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

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?

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.