Surgery Along a Link

Question

It is well known that you can construct 3-Manifolds via surgery on a link in $S^3$. Is there an analogous method to construct 2-manifolds?

Answer 1

Surgery on 2-manifolds is handlebody decomposition.

In general, surgery on an $m$-fold $M$ produces a new $m$-fold $M'$ in the following way: $$ M' = \mbox{closure}\bigg[\big(M - (S^n\times D^{m-n})\big)\cup_{S^n\times S^{m-n-1}} \big( D^{n+1}\times S^{m-n-1}\big)\bigg] $$ where the replacement is glued in by an automorphism of $S^n\times S^{m-n-1}$. See this intro book by Ranicki; here I have repeated Definition 1.2.

So the case of surgery on $S^3$ with $n=1$ involves removing the tubular neighborhood of an embedded knot $S^1\times D^2$ and replacing it with a copy of $D^2\times S^1$, glued by an automorphism of $S^1\times S^1$. If two gluings are not isotopic, then the resulting manifolds need not be homoemorphic. So surgery on the same knot can produce many different homeomorphism types, depending on choice of gluing map. (For example, the lens spaces $L(p,q)$ can all be produced by surgery on the unknot.)

Surgery on a link in $S^3$ is a sequence of surgeries on embedded knots. As you note, it is a theorem of Lickorish and Wallace that every 3-manifold is homeomorphic to a manifold produced by surgery on a link in $S^3$.

The case of surgery in $S^2$, with $n=0$, boils down to handlebody decompositions of surfaces. Remove a copy of $S^0\times D^2$. Then glue in a copy of $D^1\times S^1$ along its boundary. That is, add a handle.

Again, isotopic gluings result in homeomorphic manifolds. But the automorphism group of $S^0\times S^1$ is much simpler... Exercise: Show that the only two homeomorphism types produced by $n=0$ surgery on a two-sphere are the torus and the klein bottle.