, we have that $A^{(L)}(F)\in{\mathcal{A}}((\partial^{-1}))_{N-1}$ for every $F\in{\mathcal{A}}((\partial^{-1}))$. Moreover, if $F\in{\mathcal{A}}((\partial^{-1}))_{-N-1}$, then $(FL)_+=(LF)_+=0$, and therefore $A^{(L)}(F)=0$. In conclusion, $A^{(L)}$ induces a map $A^{(L)}:\,{\ensuremath{\raisebox{.40ex}{\ensuremath{{\mathcal{A}}((\partial^{-1}))}}
\! \big / \! \raisebox{-.40ex}{\ensuremath{{\mathcal{A}}((\partial^{-1}))_{-N-1}}}}}\longrightarrow{\mathcal{A}}((\partial^{-1}))_{N-1}$. Note that ${\mathcal{A}}((\partial^{-1}))={\mathbb{F}}[\partial,\partial^{-1}]\circ{\mathcal{A}}+{\mathcal{A}}((\partial^{-1}))_{-N-1}$, and $({\mathbb{F}}[\partial,\partial^{-1}]\circ{\mathcal{A}})\cap{\mathcal{A}}((\partial^{-1}))_{-N-1}
=\partial^{-N-1}{\mathbb{F}}[\partial^{-1}]\circ{\mathcal{A}}$. Hence, we can canonically identify $${\ensuremath{\raisebox{.40ex}{\ensuremath{{\mathcal{A}}((\partial^{-1}))}}
\! \big / \! \raisebox{-.40ex}{\ensuremath{{\mathcal{A}}((\partial^{-1}))_{-N-1}}}}}\simeq
{\ensuremath{\raisebox{.40ex}{\ensuremath{\left({\mathbb{F}}[\partial,\partial^{-1}]\circ{\mathcal{A}}\right)}}
\! \big / \! \raisebox{-.40ex}{\ensuremath{\left(\partial^{-N-1}{\mathbb{F}}[\partial^{-1}]\circ{\mathcal{A}}\right)}}}}
\,,$$ and we get the induced map $$\label{adlermap2}
A^{(L)}:
{\ensuremath{\raisebox{.40ex}{\ensuremath{\left({\mathbb{F}}[\partial,\partial^{-1}]\circ{\mathcal{A}}\right)}}
\! \big / \! \raisebox{-.40ex}{\ensuremath{\left(\partial^{-N-1}{\mathbb{F}}[\partial^{-1}]\circ{\mathcal{A}}\right)}}}}
\longrightarrow{\mathcal{A}}((\partial^{-1}))_{N-1}\,.$$

Let $I=\{-N,-N+1,-N+2,\dots\}\subset{\mathbb{Z}}$. We have the identifications $$\label{id1bis}
{\ensuremath{\raisebox{.40ex}{\ensuremath{\left({\mathbb{F}}[\partial,\partial^{-1}]\circ{\mathcal{A}}\right)}}
\! \big / \! \raisebox{-.40ex}{\ensuremath{\left(\partial^{-N-1}
{\mathbb{F}}[\partial^{-1}]\circ{\mathcal{A}}\right)}}}}\stackrel{\sim}{\longrightarrow}{\mathcal{A}}^{\oplus I}
\,\,,\,\,\,\,
\sum_{n=-N}^M\partial^n\circ F_n\mapsto (F_n)_{n\in I}\,,$$ and $$\label{id2bis}
{\mathcal{A}}((\partial^{-1}))_{N-1}\stackrel{\sim}{\longrightarrow}{\mathcal{A}}^I
\,\,,\,\,\,\,
\sum_{n=-N}^\infty P_n\partial^{-n-1}\mapsto(P_n)_{n\in I}\,.$$ Therefore the map $A^{(L)}$ in induces a map $$\label{def:H}
H^{(L)}:\, {\mathcal{A}}^{\oplus I}\longrightarrow {\mathcal{A}}^I\,.$$ This map is given by an $I\times I$ matrix differential operator $H^{(L)}=\big(H^{(L)}_{ij}(\partial)\big)_{i,j\in I}$, which we compute explicitly in terms of the generating series of its entries: $$\label{Hseries}
H^{(L)}(\partial)(z,w)=\sum_{i,j\in I}H_{ij}^{(L)}(\partial)z^{-i-1}w^{-j-1}
\,.$$

\[hseries\] We have $$\label{h}
H^{(L)}(\partial)(z,w)
=L(w)i_w(w-z-\partial)^{-1}\circ L(z)
-L(z+\partial)i_w(w-z-\partial)^{-1}\circ L^*(-w+\partial)
\,,$$ where, as usual, we expand $L(z+\partial)$ and $L^*(-w+\partial)$ in non-negative powers of $\partial$.

By definition, the matrix element $H^{(L)}_{ij}(\partial)$ is given by $$\label{20131022:eq1}
H^{(L)}_{ij}(\partial)(f)
=\operatorname{Res}_{\partial}\left(A^{(L)}\left(\partial^j\circ f\right)\partial^i\right)\,,$$ for all $f\in{\mathcal{A}}$ and $i,j\in I$. Note that, by the above observations on the Adler map $A^{(L)}$, the RHS of is zero for $i$ or $j$ less 