\[DLM3\] that if $V$ is rational then there are only finitely many inequivalent irreducible admissible $V$-modules and each irreducible admissible $V$-module is an ordinary module.

The following proposition can be found in \[L\] and \[DM2\].

\[4.19\] If $M$ is a simple weak $V-$module then $M$ is spanned by $\{u_nm|u\in V,n\in \Q\}$ where $m\in M$ is a fixed nonzero vector.

We now recall the associative algebra $A_{n}(V)$ as constructed in \[DLM4\]. Let $O_n(V)$ be the linear span of all $u\circ_n v$ and $L(-1)u+L(0)u$ where for homogeneous $u\in V$ and $v\in V,$ $$\label{g2.2}
u\circ_n v=\Res_{z}Y(u,z)v\frac{(1+z)^{\wt u+n}}{z^{2n+2}}.$$ Define the linear space $A_n(V)$ to be the quotient $V/O_{n}(V).$ We also define a second product $*_n$ on $V$ for $u$ and $v$ as above: $$\label{a5.1}
u*_nv=\sum_{m=0}^{n}(-1)^m{m+n\choose n}\Res_zY(u,z)\frac{(1+z)^{\wt\,u+n}}{z^{n+m+1}}v.$$ Extend linearly to obtain a bilinear product on $V$.

Let $M=\sum_{n\in\Z_{+}}M(n)$ be an admissible $V$-module. Following \[Z\] we define weight zero operator $o_M(v)=v_{\wt v-1}$ on $M$ for homogeneous $v$ and extend $o_M(v)$ to all $v$ by linearity. It is clear from the definition that $o_M(v)M(n)\subset M(n)$ for all $n.$

\[l2.3\] Let $V$ be a vertex operator algebra and $M$ an admissible $V$-module. Then

1\) The product $*_{n}$ induces the structure of an associative algebra on $A_{n}(V)$ with identity 