Re: [math-fun] Graduate Student Solves Decades-Old Conway Knot Problem | Quanta Magazine
I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}: (x+iy)^2 + (z+iw)^3 = 0 and x^2 + y^2 + z^2 + w^2 = 1 but which is then stereographically projected into R^3 = {(X,Y,Z)} via X = x/(1-w), Y = y/(1-w), Z = z/(1-w). If my algebra is right, the final (two) equations are 4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0 where W is short for (X^2 + Y^2 + Z^2 - 1). I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher". —Dan
Hi Dan, The quick easy depiction algorithm is to take plane cuts at equal-height intervals: https://0x0.st/ipt3.png It looks to be an overhand knot with joined ends, so it probably is an alternative trefoil. I wonder if it's possible to choose a more symmetric elliptic curve or a more symmetric projection, and then end up outputting a variety with lower degrees? --Brad On Mon, May 25, 2020 at 4:46 PM Dan Asimov <dasimov@earthlink.net> wrote:
I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}:
(x+iy)^2 + (z+iw)^3 = 0
and
x^2 + y^2 + z^2 + w^2 = 1
but which is then stereographically projected into R^3 = {(X,Y,Z)} via
X = x/(1-w), Y = y/(1-w), Z = z/(1-w).
If my algebra is right, the final (two) equations are
4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0
where W is short for (X^2 + Y^2 + Z^2 - 1).
I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher".
—Dan
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Perhaps add a z-axis color gradient to the "quick easy depiction algorithm". I don't know how much effort that would be, and of course if it turns one line of code into ten, it's not worth it. But I'm having trouble seeing the crossings. On Mon, May 25, 2020 at 7:23 PM Brad Klee <bradklee@gmail.com> wrote:
Hi Dan,
The quick easy depiction algorithm is to take plane cuts at equal-height intervals:
It looks to be an overhand knot with joined ends, so it probably is an alternative trefoil.
I wonder if it's possible to choose a more symmetric elliptic curve or a more symmetric projection, and then end up outputting a variety with lower degrees?
--Brad
On Mon, May 25, 2020 at 4:46 PM Dan Asimov <dasimov@earthlink.net> wrote:
I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}:
(x+iy)^2 + (z+iw)^3 = 0
and
x^2 + y^2 + z^2 + w^2 = 1
but which is then stereographically projected into R^3 = {(X,Y,Z)} via
X = x/(1-w), Y = y/(1-w), Z = z/(1-w).
If my algebra is right, the final (two) equations are
4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0
where W is short for (X^2 + Y^2 + Z^2 - 1).
I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher".
—Dan
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In this case at least, more trouble than its worth. There are only three crossings and they alternate. Other than that the shape is fairly normal. If I was going to do the algorithm right, I would fine grade enough to ID immediate successors, and then at the end impose a color wheel. Again, maybe later. --Brad On Mon, May 25, 2020 at 6:53 PM Allan Wechsler <acwacw@gmail.com> wrote:
Perhaps add a z-axis color gradient to the "quick easy depiction algorithm". I don't know how much effort that would be, and of course if it turns one line of code into ten, it's not worth it. But I'm having trouble seeing the crossings.
On Mon, May 25, 2020 at 7:23 PM Brad Klee <bradklee@gmail.com> wrote:
Hi Dan,
The quick easy depiction algorithm is to take plane cuts at equal-height intervals:
It looks to be an overhand knot with joined ends, so it probably is an alternative trefoil.
I wonder if it's possible to choose a more symmetric elliptic curve or a more symmetric projection, and then end up outputting a variety with lower degrees?
--Brad
On Mon, May 25, 2020 at 4:46 PM Dan Asimov <dasimov@earthlink.net> wrote:
I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}:
(x+iy)^2 + (z+iw)^3 = 0
and
x^2 + y^2 + z^2 + w^2 = 1
but which is then stereographically projected into R^3 = {(X,Y,Z)} via
X = x/(1-w), Y = y/(1-w), Z = z/(1-w).
If my algebra is right, the final (two) equations are
4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0
where W is short for (X^2 + Y^2 + Z^2 - 1).
I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher".
—Dan
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_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Okay, it was a minor quibble :) On Mon, May 25, 2020 at 8:02 PM Brad Klee <bradklee@gmail.com> wrote:
In this case at least, more trouble than its worth. There are only three crossings and they alternate. Other than that the shape is fairly normal.
If I was going to do the algorithm right, I would fine grade enough to ID immediate successors, and then at the end impose a color wheel. Again, maybe later.
--Brad
On Mon, May 25, 2020 at 6:53 PM Allan Wechsler <acwacw@gmail.com> wrote:
Perhaps add a z-axis color gradient to the "quick easy depiction algorithm". I don't know how much effort that would be, and of course if it turns one line of code into ten, it's not worth it. But I'm having trouble seeing the crossings.
On Mon, May 25, 2020 at 7:23 PM Brad Klee <bradklee@gmail.com> wrote:
Hi Dan,
The quick easy depiction algorithm is to take plane cuts at equal-height intervals:
It looks to be an overhand knot with joined ends, so it probably is an alternative trefoil.
I wonder if it's possible to choose a more symmetric elliptic curve or a more symmetric projection, and then end up outputting a variety with lower degrees?
--Brad
On Mon, May 25, 2020 at 4:46 PM Dan Asimov <dasimov@earthlink.net> wrote:
I tried to find equations in R^3 for the trefoil defined initially in R^4 = {(x,y,z,w)}:
(x+iy)^2 + (z+iw)^3 = 0
and
x^2 + y^2 + z^2 + w^2 = 1
but which is then stereographically projected into R^3 = {(X,Y,Z)} via
X = x/(1-w), Y = y/(1-w), Z = z/(1-w).
If my algebra is right, the final (two) equations are
4(X^2 - Y^2)(X^2 + Y^2 + Z^2 + 1) + 8 Z^3 - 6 Z W^2 = 0 & 8 X Y (X^2 + Y^2 + Z^2 + 1) + 12 Z^2 W - W^3 = 0
where W is short for (X^2 + Y^2 + Z^2 - 1).
I can't tell for sure if it looks right when I try to plot it in 3D using Mac "Grapher".
—Dan
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participants (3)
-
Allan Wechsler -
Brad Klee -
Dan Asimov