Why does the slope of a smooth simple closed curve have winding number one?

Question

$\def\RR{\mathbb{R}}$Let $S^1$ be the circle and let $\gamma : S^1 \to \RR^2$ be a smooth injective map with $\gamma'(t)$ everywhere nonzero. What is the easiest way to show that $t \mapsto \gamma'(t)$ has winding number $\pm 1$ around zero? I'm hoping for a proof at the level of Munkres Analysis on Manifolds, or easier.

Answer 1

This is Hopf's Umlaufsatz. There are proofs in most undergraduate differential geometry books (including my own text, on my webpage, pp. 28 ff.). The technical details are at the level of establishing path lifting for $\gamma'\colon S^1\to S^1$ and an extension of this lifting to the chord map of the curve.

Answer 2

The following is a heuristic argument in terms of complex analysis:

The given curve $\gamma$ bounds a simply connected domain $\Omega$ in the complex $w$-plane. By Riemann's mapping theorem there is a conformal map $f$ of the unit disk $D$ onto $\Omega$.

We assume that $f$ is in fact analytic in some larger disk $D_\rho$, and that $f'(z)\ne0$ in all of $D_\rho$. As $D_\rho$ is simply connected we then can write $f'(z)=e^{g(z)}$ for some analytic $g:\>D_\rho\to{\mathbb C}$.

The given curve $\gamma$ can then be parametrized as $$\gamma:\quad t\mapsto w(t)=f\bigl(e^{it}\bigr)\qquad(0\leq t\leq2\pi)\ .$$ But we are actually interested in its hodograph $\dot\gamma$ given by $$\dot\gamma:\quad t\mapsto p(t):=\dot w(t)=i\>e^{it}f'\bigl(e^{it}\bigr)=i\exp\bigl(it +g(e^{it})\bigr)\ .$$ We have to prove that $N(\dot\gamma,0)=1$. Now $$\eqalign{N(\dot\gamma,0)&={1\over2\pi i}\int_{\dot\gamma}{dp\over p} \cr &={1\over2\pi i}\int_0^{2\pi}{\ddot w(t)\over \dot w(t)}\>dt\cr &={1\over2\pi i}\int_0^{2\pi}\left(i+ {d\over dt}g(e^{it})\right)\>dt \cr &=1\ .\cr}$$

Answer 3

I think that the following paper will do: Umlaufsatz