# Trefoil knot as an algebraic curve

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?

• 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
• @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
• @Zhen: It would be great if someone can expand this is an answer (and derive some equation from it). – Martin Brandenburg May 23 '12 at 19:24
• See my answer to math.stackexchange.com/questions/3197639 . I might come over here and write a longer answer if I get a chance. – David E Speyer Jun 13 '19 at 15:47

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.

• 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!

Edit
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 $x=x_1+ix_2, y=y_1+iy_2$)
$$x_1^2+x_2^2+y_1^2+y_2^2=\epsilon^2, \quad x_1^2-x_2^2+y_1^3-3y_1y^2_2=0,\quad 2x_1x_2+3y_1^2y_2-y_2^3=0$$

• 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:

Another answer to this question, based on ideas from my answer here. I will write a trefoil knot as the transverse intersection of two smooth polynomial surfaces. I will also draw pictures!

Here is the basic idea. Let $$S^3$$ be the sphere $$|z_1|^2 + |z_2|^2 = 2$$ inside $$\mathbb{C}^2$$. I'll write each coordinate $$z_j$$ as $$x_j+i y_j$$. The trefoil knot $$K$$ is given by the equation $$z_1^3 = z_2^2$$ in $$S^2$$. Taking real and imaginary parts, we get $$x_1^3 - 3 x_1^2 y_1 = x_2^2 - y_2^2 \qquad 3 x_1^2 y_1 - y_1^3 = 2 x_2 y_2. \qquad (1)$$ I compute in my linked answer that these give smooth, transverse, surfaces in $$S^3$$.

That's $$S^3$$, but the OP asked for $$\mathbb{R}^3$$. To this end, we can remove a point from $$S^3$$. Since the OP wants a closed knot, not one that goes off to infinity, we should remove a point not on $$K$$. It is also good to remove a point where $$z_1^3-z_2^2$$ is neither purely real nor purely imaginary, so the two surfaces in $$(1)$$ will stay compact. I choose the point $$(1+i, 0)$$.

Stereographic projection away from the point $$(1+i, 0)$$ is given by the formula $$(u,v,w) \ = \ \frac{1}{2-x_1-y_1} (x_1-y_1, x_2, y_2) \qquad (2).$$ Composing $$(2)$$ with the parametrization $$(e^{2 i t}, e^{3 i t})$$ of $$K$$, we get a knot in $$\mathbb{R}^3$$ parametrized as $$\left( \frac{\cos (2 t) - \sin(2 t)}{2-\sin (2 t)-\cos (2 t)},\ \frac{\cos (3 t)}{2-\sin (2 t)-\cos (2 t)},\ \frac{\sin (3 t)}{2-\sin (2 t)-\cos(2 t)}\right).$$

To get equations for the surfaces in $$(1)$$, we need to have the inverse of $$(2)$$. That's easy to compute; the inverse map is $$(x_1, y_1, x_2, y_2) =$$ $$\frac{1}{u^2+2 v^2+2 w^2+1} \left( u^2+2 u+2 v^2+2 w^2-1,\ u^2-2 u+2 v^2+2 w^2-1,\ 4v,\ 4 w \right).$$ Plugging this into $$(1)$$ and putting everything over a common denominator, we get the equations $$2 + 12 u - 30 u^2 - 40 u^3 + 30 u^4 + 12 u^5 - 2 u^6 - 28 v^2 - 48 u v^2 + 56 u^2 v^2 + 48 u^3 v^2 - 12 u^4 v^2 - 8 v^4 + 48 u v^4 - 24 u^2 v^4 - 16 v^6 + 4 w^2 - 48 u w^2 + 88 u^2 w^2 + 48 u^3 w^2 - 12 u^4 w^2 + 48 v^2 w^2 + 96 u v^2 w^2 - 48 u^2 v^2 w^2 - 48 v^4 w^2 + 56 w^4 + 48 u w^4 - 24 u^2 w^4 - 48 v^2 w^4 - 16 w^6 =$$ $$-2 + 12 u + 30 u^2 - 40 u^3 - 30 u^4 + 12 u^5 + 2 u^6 + 12 v^2 - 48 u v^2 - 72 u^2 v^2 + 48 u^3 v^2 + 12 u^4 v^2 - 24 v^4 + 48 u v^4 + 24 u^2 v^4 + 16 v^6 - 32 v w - 32 u^2 v w - 64 v^3 w + 12 w^2 - 48 u w^2 - 72 u^2 w^2 + 48 u^3 w^2 + 12 u^4 w^2 - 48 v^2 w^2 + 96 u v^2 w^2 + 48 u^2 v^2 w^2 + 48 v^4 w^2 - 64 v w^3 - 24 w^4 + 48 u w^4 + 24 u^2 w^4 + 48 v^2 w^4 + 16 w^6=0$$ for $$K$$.

Here is the first equation, together with the knot. (Apologies for the gaps in the knot; I spent a while fighting with Mathematica and I want to go to bed. And thanks to this answer for getting me this far!)

And here are the two surfaces together. It's hard for me to actually see this, but it is two genus two surfaces, meeting along the knot.

As a final note, I spent a long time trying to get $$K$$ as the intersection of a genus $$1$$ surface with something, since $$K$$ is, after all, a torus knot. It took me a while to understand why I was failing: If $$K$$ is to be the transverse intersection of $$T$$ and $$\{ f=0 \}$$, then I want $$f$$ restricted to $$T$$ to be positive on one side of $$K$$ and negative on the other. But, if I embed $$K$$ into a torus $$T$$, then $$K$$ will not disconnect $$T$$, so this is impossible. The knot $$K$$ does disconnect the genus $$2$$ surfaces in the pictures, which is why I succeed.

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.

GF

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