represented by the same operator on $U$ for $(U,Y_{U})$ and $(U,Y^{o}_{U})$. Therefore, the assertion in Proposition \[pcontragredient\] also includes the grading information.

Let $f(x)=\sum_{n\le N}a_{n}x^{n}\in U((x^{-1}))$ for a vector space $U$. Then, for any complex number $z_{0}$, $$\begin{aligned}
f(x+z_{0}):=\sum_{n\le N}a_{n}(x+z_{0})^{n}
=\sum_{n\le N}\sum_{j\ge 0}{n\choose j}a_{n}z_{0}^{j}x^{n-j}\end{aligned}$$ exists in $U((x^{-1}))$. Let $(W,Y_{W})$ be a right weak $V$-module and $z_{0}$ be a complex number. For $v\in V$, we define $Y_{W}^{(z_{0})}(v,x)\in (\End\;W)[[x,x^{-1}]]$ by $$\begin{aligned}
Y_{W}^{(z_{0})}(v,x)w=Y_{W}(v,x+z_{0})w=(Y_{W}(v,y)w)|_{y=x+z_{0}}
=e^{z_{0}{d\over dx}}(Y_{W}(v,x)w)\end{aligned}$$ for $w\in W$.

For any right weak $V$-module $(W,Y_{W})$, the pair $(W,Y_{W}^{(z_{0})})$ is also a right weak $V$-module.

Clearly, $$\begin{aligned}
Y_{W}^{(z_{0})}({\bf 1},x)=1, \;\; Y_{W}^{(z_{0})}(v,x)w\in W((x^{-1}))\end{aligned}$$ for $v\in V,\; w\in W$. Then it remains to prove the opposite Jacobi identity.

Let $u,v\in V$. Replacing $x_{1}$ and $x_{2}$ by $x_{1}+z_{0}$ and $x_{2}+z_{0}$ in the opposite Jacobi identity (\[ehl\]) with $Y^{o}$ being replaced by $Y_{W}$, respectively, we get $$\begin{aligned}
\label{ehly}
& &x_{0}^{-1}\delta\left(\frac{x_{1}-x_{2}}{x_{0}}\right)
Y_{W}(v,x_{2}+z_{0})Y_{W}(u,x_{1}+z_{0})\nonumber\\
& &-x_{0}^{-1}\delta\left(\frac{x_{2}-x_{1}}{-x_{0}}\right)
Y_{W}(u,x_{1}+z_{0})Y_{W}(v,x_{2}+z_{0})\nonumber\\
&=&(x_{2}+z_{0})^{-1}\delta\left(\frac{x_{1}+z_{0}-x_{0}}{x_{2}+z_{0}}\right)
Y_{W}(Y(u,x_{0})v,x_{2}+z_{0}),\end{aligned}$$ noting that for $n\in {\Z}$, $$\begin{aligned}
(x_{1}+z_{0}-x_{2}-z_{0})^{n}=(x_{1}-x_{2})^{n}.\end{aligned}$$ Notice that $$\begin{aligned}
(x_{2}+z_{0})^{-1}\delta\left(\frac{x_{1}+z_{0}-x_{0}}{x_{2}+z_{0}}\right)
=e^{z_{0}({\partial\over \partial x_{1}}+{\partial\over \partial x_{2}})}
x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
=x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)\end{aligned}$$ because $$\begin{aligned}
\left({\partial \over\partial x_{1}}+{\partial \over\partial x_{2}}\right)
x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)=0\end{aligned}$$ (cf. \[Li2\]). Then we obtain $$\begin{aligned}
\label{ehlz}
& &x_{0}^{-1}\delta\left(\frac{x_{1}-x_{2}}{x_{0}}\right)
Y_{W}(v,x_{2}+z_{0})Y_{W}(u,x_{1}+z_{0})\nonumber\\
& &-x_{0}^{-1}\delta\left(\frac{x_{2}-x_{1}}{-x_{0}}\right)
Y_{W}(u,x_{1}+z_{0})Y_{W}(v,x_{2}+z_{0})\nonumber\\
&=&x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
Y_{W}(Y(u,x_{0})v,x_{2}+z_{0}).\end{aligned}$$ (This is also a generalization of the opposite Jacobi identity for $Y_{W}$ 