[math-fun] Polynomial torus knots redux
Say p, q are positive integers with gcd(p,q) = 1 and p > q. Then a (p,q) torus knot in S^3 in R^4 = C^2 = {(z,w)} is given by the intersection of the 3-sphere S^3 |z|^2 + |w|^2 = 1 and the variety z^p + w^q = 0. Via stereographic projection S^3 - {(0,0,0,1)} —> R^3 = {(X,Y,Z)}, this gives the complex polynomial equation P(X,Y,Z) = 0 where P(X,Y,Z) = 2^p (X+iY)^p + (Q+1)^(p-q) (2Z + i(Q-1))^q and Q stands for X^2 + Y^2 + Z^2, which is equivalent to the two real polynomial equations Re(P(X,Y,Z)) = 0 and Im(P(X,Y,Z) = 0, 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). Of course we could write (Re("))^2 + (Im("))^2 = 0 and get the same knot as the zero-set of just one real polynomial of degree 4p. —Dan
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).
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Brad Klee -
Dan Asimov