# Curvature and turning number of $t \mapsto (\cos(t), \sin(3t))$

Is the closed curve with period $$2 \pi$$ $$\delta(t) := (\cos(t), \sin(3t))$$ regularly homotopic to the positively traversed unit circle $$(\cos(t), \sin(t))$$?

From this question I now know that the answer is indeed yes and one answer even gives the regular homotopy.

Before finding this answer I tried to solve this task using the theorem of Whitney-Graustein, which says that two closed curves in $$\mathbb R^2$$ are regularly homotopic to each other if and only their turning number is equal.

Definition. The turning number of a closed curve $$(\gamma, \tau)$$ is $$\tag{1} n = \frac{1}{2 \pi} \int_0^{\tau} \kappa(s) ds,$$ where $$\kappa$$ is the curvature of $$\gamma$$.

For curves (not necessarily parametrised with respect to arc lenght) we derived their curvature to be $$\tag{1} \kappa(t) = \frac{\det(\gamma'(t), \gamma''(t))}{| \gamma'(t) |^3}$$ I got $$\begin{equation*} \delta'(t) = (- \sin(t), 3 \cos(3 t)) \quad \text{and} \quad \delta''(t) = - (\cos(t), 9 \sin(3 t)) \end{equation*}$$ and thus $$\begin{equation*} \kappa(t) = \frac{9 \sin(3t) \sin(t) + 3 \cos(3t) \cos(t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}} = \frac{6 \cos(2 t) - 3 \cos(4 t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}} \end{equation*}$$ As $$\kappa(t + \pi) = \kappa(t)$$ holds we have $$\int_0^{2 \pi} \kappa(s) ds = 2 \int_{0}^{\pi} \kappa(s) ds$$ and thus the turning number is $$\begin{equation*} n = \frac{2}{2 \pi} \int_{0}^{\pi} \kappa(s) ds = \frac{1}{\pi}\int_{0}^{\pi} \frac{6 \cos(2 s) - 3 \cos(4 s)}{(\sin^2(s) + 9 \cos^2(3s))^{3/2}} ds = 1.92492, \end{equation*}$$ where the last equality is given by WolframAlpha. Clearly this is incorrect as the turning number is a integer.

Is my understanding of curvature incorrect?

## 2 Answers

In the definition you're using, the turning number is given by an integral with respect to $$s$$, arclength. Your parameter $$t$$ (although you switched randomly to both $$x$$ and $$s$$) is not an arclength parameter. So, if you correct by using $$ds = \dfrac{ds}{dt}dt$$, you get the correct answer that $$n=1$$.

• I changed all $x$'s to $t$'s. So the "most general formula" for the turning number would instead have $$\int_0^{\tau} \kappa(s) | \gamma'(s) | dx$$ in (2) instead but (1) is correct as it stands? – Ramanujan Apr 18 '20 at 12:30
• Or in other words, is there an easy way to find $s(t)$? (it should be $ds$ in the integral in my comment above). – Ramanujan Apr 18 '20 at 12:42
• So you need $\displaystyle{\int_0^{2\pi} \kappa(t)\frac{ds}{dt}\,dt = \int_0^{2\pi} \kappa(t)|\gamma'(t)|\,dt}$. – Ted Shifrin Apr 18 '20 at 17:07

The winding number of a curve $$\gamma$$ parametrized as $$I\ni t \mapsto (x(t), y(t))= x(t) + i y(t) = z(t)$$ can be calculated by the formula $$\frac{1}{2 \pi i}\int_{\gamma} \frac{d z}{z}=\frac{1}{2 \pi i} \int_I \frac{z'(t)}{z(t)} dt$$ In your case the result is $$-1$$.

This can also be inferred by inspecting the curve You can also check that your curve is homotopic to the curve $$t\mapsto (\cos t, - \sin t)$$ by the obvious homotopy.

The analogous curve $$(\cos t, \sin 5t)$$ has index $$1$$. Worth looking at pictures...

$$\bf{Added:}$$ I mistook the turning number for the winding number. However, if $$\gamma\colon I \to \mathbb{C}$$ is a curve then the turning number of $$\gamma$$ equals the winding number of $$\gamma'$$, the derivative of $$\gamma$$. Indeed, one checks that this is correct for the arclength parametrization, and that it is independent on the parametrization:

Indeed, if $$\eta(s) = \gamma(\phi(s))$$, then $$\eta'(s) = \gamma'(\phi(s)) \phi'(s)$$, and $$\eta''(s)= \gamma''(\phi(s))(\phi'(s))^2 + \gamma'(\phi(s))\phi''(s)$$ so $$\int_J\frac{\eta''(s)}{\eta'(s)} = \int_J\frac{\gamma''(\phi(s))}{\gamma'(\phi(s))}\cdot \phi'(s) + \int_J \frac{\phi''(s)}{\phi'(s)}= \int_I\frac{\gamma''(s)}{\gamma'(s)}+ 0$$

$$\bf{Added}$$ It is not hard to see that $$\frac{1}{2\pi i} \int_I\frac{\gamma''(t)}{\gamma'(t)}dt = \frac{1}{2\pi} \int_I\frac{\gamma''(t) \times \gamma'(t)}{\|\gamma'(t)\|^2} dt$$ in the second integral $$\gamma$$ is considered taking values in $$\mathbb{R}^2$$.

• But the answer below and the link in my question both claim the turning number is $+1$. – Ramanujan Apr 18 '20 at 12:31
• The turning number is often not the winding number about any point. It is the total angle through which the tangent vector turns as one goes once around the curve. – Ted Shifrin Apr 18 '20 at 17:35
• @Ted Shifrin: Oh, I see. It seems that the turning number is the winding number of the tangent vector. – orangeskid Apr 20 '20 at 2:06