Is there a generalized Property P - what can we say about framed link descriptions of $S^3$?

Question

A knot $K$ is said to have Property P if every nontrivial Dehn surgery on $K$ yields a 3-manifold that is not simply connected. It is known that every knot except the unknot has Property P. I am wondering what can be said about a link that admits a nontrivial Dehn surgery that yields $S^3$.

An $n$-component link $L$ is said to have Property R if some Dehn surgery on $L$ yields $\sharp^n S^1 \times S^2$. The generalized Property R conjecture says that any such link, together with the framing used to obtain $\sharp^n S^1 \times S^2$ must be handleslide equivalent to an unlink with each component having framing 0. This conjecture is true for $n=1$ but unknown even for $n=2$ - see here. Note that by Kirby's Theorem, any two framed links that describe $\sharp^n S^1 \times S^2$ must differ by handleslides together with blowups and blowdowns - generalized Property R is asserting that the latter moves are not necessary in the case of $\sharp^n S^1 \times S^2$ for any pair of $n$-component descriptions.

Is there any sort of generalized Property P conjecture? There are certainly lots of links that that have a surgery that yields $S^3$ - for example any handlebody diagram for a 4-manifold without 1- or 3-handles. In fact, there is a conjecture that any simply connected smooth closed 4-manifold admits such a handlebody description (note: this implies S4PC) - if this is true then any such 4-manifold would yield such a framed link.

By considering the Hopf link with either $(0,0)$-framing or $(0,1)$-framing, we obtain two descriptions of $S^3$ that certainly are not handleslide equivalent - so Property P does not generalize naively like Property R. Maybe there is a bound $f(n)$, such that any two $n$-component framed link descriptions of $S^3$ require at most $f(n)$ blowups and blowdowns, together with handleslides in order to get between them?

Answer 1

I know of no analogue of property P.

In fact, there are many links in $S^3$ which admit infinitely many fillings which are also $S^3$. The simplest is probably the Whitehead link: each component is unknotted, and one can "twist" along an unknotting disk bounding one component to obtain infinitely many non isotopic links with the same complement.

The only thing that I know of is a kind of converse to this by Cameron Gordon. If a link has infinitely many Dehn fillings yielding $S^3$, then there is a collection of disjointly embedded disks and annuli in $S^3$ bounding a sublink so that all but finitely many of the fillings are obtained by $1/n$ filling along the boundaries of the disk complements, and pairs of fillings along the boundaries of the annulus components. This is not exactly stated in his paper, but I think that it follows from his proof. Applying to knots, one sees that a knot with infinitely many $S^3$ fillings is the unknot. But this is quite weak compared to property P.