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The knot must be quasipositive. You can perturb it to be transverse as in Marco Golla's answer, and then a theorem of Boileau and Orevkov (in Quasi-positivité d'une courbe analytique dans une boule pseudo-convexe, MR1836094) says that transverse knots which bound symplectic surfaces in $B^4$ must be quasipositive.


1

No, there are obstructions to this. Suppose $L$ is a Legendrian knot that bounds a symplectic surface $\Sigma$, and $T$ a (sufficiently close) transverse push-off (see Etnyre's notes or Geiges' Introduction to contact topology for references). Notice that $T$ can be chosen to be $C^\infty$-arbitrarily close to $L$. (This is clear from the construction, ...


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(I'm adding another answer, in case *cough* you have no access *cough* to the book.) In practice, you just need to crunch some numbers: you need to find integers $s_1, s_2, s_3$ such that $r_1r_2s_3 + r_2r_3s_1 + r_3r_1s_2 = b$; the surgery presentation is just obtained by doing 0-surgery on the unknot, and $r_i/s_i$-surgery on three (pairwise unlinked) ...


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One method is outlined in Montesinos' book "Classical tessellations and three-manifolds." Chapter 4 of that book deals with Seifert fibered spaces more generally, but the specifics for this question are covered in section 4.3 Constructing the manifold from the invariants and Figure 12 of that chapter. Of course, this is not the only way to get surgery ...



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