Zhu top induction, simple correspondence, and forward rationality

Write [a] for a state class in A(V), and $d=\mathrm{wt}(a)$. Translation gives $[Tc]=-\mathrm{wt}(c)[c]$, whence $[T^{j}c]=(-1{)}^j(\mathrm{wt}(c))\cdots (\mathrm{wt}(c)+j-1)[c]$. Skew-symmetry on V gives $[b\ast a]={\sum }_{l\ge 0}$ $\mathrm{binom}(d-1,l)[a_{(l-1)}b]$. Here is its finite coefficient verification. The coefficient of $[a_{(l-1)}b]$, after expanding $e^{zT}Y(a,-z)b$ and reducing T, is ${\sum }_{i=0}^{min(\mathrm{wt}(b),l)}$ $(-1{)}^i$ binom(wt(b),i) $\mathrm{binom}(d+\mathrm{wt}(b)-i-1,l-i)=\mathrm{binom}(d-1,l)$. This is the coefficient of $x^l$ in $(1+x{)}^{d+\mathrm{wt}(b)-1}(1-x\mathrm{/}(1+x){)}^{\mathrm{wt}(b)}=(1+x{)}^{d-1}$. Each state calculation is finite by truncation and the nonnegative state weights. Subtracting from [a*b] gives $[a][b]-[b][a]={\sum }_{j\ge 0}$ $\mathrm{binom}(d-1,j)[a_{(j)}b]$.

For $k\ge 1$ put $C_k(a,b)={\sum }_{i=0}^d$ $\mathrm{binom}(d,i)a_{(i-k)}b$. The actual higher-circle relation gives $[C_k]=0$ for $k\ge 2$, while $[C_1]=[a][b]$. These identities hold in the algebra itself, hence on every U, not only on already-existing module tops.

Let $D(S)={\oplus }_{n\ge 0}S(n{)}^{\ast }$ be the direct sum of the full algebraic duals, with its natural pairing against S; no S(n) is assumed finite dimensional. For homogeneous $$\begin{gathered} a \\ in{V}_d \end{gathered}$$ define $A(x)=e^{xL(1)}(-x^{-2}{)}^{L(0)}$ and the opposite field $Y_S^o(a,x)=Y_S(A(x)a,x^{-1})$. Since L(1) lowers the nonnegative state weight, for $$\begin{gathered} a \\ in{V}_d \end{gathered}$$ $Y_S^o(a,x)=(-1{)}^d{\sum }_{j=0}^d\frac{x^{-2d+j}}{j!}{Y}_S(L(1{)}^{j}a,x^{-1}),a_{(m)}^o=(-1{)}^d{\sum }_{j=0}^d\frac{(L(1{)}^{j}a{)}_{(2d-j-m-2)}^S}{j!}\mathrm{.}$ Define the candidate dual modes by $a_{(m)}^D=(a_{(m)}^o{)}^t$. An original mode of $L(1{)}^j$ a shifts S-degree by $(d-j)-(2d-j-m-2)-1=m+1-d$. Its transpose therefore sends $S(q{)}^{\ast }$ to $S(q+d-m-1{)}^{\ast }$. For fixed s∈S, $a_{(m)}^{o}s=0$ for all sufficiently negative m, since every original mode index 2d-j-m-2 then exceeds its truncation bound; thus $Y_S^o(a,x)s$ is upper-truncated. Every f∈D(S) has finite grade support, so the transpose of $a_{(m)}^o$ kills f for all sufficiently large m. The transpose modes thus define a lower-truncated candidate action on D(S), and the vacuum acts as the identity. This proves the required truncation directly, with no finite-dimensionality assumption on a grade.

We now derive the opposite-field Jacobi identity on S from the actual module Jacobi identity; this derivation is needed because the restricted-dual theorem below assumes finite-dimensional grades. Residues of the state Jacobi identity with ω give, for $j=0,1$, $[L(j),Y(a,z)]={\sum }_{i=0}^{j+1}\left(\genfrac{}{}{0px}{}{j+1}{i}\right)z^{j+1-i}Y(L(i-1)a,z)\mathrm{.}$ Thus $[L(0),Y(a,z)]=Y(L(0)a,z)+z{\mathrm{\partial }}_{zY}(a,z)$ and $[L(1),Y(a,z)]=Y(L(1)a,z)+2zY(L(0)a,z)+z^2{\mathrm{\partial }}_{zY}(a,z)$. Integrating these coefficientwise formal differential identities gives $b^{L(0)}Y(a,z)b^{-L(0)}=Y(b^{L(0)}a,bz)$ and $e^{uL(1)}Y(a,z)e^{-uL(1)}=Y\text{\hspace{0.17em}⁣}(e^{u(1-uz)L(1)}(1-uz{)}^{-2L(0)}a,\frac{z}{1-uz})\mathrm{.}$ The second identity follows by differentiating both sides in u, using the displayed L(1) bracket, and matching their values at $u=0$; all state exponentials are finite because L(1) lowers nonnegative state weight. Combining the two identities yields the exact state-coordinate formula $A(s)Y(a,z_0)b=Y\text{\hspace{0.17em}⁣}(A(s+z_0)a,-\frac{z_0}{s(s+z_0)})A(s)b,$ with $s+z_0$ expanded at $z_0=0$. Now fix homogeneous a,b, $$\begin{gathered} s_0 \\ inS(k) \end{gathered}$$, and $$\begin{gathered} f \\ inS(\mathrm{\ell }{)}^{\ast } \end{gathered}$$. By the module Jacobi identity, the two matrix coefficients of the ordered products of $Y_S(a,t_1),Y_S(b,t_2)$ and their iterate are the prescribed binomial expansions of one rational function. Here is the finiteness in this assertion: choose K with $a_{(j)}b=0$ for $j\ge K$, so the module commutator makes $(t_1-t_2{)}^K$ times the two product orders equal. In the first order, truncation on the rightmost b-field bounds $t_2$ exponents below; in the reversed order, truncation on the rightmost a-field bounds $t_1$ exponents below. The fixed input and output grades force the sum of the two exponents to be fixed, so the common numerator is a Laurent polynomial. The iterate is its expansion after $t_1=t_2+t_0$: taking the two mode residues of the original Jacobi identity gives the coefficient formula in [F2] for every integer pair of mode indices, and its rightmost-mode sums are finite. Thus the third expansion is exactly the matrix coefficient of $Y_S(Y(a,t_0)b,t_2)$. Dividing by $(t_1-t_2{)}^K$ gives the claimed rational function, with poles only at $t_1=0$, $t_2=0$, and $t_1=t_2$. This uses only mode truncation and the grade labels; it uses no sum over a basis of S(k). Replace $t_i$ by $x_i^{-1}$ and use the finite state $A(x_i)$ on the inserted states. Inversion reverses the two product expansions: the ${\iota }_{x_1,x_2}$ expansion is $Y_S^o(b,x_2)Y_S^o(a,x_1)$, and the ${\iota }_{x_2,x_1}$ expansion is $Y_S^o(a,x_1)Y_S^o(b,x_2)$. On the iterate chart $x_1=x_2+x_0$ one has $t_1-t_2=-x_0\mathrm{/}(x_1{x}_2)$; the state-coordinate formula above identifies the transformed iterate with $Y_S^o(Y(a,x_0)b,x_2)$. For every Laurent monomial $Q=x_1^q$ $x_2^r(x_1-x_2{)}^s$, the coefficientwise delta identity $x_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{x_1-x_2}{x_0}\right){\iota }_{x_1,x_2}Q-x_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{x_2-x_1}{-x_0}\right){\iota }_{x_2,x_1}Q=x_2^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{x_1-x_0}{x_2}\right){\iota }_{x_2,x_0}Q(x_0+x_2,x_2)$ follows by expanding each integer power with the designated binomial conventions and $$\begin{gathered} \delta (x)={\sum }_{n \\ inZ}{x}^n \end{gathered}$$. The transformed rational function is a finite sum of such monomials with finite pole order, so linearity proves the same identity for it and every fixed coefficient is a finite sum. Functionals in $S(\mathrm{\ell }{)}^{\ast }$ separate S(ℓ), and hence this proves the full opposite-field Jacobi identity on S. The same $s{l}_2$ brackets give $A(x^{-1})A(x)=1$: conjugating $e^{xL(1)}$ by $(-x^2{)}^{L(0)}$ gives $e^{-x^{-1}L(1)}$, which cancels $e^{x^{-1}L(1)}$.

Take the current Lie algebra generated by symbols a(m), linear in a, modulo $(Ta)(m)=-m$ a(m−1), and bracket $[a(m),b(n)]={\sum }_{j\ge 0}$ $\mathrm{binom}(m,j)(a_{(j)}b)(m+n-j)$. The sums are finite by state truncation; the coefficient identity of state Jacobi proves Lie Jacobi. Give a(m) degree d−m−1. Let P be its nonpositive-degree subalgebra. On U, negative degrees act as zero and a(d−1) acts as [a]. The displayed Zhu commutator makes this a representation of P; the translation relation agrees because $[Ta]=-d[a]$. Impose $1(-1)=1$. Form the induced module $E(U)=U(current){\otimes }_{U(P)}U$. Sorting words by degree sign using $XY=YX+[X,Y]$ gives a vector-space basis of words of positive-degree modes on U; the only sorting ambiguity of length three is Lie Jacobi, and hence has zero difference. This is the PBW argument in this particular presentation. Consequently E(U) is nonnegatively graded and $E(U)(0)=U$. Modes have their stated degree and are lower truncated on every fixed vector. Let p:E(U)→U be projection onto grade zero.

Pair the module Jacobi expression for the transpose action on D(S) with any $$\begin{gathered} s_0 \\ inS \end{gathered}$$ and f∈D(S). Transposing reverses each product, so the identity just proved for $Y_S^o$ is exactly the ordinary module Jacobi identity for D(S). All coefficient pairings are finite by the grade shifts in step 1.3 and the finite state exponential. Thus D(S) is a weak V-module. Its grade-n piece is $S(n{)}^{\ast }$, and $a_{(m)}^{D}S(n{)}^{\ast }\subset S(n+d-m-1{)}^{\ast }$; the vacuum and Jacobi identities already proved make this an admissible module. Since $\omega =L(-2)1$ and $L(1)\omega =0$ by the Virasoro relation and $T1=0$, $Y_S^o(\omega ,x)=x^{-4}{Y}_S(\omega ,x^{-1})$; comparing coefficients gives $L_D(j)=L_S(-j{)}^t$. In particular, $L_D(0)$ preserves each $S(n{)}^{\ast }$, but it need not act there by a scalar for a general admissible S. No finite-grade conclusion or double-dual isomorphism has been used.

Let $A_a$ denote the action of [a] on U. For u∈U, projection of two fields is the expansion in $\mathrm{\mid }{z}_1\mathrm{\mid }>\mathrm{\mid }{z}_2\mathrm{\mid }$ of the rational expression $p{Y}_E(a,z_1)Y_E(b,z_2)u$ $=$ $z_1^{-d}{z}_2^{-e}(A_a{A}_b+{\sum }_{j\ge 0}{A}_{a_{(j)}b}{\sum }_{r\ge 1}\mathrm{binom}(d-1+r,j)(z_2\mathrm{/}{z}_1{)}^r)u$. Indeed the rightmost b-mode has degree $r\ge 0$. At $r=0$ the coefficient is $A_a{A}_b$. For $r>0$, the a-mode lowers degree and annihilates U, so commuting it through the b-mode gives exactly the displayed coefficient. Put $w=(z_1-z_2)\mathrm{/}{z}_2$. The rational identity ${\sum }_{r\ge 1}\mathrm{binom}(d-1+r,j)(1+w{)}^{-r}={\sum }_{i=0}^j$ $\mathrm{binom}(d,i)w^{i-j-1}$ is proved by summing over j with a new variable t: the left generating function is ${\sum }_{r\ge 1}(1+t{)}^{d-1+r}(1+w{)}^{-r}=(1+t{)}^d\mathrm{/}(w-t)$. Put $H(w)={\sum }_n{A}_{a_{(n)}b}{w}^{-n-1}$ and let $F(w)=z_1^d{z}_2^{e}p{Y}_E(a,z_1)Y_E(b,z_2){\mathrm{\mid }}_U$, with $z_1=z_2(1+w)$. The generating identity gives $F(w)=A_a{A}_b+[(1+w{)}^{d}H(w){]}_{<0}$, without any further factor multiplying F: its jth contribution is $A_{a_{(j)}b}{\sum }_{i=0}^{\min (d,j)}\left(\genfrac{}{}{0px}{}{d}{i}\right)w^{i-j-1}$, exactly the strictly negative-power part. The coefficient of $w^{k-1}$ in $(1+w{)}^{d}H(w)$ is $A_{C_k(a,b)}$ for $k\ge 1$. Hence the higher-circle and star relations identify its constant coefficient with $A_a{A}_b$ and kill every positive-power coefficient. Consequently $F(w)=(1+w{)}^{d}H(w)$. On the iterate side the state weight $\mathrm{w}\mathrm{t}(a_{(n)}b)=d+e-n-1$ gives $p{Y}_E(Y(a,z_0)b,z_2){\mathrm{\mid }}_U=z_2^{-d-e}H(w)$ for $z_0=z_{2}w$. Retaining all scalar factors, $z_1^{d}p{Y}_E(a,z_1)Y_E(b,z_2){\mathrm{\mid }}_U=z_2^{-e}F(w)=z_2^{-e}(1+w{)}^{d}H(w)=z_1^{d}p{Y}_E(Y(a,z_0)b,z_2){\mathrm{\mid }}_U\mathrm{.}$ This is projected weak associativity on U with precisely the power defining the grade-zero defect. Every equality compares the specified expansions of this rational expression; it never interchanges two unrestricted infinite sums.

Here are the powers that make the propagation work without cancellation of a binomial factor in a distribution space. Put $z_1=z_0+z_2$. For homogeneous a of weight d and v∈E(U)(p), impose the coefficients of $D_{d+p}(a,b;v)=z_1^{d+p}(Y_E(a,z_1)Y_E(b,z_2)v-Y_E(Y(a,z_0)b,z_2)v)$. On the product side use its binomial substitution expansion; on the iterate side expand $z_1^{d+p}$ as the polynomial $(z_2+z_0{)}^{d+p}$. These are well-defined after the stated power because $a(m)v=0$ for $m\ge d+p$. Powers can always be increased by multiplication by the polynomial $z_1$. All coefficient statements below use these conventions and only nonnegative extra powers of $z_1$.

For an unweighted defect D, the commutator of c(m) with its two field operators is $[c(m),D(a,b)]={\sum }_{i\ge 0}\mathrm{binom}(m,i)z_1^{m-i}D(c_{(i)}a,b)+{\sum }_{i\ge 0}\mathrm{binom}(m,i)z_2^{m-i}D(a,c_{(i)}b)$. To check the iterate term, apply state Jacobi to $c(j)(Y(a,z_0)b)$. Its two contributions are $Y(a,z_0)c(j)b$ and ${\sum }_{i\ge 0}\mathrm{binom}(j,i)z_0^{j-i}Y(c(i)a,z_0)b$. Summation over j gives the asserted $z_1$ factor because ${\sum }_{j\ge i}\mathrm{binom}(m,j)\mathrm{binom}(j,i)z_2^{m-j}{z}_0^{j-i}=\mathrm{binom}(m,i)(z_2+z_0{)}^{m-i}$. This is binomial expansion, valid also for negative m. Only finitely many i remain because c(i)a and c(i)b vanish for sufficiently large i. For a fixed source degree, multiplication by the powers below makes each indicated coefficient finite, so this formal computation can be performed coefficientwise.

First fix $$\begin{gathered} v=u \\ inU \end{gathered}$$. We prove p x $D_d(a,b;u)=0$ for every current-mode word x by induction on its length. The case $x=1$ is the rational identity proved above. PBW-sort x with positive degrees left, then zero, then negative. Commutator errors have smaller length and are already treated. A positive-degree leftmost mode has zero p-projection because E has no negative grades. For a degree-zero leftmost mode, p $c(\mathrm{wt}(c)-1)=A_c$ p, so it can be removed. Otherwise every mode has negative degree, and the mode c(m) closest to the defect satisfies $m\ge \mathrm{wt}(c)$. Commute it through the defect using the displayed identity, until its action on u is zero. For the first new defect its state weight is $d^{\mathrm{\prime }}=\mathrm{wt}(c)+d-i-1$, while its $z_1$ exponent is d+m−i. The difference is $m-\mathrm{wt}(c)+1\ge 1$. Thus it is a polynomial nonnegative power of $z_1$ times the induction hypothesis with state weight d'. The second new defect still has first-state weight d and exponent d, also exactly the induction hypothesis. This decreases the number of preceding modes, proving the assertion. Negative-weight state products are zero and require no separate power convention.

Now prove p x $D_{d+p}(a,b;v)=0$ for every homogeneous v∈E(U)(p), simultaneously for all preceding words x, by induction on the length of a positive-degree PBW word generating v from U. The zero-length case was just proved. Write $v=c(m)v_0$, with $r=\mathrm{wt}(c)-m-1>0$ and $p=p_0+r$. Commuting that source mode through the defect gives a preceding c(m) times the defect on $v_0$, minus the two displayed state-product defects. The first term has power $d+p\ge d+p_0$ and is covered by the source-word induction, which already allows every preceding word. For c(i)a the exponent is $d+p+m-i=d+p_0+\mathrm{wt}(c)-i-1=d^{\mathrm{\prime }}+p_0$, exactly the required exponent for its first-state weight d'. For the other defect the exponent is $d+p\ge d+p_0$. Therefore each term is killed by every p x by the induction hypothesis. Every comparison uses equal powers or multiplication by a nonnegative power, never cancellation of $z_0+z_2$. Linearity covers every homogeneous v by PBW.

Let I be the current-submodule generated by all coefficients of these $D_{d+p}$. Each coefficient relation is homogeneous for the mode-degree grading, so I is graded. The vanishing p x D proved above shows I⊂ker p and p x $I=0$ for every x; in particular $I\cap U=0$. Put $M(U)=E(U)\mathrm{/}I$ and write F(a,z) for its induced fields. The current bracket gives weak commutativity: if $a_{(j)}b=0$ for all $j\ge K$, then $(z_1-z_2{)}^K$ $F(a,z_1)F(b,z_2)v=(z_1-z_2{)}^K$ $F(b,z_2)F(a,z_1)v$. Indeed, the coefficient commutator formula expresses the field commutator as the finite sum of the fields $F(a_{(j)}b,z_2)$ against the j-th derivatives of the formal delta; multiplication by $(z_1-z_2{)}^K$ kills every derivative with $j<K$. The relations defining I give weak associativity on each homogeneous v∈M(U)(p): $(z_0+z_2{)}^{d+p}F(a,z_0+z_2)F(b,z_2)v=(z_0+z_2{)}^{d+p}F(Y(a,z_0)b,z_2)v\mathrm{.}$ Here $d=\mathrm{wt}(a)$, and the first product is expanded after $z_1=z_0+z_2$. To see explicitly that these two relations give the full module Jacobi identity, fix homogeneous a,b, v∈M(U)(p), an output grade q, and $$\begin{gathered} \lambda \\ inM(U)(q{)}^{\ast } \end{gathered}$$. Mode-degree shifts force the sum of the $z_1,z_2$ exponents in $⟨\lambda ,F(a,z_1)F(b,z_2)v⟩$ to be fixed. Choose K as above and multiply this matrix coefficient by $(z_1-z_2{)}^K$. In the first product order, lower truncation on the rightmost b-field bounds the $z_2$ exponents below; in the reversed order, lower truncation on the rightmost a-field bounds the $z_1$ exponents below. The weak-commutativity equality therefore gives a common numerator $P(z_1,z_2)$ whose exponents are bounded below in both variables. Since their sum is fixed, P is a Laurent polynomial. Thus the two products are the expansions ${\iota }_{z_1,z_2}R$ and ${\iota }_{z_2,z_1}R$ of $R=P\mathrm{/}(z_1-z_2{)}^K$. Weak associativity identifies the third expansion, ${\iota }_{z_2,z_0}R(z_0+z_2,z_2)$, with $⟨\lambda ,F(Y(a,z_0)b,z_2)v⟩$: the equality first holds after multiplication by $(z_0+z_2{)}^{d+p}$, and cancellation is valid in the same $z_0-adic$ iterated Laurent-series space, where $(z_0+z_2{)}^{-(d+p)}=z_2^{-(d+p)}{\sum }_{r\ge 0}\left(\genfrac{}{}{0px}{}{-(d+p)}{r}\right)(z_0\mathrm{/}{z}_2{)}^r$ is the inverse expansion. For every Laurent monomial $Q=z_1^q$ $z_2^r(z_1-z_2{)}^s$, the coefficientwise delta identity $z_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{z_1-z_2}{z_0}\right){\iota }_{z_1,z_2}Q-z_0^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{z_2-z_1}{-z_0}\right){\iota }_{z_2,z_1}Q=z_2^{-1}\delta \text{\hspace{0.17em}⁣}\left(\frac{z_1-z_0}{z_2}\right){\iota }_{z_2,z_0}Q(z_0+z_2,z_2)$ holds by expanding the integer powers with their designated generalized-binomial expansions and using $$\begin{gathered} \delta (x)={\sum }_{n \\ inZ}{x}^n \end{gathered}$$. Since R is a finite linear combination of these monomials, the same identity holds with R; its numerator is finite and its pole order is K, so each fixed coefficient involves only finitely many terms. This is precisely the module Jacobi identity after pairing with λ. Such functionals separate each graded coefficient, so Jacobi holds as a vector-valued formal identity on M(U). The vacuum and lower-truncation axioms follow from the construction. A map U→Ω(W) extends uniquely to the induced current module E(U). This weak associativity also follows directly from W Jacobi: for each image vector w choose $L\ge 0$ with $a_{(m)}^{W}w=0$ for all $m\ge L$, multiply Jacobi by $z_1^L$, and take $Re{s}_{z_1}$. The second ordered term vanishes because its $z_1$-residue uses only modes $a_{(m)}^{W}w$ with $m\ge L$; the delta substitutions in the first and iterate terms give $(z_0+z_2{)}^L$ $Y_W(a,z_0+z_2)Y_W(b,z_2)w=(z_0+z_2{)}^L$ $Y_W(Y(a,z_0)b,z_2)w$. Multiplying this identity by any further nonnegative power preserves it. Since $D_{\mathrm{\ell }}=z_1^{\mathrm{\ell }-(d+p)}{D}_{d+p}$ for $\mathrm{\ell }\ge d+p$, choose ℓ also at least L; injectivity of multiplication by $z_1^{\mathrm{\ell }-(d+p)}$ then shows that the map kills every coefficient generator of I. It therefore factors uniquely through M(U), proving the universal property.

The largest graded submodule of M(U) disjoint from U is the sum of all such submodules: its grade-zero space remains zero, so the sum is still disjoint. Denote it by J(U). If U is simple, the quotient L(U) has no nonzero proper graded submodule: any nonzero graded submodule must meet U, hence contains U and generates L(U). For clarity, this is also simplicity as an ungraded weak module. A simple A(V)-module U is a quotient of its countable-dimensional algebra A(V), hence is countable dimensional. The central class [ω] acts as a scalar h. Here the scalar assertion does not assume U finite dimensional: if an endomorphism T of a countable-dimensional simple module is nonscalar, then every T−λ is invertible by Schur, and for a fixed nonzero u the vectors $(T-\lambda {)}^{-1}u$, λ∈C, are linearly independent. A finite dependence, multiplied by the product of T−λ, yields a nonzero polynomial annihilating u; its factorization and invertibility of each T−λ contradict that equation. Uncountability of C contradicts countable dimension. Thus every endomorphism is scalar. Centrality of [ω] follows either from the displayed Zhu commutator (its $j=0,1$ terms are Ta and wt(a)a, whose classes cancel), or directly from $[Ta]=-\mathrm{wt}(a)[a]$.

In M(U), the mode commutator now makes L(0) act as h+n on grade n, since it does so on U and every generating word shifts both by its degree. Distinct grades have distinct eigenvalues. Any weak submodule is therefore graded: on the finite support of each vector, polynomial interpolation in L(0) isolates every component. The graded simplicity just proved is consequently actual simplicity. Conversely, let S be a simple admissible module. If its specified grading has least nonzero grade $n_0>0$, replace it by $S_{\text{new}}(n)=S(n+n_0)$ for $n\ge 0$. For $$\begin{gathered} a \\ in{V}_r \end{gathered}$$, the mode rule gives $a_m{S}_{\text{new}}(n)\subset S_{\text{new}}(n+r-m-1)$, with negative target grades zero, so this is an admissible regrading and $S_{\text{new}}(0)\mathrm{\ne }0$. We may therefore assume $S(0)\mathrm{\ne }0$. PBW mode-degree reordering shows that the submodule generated by any zero-mode-stable subspace T⊂S(0) has grade-zero space precisely T: negative degrees kill T, and a surviving degree-zero ordered word has only zero-degree modes. Simplicity makes S(0) a simple A(V)-module. The universal map M(S(0))→S is onto, and its kernel is graded and disjoint from the top. Its kernel equals J(S(0)), since a larger top-disjoint submodule would give a nonzero proper submodule in the simple quotient. This proves S≅L(S(0)), and proves the claimed correspondence.

Apply rationality to the constructed M(A(V)). Its grading-preserving decomposition into simple admissible summands restricts on degree zero to a decomposition of its top A(V). Every nonzero summand must meet degree zero: the projection of M(A(V)) to that summand is a module map, and M(A(V)) is generated by degree zero. Each summand's degree-zero space is simple over A(V), by the degree-ordering argument in the correspondence proof. Therefore the regular left A(V)-module is semisimple, in the exact convention of the target. In fact A(V) is finite dimensional here. Write the regular module as a direct sum of simple left ideals. The vector 1 has finite support, say in $I_1\oplus \cdots \oplus I_r$. Multiplication by any a puts $a=a\cdot 1$ in this same finite sum, since each $I_i$ is a left ideal. Thus A(V) is a finite direct sum of simple left ideals. Each has at most countable dimension, since V has countably many finite-dimensional weights and A(V) is its quotient. The countable Schur argument above gives $En{d}_A(I_i)=C$. Homomorphisms between nonisomorphic simple ideals vanish, and between isomorphic ones form a one-dimensional space after choosing an isomorphism. Thus $En{d}_A(A)$ is finite dimensional, and right multiplication identifies it with $A(V{)}^{op}$. This proves finite dimensionality, hence every simple A(V)-module is finite dimensional.

Let S be a simple admissible module normalized to degree zero. Its top is simple, hence finite dimensional, and [ω] acts there as a scalar h. Cyclic generation gives $L(0)=h+n$ on S(n), so it is semisimple; S has at most countable dimension because one vector generates it under words in the countable set of state basis modes. It remains to prove finite dimensions of S(n).

Every nonzero graded submodule N⊂D(S) must meet degree zero. If $N(0)=0$, its annihilator $N^{\perp }$ in S contains S(0). It is a V-submodule: $A(z^{-1})A(z)=1$ in step 1.4 gives the exact inverse-field identity $Y_S(a,z)=Y_S^o(A(z)a,z^{-1})$. For homogeneous a, A(z)a is a finite Laurent polynomial in z with state coefficients, so every original state mode is a finite linear combination of opposite-state modes. Opposite-state invariance of the pairing then makes $N^{\perp }$ stable under the original modes. Simplicity of S implies $N^{\perp }=S$. The full grade-dual pairing separates D(S), so $N=0$, a contradiction. Apply rationality to D(S). Every simple graded summand must meet degree zero by the preceding paragraph. Since $D(S)(0)=S(0{)}^{\ast }$ is finite dimensional, there are only finitely many such summands. Each simple admissible module is countable dimensional, being cyclic under the countable state-mode algebra. Hence D(S) is countable dimensional. If some S(n) were infinite dimensional, it has a countably infinite basis. Fix a countable basis of the countable-dimensional S, let $E_j$ be the span of its first j vectors, and choose a finite basis of each finite-dimensional intersection $S(n)\cap E_j$ by row reduction. The union of these finite lists is countable and spans S(n), because each vector of S has finite support in the fixed basis; retaining each vector outside the span of its predecessors gives a countably infinite basis. Its full dual is then uncountable dimensional. Explicitly, the functionals assigning 1,λ,λ²,… to that basis are linearly independent as λ ranges over C, by the finite Vandermonde determinant. This contradicts countable dimensionality of D(S). Thus every S(n) is finite dimensional. The weights h+n are bounded below in real part, so S is ordinary. The forward implication of the original rationality criterion is now proved.