# Bitangent locus of torus knots

Anyone know how to compute the bitangent locus of a space curve, e.g. a torus knot (pick whatever parametrization you like)? Specifically, what is the set of normal vectors (in the two-sphere) of planes tangent to the curve in more than one place?

• What precisely do you mean? You could compute it in a lot of standard ways, adapting Newton's method. Do you just want to see examples or are you primarily interested in computational heuristics? – Ryan Budney Dec 31 '13 at 20:07
• I'd be pretty happy with a decent picture, even. – Eric Zaslow Jan 2 '14 at 22:06
• Given a knot in $\mathbb R^3$ its set of tangent tangent lines is a 1-manifold in $\mathbb RP^2$, and generically this is an immersed circle. So the bitangent locus is the collection of double-points of this immersed curve in $\mathbb RP^2$. I think basically every null-homologous immersion $S^1 \to \mathbb RP^2$ is realizable as the tangent lines of a knot in $\mathbb RP^2$ so there's little restriction on the bitangent locus beyond that. – Ryan Budney Jan 2 '14 at 22:55

## 1 Answer

This 2013, long (47-page) PhD-thesis paper by Niels Lubbes may give some sense of how nontrivial is this task:

"Families of bitangent planes of space curves and minimal non-fibration families," (arXiv link)

He says, "In this paper we concentrate on cone curves, but many of our methods can be used to find bitangent families of arbitrary space curves." Algorithms are described starting on p.37. p.42: An aﬃne chart of the cone curve $C$ lying on a quadric cone $Q$ and its tangent developable $Z$.

• Thanks. I had actually seen this paper and was hoping it was not the only thing out there! As you note, Lubbes studies cone curves. Also, his context is algebraic geometry, whereas general knots are not (readily) algebro-geometric objects. A bit demoralizing, but thanks! – Eric Zaslow Dec 31 '13 at 19:12
• @EricZaslow: Yes, demoralizing---It seems to be fundamentally difficult! :-/ My guess is that the avoidance of algebraic objects will only reduce your tools, rather than make it easier. – Joseph O'Rourke Jan 1 '14 at 2:00