Even-lattice chiral boson sector properties

The exponent defining $ϵ$ is bilinear in the integer coordinates of each argument, so $ϵ(\lambda ,\mu )ϵ(\lambda +\mu ,\nu )=ϵ(\mu ,\nu )ϵ(\lambda ,\mu +\nu )$ after expanding both exponents. Also $ϵ(0,\mu )=ϵ(\lambda ,0)=1$. Thus the momentum shifts are normalized and associative. Modulo two, $⟨\lambda ,\mu ⟩={\sum }_i{G}_{ii}{\lambda }_i{\mu }_i+{\sum }_{i>j}{G}_{ij}({\lambda }_i{\mu }_j+{\lambda }_j{\mu }_i)\equiv {\sum }_{i>j}{G}_{ij}({\lambda }_i{\mu }_j+{\lambda }_j{\mu }_i),$ since $G_{ii}$ is even. Therefore $ϵ(\lambda ,\mu )\mathrm{/}ϵ(\mu ,\lambda )=(-1{)}^{⟨\lambda ,\mu ⟩}$.

The explicit oscillator operators in [F1] satisfy their commutator on every polynomial momentum sector, including the zero modes. For a fixed polynomial all sufficiently large positive modes vanish. The positive exponential in $Y(e^{\lambda },z)$ differentiates that polynomial only finitely many times, so it gives a finite Laurent polynomial; the negative exponential has nonnegative powers of $z$, and each coefficient is a finite sum. Integrality of $⟨\lambda ,\mu ⟩$ then makes $Y(e^{\lambda },z)$ lower-truncated on every state. The currents are fields for the same reason. Thus these are actual fields in [F4], not merely arbitrary formal distributions.

Commuting the positive exponential of the $\lambda$ field past the negative exponential of the $\mu$ field gives the scalar contraction $\exp \text{\hspace{0.17em}⁣}(-⟨\lambda ,\mu ⟩{\sum }_{n\ge 1}\frac{(w\mathrm{/}z{)}^n}{n})={\iota }_{z,w}(1-w\mathrm{/}z{)}^{⟨\lambda ,\mu ⟩}\mathrm{.}$ Moving its zero-mode factor past $U_{\mu }$ supplies $z^{⟨\lambda ,\mu ⟩}$. The common ordered expression is consequently multiplied by $ϵ(\lambda ,\mu ){\iota }_{z,w}(z-w{)}^{⟨\lambda ,\mu ⟩}$. In the opposite order the same oscillator and momentum-shift expression is multiplied by $ϵ(\mu ,\lambda ){\iota }_{w,z}(w-z{)}^{⟨\lambda ,\mu ⟩}$. Step 1.1 identifies the signs. Multiplying by $(z-w{)}^N$ with $N\ge \max (0,-⟨\lambda ,\mu ⟩)$ makes both scalar expansions the same polynomial, proving mutual locality. The current-current commutator is $⟨v,w⟩{\mathrm{\partial }}_w\delta (z,w)$ and is killed by $(z-w{)}^2$. Directly $[a_m^v,Y(e^{\lambda },w)]=⟨v,\lambda ⟩w^{m}Y(e^{\lambda },w)$, including $m=0$; hence the current-lattice commutator is $⟨v,\lambda ⟩\delta (z,w)Y(e^{\lambda },w)$ and is killed by $z-w$. All pairs in the generating family of currents and momentum fields are local.

Let $\mathsf{T}$ be the explicit operator in [F1]. It kills $\mathbf{1}$ and satisfies $[\mathsf{T},a_m^v]=-m{a}_{m-1}^v$ for every integer $m$, so $[\mathsf{T},J^v(z)]={\mathrm{\partial }}_z{J}^v(z)$. Also $[\mathsf{T},U_{\lambda }]=a_{-1}^{\lambda }{U}_{\lambda }$. Write the two oscillator exponentials as $E_{-}(z),E_{+}(z)$, in their displayed order. Their commuting exponents give $[\mathsf{T},E_{-}]={\mathrm{\partial }}_z{E}_{-}-a_{-1}^{\lambda }{E}_{-},[\mathsf{T},E_{+}]={\mathrm{\partial }}_z{E}_{+}+z^{-1}{a}_0^{\lambda }{E}_{+}\mathrm{.}$ Here $[\mathsf{T},z^{a_0^{\lambda }}]=0$. When computing $[\mathsf{T},E_{-}{E}_{+}{U}_{\lambda }{z}^{a_0^{\lambda }}]$, use $E_{+}{a}_{-1}^{\lambda }=(a_{-1}^{\lambda }-⟨\lambda ,\lambda ⟩z^{-1})E_{+},a_0^{\lambda }{U}_{\lambda }=U_{\lambda }(a_0^{\lambda }+⟨\lambda ,\lambda ⟩)\mathrm{.}$ The two norm terms cancel, and the remaining $z^{-1}{a}_0^{\lambda }$ is precisely the derivative of the rightmost $z^{a_0^{\lambda }}$. Thus $[\mathsf{T},Y(e^{\lambda },z)]={\mathrm{\partial }}_{z}Y(e^{\lambda },z)$, with the stated zero-mode order.

The currents applied to $\mathbf{1}$ have nonnegative powers with constant state $a_{-1}^v\mathbf{1}$. Likewise $Y(e^{\lambda },z)\mathbf{1}=E_{-}(z)\mathrm{\mid }\lambda ⟩$ has nonnegative powers and constant state $\mathrm{\mid }\lambda ⟩$. Translation covariance and $\mathsf{T}\mathbf{1}=0$ give the coefficient recursion $(k+1)v_{k+1}=\mathsf{T}{v}_k$ for each such vacuum series. They are therefore exactly $e^{z\mathsf{T}}{a}_{-1}^v\mathbf{1}$ and $e^{z\mathsf{T}}\mathrm{\mid }\lambda ⟩$. The mode $Y(e^{\lambda }{)}_{(-1)}\mathbf{1}$ produces $\mathrm{\mid }\lambda ⟩$, and negative current modes then produce every oscillator monomial on that sector. These states span the full algebraic direct sum. All hypotheses of [F3] have now been verified, so it gives a unique state-field map with field truncation, vacuum/creation, translation and locality, extending the displayed generating fields.

This state-field map is exactly the descendant prescription in the definition. For a current $J^v$ and a field $B$, expand the two binomial expressions in [F4] in its $(-n)$th product, for $n\ge 1$. Their residue separates the original current modes with $m<0$ from those with $m\ge 0$ and gives $(J^v{)}_{(-n)}B=\frac{{\mathrm{\partial }}_z^{n-1}{J}_{-}^v}{(n-1)!}B+B\frac{{\mathrm{\partial }}_z^{n-1}{J}_{+}^v}{(n-1)!}\mathrm{.}$ By the state transport in [F3], its creative state is $a_{-n}^{v}b$ if $B=Y(b,z)$. Induction on the number of oscillator factors proves the displayed recursive state-field formula, and uniqueness makes it independent of how an oscillator monomial is ordered. In particular all descendant fields inherit the vertex-algebra axioms. The transported residue products give the vertex-algebra iterated products; associativity here is the formal vertex-algebra product convention, not equality of unexpanded products in incompatible radial orders.

The stress tensor is the reconstructed field of $\omega =\frac{1}{2}{\sum }_i(a_{-1}^{e_i}{)}^2\mathbf{1}$, for an orthonormal basis $e_i$. Its modes are $L_m=\frac{1}{2}{\sum }_i{\sum }_{k\in \mathbb{Z}}:a_k^{e_i}{a}_{m-k}^{e_i}:$, with every negative-index oscillator placed to the left; each sum is finite on a state. At $m=0,-1$ this gives $L_0=\frac{1}{2}{\sum }_i(a_0^{e_i}{)}^2+{\sum }_i{\sum }_{n\ge 1}{a}_{-n}^{e_i}{a}_n^{e_i},L_{-1}={\sum }_i{a}_{-1}^{e_i}{a}_0^{e_i}+{\sum }_i{\sum }_{n\ge 1}{a}_{-n-1}^{e_i}{a}_n^{e_i}=\mathsf{T}\mathrm{.}$ Thus $\mathrm{\mid }\lambda ⟩$ has weight $⟨\lambda ,\lambda ⟩\mathrm{/}2$ and $a_{-n}^v$ adds $n$. The current contraction is $J^{e_i}(z)J^{e_j}(w)\sim {\delta }_{ij}(z-w{)}^{-2}$. In $T(z)T(w)$ the two double contractions for each $i$ give $\frac{1}{2}(z-w{)}^{-4}$; the four single contractions, including their first Taylor term at $w$, give $2T(w)(z-w{)}^{-2}+{\mathrm{\partial }}_{w}T(w)(z-w{)}^{-1}$. The fourth-order coefficient is therefore $r\mathrm{/}2$, so $c=r$. This contraction calculation uses only the oscillator rank, not discreteness of the momentum labels. It gives the stated values 16 and 10 at those ranks.