# Knot theory: Converting a parametrization into an embedding

According to the Wikipedia page on torus knots a (p, q)-torus knot can be given by the parametrization:

$\begin{cases} x = \cos{pt}\cos{qt} + 2\cos{pt}\\ y = \sin{pt}\cos{qt} + 2\sin{pt}\\ z = -\sin{qt} \end{cases}$

where $t \in \left[0, 2\pi\right]$.

Unfortunately, on its own a parametrization isn't very useful I think, aside from allowing one to graph the knot in something like Mathematica. That's because a knot is an embedding of $S^1$ in $\mathbb{R}^3$, not $\left[0, 2\pi\right]$ in $\mathbb{R}^3$.

My question is can someone walk me through the process of converting a parametrization of a torus knot call it $f: \left[0, 2\pi\right] \rightarrow \mathbb{R}^3$ into an explicit embedding $g: S^1 \rightarrow \mathbb{R}^3$? More generally, is there a process for converting a parametrization of a knot into an explicit embedding of $S^1$ in $\mathbb{R}^3$?

## 1 Answer

Sure. $S^1$ actually is $[0, 2\pi] / \sim$, where $a \sim b$ iff $b - a$ is a multiple of $2\pi$.

Since the map $X$ defined by

$$X : [0, 2\pi] \to \Bbb R^3:t \mapsto (x(t), y(t), z(t))$$

satisfies $X(2\pi) = X(0)$, it passes to the quotient, $S^1$, and thus is an explicit map from $S^1$ to $\Bbb R^3$. It's not hard to check that it's an embedding, because on the quotient, it's 1-1.

Or did you have some other definition of $S^1$ that you prefer?

• Okay I understand the equivalence relation here but can you explain what you mean when you say that the map passes to the quotient? Is this just saying that the domain is actually $[0, 2\pi]/\sim$, not $[0, 2\pi]$? – EgoKilla May 15 '17 at 16:35
• There's a "projection" map $p$ from $[0, 2\pi]$ to $[0, 2\pi] / ~$, taking each item in the domain to its coset in the codomain. For instance, $0.4$ is sent to $\{0.4\}$. In general $x$ is sent to $\{x\}$; But $0$ and $2\pi$ are both sent to the coset $\{0, 2\pi\}$. To say that $X$ passes to the quotient means there's a map $Y : [0, 2\pi]/~ \to \Bbb R^3$ with the property that $X = Y \circ p$. In this case, it's easy to define. For a one-element coset $\{x\}$, we have $Y(\{x\}) = X(x)$. For $\{0, 2\pi\}$, we define $Y(\{0, 2\pi\}) = X(0)$. That's "passing to the quotient". – John Hughes May 15 '17 at 16:43