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 $n$?
Best Answer
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).