Hi Dan, Again, I think it my last message was fine as stated, but I will explain a few finer points. Call the three polynomials S(X,Y), H(X,Y,Z), and F(X,Y,Z). They all equal zero for the parametric curve {X(t),Y(t),Z(t)} after reducing terms. The linear height function, H, can be factored as H = L(X,Y)+C(X,Y)*Z with L the product of four linear factors, and C the product of two circular factors. Eight solutions of L=0 & C=0, are a solutions of H=0 & S=0 regardless of Z, see: https://0x0.st/iph1.png So the algebraic varietry { (X,Y,Z) : H=0 & S=0 } accidentally contains 8 vertical lines. If (X_0,Y_0) is any one of the eight singular points, then F(X_0,Y_0,Z)=0 implies either: 48 Z^2-45 = 0, or 144 Z^2 - 567 = 0. On the set of singularities--which corresponds to the set of knot crossings--the quadratic filter function F chooses just two points from the singular, vertical line, one on the overpass, and one on the underpass. Thus the variety: { (X,Y,Z) : H=0 & S=0 & F=0 } Is the (4,2) torus knot in the strict sense, where it contains no extra points or mirror images. --Brad On Wed, May 27, 2020 at 11:42 AM Dan Asimov <dasimov@earthlink.net> wrote:
We need three constraints to accomplish what?
—Dan
----- Version 1.0 in Mathematica: https://0x0.st/ip7E.txt Analyzing outputs, I found the following for case (3,4):
{X,Y,Z} = {Sin[t] + 2 Sin[3 t], Cos[t] - 2 Cos[3 t], 2 Sin[4 t]} 0 = -81 + 117 X^2 - 40 X^4 + 4 X^6 + 117 Y^2 - 112 X^2 Y^2 + 12 X^4 Y^2 - 40 Y^4 + 12 X^2 Y^4 + 4 Y^6; 0 = -8 X^3 Y + 8 X Y^3 + 27 Z - 24 X^2 Z + 4 X^4 Z - 24 Y^2 Z + 8 X^2 Y^2 Z + 4 Y^4 Z;
0 = 81 - 81 X^2 - 81 Y^2 + 32 X^2 Y^2 + 16 X^2 Z^2 + 16 Y^2 Z^2
Contrary to expectation, we need not two but three constraints. The first two equations define the knot + 8 vertical lines at crossing points. The third constraint filters out vertical lines. -----
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