Winding number is locally-constant for general curves (not $C^1$) using variation of argument definition

Question

I've been searching for this proof, here and on the web too, but it seems like the answer is harder to find than expected. There are many similar questions about this but they all (implicitly or explicitly) suppose continuous differentiability.

First, let me state the definition of winding number:

Let $\gamma : [0, 1] \to \mathbb{C^*}$ be a continuos function. We know that for such a path, there exists a continuous argument function, that is a function $\phi : [0, 1] \to \mathbb{R}$ such that $\gamma(t) = |\gamma(t)|e^{i\phi(t)}$, for all $t \in [0, 1]$. We define the variation of argument along $\gamma$: $${\rm vararg}(\gamma)=\phi(1)-\phi(0)$$ (We can also prove that all such argument functions are of the form $\phi + 2n\pi$, for any integer $n$, so the variation of argument is well-defined. Also, if $\gamma$ is closed, the variation of argument is a multiple of $2\pi$)

The winding number (index) of the closed curve $\gamma$ around $0$ is defined as: $${\rm I}(\gamma, 0) = \frac{\rm vararg(\gamma)}{2\pi}$$

For a closed path $\gamma : [0, 1] \to \mathbb{C}\setminus\{z_0\}$ for some $z_0 \in \mathbb{C}$, the winding number of $\gamma$ around $z_0$ is defined by translation as: $${\rm I}(\gamma, z_0) = {\rm I}(\gamma - z_0, 0)$$

I consider this as the best definition of winding number because it is intuitive, general and everything follows from it.

Now I want to prove that the winding number is constant on the connected components of the complement of the curve. This is not very hard to prove using the integral expression of the winding number, but that assumes that $\gamma$ is $C^1$.

I know that I should somehow prove that $\phi$ is continuous or just that it is constant in an open ball around any point included in the connected component.

Intuitively, I can prove it like this: consider a curve $\gamma$ that doesn't go through $0$, a point $z_0 \ne 0$ inside a ball around $0$ that is contained in the connected component of $0$. Imagine two segments from $0$ to the curve and from $z_0$ to the curve, as we move along it. Both segments follow the same point on the circle around $0$ and their movements along little circles around their respective start points have the same intervals of increase/decrease in angle (going counter-/clockwise). Therefore, whenever one completes a circle in one direction, the other does so too. Hence, the variation of the argument of $\gamma$ around $0$ and around $z_0$ is the same. By translation, we generalize the result for other points than $0$.

This argument is hard to write rigorously and gets a little too geometric, I think. I wonder if an easier take on this can be carried out.

Any proof or help is appreciated.

Answer 1

It’s a consequence of the following lemma: let $\gamma:[0,1] \rightarrow \mathbb{C}^*$ be a closed curve. There exists $\epsilon >0$ such that for every closed curve $\delta:[0,1] \rightarrow \mathbb{C}^*$ with $\|\gamma-\delta\|_{\infty} <\epsilon$, $I(\gamma,0)=I(\delta,0)$.

Indeed, you get your result by taking $\gamma$ and $\delta$ to be close translates of the same closed curve.

How to prove the lemma? Well, let $r>0$ be such that $|\gamma| \geq r$, and assume $\|\gamma-\delta\|_{\infty} < r$. Then consider $\alpha=\gamma/\delta$: then $\alpha([0,1]) \subset D=\mathbb{C} \backslash (-\infty,0]$. But it’s easy to see that there is a complex logarithm $L: D \rightarrow \mathbb{C}$, and then the imaginary part of $L \circ \alpha$ is an argument of $\alpha$. Therefore $\mathrm{vararg}(\alpha)=0$, and thus $\mathrm{vararg}(\gamma)=\mathrm{vararg}(\delta)$ so $I(\gamma,0)=I(\delta,0)$.