Algebraic graded dual of an admissible VOA module

For homogeneous $a\in V_r$, write $Y_W^o(a,z)={\sum }_{m\in \mathbb{Z}}{a}_m^o{z}^{-m-1}$. Expanding the finite state exponential in the defining opposite field gives $a_m^o=(-1{)}^r{\sum }_{j=0}^r\frac{(L(1{)}^{j}a{)}_{(2r-j-m-2)}^W}{j!}\mathrm{.}$ Each summand shifts $W(n)$ by $(r-j)-(2r-j-m-2)-1=m+1-r\mathrm{.}$ Its transpose therefore sends $W(n{)}^{\ast }$ into $W(n+r-m-1{)}^{\ast }$, exactly the admissible shift. If this target grade is negative it is zero; for a finite-support vector of $D(W)$, all sufficiently large $m$ have negative target grade. This proves field truncation on $D(W)$. The vacuum coefficient transposes to the identity.

Since $V$ is a CFT-type conformal vertex algebra by [F1] and $W$ is a left $V$-module by [F3], [F6] applies to $Y_W^o$. Pair that opposite Jacobi identity with $f\in W(p{)}^{\ast }$ and $w\in W(q)$. Every opposite mode on a homogeneous state is a finite sum because $L(1{)}^{j}a=0$ for $j>\mathrm{w}\mathrm{t}(a)$; admissibility [F2] gives the grade shift for each term, so every paired coefficient is defined without a dimension bound on any $W(n)$. The module residue identities [F4] and delta expansion convention [F5] make the resulting pairing coefficientwise formal; no analytic convergence is used. Thus the opposite Jacobi identity can be transposed coefficientwise without any finite-dimensionality assumption on $W(n)$.

Pair the opposite Jacobi identity from step 1.2 with $f$ and transpose each product. Transposition reverses the order of the two operators, so the reversed-order identity becomes the ordinary module Jacobi identity for $Y_D$. Every coefficient pairing is finite: each mode has the grade shift of step 1.1 and $f$ has one grade of support. Therefore $D(W)$ is a left $V$-module, and step 1.1 gives its admissible grading.

Evaluation $j_W(w)(f)=f(w)$ on each grade is injective because [F7], with $E=W(n)$, $L=\{w\}$ and $S=E$, extends every nonzero $w\in W(n)$ to a basis; the coordinate functional equal to $1$ on $w$ and $0$ on the other basis vectors separates it. Since $[L(0),L(1)]=-L(1)$ and $L(0)$ has integral eigenvalues on $V$, conjugation gives $(-x^2{)}^{L(0)}{e}^{xL(1)}(-x^2{)}^{-L(0)}=e^{-x^{-1}L(1)}$. Therefore $e^{x^{-1}L(1)}(-x^2{)}^{L(0)}{e}^{xL(1)}(-x^{-2}{)}^{L(0)}=1.$ Pairing twice gives $j_W(Y_W(a,z)w)=Y_{D^2(W)}(a,z)j_W(w)$, so evaluation is a module map. If each grade is finite dimensional, its gradewise evaluation map is bijective.

If $g:W\to X$ is a grading-preserving module homomorphism, then $g{Y}_W^o(a,z)=Y_X^o(a,z)g$, so its gradewise transpose $D(g)$ is a module homomorphism by the defining pairings. Precomposition gives $D(\mathrm{i}\mathrm{d})=\mathrm{i}\mathrm{d}$ and $D(h\circ g)=D(g)\circ D(h)$. A grading-preserving short exact sequence of admissible modules is exact on each grade. For each inclusion $U\subset X$ of vector spaces, apply [F7] first with $E=S=U$, $L=\mathrm{\varnothing }$, obtaining a basis $B_U$ (also for $U=0$), and then with $E=S=X$, $L=B_U$, obtaining a basis $B_X\supseteq B_U$. Extending a functional by zero on $B_X\setminus B_U$ proves that $X^{\ast }\to U^{\ast }$ is surjective. Its kernel consists exactly of the functionals factoring through $X\mathrm{/}U$, and precomposition with $X\twoheadrightarrow X\mathrm{/}U$ is injective. Thus algebraic duality is exact, and direct sums of these gradewise exact dual sequences remain exact with arrows reversed. Hence $D$ is an exact contravariant functor.