28 May
2020
28 May
'20
8:31 p.m.
For p < q, we already know at least one case (2,3) where the lowest degrees = q & q+1. It's not surprising if the degree bounds are smaller for p < q, because that choice allows symmetry around the rotational axis. One nice thing about z^p + w^q = 0 is that the parametric solution is obvious in terms of complex exponentials. It should be possible to put in an extra deformation parameter, and calculate the period annihilator, but probably more difficult due to higher degree. --Brad On Thu, May 28, 2020 at 5:05 PM Dan Asimov <dasimov@earthlink.net> wrote:
each of degree = 2p. This is the lowest degree for two real polynomials that give the (p,q) torus knot in R^3 (p > q).