Zhu quotient algebra and admissible top-level action

For homogeneous $a\in V_r$ set $C_{m,n}(a,b)={\mathrm{Res}}_z(1+z{)}^{r+n}{z}^{-m-2}Y(a,z)b$, $m\ge n\ge 0$. We prove each lies in O(V). The case $C_{0,0}$ is its defining generator. Integration by parts with F5 gives $C_{m,0}(Ta,b)=(m+2)C_{m+1,0}(a,b)+(m+1-r)C_{m,0}(a,b)$: differentiate $(1+z{)}^{r+1}{z}^{-m-2}$ and expand its extra factor 1+z. Thus if all $C_{m,0}$ lie in O(V), the same is true at m+1, since m+2 is invertible in C and Ta has weight r+1 when nonzero. Induction gives the assertion for every m. Finally $C_{m,n}={\sum }_{i=0}^n\left(\genfrac{}{}{0px}{}{n}{i}\right)C_{m-i,0}$ by expanding $(1+z{)}^n$, and $m-i\ge 0$. This proves the full higher-residue relation with no ideal property assumed.

For homogeneous $a\in V_r,b\in V_s$, put $D=T+L(0)$, a linear operator on V. Creation gives $Da=a\circ \mathbf{1}\in O(V)$ for every homogeneous a and hence $D(V)\subseteq O(V)$. The derivative property and integration by parts give $(Ta)\ast b=a\circ b-r(a\ast b)$ for $a\in V_r$, so $(Da)\ast b=a\circ b$ exactly. We also have the exact derivation identity $D(a\ast b)=(Da)\ast b+a\ast (Db)$. To verify it rather than assume it, expand $a\ast b={\sum }_i\left(\genfrac{}{}{0px}{}{r}{i}\right)a_{(i-1)}b$. The L(0)-weight of its ith term is r+s-i, while translation of the same sum is $T(a\ast b)=a\ast Tb+a\circ b-r{\mathrm{Res}}_z(1+z{)}^{r-1}{z}^{-1}Y(a,z)b$. The elementary identity $(r-i)\left(\genfrac{}{}{0px}{}{r}{i}\right)=r\left(\genfrac{}{}{0px}{}{r-1}{i}\right)$ makes the remaining weight sum cancel that last residue and gives the asserted derivation formula; for $r=0$ the cancellation reads $0=0$. Finally vacuum and creation give $\mathbf{1}\ast a=a$ and $a\ast \mathbf{1}=a$ exactly, since only their constant power-series coefficients contribute to the star residue.

Let $a\in V_r,b\in V_s$ and c be homogeneous. Weight bookkeeping in F5 and the finite star sum give $(a\ast b)\ast c={\mathrm{Res}}_y{\mathrm{Res}}_t\frac{(1+y+t{)}^r(1+y{)}^s}{ty}Y(Y(a,t)b,y)c\mathrm{.}$ Apply F5's residue identity with $m=-1$ and $P(x,y)=(1+x{)}^r(1+y{)}^s\mathrm{/}y$. Its first product expands $(x-y{)}^{-1}={\sum }_{j\ge 0}{y}^j{x}^{-j-1}$; the $j=0$ term is $a\ast (b\ast c)$. The reversed expansion is $(x-y{)}^{-1}=-{\sum }_{j\ge 0}{x}^j{y}^{-j-1}$, and the minus in the residue identity therefore makes its contribution positive. Define $E_j={\mathrm{Res}}_y(1+y{)}^s{y}^{j-1}Y(b,y)c$ and $F_j={\mathrm{Res}}_x(1+x{)}^r{x}^{j}Y(a,x)c$. Exact coefficient extraction yields $(a\ast b)\ast c-a\ast (b\ast c)={\sum }_{j\ge 1}{C}_{j-1,0}(a,E_j)+{\sum }_{j\ge 0}{C}_{j,0}(b,F_j)\mathrm{.}$ These sums are finite: $E_j={\sum }_{i=0}^s\left(\genfrac{}{}{0px}{}{s}{i}\right)b_{(j+i-1)}c$ and $F_j={\sum }_{i=0}^r\left(\genfrac{}{}{0px}{}{r}{i}\right)a_{(j+i)}c$ vanish for large j by truncation. Each summand lies in O(V) by step 1.1, even when $E_j,F_j$ are not homogeneous. Hence the associator is in O(V). Bilinearity extends this additive congruence to all inputs.

Right ideal stability follows first, without multiplying an unproved congruence: $(a\circ b)\ast c=((Da)\ast b)\ast c\equiv (Da)\ast (b\ast c)=a\circ (b\ast c)\in O(V)$, where the only congruence is the associator from step 2.1 applied to the actual inputs Da,b,c. Thus $O(V)\ast V\subseteq O(V)$. Next the exact derivation relation in step 1.2 gives $c\ast (Da)=D(c\ast a)-c\circ a\in O(V)$. Applying step 2.1 to c,Da,b yields $c\ast (a\circ b)=c\ast ((Da)\ast b)\equiv (c\ast (Da))\ast b$, which is in O(V) by the right ideal stability just proved. Thus $V\ast O(V)\subseteq O(V)$ too. Linearity across circle generators proves both ideal assertions in the Statement. This order of proof avoids assuming either ideal assertion in order to prove the other.

Step 3.1 makes the product on V/O(V) well-defined on both representatives. The associator relation of step 2.1 descends to associativity, and step 1.2 gives its two-sided unit $[\mathbf{1}]$. These conclusions use only finite formal residue identities, not a citation to the algebra-structure theorem.

For $a\in V_r$ and $m\in M(0)$, F3 implies $a_{(n)}^{M}m=0$ whenever $n\ge r$, since its target grade is negative. Hence $M(0)\subseteq \mathrm{\Omega }(M)$. F3 also gives $o_M(a)M(0)\subseteq M(0)$. F6 now supplies $o_M(O(V))M(0)=0$ and $o_M(a\ast b)m=o_M(a)o_M(b)m$, by its explicit circle/star iterate coefficient cancellations. Its vacuum identity makes the action unital. Thus $[a]\mapsto o_M(a){\mathrm{\mid }}_{M(0)}$ is the stated well-defined representation of the associative algebra from step 4.1. No dimension bound or semisimplicity condition is used.