Associative Zhu algebra for resonant induction

For $a\in V_r$, $b\in V$, and integers $m\ge n\ge 0$, put $C_{m,n}(a,b)={\mathrm{Res}}_z(1+z{)}^{r+n}{z}^{-m-2}Y(a,z)b$. We first show $C_{m,n}(a,b)\in O(V)$. The case $m=n=0$ is the defining circle relation. For $n=0$, integration by parts and translation give $C_{m,0}(Ta,b)=(m+2)C_{m+1,0}(a,b)+(m+1-r)C_{m,0}(a,b)\mathrm{.}$ Induction on $m$, simultaneously for every homogeneous state $a$, proves all $C_{m,0}$ lie in $O(V)$: the left side is a higher circle residue for $Ta$, already covered at index $m$, and $m+2\mathrm{\ne }0$. Finally $C_{m,n}={\sum }_{i=0}^n\left(\genfrac{}{}{0px}{}{n}{i}\right)C_{m-i,0}$; each $m-i\ge 0$, so every $C_{m,n}$ is in $O(V)$.

Define $D=T+L(0)$ on $V$. Creation gives $Da=a\circ \mathbf{1}\in O(V)$ for homogeneous $a$. Translation and integration by parts give $(Ta)\ast b=a\circ b-r(a\ast b)$, hence $(Da)\ast b=a\circ b$. To verify $D(a\ast b)=(Da)\ast b+a\ast (Db)$, put $C_i=\left(\genfrac{}{}{0px}{}{r}{i}\right)$ and expand the left side as ${\sum }_{i=0}^r{C}_i(a_{(i-1)}Tb-(i-1)a_{(i-2)}b+(r+s-i)a_{(i-1)}b)\mathrm{.}$ The right side has the same $Tb$ terms; its remaining terms are ${\sum }_{i=0}^r{C}_i(r+s)a_{(i-1)}b-{\sum }_{i=0}^{r+1}\left(\genfrac{}{}{0px}{}{r+1}{i}\right)(i-1)a_{(i-2)}b\mathrm{.}$ For the coefficient of $a_{(k-1)}b$, equality reduces to $k\left(\genfrac{}{}{0px}{}{r}{k}\right)+k\left(\genfrac{}{}{0px}{}{r}{k+1}\right)=k\left(\genfrac{}{}{0px}{}{r+1}{k+1}\right)$, Pascal's identity, with out-of-range binomial coefficients zero. Thus the derivation identity holds term by term, including $r=0$.

For homogeneous $a\in V_r,b\in V_s,c\in V$, set $E_j={\mathrm{Res}}_y(1+y{)}^s{y}^{j-1}Y(b,y)c$ and $F_j={\mathrm{Res}}_x(1+x{)}^r{x}^{j}Y(a,x)c$. The residue identity in [F3], applied to the two finite star sums, yields $(a\ast b)\ast c-a\ast (b\ast c)={\sum }_{j\ge 1}{C}_{j-1,0}(a,E_j)+{\sum }_{j\ge 0}{C}_{j,0}(b,F_j)\mathrm{.}$ Both sums are finite: $E_j={\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{s}{i}\right)b_{(j+i-1)}c$ and $F_j={\sum }_{i=0}^r\left(\genfrac{}{}{0px}{}{r}{i}\right)a_{(j+i)}c$, which vanish for all sufficiently large $j$ by state truncation. By step 1.1 each displayed $C$ belongs to $O(V)$, including when its second argument is inhomogeneous. Therefore the star associator lies in $O(V)$.

Since $(Da)\ast b=a\circ b$, step 2.1 gives $(a\circ b)\ast c=((Da)\ast b)\ast c\equiv (Da)\ast (b\ast c)=a\circ (b\ast c)\in O(V)$, so $O(V)\ast V\subseteq O(V)$. By step 1.2, $c\ast (Da)=D(c\ast a)-(Dc)\ast a\in O(V)$, because $D(V)\subseteq O(V)$ and $(Dc)\ast a=c\circ a$. A second use of the associator gives $c\ast (a\circ b)=c\ast ((Da)\ast b)\equiv (c\ast (Da))\ast b\in O(V)$. Hence $V\ast O(V)\subseteq O(V)$ as well. The associator descends to the quotient, so its product is associative. The finite residue formulas give $\mathbf{1}\ast a=a=a\ast \mathbf{1}$, proving the unit assertion.

Skew-symmetry $Y(b,z)a=e^{zT}Y(a,-z)b$, followed by $Ta\equiv -ra(\mathrm{m}\mathrm{o}\mathrm{d}O(V))$, gives $[b\ast a]={\sum }_{\mathrm{\ell }\ge 0}\left(\genfrac{}{}{0px}{}{r-1}{\mathrm{\ell }}\right)[a_{(\mathrm{\ell }-1)}b]\mathrm{.}$ Indeed, after expanding the exponential and replacing $T^j(a_{(n)}b)$ by $(-1{)}^j(r+s-n-1)\cdots (r+s-n+j-2)(a_{(n)}b)$, the coefficient of $a_{(\mathrm{\ell }-1)}b$ is the coefficient of $x^{\mathrm{\ell }}$ in $(1+x{)}^{r+s-1}(1-x\mathrm{/}(1+x){)}^s=(1+x{)}^{r-1}$. Subtracting the displayed formula for $[b\ast a]$ from the star sum $[a\ast b]={\sum }_{\mathrm{\ell }=0}^r\left(\genfrac{}{}{0px}{}{r}{\mathrm{\ell }}\right)[a_{(\mathrm{\ell }-1)}b]$ and reindexing $\mathrm{\ell }=j+1$ leaves the claimed commutator formula. Terms with negative binomial lower index vanish, and state truncation makes the sum finite.

Apply the commutator formula of step 4.1 with $a=\omega$, of weight two, and homogeneous $b\in V_s$. Only $j=0,1$ contribute, so $[\omega ]\ast [b]-[b]\ast [\omega ]=[Tb]+[L(0)b]=(-s+s)[b]=0,$ using $[Tb]=-s[b]$ from step 3.1 and $L(0)b=sb$. Linearity gives centrality for every Zhu class.