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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.

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  • $\begingroup$ 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. $\endgroup$
    – user10290
    Commented Mar 11, 2012 at 0:04
  • $\begingroup$ 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 $z$. $\endgroup$
    – john mangual
    Commented Mar 11, 2012 at 2:50
  • $\begingroup$ 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. $\endgroup$
    – Sungjin Kim
    Commented Mar 11, 2012 at 6:15
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    $\begingroup$ 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. $\endgroup$
    – Tom Goodwillie
    Commented Mar 13, 2012 at 0:47
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    $\begingroup$ @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. $\endgroup$
    – PVAL
    Commented Jun 21, 2014 at 11:39

2 Answers 2

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The winding number $w(z)=\frac{1}{2\pi i}\int_C\frac{d\zeta}{z-z(\zeta)}$ helps little for counting the multiple points of a curve $C=\{z(\zeta), \zeta\in S^1\}$. Indeed $w$ is an integer-valued function which is constant on the components of $\Bbb C\backslash C$. One of those components is unbounded. The other ones are called holes of $C$ and their number is indicated by $h=h(C)$. So, for using $w$ you need to know $h$. But $h$ itself already suffices to count double points. Indeed if $C$ has $h$ holes and is generic i.e. $C$ has no other multiple points than a finite number $d$ of double points, we have $$d=h-1.$$ This depends on a general property of connected graphs as I am going to explain. More in general, if $C$ is a connected graph (maybe not a curve) with $h$ holes, $n_2$ double points, $n_3$ triple points and so on, we have $$\sum_2(1-j)n_j=h-1.$$ Proof. Call singularity index of a vertex $v$ the number $s(v)$ of edges which have $v$ as vertex, reduced of $2$. For example if $v$ is smoothable we have $s(v)=0$, if it is a double point $s(v)=2$... Define the singularity index of the whole graph as the sum $s(C)$ of the indices of its vertices. For example if $C$ consists in a single vertex or has the form of a comb we have $s(C)=-2$, if it is (homeomorphic to) $S^1$ then $s(C)=0$. If $C$ is (homeomorphic to) a smooth curve with $d$ double points then $s(C)=d$. In general, if $C$ has $n_2$ double points, $n_3$ triple points ... , then $$s(C)=2\sum_2 (j-1)n_j.$$ So $s(C)$ is anyway an even number.

Now, in order to compute the Euler characteristic $\chi (C)$ ($=\sharp$ of vertices - $\sharp$ of edges) of $C$ observe that each $v$ contributes with $s(v)+2$ to the sum "$2$($\sharp$ of edges)", thus $2$($\sharp$ of edges)=$s(C)+2$($\sharp$ of vertices). This gives $\chi (C)=- s(C)/2$.

Assume now $C$ to be connected and take the $\varepsilon$-neighborhood $A$ of $C$. $A$ is an open, connected set in the plane and for $\varepsilon\ll 1$ has $h$ holes like $C$. The homology group $H_0(A)$ (any coefficients) has rank $1$ because of connectedness, $H_1(A)$ is generated by the holes and $H_2(A)=0$ because $A$ is on the plane. Hence the characteristic is $\chi (A)=1-h+0=1-h$.

But $A$ is by construction a strong deformation retract of $C$ (see wiki: "retraction"). This means that there exist a retraction $A\to C$ which is homotopic to the identity of $A$ and the points of $C$ stay fixed during the homotopy. In other words there exists $F:A\times [0,1]\to A$ continuous, with the properties $F(a,0)=a$, $F(a,1)\in C$, $\forall a\in A$ and $F(x,t)=x, \forall (x,t)\in C\times [0,1]$. Now, a strong deformation retract of a complex has the same characteristic as the complex itself (see again wiki: "retractions"), hence we have $\chi (C)=\chi (A)$. Substituting we obtain the desired formula.

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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.

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    $\begingroup$ 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. $\endgroup$
    – PVAL
    Commented Jun 21, 2014 at 11:27
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    $\begingroup$ 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... $\endgroup$
    – Dan Petersen
    Commented Jun 21, 2014 at 17:27

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