An analogue of the figure-8 knot, in four dimensions

The figure-8 knot is a lovely example of a knot in 3-dimensions. It's one of the simplest knots you can draw. The complement admits a hyperbolic structure of finite volume. And the hyperbolic structure is very nice -- it is the union of two regular ideal tetrahedra, moreover, the n-th face of tetrahedron 1 is glued to the n-th face of tetrahedron 2, with two of the gluing maps by right-handed 2π/3 twists, and two by left-handed 2π/3 twists. The complement of the figure-8 knot (thought of as a knot in the 3-sphere) fibers over the circle, and the fibre is a once-punctured torus. The monodromy is one of the simplest Anosov maps you can write down.

In a recent paper with Ben Burton and Jon Hillman, we find an analogous object in 4-dimensions. It's the complement of an embedded 2-sphere in the 4-sphere -- the knot is called a Cappell-Shaneson knot. So like the figure-8 knot, it fibers over the circle. It also has fiber a once-punctured product of circles, and the monodromy is a readily-described linear automorphism. The gluing instructions for this triangulation are a little less friendly to describe -- for those I suggest looking at the paper but they're reasonable.

The figure-8 triangulation has two edges, 4 triangles and two tetrahedra, and of course the one (ideal) vertex. The 4-manifold triangulation has similarly has one ideal vertex, but it only has one edge, 4 triangles, 5 tetrahedra and 2 pentachora. During the early drafts of the paper, I computed the edge link, it's the induced triangulation of the 2-sphere normal to the edge. It has 20 triangles. It's dual decomposition (depicted) has six hexagons and six squares.

A key point of interest about this paper is that we only prove that this triangulation is homeomorphic to the complement of the Cappell-Shaneson knot. We still hope to find a PL-equivalence, but that's a task for another day.

update: We now have a proof this manifold fibers over S1 and so it is diffeomorphic to a Cappel-Shaneson knot (exterior).   Our proof is algorithmic -- we find an explicit fibre-bundle structure on the triangulation via a somewhat tedious enumerative search.