# Knot complements with respect to Thurston's 8 geometries

We can put knot complements in three buckets: hyperbolic knots, satellite knots, and torus knots. Can this classification be made with looking at the (complete) metric on the complement, if it has curvature -1, 1 or 0?

If this is the case: what metrics (of the 8 Thurston's metics by the geometrization conjecture) can appear? hyperbolic, spherical and Euclidean?

If this is not the case: is it known which of the 8 metrics can appear as knot complements if not all?

• Does the same hold for links? – Jake B. Nov 21 '16 at 20:11
• See here for a nice survey on the subject by Ryan Budney. – Mike Miller Nov 21 '16 at 20:16
• As Mike mentions, yes this has all been worked out. Many low-dimensional topologists knew the answer for some time. I codified the answer into the survey paper Mike links to. Since knot complements are non-compact the geometry is not always unique -- in the Seifert-fibred case there are two essentially-equivalent geometries. Also, the unknot complement becomes trivial after the compression-body decomposition so it has the "empty geometry" and is exceptional among knot exteriors. – Ryan Budney Nov 21 '16 at 21:10

Geometrization of 3-manifolds has three levels.

Level 1: the compression-body decomposition. This is non-trivial only for the unknot complement, which becomes empty after this decomposition, as the unknot complement is a solid-torus.

Level 2: the connect-sum decomposition. All knot exteriors are prime so this is trivial.

Level 3: the torus decomposition. Satellite knots have this. Typically this is split into the most elementary kind "composite knots" and the more complicated satellite operations, such as Whitehead doubling (hyperbolic types) and cabling (Seifert satellite operations. The hyperbolic satellite operations have an infinite variety.

Seifert-fibred knot exteriors admit two geometries. Non-compact manifolds have a little bit of ambiguity when it comes to putting geometric structures on them. Bonahon's survey of geometric structures on 3-manifolds is a good resource for this. In short, you get both a $\mathbb R \times H^2$ and $PSL_2 \mathbb R$ structure. The $\mathbb R \times H^2$ structure is made explicit in my paper via the Milnor Fibration. Moreover, these geometric structures are not rigid in general.

This happens in satellite operations as well. The manifold that comes up via the connect-sum operation is a (trivial) punctured disc bundle over the circle. So there is a full moduli space of such $\mathbb R \times H^2$ structures.

But in short there is (1) the empty set after the compression body decomposition of the unknot exterior (2) hyperbolic knots, but there are also certain hyperbolic links that appear after decomposing satellite knots -- the exact class is all hyperbolic links of $n+1$ components that possess an $n$-component trivial sub-link and (3) the Seifert-fibred knot and link exteriors. These admit related $PSL_2 \mathbb R$ and $\mathbb R \times H^2$ structures and appear in families. The Seifert-fibred knots are the torus knots. The Seifert-fibred links that appear are the "keychain link" that generates connect-sum, and also the links I call "Seifert links" that generate the cabling operation.

• Thanks! I will try to debug your answer and read the reference. Is my statement about curvatures true? Also: if we say M has -1 (or 0 or 1) constant curvature, do we know to which geometry it belongs? – Jake B. Nov 22 '16 at 18:39
• No. The geometry is all non-positive curvature. $\mathbb R \times H^2$ is non-isotropic, there is a preferred direction of zero curvature and in the complementary directions you have negative curvature. Most Seifert-fibred spaces have this "stratified" curvature tensor, with few exceptions. – Ryan Budney Nov 22 '16 at 18:43
• Thank you, so if i get your answer right, from the list en.wikipedia.org/wiki/Geometrization_conjecture the only possible geometries are: $H^3$, $\mathbb R \times H^2$ or $PSL_2 \mathbb R$, or a combination of those when we split along tori? But we could not get lets say $\mathbb R \times S^2$? – Jake B. Nov 22 '16 at 22:01
• Correct. There are only the three geometries. It's unclear to me if there's any aesthetic difference in choosing the $\mathbb R \times H^2$ or the $SL_2 \mathbb R$ geometry on the Seifert-fibred manifolds. Depending on how the knots arise one might have a preference for using one geometry over the other, that would be an interesting thing to think about. – Ryan Budney Nov 22 '16 at 23:07