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In reading Section B of the Appendix of Chapter $1$ of Do Carmo's Differential Geometry of Curves and Surfaces, I have a few points of confusion. He writes (summarized):

Let $\alpha : [0, l] \rightarrow \mathbb{R}^2$be a closed plane curve given by $\alpha(s) = (x(s), y(s))$ parametrized by arc length. Let $t$ be the tangent indicatrix (the normalized tangent curve) given by $t(s) = (x'(s), y'(s))$. Let $\theta(s), 0 < \theta(s) < 2 \pi$ be the the angle that $t(s)$ makes with the $x$-axis, that is $x'(s) = \cos \theta(x), y'(s) = \sin \theta(s)$.

Since $\theta(s) = \arctan(\frac{y'(s)}{x'(s)})$, $\theta = \theta(s)$ is locally well defined as a differentiable function and $\frac{dt}{ds} = \frac{d}{ds} (\cos \theta, \sin \theta) = \theta'(-\sin \theta, \cos \theta) = \theta'n$ where $n$ is the normal vector.

This means that $\theta'(s) = k(s)$ and suggests defining a global differentiable function $\theta: [0,l] \rightarrow \mathbb{R}$ by $\theta(s) = \int_{0}^{s} k(s) ds$. Since $\theta' = k = x'y''-x''y' =(\arctan (\frac{y'}{x'}))'$. This global function agrees, up to constants, with the previous locally defined $\theta$. Intuitively, $\theta(s)$ measures the total rotation of the tangent vector, that is, the total angle described by the point $t(s)$ on the tangent indicatrix, as we run the curve from $0$ to $s$. Since $\alpha$ is closed, this angle is an integer multiple, $I$, of $2 \pi$; that is $\int_{0}^{l} k(s)ds = \theta(l) - \theta(0) = 2 \pi I$, where $I$ is the rotation index.

My questions are:

  1. I do not understand the statement "$\theta(s)$ measures the total rotation of the tangent vector from $0$ to $s$", to me it seems like it is simply integrating the curvature over some interval, and I cannot seem to connect these two concepts.

  2. It seems to me that $\theta(l)=\theta(0)$, and so I am not sure why this implies that the rotation index could be anything other than $0$. I am familiar with the concept from computer graphics of the winding number of a curve centered around some point, the idea being we want to know how many times the curve spins around (clockwise or anti-clockwise) some point, but I can't connect this to the equation described.

IntegrateThis
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1 Answers1

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Say $\alpha$ is a closed curve of class $C^2$. If it is parametrized by arc length, then $\alpha'(s)$ is an element of $S^1$ so it might be written as $\alpha'(s) = (\cos(\theta(s)),\sin(\theta(s))^\top$ for some angle function $\theta(s)$. The requirement that $\alpha$ is closed translates into the fact that $\alpha'(0) = \alpha'(l)$ but this does not imply that $\theta(0) = \theta(l)$ but rather that $\theta(l)-\theta(0)$ needs to be an integer muptliple of $2\pi$. This integer is called the index of the curve. I hope that this clarifies your second question. Regarding the first question, note that the signed curvature is an a priori unrelated concept: It is the projection of $\alpha''$ onto the vector $\left(\begin{smallmatrix}0 & -1 \\ 1 & 0 \end{smallmatrix}\right)\alpha'$ in the sense that

$$k(s) = \left\langle \left(\begin{smallmatrix}0 & -1 \\ 1 & 0 \end{smallmatrix}\right)\alpha'(s),\alpha''(s)\right\rangle$$

Now, using $\alpha' = (\cos(\theta),\sin(\theta))^\top$ you can compute $$ k(s) = \left\langle \left(\begin{smallmatrix}-\sin(\theta)\\ \cos(\theta)\end{smallmatrix}\right),\left(\begin{smallmatrix}-\sin(\theta)\theta'\\ \cos(\theta)\theta'\end{smallmatrix}\right)\right\rangle = \theta'(s), $$ which is how the angle $\theta$ and the signed curvature are related.

By the fundamental theorem of calculus you can compute the index of your curve by computing $$I = \frac{1}{2\pi}(\theta(l)-\theta(0)) = \frac{1}{2\pi}\int_0^l \theta'(s)\,\mathrm ds = \frac{1}{2\pi}\int_0^1 k(s)\,\mathrm ds.$$

frog
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