than $-N$. Therefore, recalling the definition of the $\delta$-function, $$\label{20131022:eq2}
\begin{array}{l}
\displaystyle{
H^{(L)}(\partial)(z,w)f
=\sum_{i,j\in{\mathbb{Z}}}\operatorname{Res}_{\partial}\left(A^{(L)}
\left(\partial^j\circ f\right)\partial^i\right)z^{-i-1}w^{-j-1}=
} \\
\displaystyle{
=\operatorname{Res}_{\partial}\!
\Big(
\big(L(\partial)\delta(w-\partial)\circ f\big)_+
L(\partial)
\delta(z-\partial)
-L(\partial)
\big(\delta(w-\partial)\circ fL(\partial)\big)_+
\delta(z-\partial)
\Big).
}
\end{array}$$ By equation , we have $$\label{20131022:eq3}
\left(L(\partial)\delta(w-\partial)\circ f\right)_+
=L(w)\delta(w-\partial)_+\circ f=L(w)i_w(w-\partial)^{-1}\circ f\,.$$ Similarly, $$\label{20131022:eq4}
\left(\delta(w-\partial)\circ fL(\partial)\right)_+
=\delta(w-\partial)_+\circ\left(L^*(-w+\partial)f\right)
=i_w(w-\partial)^{-1}\circ\left(L^*(-w+\partial)f\right)\,,$$ where $(L^*(-w+\partial)f)\in{\mathcal{A}}((w^{-1}))$ is obtained by applying the (non-negative) powers of $\partial$ to $f$ (we put parentheses to denote this). Combining equations , and , we get $$\label{20131108:eq1}
\begin{array}{c}
H^{(L)}(\partial)(z,w)f
=\operatorname{Res}_{\partial}\Big(
L(w)i_w(w-\partial)^{-1}\circ fL(\partial)\delta(z-\partial)
\\
-L(\partial)i_w(w-\partial)^{-1}\circ\left(L^*(-w+\partial)f\right)\delta(z-\partial)
\Big)\,.
\end{array}$$ By equation , inside the residue we can replace $\partial$, written on the right, by $z$ (and therefore $\partial$, written anywhere, by $z+\partial$, written in the same place). Hence, equation gives .

In the last term of equation $L^*(\partial)$ denotes the adjoint of the pseudodifferential operator $L(\partial)$, and $L^*(-w+\partial)\in{\mathcal{A}}[\partial]((w^{-1}))$ is obtained by replacing $\partial$ by $-w+\partial$ and expanding in non-negative powers of $\partial$. This is not the same as the adjoint of the differential operator $L(-w+\partial)$. In fact, we have the identity $$\label{20131022:eq5}
L(z+\partial)^*=L^*(-z+\partial)\,.$$

Note that, while $H^{(L)}(\partial)(z,w)$ lies in ${\mathcal{A}}[\partial]((z^{-1},w^{-1}))$, i.e. it has powers of $z$ and $w$ simultaneously bounded above (by construction), the two terms in the RHS of do not: they lie 