# The cyclic branched covers of "simple" knots in $S^3$

Is there a convenient place in the literature where the geometric decompositions of cyclic branched covers of $S^3$ branched over "small" knots is recorded?

By small knots, I'm referring to things like torus knots, 2-bridge knots and the Rolfsen knot table.

For 2-sheeted coverings the Bonahon-Siebenman paper is a canonical reference, and for 2-bridge knots there's the standard connection with lens spaces. But I'd like to know about $n$-sheeted coverings for larger $n$ (3,4,5,6 is a little larger) and with standard small knots.

I imagine there are a few papers out there where much of this is worked out, perhaps a Montesinos paper? But I'm not having much luck finding this kind of information.

In principle I suppose I could generate tables using SnapPy and Regina but I imagine there must be work of this kind that predates such technology.

Thanks for any insights.

Such a classification follows from the orbifold theorem. The point is that the $n$-fold cyclic branched cover over a knot is an $n$-fold orbifold cover over the orbifold whose singular locus is the knot with a cone angle of $2\pi/n$ (in fact, one only needs the theorem for orbifolds of "cyclic type"). The orbifold theorem states that a good orbifold admits a geometric decomposition. An orbifold whose underlying singular locus is a knot in $S^3$ is automatically good, so satisfies the hypothesis of the orbifold theorem (there can be no teardrop or bad football suborbifolds for straightforward reasons).

There is a version of the prime decomposition for good 3-orbifolds. In the case at hand, the only spherical orbifold that may appear is a football, which corresponds to a connect sum. So the connect sum decomposition of the knot will correspond to the prime decomposition of the cyclic orbifolds.

Thus, assume the knot is prime. Now, the only toroidal orbifolds with the same order singular points are the $2222$ pillowcase and the $333$ turnover. Clearly the turnover cannot occur for a knot orbifold, and thus one only sees the $2222$ pillowcases for the cyclic 2-orbifold. This toroidal decomposition then corresponds to the decomposition along Conway spheres discussed by Bonahon-Siebenmann. Thus, one may reduce to considering tangles with no Conway spheres. The Seifert-fibered pieces will correspond to Montesinos tangles, where one inserts rational tangles in a standard picture.

There are a few exceptional cases, corresponding to the other geometric structures, which were classified by Dunbar. One can browse his tables and discover that the only other case is a Euclidean orbifold which is the 3-fold cover of the figure 8 knot complement.

• Hi Ian, my question is about if someone has actually gone through and computed these geometric decompositions for the knot tables. Should I infer the answer is no? Dec 13, 2013 at 10:55
• @RyanBudney: well, the point is that (except for the figure 8), for $n$-fold covers where $n\geq 3$, the geometric decomposition will correspond to the geometric decomposition of the knot complement. So I think your question is essentially equivalent to whether people have worked out the satellite structure of small knots. Dec 13, 2013 at 17:35
• My question boils down to wanting to be able to quickly identify which 3-manifolds I have are some cyclic branched cover of a pair $(S^3,K)$ for (ideally) some familiar knots -- moreover, I want to know precisely which knot it is, and at least the order of the covering. By "quick" I mean done algorithmically, not by humans, as there are potentially thousands of such computations to be done. The 3-manifolds I have are all smallish (in the sense of triangulations) so my question is more looking for a correspondingly small table so that answers could be looked-up most of the time. Dec 13, 2013 at 18:31
• Ok, so given a manifold, you want to check if there is a cyclic group action, with connected fixed point set, and quotient $S^3$? First, you should try to find the prime and geometric decomposition, which can be done using Regina (I'm not sure about automation, but you're probably more familiar with this than me). Then you want to look at the symmetries of the geometric pieces which give rise to symmetries of the entire manifold. I think in principle this can be done using Snappea for the hyperbolic pieces, although I'm not sure the symmetry algorithm is rigorous. Dec 13, 2013 at 18:43
• Okay, so I think it's clear, the answer to the most hopeful version of my question is probably no. No such pre-existing table exists in the literature, at least not in the level of detail I'd like. Right, the symmetry group computation in SnapPea is susceptible to errors in computing an actual canonical simplicial refinement of the Epstein-Penner decomposition and apparently SnapPea sometimes fails to do that. I'll leave this question open for a few more days in case somebody knows of some magic tables out there... somewhere. Dec 13, 2013 at 18:52