TITLE: Surgery along knots and connected sum
QUESTION [8 upvotes]: Denote $S^3_{p/q}(K)$ by performing $p/q$-surgery along a knot $K$ in $S^3$.
Let $K$ and $J$ be two arbitrary oriented non-trivial knots in $S^3$. Is there a nice relation between surgery on the connected sum of knots and the connected sum of their surgeries?
More explicitly, let say, do we have the following homeomorphism: $$S^3_{p/q}(K \# J) \approx S^3_{p/q}(K) \ \# \ S^3_{p/q}(J).$$
It might be false. Does this homeomorphism hold for integral surgeries?

REPLY [10 votes]: If $p \neq \pm 1$ then this can't happen, because the homology groups of the two sides would be different. I suspect it can't happen at all unless one of the knots is trivial; this would follow from the well-known cabling conjecture. If both knots are non-trivial, then the right side is a reducible manifold, so you would need that at least one summand is the $3$-sphere, which doesn't happen by the resolution of the property P conjecture. The cabling conjecture asserts that a knot with a reducible surgery is a $(p,q)$ cable, the surgery would be a $pq/1$ surgery, and one of the summands would be a lens space $L(pq,1)$. But a non-trivial connected sum is not a cable of another knot.
In general, surgery doesn't commute with connected sums, so one wouldn't expect an equality of the sort you're asking about. The sphere along which the connected sum is done doesn't yield a sphere in the surgered manifold; it yields a torus.