Gordon-Luecke Theorem Revisited

3-manifoldsgt.geometric-topologyknot-linkknot-theory

$\DeclareMathOperator\Diff{Diff}\DeclareMathOperator\GL{GL}$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.

Best Answer

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.

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