📂Geometry

Proof of the Rotation Number Theorem

Theorem ¹

The winding number of a plane simple closed curve is $i_{\alpha} = \pm 1$.

Explanation

It’s a brief but very intuitive and important theorem. The proof is somewhat unique.

Proof

Let’s say $\alpha (s)$ is a curve that satisfies the condition of the theorem with length $L$. $$ 0 \le u < v \le L $$ Define two points $u, v$ appearing according to the reparameterization of arc length of the curve. Here, we aim to define the two-variable function $a (u, v)$ as a unit vector in the same direction as the vector whose starting point is $\alpha (u)$ and endpoint is $\alpha (v)$ but with size $1$. Rewriting it in formulaic terms yields the following. $$ a(u,v) := {{ \alpha (v) - \alpha (u) } \over { \left\| \alpha (v) - \alpha (u) \right\| }} $$ If $u=v$, then because the denominator becomes $0$, consider the limit when $v \to u$ as the tangent of a plane curve $t$. In other words, let’s assume $a(u,u) = t(u)$. Especially, $a(0,L)$ should be treated as having made one full rotation (in the same sense as the left and right limits being different). $$ a (0,L) = - t(0) = -t(L) $$ With this definition, $\alpha$ is a $C^{2}$ function in the following region $\Delta$.

Meanwhile, let’s define the two-variable function $\alpha (u,v)$ defined in $\Delta$ as the magnitude of the angle formed with the horizontal axis ($x$-axis) and $a(u,v)$. Care must be taken not to be confused with the given curve $\alpha (s)$, but such a definition simplifies expressions in the upcoming calculations, necessitating the reuse of $\alpha$. Remember that this definition implies $\alpha (u,u) = \theta (u)$.

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Part 1. $\displaystyle 2 \pi i_{\alpha} = \int_{\overline{AC}} d \alpha$

$$ i_{\alpha} = {{ \theta (L) - \theta (0) } \over { 2 \pi }} $$ The winding number is the integer $i_{\alpha}$ that satisfies the above.

Because it’s $\alpha (u,u) = \theta (u)$, integrating $\alpha$ along $d \theta$ to $\displaystyle \int_{\alpha} d \theta$ is the same as integrating along line segment $\overline{AC}$ to $d \alpha$ in $\displaystyle \int_{\overline{AC}} d \alpha$. Thus, we obtain the following. $$ \begin{align*} 2 \pi i_{\alpha} =& \theta (L) - \theta (0) \\ =& \int_{0}^{L} {{ d \theta } \over { d s }} ds \\ =& \int_{\alpha} d \theta \\ =& \int_{\overline{AC}} d \alpha \end{align*} $$

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Part 2. $\displaystyle \int_{\overline{AC}} d \alpha = \int_{\overline{AB}} d \alpha + \int_{\overline{BC}} d \alpha$

Green’s theorem: Given a simple plane $C^{2}$ closed curve $\mathcal{C}$ surrounding a bounded region $\mathcal{R}$ moving counterclockwise and being smooth piecewise.

If the two functions $P,Q$ defined in $\mathcal{R}$ are differentiable in $\mathcal{R}$ $$ \int_{\mathcal{C}} (Pdx + Qdy) = \iint_{\mathcal{R}} (Q_{x} - P_{y}) dx dy $$

Since $\alpha$ is a $C^{2}$ function (with its second derivative continuous), $\displaystyle {{ \partial^{2} \alpha } \over { \partial u \partial v }} = {{ \partial^{2} \alpha } \over { \partial v \partial u }}$ holds, and by Green’s theorem, $$ \begin{align*} \int_{\Delta} d \alpha =& \int_{\Delta} \left( {{ \partial \alpha } \over { \partial u }} du + {{ \partial \alpha } \over { \partial v }} dv \right) \\ =& \iint_{\blacktriangle} \left( {{ \partial^{2} \alpha } \over { \partial v \partial u }} - {{ \partial^{2} \alpha } \over { \partial u \partial v }} \right) dudv \\ =& \iint_{\blacktriangle} 0 dudv \\ =& 0 \end{align*} $$ In other words, because $\displaystyle \int_{\overline{AC} + \overline{CB} + \overline{BA}} d \alpha = 0$, we obtain the following. $$ \int_{\overline{AC}} d \alpha = \int_{\overline{AB}} d \alpha + \int_{\overline{BC}} d \alpha $$

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Part 3. $i_{\alpha} = \pm 1$

That the given curve rotates counterclockwise means integrating from $0$ to $L$ while fixing $u=0$. $$ \int_{\overline{AB}} d \alpha = \int_{\overline{BC}} d \alpha = \pi $$

That the given curve rotates clockwise means integrating from $L$ to $0$ while fixing $v=0$. Therefore, $$ \int_{\overline{AB}} d \alpha = \int_{\overline{BC}} d \alpha = - \pi $$ According to Part 2, if it’s counterclockwise, then $\displaystyle \int_{\overline{AC}} d \alpha = + 2 \pi$, if clockwise, then $\displaystyle \int_{\overline{AC}} d \alpha = - 2 \pi$. Summarizing, $$ \int_{\overline{AC}} d \alpha = \int_{\overline{AB}} d \alpha + \int_{\overline{BC}} d \alpha = \pm 2 \pi $$ and by Part 1, we obtain $i_{\alpha} = \pm 1$.

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