Knot being a subset of another knot.

Question

Can you show an example of a nontrivial knot $ K $ in $ \mathbb{R}^{4} $ such that $ K \cap \mathbb{R}^{3} $ is a nontrivial knot in $ \mathbb{R}^{3} $ ? It would be nice if you used parametric formulas.

Answer 1

Any slice knot gives an example. Consider the three sphere $S^3$ as the boundary of the four ball $B^4$. A knot $K\subset S^3$ is (smoothly) slice if it bounds a smooth disk in $B^4$ (called a slice disk).

The slice disk can be doubled to obtain a knotted $S^2$ in $S^4$ as follows. Consider the four sphere $S^4$ as two $4$-balls glued along their common $S^3$ boundary: $S^4 = B^4_1 \cup_{S^3} B^4_2$. If $K\subset S^3$ is slice, then $K$ bounds smooth disks $D_1\subset B^4_1$ and $D_2\subset B^4_2$. The union $D_1\cup D_2$ is a knotted $S^2$ in $S^4$ that contains the knot $K$ as a subset.

Just to have some visuals, the picture below depicts a knot and a slice disk. The disk looks like it intersects itself, but the colors are supposed to indicate position in the fourth dimension. Because the arc of intersection of the disk with itself is colored differently on each intersecting page, the disk is embedded in $B^4$.

I don't know enough to write down parametric formulas for the disks, but perhaps someone can improve on my answer.