Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$
If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed.
The period of a closed curve $\textbf{$\gamma$}$ is the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic.
A curve $\textbf{$\gamma$}$ is said to have a self-intersection at a point $p$ of the curve if there exist parameter values $a\neq b$ such that
- $\textbf{$\gamma$}(a) = \textbf{$\gamma$}(b) = p$, and
- if $\textbf{$\gamma$}$ is closed with period $T$, then $a-b$ is not an integer multiple of $T$.
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I am looking at the following exercise:
Show that the Cayley sextic
$$\textbf{$\gamma$}(t) = \left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right ), t \in \mathbb{R}$$
is a closed curve which has exactly one self-intersection. What is its period?
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I have done the following:
Since $\textbf{$\gamma$} (0)=(1,0)$ and $\textbf{$\gamma$}\left (\frac{\pi}{6}\right )=\left (0, \frac{\sqrt{3}}{2}\right )$, i.e., $\textbf{$\gamma$} (0)\neq \textbf{$\gamma$} \left (\frac{\pi}{6}\right )$ we have that $\textbf{$\gamma$}$ is not constant.
We have that $$\cos (t+2\pi )=\cos t , \ \ \cos (3(t+2\pi ))=\cos (3t) , \ \ \sin (3(t+2\pi ))=\sin (3t)$$
so then $$\textbf{$\gamma$}(t+2\pi ) = \left ((\cos (t+2\pi ))^3 \cos {[3(t+2\pi )]}, (\cos (t+2\pi ))^3 \sin {[3(t+2\pi )]}\right )=\left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right )=\textbf{$\gamma$}(t), \forall t$$
That means that the non-constant curve $\textbf{$\gamma$}$ is $2\pi$-periodic.
So, $\textbf{$\gamma$}$ is a closed curve.
Is this correct?
Is $2\pi$ the period? Is this the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic? Or does the period change when we have $\cos 3t$ and $\sin 3t$ instead of $\cos t$ and $\sin t$ ?
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EDIT:
We have that $$\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\left (-\frac{1}{8}, 0 \right ) \\ \textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$$
So, when we take $p=\left (-\frac{1}{8}, 0\right )$ we have that $\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$
Since the period is $\pi$, we have that $\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}=\frac{1}{3}(\pi)$, i.e., $\frac{2\pi}{3}-\frac{\pi}{3}$ is not an integer multiple of the period.
That means that $\gamma$ has a self-intersection at $p=\left (-\frac{1}{8}, 0\right )$.
Now we want to show that this self-intersection is unique.
To do that we want to find all points $x,y$ within one period of $[0,\pi]$ such that $\gamma(x)=\gamma(y)$. That is, find all solutions of the simultaneous system
$$
\begin{align*}
\cos^3(x) \cos(3x)&=\cos^3(y) \cos(3y) \\
\cos^3(x) \sin(3x)&=\cos^3(y) \sin(3y).
\end{align*}
$$
$$\frac{\cos^3(x) \cos(3x)}{\cos^3(x) \sin(3x)}=\frac{\cos^3(y) \cos(3y)}{\cos^3(y) \sin(3y)} \Rightarrow \frac{ \cos(3x)}{\sin(3x)}=\frac{ \cos(3y)}{ \sin(3y)} \\ \Rightarrow \cot (3x)=\cot (3y) \Rightarrow 3x=3y+k\pi \Rightarrow x=y+\frac{k}{3}\pi$$
Since the period is $\pi$, the only possible values for $k$ are $k=0$, $k=1$ and $k=2$.
For $k=0$ we have $3x=3y \Rightarrow x=y$. But for $x=y$ the second condition of the definition of a self-intersection is not satisfied.
So, for $k=1,2$ we have that $3x=3y+k\pi \Rightarrow x=y+\frac{k\pi}{3}$.
The first condition of the definition is satisfied with each of these values.
As for the second definition we want that $x-y=\frac{k\pi}{3}$ is not an integer multiple of $\pi$, which is also satisfied with each of these values for $k$.
So we want somehow to show that the points that we get this relation are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, or not?
But how? I don't have any idea…
Best Answer
Just using trig:
$T = \pi$
see this by computing $\gamma(0) = (1,0) = \gamma(T)$
Solve this using the second component first so that $$0 = \textrm{sin}(3T)\textrm{cos}^3(T)$$ Then $T = n \pi/3$, Do the same thing for the first component and you must have $n$ is a multiple of 3. In fact this is sufficient for all $t$, $\gamma(t+T) = \gamma(t)$.
The norm of the vector is $|\gamma(t)| = \cos^6(t)$ so to have an intersection we must have $\cos^6(t) = \cos^6(s)$ or in other words $s = \pi - t$ in this interval of $[0,\pi]$ This condition for the intersection is satisfied by the first component easily but leads to the following for the second component: $$\cos^3(t) \sin (3t) = -\cos^3(t) \sin (3t)$$ or in other words $$0 =\cos^3(t) \sin (3t) $$ The only possible intersections are then at $t = {0, \pi/3, \pi/2, 2\pi/3}$ Let $t<s$ from above then you must have either $t =0$ or $t = \pi/3$. But $t=0$ leads to a degenerate case so $t = \pi/3$ and $s = 2\pi/3$. Which you can check with substitution directly.
So $\gamma(\pi/3) = \gamma(2\pi/3)$ is the unique intersection.