# non-equivalent Riemann surfaces of genus $1$

It is well known that all the compact orientable connected Hausdorff genus $1$ surfaces are homeomorphic, but they may have different complex structures.

In fact, consider the following connected region in $\mathbb{C}$. $$G=\{ z\in\mathbb{C}\colon |z|> 1, \Im(z)>0, |\Re(z)|<\frac{1}{2} \}.$$ Consider distinct points $w_1,w_2$ in the region $G$, we can form two lattices: $$\Gamma_1=\mathbb{Z}+w_1\mathbb{Z} \,\,\,\,\mbox{ and } \,\,\,\,\Gamma_2=\mathbb{Z}+w_2\mathbb{Z}.$$ Then $\mathbb{C}/\Gamma_1$ and $\mathbb{C}/\Gamma_2$ are non-equivalent compact Riemann surfaces of genus $1$. This is quite standard fact and its proof can be found in Ahlfors' classic text.

My question is about thinking on this fact in a little different way: can we find some property of Riemann surface $\mathbb{C}/\Gamma_1$ which is not satisfied by other one, so that they are not equivalent?

In algebra or topology, it is very common practice that proving two objects to be non-equivalent involves finding some property of one object which is not satisfied by other object. I am thinking in this way for the problem of non-equivalence of Riemann surfaces of genus $1$.

• The moduli space of Riemann surfaces of genus $1$ is connected, which tells you that there is no "discrete" invariant distinguishing these surfaces. So it's unclear what would qualify as an answer. Once you know that this classification is the same as the classification of flat Riemannian tori up to scale you can appeal to tools from Riemannian geometry, e.g. the spectrum of the Laplacian. – Qiaochu Yuan Apr 27 '16 at 5:59
• @QiaochuYuan I would consider your comment "answer worthy" for this question... – gebruiker Jul 27 '17 at 11:32