When $\mathbb{C}/\Lambda$ is a Riemann surface?

Question

Let $\Lambda$ be a lattice, that is $\Lambda＝\{a\omega_1＋b\omega_2\mid a,b\in\mathbb Z\}$.

I heard that the necessarily and sufficient condition that $\mathbb{C}/\Lambda$ be a Riemann surface is $\omega_1$ and $\omega_2$ are linearly independent over $\mathbb R$.(Here, we give equivalent relation to $\mathbb{C}$ by $x\sim y$ with $x-y\in\Lambda$).

But how can I formally prove this?

Thank you in advance.

Answer 1

It is a Riemann surface iff

• $\omega_1,\omega_2$ are $\Bbb{R}$-linearly independent, in which case $\mathbb{C}/\Lambda$ is a compact Riemann surface (a complex torus, isomorphic to a projective elliptic curve),

• or $\Bbb{Q}$-linearly dependent, in which case it is isomorphic to $\Bbb{C/Z}$ or $\Bbb{C}$ depending on $\omega_j\ne 0$.

The word lattice assumes that $\omega_1,\omega_2$ are $\Bbb{R}$-linearly independent.

The non-obvious theorem is that they are often non-isomorphic as Riemann surfaces when changing the $\omega_1,\omega_2$ pair (this is proven usually with the unique space of holomorphic differential $\{c dz,c\in \Bbb{C}\}$)