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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.

Diego Hernández Rodríguez
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    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 09 '19 at 14:31
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    Such a disk always exists. You can consider the cone the over the knot. I just dropped the smoothness condition. – Diego Hernández Rodríguez Aug 09 '19 at 18:37
  • I see: this was not clear from what you wrote. – Jonny Evans Aug 09 '19 at 19:44
  • Incidentally, for the trefoil, the answer to 1 is $SL(2,R)/SL(2,Z)$, which is nice. For the figure 8 you can give a nice triangulation of the complement by two ideal hyperbolic tetrahedra: I think this is explained in Thurston's book on 3-manifolds. I think if you're interested in the cone disc then the answer to 2 should be the knot complement times R, but I'm not 100% sure I know what a "tubular neighborhood" is in that case. – Jonny Evans Aug 09 '19 at 19:56
  • Thanks. Do you have a good reference for the trefoil? – Diego Hernández Rodríguez Aug 09 '19 at 20:56
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    There seems to be a discussion with some links here: https://mathoverflow.net/questions/93942/why-s3-k-and-sl2-r-sl2-z-are-diffeomorphic-here-k-is-a-trefoil-in-s3 – Jonny Evans Aug 09 '19 at 21:12
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    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 09 '19 at 21:15

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