I'm looking for a parameterization $(x_1(u,v),x_2(u,v),x_3(u,v),x_4(u,v))$ of a knotted sphere in $\mathbb R^4$. How might one go about finding such a parameterization?

1$\begingroup$ Which knotted surface? $\endgroup$– Igor RivinDec 3, 2015 at 11:23

$\begingroup$ @IgorRivin Yeah, I left that deliberately open. I'm looking for a parameterization of any relatively basic knotted surface, that I can use to produce a visualization of the intersection of the surface with a family of 3d affine hyperplanes, and seeing how the intersections evolve over time as classical links. $\endgroup$– silvascientistDec 3, 2015 at 16:38

$\begingroup$ The question is: what do you mean by "knotted"? Which is the same question as: what do you mean by UNknotted? $\endgroup$– Igor RivinDec 3, 2015 at 16:44

$\begingroup$ @IgorRivin I suppose I should restrict to spheres. Then take unkotted to be ambient isotopic to an embedding that lies in $\mathbb R^3$ $\endgroup$– silvascientistDec 3, 2015 at 16:59

$\begingroup$ I'd understand if this question was considered too basic for this site, but I don't see how it is at all unclear what is being asked for. $\endgroup$– PVALDec 3, 2015 at 19:53
1 Answer
If you know how to parameterize a nontrivial knot in $R^3$ (say, the trefoil knot), you can use Artin's "spinning" construction to parameterize a nontrivial knot in $R^4$ via map $S^2\to R^4$ written in spherical coordinates on $S^2$. Take, say, the trefoil $T$ in $R^3$ parameterized by some smooth periodic function $f(\theta), 0\le \theta\le a$ where $a$ is the period. Let $L$ denote the straight line in $R^3$ through the points $f(0), f(c)$, $c=a \epsilon$ where $\epsilon>0$ is sufficiently small (if you want to know how small is small, ask me and I can explain). I will choose the coordinates $x_1, x_2, x_3$ so that $L$ is the $x_3$axis. Let $R_\phi$ denote the rotation by the angle $\phi$ in $R^4$ around the plane $x_1=x_2=0$ (thus, fixing $L$). Now, consider the map $F: [0, c]\times [0, 2\pi]\to R^4$, given by $$F(\theta, \phi) = R_\phi\circ f(\theta)$$ Its image is a knotted 2sphere $\Sigma$ in $R^4$. It is smooth away from the points $f(0), f(c)$. If this bothers you, it can be fixed (explicitly). Now, rescale the interval $[0, c]$ to $[0, \pi]$ via the map $s: t\mapsto \frac{\pi}{c}t$. Then the map $$G= F\circ (s^{1} \times Id): [0,\pi]\times [0,2\pi]\to R^4$$ parameterizes the same surface $\Sigma$ as before. Lastly, use the spherical coordinates on $S^2$ to convert the map $G$ to a homeomorphism $S^2\to \Sigma$.
This, I think, is as an explicit construction as it can be (provided you are not bothered by smallness issue for $\epsilon$).
Edit: Upon your request, smallness of $\epsilon$ is defined as follows: There exists a round ball $B$ (or a halfspace) in $R^3$ with points $p=f(a\epsilon), q=f(a)$ on its boundary such that the arc $\alpha=f([a\epsilon, a])$ is contained in $B$ and is unknotted in $B$, i.e., is isotopic to the line segment $pq$ in $B$ via an isotopy fixing the boundary sphere of $B$. (In practical terms, you can think of the unknottedness property of $\alpha$ as assuming that $\alpha$ is a graph of a function on the segment $pq$, with values in the plane orthogonal to $pq$.)
See for instance here for details.

$\begingroup$ Thanks for your answer. How small is sufficiently small? $\endgroup$ Dec 5, 2015 at 0:06