Reconstruction and uniqueness from creative local generating fields

For fields $A,B$ put $D_n(z)=A(z{)}_{(n)}B(z)$. Direct expansion of the two binomial series gives $(D_n{)}_{k}v={\sum }_{j\ge 0}(-1{)}^j\left(\genfrac{}{}{0px}{}{n}{j}\right)(A_{n-j}{B}_{k+j}v-(-1{)}^n{B}_{n+k-j}{A}_{j}v)$. Both sums are finite: the first by truncation of $B$ on $v$, the second by truncation of $A$ on $v$. For sufficiently large $k$, the whole first sum vanishes, while the second involves only a fixed finite set of $j$ and vanishes by truncation of $B$ on those finitely many vectors $A_{j}v$. Thus $D_n$ is a field. If $(x-z{)}^N[A(x),B(z)]=0$ and $n\ge N$, the expansions of $(x-z{)}^n$ coincide and locality gives $D_n=0$.

Here is an explicit locality calculation. Suppose $(x-z{)}^N[A(x),B(z)]=(x-w{)}^M[A(x),C(w)]=(z-w{)}^K[B(z),C(w)]=0$. For $n<N$ set $s=N-n-1\ge 0$. Multiply the residue defining $[D_n(z),C(w)]$ by $(z-w{)}^{K+M+s}$ and use $(z-w{)}^{M+s}={\sum }_{j=0}^{M+s}\left(\genfrac{}{}{0px}{}{M+s}{j}\right)(x-w{)}^{M+s-j}(-(x-z){)}^j$. In a term $j\le s$, there remain factors $(x-w{)}^M$ and $(z-w{)}^K$, so $C$ commutes through both $A$ and $B$ and the four commutator terms cancel. In a term $j>s$, the power $n+j\ge N$ is nonnegative, so its two binomial expansions coincide and $(x-z{)}^{n+j}[A(x),B(z)]=0$ cancels the same four terms before taking the residue. All outside factors are polynomials, so these manipulations are finite sums. This proves the stated exponent $K+M+s$.

Suppose in addition $A,B$ are translation covariant. Apply $[T,-]$ to their residue definition, replace commutators by ${\mathrm{\partial }}_{x}A$ and ${\mathrm{\partial }}_{z}B$, and move ${\mathrm{\partial }}_x$ from $A$ by formal integration by parts. Since ${\mathrm{\partial }}_x(x-z{)}^n=-{\mathrm{\partial }}_z(x-z{)}^n$ in each expansion, this gives $[T,D_n(z)]={\mathrm{\partial }}_z{D}_n(z)$. For any translation-covariant field $E$, the Laurent series $E(z)\mathbf{1}$ satisfies ${\mathrm{\partial }}_z(E(z)\mathbf{1})=T(E(z)\mathbf{1})$. It has no negative powers: a least negative power $q$ would have derivative coefficient $q[z^q]E(z)\mathbf{1}$ at degree $q-1$, while the right side has no term there. The coefficient recursion therefore gives $E(z)\mathbf{1}=e^{zT}{E}_{-1}\mathbf{1}$. In particular all such fields are creative. For creative $A,B$, set $a=A_{-1}\mathbf{1}$, $b=B_{-1}\mathbf{1}$. Formula 1.1 at $k=-1$ gives $(D_n{)}_{-1}\mathbf{1}=A_{n}b$: every $A_j\mathbf{1}$ with $j\ge 0$ and every $B_{-1+j}\mathbf{1}$ with $j\ge 1$ vanishes.

We prove Jacobi by coefficient extraction, including negative binomial indices. Fix local $A,B$ with exponent $N$, integers $m,n$, and $D_j=A_{(j)}B$. Then ${\mathrm{Res}}_x{x}^m({\iota }_{x,z}(x-z{)}^{n}A(x)B(z)-{\iota }_{z,x}(x-z{)}^{n}B(z)A(x))={\sum }_{i\ge 0}\left(\genfrac{}{}{0px}{}{m}{i}\right)z^{m-i}{D}_{n+i}(z)$. If $n\ge N$ both sides vanish. Otherwise set $s=N-n-1$. In the Laurent polynomial ring $\mathbb{C}[x^{\pm 1},z^{\pm 1}]$, the difference $x^m-{\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{m}{i}\right)z^{m-i}(x-z{)}^i$ is divisible by $(x-z{)}^{s+1}$: substitute $x=z+h$ and take its Taylor coefficients through order $s$; this argument works for negative $m$ because $x$ and $z$ are units. Multiplying this difference by the two $n$th-product kernels leaves a common Laurent polynomial times $(x-z{)}^N[A(x),B(z)]$, which is zero. The remaining terms give the displayed equality. Its right-hand sum is finite since $D_{n+i}=0$ for $n+i\ge N$.

Let $\mathcal{U}$ be the complex linear span of the identity field and all finite bracketed $n$th-product expressions in the generators. Every member is a translation-covariant field by 1.1 and 2.2. Every pair is mutually local: induct on the sum of the numbers of generators/identity symbols in two bracketed expressions; when one is a product, its two factors have smaller total complexity with the other expression, so 2.1 applies; the base is the given generator locality and the identity commutes with every field. Finite linear combinations remain local by taking the maximum of finitely many exponents. Define $ϵ(E)=E_{-1}\mathbf{1}$. Formula 2.2 gives $ϵ(A_{(n)}B)=A_nϵ(B)$. Repeatedly applying it shows that the image of $ϵ$ contains every spanning word from the Statement, hence is all of $V$.

The state map is injective without a dimension argument. Suppose $E\in \mathcal{U}$ has $ϵ(E)=0$. Then $E(z)\mathbf{1}=0$ by 2.2. If $E(z)v=0$ and $A^i(w)$ is a generator, locality gives $(z-w{)}^{L}E(z)A^i(w)v=(z-w{)}^L{A}^i(w)E(z)v=0$. The left-hand product $E(z)A^i(w)v$ belongs to $V((z))((w))$: its $w$ powers are bounded below, and every $w$ coefficient is a Laurent series in $z$. In this iterated-series module multiplication by $z-w$ is invertible, with inverse $z^{-1}{\sum }_{j\ge 0}(w\mathrm{/}z{)}^j$. Thus $E(z)A_n^{i}v=0$ for every $n$. Induction on word length and the spanning hypothesis give $E(z)=0$ on $V$. Consequently $ϵ:\mathcal{U}\to V$ is a linear bijection.

Define $Y(a,z)={ϵ}^{-1}(a)$. The identity field has state $\mathbf{1}$, so $Y(\mathbf{1},z)=\mathrm{id}$. The generators have states $a^i$, so $Y(a^i,z)=A^i(z)$. Truncation, translation covariance, mutual locality, and $Y(a,z)\mathbf{1}=e^{zT}a$ follow from 3.1 and 2.2. These are all the axioms in [F2]. Moreover $Y(a,z{)}_{(n)}Y(b,z)$ lies in $\mathcal{U}$ and has state $a_{(n)}b$ by 2.2; injectivity gives the claimed state-product identity. Directly in the residue definition, $A_{(-2)}\mathrm{id}={\mathrm{\partial }}_{z}A$, so this identity also gives $Y(Ta,z)={\mathrm{\partial }}_{z}Y(a,z)$.

Write $\delta (u)={\sum }_{r\in \mathbb{Z}}{u}^r$. The equality of 2.3 is exactly the result of taking ${\mathrm{Res}}_{x_0}{x}_0^n{\mathrm{Res}}_{x_1}{x}_1^m$ in the formal-distribution identity $x_0^{-1}\delta ((x_1-z)\mathrm{/}{x}_0)A(x_1)B(z)-x_0^{-1}\delta ((z-x_1)\mathrm{/}(-x_0))B(z)A(x_1)=z^{-1}\delta ((x_1-x_0)\mathrm{/}z){\sum }_j{D}_j(z)x_0^{-j-1}$. In the first two terms the binomial expansions are respectively in nonnegative powers of $z$ and $x_1$; in the third, $(x_1-x_0{)}^r$ is expanded in nonnegative powers of $x_0$. Indeed ${\mathrm{Res}}_{x_1}{x}_1^m{z}^{-1}\delta ((x_1-x_0)\mathrm{/}z)=(z+x_0{)}^m$ expanded in nonnegative powers of $x_0$, giving exactly the right side of 2.3 after the remaining residue. These moments for every integer $m,n$ determine every coefficient in $x_1,x_0$, proving the identity. The subsequent $z$ coefficients are well defined on each vector: they are finite sums by 1.1 and $D_j=0$ for $j\ge N$. Substituting $A=Y(a)$, $B=Y(b)$ and the state-product identity identifies the right side as $z^{-1}\delta ((x_1-x_0)\mathrm{/}z)Y(Y(a,x_0)b,z)$. This is the full Jacobi identity. Explicitly, its coefficients are ${\sum }_{i\ge 0}\left(\genfrac{}{}{0px}{}{m}{i}\right)(a_{(n+i)}b{)}_{(m+k-i)}={\sum }_{i\ge 0}(-1{)}^i\left(\genfrac{}{}{0px}{}{n}{i}\right)(a_{(m+n-i)}{b}_{(k+i)}-(-1{)}^n{b}_{(n+k-i)}{a}_{(m+i)})$ for all integers $m,n,k$; each side is finite after application to a vector.

For uniqueness, any other vertex algebra $Y^{\mathrm{\prime }}$ with the same vacuum, translation and generating fields has every $Y^{\mathrm{\prime }}(v)$ local with the generators and creative with state $v$. By 2.1 and induction it is local with every field in $\mathcal{U}$. For $E=Y^{\mathrm{\prime }}(v)-Y(v)$, the proof of 4.1 uses only locality with the generators, creativity, translation covariance, and the generating-word spanning condition; it therefore applies to $E$ even before knowing that $E\in \mathcal{U}$. Its state is zero, so $E=0$. Thus $Y^{\mathrm{\prime }}=Y$.

Under the additional grading hypotheses, the identity has degree zero. If fields $A,B$ have degrees $d,e$, respectively, the mode formula in 1.1 shows that $(A_{(n)}B{)}_k$ shifts degree by $d+e-n-k-2$, so $A_{(n)}B$ has degree $d+e-n-1$. Every bracketed expression is therefore homogeneous when its generators are homogeneous. Its state has that degree, since the vacuum has degree zero. The span of these homogeneous fields is all $\mathcal{U}$, and the graded bijection $ϵ$ transports their degrees to $V$. Hence the claimed graded truncation convention holds for every $Y(a)$. No analytic bound or conformal vector has been assumed or deduced. Together with 5.1, 6.1, and 6.2 this proves all assertions.