I've seen a couple papers (that I now can't find) that say that in his paper "On irreducible 3manifolds which are sufficiently large" Waldhausen proved that the data $\pi_1(\partial (S^3\setminus K)) \to \pi_1(S^3\setminus K)$ is a complete knot invariant. However, the word "knot" doesn't appear in this paper (although the phrase "to avoid an orgy of notation" does :). Is the claimed result a straightforward corollary of his main results? Or am I looking at the wrong paper?
3 Answers
As Ryan says, this follows from Waldhausen's paper, when appropriately interpreted. Sufficiently large 3manifolds are usually called "Haken" in the literature, and as Ryan says, they are irreducible and contain an incompressible surface (which means that the surface is incompressible and boundary incompressible). An irreducible manifold with nonempty boundary and not a ball (ie no 2sphere boundary components) is always sufficiently large, by a homology and surgery argument. By Alexander's Lemma, knot complements are irreducible, and therefore sufficiently large (the sphere theorem implies that they are aspherical).
Waldhausen's theorem implies that if one has two sufficiently large 3manifolds $M_1, M_2$ with connected boundary components, and an isomorphism $\pi_1(M_1) \to \pi_1(M_2)$ inducing an isomorphism $\pi_1(\partial M_1) \to \pi_1(\partial M_2)$, then $M_1$ is homeomorphic to $M_2$. This is proven by first showing that there is a homotopy equivalence $M_1\simeq M_2$ which restricts to a homotopy equivalence $\partial M_1\simeq \partial M_2$. Then Waldhausen shows that this relative homotopy equivalence is homotopic to a homeomorphism by induction on a hierarchy. The peripheral data is necessary if the manifold has essential annuli, for example the square and granny knots have homotopy equivalent complements.
If $K_1, K_2\subset S^3$ are (tame) knots, and $M_1=S^3\mathcal{N}(K_1), M_2=S^3\mathcal{N}(K_2)$ are two knot complements, then Waldhausen's theorem applies. However, one must also cite the knot complement problem solved by Gordon and Luecke, in order to conclude that $K_1$ and $K_2$ are isotopic knots. Otherwise, one must also hypothesize that the isomorphism $\partial M_1 \to \partial M_2$ takes the meridian to the meridian (the longitudes are determined homologically). This extra data is necessary to solve the isotopy problem for knots in a general 3manifold $M$, to guarantee that the homeomorphism $(M_1,\partial M_1)\to (M_2,\partial M_2)$ extends to a homeomorphism $(M,K_1)\to (M,K_2)$, since for example there are knots in lens spaces which have homeomorphic complements by a result of BleilerHodgsonWeeks.

$\begingroup$ @Samuelson: the "filling slope" construction I gave is basically the way of avoiding dealing with the GordonLuecke knot complement problem. In particular, you can't use GordonLuecke for link complements, but the fillingslope technique does generalize. $\endgroup$ Aug 15, 2010 at 20:47

1$\begingroup$ I think it's still an open problem as which links have the same link complement. I believe Gordon has some results on this but as far as I know these results are not known to be complete? $\endgroup$ Aug 15, 2010 at 20:53

$\begingroup$ Since knots are essentially classified perhaps I should have phrased that as, it's an open problem to find an efficient procedure to go from one link and construct all the links whose complements are homeo/diffeomorphic to your original link complement. $\endgroup$ Aug 15, 2010 at 21:02

1$\begingroup$ @Minkowski: if the knots admit an orientation, then the orientation of $S^3$ induces an orientation of the meridians. If the map on peripheral elements preserves the orientation of meridian and longitude, then the map of knots will also be orientation preserving. And orientation preserving homeomorphisms of $S^3$ are connected, so the knots are isotopic. If the knots are unoriented, just try both orientations to compare them. So eg one can also tell if a knot is invertible or amphichiral. $\endgroup$– Ian AgolMar 28 at 2:31

1$\begingroup$ Yes, I think that’s right. In principle one could incorporate the orientation into the group theory, by taking an orientation of the relative fundamental class of the fundamental group (since the knot complement is a $K(\pi,1)$). This orientation of the peripheral subgroup is moreorless equivalent since it orients the peripheral torus, and hence the 3manifold by the long exact sequence in homology. $\endgroup$– Ian AgolMar 28 at 20:12
You're looking at the right paper. His results apply to a broad class of 3manifolds, which knot complements happen to be a part of. I don't have the paper here with me but I believe the class was then called "sufficiently large". Which I believe in this case means irreducible and containing an incompressible surface.
edit: Technically what he's describing is a "complete 3manifold invariant". To turn it into a complete knot invariant you need the following observation. Given a knot complement you can turn it into a knot (in some $3$manifold) by filling in the boundary $S^1 \times S^1$ with a $S^1 \times D^2$. To do that you need a gluing map, which amounts to specifying the slope of the $D^2$factor in $S^1 \times S^1$. Given a knot, the invariant of the knot is the knot complement together with the filling slope that recovers $S^3$. The knot complement together with this natural filling slope is the complete invariant of the knot (up to mirror inverse). Waldhausen's paper shows you how if you reduce that information to $\pi_1 \partial M \to \pi_1 M$ together with a the filling slope (thought of as an element of $\pi_1 \partial M$), that is also a complete invariant of the knot.

$\begingroup$ Oh, so different gluing maps (of $\partial S^1 \times D^2$ to $\partial M$ can give different 3manifolds? That's good to know. Also, it would be nice if there were a way to indicate "both these answers are very useful." $\endgroup$ Aug 17, 2010 at 3:29

$\begingroup$ Oh, what did you mean by "knots are essentially classified"? Did you mean the Reidemeister moves? $\endgroup$ Aug 17, 2010 at 3:32

1$\begingroup$ Regarding your 1st question, the answer is yes. Details: en.wikipedia.org/wiki/Dehn_surgery Regarding your 2nd question, no. As far as I know, if two knots are not isotopic, you don't know how many Reidemeister moves you have to make to come to that conclusion. I mean an efficient algorithm  one you could consider implementing. $\endgroup$ Aug 17, 2010 at 7:45

1$\begingroup$ The algorithm I have in mind goes more like this: take your knot/link complement, triangulate it. Perform the connectsum and JSJdecomposition (Jaco, Rubinstein, Oertel, Burton, etc). Recognise the Seifertfibered parts (same credits). Geometrize the hyperbolic parts (this is the cusped version of the Manning algorithm, I believe due to Tillman and perhaps others), then you have to compare the hyperbolic manifolds. Ideally you'd do this by an EpsteinPenner canonical polyhedral decomposition but perhaps there are more efficient ways. $\endgroup$ Aug 17, 2010 at 7:49
This topic is treated in G. Burde & H. Zieschang's book, Knots, 2nd edition, p. 40.