===== recycled on Sat Aug 10 11:22:23 EDT 2019 by drorbn on Ubuntu-on-X2 ======

Using the pairing
\[
  \langle z_i^m,\zeta_j^n \rangle
  = \left.\partial_{\zeta_i}^m\zeta_j^n\right|_{\zeta_A\to 0}
  = \delta_{ij}\delta_{mn}n!,
\]
we get a map
\begin{multline*}
   \calD\colon\Hom(\bbQ\llbracket z_A\rrbracket\to\bbQ\llbracket z_B\rrbracket)
  \cong\bbQ\llbracket z_A\rrbracket^*\otimes\bbQ\llbracket z_B\rrbracket \\
  \cong\bbQ\llbracket \zeta_A\rrbracket\otimes\bbQ\llbracket z_B\rrbracket
  \cong\bbQ\llbracket \zeta_A,z_B \rrbracket
\end{multline*}

===== recycled on Sat Aug 10 11:22:27 EDT 2019 by drorbn on Ubuntu-on-X2 ======


{\bf Example.} $\calD(id\colon\bbQ\llbracket z\rrbracket\to\bbQ\llbracket
z\rrbracket)=\bbe^{\zeta z}$. Indeed,
\[ \langle z^n, \bbe^{\zeta z}\rangle
  = \left\langle z^n, \sum_m\frac{(\zeta z)^m}{m!}\right\rangle
  = \sum_m\frac{z^m}{m!}\delta_{mn}n! = z^n.
\]

===== recycled on Sat Aug 10 11:23:48 EDT 2019 by drorbn on Ubuntu-on-X2 ======


{\bf\red Other GDOs.} {\bf Claim.} If $L\colon\bbQ\llbracket
z_A\rrbracket\to\bbQ\llbracket z_B\rrbracket$ is linear, then
$\calD(L)=L\left(\bbe^{\sum_{i\in A}\zeta_i z_i}\right)$.\hfill\text{{\bf
Proof.} Exercise.}