In what topology, elliptic curve is homeomorphic to torus?

Question

I often hear elliptic curve is homeomorphic to torus as topology.

In this context, what topology we are dealing?

Euclid topology on C and it's quotient topology? Or Zariski topology?

Thank you in advance.

Answer 1

Question: "I often hear elliptic curve is homeomorphic to torus as topology. In this context, what topology we are dealing? Euclid topology on C and it's quotient topology? Or Zariski topology? Thank you in advance."

Answer: There is a section in Hartshorne (Section IV.4) on elliptic curves that treats elliptic curves over the complex numbers $K$. If $\Gamma \subseteq K$ is a "lattice" (a subgroup on the form $\Gamma\cong \mathbb{Z}\oplus \mathbb{Z}\tau$) with $\tau \notin \mathbb{R} \subseteq K$, you may define the "quotient" $E(\Gamma):=K/\Gamma$. When defined like this, it follows $E(\Gamma)$ is a topological space with a continuous "quotient map" $p: K \rightarrow E(\Gamma)$. In HH.IV.4 they prove that $E(\Gamma)$ may be given the structure of an algebraic variety over $K$. This algebraic variety has an embedding $E(\Gamma)\cong V(F) \subseteq \mathbb{P}^2_K$, where $F(x,y,z) \in K[x,y,z]$ is an explicit polynomial in $x, y$ and $z$ - they construct in HH a 2-variable polynomial in $x,y$.

As a topological space (in the "strong" topology inherited from $K$) there is for any lattice $\Gamma$ a homeomorphism

$$E(\Gamma)^{top} \cong \mathbb{R}/\mathbb{Z}\times \mathbb{R}/\mathbb{Z} \cong S^1 \times S^1.$$

Theorem IV.4.15B proves that for different choices of $\tau,\tau'$ you get lattices $\Gamma, \Gamma'$ and curves $E:=E(\Gamma), E':=E(\Gamma')$. The curves $E,E'$ are isomorphic as algebraic varieties iff there are integers $a,b,c,d$ with $ad-bc \neq +1,-1$ and

$$E1.\text{ } \tau'=\frac{a\tau+b}{c\tau+d}.$$

There is always a homeomorphism $E(\Gamma)^{top} \cong E(\Gamma')^{top}$.

Question: "Euclid topology on C and it's quotient topology? Or Zariski topology?"

Answer: The curves $E(\Gamma), E(\Gamma')$ are always "isomorphic" as topological spaces in the "strong" topology. They are isomorphic as algebraic varieties (equipped with the Zariski topology) iff condition $E1$ holds. Moreover, if we equip $E(\Gamma),E(\Gamma')$ with the structure of complex projective manifolds to get manifolds $E(\Gamma)_s, E(\Gamma')_s$, it follows $E(\Gamma)_s \cong E(\Gamma')_s$ are isomorphic as complex manifolds (in the "strong" topology) iff $E1$ holds.

Note: If

\begin{align*} A= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*}

with $a,b,c,d\in \mathbb{Z}$ and $det(A) \neq 1,-1$ you get a set of matrices

$$M \subseteq Mat(2,\mathbb{Z})$$

and $M$ is not a subgroup of $G:=SL(2,\mathbb{Z})$. Hence the relation in $E1$ is not induced by an action of a subgroup $H \subseteq G$. Still you get a "parameter space" of elliptic curves, parametrized by $\tau \in K-\mathbb{R}$, where two complex numbers $\tau,\tau'$ are equivalent iff $E1$ holds for some $A\in M$.

Answer 2

It depends what you mean by "torus" - if you mean the manifold $S^1\times S^1$, the only possible choice is the Euclidean/analytic topology, because manifolds are Hausdorff and the Zariski topology is not Hausdorff except on finite sets. There are other ways to consider torii, though.

If that is indeed what you mean, this is talked about basically all over the place - for instance, here's a description at wikipedia.