Complete knot invariant?

Question

I've seen a couple papers (that I now can't find) that say that in his paper "On irreducible 3-manifolds 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?

Answer 1

As Ryan says, this follows from Waldhausen's paper, when appropriately interpreted. Sufficiently large 3-manifolds 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 non-empty boundary and not a ball (ie no 2-sphere 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 3-manifolds $M_1, M_2$ 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 3-manifold $M$, 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 Bleiler-Hodgson-Weeks.

Answer 2

You're looking at the right paper. His results apply to a broad class of 3-manifolds, 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 3-manifold 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.

Answer 3

This topic is treated in G. Burde & H. Zieschang's book, Knots, 2nd edition, p. 40.