Winding number is locally-constant for general curves (not C1) using variation of argument definition
autumnunfound91
Open question
2022-08-30
Winding number is locally-constant for general curves (not ) using variation of argument definition 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 be a continuos function. We know that for such a path, there exists a continuous argument function, that is a function such that , for all . We define the variation of argument along :
(We can also prove that all such argument functions are of the form , for any integer n, so the variation of argument is well-defined. Also, if is closed, the variation of argument is a multiple of ) The winding number (index) of the closed curve around 0 is defined as:
For a closed path for some , the winding number of around is defined by translation as:
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 is . I know that I should somehow prove that 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 γ that doesn't go through 0, a point inside a ball around 0 that is contained in the connected component of 0. Imagine two segments from 0 to the curve and from 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 around 0 and around 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.
Answer & Explanation
lywyk0
Beginner2022-08-31Added 12 answers
Step 1 It’s a consequence of the following lemma: let be a closed curve. There exists such that for every closed curve with , . Indeed, you get your result by taking and to be close translates of the same closed curve. Step 2 How to prove the lemma? Well, let be such that , and assume . Then consider : then . But it’s easy to see that there is a complex logarithm , and then the imaginary part of is an argument of . Therefore , and thus so .