# Small examples of exceptional hyperbolic Dehn Filling of hyperbolic manifolds

For experimental purposes, I would like to have a small (i.e. triangulated with few tetrahedra) example of a manifold $$M$$ with the following properties:

1. $$M$$ is a hyperbolic manifold with finite volume (and possibly with only one cusp);

2. $$M$$ has a so called "exceptional hyperbolic Dehn filling" $$N$$, that is to say, a Dehn filling $$N$$ which is hyperbolic, but where the geodesic representative of the core $$c$$ of the filling torus, is not isotopic to $$c$$.

See also this question for a discussion of exceptional hyperbolic Dehn fillings.

## 3 Answers

Here are two examples of exceptional hyperbolic fillings. In the first the core curve becomes parabolic after filling (so cannot be isotopic to a geodesic). In the second, the core curve is homotopic (but not isotopic) to a geodesic.

Let $$L$$ be the two-component link L11n127. The complement $$S^3 - L$$ is hyperbolic and can be triangulated with 11 tetrahedra. Consulting the diagram, we see that $$L$$ has one component $$K$$ which is a copy of the figure eight. The other component $$U$$ is a copy of the unknot. So the meridional filling of $$U$$ yields $$S^3 - K$$, which is hyperbolic. Let $$\mu$$ be the core curve of the filling solid torus. Then $$\mu$$ is homotopic to a meridian of $$K$$; thus $$\mu$$ is not isotopic to a geodesic.

The two-component link L9a8 is similar, but a bit more complicated. Here the link complement requires 13 tetrahedra. But now the core curve of the filling is homotopic (but not isotopic) to an "augmenting loop". It requires a bit of work to show that the augmenting loop is a geodesic; in fact it is a systole of the figure-eight knot complement.

It would be interesting to know the first exceptional hyperbolic filling appearing in the various censuses coming with SnapPy.

Another sort of example is a geodesic that is not embedded. In the Whitehead link complement, there is a thrice-punctured sphere (pants) which is totally geodesic. A figure 8 curve in such a surface has a geodesic realization which is not embedded. Thus one can take a curve which is homotopic to this immersed geodesic. The example below has a triangulation with 12 tetrahedra according to SnapPy, where the green curve upon meridian filling is homotopic to the immersed geodesic. There is a thrice-punctured sphere in the figure 8 complement which is immersed, so one could also find such an example there.

• Ok, I played around a bit with Matthias's new version of "inside view" in snappy. It seems that there are two shortest immersed geodesics in the figure eight-knot complement. They have complex length $2.63391579384963 + 3.14159265358979i$ and, in unsimplified generators, they are represented by the words $abaB$ and $bbCC$. Instead of living in the immersed geodesic surface, they run from edge midpoint to edge midpoint inside the tetrahedra. So staring at the tetrahedra, as mapped into the diagram, might let us draw the desired diagram. Jul 1 at 8:04
• Or: we could look at the two-component links, having the figure-eight as one component, and hope to spot this example among those... A symmetry argument could prove that the core of the other component self intersects? Jul 1 at 8:06
• More messing about with SnapPy suggests that L8a2 (katlas.org/wiki/L8a2) works. This makes sense, as it is a fairly nice filling of your example. I'll draw a picture and add this as another answer. Jul 1 at 8:34

Here is a variant of Ian's answer. We cut along the round component in his diagram, perform a full twist (with the correct handedness), and reglue. This gives the following three-component link.

If you are seeing this in black and white, then below we call the figure-eight component "black", the round unknot "purple", and the infinity sign "green". The purple loop is a geodesic (thought of as living in the figure eight knot complement) so still bounds an embedded pair of pants $$P$$ with one material boundary (on purple) and two ideal boundaries (on black). The green is a geodesic immersed in $$P$$. So, we delete the purple loop from the link, perturb the green loop (as shown), and obtain the desired link.

Ok, checking my work in SnapPy, it seems that there is something interesting going on. Depending on how we choose the central crossing for green we get the links L8a3 or L6a1. The latter has only six tetrahedra (so that is good for the original poster's question). Also, this suggests that there is a smooth path in the space of cone manifolds that (a) connects L8a3 to L6a1 and (b) passes through the figure-eight.

Amusingly enough, this last sentence points to a much lighter proof that an example, as asked for in the original post, exists. Switching the central crossing on the green loop (and deleting purple) moves us between a pair of links with distinct complements. Performing a meridional filling on each gives us two core curves in the figure-eight knot complement. These core curves are homotopic, but not isotopic (because their complements are not homeomorphic), to each other. Thus at most one of the cores can be isotopic to a geodesic. (In fact, neither is, but that requires more work to see.).