-
L(z_3+x+y)
\big(\big|_{x=\lambda+\partial}L^*(-z_1+\lambda)\big)
\big(\big|_{y=\mu+\partial}L^*(-z_2+\mu)\big)
\Big)
\,.}
\end{array}$$ Combining the second term in the RHS of and the first term in the RHS of we get $0$, and combining the first term in the RHS of and the second term in the RHS of we also get $0$. Next, combining the first terms in the RHS of , and , we get $$\begin{array}{l}
\vphantom{\Big(}
\displaystyle{
\Big(
i_{z_1}(z_1-z_2-x)^{-1}
i_{z_2}(z_2-z_3-y)^{-1}
-
i_{z_1}(z_1-z_3-x-y)^{-1}
i_{z_2}(z_2-z_3-y)^{-1}
} \\
\vphantom{\Big(}
\displaystyle{
-
i_{z_1}(z_1-z_3-x-y)^{-1}
i_{z_1}(z_1-z_2-x)^{-1}
\Big)
L(z_1)
\big(\big|_{x=\lambda+\partial}L(z_2)\big)
\big(\big|_{y=\mu+\partial}L(z_3)\big) 
\,,
}
\end{array}$$ and this is zero by the obvious identity $$\label{obvious}
a^{-1}b^{-1}-i_a(a+b)^{-1}b^{-1}-i_a(a+b)^{-1}a^{-1}=0\,.$$ Next, combining the second terms in the RHS of , and , we get $$\begin{array}{c}
\vphantom{\Big(}
\displaystyle{
\Big(
i_{z_1}(z_1-z_3-x-y)^{-1}
i_{z_2}(z_2-z_3-y)^{-1}
+
i_{z_1}(z_1-z_3-x-y)^{-1}
\times
} \\
\vphantom{\Big(}
\displaystyle{
\times
i_{z_2}(z_1-z_2-x)^{-1}
-
i_{z_1}(z_1-z_2-x)^{-1} 
i_{z_2}(z_2-z_3-y)^{-1}
\Big)
\times
} \\
\vphantom{\Big(}
\displaystyle{
\times
L(z_3+x+y)
\big(\big|_{x=\lambda+\partial}L^*(-z_1+\lambda)\big)
\big(\big|_{y=\mu+\partial}L^*(-z_2+\mu)\big)
\,,}
\end{array}$$ which, by the identity $i_{z_2}(z_1-z_2-x)^{-1}=i_{z_1}(z_1-z_2-x)^{-1}-\delta(z_1-z_2-x)$ and equation , can be rewritten as follow: $$\label{eq7}
\begin{array}{c}
\vphantom{\Big(}
\displaystyle{
-i_{z_1}(z_1-z_3-x-y)^{-1}
\delta(z_1-z_2-x)
L(z_3+x+y)
\times
} \\
\vphantom{\Big(}
\displaystyle{
\times
\big(\big|_{x=\lambda+\partial}L^*(-z_1+\lambda)\big)
\big(\big|_{y=\mu+\partial}L^*(-z_2+\mu)\big)
\,,}
\end{array}$$ Next, combining the first term in the RHS of and the second term in the RHS of , we get the opposite of , proving the claim.

\[20131029:lem\] If $L(\partial)\in{\mathcal{V}}[\partial]$, then $H^{(L)}(\partial)(z,w)\in{\mathcal{V}}[\partial][z,w]$.

