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?
Complete Knot Invariant – Knot Theory and Geometric Topology
gt.geometric-topologyknot-theory
Related Solutions
This showed up in my snail-mail today, so I'm sharing the wealth:
I'm very curious where this came up. In any case, the answer to the first question is yes, it does distinguish these trefoils; you found the minimal representatives.
Let $a_0,\dots,a_{N-1}$ be the roots of unity that are visited along the knot, in (cyclic) order. Suppose we have a minimal representative for some non-trivial knot. Then we cannot have $|a_k - a_{k+1}| = 1$ for any $k$, as otherwise we could replace this pair $a_k, a_{k+1}$ by a single root of unity (for $N-1$), adjusting the other roots of unity as appropriate. A little more subtly, we cannot have $|a_{k-1} - a_{k+1}| = 1$ either, as then we could again delete $a_k$ from the sequence to get a smaller representation. With these simple constraints, the smallest possible sequence for a non-trivial knot is the one you found for one of the trefoils with $N=7$. There are several possibilities for $N=8$, including the one you found for the other trefoil. I've included a very short Haskell program below that computes this. The possibilities for $N=8$ are $$ (2,7,5,3,1,6,4,0)\quad (2,5,7,3,1,6,4,0)\quad (3,6,1,4,7,2,5,0)\quad (2,6,4,1,7,3,5,0) $$ $$ (3,1,6,4,2,7,5,0)\quad (2,4,6,1,3,7,5,0)\quad (3,5,1,7,4,2,6,0)\quad (4,2,7,5,1,3,6,0) $$ $$ (3,1,5,7,2,4,6,0)\quad (5,3,1,6,4,2,7,0)\quad (2,4,6,1,3,5,7,0) $$
For the second question, I have never heard of this representation before.
Here is the code, for anyone interested.
-- A (partial) circular stick representation is a list of integers,
-- the order of the roots of unity to visit in order
type CircStick = [Int]
-- The next element ak after a partial representation a1, ..., a{k-1}
-- must satisfy
-- (a) ak has not already been seen
-- (b) |ak - a{k-1}| > 1
-- (c) |ak - a{k-2}| > 1
-- There are a few more "easy" constraint, eg the first and last entries
-- cannot differ by one. We do not impose those constraint here.
nexts :: Int -> CircStick -> [Int]
nexts n [] = [0]
nexts n [a1] = filter (\a -> abs (a-a1) > 1) [0..n-1]
nexts n (a1:a2:as) =
filter (\a -> not (elem a as)) $
filter (\a -> abs (a-a1) > 1) $
filter (\a -> abs (a-a2) > 1) $
[1..n-1]
completions :: Int -> CircStick -> [CircStick]
completions n as | length as >= n = [as]
completions n as =
concat [completions n (a:as) | a <- nexts n as]
-- Impose final constraints:
-- (a) Last entry cannot be 1
-- (b) Take entry that is lexicographically less than its reverse
-- (c) first and next-to-last entries cannot differ by one
circSticks :: Int -> [CircStick]
circSticks n =
filter (\as -> abs ((as!!0) - (as!!(n-2))) > 1) $
filter (\as -> as < tail (reverse as)) $
filter (\as -> head as /= 1) $
(completions n [])
Edit: For those interested, here are the 108 possibilities for $N=9$. I hope there's some way of checking what these are more efficiently than just going through them by hand.
[[2,7,5,3,8,1,6,4,0],[2,7,5,3,1,8,6,4,0],[2,5,7,3,1,8,6,4,0],[2,7,5,1,3,8,6,4,0],[2,5,7,1,3,8,6,4,0],[2,6,8,3,5,1,7,4,0],[2,7,5,3,1,6,8,4,0],[2,5,7,3,1,6,8,4,0],[2,7,5,1,3,6,8,4,0],[2,5,7,1,3,6,8,4,0],[3,8,6,1,4,7,2,5,0],[3,6,8,1,4,7,2,5,0],[3,7,1,4,6,8,2,5,0],[2,8,6,4,1,7,3,5,0],[2,6,8,4,1,7,3,5,0],[2,7,4,1,6,8,3,5,0],[2,4,8,6,3,1,7,5,0],[2,4,6,8,3,1,7,5,0],[3,8,1,6,4,2,7,5,0],[3,1,8,6,4,2,7,5,0],[3,6,1,8,4,2,7,5,0],[3,1,6,8,4,2,7,5,0],[2,8,4,6,1,3,7,5,0],[2,4,8,6,1,3,7,5,0],[2,6,4,8,1,3,7,5,0],[2,4,6,8,1,3,7,5,0],[3,7,1,4,6,2,8,5,0],[2,7,4,1,6,3,8,5,0],[3,8,5,1,7,4,2,6,0],[3,7,5,1,8,4,2,6,0],[3,5,7,1,4,8,2,6,0],[4,7,1,3,5,8,2,6,0],[4,1,7,3,5,8,2,6,0],[4,8,2,5,7,1,3,6,0],[4,7,2,5,8,1,3,6,0],[4,2,7,5,1,8,3,6,0],[2,4,7,1,5,8,3,6,0],[2,8,5,3,7,1,4,6,0],[2,7,5,3,8,1,4,6,0],[3,8,1,5,7,2,4,6,0],[3,7,1,5,8,2,4,6,0],[2,7,5,3,1,8,4,6,0],[2,5,7,3,1,8,4,6,0],[3,1,7,5,2,8,4,6,0],[4,2,7,5,3,1,8,6,0],[3,5,1,7,4,2,8,6,0],[4,2,7,5,1,3,8,6,0],[2,4,7,1,5,3,8,6,0],[3,1,5,7,2,4,8,6,0],[5,8,3,1,6,4,2,7,0],[5,3,8,1,6,4,2,7,0],[3,5,8,1,6,4,2,7,0],[5,3,1,8,6,4,2,7,0],[3,5,1,8,6,4,2,7,0],[5,1,3,8,6,4,2,7,0],[5,3,1,6,8,4,2,7,0],[5,1,3,6,8,4,2,7,0],[4,6,1,3,8,5,2,7,0],[4,1,6,3,8,5,2,7,0],[4,6,2,8,5,1,3,7,0],[5,8,2,4,6,1,3,7,0],[5,2,8,4,6,1,3,7,0],[2,6,4,8,1,5,3,7,0],[4,2,6,8,1,5,3,7,0],[2,4,6,8,1,5,3,7,0],[2,6,4,1,8,5,3,7,0],[2,4,6,1,8,5,3,7,0],[2,5,8,3,6,1,4,7,0],[5,3,1,8,6,2,4,7,0],[3,5,1,8,6,2,4,7,0],[5,1,3,8,6,2,4,7,0],[5,3,1,6,8,2,4,7,0],[5,1,3,6,8,2,4,7,0],[4,2,8,6,3,1,5,7,0],[2,4,8,6,3,1,5,7,0],[4,2,6,8,3,1,5,7,0],[2,4,6,8,3,1,5,7,0],[3,6,1,4,8,2,5,7,0],[3,1,6,4,8,2,5,7,0],[2,8,4,6,1,3,5,7,0],[4,2,8,6,1,3,5,7,0],[2,4,8,6,1,3,5,7,0],[2,6,4,8,1,3,5,7,0],[4,2,6,8,1,3,5,7,0],[2,4,6,8,1,3,5,7,0],[2,6,4,1,8,3,5,7,0],[2,4,6,1,8,3,5,7,0],[5,7,3,1,6,4,2,8,0],[4,6,1,3,7,5,2,8,0],[5,3,7,1,4,6,2,8,0],[3,5,7,1,4,6,2,8,0],[6,4,2,7,5,1,3,8,0],[5,7,2,4,6,1,3,8,0],[6,2,4,7,1,5,3,8,0],[2,6,4,1,7,5,3,8,0],[5,2,7,4,1,6,3,8,0],[6,3,1,5,7,2,4,8,0],[6,1,3,5,7,2,4,8,0],[2,7,5,3,1,6,4,8,0],[2,5,7,3,1,6,4,8,0],[3,6,1,4,7,2,5,8,0],[6,2,4,7,1,3,5,8,0],[2,6,4,1,7,3,5,8,0],[4,2,7,5,3,1,6,8,0],[5,3,1,7,4,2,6,8,0],[3,5,1,7,4,2,6,8,0],[4,2,7,5,1,3,6,8,0],[3,1,5,7,2,4,6,8,0]]
Best Answer
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.