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In Lawrence Washington's book Elliptic Curves: Number Theory and Criptography I read that if $E$ is an elliptic curve defined over the real numbers $\mathbb{R}$ then the set of real points $E(\mathbb{R})$ can be obtained as the intersection of the torus of complex points $E(\mathbb{C})$ and a plane. The following is the relevant page of the book.

Real Locus of Elliptic Curve

Basically it says that if the plane passes through the hole in the torus, then the real locus of the elliptic curve looks like the following

Elliptic Curve over the reals, with two components

and if not, it looks like this one

enter image description here

So passing through the hole in the middle seems to determine whether the graph of the elliptic curve has two real components or just one. I have never seen such a claim before in other books about elliptic curves, and I'm really curious about it. Thus I have a couple of questions about this.

  1. Can this claim about the real locus being the intersection of a torus with a plane be made precise somehow, and if so can anyone please provide an explanation about it?
  2. If this is possible, can an explicit example be given?

Notes:

  1. I have to say that I'm really confused about this because the correspondence of an elliptic curve with a torus (when considering the complex points $E(\mathbb{C})$) is given by an isomorphism with the complex modulo a lattice, so even thinking about intersecting "the torus" with a plane seems rather odd to me.
  2. The reference to section 9.3 in the book does not seem to clarify this, it basically deals with the identification of the elliptic curve with the torus.

Thank you very much for any help with these questions.

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FWIW: toric sections are quartics, not cubics. So, there isn't a direct relation between slices of a doughnut and elliptic curves, unless you count the idea of being able to projectively transform any quartic into a cubic... –  J. M. Apr 2 '13 at 17:43
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2 Answers

up vote 12 down vote accepted

The complex points of the elliptic curve lie in the projective plane over $\mathbb C$. Supposing that the coefficients of the elliptic curve are in $\mathbb R$, it then makes perfect sense to intersect the curve with the projective plane over $\mathbb R$. If we forget the point at infinity, then we can just think that we are intersecting a certain punctured torus in $\mathbb C^2$ with $\mathbb R^2$.

Of course, the torus is not embedded as neatly as in the picture you posted (or, more accurately, it is not so much a question of "neatness" of the embedding, but rather the fact that the torus is embedded holomorphically in $\mathbb C^2$, which as a real manifold is of dimension four), but the description of the intersection is nevertheless correct.

Added: David Speyer's answer here makes an important point about the topology of the intersection, which I didn't address in my answer.

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I see, that makes sense. I was thinking naively of a plane intersecting a torus like the one in the picture and that was really confusing me. Thank you very much for clearing up the issue. –  Adrián Barquero May 22 '11 at 4:29
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It seems to me that the passage in Washington is misleading in one respect. If you have a torus $T$ in $\mathbb{R}^3$ and intersect it with a plane which misses the hole, the resulting intersection will be contractible in $T$. However, the real locus of an elliptic curve is NOT contractible within that curve.

In the case where the real locus has one component, the elliptic curve looks like $\mathbb{C}/\Lambda$ where $\Lambda$ is the lattice generated by $1$ and $1/2 + i \tau$ for some real $\tau$. A fundamental domain for this lattice is a rhombus, with vertices $0$, $1/2 + i \tau$, $1$ and $1/2 - i \tau$. The real locus of the curve is the diagonal of this rhombus running from $0$ to $1$. In particular, it is not contractible.

Other than that, I of course agree with Matt E.'s answer.

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