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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