MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
btw, Lackenby has some notes which partially discuss the proof of Waldhausen's theorm: see "Three-dimensional manifolds (Graduate Course, Michaelmas 1999)" – Ian Agol Aug 26 '10 at 15:20
up vote 22 down vote accepted

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.

share|cite|improve this answer
@Samuelson: the "filling slope" construction I gave is basically the way of avoiding dealing with the Gordon-Luecke knot complement problem. In particular, you can't use Gordon-Luecke for link complements, but the filling-slope technique does generalize. – Ryan Budney Aug 15 '10 at 20:47
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? – Ryan Budney Aug 15 '10 at 20:53
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. – Ryan Budney Aug 15 '10 at 21:02
Thanks for the detailed answer! – Peter Samuelson Aug 17 '10 at 3:33
@Ryan: your summary of what's known about link complements is spot on. – Dave Futer Aug 20 '10 at 14:22

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.

share|cite|improve this answer
Oh, so different gluing maps (of $\partial S^1 \times D^2$ to $\partial M$ can give different 3-manifolds? That's good to know. Also, it would be nice if there were a way to indicate "both these answers are very useful." – Peter Samuelson Aug 17 '10 at 3:29
Oh, what did you mean by "knots are essentially classified"? Did you mean the Reidemeister moves? – Peter Samuelson Aug 17 '10 at 3:32
Regarding your 1st question, the answer is yes. Details: 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. – Ryan Budney Aug 17 '10 at 7:45
The algorithm I have in mind goes more like this: take your knot/link complement, triangulate it. Perform the connect-sum and JSJ-decomposition (Jaco, Rubinstein, Oertel, Burton, etc). Recognise the Seifert-fibered 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 Epstein-Penner canonical polyhedral decomposition but perhaps there are more efficient ways. – Ryan Budney Aug 17 '10 at 7:49

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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