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

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The three triplet primary fields satisfy a level-three null relation

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

For the actual rank-one square-zero lattice fields at every integer $p \geq 2$ , put $F = e^{- α}$ , $H = Q F$ , $E = Q^{2} F$ and $Q = (e^{α})_{(0)}$ . Each of $X = F, H, E$ is primary of weight $2 p - 1$ and satisfies the state identity $(L (- 1)^{3} - 4 p L (- 2) L (- 1) + 2 p (2 p - 1) L (- 3)) X = 0.$ Consequently, on any actual lattice module $M$ , for a Virasoro-invariant subspace $B \subseteq M$ and a Virasoro-module quotient map $π : M \to C$ , the assignment $w \mapsto π Y (w, z) ∣_{B}$ on the Virasoro submodule of the state space generated by any one of $F, H, E$ satisfies the derivative, primary commutator, and descendant-residue identities of a formal Virasoro intertwining field. These assertions concern the actual vertex algebra and module fields, not merely the oscillator Lie algebra representation.

Statement references

Exact lengths of boundary long-screening strings

Explicit rank-one square-zero lattice fields

Jacobi identities for the rank-one square-zero lattice extension

Conformal subalgebra in the kernel of $v_{(0)}$

Facts & Assumptions

Given: The integer $p \geq 2$ , the rank-one oscillator formulas, and the proved supplier interfaces below.

[A1]

Current Explicit rank-one square-zero lattice fields fixes $⟨ α, α ⟩ = 2 p$ , $F_{t} = C [x_{1}, x_{2}, \dots] e^{t α}$ , $α_{- n} = x_{n}$ , $α_{n} = 2 p n \partial_{x_{n}}$ , $α_{0} = 2 p t$ , the exponential fields, and the normal-order prescription. Its exact definition is used in the calculation below.

[A2]

Current Jacobi identities for the rank-one square-zero lattice extension asserts the ordinary vertex algebra on $A$ , its ordinary module $M$ , the full adjoint and module Jacobi identities, and the displayed translation operator. Its actual-field Jacobi identities and translation formula are used below.

[A3]

Current Conformal subalgebra in the kernel of $v_{(0)}$ supplies the conformal modes $L (n) = \frac{1}{4 p} \sum_{j \in Z} : α_{j} α_{n - j} : - \frac{p - 1}{2 p} (n + 1) α_{n},$ the Virasoro action and weight formula, the derivation $Q = (e^{α})_{(0)}$ commuting with every $L (n)$ , the table $Q F = H$ , $Q H = E$ , $Q E = 0$ , and primarity of $F, H, E$ of weight $2 p - 1$ . Its complete current proof supplies these identities for the actual fields.

[A4]

Exact lengths of boundary long-screening strings with $i = 0$ , $n = 1$ gives $Q^{2} F \neq 0$ . The coefficient maps are the same as the present lattice screening by the exponential formula and scalar cocycle one in [A1]. Consequently $E \neq 0$ , $H \neq 0$ (otherwise $Q H = E$ would vanish), and $F \neq 0$ as a labelled constant polynomial.

Proof

1.1givenA1A2A3algebra

Put $q = (p - 1) / (2 p)$ . The translation formula in [A2] gives $L (- 1) (f e^{t α}) = (\sum_{k \geq 1} k x_{k + 1} \partial_{x_{k}} f + t x_{1} f) e^{t α}$ . The conformal modes in [A3] give $L (- 2) e^{t α} = ((t + q) x_{2} + x_{1}^{2} / (4 p)) e^{t α}$ , $[L (- 2), x_{1}] = x_{3}$ , and $L (- 3) e^{t α} = ((t + 2 q) x_{3} + x_{1} x_{2} / (2 p)) e^{t α}$ . Starting with $F = e^{- α}$ and applying these operators successively yields $L (- 1)^{3} F = (- 2 x_{3} + 3 x_{1} x_{2} - x_{1}^{3}) e^{- α},$ $L (- 2) L (- 1) F = (- x_{3} + (1 - q) x_{1} x_{2} - \frac{x_{1}^{3}}{4 p}) e^{- α}, L (- 3) F = (- \frac{x_{3}}{p} + \frac{x_{1} x_{2}}{2 p}) e^{- α} .$ The first equality follows by applying the displayed $L (- 1)$ formula three times; the second uses $L (- 1) F = - x_{1} e^{- α}$ and the commutator with $x_{1}$ ; the third substitutes $t = - 1$ .

2.1givenA3step 1.1algebra

In the asserted linear combination of these three states, the coefficients of $x_{1}^{3}$ , $x_{3}$ , and $x_{1} x_{2}$ are respectively $- 1 + 1 = 0$ , $- 2 + 4 p - 2 (2 p - 1) = 0$ , and $3 - 4 p (1 - q) + (2 p - 1) = 0$ because $q = (p - 1) / (2 p)$ . These monomials exhaust the three displayed expressions, so the null relation holds for $F$ .

3.1givenA3A4step 2.1algebra

The differential operator $P (L) = L (- 1)^{3} - 4 p L (- 2) L (- 1) + 2 p (2 p - 1) L (- 3)$ commutes with $Q$ by [A3]. Thus $P (L) H = Q P (L) F = 0$ and $P (L) E = Q^{2} P (L) F = 0$ . The same supplier gives their weight and positive-mode equations; [A4] ensures that $F, H, E$ are all nonzero, so these are primaries in its stated convention.

4.1givenA2A3step 3.1algebra

Write $Y_{M} (w, z)$ for the actual module field and $T_{M} (y) = Y_{M} (ω, y) = \sum_{n \in Z} L_{M} (n) y^{- n - 2}$ . For every state $w$ , module translation gives $Y_{M} (L (- 1) w, z) = \partial_{z} Y_{M} (w, z)$ . Taking first the residue in the iterate variable and then ${Res}_{y} y^{n + 1}$ in module Jacobi gives, for every integer $n$ , $[L_{M} (n), Y_{M} (w, z)] = \sum_{j \geq 0} (\binom{n + 1}{j}) z^{n + 1 - j} Y_{M} (L (j - 1) w, z) .$ Indeed the delta substitution replaces $y^{n + 1}$ by $(z + x)^{n + 1} = \sum_{j \geq 0} (\binom{n + 1}{j}) z^{n + 1 - j} x^{j}$ , where $x$ is the iterate variable, and ${Res}_{x} x^{j} Y (ω, x) w = L (j - 1) w$ . Truncation makes the sum finite on a fixed $w$ . For $w = X$ , primarity from step 3.1 leaves exactly $j = 0, 1$ , giving $[L_{M} (n), Y_{M} (X, z)] = z^{n} (z \partial_{z} + (n + 1) (2 p - 1)) Y_{M} (X, z)$ .

5.1givenA2A3step 4.1algebra

For $r \geq 2$ , take ${Res}_{x} x^{1 - r} {Res}_{y}$ of the same module Jacobi identity. Since $ω_{(1 - r)} = L (- r)$ , this yields $Y_{M} (L (- r) w, z) = {Res}_{y} (ι_{y, z} (y - z)^{1 - r} T_{M} (y) Y_{M} (w, z) - ι_{z, y} (y - z)^{1 - r} Y_{M} (w, z) T_{M} (y)) .$ Here ${Res}_{y}$ means the coefficient of $y^{- 1}$ , and the two prescribed expansions are $ι_{y, z} (y - z)^{1 - r} = \sum_{j \geq 0} (\binom{r + j - 2}{j}) z^{j} y^{1 - r - j}, ι_{z, y} (y - z)^{1 - r} = (- 1)^{r - 1} \sum_{j \geq 0} (\binom{r + j - 2}{j}) y^{j} z^{1 - r - j} .$ The first residue contains $L_{M} (- r - j) Y_{M} (w, z)$ multiplied by $z^{j}$ ; each coefficient on a fixed input is a finite sum by the lower bound on powers in $Y_{M} (w, z)$ . The second contains $Y_{M} (w, z) L_{M} (j - 1)$ , and truncation of $T_{M} (y)$ makes all sufficiently large $j$ vanish on that input. Thus both residues are well-defined coefficientwise, with no analytic convergence assumption.

6.1givenstep 2.1step 3.1step 4.1step 5.1algebra∎

Let $A_{X}$ be the linear span of finite Virasoro mode words on $X$ , let $ι : B ↪ M$ be the given inclusion, and define the complex-linear assignment $I (w, z) = π Y_{M} (w, z) ι$ for $w \in A_{X}$ ; its coefficients are maps $B \to C$ . Put $T_{B} (y) = \sum_{n} L_{B} (n) y^{- n - 2}$ and $T_{C} (y) = \sum_{n} L_{C} (n) y^{- n - 2}$ . Virasoro invariance and equivariance give $L_{M} (n) ι = ι L_{B} (n)$ and $π L_{M} (n) = L_{C} (n) π$ . Composing step 4.1 with $π$ and $ι$ therefore gives $I (L (- 1) w, z) = \partial_{z} I (w, z), L_{C} (n) I (w, z) - I (w, z) L_{B} (n) = \sum_{j \geq 0} (\binom{n + 1}{j}) z^{n + 1 - j} I (L (j - 1) w, z) .$ Composing step 5.1 gives precisely $I (L (- r) w, z) = {Res}_{y} (ι_{y, z} (y - z)^{1 - r} T_{C} (y) I (w, z) - ι_{z, y} (y - z)^{1 - r} I (w, z) T_{B} (y)), r \geq 2.$ These are the derivative, commutator (primary at $w = X$ ), and descendant-residue identities asserted in the Statement. They hold even when $I = 0$ ; no simplicity or highest-weight grading of arbitrary $B, C$ is asserted. Finally the null state from steps 2.1–3.1 maps to zero by linearity of $I$ .

Depends on

Dependency tree · 2 levels

Sources

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