for the trefoil, I had in mind r[t_] := 2 + Sin[3 t] z[t_] := Cos[3 t] x[t_] := r[t] Cos[2 t] y[t_] := r[t] Sin[2 t] ParametricPlot3D[ {x[t], y[t], z[t]}, {t, 0, 3 Pi}] then we have two equations (sqrt(x^2+y^2) - 2)^2 + z^2 = 1 This just says we’re on a torus, and we can turn it into a 4th-degree polynomial: (3 + x^2 + y^2 + z^2)^2 - 16 (x^2 + y^2) = 0 . Then we have (x/r) = cos 2t and z = cos 3t which with trig identities leads to 2z^2 - 4 (c/r)^3 + 3 (c/r) - 1 = 0 then eliminating r is a bit of a pain. - Cris
On May 25, 2020, at 10:28 AM, Brad Klee <bradklee@gmail.com> wrote:
Hi Dan,
I think it's obvious, but I will explain a few finer points.
Call the polynomial equations P1(X,Y) and P2(X,Y,Z). The (3,4) torus knot can be represented as an algebraic variety: K(3,4)={(X,Y,Z) in R^3: P1(X,Y)=0 & P2(X,Y,Z)=0}, and K(3,4)={ (Sin[t] + 2 Sin[3t], Cos[t] - 2 Cos[3t], Sin[4t] ) in R^3 : t in [0,2*pi] }.
The proof is to observe that the map to trigonometric polynomial achieves identity after reducing terms. You must also show that the two polynomials are independent, but that follows from deg(P1) > deg(P2), as P2 includes more variables, so cannot be of the form f(Z)*P1. (Also, P1 does not factor to a lower-degree invariant.)
The polynomial P1 has a special interpretation. Curve C={(X,Y): P1(X,Y)=0} is the projection of K onto the pane orthogonal to the Z axis. We can then solve P1(X,Y,Z)=0 for Z(X,Y), the height function of knot K, which has two branches.
I am fairly sure this is what Cris Moore originally had in mind; though, I don't know the details of his encoding of the trefoil. It depends slightly on the choice of parametric form (but the degrees will be the same, because they must come from Chebyshev polynomials).
Per intuition I could guess the existence of an algorithm from any knot parameterized by trig. polynomials to an algebraic variety, but have not worked out details.
--Brad
On Mon, May 25, 2020 at 10:52 AM Dan Asimov <dasimov@earthlink.net> wrote:
What do these equations represent?
What are X, Y, and Z ?
—Dan
----- For the (3,4) knot I calculated that,
0 = -81 X^2 + 117 X^4 - 40 X^6 + 4 X^8 - 81 Y^2 + 234 X^2 Y^2 - 152 X^4 Y^2 + 16 X^6 Y^2 + 117 Y^4 - 152 X^2 Y^4 + 24 X^4 Y^4 - 40 Y^6 + 16 X^2 Y^6 + 4 Y^8
0 = 9 X^2 - 9 X^4 + 9 Y^2 + 14 X^2 Y^2 - 9 Y^4 - 144 Z^2 + 64 X^2 Z^2 + 64 Y^2 Z^2
{X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], Sin[4 t]} .
The total number of (X,Y) monomials increases quadratically and total (X,Y,Z) monomials increase cubically, as functions of degree. These equations can be solved to give an upper degree bound.
For integral calculations, this is a really nice way to represent the knot. We already have a [0,2*pi] parameterization built in, and trigonometric polynomials are easy to integrate.
Nice Idea! -----
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Cris Moore moore@santafe.edu “Looking after the numerous pieces of routine work, attending to scientific correspondence, getting material in form for publication, etc., has consumed so much of my time... If one could only go on and on with original work, looking for new stars, variables, classifying spectra and studying their peculiarities and changes, life would be a most beautiful dream.” — Williamina Fleming