Mutually local creative generators determine a state–field map

For fields $A(z),B(z)$ and any integer $n$, the residue definition in [F1] gives $(A_{(n)}B{)}_{(m)}v={\sum }_{j\ge 0}(-1{)}^j\left(\genfrac{}{}{0px}{}{n}{j}\right)A_{(n-j)}{B}_{(m+j)}v-(-1{)}^n{\sum }_{j\ge 0}(-1{)}^j\left(\genfrac{}{}{0px}{}{n}{j}\right)B_{(m+n-j)}{A}_{(j)}v$ by expanding ${\iota }_{z,w}(z-w{)}^n$ and ${\iota }_{w,z}(z-w{)}^n$ and taking the $z^{-1}{w}^{-m-1}$ coefficient. For fixed $m,v$, the first sum is finite because $B_{(m+j)}v=0$ for large $j$, and the second is finite because $A_{(j)}v=0$ for large $j$. For lower truncation, fix $n,v$ and choose $J\ge 0$ with $A_{(j)}v=0$ for $j\ge J$. Choose $b_v$ with $B_{(r)}v=0$ for $r\ge b_v$, and for each $0\le j<J$ choose $b_j$ with $B_{(r)}{A}_{(j)}v=0$ for $r\ge b_j$. Whenever $m\ge b_v$ and $m\ge b_j-n+j$ for every $j<J$, both sums vanish termwise. These finitely many conditions give a cutoff, including $J=0$ when the second family is empty. Thus every $n$th product is again a field.

(Dong locality.) Suppose $A(z),B(x),C(w)$ are pairwise local, and choose a common positive integer locality order $r$ large enough that $r+n>0$. Put $D_n(x,w)={\iota }_{x,w}(x-w{)}^{n}B(x)C(w)-{\iota }_{w,x}(x-w{)}^{n}C(w)B(x)$, so $B_{(n)}C={\mathrm{Res}}_x{D}_n(x,w)$ by [F1]. Expand the commutator residue using $(z-w{)}^{4r}=(z-w{)}^r{\sum }_{i=0}^{3r}\left(\genfrac{}{}{0px}{}{3r}{i}\right)(z-x{)}^i(x-w{)}^{3r-i}$. For each $i$, set $Q_i=(z-w{)}^r(z-x{)}^i(x-w{)}^{3r-i}{D}_n(x,w)$. If $i\le r$, then $3r-i+n\ge 2r+n\ge r+1$. The two binomial expansions of $(x-w{)}^{3r-i+n}$ are therefore the same polynomial, so $Q_i$ is that polynomial times $(z-w{)}^r(z-x{)}^i[B(x),C(w)]$ and is zero by locality of $B,C$. If $i>r$, $(z-x{)}^i$ contains a locality factor for $A,B$ and $(z-w{)}^r$ contains one for $A,C$. These scalar factors commute with the fields; moving $A(z)$ past $B(x)$ and then $C(w)$, or past $C(w)$ and then $B(x)$, shows $[A(z),Q_i]=0$ for each of the two ordered products in $D_n$. Thus $(z-w{)}^{4r}[A(z),(B_{(n)}C)(w)]={\sum }_{i=0}^{3r}\left(\genfrac{}{}{0px}{}{3r}{i}\right){\mathrm{Res}}_x[A(z),Q_i]=0$. If $(z-w{)}^r[A(z),B(w)]=0$, differentiating and multiplying by $z-w$ gives $(z-w{)}^{r+1}[{\mathrm{\partial }}_{z}A(z),B(w)]=0$; differentiating in $w$ gives the same assertion for a derivative of the second field. A derivative of a field remains a field because its mode coefficients are scalar multiples of shifted modes, so derivatives and linear combinations preserve pairwise locality. Hence the closure of $\mathcal{S}$ under linear combinations, derivatives, and all $n$th products is pairwise local.

If $A(z)\mathbf{1}=e^{zT}a$ and $B(z)\mathbf{1}=e^{zT}b$, then $[T,A(z)]={\mathrm{\partial }}_{z}A(z)$ implies $A(z)e^{wT}=e^{wT}A(z-w)$ by differentiating both sides as formal power series in $w$. In the residue defining $A(z{)}_{(n)}B(w)\mathbf{1}$, the second term has no $z^{-1}$ coefficient because ${\iota }_{w,z}(z-w{)}^n$ and $e^{zT}$ contain only nonnegative powers of $z$. In the first term write $A(u)b={\sum }_q{A}_{(q)}b{u}^{-q-1}$, where only finitely many $q\ge 0$ contribute by field truncation. For every integer $s$, ${\mathrm{Res}}_z{\iota }_{z,w}(z-w{)}^s$ is $1$ when $s=-1$ and $0$ otherwise, since its powers of $z$ are $s-j$ with $j\ge 0$. Thus ${\mathrm{Res}}_z{\iota }_{z,w}(z-w{)}^{n}A(z-w)b={\sum }_q{A}_{(q)}b{\mathrm{Res}}_z{\iota }_{z,w}(z-w{)}^{n-q-1}=A_{(n)}b$, because only $q=n$ contributes. The coefficientwise residue is therefore finite and $(A(z{)}_{(n)}B(w))\mathbf{1}=e^{wT}{A}_{(n)}b$. Likewise, if $A$ and $B$ are translation-covariant, then $A_{(n)}B$ is translation-covariant: in its residue, $[T,-]$ differentiates both fields, and ${\mathrm{\partial }}_z+{\mathrm{\partial }}_w$ kills $(z-w{)}^n$ in either binomial expansion; the remaining total $z$-derivative has zero residue. Derivatives preserve both creativity and covariance.

Let $\mathcal{F}$ be the linear span of ${\mathrm{id}}_V$ and all fields obtained from $\mathcal{S}$ by finitely many derivatives and $n$th products. By steps 1.1–1.2 its elements are fields and are pairwise local; by step 2.1 they are translation-covariant and creative, so each has a unique state $s(F)\in V$ with $F(z)\mathbf{1}=e^{zT}s(F)$. Step 2.1 gives $s(A_{(n)}B)=A_{(n)}s(B)$. Starting with $s({\mathrm{id}}_V)=\mathbf{1}$, this identity and the assumed negative-mode spanning condition show that $s:\mathcal{F}\to V$ is surjective.

The state map $s$ is injective. If $H\in \mathcal{F}$ has $s(H)=0$, then $H(z)\mathbf{1}=0$. For arbitrary $b\in V$, choose $G\in \mathcal{F}$ with $s(G)=b$ by step 3.1. Locality gives $(z-w{)}^M[H(z),G(w)]=0$ for some $M$. Apply it to $\mathbf{1}$ and use $G(w)\mathbf{1}=e^{wT}b$; the reversed term vanishes because $H(z)\mathbf{1}=0$. Setting $w=0$ yields $z^{M}H(z)b=0$, and multiplication by $z^M$ is injective on formal distributions. Thus $H(z)b=0$ for every $b$, so $H=0$.

Since $s$ is bijective, define $Y(a,z)=s^{-1}(a)(z)$. Step 1.1 gives lower truncation; the state of ${\mathrm{id}}_V$ is $\mathbf{1}$, so injectivity gives $Y(\mathbf{1},z)={\mathrm{id}}_V$; creativity gives $Y(a,z)\mathbf{1}=e^{zT}a$; translation covariance and locality follow from the defining properties of $\mathcal{F}$. These are exactly the axioms in [F2], so this is a vertex-algebra state–field map extending $\mathcal{S}$.

For any other state–field map extending $\mathcal{S}$, the difference $H(z)$ from $Y(a,z)$ is a field with $H(z)\mathbf{1}=0$ and is local with every element of $\mathcal{F}$: it is local with generators by the vertex-algebra locality axiom, and repeated application of step 1.2 preserves locality under derivatives and $n$th products. For any $b\in V$, choose $G\in \mathcal{F}$ with $G(w)\mathbf{1}=e^{wT}b$ by step 3.1. Applying the locality identity $[H(z),G(w)]=0$ after multiplication by a sufficient power of $z-w$ to $\mathbf{1}$, then setting $w=0$, gives $z^{M}H(z)b=0$; injectivity of multiplication by $z^M$ gives $H(z)b=0$. This holds for every $b$, so $H=0$ and the extension is unique.