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Winding Numbers

Suppose that I have a loop in the plane, and I want to know how many times it winds around, say, the origin. (While the given loop doubles back on

8 months ago

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Suppose that I have a loop in the plane, and I want to know how many times it winds around, say, the origin.

(While the given loop doubles back on itself at certain points, you can see that at the end of the day it goes around the origin exactly once, counterclockwise.)

This is easy if your loop is, say, circular. But if you have some crazily complicated path, this can become hard to see. It turns out there is a straightforward way to make sense of these things, though. If your curve is  𝐶 , then all you need to do is to compute the integral

This can be shown to be precisely equal to the winding number! It also allows us to prove some really rather advanced mathematical statements. For example, I shall now show that  1+1=2 .

We’ll accomplish this by reducing the rather complicated language of arithmetic down to something easier to understand: namely, paths and integrals. To do this, we shall need some way of adding paths. Thankfully, there is a reasonably straightforward way to do this.

The idea is to choose some starting point for all your loops. Then, given two loops  𝐶1  and  𝐶2 , we can define a third loop  𝐶3  as the loop made by first traveling around  𝐶1  and then traveling around  𝐶2 . Of course, we need to specify the parametric equations to really be sure that this is well-defined. If  𝐶1  is given by the parametric equation

𝑝1(𝑡)=(𝑥1(𝑡),𝑦1(𝑡)),
and  𝐶2  is given by the parametric equation
𝑝2(𝑡)=(𝑥2(𝑡),𝑦2(𝑡)),

where  𝑡  goes from  0  to  1 , then  𝐶3  will have the parametric equation

Intuitively, this is exactly what we mean by saying that 𝐶3 goes around 𝐶1 and then 𝐶2.

Now, here is the beautiful thing: if  𝐶1  loops around the origin  𝑚  times counterclockwise, and  𝐶2  loops around the origin  𝑛  times counterclockwise, then  𝐶3=𝐶1+𝐶2  loops around the origin  𝑚+𝑛  times! So, to compute  1+1 , all we need to do is to choose a simple loop that goes around the origin once, compute its sum with itself, and then use the integral formula for the winding number to obtain  1+1. Let’s do it!

We will just use the unit circle  𝐶  going around the origin counterclockwise. We can write this path parametrically as:

𝑝(𝑡)=𝑒2𝜋𝑖𝑡=(cos(2𝜋𝑡),sin(2𝜋𝑡)). Now, what is  𝐶+𝐶 ? Well, per our parametric coordinates, this will be given by

where the middle line follows from basic trigonometric identities. Now, it just remains to calculate

To do this, we use the u-substitution

now, yielding

It’s a remarkably technical result. You need to construct the universal covering of the circle—this turns out to be the real line. This induces a map between the fundamental groups of the real line and the circle. The fundamental group of the real line is the trivial group because it is easy to show that the real line contracts to a point.

Now, there is a nice trick: both the real line and the circle can be thought of as groups—Lie groups, specifically. But then we can set up a correspondence between the fundamental groups of these Lie groups and the groups themselves, and at the end of the day, we see that the kernel of the homomorphism from the real line to the circle is exactly the fundamental group of the circle. However, it is easy to see that the pre-image of 0 under this homomorphism is the integers, and therefore the fundamental group of the circle is isomorphic to the integers.

Thinking about specific paths (you can use the winding number formula, and the fact that changing paths by homotopy doesn’t change the value of the integral), you can work out that the geometric interpretation is that the winding number is the integer to which a particular class of paths is sent by the aforementioned isomorphism. We conclude that, yes, if we concatenate two paths (which is the geometric meaning of multiplying elements in the fundamental group), then that corresponds to adding their winding numbers.

This type of mind-bending stuff can come very handy when you want to know if say a robot has circulated back from its trajectory. Of how many ways your kids find to change subject and avoid discussing their math grades LOL

Mario Esposito

Published 8 months ago