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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 $\frac{1}{2\pi}\int_{S^1}f^*(d\theta).$

I have no clue, except for the winding number of $f$ around 0 is just the degree of $u$, where $$u(x) = \frac{f(x)}{|f(x)|}.$$

Degree. $Y$ is connected and has the same dimension as $X$. We define the degree of an arbitrary smooth map $f: X \to Y$ to be the intersection number of $f$ with any point, $\deg(f) = I(f,\{y\}).$

Intersection number. If $f: X \to Z$ is transversal to $Z$, $f, Z$ are appropriate for intersection theory, then $f^{-1}(Z)$ is a finite number of points, each with an orientation number $\pm 1$ provided by the preimage orientation.

Thank you~

Community
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Tumbleweed
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2 Answers2

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Hints: (1) The degree theorem tells you that if $f\colon M \to N$ is a smooth map between compact oriented $n$-manifolds and $\omega$ is an $n$-form on $N$, then $$\int_M f^*\omega = \text{deg}(f)\int_N\omega\,.$$ (2) Show that $\displaystyle\int_{S^1} f^*d\theta = \displaystyle\int_{S^1} \Big(\frac f{|f|}\Big)^*d\theta $.

Ted Shifrin
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  • Thanks Ted. I am almost there except for I need to prove $\int_\Omega u^* d\theta = \int_{S^1} f^* d\theta$. Can I just claim they are equal? Or how can I justify it? I've been thinking of applying change of variable twice, but feel shooting with cannon. Thank you. – Tumbleweed Aug 03 '13 at 18:28
  • @Jellybean: ($\Omega$ should be $S^1$ above.) The easiest argument is a direct calculation. Note $d\theta=\dfrac{-y,dx+x,dy}{x^2+y^2}$ is invariant under scaling. – Ted Shifrin Aug 03 '13 at 20:59
  • Thank you Ted. I've trying to think very hard, and I see if we scale $x,y$ then $d\theta$ is invariant. I also see that we can take the demoninator $|f|$ out from pull-pack and put in $\theta$ as $d \frac{\theta}{|f|}$. But this is different from scaling $x,y$, and this will yields the change of $d\theta$. Where did I understand you incorrectly? – Tumbleweed Aug 04 '13 at 23:09
  • Sadly, what you wrote doesn't make sense. There are fancier ways to do this, but just write it out carefully, using the product rule. If $f=(u,v)$, when you use the explicit formula for $d\theta$, when you pull back you should get $(-v,du+u,dv)/(u^2+v^2)$ plus a term with $d(1/\sqrt{u^2+v^2})$. Show that its coefficient is actually $0$. – Ted Shifrin Aug 05 '13 at 06:26
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I am sorry for having no time to write a complete answer. But I suggest you to read the definition of degree in more detail (what does it tell you about the behavior of $u$?). The wikipedia article on winding number and degree may be a good start.

Bombyx mori
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  • I read the wiki article. I don't have much trouble when I learned winding number in complex analysis. But my question is from a differentiable viewpoint. I'll add the definition in my question, which does not make much sense to me solving this question.. thanks. – Tumbleweed Aug 03 '13 at 07:24
  • @Jellybean: The degree defined this way from differential topology is not entirely enlightening. Try to think about the homotopy type of the map $\mathbb{S}^{n}\rightarrow \mathbb{S}^{n}$ involved. – Bombyx mori Aug 04 '13 at 06:01