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Suppose $B$ is the binding (with more than one component) of a planar open book on a 3-manifold $Y$ and let $L\subset B$ be the complement of a single component of the binding. Now we perform page-framed surgery on $L$, the resulting 3-manifold is $S^3$ since the induced open book has disk pages. Then $L$ is a fibered link in $S^3$ and we can consider the fibration $f:S^3-L\to S^1$. The question is how does $df$ look like near each component of $L$? I think we can take coordinates $(r,\theta, \phi)$ near each component (with $\theta$ representing the meridian and $\phi$ the longitude from page framing), and then $f=\theta$ and $df=d\theta$.

I am asking this question because in lemma 2.4 of this paper of Gay and Stipsicz: http://arxiv.org/abs/0908.3774, they are claiming that $df=md\theta+ld\phi$ near each component for constants $m$ and $l$ depending on the component. I cannot figure out what specific role $S^3$ is playing in their argument.

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  • $\begingroup$ The link $L'$ in $S^3$ (the surgery dual of the link $L$ in $Y$) is not necessarily a fibered link. If the knot $K$ is the remaining component of the binding $B$ in $Y$, then after surgery on $L$, in $S^3$ we see that $L'$ is a closed pure braid with $K$ as a braid axis. Generically a link that is the closure of a pure braid is not fibered. But Gay and Stipsicz do not claim $L'$ is fibered. From a glance, it appears that the function of $S^3$ here is that it is a homology sphere so that $L'$ will be null-homologous. $\endgroup$
    – Ken Baker
    Jan 15, 2015 at 14:46
  • $\begingroup$ I see what you are saying, but then what could be the map $\sigma: S^3-L'\to S^1$ they are talking about? $\endgroup$
    – nikita
    Jan 16, 2015 at 16:57
  • $\begingroup$ Any map dual to a Seifert surface for $L'$. They just ask that the preimage of a regular value is a Seifert surface. $\endgroup$
    – Ken Baker
    Jan 18, 2015 at 14:18

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