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. -----
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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Ha ha, typo, should be (4,3) at the end. —Brad
On May 27, 2020, at 12:30 PM, Brad Klee <bradklee@gmail.com> wrote:
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:
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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I believe there’s a theorem in algebraic geometry that implies that those straight lines are factors of the system of polynomials, and can be removed rather than having to add a third equation. Varieties like e.g. the union of a line and a circle are not “irreducible”, and they can be factored into irreducible ones.
On May 27, 2020, at 11:30 AM, Brad Klee <bradklee@gmail.com> wrote:
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:
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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Cristopher Moore Professor, Santa Fe Institute I can think of nothing more dangerous, more divisive, or more self-destructive than the effort to prey on what is called 'white backlash'... it is dangerous because it threatens to vest power in the hands of second-rate men whose only qualification is their ability to pander to other men's fears. I think it divides this nation at a very critical time—and therefore it weakens us as a united country. — Lyndon Johnson, 1966
Certainly if you have F1*F2=0 then you can take it apart to F1=0, F2=0, but that isn't exactly what's happening here. You'd have to be more specific... In the meantime, we can make use of F1=0 and F2=0 if and only if F1^2+F2^2=0, when F1 & F2 real. For trefoil, if we write that: 0 = S(X,Y) = 27 - 108 X^2 + 64 X^4 - 96 X^2 Y - 108 Y^2 + 128 X^2 Y^2 + 32 Y^3 + 64 Y^4; 0 = H(X,Y,Z) = -8 X^3 + 24 X Y^2 - 9 Z + 16 X^2 Z + 16 Y^2 Z; 0 = F(X,Y,Z) = -9 + 52 X^2 - 96 X^2 Y + 52 Y^2 + 32 Y^3 - 64 Z^2; We have deg(S) = 4, deg(H^2+F^2) = 6. And we can compare with Dan's model that, 6 + 4 < 6 + 5. However, in invariant theory we probably would want to compare the sum of degrees of the Groebner basis, for which I get 29 to 220. The symmetric model wins by an order of magnitude. I don't think it's a good idea to try and force the variety definition to have only two terms, and for the purpose of scoring, we might even want to have as many polynomials as possible. The SHF idea has potential to generalize, maybe even to all sufficiently tame knots. One thing I really like about SHF is that we can give invariant descriptions. For the trefoil: V(S) : degree four shadow. V(H) : (3 lines) and a circle V(F) : over/under points at singularities V(S) All similar choices of coordinates and parameters should still have the same description as above. We could guess the topological description determines degree invariants. More later... --Brad On Wed, May 27, 2020 at 12:40 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
I believe there’s a theorem in algebraic geometry that implies that those straight lines are factors of the system of polynomials, and can be removed rather than having to add a third equation. Varieties like e.g. the union of a line and a circle are not “irreducible”, and they can be factored into irreducible ones.
On Wed, May 27, 2020 at 12:40 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
I believe there’s a theorem in algebraic geometry that implies that those straight lines are factors of the system of polynomials, and can be removed rather than having to add a third equation. Varieties like e.g. the union of a line and a circle are not “irreducible”, and they can be factored into irreducible ones.
On May 27, 2020, at 11:30 AM, Brad Klee <bradklee@gmail.com> wrote:
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:
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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Cristopher Moore Professor, Santa Fe Institute
I can think of nothing more dangerous, more divisive, or more self-destructive than the effort to prey on what is called 'white backlash'... it is dangerous because it threatens to vest power in the hands of second-rate men whose only qualification is their ability to pander to other men's fears. I think it divides this nation at a very critical time—and therefore it weakens us as a united country. — Lyndon Johnson, 1966
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participants (3)
-
Brad Klee -
Cris Moore -
Dan Asimov