On one hand the real locus of a complex elliptic curve is the intersection of a plane with a torus (i.e. a torus embedded in $\mathbb{C}^2$ plus infinity). And an elliptic curve has no cusps or self-intersections. But the intersection of a plane with a non-singular torus can have those.

I suppose those intersections of a complex cubic curve with a plane which can (up to some complex projective change of coordinates) give a real locus cannot have cusps or self intersections unless the cubic does too.

But is that right? Is it easy to see, once you see it? If it is not right what is?

  • $\begingroup$ No. Elliptic curves by definition are smooth. The real locus is homeomorphic to one or two circles. This is not MO level. $\endgroup$ – Felipe Voloch Aug 18 '13 at 12:52
  • $\begingroup$ @Felipe I suspect you are reading the title but not the question. I mentioned that an elliptic curve has no cusps or self-intersections -- a geometric way of saying it is smooth. I do not think you mean to deny a real-linear subspace of $\mathbb{C}^2$ of real dimension 2 can meet an elliptic curve in a singular real curve. $\endgroup$ – Colin McLarty Aug 18 '13 at 14:47
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    $\begingroup$ Sorry, did you mean a random real 2-dim subspace of $\mathbb{C}^2$ as a 4-dim real v.s.? Such a thing can have singularities but won't be the real points of an elliptic curve. Smooth in the algebraic category gives smooth in the usual analysis sense, so a real elliptic curve is a smooth 1-dim'l compact manifold, so a union of circles. Is that what you are asking? $\endgroup$ – Felipe Voloch Aug 18 '13 at 14:57
  • $\begingroup$ I meant how do the planes that can be planes of reality differ from random real 2-dim subspaces of $\mathbb{C}^2$ so that their intersections with an elliptic curve cannot be singular. The bad ambiguity is that I did not say the curve is defined over $\mathbb{R}$, though I should have since otherwise the claim is false. I got too caught up seeking a geometric picture of planes of reality in $\mathbb{C}^2$. The point is easy to see algebraically -- as you say. $\endgroup$ – Colin McLarty Aug 18 '13 at 15:40
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    $\begingroup$ It sounds like you are evolving your question into something like "what can you get by intersecting the Weil restriction of a complex elliptic curve embedded in the plane, with a real plane?". As it happens, generic intersections will have dimension zero, not one. $\endgroup$ – S. Carnahan Aug 18 '13 at 17:27

The tangent plane to an elliptic curve at a point is a holomorphic curve, meaning it is closed under multiplication by $i$. The real plane does not intersect its product with $i$, Which is the imaginary plane. Hence they cannot be identical, which would happen at a cusp or self-intersection point.

  • $\begingroup$ You assuming the point of tangency is $\langle 0,0\rangle$ (which we can arrange by a translation). Right? If so then I understand this and like it. $\endgroup$ – Colin McLarty Aug 18 '13 at 19:38
  • $\begingroup$ I was thinking of something slightly different, but equivalent: the tangent plane lives in the tangent space, which is a two-dimensional vector space that happens to be isomorphic to the affine space the elliptic curve is in. $\endgroup$ – Will Sawin Aug 19 '13 at 1:52

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