$\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\GL{GL}$Here is an proof-sketch of a strengthened Gordon-Luecke theorem. This is presumably known, but is it written down somewhere? I am also curious if there is a pre-geometrization proof of this statement, in roughly this generality.

Theorem: Let $L$ be an $n$-component link in the 3-sphere. Let $M$ be the exterior of the link, i.e. $S^3 \setminus \nu L$ where $\nu L$ is an open tubular neighbourhood of $L$. Then there is a natural representation given by restricting diffeomorphisms to the boundary

$$\pi_0 \Diff(M) \to \pi_0 \Diff(\partial M).$$

This representation has infinite image if and only if the link $L$ contains a split unknot, or a split Hopf link.

Notation: $\Diff(M)$ means to the full group of diffeomorphisms of the knot exterior, so $\pi_0 \Diff(M)$ means isotopy-classes, i.e. the mapping class group. Similarly $\pi_0 \Diff(\partial M)$ is the mapping class group of the boundary, i.e. a disjoint union of tori. So this group is the wreath product $\GL_2 \mathbb Z\wr\Sigma_n$, i.e. one can permute tori and act by automorphisms of tori.

Proof: The unknot exterior is $S^1 \times D^2$ and the Hopf link exterior is $S^1 \times S^1 \times I$, and both admit mapping-classes of infinite order — generalized Dehn twists.

So what remains to be shown is that if a link does not have one of these as split sublinks, then the representation is of finite order.

The idea is to prove it using geometrization.

For links whose exteriors are Seifert-fibered manifolds this is a direct computation using the Seifert data and their mapping class groups.

For hyperbolic links the mapping class group is finite, so the image is finite.

For satellite links, we can take the subgroup of finite index that preserves all the 3-manifolds in the JSJ-decomposition, and these are Seifert-fibered or hyperbolic link exteriors, so the result follows from the previous two steps. Lastly, unknot and Hopf link exteriors do not arise via torus splittings i.e. the JSJ decomposition.

Off the top of my head the only extensions to Gordon-Luecke that I know of concern the question of if two distinct links can have diffeomorphic complements. While this is related to my question, it only would appear to be "in the neighbourhood" and not really on exactly the same topic.

  • $\begingroup$ I suppose one could avoid geometrization by using Johansson's finiteness theorem, for 3-manifold mapping class groups. $\endgroup$ Commented Sep 15, 2021 at 0:18
  • $\begingroup$ Is it obvious how this implies the Gordon-Luecke theorem? I do not see this. Also, it seems to me that the same argument should work for links in general 3-manifolds (with the statement slightly changed in the obvious way). Would this then imply that any knot in a 3-manifold (except for the obvious counterexamples: rational unknots in lens spaces, whose exteriors are solid tori) is determined by its complement? $\endgroup$
    – Marc Kegel
    Commented Sep 22, 2021 at 8:40
  • $\begingroup$ The full Gordon-Luecke theorem concerns embeddings of knot exteriors in $S^3$, i.e. it states a non-trivial knot exterior has at most two smooth embeddings in $S^3$ modulo parametrization: the standard embedding and its mirror-image. i.e. you can't embed one knot exterior to be the exterior of a different knot. This question is about the parametrizations of the exteriors, and how they can change the peripheral system. Someone changed the title of my question, making the intent perhaps a little more confusing. So this is close to the Gordon-Luecke theorem in spirit, but not exactly it. $\endgroup$ Commented Sep 22, 2021 at 10:07
  • $\begingroup$ Ok, now I understood it. Thanks for the clarification. $\endgroup$
    – Marc Kegel
    Commented Sep 22, 2021 at 15:12

1 Answer 1


I think that the correct theorem should be the following:

If $L$ does not contain any split unknot or any two coaxial components, the image of the map is finite

Following Cameron Gordon, two coaxial components are two components $K_1$ and $K_2$ such that $K_1$ is an essential curve in the boundary $\partial N(K_2)$ of a solid torus regular neighbourhood of $K_2$. We also suppose that no other component of $L$ intersects $N(K_2)$.

Here is one example:

Two coaxial components

If you have two coaxial components, there is an incompressible annulus in $S^3\setminus L$ which is contained in $N(K_2)$, which winds once along $K_1$ and some $n$ times along $K_2$. The Hopf link corresponds to the case where $K_2$ is the unknot and $n=0$.

Every time you have an incompressible annulus that connects two different components of the link you can do a Dehn twist along it and act non-trivially on the boundary. It has infinite order.

I think that the proof should go as follows: the only infinite-order automorphisms in a 3-manifold with boundary that arise as a link complement should be twists along discs or annuli. So $S^3\setminus L$ contains an essential disc or annulus. In the second case, since we are in $S^3$, the annulus arises necessarily from two coaxial components.

  • $\begingroup$ Your argument seems isomorphic to my argument, but I missed a case in the statement of the theorem. Coaxial components are a type of cabelling in the satellite decomposition. I suppose I would call this a splice with a keychain link -- splicing on the keyring. I missed this case likely because I'm used to only splicing on the keys. Thanks. $\endgroup$ Commented Sep 16, 2021 at 17:35
  • $\begingroup$ I suppose this comes up in one other way in the JSJ decomposition -- splicing with a Seifert-fibered link where one component is a singular fiber and $n \geq 2$ components are regular fibers of a Seifert-fibering of $S^3$. $\endgroup$ Commented Sep 16, 2021 at 18:25

Your Answer

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

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