I've been working through some problems in complex analysis (Hille's Analytic Function Theory) and I came across the Weierstrass $\mathscr{P}$ function defined as follows:
$\mathscr{P}_{\Lambda}(z)=\frac{1}{z^2}+\sum\limits_{\omega \in \Lambda}{}^{'}\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}$,
and it can be shown that it satisfies the differential equation
$(\mathscr{P}_{\Lambda}'(z))^2=4\mathscr{P}_{\Lambda}(z)-60G_4(\Lambda)\mathscr{P}_{\Lambda}(z)-140G_6(\Lambda)$.
Applying the map $(\mathscr{P}_{\Lambda}(z),\mathscr{P}_{\Lambda}'(z))\rightarrow(x,y)$ gives rise to the elliptic curve $y^2-4x^3-g_2x-g_3$. Thus, every lattice has a corresponding elliptic curve. By using the surjectivity of the modular invariant (see Diamond & Shurman), we can show the other direction - every elliptic curve has a corresponding lattice.
Hille later shows that for any doubly-periodic ("elliptic") function of order $m$ satisfies a certain differential equation
$\left(\frac{df}{dz}\right)^m+\sum\limits_{k=1}^{m}P_k(f)\left(\frac{df}{dz}\right)^{m-k}=0, \quad \deg P_k(z) \leq 2k$.
Applying a similar map $(f(z),\frac{df}{dz})\rightarrow (x,y)$ results in the algebraic curve
$y^m+\sum\limits_{k=1}^{m}P_k(y)x^{m-k}=0, \quad \deg P_k(z) \leq 2k$.
Thus, every doubly-periodic function has a corresponding algebraic curve. However, his proof is by contradiction and non-constructive. My question is if, as before, we can go backwards: given an algebraic curve, is there a corresponding-doubly periodic function?
Any function of genus $0$ cannot corresponds to a doubly periodic function because each of them or a rationally equivalent curve are parameterized by a rational parameter what is not possible for an elliptic curve (which is of genus $1$). Curves of genus $1$ are all them rationally equivalent to a non-singular cubic and this elliptic cubic is rationally parameterized not by a rational parameter but yes precisely by a Weierstrass $\mathscr{P}$ function.