When are two tori biholomorphic?

Question

If $\Lambda \subset \mathbb{C}$ is a lattice, let $T_{\Lambda}$ be the torus $\mathbb{C}/\Lambda$. My question is:

If $\Lambda_1, \Lambda_2 \subset \mathbb{C}$ are two lattices, when are $T_{\Lambda_1}$ and $T_{\Lambda_2}$ biholomorphic?

I found that if there exists a biholomorphism $\varphi : \mathbb{C} \to \mathbb{C}$ sending $\Lambda_1$ to $\Lambda_2$ (ie. $\varphi(\Lambda_1)= \Lambda_2$), then $\varphi$ induces a biholomorphism from $T_{\Lambda_1}$ to $T_{\Lambda_2}$. So I looked for biholomorphic matrices and I found $G=\left\{ \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) : (a,b) \in \mathbb{R}^2 \backslash \{(0,0)\} \right\}$; as a subgroup of $\text{GL}_2(\mathbb{R})$, $G \simeq \mathbb{C}^*$ hence:

If $(\omega_1,\omega_2)$ and $(\alpha_1,\alpha_2)$ are colinear in $\mathbb{C}^2$ (as a $\mathbb{C}$-vector space), then $T_{\Lambda_1}$ and $T_{\Lambda_2}$ are biholomorphic, with $\Lambda_1=\omega_1 \mathbb{Z}+ \omega_2 \mathbb{Z}$ and $\Lambda_2= \alpha_1 \mathbb{Z}+ \alpha_2 \mathbb{Z}$.

However, it is far from being complete. Also, I don't know how to prove that two such tori are not biholomorphic, is there a nice invariant for that? (In fact, I realised I only know invariant by homeomorphism.)

Answer 1

Two lattices determine the same torus (or elliptic curve) if and only if the lattices $\Lambda_1, \Lambda_2$ are homotethic, i.e. $\Lambda_1 = \alpha \Lambda_2$ for some $\alpha \in \mathbf C^\times$. You have been able to prove one direction. For the other direction, suppose you are given an isomorphism $\mathbf C/\Lambda_1 \to \mathbf C/\Lambda_2$. This isomorphism lifts to an isomorphism between their universal covering spaces, $\mathbf C \to \mathbf C$, such that $\varphi(\Lambda_1) = \Lambda_2$. Then we are reduced to the case which you have already shown.

To answer your question regarding invariants: the set $\{\Lambda : \Lambda \subseteq \mathbf C \text{ is a lattice}\}/\sim$, where $\sim$ is the equivalence relation $\Lambda \sim \alpha \Lambda$, classifies elliptic curves, by what we have just said. It turns out that this set has a very important geometric description. If $\Lambda = \left<\omega_1, \omega_2\right >$, then we have $\Lambda \sim \left< \omega_1/\omega_2, 1\right >$, and we can always make sure that $\tau = \omega_1/\omega_2$ has positive imaginary part (by switching $\omega_1$ and $\omega_2$). Moreover, if we change the basis $\{\omega_1, \omega_2\}$ by an element of $\text{SL}_2(\mathbf Z)$, so we get the basis $\{a\omega_1 + b\omega_2, c\omega_1 + d\omega_2\}$, then $\tau$ is changed into $\frac{a\tau + b}{c\tau + d}$. Thus, we have arrived at the following observation:

The set $\{\Lambda : \Lambda \subseteq \mathbf C \text{ is a lattice}\}/\sim$ is in bijection with the orbits of $\text{SL}_2(\mathbf Z)$ acting as Möbius transformations on the upper half-plane $\mathbf H$.

In turns out that $\mathbf H/\text{SL}_2(\mathbf Z)$ has the structure of a noncompact Riemann surface isomorphic to $\mathbf C$. The function which usually supplies this isomorphism is called the elliptic modular invariant, or the Klein j-function. It is the simplest (nontrivial) example of a modular form. Thus, every elliptic curve over $\mathbf C$ has a $j$-invariant, which is a complex number, and two elliptic curves are isomorphic if and only if they have the same $j$-invariant.

If you'd like to learn more, any book which discusses elliptic curves over $\mathbf C$ should satisfy your curiosity.