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Is the trefoil knot with its usual embedding into affine $3$-space

an algebraic curve (maybe after extending scalars to $\mathbb{C}$)? Is there even some thickening to some algebraic surface? If not, is there at least some similar algebraic curve which describes this type of knot? I hope that this question is not silly, I know almost nothing about this classical stuff on algebraic curves. A google research indicates that there is some connection with the cusp $y^2=x^3$, but I don't really get it.

PS: I am interested in explicit equations. Specifically, is the trefoil cut out by two equations in affine $3$-space?

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Some discussion in MO. My answer there is probably mostly useless because it is about a parametrization of a tubular surface around the trefoil. But the Milnor fibration may be useful to you (but goes over my head). – Jyrki Lahtonen May 23 '12 at 18:59
Analytic parametrizations can be also found on wikipedia; I am looking for an implicit definition given by a polynomial ( = algebraic curve). Agol's answer refers to the cusp, hm ... – Martin Brandenburg May 23 '12 at 19:01
But are you thinking about the curve in the "center" of the tube, or the surface of the tube? Are you thinking about it as a real curve, a real surface or a complex curve? – Jyrki Lahtonen May 23 '12 at 19:09
@Martin: Wikipedia says, "In algebraic geometry, the trefoil can also be obtained as the intersection in $\mathbb{C}^2$ of the unit 3-sphere $S^3$ with the complex plane curve of zeroes of the complex polynomial $z^2 + w^3$ (a cuspidal cubic)." – Zhen Lin May 23 '12 at 19:16
For the curve, it seems like you should be able to find an easy algebraization from its characterization as the (3,2) torus knot; if you're interested in an implicit characterization for a thickened surface then it seems at least at first glance like level sets of that algebraic version should get the job done, though admittedly I haven't plugged through the math to try this... – Steven Stadnicki May 23 '12 at 19:17
up vote 6 down vote accepted

There is a paper of Stephan Klaus that gives an explicit algebraic surface construction of the solid trefoil.

In addition, I found part of a solution to finding an algebraic curve isotopic to the trefoil by Michael Trott, but unfortunately the final pages are missing from the Google Books preview.

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Perfect! Thanks. Perhaps someone can add the formula which is only partially visible in the google books preview. – Martin Brandenburg May 23 '12 at 19:23
Here is the equation of the terfoil surface for copy & paste in SURFER or other visualization software: (-8*(x^2 + y^2)^2*(x^2 + y^2 + 1 + z^2 + a^2 - b^2) + 4*a^2*(2*(x^2 + y^2)^2 -(x^3 - 3*xy^2)*(x^2 + y^2 + 1)) + 8*a^2*(3*x^2*y - y^3)*z + 4*a^2*(x^3 - 3*xy^2)*z^2)^2 -(x^2 + y^2)*(2*(x^2 + y^2)*(x^2 + y^2 + 1 + z^2 + a^2 - b^2)^2 + 8*(x^2 + y^2)^2 +4*a^2*(2*(x^3-3*x*y^2)-(x^2+y^2)*(x^2+y^2+1))-8*a^2*(3*x^2*y-y^3)*z-4*(x^2+y^2)‌​*a^2*z^2)^2 – Martin Brandenburg May 23 '12 at 19:48

Consider two relatively prime integers $p,q \geq 2$ and the complex affine curve $C\subset \mathbb C^2 $given by $x^p+y^q=0$, which has an isolated singularity at the origin $O=(0,0)$.

If $C$ is intersected with the real $3$-sphere $S_\epsilon$ of equation $\mid x\mid ^2+\mid y \mid^2=\epsilon ^2$, Brauner proved in 1928 that the resulting real algebraic curve $C\cap S_\epsilon$ is a $(p,q)$ knot.
For $p=2, q=3$ , you obtain the trefoil knot, as you correctly conjectured.

Milnor has written a fantastic booklet (122 pages...) on the subject: Singular points of complex hypersurfaces, published by Princeton in 1968, with Brauner's theorem proved on the second page!

The intersection $C\cap S_\epsilon\subset \mathbb C^2=\mathbb R^4$ is a real algebraic curve with completely explicit polynomial equations .
For the trefoil knot with $p=2,q=3 $ for example, we get (writing $x=x_1+ix_2, y=y_1+iy_2$)
$$ x_1^2+x_2^2+y_1^2+y_2^2=\epsilon^2, \quad x_1^2-x_2^2+y_1^3-3y_1y^2_2=0,\quad 2x_1x_2+3y_1^2y_2-y_2^3=0 $$

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Thanks. Maybe I am mistaken, but $S_{\epsilon}$ doesn't seem to be an algebraic subset of $\mathbb{C}^2$. Or is the ground field still $\mathbb{R}$? Also, does this yield an explicit equation? – Martin Brandenburg May 23 '12 at 20:04
Dear Martin, yes, I meant real algebraic subset. I have written an edit. – Georges Elencwajg May 23 '12 at 20:42
Alright. So this cuts out the trefoil in $\mathbb{R}^4$ using three equations. I would like to know if it is possible to cut it out in $\mathbb{R}^3$ via one equation. – Martin Brandenburg May 24 '12 at 6:20
Dear Martin, sure, you can sum the squares of any number of equations defining a curve (or, for that matter, any real algebraic variety) and get one equation, but this is rather meaningless. You can't hope to define a curve in $\mathbb R^3$ in a serious way by just one equation. – Georges Elencwajg May 24 '12 at 7:35
Dear Martin, I have tried to answer your original question and some in the comments as best I could. I am sure you have many other interesting questions, but I suggest you ask them in new posts. – Georges Elencwajg May 24 '12 at 9:24

I know I am a bit late, but I think it is pretty easy to see that the trefoil can be described in terms of algebraic equations. Consider the parametrization from Wikipedia: $$ x = \sin t + 2 \sin 2t, \quad y=\cos t - 2 \cos 2t, \quad z=-\sin 3t. $$ Using the rules for double and triple angle this is equal to $$ x = \sin t + 4 \sin t\cos t, \quad y=\cos t - 2 +4\sin^2 t, \quad z=-\sin t (4\cos^2 t -1). $$ Denoting $s=\sin t$ and $c=\cos t$ you have the ideal $$ I=\langle s + 4 sc-x,c - 2 +4s-y,-s (4c -1)-z,s^2+c^2-1\rangle $$ describing the trefoil. The equation $s^2+c^2-1=0$ makes the connection between $\sin t$ and $\cos t$ while eliminating the parameter $t$. Now you can eliminate $s$ and $c$ in order to obtain a representation in the variables $x,y$ and $z$. Here is the SINGULAR code:

ring R = 0,(x,y,z,s,c),dp;
ideal I = 4*s*c-x+s, 4*s^2-y+c-2, -4*s*c^2-z+s, s^2+c^2-1;
ideal J = eliminate(I,sc);

Now $J=\langle P_1,P_2,P_3,P_4,P_5,P_6\rangle$ with $$ \begin{aligned} P_1&=12x^2y-4y^3-13x^2-13y^2+64z^2+9,\\ P_2&=x^3-3xy^2+4x^2z+4y^2z-9z,\\ P_3&=64y^3z+9x^3+9xy^2+4x^2z+100y^2z-192xz^2\\ &\phantom{=}-256z^3-72xy-180yz+27x-36z,\\ P_4&=256xy^2z-256x^2z^2-256y^2z^2-16y^3-128xyz\\ &\phantom{=}+256yz^2+35x^2-61y^2-240xz+640z^2-36y+9,\\ P_5&=48y^4+112y^3-144xyz-192yz^2-35x^2\\ &\phantom{=}-143y^2-108xz-160z^2-108y+99,\\ P_6&=8xy^3-4x^3+8xy^2-24x^2z-32xz^2-18xy-18yz+9x+27z. \end{aligned} $$ Maybe there are also generators with smaller degree. I haven't tried to find them yet.

There is also the famous trick to generate a single equation for the space curve: $$ T=\sum_{i=1}^6 P_i^2. $$ Over the real numbers you have $V_\mathbb{R}(J)=V_\mathbb{R}(T)$.

I'd like to post an image of a visualization based on the algebraic equations. But unfortunately, I don't have enough reputation yet.

Edit: Now I have enough reputation to post an image:

enter image description here

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See the following paper for lots of examples:

G. Freudenburg, "Bivariate analogues of Chebyshev polynomials with application to embeddings of affine spaces", CRM Proceedings and Lecture Notes, vol. 54 (2011), American Math. Society, 39-56.


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It all seems rather complicated to me, but no doubt I've missed something.

I just used polar coordinates $1/r = 1 + 0.5 \cos (3A/2)$.

This gives a lovely plane trefoil. The cartesian equation has 11 terms & is of the sixth degree in x & y - can give it if anyone interested

David R Roberts

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