Intersection of Satellite, Torus, and Hyperbolic Knots

definitionknot-theorylow-dimensional-topology

Thurston famously proved every knot that is neither a satellite nor a torus knot is hyperbolic.

I was thinking about the definition of satellite knots, in particular which cases we should eliminate to make Thurston’s result make sense. Otherwise it seems to me that all knots are satellite.

Here’s a definition: take a pattern knot $K$, which is a knot embedded in an unknotted solid torus, and a companion knot, which is another knot $K’$. Then we map the solid torus where $K$ lives into the tubular neighborhood of $K’$, in an untwisted way, i.e., meridian goes to meridian, and preferred longitude goes to preferred longitude (the one with framing $0$). The image of $K$ is $K’’$, a satellite knot.

Approach:

First, we should eliminate the knots $K$ that fit inside a 3-ball in the torus. Rolfsen call this a geometrically essential knot, and Peter Cromwell a pattern with wrapping number $0$.

The reason is that if this were the case then every knot is satellite, because $K’’=K$ and every knot fit inside a 3-ball anyway.

Second, we should eliminate the knots $K$ that coincide with a longitude, because then $K’’ = K’$ and again every knot is satellite. In Cromwell language these are patterns $K$ which are torus knots with wrapping number $1$.

Third, we should eliminate the option of $K’$ being the unknot, else $K’’ = K$ again, and since every knot is a connected sum with the unknot, we can fit any knot inside a solid torus, and this would mean every knot is satellite.

Question: Are there more cases we should eliminate? After eliminating these cases, how do we see no knot can be both a torus knot and a satellite?

Best Answer

From the perspective of geometrization, satellite knots are knots with a nontrivial JSJ decomposition. This means that there exists an incompressible torus $T$ in the exterior $X$ of the knot $K$ that isn't boundary parallel (i.e., it's not isotopic to the boundary of the knot exterior).

Since $T$ is a torus, a consequence of Dehn's lemma is that $S^3-T$ consists of two components: a solid torus and a knot complement. If $K$ were contained in the knot complement component, then that would mean $T$ bounds a solid torus disjoint from $K$. This contradicts the fact $T$ is incompressible since any meridian disk of the solid torus would serve as a compression disk for $T$. Hence, $K$ is in a solid torus component of $S^3-T$.

If $K$ is contained in a 3-ball in the solid torus component, then one can find a compression disk for $T$ on that side, hence the "geometrically essential knot" condition. Conversely, if there's a compression disk on the solid torus side, one can take the complement of it in the solid torus to get a 3-ball that $K$ is contained within.

If $K$ is a longitude of the solid torus component, then $T$ is boundary parallel, and conversely boundary parallel implies $K$ is a longitude for the solid torus.

If the longitude of the solid torus is an unknot, then there exists a compression disk for $T$ outside the solid torus component (from the disk that the unknot bounds). Conversely, if there's a disk out there then the longitude of the solid torus is an unknot.

Since compression disks are either on one side of $T$ or the other due to orientability of $T$, the three conditions you mention are equivalent to the conditions that $T$ be an incompressible torus that isn't boundary parallel. So, the three conditions you mention are all you need for the definition of a satellite knot for Thurston's classification.

To see torus knots aren't satellite knots, the idea is that the exterior of a torus knot can be given the structure of a Seifert-fibered space. Incompressible non-boundary-parallel tori have to be isotopic to either horizontal surfaces (transverse to each fibers it intersects) or vertical surfaces (unions of fibers). Since $X$ has nonempty boundary, the torus must be a vertical surface, but every vertical torus for a torus knot must either be boundary parallel or bound a solid torus in $X$ due to the fact that the base orbifold is a disk with only two cone points.

Related Question