Is the real locus of an elliptic curve the intersection of a torus with a plane?

Solution 1:

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.

Solution 2:

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.