Is it a Closed Curve?

Question

In the figure below, the black curve is an intersection of two cylinders in 3D. So according to Kristopher Tapp's definition of a closed curve (see below), is this a closed curve?

According to Kristopher Tapp, some definitions and propositions are as follows:

Definition 1:
A $\textbf{parametrized curve}$ in $\mathbb{R}^n$ is a smooth function $\gamma: I \rightarrow \mathbb{R}^n$, where $I \subset \mathbb{R}$ is an interval.

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Definition 2:
Let $\gamma: I \rightarrow \mathbb{R}^n$ be a curve. It is called $\textbf{regular}$ if its speed is always nonzero ($|\gamma'(t)| \ne 0$ for all $t \in I$).

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Definition 3:
A $\textbf{closed curve}$ means a regular curve of the form $\gamma:[a,b] \rightarrow \mathbb{R}^n$ such that $\gamma(a)=\gamma(b)$ and all derivatives match:
\begin{equation*} \gamma'(a)=\gamma'(b), \quad \gamma''(a)=\gamma''(b), etc. \end{equation*}
If additionally $\gamma$ is one-to-one on the domain $[a,b)$, then it is called a $\textbf{simple closed curve}$.

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Proposition 1:
A regular curve $\gamma:[a,b] \rightarrow \mathbb{R}^n$ is a closed curve if and only if there exists a periodic regular curve $\hat{\gamma}: \mathbb{R} \rightarrow \mathbb{R}^n$ with period $b-a$ such that $\hat{\gamma}(t)=\gamma(t)$ for all $t \in [a,b]$.

Answer 1

Your black object $C$ is the intersection of two cylinders, hence a set. This set is closed and bounded, hence compact. The question is whether this set $C$ can be parametrized as a regular closed curve.

The projection of $C$ to the $(x,y)$-plane is the circle $\ C'\colon\> x^2+(y-1)^2=1$, but $C$ covers this circle twice. We therefore parametrize the doubled circle $C'$ as $$t\mapsto\bigl(\sin(2t), 1+\cos(2t)\bigr)=\bigl(\sin(2t),2\cos^2 t\bigr)\qquad(0\leq t\leq 2\pi)\ .$$ We now have to take care of the $z$-coordinate. Since the point ${\bf r}(t)$ has to lie on the red cylinder we necessarily have $$z^2(t)=4-y^2(t)=4(1-\cos^4 t)=4\sin^2 t\>(1+\cos^2 t)\ .$$ We now choose the square root such that for $0\leq t\leq\pi$ we obtain $z(t)\geq0$, and for $\pi\leq t\leq 2\pi$ we obtain $z(t)\leq 0$. This is smoothly possible by letting $$z(t)=2\sin t\>\sqrt{1+\cos^2 t}\ .$$ It follows that $$\gamma:\quad t\mapsto{\bf r}(t):=2\bigl(\sin t\cos t,\cos^2 t, \sin t\>\sqrt{1+\cos^2 t}\bigr)\qquad\bigl(t\in{\mathbb R}/(2\pi)\bigr)$$ is a representation of $C$ of the required kind, whereby it remains to check that $\dot{\bf r}(t)\ne{\bf 0}$ for all $t$. I may leave this to you.

In this example we were lucky, since we could give a presentation of $C$ in terms of elementary functions. If such a thing is not possible a more theoretical approach is necessary. One could try to use the implicit function theorem which tells us that $C$ is "locally" a curve. It then would be a compact one-dimensional manifold, hence a union of simple closed curves. But unfortunately a crucial assumption of the implicit function theorem is not fulfilled at the point $(0,2,0)\in C$. The two cylinders do not intersect transversally there – that's why $(0,2,0)$ is a double point of $C$. Such a singularity would require additional analysis.

Answer 2

Let $I$ be an interval, and $\gamma:I\rightarrow \mathbb{R}^3$ a smooth function.

First you must differentiate between $\gamma$ and $\gamma(I) = \{\gamma(x):x\in I\}$. Usually $\gamma$ is referred to as a curve, and $\gamma(I)$ is the trajectory. (Occasionally the trajectory is referred to as a curve, but it's uncommon.)

In your picture, the intersection between the cylinders is a set. It can be viewed as the trajectory of a curve, but is not a curve itself. Consider this curve:

$$\gamma:[0,2\pi]\rightarrow \mathbb{R}^3,\qquad\gamma(x) = (\sin(2x), 1+\cos(2x), 2\sin(x))$$

This is a closed curve using the definition given, and its trajectory is the intersection of the two cylinders. However, you can just as easily define curves whose images are the set you described which are not closed curves. For example: $$f:[0,1]\rightarrow I,\qquad f(x) = 2\pi x^2$$ $$\gamma_2:[0,1]\rightarrow \mathbb{R}^3,\qquad\gamma_2(x) = (\gamma\circ f)(x)$$

$\gamma_2$ is smooth, as it is a composition of smooth functions, so it is a curve using your given definition. However, the derivatives do not match at the endpoints: $$\sin(4\pi x^2)' = 8\pi x\cos(4\pi x^2), \mathrm{but}\ 0 \neq 8\pi$$ This means it isn't a closed curve using your definition.

Basically you can't talk about whether or not it is a closed curve without the curve itself. The trajectory doesn't provide enough information.