Jacobi identities for the rank-one square-zero lattice extension

Give $P$ oscillator degree $\mathrm{deg}{x}_n=n$. On a fixed polynomial, annihilation operators have bounded total index, and, for a fixed output degree, so do creation operators. Hence every coefficient in each normal field is a finite sum. In a fixed state charge and input charge, the exponential has a fixed integer initial power of $z$ and only a polynomial in $z^{-1}$ from annihilation; negative oscillator modes add only nonnegative powers after their divided derivatives. Nonnegative oscillator parts on the right have indices bounded by the input degree. This gives a lower bound on all powers of $z$ and proves truncation on both spaces, as required in [F4]. The empty oscillator monomial of charge zero gives $Y(\mathbf{1},z)=I$. For $T$ as displayed, direct differentiation gives $[T,{\alpha }_n]=-n{\alpha }_{n-1}$, including $[T,{\alpha }_1]=-{\alpha }_0$ and $[T,{\alpha }_0]=0$. Applying the product rule to the two exponential factors and the charge shift gives $[T,{\mathcal{E}}_r(z)]={\mathrm{\partial }}_z{\mathcal{E}}_r(z)$ on $\mathcal{A}$; the zero-mode term follows from the difference $(t+r)x_1-t{x}_1=r{x}_1$ across the shift. To check translation on every normal field, split $\alpha (z)={\alpha }^{-}(z)+{\alpha }^{+}(z)$ into negative and nonnegative oscillator modes. Reindexing $[T,{\alpha }_n]=-n{\alpha }_{n-1}$ separately on $n<0$ and $n\ge 0$ gives $[T,{\alpha }^{\pm }(z)]={\mathrm{\partial }}_z{\alpha }^{\pm }(z)$; differentiation preserves each split, so the same holds for every divided derivative $b_n^{\pm }$. The defining normal-order sum is finite and remains in the same order under these commutators. The commutator derivation rule and the ordinary product rule therefore give $[T,:b_{n_1}\cdots b_{n_{\mathrm{\ell }}}{\mathcal{E}}_r:]={\mathrm{\partial }}_z:b_{n_1}\cdots b_{n_{\mathrm{\ell }}}{\mathcal{E}}_r:$. On the vacuum each field has only nonnegative powers and constant term equal to its state: the creation exponential has constant term one, and the negative part of $b_n(z)$ is ${\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{n+k-1}{k}\right)x_{n+k}{z}^k$. With $T\mathbf{1}=0$, translation covariance gives the coefficient recursion $[z^{k+1}]Y(a,z)\mathbf{1}=T[z^k]Y(a,z)\mathbf{1}\mathrm{/}(k+1)$, so creation is $Y(a,z)\mathbf{1}=e^{zT}a$. These are the state-field axioms in [F4].

The oscillator commutator is $[{\alpha }_m,{\alpha }_n]=2pm{\delta }_{m,-n}I$. Moving annihilation exponentials across creation exponentials therefore contributes $\exp (-2prs{\sum }_{n\ge 1}(w\mathrm{/}z{)}^n\mathrm{/}n)=(1-w\mathrm{/}z{)}^{2prs}$. Moving the charge shift across $z^{r{\alpha }_0}$ contributes $z^{2prs}$. Thus ${\mathcal{E}}_r(z){\mathcal{E}}_s(w)={\iota }_{z,w}(z-w{)}^{2prs}{B}_{r,s}(z,w)$, where $B_{r,s}(z,w)=\exp \text{\hspace{0.17em}⁣}({\sum }_{n\ge 1}\frac{{\alpha }_{-n}}{n}(r{z}^n+s{w}^n))\exp \text{\hspace{0.17em}⁣}(-{\sum }_{n\ge 1}\frac{{\alpha }_n}{n}(r{z}^{-n}+s{w}^{-n}))e_{r+s}{z}^{r{\alpha }_0}{w}^{s{\alpha }_0}\mathrm{.}$ Here $e_{r+s}$ includes the sum of the two state-sector indices and is zero if that sum plus the input-sector index exceeds one. The scalar cocycle is one. Reversing the product gives the other expansion of the same expression, since $2prs$ is even. If both states have sector one, both ordered products and all products of their states are zero; the same is true if the input sector makes the total sector exceed one. Thus all subsequent formulas may be checked on allowed sectors only.

For states $u={\prod }_{i=1}^k{x}_{a_i}{e}^{r\alpha }$ and $b={\prod }_{j=1}^l{x}_{d_j}{e}^{s\alpha }$, form the following finite sum $G(z,w)$. Start with $(z-w{)}^{2prs}{B}_{r,s}(z,w)$ and the oscillator factors $b_{a_i}(z),b_{d_j}(w)$. Each oscillator may remain unmatched, may be paired with one oscillator on the other side, or may be paired with the exponential on the other side. No oscillator may be used twice; several distinct oscillators may pair with the same exponential. A left-right oscillator pair contributes $2p(-1{)}^{a_i-1}(a_i+d_j-1)!\mathrm{/}((a_i-1)!(d_j-1)!)(z-w{)}^{-a_i-d_j}$. A left oscillator paired with the right exponential contributes $2ps(-1{)}^{a_i-1}(z-w{)}^{-a_i}$; a right oscillator paired with the left exponential contributes $-2pr(z-w{)}^{-d_j}$. Normally order all unmatched oscillators with $B_{r,s}$. Sum these products over all choices, counting the oscillator factors by their original indices even if their orders coincide. These contraction factors follow respectively by differentiating $2p\mathrm{/}(z-w{)}^2$, $2ps\mathrm{/}(z-w)$ and $-2pr\mathrm{/}(z-w)$ with the prescribed divided derivatives. Repeatedly applying $XY=YX+[X,Y]$ when moving an annihilator across a creator or an exponential gives exactly these choices: an oscillator-oscillator commutator removes the two oscillators, while commuting with an exponential removes only that oscillator and leaves the exponential intact. All contractions are scalar. Hence the two ordered products are precisely ${\iota }_{z,w}G(z,w)$ and ${\iota }_{w,z}G(z,w)$. This also proves locality, since only finitely many contraction pole orders occur and the unmatched normal expression is the same in the two orders.

Apply the same expression to the vacuum on $\mathcal{A}$ and set $w=0$ to compute $Y_{\mathcal{A}}(u,\zeta )b$. The scalar part is the contraction product of step 3.1 times ${\zeta }^{2prs}$. The unmatched left oscillator $b_a$ becomes ${\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{a+k-1}{k}\right)x_{a+k}{\zeta }^k$, the unmatched right oscillator becomes $x_d$, and the exponential part becomes $\exp ({\sum }_{n\ge 1}r{x}_n{\zeta }^n\mathrm{/}n)e^{(r+s)\alpha }$. This gives every state coefficient, including regular powers; each coefficient is finite. Apply $Y_{\mathcal{M}}(-,w)$ coefficientwise. The oscillator substitution is $b_a(w+\zeta )={\sum }_{k\ge 0}\left(\genfrac{}{}{0px}{}{a+k-1}{k}\right){\zeta }^k{b}_{a+k}(w)$. For the exponential substitution the required identity is $B_{r,s}(w+\zeta ,w)=:\exp \text{\hspace{0.17em}⁣}({\sum }_{n\ge 1}\frac{r{\zeta }^n}{n}{b}_n(w)){\mathcal{E}}_{r+s}(w):\mathrm{.}$ To verify it, expand each oscillator mode in $b_n(w)={\mathrm{\partial }}_w^{n-1}\alpha (w)\mathrm{/}(n-1)!$: summing ${\zeta }^n\mathrm{/}n!$ times its $(n-1)$st derivative is its formal integral from $w$ to $w+\zeta$. A negative mode gives ${\alpha }_{-k}((w+\zeta {)}^k-w^k)\mathrm{/}k$, a positive mode gives $-{\alpha }_k((w+\zeta {)}^{-k}-w^{-k})\mathrm{/}k$, and the zero mode gives ${\alpha }_0\log (1+\zeta \mathrm{/}w)$. Adding the exponent of ${\mathcal{E}}_{r+s}(w)$ yields exactly the two exponentials and zero-mode powers in $B_{r,s}(w+\zeta ,w)$. The logarithm here is its formal power series; exponentiation gives the ordinary binomial power $(1+\zeta \mathrm{/}w{)}^{r{\alpha }_0}$, so no logarithmic field is introduced. Consequently the iterate is exactly $Y_{\mathcal{M}}(Y_{\mathcal{A}}(u,\zeta )b,w)={\iota }_{w,\zeta }G(w+\zeta ,w)$. The calculation with $\mathcal{M}$ replaced by $\mathcal{A}$ is identical.

Fix an input polynomial, its charge and sector, and project the output polynomial to oscillator degrees at most $D$. Every normal expression in step 3.1 then has finitely many terms: annihilation indices are bounded by the input degree and creation indices by $D$. The zero-mode factors are integer powers, so the projected $G(z,w)$ is a finite sum of operator coefficients times $z^a{w}^b(z-w{)}^k$ with $a,b,k\in \mathbb{Z}$. For each such monomial coefficient, the formal delta substitution is $z_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{z-w}{z_0}\right){\iota }_{z,w}\text{\hspace{0.17em}⁣}(z^a{w}^b(z-w{)}^k)-z_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{w-z}{-z_0}\right){\iota }_{w,z}\text{\hspace{0.17em}⁣}(z^a{w}^b(z-w{)}^k)=w^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{z-z_0}{w}\right)(w+z_0{)}^a{w}^b{z}_0^k\mathrm{.}$ To verify it, first set $a=b=0$. In the defining delta sums, multiplication by $(z-w{)}^k$ shifts the summation index by $k$ in both ordered expansions, including when $k<0$ because each product retains its prescribed $\iota$ expansion; on the third expansion the substitution $z=w+z_0$ gives $z_0^k$. The remaining factor $w^b$ is common to all three terms. For $z^a$, use $(z-w-z_0)\delta ((z-w)\mathrm{/}{z}_0)=0$ in the first expansion and the same relation in the reversed expansion; on the third expansion it is $(z-z_0-w)\delta ((z-z_0)\mathrm{/}w)=0$. This gives $z^a=(w+z_0{)}^a$ on the delta support for nonnegative $a$ by multiplication, and for negative $a$ by multiplying by the formal inverses, expanded in nonnegative powers of $z_0\mathrm{/}w$. Thus the displayed identity holds coefficientwise for every integer $a,b,k$. Apply it to each finite term of $G$. The two product terms are those of step 3.1, and the iterate is the substitution $G(w+z_0,w)$ established in step 4.1, so the full Jacobi identity holds on the degree projection. Since $D$ was arbitrary, it holds coefficientwise before projection. This proves the module Jacobi identity on $\mathcal{M}$ and the adjoint Jacobi identity on $\mathcal{A}$, for monomial states and hence all states. Together with step 1.1 and locality in step 3.1, the latter establishes the ordinary vertex-algebra axioms in [F4]. The displayed formula for $T$ preserves every charge and sector label, so $T(I)\subseteq I$. The sector rule makes every mode product of an $I$-state with an $\mathcal{A}$-state lie in $I$ and every mode product of two $I$-states vanish. These are exactly the square-zero-ideal conditions in the Statement.