Suppose I have two distinct fibered knots in a homology sphere. Is it possible for them to have (orientation-preservingly) homeomorphic exteriors?

See Oriented knot complement conjecture for fibered knots for another version of this question.

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    $\begingroup$ I am certain that such examples exist. Pick a hyperbolic homology sphere $M$, chosen so that $M$ admits an orientation preserving involution $\phi$. (For example, generic elements of the genus two Torelli group will build such manifolds with Heegaard genus two.) Now, there is a classical result that all three-manifolds admit fibered knots. Pick a generic such knot $K \subset M$. Then $K$ and $\phi(K)$ are the desired knots. $\endgroup$
    – Sam Nead
    Feb 8 '15 at 21:11

Here we are assuming "distinct" means "non-isotopic".

Here is an example, found using SnapPy: http://www.math.uic.edu/t3m/SnapPy/

Consider the SnapPea manifold M = Manifold('m390'). This has first homology group Z. We Dehn fill M along the slope (2,-3), via the command M.dehn_fill((2,-3)). Note that the filled manifold is still called M. We check M.homology() to see M is a homology sphere.

By examining M.length_spectrum(2) we find M has a pair of geodesics of length 1.169 + 1.937*I. Let's call these the "twins" K and K'. The twins are tied for second and third shortest curves in the manifold.

We now look at the list M.dual_curves(). These are the curves that SnapPy can drill for us. Luckily for us, the zeroth dual curve is one of the twins, K. So we drill K via N = M.drill(0). Just to be on the safe side, we take P = N.filled_triangulation(). Now, P.identify() tells us that P is the SnapPea manifold v2869. If we check Nathan Dunfield's list


we find that v2869 is fibered. So, K is a fibered knot in the homology sphere M.

It is left to show that the complement of K' in M is homeomorphic to that of K. We examine P.length_spectrum(). The second shortest curve has length 0.870 + 2.070*I. This will be K'; its length has changed because we drilled K. Again we check the dual curves to find that K' can be drilled. [There is some junk in the collection of dual curves. I don't know why that happens?] So form Q = P.drill(2).

Now, if we fill either cusp of Q using the meridional slope (1,0), we get manifolds isomorphic to P. Here is the list of commands:

  • Q.dehn_fill((1,0),0) # fill K
  • P.is_isometric_to(Q) # check
  • Q.dehn_fill((0,0),0) # drill K
  • Q.dehn_fill((1,0),1) # fill K'
  • P.is_isometric_to(Q) # check

Since the isometry checker returned true in both cases, we are done.

Just as a final remark - I found m390 by looking in the cusped census for homology solid tori, with symmetries, with fundamental group of rank at least three (according to SnapPy). The first requirement is because we want to fill to get a homology sphere. The second and the third help the twins appear.

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    $\begingroup$ This is an amazing SnapPea-expert answer! $\endgroup$ Feb 9 '15 at 0:52
  • $\begingroup$ It would be more interesting to find an example for which the homeomorphism does not extend over the Dehn fillings - this is usually what is meant by "cosmetic surgery" I think. $\endgroup$
    – Ian Agol
    Feb 11 '15 at 4:24
  • $\begingroup$ @Ian - After I wrote my answer, the poster realized that: see mathoverflow.net/questions/196088/… I have written the obvious answer there, stating that the cosmetic surgery conjecture is still open. Do you see a way to use the extra hypotheses? $\endgroup$
    – Sam Nead
    Feb 11 '15 at 9:18

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