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A curve dividing the plane into 4 regions.

I am trying to understand the concept of the winding number of a curve.

As the title suggests, I would like to figure out the winding number of C around points in the regions 1, 2, 3 and 4 (It will be the best if I can do this simply by inspection with no computations).

I denote the first loop curve $C_{1}$ then $C_{2}$ and $C_{3}$ respectively. By definition, the winding number of the curve, $C$ around $a$ is defined by: $$n(C,a)=\frac{1}{2\pi i}\int_{C} \frac{1}{z-a}dz.$$ However, I did not find the formula useful. Rather, based on https://en.wikipedia.org/wiki/Winding_number, my findings are presented below.

My answer is as follows:
Region 1: $n(C,a)=-1$.
Reason: Suppose a point, $z$ is in region 1 (being enclosed by loop 1). Since the curve is clockwise, by visual inspection, we have $n(C,a)=-1$.

Region 3: $n(C,a)=-1$.
Reason: Same explanation as above.

Region 2: $n(C,a)=1$.
Reason: This time, the loop is anti-clockwise.

Region 4: This is slightly tricky and I am unsure.

Can someone help me to verify if my winding number for regions 1, 2 and 3 are correct and is my way of determining winding number correct?

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You are right about regions $1$, $2$, and $3$.

About region $4$: from each $p$ in that region you can draw a ray starting at $p$ and going to $\infty$ which doesn't even touch the line. Therefore, yes, the answer is $0$.

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  • $\begingroup$ Hi @José Carlos Santos, thank you for the verification. May I Kindly clarify what is the line you are referring to when we draw a ray at any $p$ in region 4 going to $\infty$? $\endgroup$ – Cleytus Mar 26 '18 at 14:54
  • $\begingroup$ @Cleytus For me, the geometric (sort of) way of determining the winding number of $C$ with respect to $p$ is this: I consider a ray starting at $p$. For each point of intersection with $C$ I see if $C$ makes the ray turn around $p$ in a counter-clockwise or in a clockwise way and I count $+1$ in the first case and $-1$ in the second case (and $0$ if I was silly enough to pick a ray tangent to $C$). Then I add all these numbers and the sum is the winding number (it depends only on $p$, not on the direction of the ray). $\endgroup$ – José Carlos Santos Mar 26 '18 at 15:00
  • $\begingroup$ Thank you Sir, have a great day! $\endgroup$ – Cleytus Mar 26 '18 at 15:03
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Here's a thought experiment for you to carry out, or you can actually carry it out physically if you are so inclined.

Make a long, flexible loop of beads which cycles among three colors $RBGRBGRBG\ldots$ where $R$ is red, $B$ is blue, $G$ is green. The purpose of the color cycling is so that you can tell the orientation anywhere on the loop: the positive direction is as written above, and the negative direction is $GBRGBRGBR\ldots$.

Lay the loop down on a table top as your picture indicates, so that the positive orientation on the loop points in the same direction as the arrows on your picture.

To figure out the winding number around region 1 of the tabletop, first place the tip of one finger in region 1. Next, using the other hand, and keeping your fingertip pressed steadily on the table top, try to rearrange the loop of beads so that one of three outcomes occurs, and thereby define an integer $n \in \mathbb{Z}$, as follows:

  • The loop stays on a circle of constant radius centered at your finger, going counterclockwise around that circle. The integer $n$ is positive, and it counts the number of counterclockwise rotations.
  • The loop can be pulled away from your finger and bunched all up in a tiny pile far from your finger. The number $n$ is zero.
  • The loop stays on a circle of constant radius centered at your finger, going clockwise around that circle. The integer $n$ is a negative integer, and its absolute value counts the number of clockwise rotations.

That integer $n$ is equal to $n(C,a)$ where $a$ is the point where your fingertip touches the top of the table.

This explanation exploits an important fact about winding number, namely that assuming $C$ misses $a$, the integer $n(C,a)$ is unchanged when $C$ is altered by a homotopy (that misses $a$).

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  • $\begingroup$ That is a very interesting experiment indeed! I will definitely try it out, say, during my upcoming summer break? Haha! Thank you and have a great day. $\endgroup$ – Cleytus Mar 26 '18 at 16:06

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