Proposition 8.9.7.26EFAEC0B76D Write $R_{p-1}^{\widehat a}$ for the shifted current obtained by omitting $u_ a$, in the inherited order, and use two hats for two omissions. Put $R_0=1$. For $1\leq p\leq m$,
\[ \begin{aligned} \mathrm dR_ p& =\mathrm d\omega \wedge \sum _{a=1}^ pR_{p-1}^{\widehat a},\\ \mathrm d^{\mathrm c}R_ p& =\mathrm d^{\mathrm c}\omega \wedge \sum _{a=1}^ pR_{p-1}^{\widehat a},\\ \mathrm{dd}^{\mathrm{c}}R_ p& =\mathrm{dd}^{\mathrm{c}}\omega \wedge \sum _{a=1}^ pR_{p-1}^{\widehat a} +2\, \mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega \wedge \sum _{a<b}R_{p-2}^{\widehat a\widehat b}. \end{aligned} \]
The last sum is absent for $p=1$. All currents on the right have order zero. If all but at most one potential are smooth, $R_ p$ is independent of the ordering and is the ordinary product of the smooth factors with $\omega +\mathrm{dd}^{\mathrm{c}}u$.
Proof.
Every $D_ I$ is both $\mathrm d$- and $\mathrm d^{\mathrm c}$-closed. Differentiating (8.9.5) therefore gives
\[ \mathrm dR_ p=\mathrm d\omega \wedge \sum _ I(p-|I|)\omega ^{p-|I|-1}\wedge D_ I, \]
with zero terms omitted. The coefficient $p-|I|$ counts the omitted indices outside $I$, which proves the first formula; the second is identical with $\mathrm d^{\mathrm c}$ in place of $\mathrm d$. Apply $\mathrm d$ to the second formula. Each pair of omitted indices occurs twice, and $-\mathrm d^{\mathrm c}\omega \wedge \mathrm d\omega =\mathrm d\omega \wedge \mathrm d^{\mathrm c}\omega $ because both forms have odd degree three. This gives the displayed $\mathrm{dd}^{\mathrm{c}}$ formula, with every background derivative retained.
If all but one potential are smooth, applying $\mathrm{dd}^{\mathrm{c}}$ to the bounded potential times a smooth closed product gives its $\mathrm{dd}^{\mathrm{c}}$ wedged with that product. At later smooth steps, closedness again makes the recursion ordinary smooth multiplication. This is independent of the position of the bounded potential, and the subset expansion gives the last assertion.
$\square$
Pome
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