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In Alain Roberts "Elliptic curves: notes from postgraduate lectures given in Lausanne 1971/72" page 11 (available on google books unless you already tried to read another chapter), there is a hand drawn picture of a real 2-dimensional torus and a real plane, which topologically represent the way a complex cubic (with two real components) and the real projective plane sit in the complex projective plane. Taking the picture on face value, one should be able to project an open subset of the complex projective plane to $\mathbb{R}^3$, so that there is some real line $L$ that passes through the "doughnut" defined by the image of the complex cubic.

I tried to reproduce this picture on a computer, using the map $\mathbb{CP}^2\to\mathbb{R}^7$ given by


projecting to various $\mathbb{R}^3$s, and looking for $L$ by trial and error; all in vain. Which brings me to....


  • Is there such a line (the map I used does not send the real projective plane to a plane, so it does not have to be the case even if Roberts picture is correct) ?

  • Is there an algorithm to find such a line ?

  • Is there a "better" way to project an open part of the complex projective plane to $\mathbb{R}^3$ ?

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I guess you're rejecting the obvious map to $\mathbb{C}^2 = \mathbb{R}^4$ (normalizing out $z_3$) because the elliptic curve necessarily intersects the $\mathbb{CP}^1$ you removed to make that picture? I suspect that's what Roberts is trying to suggest anyway, since the intersection is just at most a few points. – Dylan Thurston Apr 2 '10 at 4:41
@Dylan: yup - I want to see the entire torus, not the one with the infinity part thrown away. Note that there is one obvious cheat: embedding the elliptic curve in CP^1 times CP^1 which sits nicely in R^6, and projecting; the problem with this approach is that you are loosing all hope of seeing the Fubini Studi metric on CP^2. – David Lehavi Apr 2 '10 at 21:22
up vote 3 down vote accepted

I found this article: "Visualizing Elliptic Curves" by Donu Arapura it is available at the following URL: In it he discusses a projection that sends sends the real part of $x$ to $x_1$ and the real part of $y$ to $x_3$ thus it would seem to preserve the entire real plane and any line in it. So this might be useful to you.

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Thanks - this is exactly what I looked for ! – David Lehavi May 4 '10 at 7:26

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