Plane Curve invariants via Contour Integrals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:23:12Z http://mathoverflow.net/feeds/question/98575 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98575/plane-curve-invariants-via-contour-integrals Plane Curve invariants via Contour Integrals John Mangual 2012-06-01T14:56:48Z 2012-06-01T15:11:13Z <p>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 <a href="http://mathoverflow.net/questions/90856/computing-self-intersections-with-complex-analysis" rel="nofollow">self-intersections of curves</a>. 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. $\frac{1}{2\pi i} \oint \oint \frac{dz_1 dz_2}{z_1 - z_2}$</p> <p>Inspired by <a href="http://www.ihes.fr/~maxim/TEXTS/VassilievKnot.pdf" rel="nofollow">Kontsevich's integral for knots</a> (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 <em>signed</em> self-intersections.</p> <p><img src="http://oi47.tinypic.com/15i43mo.jpg" alt="alt text"></p> <p>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.</p>