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

We learn in complex analysis class how to find the winding number of the contour $\Gamma$ around the origin. \[ n = \frac{1}{2\pi i} \oint \frac{dz}{z} = \frac{1}{2\pi i} \oint d(\log z) = \frac{1}{2\pi } \oint d\theta \] My goal had been to count the number of self-intersections of curves. I guessed some integral over a torus $\Gamma \times \Gamma$ which would have a pole-like object whenever $z_1 = z_2$. In the back of my mind, I worried maybe integrating along $t_1 = t_2$ would have zero contribution. \[ \frac{1}{2\pi i} \oint \oint \frac{dz_1 dz_2}{z_1 - z_2} \]

Inspired by Kontsevich's integral for knots (and some more recent papers), I learned of something that comes close \[ \frac{1}{2\pi i} \oint \oint \frac{dz_1 - dz_2}{z_1 - z_2} = \frac{1}{2\pi i} \oint \oint d \log(z_1 - z_2) = \frac{1}{2\pi i} \oint \oint \arg (z_1 - z_2) \] So formulas like these are measuring how the chords of curves wind around each other. This seems to be known as the Whitney invariant for plane curves, counting signed self-intersections.

alt text

Is there a way to get self-intersections all of the same sign? This must be related the Vassiliev invariants as well, but I'd like to focus on plane curves.

share|improve this question
    
This measure blows up when the curve is self-tangent, since if you go one way it has 2 intersection and if you go the other way it has 0. Therefore, you need to integrate something that has a pole or is otherwise not well-defined when the curve is self-tangent. This seems hard, because for it to be self-tangent $(z_1-z_2)$ and $d(z_1-z_2)$ both have to be zero. –  Will Sawin Jun 1 '12 at 17:02
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.