# In what topology, elliptic curve is homeomorphic to torus?

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.

• Please make the question clearer. What does an elliptic curve mean to you? If you approach them from the direction of complex analysis the answer is almost obvious, because an elliptic curve is constructed as $\Bbb{C}/\Lambda$ where $\Lambda$ is a lattice (rank two discrete free abelian group in the complex plane). If your background is an cryptography, and you have only seen elliptic curves over finite fields (or their algebraic closure), then a good answer would need a lot more. If you have seen sets of real points, then we need yet another. Jul 4 '21 at 6:54
• I define elliptic curve over field K as genus 1 cuver over K with base point. I'm familiar with this elliptic curve over any field. Specially if K＝C, I met the fact that elliptic curve is homeomorphic to torus.
– user925204
Jul 4 '21 at 6:59
• Neronoggshafarevich, that should a part of the question. Not a comment. See the edit button under the question? Also, if you want elliptic curves compared to torii, may be you should also explain what a torus is for you. That also means different things in different places. Jul 5 '21 at 9:15

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.

• What is the definition of ''torus'' in your latter argument?(In that meaning, can the apology be different from euclidian?)
– user925204
Jul 4 '21 at 7:02
• I've never met a Euclidean apology before! The second paragraph is supposed to refer to taking the torus to be the manifold $S^1\times S^1$ as described in the first paragraph - I don't describe any other torii or make use of any other definitions, I just know that there are a lot of other ways one can use that word in math. Jul 4 '21 at 7:09

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 $$A\in M$$.