## Elliptic Curves, Lattices, Lie Algebras

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I've recently started to look at elliptic curves and have three basic questions:

1. Is it correct to say that elliptic curves $E$ in the projective plane are in bijective correspondence with lattices $L$ in the complex plane via $E$ <--> $C/L$.

2. If so, is there an explicit expression of the lattice generators in terms of the equation defining the curve? Or, at least, is there a simple example of a curve and its corresponding lattice?

3. Since every elliptic curve is a Lie group, it must have a corresponding Lie algebra. Is there an explicit expression of the Lie algebra in terms of the equation or lattice? Or, again, a simple example of a curve and its Lie algebra (or, even better, an example of a curve, its lattice, and its Lie algebra).

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## 4 Answers

What you want in terms of the relation between lattices and elliptic curves over C is proposition I.4.4 of Silverman's Advanced topics in the arithmetic of elliptic curves. Additionally, to go from a lattice to the equation of the elliptic curve (explicitly), you use Eisenstein series as in Corollary I.4.3 of that book. To go from the elliptic curve to the lattice is described in the proof of proposition I.4.4: you look at the homology of the curve and compute period integrals (incidentally, this is how you go from an abelian variety over C to a complex torus).

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For a bit more info on question 3: if you are interested in the elliptic curve only as a complex Lie group, then when you identify it with C/L for C the complex plane and L a lattice, the Lie algebra is canonically C and the exponential map is the reduction mod L.

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1. No, for any cubic curve in the plane, there is a family of cubic plane curves (3 or 4 dimensional - I forget Edit: 8-dimensional, with a transitive action of PGL3) that are isomorphic as curves. For any lattice in C, there is a 1-dimensional family of lattices that form isomorphic curves. Over C, the curves are classified by their j-invariant (a complex number).
2. You can fix the lattice and choose generators so that one of the generators is 1 and the other lies in the upper half plane. There is then an action of SL2(Z) on the choices of generators, yielding an action on the upper half plane. If you choose a fundamental domain for this action, you get a "canonical" choice of lattice for each elliptic curve.
3. The Lie algebra is the unique 1-dimensional Lie algebra, whose bracket is zero. One can sometimes find more interesting information using the formal group law, but that mostly applies when you work in characteristic p.
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Thanks alot for your answers. One more question then: How much of the structure of an elliptic curve does this correspondence preserve? That is, is it homeomorphic, diffeomorphic, or biholomorphic? – John McCarthy Oct 24 2009 at 15:55
Corollary I.4.3 of Silverman that I speak of in my answer tells you that the map from C/lattice to E is a complex analytic isomorphism (it's given by the Weierstrass P-function and its derivative). – Rob Harron Oct 24 2009 at 16:10
Great, that's what I wanted to hear. Thanks alot. – John McCarthy Oct 24 2009 at 16:15

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Scott, I am confused by point 1. There is definitely a typo here, but I think there may be a larger error. {Smooth cubic curves}/PGL_3 is in bijection with {genus one curves}/isomorphism. To see this, notice that, given a genus one curve C, the set of embeddings of C into P^2, modulo PGL_3, is in bijection Pic^3(C). The automorphisms of C act transitively on Pic^3(C). I can think of various ways to vary my answer by talking about genus 1 curves with a specified origin or by not quotienting by PGL_3, but I can't think of any question whose answer should be "3 or 4 dimensional". – David Speyer Nov 3 2009 at 20:17
I was looking at free parameters in the full Weierstrass cubic form, but you're right that it is a poor choice. I will try to change it appropriately. – S. Carnahan♦ Nov 3 2009 at 21:39

Depending on what you mean, then no. Elliptic curves (abstractly) are in bijection with lattices modulo homothety (also, the upper half plane modulo SL(2,Z)). As for the Lie algebra of an elliptic curve, an elliptic curve is abelian, so the lie algebra is a one dimensional abelian complex Lie algebra, so it has trivial bracket.

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you mean "...with lattices modulo homothety" or "...with the upper half plane modulo SL(2,Z)" in that first sentence. – Kevin Buzzard Nov 3 2009 at 21:16
Yeah, good point.  I don't do much with elliptic curves (most of the problems I think of are only interesting in the hyperbolic case) so I mix the language up at times.  Corrected. – Charles Siegel Nov 3 2009 at 22:22
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