was mentioned in \[HL1\] one should think the pair $(W,Y^{o})$ as a right weak $V$-module. Since we shall discuss a little more further, let us make a formal definition here: A [*right* ]{} weak $V$-module is a vector space $W$ equipped with a linear map $Y_{W}$ from $V$ to $(\End W)[[x,x^{-1}]]$ such that for $u,v\in V$, $w\in W$, $$\begin{aligned}
& &Y_{W}(v,x)w\in W((x^{-1})),\\
& &Y_{W}({\bf 1},x)=1,\end{aligned}$$ and such that the opposite Jacobi identity (\[ehl\]) with $Y^{o}$ being replaced by $Y_{W}$ holds.

Let $U$ be a vector space together with a linear map $Y_{U}$ from the vertex operator algebra $V$ to $(\End\; U)[[x,x^{-1}]]$, e.g., $(U,Y_{U})$ is a left or right (weak) $V$-module. For $v\in V$, we define (cf. (\[eyovx\])) $$\begin{aligned}
Y_{U}^{o}(v,x)=Y_{U}(e^{xL(1)}(-x^{-2})^{L(0)}v,x^{-1})
\in (\End\; U)[[x,x^{-1}]].\end{aligned}$$ Then FHL’s proof in \[FHL\] for Proposition \[pfhl0\] again in fact proves the following result, which gives the equivalence between the notion of left $V$-module and the notion of right $V$-module:

Let $V$ be a vertex operator algebra and $U$ a vector space together with a linear map $Y_{U}$ from $V$ to $(\End\; U)[[x,x^{-1}]]$. Then $(U,Y_{U})$ is a left (weak) $V$-module if and only if $(U,Y^{o}_{U})$ is a right (weak) $V$-module.

Notice that $$\begin{aligned}
\Res_{x}xY^{o}(\omega,x)=\Res_{x}x^{-3}Y(\omega,x^{-1})=\Res_{x}xY(\omega,x).\end{aligned}$$ Then $\omega_{1}=L(0)$ is 