Vertex Algebra Axioms and State–Field Constructions

For $m,n\ge 0$, the loop term $[x,y{]}_{m+n}$ belongs to $\mathfrak{g}[t]$. The central term in [F1] can be nonzero only if $m+n=0$, which for nonnegative $m,n$ forces $m=n=0$ and hence its coefficient $m$ is zero. Thus $\mathfrak{g}[t]$ is closed under brackets in the central extension, and $K$ commutes with it. Bilinearity extends this to all finite sums in $\mathfrak{p}$.

The proposed scalar action sends every bracket in $\mathfrak{g}[t]$ to zero, agrees with the zero bracket between $K$ and $\mathfrak{g}[t]$, and sends $[K,K]=0$ to zero. Hence it is a Lie-algebra representation on the one-dimensional space ${\mathbb{C}}_k$; the scalar $k$ is unrestricted.

Set $\delta (z,w)={\sum }_{r\in \mathbb{Z}}{z}^{-r-1}{w}^r$. Comparing coefficients gives $(z-w)\delta =0$. Differentiating this identity $j$ times in $w$ gives $(z-w){\mathrm{\partial }}_w^j\delta =j{\mathrm{\partial }}_w^{j-1}\delta$ for $j\ge 1$. Also, direct residue extraction gives ${\mathrm{Res}}_z((z-w{)}^r{\mathrm{\partial }}_w^j\delta (z,w)\mathrm{/}j!)=1$ if $r=j$ and $0$ otherwise for $r,j\ge 0$: repeated use of the differentiated identity reduces $r<j$ to a positive derivative of ${\mathrm{Res}}_z\delta =1$, and $r>j$ to a positive power of $(z-w)$ times $\delta$.

If $N=0$, locality says $C=0$. If $N=1$, writing $C={\sum }_{p,q}{c}_{p,q}{z}^{-p-1}{w}^{-q-1}$, the equation $(z-w)C=0$ gives $c_{p+1,q}=c_{p,q+1}$. Hence $c_{p,q}$ depends only on $p+q$, so $C=({\mathrm{Res}}_{z}C)\delta$. This is the required expansion for order one.

Assume $N\ge 2$ and that the expansion is proved for order $N-1$. Since $(z-w{)}^{N-1}((z-w)C)=0$, write $(z-w)C={\sum }_{j=0}^{N-2}{d}_j(w){\mathrm{\partial }}_w^j\delta \mathrm{/}j!$ by induction. The identity in step 1.1 shows that $C-{\sum }_{j=0}^{N-2}{d}_j(w)\frac{{\mathrm{\partial }}_w^{j+1}\delta }{(j+1)!}$ is killed by $z-w$, hence is a multiple $d_0^{\mathrm{\prime }}(w)\delta$ by step 2.1. This gives a finite delta-derivative expansion for $C$. Taking residues after multiplying by $(z-w{)}^r$ and using the residue calculation in step 1.1 identifies its $r$th coefficient as ${\mathrm{Res}}_z((z-w{)}^{r}C)$ and proves uniqueness.

The geometric-series identity $\delta =i_{z,w}(z-w{)}^{-1}-i_{w,z}(z-w{)}^{-1}$ follows by writing the first expansion in nonnegative powers of $w\mathrm{/}z$ and the second in nonnegative powers of $z\mathrm{/}w$. Differentiating $j$ times gives ${\mathrm{\partial }}_w^j\delta \mathrm{/}j!=i_{z,w}(z-w{)}^{-j-1}-i_{w,z}(z-w{)}^{-j-1}$. Substitution in step 2.2 yields the stated singular OPE coefficients and fixes both expansion conventions.

Since ${\mathrm{Res}}_z(z^m\delta (z,w))=w^m$, differentiation gives ${\mathrm{Res}}_z(z^m{\mathrm{\partial }}_w^j\delta \mathrm{/}j!)=\left(\genfrac{}{}{0px}{}{m}{j}\right)w^{m-j}$ for every integer $m$. Apply ${\mathrm{Res}}_z{z}^m{\mathrm{Res}}_w{w}^n$ to step 2.2. The result is $[A_{(m)},B_{(n)}]={\sum }_{j=0}^{N-1}\left(\genfrac{}{}{0px}{}{m}{j}\right)(C_j{)}_{(m+n-j)}$. The sum is finite; the $N=0$ case was handled in step 2.1.

Fix a finite ordered basis $e_1,\dots ,e_d$ of $\mathfrak{g}$ and the countable basis $e_{i,n}=e_i\otimes t^n$ of the loop modes, together with $K$. Fix a total order on this basis that places every mode with $n<0$ before every basis element of $\mathfrak{p}$; within each part use any fixed enumeration. For basis letters $b_j>b_i$, apply the defining rewrite $b_j{b}_i\mapsto b_i{b}_j+[b_j,b_i]$. Order words by length and then inversion number. The swapped word has one fewer inversion and every bracket term has smaller length; each bracket is a finite linear combination of basis letters. Thus every reduction tree is finitely branching and strictly decreases this well-founded pair, so it terminates in a finite linear combination of ordered words.

Reductions at disjoint adjacent pairs commute. The only overlapping reductions occur in a descending triple $xyz$ of basis letters with $x>y>z$. Reducing the left pair first and then swapping the remaining inversions gives $xyz⟶zyx+[y,z]x+y[x,z]+[x,y]z,$ whereas reducing the right pair first gives $xyz⟶zyx+z[x,y]+[x,z]y+x[y,z]\mathrm{.}$ Their difference is $([y,z]x-x[y,z])+(y[x,z]-[x,z]y)+([x,y]z-z[x,y])$. By bilinearity, the two-letter defining relations reduce these three differences respectively to $[[y,z],x]$, $[y,[x,z]]$, and $[[x,y],z]$; their sum is zero by Jacobi. Each bracket expands into finitely many basis letters, and every comparison term has smaller word length than the original triple. Induction on (length, inversion number) therefore joins every overlapping reduction, also inside any fixed left and right word context, so all reduction paths from each word agree.

The resulting linear normal-form map fixes every ordered word and sends each contextual relation $u(b_j{b}_i-b_i{b}_j-[b_j,b_i])v$ to zero, by the uniqueness in step 1.2. This checks all basis pairs: a reversed pair is the negative of its oriented relation by antisymmetry, and an equal pair has bracket zero. Bilinearity then gives the same for arbitrary $x,y\in \widehat{\mathfrak{g}}$, so the map kills the full two-sided ideal $I$ of [F2]. Conversely each reduction changes a word by an element of $I$. Hence ordered monomials span the quotient, and a linear combination of distinct ordered monomials in $I$ would be fixed by the normal-form map and also sent to zero; it must be zero. Ordered monomials are therefore a basis of $U(\widehat{\mathfrak{g}})$.

The negative-mode span ${\mathfrak{n}}_{-}:=\mathfrak{g}\otimes t^{-1}\mathbb{C}[t^{-1}]$ is a Lie subalgebra because sums of two negative indices stay negative and [F1] has no central term for them. The subalgebra $\mathfrak{p}=\mathfrak{g}[t]\oplus \mathbb{C}K$ is established in [F4]. With the order fixed in step 1.1, each PBW word is a negative-mode monomial followed by a $U(\mathfrak{p})$-monomial. Thus multiplication identifies $U({\mathfrak{n}}_{-})\otimes U(\mathfrak{p})$ with $U(\widehat{\mathfrak{g}})$ as a vector space and as a right $U(\mathfrak{p})$-module.

Tensor the right-module decomposition in step 3.1 over $U(\mathfrak{p})$ with ${\mathbb{C}}_k$. It identifies [F3] with $U({\mathfrak{n}}_{-})\otimes {\mathbb{C}}_k$; the negative-mode PBW monomials tensor the nonzero generator of ${\mathbb{C}}_k$ to a basis. The empty monomial is nonzero, proving the claimed basis and vacuum assertion.

For current modes, $D([x_m,y_n])=-(m+n)[x,y{]}_{m+n-1}$, whereas $[D(x_m),y_n]+[x_m,D(y_n)]=-(m+n)[x,y{]}_{m+n-1}-m((m-1)+n)B(x,y){\delta }_{m+n-1,0}K$. If the Kronecker factor is nonzero then $m+n=1$, so its central coefficient $m(m+n-1)$ is zero. Thus the two expressions agree. Brackets with $K$ also satisfy the derivation identity because $K$ is central and $D(K)=0$. Bilinearity proves that $D$ is a Lie derivation on all finite sums.

If $n\ge 1$, then $D(x_n)=-n{x}_{n-1}\in \mathfrak{g}[t]$, while $D(x_0)=0$ and $D(K)=0$. Hence $D(\mathfrak{p})\subseteq \mathfrak{p}$. Extend $D$ to the tensor algebra by the Leibniz rule. It preserves the ideal generated by $uv-vu-[u,v]$, since its derivative is a sum of the same defining relations with $u$ or $v$ replaced by $D(u)$ or $D(v)$. Therefore it descends to a derivation $D_U$ of $U(\widehat{\mathfrak{g}})$.

Define $T(u\otimes 1)=D_U(u)\otimes 1$. This respects the induced-module relation $ua\otimes 1=u\otimes a1$ for $a\in U(\mathfrak{p})$: the Leibniz rule gives the same $D_U(u)a$ term on both sides, and the remaining term $u{D}_U(a)\otimes 1$ equals $u\otimes D_U(a)1=0$ because $D(\mathfrak{p})\subseteq \mathfrak{p}$ and the vacuum character kills $\mathfrak{g}[t]$ and $D(K)=0$. Thus $T$ is well-defined. Since $D_U(1)=0$, $T\mathbf{1}=0$, and for every current mode the defining derivation action gives $[T,x_n]=D(x_n)=-n{x}_{n-1}$.

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.

We prove by induction on the length of a negative-mode monomial $W\mathbf{1}$ that, for fixed $x\in \mathfrak{g}$, $x_{m}W\mathbf{1}=0$ for all sufficiently large $m$. For the empty monomial this holds for every $m\ge 0$, since $\mathfrak{g}[t]$ kills $\mathbf{1}$.

For the induction step write $W\mathbf{1}=y_{-r}{W}^{\mathrm{\prime }}\mathbf{1}$ with $r\ge 1$. By induction choose bounds for $x_m{W}^{\mathrm{\prime }}\mathbf{1}$ and $[x,y{]}_{m-r}{W}^{\mathrm{\prime }}\mathbf{1}$. Take $m$ larger than both bounds and larger than $r$. Then [F2] and [F3] give $x_m{y}_{-r}{W}^{\mathrm{\prime }}\mathbf{1}=y_{-r}{x}_m{W}^{\mathrm{\prime }}\mathbf{1}+[x,y{]}_{m-r}{W}^{\mathrm{\prime }}\mathbf{1}+mkB(x,y){\delta }_{m,r}{W}^{\mathrm{\prime }}\mathbf{1}=0$. The first two terms vanish by the chosen bounds and the central term vanishes because $m>r$. This proves the induction step.

By [F1], any $v$ is a finite linear combination of such monomials. For each summand step 2.1 supplies a bound; the maximum of these finitely many bounds works for $v$. This is exactly the restrictedness condition in [F4], with no energy grading or unitarity assumed.