Algebraic equations for knots

Question

Is there any general method to find out algebraic equations rather than parametric equations for a knot? That is, I want a set of two equations in $\mathbb{R}^3$ whose common solution is the given knot.

Answer 1

Someone already pointed out Akbulut and King, "All knots are algebraic," 1981. This shows existence, but I don't believe it gives an algorithm. (I think there is a John Nash paper about the general case of closed subsets of $\mathbb{R}^n$.)

Bode and Dennis, "Constructing a polynomial whose nodal set is any prescribed knot or link" in JKTR 2019 shows how every knot is isotopic to the zero set of a single complex polynomial. This is using $S^3$ as the unit sphere in $\mathbb{C}^2$, and their polynomial ends up being only in $z$ and $\overline{z}$ and $w$ (no need for $\overline{w}$!). In other words, every knot is the zero set of a pair of three-variable polynomials in $\mathbb{R}[x,y,z,w]$ that do not use $w$. I do not think the solutions sets need to be non-singular or even need to intersect transversely. However, the intersection ends up being a smooth algebraic curve. The idea is sort of to do a Fourier transform of the knot in braid position.

Osamu Saeki and Harold Levine have a number of papers on stable maps, which is an obliquely related concept. Saeki in "Stable maps and links in 3-manifolds," 1994, shows every link is the fiber of a stable map $S^3\to \mathbb{R}^2$. Kalmár and Stipsicz, "Maps on 3-manifolds given by surgery," 2012, I believe contains some bounds on the complexity of such a map.

The $C^\infty$ case is relatively easy: given an oriented link $L$ in $S^3$, take the Alexander dual $\alpha\in H^1(S^3-L)$ corresponding to the orientation of $L$. This gives a map $S^3-\nu(L)\to S^1$ representing it, which we may presume to be smooth, and furthermore we may assume the map is a homeomorphism on meridians. Include $S^1\hookrightarrow \mathbb{R}^2$ as the unit circle. Then, we may extend the map to $f:S^3\to \mathbb{R}^2$ by "collapsing" the solid tori. Lastly, take a smooth approximation, which only needs to modify the map outside $f^{-1}(0)=L$. This induces the orientation of $L$, too. This is hardly computable, however.

If you want, from here you can take a stable representative for $f$ that similarly keeps $f|_L$ the same. If you choose a smooth line in $\mathbb{R}^2$ running through $0$ whose endpoints are at infinity while intersecting the image of the singular set only transversely (avoiding cusps!), then the preimage of this is a smooth closed surface in $S^3$ containing that link. If you choose a second line with the same properties except that it intersects the first line only at the origin, transversely, then the intersection of these two closed surfaces in $S^3$ is the link. (If this seems unbelievable, consider fibered knots. If you take four Seifert surfaces from the pages of the open book decomposition, then interleaved pairs form these closed surfaces. For non-fibered knots, as rays from the origin rotate, the surfaces undergo various surgeries, but they are all non-intersecting except at the fiber by construction.)