4
$\begingroup$

Suppose I have two inequivalent fibered knots in a homology sphere. When I say 'inequivalent', I mean that there is no orientation-preserving homeomorphism of the space that takes one to the other. Can they have orientation-preservingly homeomorphic complements?

I guess in general the answer is 'we don't know, but we think not', which is the oriented knot complement conjecture. What I'm asking here is whether fiberedness makes things much easier. It intuitively feels to me, that fibered knot complements are quite 'rigid', so perhaps it should somehow be easier to resolve the conjecture in this case.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I suppose then one knows that 0-framed surgery is fibered (with non-zero genus), which implies it has non-trivial Floer homology (as opposed to $S^2\times S^1$. So the Floer exact triangle might impose some restrictions. $\endgroup$
    – Ian Agol
    Feb 27, 2015 at 4:36

1 Answer 1

Reset to default
2
$\begingroup$

If we remove the hypotheses of fibered knots lying in a homology sphere your question becomes the "oriented knot complement conjecture". See Problem 1.81(D) of Kirby's problem list. Problem 1.81(D) is equivalent to the somewhat better known Problem 1.81(A), the "cosmetic surgery conjecture".

With your additional requirements, I'll guess that no such triple $(M, K, K')$ exists. Some work has been done under the additional assumption that $M$ is an $L$-space. A google search on the cosmetic surgery conjecture will give the state-of-the-art in this direction. (See also this question.) I don't know how to use the fibered hypothesis - perhaps an expert in Heegaard Floer theory will tell us more.

Improve this answer
$\endgroup$

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.