Elliptic curve with prescribed lattice

Question

It's well known that there is a connection between elliptic curves and lattices. To establish such a connection one needs to use Eisenstein series.

How one can one write down the explicit equation of an elliptic curve knowing its lattice?

For example, what is the equation of the elliptic curve with lattice $<1,i>$?

Answer 1

Let $L$ be a lattice. Then the explicit equation of the associated elliptic curve is given by $$ E_L\colon y^2=4x^3-g_2(L)x-g_3(L). $$ Here the Eisenstein series are given by $$ g_2(L)=60\sum_{L^{\times}}\frac{1}{\omega^4},\quad g_3(L)=140\sum_{L^{\times}}\frac{1}{\omega^6}. $$ If we can compute these series explicitly, then we have an explicit equation of the elliptic curve. For standard lattices, like the Gaussian lattice $L=\mathbb{Z}[i]$, or the hexagonal lattice these series are well known and have been computed. For the Gaussian lattice with basis $1,i$ the result is $$ g_2(L)=\frac{\Gamma(1/4)^8}{16\pi^2}>0,\quad g_3(L)=0. $$ For the lattice $L=\mathbb{Z}[\frac{-1-\sqrt{-3}}{2}]$ we obtain the curve $y^2=4x^3-g_3(L)$.

Answer 2

I think that, in general, this is a tough question. If $\Lambda$ is a lattice in $\mathbb C$, then the elliptic curve $\mathbb C/\Lambda$ has an equation of the form $y^2=x^3+g_2(\Lambda)x+g_3(\Lambda)$ where $$g_2(\Lambda)=60\sum_{0\neq \omega\in \Lambda}\frac{1}{\omega^4}$$ and $$g_3(\Lambda)=140\sum_{0\neq \omega\in \Lambda}\frac{1}{\omega^6}$$

If $\Lambda=\langle 1,i\rangle$, it is an easy exercise to check that $g_3(\Lambda)=0$. Therefore the elliptic curve $\mathbb C/\Lambda$ has j-invariant $1728$ and thus is isomorphic to the curve $y^2=x^3+x$.