Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

One may think of a plane triangle as a kind of degenerate cubic curve. Many Euclidean geometry theorems posit a map from plane triangles to plane triangles (e.g., take all the angle bisectors) and then predict that one gets an even more degenerate output.

Can one lift these theorems (systematically?) to general (or at least less degenerate) cubic curves?

share|improve this question

1 Answer 1

Any triangle is equivalent under projective transformations to the standard triangle whose sides are on the lines $x=0,y=0,z=0$ of the projective plane. If you select a point $(0:y_1:z_1),(x_2:y_2:0),(x_3:0:z_3)$ on each of these lines not at the vertices, then they are collinear if and only if $(y_1/z_1)(z_2/x_2)(x_3/y_3)=-1$ as you can check. This is Menelaus's theorem of classical plane geometry. You can think of the lines as copies of $\mathbb{G}_m$ and make a group out of $\mathbb{G}_m \times \mathbb{Z}/3$ in a suitable way and interpret Menelaus's theorem as the three points are collinear if and only if they add to zero in this group. Once you do that, it is easy to view this statement as a degenerate case of the fact that three points on a cubic are collinear if and only if they add to zero in the group law.

Ceva's theorem, which guarantees that certain lines from the vertices of a triangle (e.g. bisectors) are collinear, is the dual of Menelaus and thus admits a similar, but less obvious, interpretation.

share|improve this answer
    
Your first observation suggests that I modify my question. Since notions of angle aren't projectively invariant, I suppose my original maps ought to receive two triangles, the actual one in play (whose angles might get bisected, say) and a second reference triangle there just to rigidify the projective plane. So I might ask instead for an appropriate lifting to maps from a pair of cubic surfaces to a cubic surface. –  David Feldman Apr 15 '12 at 1:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.