We consider the Laplacian

$$\begin{split}\Delta_{a}(\frac{\partial w^{p\bar{q}}}{\partial\eta_{r}}-\frac{\partial w^{rq}}{\partial\eta_{p}})=&\frac{\partial}{\partial\eta_{r}}\Delta_{a}w^{p\bar{q}}-\frac{\partial}{\partial\eta_{p}}\Delta_{a}w^{rq}\\ =&-4\pi A^{-1}\{\frac{\partial}{\partial\eta_{r}}(\delta(f_{S})z_{p}\bar{z}_{q})-\frac{\partial}{\partial\eta_{p}}(\delta(f_{S})z_{r}\bar{z}_{q})\}\\ =&-4\pi A^{-1}\delta(f_{S})\{\frac{\partial}{\partial\eta_{r}}(z_{p}\bar{z}_{q})-\frac{\partial}{\partial\eta_{p}}(z_{r}\bar{z}_{q})\}=0,\end{split}$$

where we have crucially used that $S$ is an algebraic cycle to deduce $\partial\delta(f_{S})=0$. Thus $\frac{\partial w^{p\bar{q}}}{\partial\eta_{r}}-\frac{\partial w^{rq}}{\partial\eta_{p}}=0$ would follow from a Liouville theorem argument, by checking some a priori growth condition

$$|\frac{\partial w^{p\bar{q}}}{\partial\eta_{r}}|\lesssim\begin{cases}R^{-1},\quad&\varrho\gtrsim A^{1/4},\\ A^{1/4}R^{-2},\quad&\varrho\lesssim A^{1/4},\end{cases}$$

which is easy to derive using the techniques in the previous lemmas in this Section. The $\bar{\eta}$ derivatives can be treated similarly. ∎