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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.

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    $\begingroup$ 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. $\endgroup$ – Jonny Evans Aug 9 '19 at 14:31
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    $\begingroup$ Such a disk always exists. You can consider the cone the over the knot. I just dropped the smoothness condition. $\endgroup$ – Diego Hernández Rodríguez Aug 9 '19 at 18:37
  • $\begingroup$ I see: this was not clear from what you wrote. $\endgroup$ – Jonny Evans Aug 9 '19 at 19:44
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    $\begingroup$ There seems to be a discussion with some links here: mathoverflow.net/questions/93942/… $\endgroup$ – Jonny Evans Aug 9 '19 at 21:12
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    $\begingroup$ 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. $\endgroup$ – Jonny Evans Aug 9 '19 at 21:15

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