Explicit rank-one square-zero lattice fields

Fix an integer $p\ge 2$. Let $P=\mathbb{C}[x_1,x_2,\dots ]$ with finite-variable polynomials, and for each $t\in \mathbb{Q}$ let ${\mathcal{F}}_t=P{e}^{t\alpha }$ be a labelled copy of $P$. On ${\mathcal{F}}_t$ define ${\alpha }_{-n}=x_n$, ${\alpha }_n=2pn{\mathrm{\partial }}_{x_n}$ for $n\ge 1$, and ${\alpha }_0=2ptI$. Let $L=\mathbb{Z}\alpha$ have bilinear form $⟨r\alpha ,s\alpha ⟩=2prs$. For $ϵ\in \{0,1\}$ define the sector of state charge $r=m-ϵ\mathrm{/}p$ by ${\mathcal{A}}_{ϵ}={⨁}_{m\in \mathbb{Z}}{\mathcal{F}}_{m-ϵ\mathrm{/}p},\mathcal{A}={\mathcal{A}}_0\oplus {\mathcal{A}}_1\mathrm{.}$ Write $V_L={\mathcal{A}}_0$ and $I={\mathcal{A}}_1$. Define two module sectors by ${\mathcal{M}}_{ϵ}={⨁}_{m\in \mathbb{Z}}{\mathcal{F}}_{m+1\mathrm{/}2-ϵ\mathrm{/}p},\mathcal{M}={\mathcal{M}}_0\oplus {\mathcal{M}}_1\mathrm{.}$

For a state charge $r=m-\delta \mathrm{/}p$ with $\delta \in \{0,1\}$, let $e_r$ shift the input charge $t$ to $t+r$ when the input sector label $ϵ$ satisfies $ϵ+\delta \le 1$; set the action to zero if $ϵ+\delta =2$. Use the same rule on $\mathcal{A}$ and $\mathcal{M}$. The scalar cocycle for every allowed shift is $1$. If two state charges have sector labels whose sum is at most one, their pairing $2prs$ is an even integer: at least one of the two labels is zero, and the pairing is then an even integer by the displayed charge formulas. The product of two sector-one states is defined to be zero.

On either $\mathcal{A}$ or $\mathcal{M}$, define the exponential field of state charge $r$ by ${\mathcal{E}}_r(z)=\exp \text{\hspace{0.17em}⁣}\left({\sum }_{n\ge 1}\frac{r{\alpha }_{-n}}{n}{z}^n\right)\exp \text{\hspace{0.17em}⁣}(-{\sum }_{n\ge 1}\frac{r{\alpha }_n}{n}{z}^{-n})e_r{z}^{r{\alpha }_0},$ where $z^{r{\alpha }_0}f{e}^{t\alpha }=z^{2prt}f{e}^{t\alpha }$. Every allowed action has integer exponent $2prt$. On a forbidden action the field is zero. Each coefficient is algebraic: the annihilation exponential acts by a finite sum on a polynomial, and the creation exponential contributes finitely to any fixed power of $z$.

Put $b_n(z)={\mathrm{\partial }}_z^{n-1}\alpha (z)\mathrm{/}(n-1)!$ for $n\ge 1$, where $\alpha (z)={\sum }_{k\in \mathbb{Z}}{\alpha }_k{z}^{-k-1}$. For a monomial state $x_{n_1}\cdots x_{n_{\mathrm{\ell }}}{e}^{r\alpha }$, define its candidate field by $Y(x_{n_1}\cdots x_{n_{\mathrm{\ell }}}{e}^{r\alpha },z)=:b_{n_1}(z)\cdots b_{n_{\mathrm{\ell }}}(z){\mathcal{E}}_r(z):\mathrm{.}$ Here normal ordering is the finite sum over all choices of negative and nonnegative oscillator parts of the $\mathrm{\ell }$ factors $b_{n_i}$, placing every negative part left of ${\mathcal{E}}_r$ and every nonnegative part right of it; factors in each group commute. Extend linearly and denote the resulting actions by $Y_{\mathcal{A}}$ and $Y_{\mathcal{M}}$ according to their target space. This definition gives candidate fields only; the vertex-algebra and module axioms are separate conclusions.

Set $\mathbf{1}=e^0$. On $\mathcal{A}$ define $T(f{e}^{t\alpha })=({\sum }_{n\ge 1}n{x}_{n+1}{\mathrm{\partial }}_{x_n}f+t{x}_{1}f)e^{t\alpha }\mathrm{.}$

Rank-one Feigin–Fuchs oscillator spaces and long-screening coefficients

Formal distributions and fields on a vector space

Adamović–Milas, lattice construction, Sections 3–4

Adamović–Milas, triplet algebra, Section 1

The charge labels are algebraic copies; fractional labels do not denote analytic exponentials. The sector rule is precisely the square-zero extension convention: the sector-one part is an ideal candidate, and any product of two sector-one states is zero. The Jacobi, vacuum, creation, translation, and module properties of these fields are proved in a separate lemma. No screening-kernel equality, triplet-generation claim, or endpoint-module claim is part of this definition.