Suppose we have a fibred knot $K$ with a fiber surface $F$ and let $c$ be an unknot disjoint from $F$ (but not homotopically trivial in the complement of $F$). Is it possible that every twist along $c$ leaves $K$ fibred with $F$ still being the fiber surface?

  • 3
    $\begingroup$ What do you mean by "twist along $c$"? $\endgroup$ – Marco Golla Sep 14 '15 at 10:06
  • $\begingroup$ @MarcoGolla technically it means 'perform a $1/n$ surgery on $c$ for some non-zero integer $n$'. I imagine it as cutting the solid torus (exterior of $c$) along a meridional disc, twisting and regluing. $\endgroup$ – shestipalov Sep 21 '15 at 10:31

Such examples were constructed by Morton. He showed that one can find unknotted curves lying on fiber surfaces with zero framing. Twisting about them preserves fiberedness and the fact that it is a knot in $S^3$. Also, curves on the fiber can be pushed disjoint from the fiber meeting your requirement.

Edit: As Danny points out in the comments John Stallings originally came up with this construction.

  • 3
    $\begingroup$ I think this construction is due to Stallings (Constructions of fibred knots and links, Proceedings of Symposia in Pure Mathematics, 1978) and is called a Stallings twist. You need to require that the framing on the curve from the normal to the Seifert surface is the untwisted framing. $\endgroup$ – Danny Ruberman Sep 15 '15 at 0:50
  • $\begingroup$ Good point, I forgot about that. $\endgroup$ – Ian Agol Sep 15 '15 at 1:02
  • $\begingroup$ @IanAgol I am probably being stupid, but I can't figure something out. Stallings describes his unknot as lying in the fiber surface. So if I push it away from the surface and then do a twist, does the surface still remain to be a fiber surface? $\endgroup$ – shestipalov Sep 21 '15 at 10:34
  • $\begingroup$ @shestipalov: Yes, the surface will remain a fiber, because the twist acts as a Dehn twist of the monodromy of the fibration. $\endgroup$ – Ian Agol Sep 21 '15 at 14:10

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.