Associative structure of the higher Zhu quotients

For homogeneous $u\in V_r$ and arbitrary $v\in V$ define $R_{m,k}(u,v)={\mathrm{Res}}_z\frac{(1+z{)}^{r+n+k}}{z^{2n+2+m}}Y(u,z)v(m\ge k\ge 0)\mathrm{.}$ Every such vector lies in the span of circle generators, without using any ideal property. Indeed $R_{0,0}=u{\circ }_{n}v$. Integration by parts and [F3] give $R_{m,0}(Tu,v)=(2n+m+2)R_{m+1,0}(u,v)+(n+m+1-r)R_{m,0}(u,v)\mathrm{.}$ If $Tu=0$, the same equality holds with left side zero. Induction on $m$, simultaneously for all homogeneous first inputs, therefore gives $R_{m,0}\in O$, since $2n+m+2$ is nonzero. Expanding the extra factor $(1+z{)}^k$ then gives $R_{m,k}={\sum }_{i=0}^k\left(\genfrac{}{}{0px}{}{k}{i}\right)R_{m-i,0}$, with every $m-i\ge 0$. All these residues are finite sums by truncation and the nonnegative exponents.

Two exact identities will later convert this residue information into both ideal properties. For homogeneous $u\in V_r$, $(Du){\ast }_{n}v=K_n(u{\circ }_{n}v),D(u{\ast }_{n}v)=(Du){\ast }_{n}v+u{\ast }_n(Dv)\mathrm{.}$ For the first, integrate $Y(Tu,z)={\mathrm{\partial }}_{z}Y(u,z)$ against $(1+z{)}^{r+n+1}{P}_n(z)$ and add $r(u{\ast }_{n}v)$. The result is ${\mathrm{Res}}_z(1+z{)}^{r+n}Y(u,z)v{\sum }_{p=0}^n{a}_p((n+p+1)z^{-n-p-2}+p{z}^{-n-p-1})\mathrm{.}$ For $1\le p\le n$, $p{a}_p=-(n+p)a_{p-1}$; all adjacent coefficients cancel and the sum is exactly $K_n{z}^{-2n-2}$. This proves the first identity. For the second let $v\in V_s$, use $T(Y(u,z)v)={\mathrm{\partial }}_{z}Y(u,z)v+Y(u,z)Tv$ and $L(0)Y(u,z)v=(r+s+z{\mathrm{\partial }}_z)Y(u,z)v$, the latter being the weight rule of [F3]. Integration by parts gives $D(u{\ast }_{n}v)=u{\ast }_{n}Tv+{\mathrm{Res}}_z((r+s-1)f_r(z)-(1+z)f_r^{\mathrm{\prime }}(z))Y(u,z)v\mathrm{.}$ On the other hand $f_{r+1}=(1+z)f_r$ shows that $(Du){\ast }_{n}v$ is the same residue with $r+s-1$ replaced by $r-1$; adding $u{\ast }_n(Dv)=u{\ast }_{n}Tv+s(u{\ast }_{n}v)$ proves the identity. Linearity extends both formulas to arbitrary inputs, with the circle operation extended linearly in its first input.

We prove the associator belongs to $O$ before using any product of congruence classes. Let $u\in V_r$, $v\in V_s$ and $w\in V$ be homogeneous. Expanding the two star products and using the actual weight of each $u_{(i-n-p-1)}v$ gives the exact expression $(u{\ast }_{n}v){\ast }_{n}w={\sum }_{p,q=0}^n{a}_p{a}_q{\mathrm{Res}}_y{\mathrm{Res}}_t\frac{(1+y+t{)}^{r+n}(1+y{)}^{s+n+p}}{t^{n+p+1}{y}^{n+q+1}}Y(Y(u,t)v,y)w\mathrm{.}$ Apply [F3] with $m=-n-p-1$ and $P(x,y)=(1+x{)}^{r+n}(1+y{)}^{s+n+p}{y}^{-n-q-1}$. The forward expansion uses ${\iota }_{x,y}(x-y{)}^{-n-p-1}={\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{n+p+j}{j}\right)x^{-n-p-j-1}{y}^j,$ and the reversed expansion uses ${\iota }_{y,x}(x-y{)}^{-n-p-1}=(-1{)}^{n+p+1}{\sum }_{j\ge 0}\left(\genfrac{}{}{0px}{}{n+p+j}{j}\right)x^j{y}^{-n-p-j-1}\mathrm{.}$ Every reversed-product residue lies in $O$: after extracting the $x$ residue its outer $v$-residue has numerator $(1+y{)}^{s+n+p}$ and denominator $y^{2n+2+p+q+j}$, so it is $R_{p+q+j,p}(v,F_j)$, where $F_j={\mathrm{Res}}_x(1+x{)}^{r+n}{x}^{j}Y(u,x)w$. Here $p+q+j\ge p$. Every forward residue with $j>n-p$ also lies in $O$: after extracting its $y$ residue its outer $u$-residue is $R_{p+j-n-1,0}(u,E_j)$, where $E_j={\mathrm{Res}}_y(1+y{)}^{s+n+p}{y}^{-n-q-1+j}Y(v,y)w$. Both conclusions use only step 1.1. All $j$ sums are actually finite on the fixed $w$: the finite polynomial numerators make $F_j$ and $E_j$ sums of modes whose indices tend to $+\mathrm{\infty }$ with $j$, so truncation kills them for large $j$. Thus the additive congruence left after discarding these residues is $(u{\ast }_{n}v){\ast }_{n}w\equiv {\sum }_{p,q=0}^n{\sum }_{j=0}^{n-p}{a}_p{a}_q\left(\genfrac{}{}{0px}{}{n+p+j}{j}\right){\mathrm{Res}}_x{\mathrm{Res}}_y\frac{(1+x{)}^{r+n}(1+y{)}^{s+n}(1+y{)}^p{y}^j}{x^{n+p+j+1}{y}^{n+q+1}}Y(u,x)Y(v,y)w(\mathrm{m}\mathrm{o}\mathrm{d}O)\mathrm{.}$ No ideal assertion was used in discarding any term.

The finite kernel in step 2.1 equals the kernel of $u{\ast }_n(v{\ast }_{n}w)$. Explicitly, as a Laurent polynomial in $x$ with coefficients in $\mathbb{C}[y]$, ${\sum }_{p=0}^n{a}_p(1+y{)}^p{\sum }_{j=0}^{n-p}\left(\genfrac{}{}{0px}{}{n+p+j}{j}\right)y^j{x}^{-p-j}={\sum }_{h=0}^n{a}_h{x}^{-h}\mathrm{.}$ For a fixed $0\le h\le n$, its left $x^{-h}$ coefficient is ${\sum }_{p=0}^h(-1{)}^p\left(\genfrac{}{}{0px}{}{n+p}{n}\right)\left(\genfrac{}{}{0px}{}{n+h}{h-p}\right)(1+y{)}^p{y}^{h-p}=\left(\genfrac{}{}{0px}{}{n+h}{n}\right){\sum }_{p=0}^h(-1{)}^p\left(\genfrac{}{}{0px}{}{h}{p}\right)(1+y{)}^p{y}^{h-p}\mathrm{.}$ The factorial identity giving this equality is valid at every displayed index. The binomial theorem makes the final sum $(y-(1+y){)}^h=(-1{)}^h$, so the coefficient is exactly $a_h$. Substitution in step 2.1 proves $(u{\ast }_{n}v){\ast }_{n}w-u{\ast }_n(v{\ast }_{n}w)\in O$. Bilinearity extends this independently proved associator congruence to arbitrary triples.

Right ideal stability now follows without multiplying a prior congruence by an arbitrary state. The generators $Du$ satisfy $(Du){\ast }_{n}w=K_n(u{\circ }_{n}w)\in O$ by step 1.2. The other generators satisfy $(u{\circ }_{n}v){\ast }_{n}w=K_n^{-1}((Du){\ast }_{n}v){\ast }_{n}w\equiv K_n^{-1}(Du){\ast }_n(v{\ast }_{n}w)=u{\circ }_n(v{\ast }_{n}w)\in O,$ where the only congruence is the associator of step 3.1 applied to the actual states $Du,v,w$. Hence $O{\ast }_{n}V\subset O$. The exact derivation formula of step 1.2 next gives $u{\ast }_n(Dv)=D(u{\ast }_{n}v)-(Du){\ast }_{n}v\in O$. Finally $u{\ast }_n(v{\circ }_{n}w)=K_n^{-1}u{\ast }_n((Dv){\ast }_{n}w)\equiv K_n^{-1}(u{\ast }_{n}Dv){\ast }_{n}w\in O\mathrm{.}$ The last inclusion uses the already proved right ideal stability on the actual element $u{\ast }_{n}Dv\in O$. These statements establish $V{\ast }_{n}O\subset O$ as well. This order avoids circular use of left and right ideal properties.

The vacuum is a two-sided identity modulo $O$. On the left, $Y(\mathbf{1},z)=I$ and $\mathrm{wt}\mathbf{1}=0$ give $\mathbf{1}{\ast }_{n}u={\sum }_{p=0}^n{a}_p{\mathrm{Res}}_z(1+z{)}^n{z}^{-n-p-1}u=u,$ since the polynomial $(1+z{)}^n$ has no coefficient in degree $n+p$ for $p>0$, and its degree-$n$ coefficient is $1$. For the right identity let $u\in V_r$. The relations $D(T^{k-1}u)\in O$ give, by induction, $T^{k}u\equiv (-1{)}^{k}r(r+1)\cdots (r+k-1)u(\mathrm{m}\mathrm{o}\mathrm{d}O)(k\ge 1)\mathrm{.}$ Thus coefficientwise $e^{zT}u\equiv (1+z{)}^{-r}u$ modulo $O$. Only finitely many creation coefficients enter any of the star residues, so substitution into $u{\ast }_n\mathbf{1}$ is an ordinary finite linear congruence. It gives the same scalar residue as on the left, hence $u{\ast }_n\mathbf{1}\equiv u$. Step 4.1 makes the quotient product independent of both representatives; step 3.1 then gives associativity and the two vacuum identities give unit $[\mathbf{1}]$. All calculations hold also for $n=0$ and weight $r=0$. No dimension assertion or representation-theoretic classification is used.