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?
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3$\begingroup$ What do you mean by "twist along $c$"? $\endgroup$– Marco GollaSep 14, 2015 at 10:06
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$\begingroup$ @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. $\endgroup$– shestipalovSep 21, 2015 at 10:31
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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.
answered Sep 14, 2015 at 18:25
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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. $\endgroup$ Sep 15, 2015 at 0:50
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$\begingroup$ @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? $\endgroup$ Sep 21, 2015 at 10:34
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$\begingroup$ @shestipalov: Yes, the surface will remain a fiber, because the twist acts as a Dehn twist of the monodromy of the fibration. $\endgroup$– Ian AgolSep 21, 2015 at 14:10