# Torus knots on tori

The T(3,2) torus knot shown on the torus.

My summer research student Kyle Patterson was interested in printing torus knots together with the tori that they lie on. Recall that torus knots are knots (and links) that can be moved so that they are embedded on a torus (or doughnut shape surface). Kyle constructed torus knots and links using the techniques described in this earlier blog post and this post too.

T(3,2) torus knot with out the torus.

If we focus our attention on the T(3,2) torus knot, then this knot winds 3 times along the long way and 2 times along the short way around the torus. This knot has the following parametrization: $x(t)=\cos(3t)(3+\cos(2t)), \ y(t)= \sin(3t)(3+\cos(2t)), \ z(t) = \sin(2t)$ for t in [0,2π]. We note that the torus this knot lies on is defined by two circles.  We take a large/ring circle of radius 3 (given by equations x2+y2=9, z=0). We make each point of the ring circle the center of a small/pipe circle of radius 1 so that the plane of the pipe circle is perpendicular to the plane of the ring circle.  To construct this torus in Cinema 4D, simply add the Torus shape, then go to the Attributes menu and set the Ring Radius to 3cm and the Pipe Radius to 1cm. In order to have a smooth looking torus we need to change the number of Ring Segments to 150 and the number of Pipe Segments to 50.

The torus minus the T(3,2) torus knot.

To finish the construction we used the Boole tool in two different ways. First, we took the union of the knot and the torus to create a new surface which reveals the knot lying on the torus. Second, we took the difference of the torus and the knot. This created a torus surface with a smooth groove in the surface following the path of the knot.

We now have three different kinds of files that can be printed. We could print the knot, the torus with the knot, and a torus with a smooth groove following the knot.  We printed the torus with the smooth groove following the T(3,2) torus knot. We also created a 3D print where the knot is one color and the torus is a different color. To do this we use the torus with groove file and the knot file and the Ultimaker5 3D printer in the IQ center at WLU.

We repeated this construction for the T(2,3), T(3,3) and T(3,6) torus knots and links.

A red T(3,2) torus knot on a white torus.

# Torus knots on hyperboloids (part 2)

In the post Hyperboloidal Representation of Torus Knots we gave an explicit construction of T(p,q) torus knots and links where the knots are polygonal and the edges lie on two distinct hyperboloids of one sheet. As the knot is traversed, the edges alternate between lying on the first then the second hyperboloid. Each hyperboloid of one sheet is a doubly ruled surface (as described in the post Doubly ruled surfaces). For each of the hyperboloids the edges of the torus knot lie in one of the rulings.

The T(3,5) torus knot lying on two hyperboloids of one sheet.

I wanted to be able to create a 3D-printable model that showed both the T(p,q) knot and the two hyperboloids of one sheet. I used Cinema 4D to do this, and ended up making several attempts to get things just right.

The T(3,5) torus knot was constructed as described in the post Hyperboloidal Representation of Torus Knots. To sum up, the vertices were computed, then the knot was constructed in the standard way.  There are two relatively easy ways to construct the two hyperboloids of one sheet. In the computation for the vertices, we can find an explicit formula for the hyperboloids. We could parametrize a generating hyperbola curve for the hyperboloid, put it into the Formula tool, then use a Lathe tool to rotate the curve around the y-axis. (Remember that in Cinema 4D, the y-axis points up.)

The edges of the T(3,5) torus knot lie on these two hyperboloids of one sheet.

Rather than do this, I kept things simple. I added an Empty Spline, opened the Structure Manager, then added in the first three vertices of the T(3,5) torus knot. This created the first two edges of the knot. However, these had the wrong orientation – the vertices were entered for the standard xyz-orientation, not the orientation in Cinema 4D. I thus went into Model mode, and rotated the vertices 90o in the y-direction. I then added a Lathe and moved the Spline under the Lathe. (The reason why this works can be found in the post Doubly ruled surfaces – the formulas explicitly show that you can take one line segment in a ruling and rotate it around the z-axis to create the surface.) At this point the surface had many obvious edges. I went back into the Spline and edited the Object Properties: the Type is Linear, the Intermediate Points is Uniform, the Number is 50. (For some reason Bezier gives a bad looking surface.) I went back into the Lathe and edited the Object Properties. Here I increased the Subdivision to 50.

One of the errors I first made when constructing this surface was to take the coordinates of the vertices of the T(3,5) knot from the 3D printable model. This was a mistake, as the original vertices had been replaced by two nearby vertices for the Chamfer at the corners. This meant the two hyperboloids of one sheet where not correct.

The T(3,5) knot on the two hyperboloids of one sheet.

At this point, we can put the object in a better form for 3D printing. One of the adjustments I made was to scale the hyperboloid surface up slightly. When 3D printed, the top and bottom edges of this surface end up each print as one single filament. This can be problematic to print. So increasing the size of the surface slightly gives a bit of wiggle room.  I used the Boole tool (A union B)  to join the hyperboloid surface to the knot. In order to even see the union of these two surfaces, I needed to turn off High Quality in the Object Properties of the Boole. I also created one object from the Boole by going to the Object menu of Objects and then “Current State to Object”. On the new Boole that was created, I selected  “Connect objects + Delete”. I’m not sure these last two steps are entirely necessary. I’ve noticed that the slicing programs for 3D printers are increasingly sophisticated and are able to handle objects that aren’t completely correctly triangulated much more easily than in the past.

The hyperboloid surface without the T(3,5) torus knot.

Finally, I wanted to be able to 3D print the knot in one color and the surface in a different color. I went back into Cinema 4D and created several files. The first was the knot, the second was the surface without the knot. I created this using the Boole tool and the “A subtract B” option. I then printed the surface and the knot on an Ultimaker 5 printer. I had trouble printing this surface as the printer added in unnecessary supports. This is a work in progress! I’ll post an update with a correctly printed knot when we make it happen.

# Constructing more equilateral stick knots

A polygonal knot is a knot that is built from a finite number of straight line segments or edges. Two edges meet at a vertex of the knot. A polygonal knot can thus be described by listing the vertices v1, …, vn. The edges join each pair of vertices: e1 = v1v2, e2 = v2v3, …, en = vnv0. An equilateral stick knot is a polygonal knot where each of the edges has the same length.

A collection of 3D printed equilateral stick knots. Top row: 9:29 and 9_36. Bottom row: 8_13, 7_4 and 8_17.

One of the many open problems in knot theory is to find the least number of edges or sticks needed to build a particular knot. This number is referred to the stick number s(K) of a knot K. The stick number of a few of the simplest knots is known. For example: s(unknot)=0, s(trefoil)=6, s(figure 8)=7. There are a number of results giving upper and lower bounds of the stick number in terms of the crossing number of a knot. These inequalities can be combined with physical constructions of a particular knot to prove a given stick number.   In a similar fashion, the equilateral stick number can be found for a knot. This is the least number of equal length sticks needed to build a particular knot. It is an open question as to whether the equilateral and regular stick numbers are the same. In the not too distant past, Eric Rawdon and Rob Scharein (Knot Plot) produced a list of knots through 10 crossings for which they could not match the stick number and equilateral stick numbers.

The 9_36 equilateral stick knot.

Clayton Shonkwiler, and Thomas Eddy have a method which explores the space of equilateral stick knots. They use this method to find realizations of all knots up to 12 crossings. The coordinates of these knots can be found in following GitHub repository: https://github.com/thomaseddy/stick-knot-gen Since these coordinates give explicit constructions for equilateral stick knots, they give an upper bound on the equilateral stick number for the knots in question. The GitHub site also lists knots where the exact equilateral stick number is known.  (This GitHub site also gives references to numerous papers of Clay and his coauthors on both equilateral stick knots on the superbridge index of knots.) This data was used to confirm that all knots through 10 crossings except for 929 have the same stick number and equilateral stick number. This nicely handled the questions raised by Eric and Rob.

At the start of summer I asked Clay if there were any equilateral stick knots that he’d like to see 3D printed. He listed the 936 knot and the 929 knot. The 936 knot has stick number ≤ 10. However the coordinates here are not a minimal stick realization. Instead the realization is the only one he knows that has superbridge number equal to 4. Indeed, this realization proves that the superbridge index of 936 is equal to 4. An image of the completed 3D printable version of the 936 knot is found above. Note that while the thickened tube around the knot has self intersections, the knot itself does not. The self-intersections are a quirk of how close the edges come to each other in that particular realization.

The polygonal 9_29 knot imported into Cinema4D

The 929 knot realization is a very recent discovery by Clay, Jason Cantarella and Andrew Rechnitzer. They sent me the data for the vertices (which will be eventually become available on arXiv).  The realization is incredibly close to being singular, so it was a real challenge to create a nice 3D print for it. It turns out that this example shows that both the equilateral and regular stick numbers of 929 are equal to 9. Thus this recent work shows that all knots through 10 crossings have the same stick number and equilateral stick number!  The general consensus is that the stick number and equilateral stick number are distinct knot invariants. However, an example illustrating this still needs to be found.

I used the 3D editing program Cinema 4D to create a 3D-printable model of these knots. The method is pretty straightforward. (This is the same technique that was described in a previous post by my Spring term Knot Theory students.)

2. In Cinema 4D open an Empty Spline curve.
3. Go to Structure Manager and then import the vertices into the spline.
4. Go into Point Mode, select the Spline Pen.
5. Join the first and last vertices together.
1. Scale the model until it is 6-8cm in each dimension
2. Go to the Object Properties. Make sure the Type is Linear, the Intermediate Points is Uniform and the Number is 50.
8. Add a Sweep. Make the Circle and the Spline the “children” of the Sweep.

The 9_29 knot first attempt.

At this point the model has some real challenges. The image is the exact same knot as displayed above. However, the addition of the tube means the edges of the knot have been extended far beyond their actual length. What is needed is to put in a Chamfer on the vertices. This replaces one vertex, with two vertices on the edges very close to the original vertex. This has the effect of smoothing out the corner. Cinema4D gives the radius of the Chamfer as the radius of an imaginary circle smoothly joining the two edges at the vertex.

The 9_29 knot with trimmed vertices. The Chamfer is too big, so the knot type has changed.

A different view of the trimmed 9_29 knot – the knot type has changed.

I first tried a Chamfer of radius 0.1cm. However this was problematic. The angle between some of the edges of the knot are very close to zero. This means that the Chamfer cuts of a great deal of length of the edges as shown in the figures above. This has the dual effect of both changing the knot type and breaking the equilateral property. I then tried a Chamfer of radius 0.01cm. I also had to increase the number of points on the spline to 100. The end result was still a bit “spiky”. To solve the problem meant that I needed to go back into the Spline and change the Object Properties: the Type became Bezier and the Intermediate points became Adaptive. The final version is found below.

The finished 9_29 knot.

My summer research student Timi Patterson also constructed the 74, 813, and 817 knots this summer. We’ve uploaded all of these knots to Thingiverse.

# Doubly-ruled surfaces

An example of a hyperboloid of one sheet.

Some of the most interesting quadratic surfaces are the doubly-ruled surfaces: the hyperboloid of one sheet (like x2+y2-z2=1)  and the hyperbolic paraboloid (like z=x2-y2). These surfaces have made an appearance in this blog previously when we discussed how to create good 3D printable models of these surfaces.

An example of a hyperbolic paraboloid.

Ruled surfaces have been studied extensively in geometry. Formally, a ruled surface is a surface which is a union of straight lines. The straight lines are called rulings (or the generators) of the surface. An example to think about is a cylinder: this is a union of straight lines, each of which intersects a circle. Just as with the circle and the cylinder, in a general ruled surface there is a curve in space which intersects each of the lines. Thus the surface can easily be parametrized as follows. Let g:(a,b)->R3 be a parametrization of the curve on the surface, let vt be a vector at point g(t) which points in the direction of the line passing through g(t). Then the surface is parametrized by f:(a,b)x(c,d)->R3 defined by f(t,s) = g(t)+svt.

A helicoid. Image from https://minimalsurfaces.blog/

A nice example of such a surface is the helicoid where lines parallel to the xy-place pass through each point of a helix (for example parametrized by g(t)=(cos(t), sin(t),t)). One possible parametrization of a helicoid is f(t,s) = (s.cos(t), s.sin(t), t).  Ruled surfaces are very familiar to us, if the rulings are all parallel to each other, then the ruled surface is a generalized cylinder. For example, the surface given by y=x2 in R3 is a generalized cylinder parametrized by f(t,s)=(t,t2,s).

The two rulings on a hyperbolic paraboloid surface.

Now there are surfaces which are doubly-ruled surfaces, meaning that each point of this surface belongs to two distinct lines. In other words, this surface has two distinct rulings. You can use projective geometry to show that this surface, the plane, and the hyperboloic paraboloid are the only surfaces that have this property. Robert Byrant has a nice proof of this fact using differential geometry in a 2012 Math Overflow post

Let us focus our attention on the hyperboloid one sheet given by  x2+y2-z2=1.  To give explicit equations for the two rulings, I will follow the text Elementary Differential Geometry by Andrew Pressley. For every t, we can show the straight line Lt given by $(x-z)\cos t=(1-y)\sin t, \ \ \ (x+z)\sin t = (1+y)\cos t$ lies on the surface. To see this, multiply the two equations together to get $(x^2-z^2)\sin t\cos t =(1-y^2)\sin t\cos t.$ in other words x2+y2-z2=1 unless cos(t)=0 or sin(t)=0. If cos(t)=0, then x=-z and y=1, and if sin(t)=0 then x=z and y=-1, and both of these lines lie on the surface. A short computation reveals that for each t,  line Lt contains the point (sin(2t),-cos(2t), 0) and is parallel to the vector (cos(2t), sin(2t), 1). We thus get all lines in one ruling for t in [0,π).

One set of rulings on a hyperboloid of one sheet.

We now show that any point on the hyperboloid lies on one of these lines. Take a point (x,y,z) on the surface x2+y2-z2=1. If x does not equal z, then let t be such that cot(t) = (1-y)/(x-z). If x does not equal -z, then let t be such that tan(t) = (1+y)(x+z). In both cases the point is on Lt. The only cases left are the points (0,1,0) and (0,-1,0). But these points lie on the lines Lt when t=π/2 and t=0 respectively.  In addition, we can check that the lines in this ruling do not intersect. Suppose (x,y,z) lies on Lt and Ls and t does not equal s. Then $(1-y)\tan t = (1-y)\tan s, \ \ \text{and} \ \ (1+y)\cot t = (1+y)\cot s.$ Assuming the tan and cot functions are not zero or undefined gives both y=1 and y=-1, a contradiction. The case where t=0 and s=π/2 give disjoint lines too: $L_0(t) = (t,-1,t) \ \ \text{and} \ \ L_{\pi/2}(s)=(-s,1,s).$

What is the other ruling? For every t, we can show the straight line Ms given by $(x-z)\cos s=(1+y)\sin s, \ \ \ (x+z)\sin s = (1-y)\cos s$ lies on the surface. The computations are almost identical, so we omit them. Each point of the surface lies on a line in Ms and the lines in this ruling do not intersect.

The second set of rulings on a hyperboloid of one sheet.

Another computation shows that if t+s is not a multiple of π, then Lt and Ms intersect in the point $\left( \frac{\cos(t-s)}{\sin(t+s)}, \frac{\sin(t-s)}{\sin(t+s)}, \frac{\cos(t+s)}{\sin(t+s)} \right).$  For each t in [0,π), there is one s in [0,π) such that t+s is a multiple of π, and the lines Lt and Ms do not intersect. Intuitively, the two lines are on opposite sides of the hyperboloid of one sheet.

There are many other interesting observations to be made about these surfaces. For example, any set of three skew lines generates such a surface and the three skew lines lie in one of the rulings. Ian Agol posted a nice proof of this in the same 2012 Math Overflow post.

# Hyperboloidal Representation of Torus Knots

Written by Timi Patterson (2024 Summer Research Scholar), added to by Elizabeth Denne.

A T(3,5) torus knot arranged on two hyperboloids of one sheet.

In his book Knots and Links, Peter Cromwell details a representation of torus knots embedded in a parameterization on the union of two hyperboloids. He provides these instructions in Section 1.5:

Choose an angle θ in [0, π/2] and construct two points: $A=(\cos\theta, -\sin\theta, -1), \ \ B=(\cos\theta, \sin\theta, 1).$ The straight line through A and B is defined by $x=\cos\theta, y=z\sin\theta.$ Rotating this line about the z-axis gives the hyperboloid $x^2+y^2-z^2\sin^2\theta = \cos^2\theta.$ Let Ht denote the annulus obtained by restricting z between the interval from -1 to 1. The boundary curves of the annulus are unit circles: $x^2+y^2=1, \ \ z=\pm 1.$ Take the union of two of these Ht annuli with different values of t=theta. This new surface is topologically a torus.

The (p,q) torus knot with p strictly greater than q (and q greater than or equal 2) can be embedded in one of these “hyperboloidal” tori as follows. Choose t=theta and s=phi such that $\frac{q}{p}\cdot \frac{\pi}{2} <\theta < \min\{ \frac{\pi}{2}, \frac{q}{p} \pi\} \ \ \ \text{and} \ \ \ \phi = \frac{q}{p}\pi -\theta .$  The knot will lie in the torus which is the union of Ht and Hs.

Now take i in {0, 1, 2, … , 2p}. If i is odd, the vertices of the knot are $v_i=( \cos((i-1)\pi\frac{q}{p} + 2\theta), \sin((i-1)\pi\frac{q}{p}+2\theta), 1),$ and if i is even, the vertices of the knot are $v_i = (\cos((i\pi\frac{q}{p}), \sin(i\pi\frac{q}{p}), -1) .$

Polygonal (3,2) torus knot whose edges lie on hyperboloids of one sheet.

Following these instructions for the trefoil knot viewed as a T(3,2) torus knot, with$\theta = \frac{2\pi}{5} \ \ \text{and} \ \ \phi = \frac{4\pi}{15},$

I constructed the following vertices:
 x y z 1 -0.8090169944 0.5877852523 1.0 2 -0.5 -0.8660254038 -1.0 3 0.9135454576 0.4067366431 1.0 4 -0.5 0.8660254038 -1.0 5 -0.1045284633 -0.9945218954 1.0 6 1.0 0.0 -1.0

One component of the T(4,2) torus link.

I first constructed the T(5,3) torus knot in Cinema 4D, as the vertices were detailed in the book. I did this by creating splines using the vertices created by the functions, and using Cinema 4D’s sweep function to create a model with a thickness. I used the Chamfer tool to smooth out the corners. I then went on to create the trefoil knot, the T(10,8) torus knot, the T(4,2), T(12,3), and T(10,8) torus links all in Cinema 4D with the same technique.

The T(4,2) torus link. Note the two components differ by a 90 degree rotation.

To create the links, I had to separate the functions into the different components. Take for example the T(4,2) torus link. When evaluating the formulas, the q/p reduces to 1/2. To then create the two different components of T(2,1), the first component uses the vertices constructed as described above. To construct the vertices of the second component, simply add π/2 to the inside of the trig function in each component. (For example cos(iπ/2 +π/2) for the first term in the even index vertex.) Therefore, I had two components with these vertices (for theta=3π/8 and phi=π/8).

Component 1:

 x y z 1 -0.7071067812 0.7071067812 1.0 2 -1.0 0.0 -1.0 3 0.7071067812 -0.7071067812 1.0 4 1.0 0.0 -1.0

Component 2:

 x y z 1 0.7071067812 0.7071067812 1.0 2 0.0 -1.0 -1.0 3 -0.7071067812 -0.7071067812 1.0 4 0.0 1.0 -1.0

One version of the T(10,8) torus link.

A different version of the T(10,8) torus link.

One thing that I experimented with some when working with the T(10,8) torus link is manipulating the theta value to try and reduce any intersections of the model. I created two different models, one with theta=5π/11, the other with theta=5π/12. They varied a lot with where the self-intersections of the tubes were, but alas both of the tubes did self intersect. That will most likely happen with a lot of torus knots or links with p’s and q’s of closer value, but some self-intersections may be able to be avoided by manipulating the theta value,

The T(10,3) torus knot.

I then went on to print out most of these 3D models. The T(10,3) torus knot and T(10,8) torus link are shown above. It turned turns out that these models are surprisingly difficult to print. Take a look at the models. There is only a small area on the base of each V shape. The edges of the knots have to “grow” from this small base. This means the models are unstable. So even though the angle of the edges is high with respect to the ground, the models still need support. We printed several without supports and had some spectacular failures, as shown below. After the edges of the knots fell over on the print bed, the printer kept going leaving a squiggly mess of filament.  The solution to this problem was to increase the angle for the supports from 43 to 55 degrees.

Several of the builds failed as the edges of the knots fell over during 3D printing.

# Parametrizing Petal Projections

Written by Keally Rohrbacher and Sawyer Dunn-Matrullo (students in Math 383D Knot Theory Spring 2023).

Figure 1: Trefoil knot petal projection.

Our plan was to create models of knots with petal projections. This is a particular type of diagram of knots that has only one crossing, and a certain odd number of arcs which cross each other there. For example, Figure 1 shows the petal projection of the Trefoil Knot (image from Wikipedia). The numbers written on the arcs in Figure 1 are important, as the order in which the knot crosses through itself is what distinguishes different petal projections of knots from each other.

We wanted to produce 3D models of the 41, 52, and 61 knots, all of which have petal projections (shown in the figures below).

Figure 2: Side view of the petal projection of the 4_1 knot.

Since it is not important what the actual picture of the petal projection is, just that it crosses through the center in a particular order, we had some choice as to how we were doing to construct these knots. We decided that it would be interesting to construct these as parameterized functions. We knew that we could make the x and y-components of this function fairly easily using a rose curve, a type of polar function which produces a cool rose that looks just like the petal projection in 2D. So, we knew that all we had to do was come up with a function for the z-axis which would parametrize the rose curve to go around and hit the center at certain heights to produce the petal we wanted. We knew which order the strand should go through the crossing in for each knot from the paper “Knot Projections with a Single Multi-Cossing” by Colin Adams and his coauthors.

Figure 3: Top view of the petal projection of the 4_1 knot.

We considered a couple ideas for finding a function that would hit these particular heights, but ultimately decided to try to find a polynomial function.

Figure 4: Side view of the petal projection of the 5_2 knot.

We did this by creating points with the heights we needed to hit at even intervals and plugged these points into an online calculator which uses Lagrangian interpolation to produce a polynomial which hit all of these points. Once we defined this as the z-component for our function, and used the rose curve for the x- and y-components, we plugged our curve into a 3D graphing calculator called GeoGebra. The side views, in figures 2, 4 and 6 (above and below), depict these heights being hit in proper order in accordance with the Adams’ paper. The top views in figures 3, 5 and 7 show the petal projection shape of the rose curve projected to the xy-plane.

Figure 5: Top view of the petal projection of the 5_2 knot.

We ran into a couple issues during this project. The most annoying of which was trying to define our curve as a parametrized function in Cinema 4D.

Figure 6: Side view of the petal projection of the 6_1 knot.

Because we had to hit so many points, the function on the z-axis ended up being a degree 11 polynomial for both knots. And while we were able to produce this on GeoGebra, a powerful calculator, we could not make a satisfactory spline of our function in Cinema 4D. We spent a long time trying to fix this problem by manipulating points on the software. We were exasperated to find that we could simply download an .stl file directly from GeoGebra, where we already had constructed the knot, circumventing our entire issue. (Note that 3D printers use .stl files as their start point.) This made its own set of issues, however, as the shapes from GeoGebra were not smooth. But, for our final print of the knots, we also graphed the curves on Mathematica like we did in GeoGebra, again avoiding the issue of plotting a curve in Cinema 4D and Mathematica gave us models with much smoother curves.

Figure 7: Top view of the petal projection of the 6_1 knot.

The most challenging part of coming up with our functions though was creating and working with high order polynomials, but this mostly just involved us typing out long equations many times. We found that WolframAlpha easily came up with the required higher order polynomials for the 61 knot (which ended up being a degree 14 polynomial).

# Tying Knots on the Shortest Lattice Walks

Written by Chadrack Bantange and JCW (students in Math 383D Knot Theory Spring 2023).

# Background:

What is a minimal cubic lattice knot?

The cubic lattice.

Let’s start with: what is the cubic lattice? The cubic lattice is all the points (a, b, c) in R3 such that a, b, c are integers. That is, the cubic lattice is composed of all the points in three dimensions that have integer entries. The image to the right is a depiction of the cubic lattice (from Wolfram MathWorld).

Next, what would it mean for a knot to be in the cubic lattice? Cubic lattice knots are knots whose vertices lie in the cubic lattice. That is, the vertices of the knot can be represented by (a, b, c) where a, b, c are integers.

Another way to think about cubic lattice knots would be to imagine you start your knot on the point (0, 0, 0). To take a step tracing the knot one must take one of the steps as depicted below. Observe that on this first step from the origin one has six options. One could go: left, right, up, down, forward, or back. Consider the diagram below which presents all six of the options. The step which traces the knot is itself not a part of the cubic lattice. If we move up from the origin to point (0, 0, 1) all the points between these vertices makeup the knot, but are not part of the cubic lattice per se.

Figure showing the 6 points in the cubic lattice one unit from the origin.

In each of the subsequent steps you have similar options. Because you are walking within the cubic lattice, on each step you can only ever add +1 or -1 to just one of your x, y, or z coordinates to get your next vertex. Note: since we are drawing a knot, you will have one less option than when starting at the origin, because you do not want to walk back on the knot you have already drawn. If you go up from (0, 0, 0) to (0, 0, 1), then you could not go back down to (0, 0, 0) when tracing your knot.

Lastly, what is a minimal cubic lattice knot? Each knot has multiple equivalent representations and diagrams that can be reached via planar isotopies and Reidemeister moves. This means there are many versions of each kind of knot in the cubic lattice.

Consider the trefoil. One could trace a trefoil knot in the cubic lattice that globally looked somewhat smooth and like a common depiction, as below on the left. This, however, is not particularly mathematically interesting or remarkable. Rather, to ask the more interesting question we ask: what is the minimal cubic lattice knot? That is, for each of the knot equivalence classes what is the shortest walk we could take in the cubic lattice to trace out a knot in the equivalence class? On the right, observe the minimal cubic lattice knot for the trefoil. In the case of the trefoil (31), the minimal walk in the cubic lattice is 24 steps. (Left image from the Rolfsen Knot Table Mosaic.)

Figure showing the standard image of the trefoil knot (left), and the image of the minimal length trefoil in the cubic lattice (right).

# Construction

To build the cubic lattice knots, we used Cinema 4D. As part of this process, we referred to Andrew Rechnitzer’s website on plotting minimal cubic lattices). For all the cubic lattices, we noticed that when importing the data from Notepad into Cinema 4D, one line of the coordinates was missing, making the object look incomplete. This was a challenge we experienced in constructing our knots. Correcting this issue involved manually attaching numbers to each line of the coordinates while also inserting a row of zeros in the first line of the dataset. As we went on building more knots, we noticed that this process was time-consuming. Alternatively, we used Microsoft Excel to copy our coordinates from Notepad, pasted into Excel using the “paste special” option, and we were able to use Excel to automatically attach numbers to the individual coordinates. We then pasted these coordinates back into Notepad and saved it as a “txt” format.

Once the data was imported into Cinema 4D, we went to “view” in the left menu, selected “Frame Geometry” to get a better geometric view of the knot. The spline automatically connected all the points except the last to the first. Before connecting them, we first made sure that under Object manager, we have “Rectangle—Spline” selected; under “Attributes—Object—Type” ,  choose “bezier or linear” depending on how smooth we want the corners of the knot to look like. For some of the knots, “bezier” was the best smoother of the corners, while for others, “linear” was best. To connect the vertices, we selected “spline pen” and were able to click on the two, unconnected points to add the last edge to the spline.

At this point all we had was a spline made of 1-dimensional lines. In order to make the knot 3-dimensional for printing and visualization, we added “circle” and “sweep” to our Cinema 4D workspace. The circle would serve as the cross section for our knot and the sweep is what swept the circle cross section along the spline we created above. We then adjusted the radius of the knot, which in most cases was 0.25 cm. In order to smooth out the corners, we added a Chamfer to the knots and set its radius at 0.25 cm for the 3, 4, 5, and 6 crossing knots and 0.3 cm for the 7 crossing cubic knots.  We then scaled the object to be hand sized. There were a lot of variations in this regard depending on the knot, especially since some of the minimal cubic lattice knots occupied the space of a cube, while others occupied the space of a rectangular prism. In general, however, the scale ranged from 4 to 8 cm in all the three axes (x,y,z). We saved both .stl and .c4d files, ready to be printed in the IQ center at WLU.

Here are some screenshots of our work. In both cases, the standard image of the knot came from the Rolfsen Knot Table Mosaic.

On the left, the standard view of 5_2 knot. On the right, the minimal cubic lattice knot of 5_2.

On the left the standard view of the knot 6_1. On the right the minimal cubic lattice knot of 6_1.

# Flowering 3D Models

Written by Claire Gilreath. Joanne Wang, and Selihom Gobeze (students in Math 383D Knot Theory Spring 2023).

# Building Petal Knots in Cinema 4D

Once we had found all of our coordinates for our petal knots, we were excited to actually build the knots in Cinema4D! Of course, this was not without its challenges. We started with 3_1. Our first step was to to import our points to a spline. Naturally, we ran into issues here because we were not aware that the points from the first row are read as the x, y, and z labels, so the data is not actually imported. We fixed this by editing our .txt file to include a row of 0s at the top. Also, one of the points was incorrect and we had to go back to WolframAlpha for a quick fix (this happened a couple of times throughout the process).

This is the petal trefoil knot made in Cinema 4D.

At this point, we were not aware that we needed to connect the last point to the first point, so we skipped that step and decided to try the Spline Smooth tool to round out our polygonal edges. We found that the result was not as uniform as we had expected but we decided to keep going to see what would happen after the sweep. We made a circle and decided to set the radius at 0.25cm in order to avoid self-intersections but give the model enough structure to support itself. Then, we made the sweep and looked at our slightly wonky 3_1 knot. We realized that the ends were not connected so we used stitch and sew to fix that (this was a result of failing to connect the last two points of the spline). To the right is our final improved 3_1 knot.

This is the petal 5_1 knot made in Cinema 4D.

We moved on to the 5_1 knot and followed the same procedure. Once we had built it, Professor Denne looked at our slightly wonky 5_1, and suggested we try to use Chamfer tool instead of the Spline Smooth tool to make the model look smoother and more uniform, with less harsh edges of the petals. In doing this, we discovered that our last point vanished when we tried to Chamfer. We realized that we needed to close the spline by deleting our last point at (0,0,0) and using the Spline Pen to connect the first point to the last point to close the gap. We set the radius of the Chamfer to 3cm and compared our new 5_1 to the wonky one. We decided we liked the Chamfer version better and fixed our 3_1 the same way. This time we set the radius to 5cm, which we found looked much smoother and decided to stick with that for all knots. Our completed 5_1 knot is pictured above.

This is the petal 6_2 knot made in Cinema 4D.

We repeated this same process with 6_2, expecting everything to be a lot easier now that we had solidified our methods. However, this was not the case. As we rotated around the knot after sweeping, we noticed that we had a self-intersection at one point and two strands that were concerningly close together. Though we’re not entirely sure why this happened, we think the Chamfer forced the two strands too close together. We first tried to make the radius of the circle inside the sweep smaller, but the strands were still touching at 0.17cm, so we decided to manually pull the spline points apart after the Chamfer but before the sweep. This fixed our problem and everything looked good after the sweep this time. Our completed 6_2 knot is pictured above.

As we built 6_3 we were worried about self-intersections, since the (x,y) coordinates were the same as in 6_2. We did not run into any problems this time, perhaps because of the different heights. Below is a picture of 6_3 from four angles, showing the “nice” petal view as well as the less attractive side views.

Different views of the petal 6_3 knot made in Cinema 4D.

# Trig Trials

Written by Claire Gilreath. Joanne Wang, and Selihom Gobeze (students in Math 383D Knot Theory Spring 2023).

# Finding the Coordinates of Petal Knots

We decided we wanted to try to 3D print the petal projections of knots for our project because they looked pretty and seemed like a challenge.

A petal projection of the trefoil knot.

Petal projections of knots are projections where all of the crossings are aligned through a single line in 3d space, giving the knot a flower-like appearance. A two-dimensional petal projection of the trefoil (3_1) is pictured to the right (via this image from Wikipedia). We decided to work on creating petal versions of 3_1, 5_1, 6_2, and 6_3 knots. We relied heavily on the work described in Colin Adams’ “Knot Projections with a Single Multi-Crossing” paper. In the appendix of this paper, Adams and his coauthors described the order of each petal for the knots we hoped to construct. Also, we utilized the Wikipedia page  about petal projections of knots to better understand their structure.

We were a bit uncertain of the best method to create these knots, since as far as we knew, there were no equations we could plug into Cinema 4D, nor were there coordinates constructed by someone else. After talking to Professor Denne, we felt that our best bet would be creating our own sets of coordinates using trig and polar coordinates. We decided to find four points per petal: the center, the top of the arc of each petal, and the end of both edges. To do this we constructed a circle through the ends of the edges, shown in red in the diagram of the trefoil below.

This figure shows the details behind the computations needed to find the coordinates of points along the petal trefoil knot.

We then found the angle between each of the edges using the formula Pi/(number of petals), so for the trefoil, we had Pi/5. The length of each red edge was the radius of the circle, r, which we determined in a later step after finding the heights. Next, we added lines from the top of each petal to the center of our circle, shown in green, and found the angle between these lines using the formula 2Pi/(the number of petals), this was 2Pi/5 for the trefoil. To find the length of the green lines, we treated the curved part of the petal as a semicircle and found its radius by finding the length of the chord between the two red edges and dividing by 2. We then found the length of the other piece by using the Pythagorean theorem and added both measurements together. These calculations are shown in purple in the diagram above, where the length of the green line is given by the following formula: the square root of (r2.-(r*sin(Pi/5))2) plus r*sin(Pi/5). Using trig, this is r(cos(Pi/5)+sin(Pi/5)). We called this quantity x as we labeled the points of the trefoil in the diagram above.

The next step was to find the heights of each point and then convert everything into Cartesian coordinates in order to have points that Cinema4D could understand.

This figure shows the heights given to the points on the petal trefoil knot.

We used the orders from the last page of Adams’ paper which gave us the heights for the edges at (0,0). We also assigned heights to the points we found by averaging the heights of the edges to find the height for the top of the petal. We then averaged the height of the top of the petal and the edges at (0,0) to find the height of the end of each edge. For the trefoil, we came up with the heights to the right and then scaled them up by 1.5, so our model would be a bit larger after printing. This also guided our decision to make the radius of our circle (r) be 3 for 3_1 and 5_1. We used r = 3.5 for 6_2 and 6_3.

At this point, we had found the polar coordinates for all of the points we wished to plot in Cinema4D, so we converted them to decimal approximations with the help of WolframAlpha. The Cartesian coordinates of 3_1 are shown in the table below. We followed a similar process for 5_1, 6_2, and 6_3 knots. Points completed!

 Vertex x y z 1 0 0 6 2 -0.927050983 2.853169549 4.875 3 0 4.19040674 3.75 4 0.927050983 2.853169549 2.625 5 0 0 1.5 6 -0.927050983 -2.853169549 2.25 7 -2.463059283 -3.390110266 3 8 -2.427050983 -1.763355757 3.75 9 0 0 4.5 10 2.427050983 1.763355757 3.375 11 3.985313636 1.294906896 2.25 12 3 0 1.125 13 0 0 0 14 -3 0 0.75 15 -3.985313636 1.294906896 1.5 16 -2.427050983 1.763355757 2.25 17 0 0 3 18 2.427050983 -1.763355757 3.75 19 2.463059283 -3.390110266 4.5 20 0.927050983 -2.853169549 5.25 21 0 0 6

# The Figure-8 Knot and its Mirror Image

Written by Elizabeth Marshall, Mason Shelley, and Libby Kerr (students in Math 383D Knot Theory Spring 2023).

We built different models depicting how the figure-8 knot 41 is achiral, meaning the knot is equivalent to its mirror image. In this case, if we were given a knot diagram, its mirror image would be the same diagram with swapped crossings. We observe some knot theory facts: the figure-8 knot (41) is a 1 component alternating knot with a crossing number of 4 and an unknotting number of 1. The crossing number of a link is the minimum number of crossings needed in a diagram, and the unknotting number is the minimum number of times the knot must pass through itself before it becomes the unknot. While the knot looks trivial, its composition is surprisingly difficult to model.

Starting with a mirror image of the figure-8 knot (Step 1 below), each of the subsequent steps shows one or two R-moves that lead the knot back to its original state. An R-move is a simple manipulation of a piece of the knot in space, where a series of them can result in significant alteration of the knot. The steps are as follows:

Step2: Make an R-2 move that forms two “over” crossings on top of the figure-8 shape of the knot.

Step 3: Make an R-3 move that brings the arc on the left of the figure-8 shape to the center of the diagram.

Step 4: Make an R-2 move that pulls the now central arc over to the right of the main shape.

Step 5: Complete an R-3 and an R-1 move that form an outer arc to the top right.

Step 6: Finally, make an R-1 move that results in the figure-8 knot.

Step 7: Flip the figure in Step 6 180 degrees. This move reveals the figure-8 knot with crossings that are the opposite of those seen in Figure 1.

We used Cinema 4D to develop our models. On this software, we used the drawing tool to build our original knot shown in Step 1. To draw the subsequent steps representative of the R-moves we manipulated the knot from the previous step, scaled it appropriately, and checked to make sure all crossings were correct.

Using new software presented numerous challenges that we faced throughout the project. We initially found parametric equations on Wikipedia for the figure-8 knot that resulted in the wrong knot when visualized in Mathematica and then resulted in a simple line in space when entered in Cinema 4D. This meant that the this parameterization was not going to work. Without these equations, we were forced to draw the knot using the spline pen. It took nearly 10 attempts to draw the first image of the knot but eventually, we made a cohesive, recognizable figure-8 knot.

Using this tool, created one big challenge with regards to placing the points for future manipulation during the design process. Our most successful strategy was to draw the structure of the knot in 2D (holding the z-axis flat) and manipulating the points to create the over/under crossings after the knot was drawn.

Another challenge was figuring out how to properly demonstrate the transition of the knot’s mirror image back to its original state. In order to make the manipulations clear, we made a video showing the use of R-moves to visualize how the diagrams were interconnected.

Towards the end of our design process, we decided to take out the 6th step (shown above) because there was too much similarity between steps 6 and 7 (final model); that is why you’ll see 7 steps and only 6 models.

Overall, the main challenge that we ran into was using the software and properly putting our ideas into the diagrams. A massive help to our group was Dave Pfaff (IQ Center WLU), who is the expert that helped us navigate Cinema 4D. Thank you so much Dave, all of your help is greatly appreciated.