^{j} f_{n}$ is a highest-weight vector and that $K$ is the linear span of all $L (- r_{1}) \dots L (- r_{s}) Q^{j} f_{n}$ with $s, n, j \geq 0$ and $r_{i} \geq 1$ , where $f_{0} = 1$ , $f_{1} = F$ , and for each $n \geq 1$ an integer $k_{n} \geq 1$ satisfies $f_{n + 1} = F_{(- k_{n})} f_{n}$ . Let $W$ be the smallest vertex subalgebra of $V$ containing $ω, F, H, E$ . Then $W = K$ and is strongly generated by $ω, F, H, E$ , meaning it is spanned by finite words of strictly negative modes of these four states applied to $1$ .

Statement references

Conformal vertex algebra of CFT type

Vertex algebra in the state–field convention

Facts & Assumptions

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

[A1]

The grading, weight preservation, mode-derivation identity, commutation with all $L (n)$ , four-state $Q$ table, highest-weight conditions on nonzero $Q^{j} f_{n}$ , subalgebra containment, recursive sequence $(f_{n})$ , and descendant-spanning property in the Statement are assumptions.

[F1]

Conformal vertex algebra of CFT type and Vertex algebra in the state–field convention fix $T = L (- 1)$ , the vacuum creation identity, and $[T, a_{(m)}] = - m a_{(m - 1)}$ .

Proof

1.1givenA1F1algebra

Let $U$ be the span of finite words of strictly negative modes of $ω, F, H, E$ on $1$ . The derivation identity gives $[Q, a_{(m)}] = (Q a)_{(m)}$ . Since $1_{(- 1)} 1 = 1$ , applying it to this product gives $Q 1 = (Q 1)_{(- 1)} 1 + 1_{(- 1)} Q 1 = 2 Q 1$ , so $Q 1 = 0$ . Applying the mode commutator successively to a word and using $Q ω = 0$ , $Q F = H$ , $Q H = E$ , and $Q E = 0$ shows that every term of its image is again such a word; hence $Q U \subset U$ . Creation gives $F = F_{(- 1)} 1 \in U$ . Induction on $n$ using $f_{n + 1} = F_{(- k_{n})} f_{n}$ with $k_{n} \geq 1$ gives $f_{n} \in U$ , and then $Q^{j} f_{n} \in U$ for all $j$ .

2.1givenA1F1step 1.1algebra

For $r \geq 2$ , $L (- r) = ω_{(- r + 1)}$ is a strictly negative mode of a displayed generator. Also $T 1 = 0$ , and $[T, a_{(- k)}] = k a_{(- k - 1)}$ for $k \geq 1$ ; commuting $T$ through a finite word therefore expresses its translate as a sum of words in $U$ . Thus every Virasoro descendant of every $Q^{j} f_{n}$ lies in $U$ , and the assumed spanning property gives $K \subset U$ .

3.1givenA1step 2.1∎

By definition $U \subset W$ . Since $K$ is a vertex subalgebra containing the four generators, $W \subset K$ . Hence $W = K = U$ , which is precisely strong generation by the four states.

Depends on

Dependency tree · 2 levels

Sources

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