How can I prove that this defintion of a winding number is valid?

Question

I have read a definition of a winding number on wikipedia and it involves finding the continuous polar parametrization of the curve, but then the question arises, why does such a parametriation always exist ? This question was once mentioned in the accepted answer to this question and it says there that it is a non-trivial fact from topology. Is it a really hard statement to prove ? If so could you give some outline of the proof or a source to read it ? And if it is not that hard could you give a hint as to why can I prove this statement ?

Answer 1

This is a modification of a standard proof that appears in the differential geometry of plane curves (if you're interested, see Lemma 3.6 on p. 27 of my differential geometry text, freely linked in my profile). One can appeal to the theory of covering spaces in elementary algebraic topology, but here is a self-contained treatment.

We assume that you are starting with a continuous curve $\alpha\colon [0,1]\to\Bbb R^2-\{0\}$ with $\alpha(0)=\alpha(1)$. Then the map $\beta =\alpha/\|\alpha\|$ is a continuous map from $[0,1]$ to the unit circle.

Consider the four open semicircles $U_1=\{(x,y)\in S^1: x>0\}$, $U_2=\{(x,y)\in S^1: x<0\}$, $U_3=\{(x,y)\in S^1: y>0\}$, and $U_4=\{(x,y)\in S^1: y<0\}$. Then the functions \begin{align*} \psi_{1,n}(x,y) &= \arctan(y/x) + 2n\pi \\ \psi_{2,n}(x,y) &= \arctan(y/x) + (2n+1)\pi \\ \psi_{3,n}(x,y) &= -\arctan(x/y) + (2n+\tfrac12)\pi \\ \psi_{4,n}(x,y) &= -\arctan(x/y) + (2n-\tfrac12)\pi \end{align*} are continuous (indeed, smooth) maps $\psi_{i,n}\:U_i\to\Bbb R$ with the property that $\big(\!\cos(\psi_{i,n}(x,y)),\sin(\psi_{i,n}(x,y))\big) = (x,y)$ for every $i=1,2,3,4$ and $n\in\Bbb Z$.

Now define $\theta(0)$ any way you want so that $\beta(0)=\big (\cos\theta(0),\sin\theta(0)\big)$. Define $S=\{t\in [0,1]: \theta \text{ is defined and continuous on } [0,t]\}$ and let $t_0=\sup S$. Suppose first that $s_0<1$. Choose $i$ so that $\beta(t_0)\in U_i$ and choose $n\in\Bbb Z$ so that $\psi_{i,n}(\beta(t_0)) = \lim\limits_{t\to t_0^-}\theta(t)$. Because $\beta$ is continuous at $t_0$, there is $\delta>0$ so that $\beta(t)\in U_i$ for all $t$ with $|t-t_0|<\delta$. Then setting $\theta(t)=\psi_{i,n}(\beta(t))$ for all $t_0\le t<t_0+\delta$ gives us a continuous function defined on $[0,t_0+\delta/2]$, contradicting our definition of $t_0$. Thus, we cannot have $t_0<1$. But the same argument shows that when $t_0=1$, the function $\theta$ is continuous on all of $[0,1]$.

The winding number of $\alpha$ around the origin is, of course, $\frac1{2\pi}\big(\theta(1)-\theta(0)\big)$, and this is independent of your original choice of the value of $\theta(0)$. (Since $\beta(1)=\beta(0)$, it follows that $\theta(1)-\theta(0)$ must be a multiple of $2\pi$. Why?)

Answer 2

This is going to require a bit of topology.

First, we are taking some curve $\gamma : [0, 1] \to \mathbb{R}^2 \setminus \{(0, 0)\}$ such that $\gamma(0) = \gamma(1)$. Let $\mathbb{S}^1$ be the circle, defined by $\mathbb{S}^1 = \{(x, y) \mid x^2 + y^2 = 1\}$.

Consider the map $f : \mathbb{R}^2 \setminus \{(0, 0)\} \to \mathbb{S}^1$ defined by $f(x) = \frac{x}{|x|}$. Note that $f$ is continuous.

We define the curve $\delta : [0, 1] \to \mathbb{S}^1$ by $\delta = f \circ \gamma$. Since $\delta$ is the composition of two continuous maps, it is continuous.

Now, we consider the function $g : \mathbb{R} \to \mathbb{S}^1$ defined by $g(t) = (\cos t, \sin t)$.

We can prove that $g$ is a covering space map. This means that for all $x \in \mathbb{S}^1$, there is some neighbourhood $U$ of $x$ such that $g^{-1}(U)$ can be written as $\bigcup\limits_{i \in I} U_i$, where the $U_i$ are pairwise disjoint and the map $g|_{U_i} : U_i \to U$ is a homeomorphism.

This is rather straightforward. For any point $p$ on the circle, we can find some $\theta$ such that $p = g(\theta)$. We can take some small $\epsilon$ (say, $\epsilon = 1$) and some neighbourhood $U$ of $p$ such that $g^{-1}(U) = \bigcup\limits_{n \in \mathbb{Z}} (\theta + 2 \pi n - \epsilon, \theta + 2 \pi n + \epsilon)$.

We then apply the following theorem:

Thm. Let $g : A \to B$ be a covering space map. Consider some curve $\delta : [0, 1] \to B$, and consider some $a \in A$ such that $g(a) = \delta(0)$. Then there exists a unique curve $\delta' : [0, 1] \to A$ such that $\delta'(0) = a$ and $g \circ \delta' = \delta$.

This is known as the "lifting property of covering spaces".

So in particular, take some $\theta_0 \in \mathbb{R}$ such that $g(\theta_0) = \delta(0)$. Then consider the unique curve $\delta' : [0, 1] \to \mathbb{R}$ such that $\delta'(0) = \theta_0$ and $g \circ \delta' = \delta$. Then the winding number of the curve $\gamma$ is defined to be $\frac{\delta'(1) - \delta'(0)}{2 \pi}$.

The astute reader will note that there is a single point in this proof where we have some discretion in making a choice. In particular, we could have chosen, instead of $\theta_0$, a different quantity $\theta_0 + 2 \pi n$ (noting that $g(\theta_0 + 2 \pi n) = g(\theta_0) = \delta(0)$). This would have led us to pick a different curve $\delta'' : [0, 1]$ such that $\delta''(0) = \theta_0 + 2 \pi n$ and $g \circ \delta'' = \delta$. Let us note that the function $\delta''(t) = \delta'(t) + 2 \pi n$ satisfies this property; therefore, by uniqueness, this is the equation for $\delta''$. Therefore, we have $\delta''(1) - \delta''(0) = \delta'(1) - \delta'(0)$, so we would have gotten the same answer for the winding number no matter what we pick for $\theta_0$.

Note that this is not the only way to go about defining the winding number. Any method which can be used to prove that $\pi_1(\mathbb{S}^1) \cong \mathbb{Z}$ can be used to define the Winding Number.

In particular, there is a very combinatorial proof which uses Van Kampen's Theorem for fundamental groupoids. But this is rather advanced material for someone who is likely seeing this for the first time.