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It seems like there is an algorithm to find the Heegard diagram of a 3 manifold obtained by surgery on a link. Also someone told me I can find it in the Gompf and Stipciz's book. But I could not find it. Can anyone help?

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What do you mean by "the Heegaard diagram"? There are many. Do you only want to find one, or do you want to find a special one? – Ryan Budney Oct 10 '12 at 3:10
For the record, the correct spelling is "Stipsicz", and not "Stipciz". – Marco Golla Oct 10 '12 at 13:37
I just want to find any heegard diagram with any genus and and any attaching curves. – mark Oct 10 '12 at 14:31
I guess this person read Exercise 6.2.2 on page 210. – Turion Feb 10 '15 at 16:06
up vote 13 down vote accepted

(1) Choose a planar presentation of your link, approximately in the plane of the blackboard. Let $N$ be a tubular neighborhood of the link in this position.

(2) For each crossing $c$ of the presentation, add a 1-handle $T_c$ to $N$ which is perpendicular to the blackboard and connects the upper and lower parts of the crossing. Let $H$ be the union of $N$ and all the 1-handles $T_c$.

(3) $S^3 \setminus H$ is a handlebody. The corresponding set of Heegaard curves on $\partial H$ bijects with the complementary regions of the (flattened) planar diagram of the link.

(4) $H$ is, of course, also a handlebody. For the corresponding Heegaard curves take the surgery curves on $N$ (this involves a choice to make them disjoint from the attaching disks of the $T_c$) union the obvious small, disk-bounding curves on the boundary of each $T_c$. (Correction added later: Actually, if there are $k$ components of the link then one should omit $k-1$ of the $T_c$ curves. The omitted crossings should be minimal with respect to connecting the components of the link. Also, I'm assuming that the planar projection is a connected graph.)

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This is a nice very-rough-outline! There's the "digging a trench" construction needed to prove that surgery along the curve gives rise to Dehn twists on the handlebody; you also have to do something non-trivial to deal with non-unit framings. – Daniel Moskovich Oct 11 '12 at 8:32
I agree that the outline is rough, but on the other hand the question asked for an algorithm, not a proof that the algorithm was correct. ... Actually, I don't understand your comment. I don't think I ever have to mention or think about Dehn twists, and I think what I wrote works fine for non-unit and even non-integer surgeries. – Kevin Walker Oct 11 '12 at 14:13
... Put another way, the output is a surface with two sets of curves, not two handlebodies with a homeomorphism between their boundaries. – Kevin Walker Oct 11 '12 at 14:18
How does this work for example for a Lens space $L(p,q)$? That's just $-p/q$-surgery on the unknot. (1) So we take $N = S^1 \times D^2 \stackrel{\text{unknot }-p/q}{\hookrightarrow} S^3$. (2) We don't need any 1-handles since there are no crossings. $H = N$. (3) I can only guess what you mean by this step. I'm drawing the circle ${0} \times \partial D^2 \subset \partial H$. (4) What exactly are the surgery curves on $N$? In particular, for general $p,q$, the curve must go around $H$ several times? I take it this is just the torus knot? – Turion Sep 9 '14 at 16:31
In the lens space example, the two curves of the Heegaard diagram are (1) a $-p/q$ curve on $\partial N$, and (2) a curve which bounds a disk in $S^3\setminus N$, i.e. a longitude. – Kevin Walker Sep 9 '14 at 19:11

I don't know about Gompf and Stipsicz, but the algorithm to obtain a Heegaard diagram from a surgery presentation may be found for example in A Simple Proof of the Fundamental Theorem of Kirby Calculus on Links by Ning Lu, Trans. Amer. Math. Soc. Vol 331(1) pp. 143-156 (1992). It's not difficult, and it's worth knowing.

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