Computing Self-Intersections with Complex Analysis

It is possible to find the winding number of a path $C \subset \mathbb{C}$ using complex analysis:

$$n = \oint_C\frac{dz}{z}.$$ You can also count the number of roots of $f(z) = 0$ inside a close curved $C$:

$$n = \oint_C dz\,\frac{f'(z)}{f(z)}.$$

How do we count the number of self-intersections or self-tangencies of $C$ with itself?

$$\#[C \cap C] = \oint \frac{dz}{z}\frac{dw}{w} f(z,w)?$$ Self-intersection is a global event depending on two points - probably it is some kind of double-integral. • If you know that your path is closed and does not completely intersect itself, like a circle which just keeps winding around itself, then it seems to me that the winding number minus one will count the number of self-intersections or self-tangencies since every time it winds it must pass through itself or touch itself tangentially. This might just work for the examples I am visualizing and the ones in your nice picture. – user10290 Mar 11 '12 at 0:04
• i believe Arnold showed the self-intersections + self-tangencies is conserved or something like that. To only count self-intersections seems harder -- try telling apart the cardioid (with a loop) and a trefoil. Maybe the combinatorial answer involves $f:\mathbb{C} \to \mathbb{N}$ by $f(z)=$ winding number of $C$ around $z$. – john mangual Mar 11 '12 at 2:50
• I think the winding number has no relation to self-intersections even if the curve is not completely intersecting itself. Consider a curve winds origin 2 times, with the first time, following unit circle, and second time, following hypocycloid(you can make the inner circle arbitrarily small so that the curve intersects the unit circle arbitrarily many times). Maybe we can give the lower bound of self-intersection number. – Sungjin Kim Mar 11 '12 at 6:15
• If the curve is immersed, and if its self-intersections occur transversely, then the number of these is of the opposite parity from the winding number. I don't see a way of specifying the multiplicity of a self-intersections as an integer, as opposed to a mod $2$ integer. – Tom Goodwillie Mar 13 '12 at 0:47
• @TomGoodwillie Whitney gives an explicit definiton of "self-intersection number" in his paper "The self-intersections of a smooth n-manifold in 2n-space". For n even this is an integer invariant. For n odd (i.e. n=1) , this is only a mod $2$ invariant, and this can be seen by picking the opposite orientation on your manifold when calculating the intersection number. – PVAL Jun 21 '14 at 11:39

The intersection number of two closed curves in $\mathbb{C}$ can be calculated by means of an integral (http://en.m.wikipedia.org/wiki/Intersection_number#Definition_for_Riemann_surfaces). I would assume this could be applied to calculating the number of selfintersection (e.g. if need be by appropriate subdividing the curve), but I have not checked the details.
• The intersection number of two closed curves in $\Bbb C$ is always even, and the self intersection number does not have to be. So while a similar method probably works its not clear how to directly translate that result to one on self-intersection. – PVAL Jun 21 '14 at 11:27
• The integral you link to doesn't actually compute the number of intersections between two curves; rather, it computes the intersection number between their respective cohomology classes. For curves on $\mathbb P^1$, which is simply connected, this is of course always zero... – Dan Petersen Jun 21 '14 at 17:27