in ${\mathcal{A}}[\partial]((z^{-1}))((w^{-1}))$ (and the powers of $z$ are not bounded above).

Preliminary properties of the Adler map {#sec:2.2}
---------------------------------------

Let ${\mathcal{V}}$ be a differential algebra, let $L(\partial)\in{\mathcal{V}}((\partial^{-1}))$ be a pseudodifferential operator of order $N$, and let $H^{(L)}(\partial)(z,w)$ be as in .

\[20130925:lem1\]

(a) $H^{(L)}(\partial)(z,w)
    =-H^{(L)}(\partial)^*(w,z)$.

(b) The following identity holds: $$\label{20130925:eq_jacobi}
    \begin{array}{l}
    \displaystyle{
    H^{(L)}(\lambda)(z_2,z_1)
    i_{z_2}(z_2-z_3-\mu-\partial)^{-1}
    L(z_3)
    } \\
    \displaystyle{
    -H^{(L)}(\lambda)(z_3+\mu+\partial,z_1)
    i_{z_2}(z_2-z_3-\mu-\partial)^{-1}
    L^*(-z_2+\mu)
    } \\
    \displaystyle{
    -L(z_1)
    i_{z_1}(z_1-z_3-\lambda-\mu-\partial)^{-1}
    H^{(L)}(\mu)(z_3,z_2)
    } \\
    \displaystyle{
    +L(z_3+\lambda+\mu+\partial)
    i_{z_1}(z_1-z_3-\lambda-\mu-\partial)^{-1}
    \Big(\Big|_{x=\lambda+\mu+\partial}
    \!\!\!
    H^{(L)}(\mu)(z_1-x,z_2)
    \Big)
    } \\
    \displaystyle{
    =H^{(L)}(\lambda+\mu+\partial)(z_3,z_1)
    i_{z_1}(z_1-z_2-\lambda-\partial)^{-1}
    L(z_2)
    } \\
    \displaystyle{
    -H^{(L)}(\lambda+\mu+\partial)(z_3,z_2+\lambda+\partial)
    i_{z_1}(z_1-z_2-\lambda-\partial)^{-1}L^*(-z_1+\lambda)\,.
    }
    \end{array}$$

In the fourth term of the RHS of equation we used the following notation: for a Laurent series $P(z)=\sum_{n=-\infty}^Nc_nz^n\in{\mathcal{A}}((z^{-1}))$ and elements $a,b\in{\mathcal{A}}$, we let $$\label{20131024:eq1}
a\Big(\Big|_{x=\nu+\partial}P(z+x)b\Big)
=
\sum_{n=-\infty}^Nai_z(z+\nu+\partial)^n(c_nb)
\,.$$

Taking the adjoint of and using equation , we get $$\label{h2}
H^{(L)}(\partial)^*(z,w)
=L(z)i_w(w-z+\partial)^{-1}L(w)
-L(w+\partial)i_w(w-z+\partial)^{-1}L^*(-z+\partial)
\,.$$ Combining equations and , and using equation , we get, $$\label{20131108:eq2}
\begin{array}{l}
H^{(L)}(\partial)(z,w)
+H^{(L)}(\partial)^*(w,z)
\\
=L(w)\delta(w-z-\partial)L(z)
-L(z+\partial)\delta(w-z-\partial)L^*(-w+\partial)
\,.
\end{array}$$ By equation , the first term in the RHS of is equal to $L(z+\partial)\delta(w-z-\partial)L(z)$. Moreover, if $L(z)=\sum_na_nz^n$, then $L^*(-w+\partial)=\sum_n(w-\partial)^na_n$. Therefore, by equation , $\delta(w-z-\partial)L^*(-w+\partial)=\delta(w-z-\partial)\sum_nz^na_n=\delta(w-z-\partial)L(z)$. Hence, the second term in the RHS of is equal to $L(z+\partial)\delta(w-z-\partial)L(z)$ as well, proving part (a).

Using , we can rewrite each 