Let $X$ be a smooth closed connected 4-manifold. It admits a handlebody structure, having a unique 0- and a unique 4-handle. We can express the handlebody structure as a Kirby diagram (https://en.wikipedia.org/wiki/Kirby_calculus). Suppose $Y$ is a subhandlebody of $X$, which contains the 0-handle of $X$ and some 1,2,3-handles of $X$. (So $Y$ has nonempty boundary.) Then we can view the handlebody structure of $Y$ inside the Kirby diagram of $X$. Is there a method or algorithm to draw the Kirby diagram of the complement $X-\operatorname{int}(Y)$? If not, how about in the simpler case where $X$ has no 1- or 3-handles, so that the Kirby diagram of $X$ is just a link in $S^3$ with integer framings?
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$\begingroup$ In your first question presumably you meant to say "a Kirby diagram" not "the" diagram. Handle structures on the complement of submanifolds is an old topic, in dimension 3 the standard is the one that generates the Wirthinger presentation of knot complements. You can do the same kind of construction in 4-manifolds but I suspect the level of detail in the resulting handle decomposition is heavily dependent on how precise your original (manifold, submanifold) pair is described. $\endgroup$– Ryan BudneyCommented Dec 2 at 15:47
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$\begingroup$ There is a general prescription for this when the handle structures come from a Morse function on $X$ (which you then restrict to the submanifold, and to the exterior of the submanifold, respectively). The prescription looks much like the Wirthinger presentation. $\endgroup$– Ryan BudneyCommented Dec 2 at 16:04
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When the decomposition of $X$ has no 3-handles, this is often feasible. The trick is to turn the handle decomposition of $X$ upside down (see Example 5.5.5 in Gompf and Stipsicz's 4-manifolds and Kirby calculus book) and to view $Y$ as the union of 2-, 3-, and 4-handles.
To give you a better picture, by "turning $X$ upside down" I mean swapping the roles of 1- and 3-handles, and viewing 2-handles as attached to the union of the 4- and 3-handle. (In Morse theory terms, you're looking at the function $-f$ instead of looking at $f$.)
Let me not describe exactly how to turn things upside down, which does get tricky in practice, but rather tell you the punchline: if $Y$ is a union of 4-, 3-, and 2-handles, then you just do not attach them. That is, just remove the link components corresponding to the 2-handles of $Y$, and specify how many 3-handles you want to attach at the end (and how you attach them, but that's hardly ever easy to make explicit).
If the handle decomposition has no 3-handles either, then you can really just forget about the 2-handles. You still have to do the work of turning the diagram upside down, though, but without 1- and 3-handles things are usually easier. What you need to do is to find a sequence of Kirby (3D) moves that take your 2-handle presentation to the empty presentation of $S^3$, and drag the 0-framed meridians of the link with you all along, and finally take the mirror of everything. (Sorry for not being very explicit, but to me this is more dark magic art than science.) You also keep track of which of the components of the link you obtain were meridians of the components corresponding to the 2-handles of $Y$, and you just remove those from the diagram.
As a concrete example, maybe, let us look at $X = \mathbb{CP}^2$ and $Y$ the neighbourhood of a complex line. Then we're in your setup: $X$ has a handle decomposition with one 1-handle (a +1-framed unknot $O$) and $Y$ is the union of the 0- handle and of the 2-handle. The procedure now says: take $O$, and a (0-framed) meridian $\mu$ and do Kirby moves on this link until $O$ disappears. In this case, this is a single blow-down, which turns $\mu$ into a $-1$-framed unknot (which was expected from Property R). Now you mirror everything, so you get a $+1$-framed unknot. To get the complement of $Y$, just remove this component, so you get a 4-ball (ok, not too surprising).
