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Let $K \subset S^3$ be a slice knot. Then it bounds a smooth embedded disk $D \subset B^4$. Let $S^3_{p/q}(K)$ denote a $3$-manifold obtained by $p/q$-surgery on $K \subset S^3$.

The following theorem is due to Gordon:

Gordon, C. M. (1975). Knots, homology spheres, and contractible 4-manifolds. Topology, 14(2), 151-172.

Theorem: For a slice knot $K \subset S^3$, $S^3_{\pm 1}(K)$ bounds a contractible $4$-manifold.

I wonder that Gordon's theorem can be generalized to $1/n$ surgeries on slice knots for all $n$?

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2 Answers 2

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Yes, the generalisation is also true. This must be written somewhere, but I don't know where (any help from other users?), and finding such a statement is often hard.

So, here's the idea instead. Turn the surgery into an integral surgery, i.e. do 0-surgery on $K$ and $-n$-surgery on a meridian $L$ of $K$. 4-dimensionally, you're constructing a 4-manifold $X$ by attaching two 2-handles to $B^4$. $X$ contains a 0-sphere (the capped-off slice disc of $K$) and a $-n$-sphere (the capped off meridian disc of $L$) intersecting transversely once. A regular neighbourhood $N$ of the union of these two spheres is a plumbing (by definition) whose boundary is $S^3$ (this is certainly done in Gompf and Stipsicz's book). Now surger out $N$ and replace it with a 4-ball $B$, to get $W$. $W$ is an integral homology ball by excision ($\tilde H_*(W) = H_*(W,B) = H_*(X,N)$) and the long exact sequence of the pair $(X,N)$. The fundamental group hasn't changed with the surgery either: $X\setminus N$ is simply-connected because both $N$, $\partial N$ and $X$ are, and then $W$ is also simply-connected (both steps use Seifert-van Kampen).

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    $\begingroup$ You do a "slam-dunk" and put a "dot" on a given slice knot. Then they cancel each other "algebraically" on homotopy/homology level. $\endgroup$
    – Oğuz Şavk
    Mar 8, 2022 at 11:08
  • $\begingroup$ Do we know the handle decomposition of the resulting contractible $4$-manifold? $\endgroup$
    – Max Schumann
    Mar 9, 2022 at 10:42
  • $\begingroup$ That's more delicate. Short answer is I don't know. My guess is that you need to have a handle decomposition of the complement of the slice disc, and then Akbulut has probably done the construction explicitly for you. I would look at his book on 4-manifolds (maybe after peeking at Gompf and Stipsicz's book). $\endgroup$
    – Marco Golla
    Mar 9, 2022 at 11:11
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There is a more general result that this follows from. This is in Gordon's paper cited above; see Theorem 4. In any event, you don't really need the handle arguments in Marco's answer.

Here is a sketch of Gordon's construction, in slightly different language. Of $K$ and $K'$ are concordant, then for any $p/q$ there is a homology cobordism between $S^3_{p/q}(K)$ and $S^3_{p/q}(K')$ with fundamental group normally generated by the meridian of $K$ or the meridian of $K'$.

To see this, consider a tubular neighborhood $S^1 \times D^2 \times I$ of the concordance, and do $p/q$ surgery at each level. The meridian of the concordance is conjugate to the meridian of either end, and normally generates the fundamental group of the complement, and hence the fundamental group of the homology cobordism.

When $K$ is slice, apply this to $K' = $ the unknot to get a homology cobordism to a lens space $L(p,q)$ with fundamental group $\mathbb{Z}/p$. In particular when $p=1$, you get a simply connected homology cobordism to $S^3$, which gives a contractible $4$-manifold with boundary $S^3_{1/q}(K)$ as desired.

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