hold, which sum up to the desired equality. Recall the expression
$\widetilde{\exp}$ used in the proof of Lemma A.6. We denote the series

(A.38)
$\begin{gathered}A(Z)_{+}=-\frac{1}{\nu}\sum_{n<0}Z^{-n-1|1}\alpha_{(n|1)},\quad
A(Z)_{-}=-\frac{1}{\nu}\sum_{n\geq 0}Z^{-n-1|1}\alpha_{(n|1)},\\\
B(Z)_{+}=\frac{1}{\nu}\sum_{n<0}\frac{Z^{-n|0}}{n}\alpha_{(n|0)},\quad
B(Z)_{-}=\frac{1}{\nu}\sum_{n>0}\frac{Z^{-n|0}}{n}\alpha_{(n|0)}\end{gathered}$

so that
$\widetilde{\exp}=\exp(A(Z)_{+})\exp(B(Z)_{+})\exp(B(Z)_{-})\exp(A(Z)_{-})Z^{-\frac{1}{\nu}\alpha_{(0|0)}|0}$.
Since $\partial_{z}$ acts as an even derivation for the exponential components
of $\widetilde{\exp}$, one can compute that

$\displaystyle\theta\partial_{z}\big{(}e^{-\frac{1}{\nu}\int\alpha(Z)}\big{)}=\theta\,s_{-\frac{\alpha}{\nu}}\,(\partial_{z}B(Z)_{+})\,\widetilde{\exp}+\theta\,s_{-\frac{\alpha}{\nu}}\,\widetilde{\exp}\,(\partial_{z}B(Z)_{-})+\theta\,s_{-\frac{\alpha}{\nu}}\,\widetilde{\exp}\,\Big{(}-\frac{1}{\nu}Z^{-1|0}\alpha_{(0|0)}\Big{)}$
$\displaystyle=\Big{(}-\frac{1}{\nu}\sum_{n<0}Z^{-n-1|1}\alpha_{(n|0)}\Big{)}e^{-\frac{1}{\nu}\int\alpha(Z)}+(-1)^{p(\alpha)+1}e^{-\frac{1}{\nu}\int\alpha(Z)}\Big{(}-\frac{1}{\nu}\sum_{n\geq
0}Z^{-n-1|1}\alpha_{(n|0)}\Big{)}$
$\displaystyle=-\frac{1}{\nu}:\\!\Big{(}\sum_{n\in\mathbb{Z}}Z^{-n-1|1}\alpha_{(n|0)}\Big{)}e^{-\frac{1}{\nu}\int\alpha(Z)}\\!:,$

which proves (A.36). Note that the differential $\partial_{\theta}$ acts as an
odd differential on the space of formal power series in $Z$ and $Z^{-1}$, and
for any even series $X(Z)$,

(A.39)
$\partial_{\theta}\big{(}\exp(X(Z))\big{)}=(\partial_{\theta}X(Z))\big{(}\\!\exp(X(Z))\big{)}=\big{(}\\!\exp(X(Z))\big{)}(\partial_{\theta}X(Z)),$

when $\partial_{\theta}X(Z)$ and $X(Z)$ supercommute with each other. By the
commutator relation (A.33), one can apply (A.39) for each exponential
component made up of (A.38), when computing (A.37). Therefore, the similar
computation

$\displaystyle\partial_{\theta}\big{(}e^{-\frac{1}{\nu}\int\alpha(Z)}\big{)}=(-1)^{p(\alpha)+1}s_{-\frac{\alpha}{\nu}}\,\partial_{\theta}\big{(}A(Z)_{+}\big{)}\,\widetilde{\exp}+(-1)^{p(\alpha)+1}s_{-\frac{\alpha}{\nu}}\,\widetilde{\exp}\,\partial_{\theta}(A(Z)_{-})$
$\displaystyle=\Big{(}-\frac{1}{\nu}\sum_{n<0}Z^{-n-1|0}\alpha_{(n|1)}\Big{)}e^{-\frac{1}{\nu}\int\alpha(Z)}+(-1)^{p(\alpha)+1}e^{-\frac{1}{\nu}\int\alpha(Z)}\Big{(}-\frac{1}{\nu}\sum_{n\geq
0}Z^{-n-1|0}\alpha_{(n|1)}\Big{)}$
$\displaystyle=-\frac{1}{\nu}:\\!\Big{(}\sum_{n\in\mathbb{Z}}Z^{-n-1|0}\alpha_{(n|1)}\Big{)}e^{-\frac{1}{\nu}\int\alpha(Z)}\\!:$

shows the equality (A.37). ∎

## Acknowledgement

I would like to express my gratitude to my supervisor, Uhi Rinn Suh, for her
invaluable support and guidance in enhancing this paper. Additionally, I am
deeply thankful to Naoki Genra for the enriching discussions we had during his
visit to Seoul National University in 2023. My appreciation also goes to Gahng
Sahn Lee for drawing my attention to the concept of coherent modules.

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