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Every closed, compact, orientable 3-manifold can be represented by a Heegaard diagram. Similarly every such 3-manifold can be represented by a framed link diagram. Is there any general procedure for going between these? Specifically, given a Heegaard diagram is there an algorithmic way to obtain a framed link diagram of the same manifold, and vice versa?

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  • $\begingroup$ Yes, this is how you prove the famous theorem of Lickorish and Wallace. (One writes any element of the mapping class group as a sequence of dehn twists and does surgery along the twisting curves). $\endgroup$ Feb 11, 2017 at 2:42

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As PVAL already said in the comments Lickorish's proof of Lickerish-Wallace theorem somehow tells you how to go from Heegaard diagrams to surgery presentations.

To go from surgery diagrams to Heegaard decomposition there is an algorithm. I tell to you how it works for surgeries on knots, then you can easily generalise to the case of links.

  • Step 1: arrange your knot $K \subset S^3$ so that the projection on the $xy$-plane is regular (Fig. 1)
  • Step 2: pick a tubular neighbourhood of $K$ and denote by $T \subset S^3$ its boundary torus (Fig. 2)
  • Step 3: picture on $T$ the surgery longitude (call it $\gamma$)
  • Step 4: in a neighbourhood of each crossing add a pipe as pictured in Figure 3. The resulting surface is going to be the surface underlying the Heegaard diagram of the surgery, denote it by $\Sigma$.
  • Step 5: take as $\beta$-curves the curve $\gamma$ together with the boundary of the compressing disks of the pipes we introduced at the crossings (the blu curve pictured in Fig. 3)
  • Step 6: take as $\alpha$-curves the boundary of the bounded regions of the diagram as shown in Figure 4.

The reason why this algorithm produce the right answer is not that deep: after gluing three-dimensional 2-handles along the $\alpha$- and the $\beta$-curves $\not=\gamma$, we get a three-manifold $Y$ with $2$ boundary components (a torus and a sphere). If we fill the sphere boundary component of $Y$ with a three-ball we get the complement of $K$ in $S^3$, and the Dhen filling operation along $\gamma$ prescribes to attach a 2-handle along $\gamma$ and fill the only sphere boundary component of the resulting three-manifold with a three-ball. This is the same as attaching 2-handles along all the $\alpha$- and the $\beta$-curves and fill the two boundary components with three-balls as prescribed by the Heegaard diagram $(\Sigma, \alpha, \beta)$.

enter image description here

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  • $\begingroup$ Thank you very much this was very clear and helpful!! $\endgroup$
    – user101010
    Feb 13, 2017 at 5:01
  • $\begingroup$ Very nice. So if I understand correctly, the handle body of your example is genus 4? $\endgroup$
    – Bob
    Jul 31, 2018 at 7:52
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    $\begingroup$ Thank you Bob, I'm glad you liked my explanation. Indeed, it has genus four. Actually, try to convince yourself that in the diagram there are always gonna be as many $\alpha$-curves as many $\beta$-curves as much as the genus of $\Sigma$ (I think this is a funny exercise). $\endgroup$ Jul 31, 2018 at 10:35
  • $\begingroup$ Sorry to keep bugging you but I'm very interested in this material. Is there a book or a paper that discusses your algorithm? $\endgroup$
    – Bob
    Aug 15, 2018 at 1:03
  • $\begingroup$ If the surgery coefficient is varied, how does this link+surgery -> surgery on Heegaard splitting prescription change? $\endgroup$ Nov 8, 2018 at 6:03

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