# Surgery on trivial knots

I know a theorem that any closed orientable 3 manifold can be obtained from the sphere $S^3$ by surgery along a framed knot.

I think I read or heard somewhere that as a surgery link, we can take unknots with integer framing.

I am confused that whether this means that we can find a surgery link whose connected components are unknots with integer framing AND each components are "separated" (there exist disjoint balls containing each components. Is this same as linking number is zero?)

Or does it mean that each component is unknot but it might be they are linked?

If you take the Lickorish-Wallace theorem for granted (every 3-manifold is obtained by integral surgery on a link in $S^3$), it is not too hard to see how to make every component of such a link an unknot using Kirby moves. You can blow up and then do handle slides to change the crossings of any component of your link without changing the resulting $3$-manifold. It is a classical fact that every knot can be turned into the unknot by changing some of its crossings. Hence we can always alter a framed link in $S^3$ in such a way that we get a new framed link with all components unknots.
Here's a picture of the "Kirby move" that changes crossings (from Gompf and Stipsicz's $4$-Manifolds and Kirby Calculus):
If $M$ is a $3$-manifold obtained by surgery on a framed unknot, then $M$ is $S^3$, $S^2 \times S^1$, or a lens space. If $M$ is obtained by surgery on a framed unlink, then $M$ is a connected sum of copies of $S^3$, $S^2 \times S^1$, and lens spaces. There are many $3$-manifolds that are not of this type. One example is the Poincaré homology sphere $\Sigma(2,3,5)$. Hence in general a $3$-manifold can only be obtained by integral surgery on a nontrivial link with all components unknots, not necessarily an unlink.
• See the edit to my answer. Basically any $3$-manifold whose fundamental group cannot be written as a free product of cyclic groups will not be obtained from surgery on any unlink. – Henry T. Horton Oct 22 '13 at 4:37