TITLE: Classification of knots in solid torus
QUESTION [7 upvotes]: What is known about the classification of knots in a solid torus $S^1 \times D^2$?  Is enumerating them a reasonable problem? Do we get a similar classification as for knots in $S^3$? Ideally there would be a simple description of the Seifert fibered knots in $S^1 \times D^2$ like for prime non-satellite knots in $S^3$ (they're exactly the torus knots).
The motivation is to better understand the classification of prime knots in $S^3$ as

torus knots,
hyperbolic knots, or
nontrivial satellites,

because satellites come from combining a knot in $S^3$ and a knot in $S^1 \times D^2$.
One issue is that we would want to exclude knots that are only knotted in a ball in $S^1 \times D^2$ because these aren't really new: they come from a knot in $S^3$, and they give composite knots in the satellite construction. Is there a relatively simple way to exclude these?
More formally: one class of knots to exclude are those obtained by taking a $(1,1)$-tangle in a $3$-ball, then closing up the ends so that the knot is not null-homologus in $S^1 \times D^2$. These are nontrivial knots in $S^1 \times D^2$ but they don't really use the topology of the solid torus in an interesting way, so we want to exclude them from our classification.

REPLY [9 votes]: Up to Dehn twists, the class of knots in the solid torus is identical to the class of two-component links in the three-sphere, where the first component is an unknot.
For example, Seifert fibered knots in the solid torus give Seifert fibered links in the three-sphere.  The base space is the orbifold $S^2(p,\infty,\infty)$.
There is a similar theory of “satellite” links.  The knots you wish to exclude (knotted in a three-ball inside the solid torus) are a special case of these.  When thought of as links, they are exactly the split links in the three-sphere (where again the first component is the unknot).
Finally, the knot complement in the solid torus is hyperbolic  exactly when the link complement in the three-sphere is.