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Let, $\Gamma, \Gamma'$ be lattices of $\mathbb C$, define ellicptic curves by $\mathbb C/\Gamma , \mathbb C/\Gamma'$, then

$\mathbb C/\Gamma , \mathbb C/\Gamma'$ are isomorphic $\Leftrightarrow$ $\Gamma=\lambda\Gamma'.$

The $\Leftarrow$ part is easy, but how to prove the $\Rightarrow$ part?

(Rmk:

I am reading Serre's A Course In Arithmetic, it doesn't particularly treat elliptic curves, and just writes:

"Let us associate to a lattice $\Gamma$ of $\mathbb C$ the elliptic curve $E_\Gamma =\mathbb C/ \Gamma$. It is easy to see that two lattices $\Gamma, \Gamma'$ define isomorphic elliptic curves if and only if they are homothety."

That's all I know about elliptic curves now, so I am not sure what does "isomorphic" mean in original question. ... I thought it means group isomorphism, but I am not sure now. It is helpful that anyone clarifies what is the author talking about.)

Cornman
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CYC
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  • What's your definition of isomorphism of elliptic curves? – Amitai Yuval Feb 10 '15 at 10:22
  • @AmitaiYuval Group isomorphism in my mind. (I encounter elliptic curves the first time, and the book($A$ $Course$ $In$ $Arithmetic$) I am reading doesn't specifically define it.) – CYC Feb 10 '15 at 10:40
  • Every $\Bbb C/\Lambda$ is obviously isomorphic to the torus $(\Bbb R/\Bbb Z)^2$ as a real Lie group. (Not only does this imply a group isomorphism, but a smooth group isomorphism.) The statement is true however if isomorphism is taken to mean biholomorphism - with no reference to the group structure necessary, only the complex geometry. – anon Feb 10 '15 at 11:04
  • @anon So in order to let the statement true, such $isomorphism$ should mean $biholomorphism$? – CYC Feb 10 '15 at 11:08
  • In Husemöller's "Elliptic Curves" that is precisely the definition of "isomorphic Complex Tori" (page 211, chapter 11, 1.4) . This can be seen as preceeding the fact that complex torus is the same as (complex) elliptic curve. The question assumes already that $;C/\Gamma;$ is an elliptic curve, so that things have to be done, imo, in another order. Are you reading some particular book on this? – Timbuc Feb 10 '15 at 11:34
  • @Timbuc I am reading Serre's $A$ $Course$ $In$ $Arithmetic$, it doesn't particular treat elliptic curves, and just writes "Let us associate to a lattice $\Gamma$ of $\mathbb C$ the $elliptic$ $curve$ $E_{\Gamma} =\mathbb C/ \Gamma$. It is easy to see that two lattices $\Gamma, \Gamma'$ define isomorphic elliptic curves if and only if they are homothety." That's all I know about elliptic curves now, so I am not sure what does "isomorphic" means ... I thought it means $group$ $isomorphism$, but I am not sure now. – CYC Feb 10 '15 at 12:04
  • @CYC I see. That book is one of the most beautiful ones I know in higher mathematics, but it can be tough to read trying to learn something from scratch. I'll try to come up later with some basic, algebraic approach to this. – Timbuc Feb 10 '15 at 12:30
  • @CYC If you've read the book then perhaps you can see two elliptic curves as defined there are isomorphic iff the lattices are congruent modulo the modular group...? Because with this then I think the direction $;\implies;$ can be proved with reasonable ease. – Timbuc Feb 10 '15 at 13:16

1 Answers1

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Hint:

A map $\mathbb{C}/\Gamma \to \mathbb{C}/\Gamma'$ comes from a map $\phi\colon \mathbb{C} \to \mathbb{C}$ such that $\phi( z + \gamma) = \phi(z) + f(\gamma)$, for some function $f$. Hence for the derivative we have $\phi'(z+ \gamma) = \phi'(z)$, and so $\phi'(z)$ is a $\Gamma$ periodic functions and thus constant $\equiv \alpha$. Therefore $\phi(z) = \alpha z + \beta$ with the extra condition $\alpha(\Gamma) \subset \Gamma'$. We have found all the maps between tori. Now it's easy.

orangeskid
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  • I am not familiar with this, isn't isomorphism of $\mathbb C /\Gamma$ a group isomorphism? If it does, then what's the definition of $\phi$ and $f$? – CYC Feb 10 '15 at 10:53
  • Yes, you can extra add the condition that $\phi$ is additive. But you can prove a more general result for maps ( regular or holomorphic). – orangeskid Feb 10 '15 at 10:55
  • Will a $homomorphism$ on $\mathbb C$ as additive abelian group automatically become a $holomorphic$ function? – CYC Feb 10 '15 at 11:03
  • Or the $isomorphism$ of elliptic curves defined by $\mathbb C / \Gamma$ means existence of $biholomorphism$ at first place? – CYC Feb 10 '15 at 11:11
  • @CYC: You have to mind the Galois action. You could consider an elliptic curve defined over $\mathbb{R}$ ( say with a lattice $\mathbb{Z}[i]$ ) and consider $(x,y) \mapsto (\bar x, \bar y)$. Yeah, not algebraic. – orangeskid Feb 10 '15 at 11:13
  • So what does "map $\mathbb C/\Gamma \rightarrow \mathbb C/\Gamma'$" means exactly? And how does $\phi$ and $f$ comes from that map? I think I need to clarify some confusion. – CYC Feb 10 '15 at 11:29
  • $\mathbb{C}/\Gamma$ has a natural structure of a complex manifold of dimension $1$. It is in fact a complex compact Lie group. It turns out that holomorphic maps between two such manifolds are translates of morphisms of Lie groups. ( basically, $\alpha z + \beta$ modulo the lattices.) – orangeskid Feb 10 '15 at 11:36
  • @CYC: you need to read a bit about lifts for covering maps. – orangeskid Feb 10 '15 at 11:37
  • @CYC: This approach is the complex analytic one, not the algebraic one. They do give the same conclusions. I suggest trying to understand continuous maps from $S^1$ to $S^1$, that is, maps from $\mathbb{R}/\mathbb{Z}$ to itself, continuous. It won't be identical to the elliptic curve case but the start is similar. – orangeskid Feb 10 '15 at 22:55