Re: [math-fun] Knot Annihilator
Actually, I'm just trying to understand the comment "we need three constraints". I'm not asking about the details of your calculation. I'm asking about what is the final goal that requires three constraints? I ask because, if the final goal is to represent the (3,4) torus knot in 3-space as the intersection of the zero-loci of each of several real polynomials, I would have expected that only two are required. —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. -----
I was also surprised by needing three constraints, but consider that the algorithm outputs four, with one syzygy. Any construction with a linear height function H and a Shadow function S will need a filter function F to reduce the line-like singularity to two crossing points on the curve. It is a question whether the three polynomials, H, S, F can be combined to two without changing the set of solutions. I don’t know just yet, seems probable. —Brad PS. I am not distinguishing between (3,4) and (4,3) for now.
On May 27, 2020, at 12:32 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Actually, I'm just trying to understand the comment "we need three constraints". I'm not asking about the details of your calculation. I'm asking about what is the final goal that requires three constraints?
I ask because, if the final goal is to represent the (3,4) torus knot in 3-space as the intersection of the zero-loci of each of several real polynomials, I would have expected that only two are required.
—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:
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. -----
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participants (2)
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Brad Klee -
Dan Asimov