We need to prove the converse to Lemma 2.12. The $T^{2}$-symmetry acts on functions via

$$(e^{i\theta_{1}},e^{i\theta_{2}})\cdot f=f(e^{i(\theta_{1}+\theta_{2})}z_{0},e^{-i\theta_{1}}z_{1},e^{-i\theta_{2}}z_{2}).$$

This action allows us to expand any holomorphic function $f$ as a Fourier series on every $T^{2}$-fibre:

$$f=\sum_{n,m\in\mathbb{Z}^{2}}f_{n,m},$$

where $f_{n,m}$ has weight $(n,m)$ with respect to the $T^{2}$ action. Since the $T^{2}$ action is holomorphic, the Fourier components $f_{n,m}$ are also holomorphic. Furthermore, these $f_{n,m}$ satisfy the same growth condition as $f$ after perhaps increasing $K_{1}$.

We claim every $f_{n,m}$ is algebraic. To see this, we can find a suitable monomial of $z_{0},z_{1},z_{2}$ which has the same weight as $f_{n,m}$, such that $f_{n,m}$ divided by this monomial has no pole along $\mathfrak{D}_{i}$. But this quotient function is $T^{2}$-invariant and holomorphic, so depends only on $\eta$, and in fact has to be a polynomial of $\eta$ by the growth condition.

By applying the Parseval identify to every $T^{2}$-fibre, we obtain

$$\frac{1}{4\pi^{2}}\int_{T^{2}}|f|^{2}\vartheta_{1}\wedge\vartheta_{2}=\sum_{n,m}|f_{n,m}|^{2}.$$

Both LHS and RHS are functions of $\mu_{1},\mu_{2},\eta$, and LHS has a bound of type (2.18) by assumption. But for any given $K_{2},K_{3}$, only finitely many monomials of $z_{0},z_{1},z_{2}$ satisfy the growth bound (2.18) globally, so only finitely many $f_{n,m}$ can appear as summands. Hence $f$ is algebraic as required. ∎