Full module Jacobi identity for a square-zero screening deformation

Truncation in $A$ makes $v_{(n)}a=0$ for sufficiently large $n$ for each fixed $a$, so $D(x)a\in I[x^{-1}]$. Every mode of $v$ annihilates $I$ by $Y(I,z)I=0$, and therefore $D(x)I=0$ and $D(x)D(y)a=0$. The vacuum creation identity $Y(v,z)\mathbf{1}\in A\text{\hspace{0.17em}⁣}\left[\left[z\right]\right]$ gives $v_{(n)}\mathbf{1}=0$ for $n\ge 0$, whence $\mathrm{\Delta }(x)\mathbf{1}=\mathbf{1}$. Each deformed field is a finite sum of monomials in $x^{-1}$ times module fields; it is thus linear in $a$ and lower truncated on every vector of $M$. The vacuum acts as the identity by [F1].

Taking ${\mathrm{Res}}_{z_0}$ of the Given adjoint Jacobi identity and then ${\mathrm{Res}}_{z_1}{z}_1^n$, for an integer $n\ge 0$, gives $[v_{(n)},Y(a,z)]={\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)z^{n-j}Y(v_{(j)}a,z)$. Indeed, in the right-hand formal delta one substitutes $z_1=z+z_0$, whose $n$th power is the finite binomial sum; the residue in $z_0$ selects $v_{(j)}a$. For $a\in S$ the $j=0$ summand is zero. Thus, with $c_n=(-1{)}^{n+1}\mathrm{/}n$, coefficientwise in $x^{-1}$ one obtains $[D(x),Y(a,z)]=Y(D(x+z)a,z)$, because for each $j\ge 1$ the scalar coefficient is ${\sum }_{n\ge j}{c}_n\left(\genfrac{}{}{0px}{}{n}{j}\right)z^{n-j}{x}^{-n}=\frac{(-1{)}^{j+1}}{j}(x+z{)}^{-j}$. To verify this last equality, put $n=j+k$, use $\left(\genfrac{}{}{0px}{}{j+k}{j}\right)\mathrm{/}(j+k)=\left(\genfrac{}{}{0px}{}{j+k-1}{k}\right)\mathrm{/}j$, and expand $(1+z\mathrm{/}x{)}^{-j}={\sum }_{k\ge 0}(-1{)}^k\left(\genfrac{}{}{0px}{}{j+k-1}{k}\right)(z\mathrm{/}x{)}^k$.

Apply step 2.1 to $b\in A$. Since $D(x+z)a\in I((x^{-1}))\left[\left[z\right]\right]$ and $D(x)b\in I[x^{-1}]$, square-zero multiplication gives $Y(D(x+z)a,z)D(x)b=0$. Consequently the commutator identity rearranges to the exact intertwining identity $\mathrm{\Delta }(x)Y(a,z)b=Y(\mathrm{\Delta }(x+z)a,z)\mathrm{\Delta }(x)b$, for $a\in S$. Every coefficient in $x^{-1}$ is a finite sum in step 2.1, so this equality has the stated binomial expansion and does not require evaluating convergent series.

Fix $a,b\in S$ and write $\delta (u)={\sum }_{r\in \mathbb{Z}}{u}^r$. Apply the original module Jacobi identity [F1] coefficientwise to the finite Laurent-polynomial inputs $\mathrm{\Delta }(x_1)a$ and $\mathrm{\Delta }(x_2)b$. Its left-hand side is exactly $x_0^{-1}\delta ((x_1-x_2)\mathrm{/}{x}_0){\tilde{Y}}_M(a,x_1){\tilde{Y}}_M(b,x_2)-x_0^{-1}\delta ((x_2-x_1)\mathrm{/}(-x_0)){\tilde{Y}}_M(b,x_2){\tilde{Y}}_M(a,x_1)$. Its right-hand side is $x_2^{-1}\delta ((x_1-x_0)\mathrm{/}{x}_2)Y_M(Y(\mathrm{\Delta }(x_1)a,x_0)\mathrm{\Delta }(x_2)b,x_2)$. Formal-delta substitution replaces $x_1$ here by $x_2+x_0$, expanded in nonnegative powers of $x_0$; for a Laurent monomial this follows from $x_1^r\delta ((x_1-x_0)\mathrm{/}{x}_2)=(x_2+x_0{)}^r\delta ((x_1-x_0)\mathrm{/}{x}_2)$, for every integer $r$. Step 3.1 then changes the inner state to $\mathrm{\Delta }(x_2)Y(a,x_0)b$. Since $S$ is a vertex subalgebra, the coefficients of $Y(a,x_0)b$ lie in $S$, so the resulting right-hand side is precisely $x_2^{-1}\delta ((x_1-x_0)\mathrm{/}{x}_2){\tilde{Y}}_M(Y(a,x_0)b,x_2)$. This is the full module Jacobi identity. Together with the truncation and vacuum verification in step 1.1 it proves every required module axiom.