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Given a compact connected 3-manifold $M$ with non-empty boundary, and a link $L \subset M$, is there a handlebody decomposition of $M = H^0 \cup (\cup_i H^1_i) \cup \{\text{2-handles}\}$ such that:

  1. $L \cap H^0$ is a set of trivial arcs;
  2. $L \cap H^1_i$ is either empty or the core of $H^1_i$ for all $i$;
  3. $L$ does not intersect the 2-handles?
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Take a Heegaard splitting of the link exterior and then glue in solid tori with your link components as cores. The Heegaard splitting survives as a Heegaard splitting of the filled manifold and since the link is disjoint from the Heegaard surface it is disjoint from the 2-handles. To see that the other properties hold, you just have to consider the compressionbody obtained by Dehn filling torus boundary components of a compressionbody.

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  • $\begingroup$ This is much cleaner than my answer! Very nice! $\endgroup$ Feb 4 '14 at 3:32
  • $\begingroup$ So, in other words you construct a surface in $M$ which divides $M$ into two compressionbodies, and such that the whole link is contained in one side, is this the point? Also, when you make these Dehn fillings on the compressionbody, what do you assume in order to get another compression body (or maybe a handlebody)? $\endgroup$ Feb 4 '14 at 12:02
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    $\begingroup$ You actually need the Heegaard splitting to be a Heegaard splitting for the link exterior, not just "contain the link to one side". When you Dehn fill toroidal boundary components of the negative boundary of a compressionbody, you will always get a compressionbody (or handlebody). To see this, recall that compressionbodies are characterized by the presence of a complete set of discs cutting the compressionbody into balls and products. In the product regions there are annuli which can be completed to discs upon Dehn filling to create a complete set of discs for the compressionbody post filling. $\endgroup$ Feb 4 '14 at 17:40
  • $\begingroup$ So, these annuli determine also the framings for the Dehn fillings. Thank you for your answer. $\endgroup$ Feb 5 '14 at 16:16
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Here is a rough construction of such a decomposition. First, glue another copy of $M$ to $M$ along their joint boundaries to get rid of the boundary, and then reduce your question to $S^3$ by performing surgery on a framed link $K\subset M$ that lives inside a ball $B$ that is disjoint from $L$.

Next, form a handlebody by starting with a tubular neighbourhood of $L$, choosing a diagram for $L$, and adding a small `bridging cylinder' between each overcrossing cylinder and its corresponding undercrossing cylinder ("overcrossing" and "undercrossing" determined by the diagram). Note that this handlebody lies entirely outside $B$. If needed, add further bridging cylinders to make the complement also a handlebody. The handlebody thus constructed has a handle decomposition with only $0$--handles and $1$--handles which we complete to a handle decomposition of $S^3$ that satisfies all three conditions which you listed.

Finally, to obtain a Heegaard splitting of $M$ which induces a handle decomposition of $M$ satisfying your conditions, choose a Heegaard splitting of $M$ so that $B$ is inside one of the handlebodies, and connect-sum your handlebody outside $B$ (the one containing $L$) with the other handlebody inside $M$, adding bridging cylinders as required.

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  • $\begingroup$ The framed link $K$ unlikely is contained in a ball in $M$. In my understanding, the addition of solid cylinders makes the complement of this thing a handlebody (maybe, by destroying the cores of the 2-handles). Is this correct? In this case probably your construction can be applied directly to $M$, isn't? $\endgroup$ Feb 3 '14 at 14:51
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    $\begingroup$ Yes- I was careless; K won't be in a ball in general, and the construction should of course be applied directly to M. Alternatively, you can realize surgery on K as Dehn twists on a Heegaard surface via a similar construction, and you can choose this surface so that L is inside one of the handlebodies (by taking the radius of the tubular neighbourhood of K to be `small'). Then apply the construction as mentioned. $\endgroup$ Feb 3 '14 at 15:59

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