What is the lowest degree polynomial that turns a circle into a nontrivial knot?

Question

The polynomial

$$f(x,y) = \begin{bmatrix} (x^2-y^2)(y(4x^2-1)+2) \\ 2xy(y(4x^2-1)+2) \\ x^3-3xy^2 \end{bmatrix} $$

maps the circle $\left\{ (x, y) \mid x^2 + y^2 = 1 \right\}$ to a trefoil knot, and I suspect that this is the lowest degree polynomial that can map the circle to a nontrivial knot- i.e. that there are no degree $4$ polynomials that can map the circle to a knot. However after significant thought I have no idea how to approach proving this. With the stick number, which feels analogous, it is possible to reduce the problem of $n=4$ or $5$ to finitely many cases, but that doesn't seem possible here.

I would like either a proof or a method of approaching a proof for this. I would also appreciate a way of proving that this is the lowest degree polynomial that produces a trefoil knot, which I am equally stumped by.

It seems natural that degree $5$ would be minimal as the trefoil knot is the simplest knot and is the $(3,2)$ torus knot, but I have no deeper reasoning besides that it feels like with lower degree polynomials you can't fit enough bends in.

The way I produced this polynomial was starting with a this parameterisation of the trefoil knot and expanding for example $\sin 3\theta = 4 \sin \theta \cos^2 \theta - \sin \theta$ so the formula is all in terms of $\cos \theta$ and $\sin \theta$, which for the circle are just $x$ and $y$.

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Edit: apparently there is a degree $3$ polynomial that works, so now the objective is to prove no degree $2$ polynomials work- hopefully that's more tractable.

Answer 1

Actually, the following is a parametrisation of the trefoil

\begin{align*} x&=2\cos 2t+\sin t\\ y&=2\sin 2t+\cos t\\ z&=\sin(3t-1.5) \end{align*} all of which can be written as polynomials in $\sin$ and $\cos$ via the multiple angle formulae. Hence, it can be done with degree 3.

Explicitly,

\begin{equation*} f(x,y)= \begin{pmatrix} 2x^2-2y^2+y\\ 4xy+x\\ (3x^2y-y^3)\cos(1.5)-(x^3-3xy^2)\sin(1.5) \end{pmatrix} \end{equation*}

Answer 2

I will prove that there is no degree $2$ embedding of a nontrivial knot.

Assume the contrary. By applying the substitution $x = \cos t, y = \sin t$, we get a periodic parametrization $g \colon [0, 2\pi) \to \mathbb R^3$, which can be expressed in the form $$ g(t) := \vec{v_1}\cos 2t + \vec{v_2} \sin 2t + \vec{v_3} \cos t + \vec{v_4} \sin t + \vec{v_5}. $$ Up to an affine transformation, we can assume $$ g(t) := \vec{a}\cos 2t + \vec{b} \sin 2t + \hat{x}\cos t + \hat{y}\sin t. $$ Let $W:=\operatorname{span}(\hat{x}, \hat{y})$ and $Z:=\operatorname{span}(\vec{a}, \vec{b})$. If $Z \subseteq W$, then the knot is a (non-self-intersecting) plane curve, i.e. the unknot. Hence, we can assume $Z \not\subseteq W$.

Let $L := Z \cap W$. Up to rotating the knot around $\hat z$ and shifting the parameter $t$, we may further assume that $L = \operatorname{span}(\hat y)$.

Take $\vec v \in Z\setminus L$ and consider a projection $p\colon \mathbb R^3 \twoheadrightarrow \mathbb R^2$ which fixes $W$ and sends $\vec v$ to zero. Then the composition $p\circ g$ has the form $$ p(g(t)) = \hat{x} \cos t + \hat{y}\,(\sin t + \lambda \cos(2t + \phi))). $$ Let us find the self-intersections of this curve on $[0, 2\pi)$. If $p(g(t)) = p(g(t'))$ for $t < t'$, since $\cos t$ is $2$-to-$1$, we have $0 < t < \pi$ and $t' = 2\pi-t$. But then \begin{align} \sin t + \lambda \cos(2t + \phi) &= \sin (2\pi-t) + \lambda \cos(2(2\pi-t) + \phi) \\ &= -\sin t + \lambda \cos(2t - \phi) \\ \end{align} that is, $2\sin t = \lambda \cos(2t - \phi) - \lambda \cos(2t + \phi) = 4\lambda \sin \phi \sin t\cos t$. But this equality can hold at most once, i.e. when $\cos t = 1/(2\lambda\sin\phi)$. Hence, generically in $\vec{v}$, we get a diagram with one crossing, which cannot represent a nontrivial knot.

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EDIT: There is also a degenerate case when $\dim Z = 1$ and $\dim L = 0$. However, if we take a projection onto $W$ with kernel $Z$, we get a circle, which cannot be the diagram of a nontrivial knot.

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EDIT: If $\vec{v_3}$ and $\vec{v_4}$ are linearly dependent (thanks @Joshua), and $\operatorname{span}(\vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4}) = \mathbb R^3$, we can assume up to affine transformations and parameter shifts $$ g(t) = \hat x \cos 2t + \hat y \sin 2t + \hat z \sin t. $$ Projecting onto $\operatorname{span}(\hat y, \hat z)$, we get a curve $(\sin 2t, \sin t)$. It is not hard to see that this is a smooth curve with only one self-intersection at $(0,0)$, which gives a diagram with only one crossing.