Knot and its embedded disk

Let $$K \subset S^3$$ be an arbitrary knot. Let $$D$$ denote the embedded disk in $$B^4$$ bounded by $$K$$.

Up to diffeomorphism, is it possible to describe the followings (at least for some trivial knots, such as trefoil, figure-eight knot,etc.):

1) $$S^3 \setminus \nu(K)$$

2) $$B^4 \setminus \nu(D)$$

3) $$\partial (B^4 \setminus \nu(D))$$

where $$\nu(*)$$ denotes the tubular neighborhood. I understand that the answer is yes for slice knots.

• Such a D only exists if K is slice (by definition). Even then, I would imagine that the diffeomorphism type of the complement of a slice disc depends on which slice disc you pick. – Jonny Evans Aug 9 '19 at 14:31
• Such a disk always exists. You can consider the cone the over the knot. I just dropped the smoothness condition. – Diego Hernández Rodríguez Aug 9 '19 at 18:37
• I see: this was not clear from what you wrote. – Jonny Evans Aug 9 '19 at 19:44
• There seems to be a discussion with some links here: mathoverflow.net/questions/93942/… – Jonny Evans Aug 9 '19 at 21:12
• The cone disc you're interested in (basically) the algebraic curve y^2=x^3 in C^2 when K is a trefoil knot (and has such an algebraic description whenever K is an iterated torus knot). This fact is used in some proofs of the identification of the complement with the homogeneous space of SL(2,R) for the trefoil. – Jonny Evans Aug 9 '19 at 21:15