Jacobi and residue calculus from the state-field locality axioms

The family $\mathcal{S}=\{Y(a,z):a\in V\}$ is a family of fields with the properties in [F1]. The creation identity gives $a=Y(a,z{)}_{(-1)}\mathbf{1}$, so its negative modes span $V$; it also makes the state-field map injective, since equality of two fields gives equality of their constant terms on $\mathbf{1}$. Apply [F4] to this indexed family. The resulting state-field map agrees with $Y(a,z)$ on every state $a$, so it is exactly the original map. We derive Jacobi from the following local product and residue calculations. If $A,B$ are fields and $C=A_{(n)}B$ is their $n$th product from [F5], expansion gives $C_{(m)}v={\sum }_{j\ge 0}(-1{)}^j\left(\genfrac{}{}{0px}{}{n}{j}\right)(A_{(n-j)}{B}_{(m+j)}v-(-1{)}^n{B}_{(m+n-j)}{A}_{(j)}v)\mathrm{.}$ For fixed $v$ both sums are finite by lower truncation. Moreover $C$ is a field: choose $J$ with $A_{(j)}v=0$ for $j\ge J$, choose $b$ with $B_{(q)}v=0$ for $q\ge b$, and for each $0\le j<J$ choose $b_j$ with $B_{(q)}{A}_{(j)}v=0$ for $q\ge b_j$. For all $m\ge b$ and $m\ge b_j-n+j$, every term vanishes. These are finitely many cutoffs; if $J=0$, the second family is empty.

Let $A(z),B(x),C(y)$ be pairwise local fields, and choose a common positive locality order $r$ with $r+n>0$. Put $D_n(x,y)={\iota }_{x,y}(x-y{)}^{n}B(x)C(y)-{\iota }_{y,x}(x-y{)}^{n}C(y)B(x)\mathrm{.}$ Use $(z-y{)}^{4r}=(z-y{)}^r{\sum }_{i=0}^{3r}\left(\genfrac{}{}{0px}{}{3r}{i}\right)(z-x{)}^i(x-y{)}^{3r-i}\mathrm{.}$ If $i\le r$, then $3r-i+n\ge r+1$, so the two expansions of $(x-y{)}^{3r-i+n}$ are the same polynomial and the corresponding summand vanishes by the locality of $B,C$. If $i>r$, the factors $(z-x{)}^i$ and $(z-y{)}^r$ permit $A$ to commute past $B$ and $C$, respectively, in either ordered product. Taking ${\mathrm{Res}}_x$ therefore proves $(z-y{)}^{4r}[A(z),(B_{(n)}C)(y)]=0$. Differentiating a locality identity and multiplying once more by its variable difference shows that derivatives of local fields remain local. Thus every derivative and every iterated $n$th product of state fields is local with each state field.

Suppose $A,B$ are creative and translation-covariant with states $a,b$, so $A(x)\mathbf{1}=e^{xT}a$ and $B(y)\mathbf{1}=e^{yT}b$. Translation covariance gives $A(x)e^{yT}=e^{yT}A(x-y)$ by expanding $e^{-yT}A(x)e^{yT}$ in powers of $y$. In the reversed term defining $A_{(n)}B$, both ${\iota }_{y,x}(x-y{)}^n$ and $A(x)\mathbf{1}$ have nonnegative powers of $x$, so its $x$-residue is zero. In the first term the residue is $e^{yT}{\mathrm{Res}}_x{\iota }_{x,y}(x-y{)}^{n}A(x-y)b$. For any integer $d$, the coefficient of $x^{-1}$ in ${\iota }_{x,y}(x-y{)}^d$ is $1$ for $d=-1$ and $0$ otherwise. Expanding $A(x-y)b$ therefore selects precisely $A_{(n)}b$ and proves $(A_{(n)}B)(y)\mathbf{1}=e^{yT}{A}_{(n)}b$. The sum is finite: large positive modes annihilate $b$, and sufficiently negative modes yield nonnegative powers of $x-y$. The same calculation gives $({\mathrm{\partial }}_{y}Y(a,y))\mathbf{1}=e^{yT}Ta$.

A field $H(z)$ local with every state field and satisfying $H(z)\mathbf{1}=0$ must vanish. Indeed, locality with $Y(c,w)$ applied to $\mathbf{1}$ gives $(z-w{)}^{N}H(z)e^{wT}c=0$; the expression has only nonnegative powers of $w$, so setting $w=0$ gives $z^{N}H(z)c=0$. Multiplication by $z^N$ is injective on formal distributions, hence $H(z)c=0$ for every $c$. Apply this to $Y(a{)}_{(m)}Y(b)-Y(a_{(m)}b)$, whose creative state is zero by step 2.2, to obtain $Y(a{)}_{(m)}Y(b)=Y(a_{(m)}b)$. Apply it to ${\mathrm{\partial }}_{z}Y(a,z)-Y(Ta,z)$, using the same locality and creative-state calculation, to obtain $Y(Ta,z)={\mathrm{\partial }}_{z}Y(a,z)$.

Put ${\delta }_y(x)=x^{-1}\delta (y\mathrm{/}x)$. If a formal distribution $R(x,y)$ is killed by $(x-y{)}^L$, its finite delta expansion is $R(x,y)={\sum }_{i=0}^{L-1}{C}_i(y)\frac{{\mathrm{\partial }}_y^i{\delta }_y(x)}{i!},C_i(y)={\mathrm{Res}}_x(x-y{)}^{i}R(x,y)\mathrm{.}$ For $L=1$, write $R={\sum }_{q\in \mathbb{Z}}{r}_q(y)x^q$. The equation $(x-y)R=0$ gives $r_{q-1}(y)=y{r}_q(y)$ for every $q$, so $r_q(y)=y^{-q-1}{r}_{-1}(y)$ and $R={\delta }_y(x){\mathrm{Res}}_{x}R$. For larger $L$, apply this case to $(x-y{)}^{L-1}R$, subtract its top term, and induct. The coefficient formula follows from ${\mathrm{Res}}_x(x-y{)}^j{\mathrm{\partial }}_y^i{\delta }_y(x)=i!{\delta }_{ij}$, which is obtained by differentiating the delta recurrence. Now choose a locality order $N$ for $Y(a,x),Y(b,y)$. For each integer $m$, set $S_m={\iota }_{x,y}(x-y{)}^{m}Y(a,x)Y(b,y)-{\iota }_{y,x}(x-y{)}^{m}Y(b,y)Y(a,x)\mathrm{.}$ For any positive $L$ with $m+L\ge N$, $(x-y{)}^L{S}_m=0$. The coefficient $C_i$ in its delta expansion is the $m+i$th product field, hence $Y(a_{(m+i)}b,y)$ by step 3.1. Therefore $S_m={\sum }_{i\ge 0}Y(a_{(m+i)}b,y)\frac{{\mathrm{\partial }}_y^i{\delta }_y(x)}{i!}\mathrm{.}$ The sum is finite because $a_{(m+i)}b=0$ once $m+i\ge N$. Taylor expansion of the formal delta gives $y^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{x-t}{y}\right)={\sum }_{i\ge 0}{t}^i\frac{{\mathrm{\partial }}_y^i{\delta }_y(x)}{i!},$ so the displayed expression is exactly the $t$-residue of $t^m{y}^{-1}\delta ((x-t)\mathrm{/}y)Y(Y(a,t)b,y)$. The two other $t$-residues are the two ordered products in $S_m$. Thus their coefficients agree for every integer $m$, proving the full Jacobi identity on $V$.

On $M$, use the supplied module Jacobi identity [F3]; on $V$, use step 4.1. Multiplying by $t^m{y}^n$ and taking residues gives the iterate identity, since ${\iota }_{x,y}(x-y{)}^m={\sum }_{j\ge 0}(-1{)}^j\left(\genfrac{}{}{0px}{}{m}{j}\right)x^{m-j}{y}^j,{\iota }_{y,x}(x-y{)}^m={\sum }_{j\ge 0}(-1{)}^{m+j}\left(\genfrac{}{}{0px}{}{m}{j}\right)y^{m-j}{x}^j\mathrm{.}$ The rightmost modes annihilate each fixed vector for large $j$, so the sum is finite. Taking the commutator residues gives the commutator identity; its terms vanish for large $j$ because $a_{(j)}b=0$. Finally write $P(x,y)$ as a finite sum of Laurent monomials $x^a{y}^b$. For each integer $a$, the formal delta substitution is $x^a{y}^{-1}\delta ((x-t)\mathrm{/}y)=(y+t{)}^a{y}^{-1}\delta ((x-t)\mathrm{/}y)$, where $(y+t{)}^a={\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{a}{j}\right)y^{a-j}{t}^j$ for negative $a$ as well as nonnegative $a$. This identity follows coefficientwise from the defining delta series. Multiplying Jacobi by each $x^a{y}^b$ and taking residues therefore gives the stated Laurent-polynomial residue identity.

The commutator formula in step 5.1, applied to ${\omega }_{(1)}=L(0)$, has only the $j=0,1$ terms because $\left(\genfrac{}{}{0px}{}{1}{j}\right)=0$ for $j\ge 2$. If $L(0)a=ra$, then $[L(0),Y(a,z)]=Y(L(0)a,z)+zY(L(-1)a,z)=rY(a,z)+z{\mathrm{\partial }}_{z}Y(a,z),$ using $L(-1)=T$ and step 3.1. Comparing the coefficient of $z^{-p-1}$ gives $[L(0),a_{(p)}]=(r-p-1)a_{(p)}$. Acting on $b\in V_s$ proves the weight rule.