Re: [math-fun] [xbbn] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
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! -----
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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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
Comparing parametric definitions, Cris has cos(5t) when the minimal possible choice is cos(3t). The choice 5t makes the knot curvier than necessary, and makes Chebyshev map higher degree. The algorithm I mentioned has an even / odd case split, but it did manage to calculate the following weak-invariant data: {X,Y,Z}={Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t], Sin[3 t]} 0 = 27 - 27 X^2 + 4 X^4 - 12 X^2 Y - 27 Y^2 + 8 X^2 Y^2 + 4 Y^3 + 4 Y^4 0 = -X^3 + 3 X Y^2 - 9 Z + 4 X^2 Z + 4 Y^2 Z So the Trefoil can be found using polynomials degree 4 in (X,Y) and degree 3 in (X,Y,Z). In particular, the height function Z(X,Y) is a ratio of polynomials in the (X,Y) variables. I will see if I can fix the algorithm later, and print off more polynomials for the first few cases. --Brad On Mon, May 25, 2020 at 12:50 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
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}]
nice. your 2d shadow is degree 4, while mine was degree 6. my construction went around a circle, with a fixed-length “elbow” tracing the knot out on the torus. C
On May 25, 2020, at 1:28 PM, Brad Klee <bradklee@gmail.com> wrote:
Comparing parametric definitions, Cris has cos(5t) when the minimal possible choice is cos(3t). The choice 5t makes the knot curvier than necessary, and makes Chebyshev map higher degree.
The algorithm I mentioned has an even / odd case split, but it did manage to calculate the following weak-invariant data:
{X,Y,Z}={Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t], Sin[3 t]} 0 = 27 - 27 X^2 + 4 X^4 - 12 X^2 Y - 27 Y^2 + 8 X^2 Y^2 + 4 Y^3 + 4 Y^4 0 = -X^3 + 3 X Y^2 - 9 Z + 4 X^2 Z + 4 Y^2 Z
So the Trefoil can be found using polynomials degree 4 in (X,Y) and degree 3 in (X,Y,Z). In particular, the height function Z(X,Y) is a ratio of polynomials in the (X,Y) variables.
I will see if I can fix the algorithm later, and print off more polynomials for the first few cases.
--Brad
On Mon, May 25, 2020 at 12:50 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
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}]
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I don't know if this is what Dan was getting at, but depending on rigor requirements, double-valuedness of the height function could be a problem. That is, the variety might actually be a superposition of chiral-opposite knots. I will look at it more later, maybe this weekend... --Brad On Mon, May 25, 2020 at 3:09 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
nice. your 2d shadow is degree 4, while mine was degree 6. my construction went around a circle, with a fixed-length “elbow” tracing the knot out on the torus.
C
On May 25, 2020, at 1:28 PM, Brad Klee <bradklee@gmail.com> wrote:
Comparing parametric definitions, Cris has cos(5t) when the minimal possible choice is cos(3t). The choice 5t makes the knot curvier than necessary, and makes Chebyshev map higher degree.
The algorithm I mentioned has an even / odd case split, but it did manage to calculate the following weak-invariant data:
{X,Y,Z}={Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t], Sin[3 t]} 0 = 27 - 27 X^2 + 4 X^4 - 12 X^2 Y - 27 Y^2 + 8 X^2 Y^2 + 4 Y^3 + 4 Y^4 0 = -X^3 + 3 X Y^2 - 9 Z + 4 X^2 Z + 4 Y^2 Z
So the Trefoil can be found using polynomials degree 4 in (X,Y) and degree 3 in (X,Y,Z). In particular, the height function Z(X,Y) is a ratio of polynomials in the (X,Y) variables.
I will see if I can fix the algorithm later, and print off more polynomials for the first few cases.
--Brad
On Mon, May 25, 2020 at 12:50 PM Cris Moore via math-fun < math-fun@mailman.xmission.com> wrote:
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}]
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participants (3)
-
Brad Klee -
Cris Moore -
Dan Asimov