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?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.