# Given a link $L\subset S^3$ how to construct a link $L'$ whose complement have hyperbolic structure?

Thurston claimed that almost all closed 3-manifolds are hyperbolic. To support this, he said that every closed 3-manifold is obtained by Dehn surgery along some link whose complement is hyperbolic. And he proved that most surgeries along this link have hyperbolic structure.

The Lickorish–Wallace theorem tells us that every closed orientable 3 manifold can be described via integral Dehn surgery along a link in $S^3$. So my Question is, given a link $L\subset S^3$, how to construct (modify) a link $L'$ whose complement is hyperbolic? Because we know that $\infty$-sugery along any knot does not change the manifold. Thus we can hope to prove Thurston's claim.

We know that if a knot $K\subset S^3$ is not satellite or torus, then it is hyperbolic. So let's say, if we start with a sattelite knot, then we can add an unknot which intersect the indecomposable torus non-trivially. Then in this process we can kill off the indecomposable torus. But I am unable to prove that in this process we have not constructed any new indecomposable torus. Also I don't know whether this complememnt is Seifert or hyperbolic.

EDIT: (Question 2) Can we construct $L'$ from $L$ by adding only trivial knots?

• This would be a better question if you said how L' and L are intended to be related. – Danny Ruberman Dec 5 '17 at 18:11
• @DannyRuberman I have edited my question. Please have a look. Thanks – Anubhav Mukherjee Dec 5 '17 at 18:46

Prime augmented alternating links are hyperbolic. These are obtained from an alternating link by adding little loops about each twist region (so in the figure, each crossing).

Any link with a prime projection can be obtained by Dehn filling on an augmented alternating link ($\pm 1$ surgery along crossing loops changes the crossing). So this gives an elementary way of embedding $L\subset L'$ where $L$ is a link and $L'-L$ is an unlink, and $L'$ has hyperbolic complement.

Robert Myers, in his paper "Simple knots in compact, orientable 3-manifolds", proved that every compact, orientable three-manifold (without sphere boundary components) contains a hyperbolic knot.

It follows that every link $L$ in the three-sphere is a sublink of a hyperbolic link $L'$. This answers your first question. It is also possible to arrange that the components of $L' - L$ are unknots in $S^3$, but I do not have a reference at hand.

Not quite by adding trivial knots, but in a lot of situations it is possible to homotope a link to get a new link with hyperbolic complement. See "Excellent 1-manifolds in compact 3-manifolds" (Myers). Some of these results work for when there are additional boundary components too and allow some control over the extra boundary - see "Totally geodesic boundaries of knot complements" (Kent).