2
$\begingroup$

The Gordon–Luecke theorem states that if two (tame) knots have homeomorphic complements in $S^3$, then they are equivalent (up to isotopy and reflections).

It is not the case, however, that distinct knots have complements in $S^3$ with distinct fundamental groups. The standard counterexample is that the square knot and the granny knot have the same knot group. (Though prime knots are distinguished by knot groups).

My question lies in the middle: if two tame knots have homotopy equivalent (but not homeomorphic) knot complements, must they be the same knot up to isotopy and reflection?

Are there any results in this direction?

$\endgroup$
4
$\begingroup$

A knot complement is an Eilenberg–MacLane space (see here, for example), and so its homotopy type is determined by the knot group. Hence your (and every) counterexample for the knot group is also a counterexample for the homotopy type.

$\endgroup$
4
$\begingroup$

Knot complements are aspherical (Papakyriakopoulos), hence, they are homotopy-equivalent if and only if they have isomorphic fundamental groups. The right notion is a relative homotopy-equivalence $$ (X(K_1), \partial X(K_1))\to (X(K_2), \partial X(K_2)), $$ where $X(K)$ denotes the knot exterior (the complement to an open tubular neighborhood of $K$ in $S^3$). Every such relative homotopy-equivalence is (relatively) homotopic to a homeomorphism (this is due to Waldhausen, in greater generality), thus, the Gordon-Luecke theorem applies.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.