Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f: S^1 \to \mathbb{R}^2 - \{0\}$ be a smooth map. Define the winding number of $f$ about 0 and prove that it equals $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

According to the definition, we define the winding number of $f$ about $0$ to be $$W(f,0) = \deg(u),$$ where $$u(x) = \frac{f(x) - 0}{|f(x) - 0|} = \frac{f(x)}{|f(x)|}.$$

According to the degree formula, we have that $$\deg(f) = \frac{\int_{S^1} f^* d\theta }{\int_{\mathbb{R}^2 - \{0\}} \theta}.$$

I am particularly uncertain about this claim I made. Could someone help me take a look at it?

But the integration around $\mathbb{R}^2 - \{0\}$ is just $2 \pi$, hence we get the winding number of $f$ about 0 is $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

Winding number If $X$ is a compact, oriented, $l$-dimensional manifold and $f: X \to \mathbb{R}^{l+1}$ is a smooth map, the winding number of $f$ around any point $z \in \mathbb{R}^{l+1} - f(X)$ is defined by $$u(x) = \frac{f(x) - z}{|f(x) - z|},$$ and set $W(f,z) = \deg(u).$

Degree Formula Let $f: X \to Y$ be an arbitrary smooth map of two compact, oriented manifolds of dimension $k$, and let $\omega$ be a $k$-form on $Y$. Then $$\int_X f^*\omega = \deg(f)\int_Y \omega.$$

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.