Triplex Numbers

A couple of years ago I stumbled upon this video by someone calling himself "Casual Graphman":

You really need to watch the video first before continuing, but basically the author constructed a unital algebra called the "triplex numbers". It's reminiscent of the split-complex numbers, except instead of \(j^2=1\) the author defines \(j^3=i^3=1,j^2=i,i^2=j\). This triplex number system is three-dimensional with basis \(\{1,i,j\}\) and each triplex number has the form \(a + bi + cj\), where \(a,b,c\) are real numbers.

Around the time I discovered this video, I was in the early stages of my abstract algebra adventures. By then the seeds of what would become the polytopic groups and polytopic algebras were just starting to germinate in my mind. I remember getting stuck trying to make sense of a group whose convex hull was a \(\bf{3}\)-simplex and being intrigued by this video because of the similarity between the third roots of unity (that form a triangle in the complex plane) and the basis of triplex space: both were cyclic groups of order three under multiplication.

The connection went even deeper when, to my utter astonishment, I learned that a polytopic group of a regular tetrahedron exists in triplex space, generated by \(\left\langle{-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}i-\frac{\sqrt{3}+1}{3}j}\right\rangle\):

The person I learned this from (anixx from the Math Stack Exchange website) used a slightly different notation: He preferred \(\bf{j}\) and \(\bf{k}\) instead of \(i\) and \(j\), probably to not cause ambiguity with complex numbers, so what he actually showed me was more like \(-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{\sqrt{3}+1}{3}k\). Here's what the group looks like:

\[\left\{-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{(\sqrt{3}+1)}{3}k,\quad-\frac{1}{3}+\frac{2}{3}j+\frac{2}{3}k,\quad-\frac{1}{3}-\frac{(\sqrt{3}+1)}{3}j+\frac{(\sqrt{3}-1)}{3}k,\quad{1}\right\}\]

I will use \(j\) and \(k\) for the rest of this post to make comparisons with complex numbers easier. And since it is a three-dimensional polytopic algebra based on a \(3\)-orthoplexic group (generated by  \(\left\langle{-j}\right\rangle\)), I will use the following notation for the triplex algebra: \(\bf{\Bbb{X}_3}\)

Casual Graphman was very thorough in analyzing the algebra he constructed. Not only did he show that \(\Bbb{X}_3\) contains zero divisors, but he also showed that the concept of the conjugate in this algebra is not an involution, making this a very weird algebra indeed. Not only that, but he also gave matrix representations of the basis elements, as well as closed form exponential functions for both the standard basis and the "diagonal" basis.

The Circle Group and the Cylinder Group of \(\Bbb{X}_3\) 

The exponential function he derived became very useful for me later on, when I discovered that just as how \(\exp({\theta{i}})\) always lands on the unit circle in the complex plane, the expression \(\exp\left(\theta\frac{j-k}{\sqrt{3}}\right)\) always lands on the specific circle that touches the basis elements \(\{1,j,k\}\). This is the circle group of \(\Bbb{X}_3\) and it has the following exponential map:

\[\exp\left(\theta\frac{j-k}{\sqrt{3}}\right)=\frac{1+j+k}{3}+\left(\frac{2-j-k}{3}\right)\cos(\theta)+\left(\frac{j-k}{\sqrt{3}}\right)\sin(\theta)\]

(Take note that \(\frac{j-k}{\sqrt{3}}\) is part of the "diagonal basis".)

Unlike the circle group of \(\Bbb{C}\), the circle group of \(\Bbb{X}_3\) is not centered at the origin and its radius is \(\sqrt{\frac{2}{3}}\) instead of \(1\), although both circle groups do share the property that all their elements are a unit length away from the origin. Also, both circle groups consist of generators of cyclic groups with a two-dimensional span (so they are 2-polytopic groups).

The circle group of \(\Bbb{X}_3\) is actually just half of a bigger multiplicative group which I will call the "cylinder group". This cylinder group contains both the circle group as well as the additive inverse of all the points of the circle group, which of course forms another circle on the opposite side of the space. This opposite circle generates polytopic groups as well, but their span is three-dimensional; in fact they generate a specific kind of polyhedron called an antiprism. The convex hull of all these antiprisms is a cylinder, hence the group's name.

Take note that both the \(3\)-orthoplex and the \(3\)-simplex are antiprisms, and they are indeed generated by points on this "additive inverse" circle: \(-j\) generates the \(3\)-orthoplexic group and \(-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{\sqrt{3}+1}{3}k\) generates the \(3\)-simplexic group of the space.

As for the other regular polyhedra, the cube, the dodecahedron, and the icosahedron, they are not antiprisms and so they don't have polytopic groups in \(\Bbb{X}_3\). Which is sad.

Cubic Coincidences

Interestingly enough, if you take the vertices of the \(3\)-simplex described above, as well as their additive inverses, the convex hull of that set is a cube. This is of course just basic 3D geometry: a cube's vertices can be seen as the vertices of two opposite tetrahedra. This multiplicative group of the cube's vertices is not even a cyclic group. Which, again, is sad.

On the other hand, there are interesting coincidences related to this cube. For one, \(\frac{j-k}{\sqrt{3}}\) is the midpoint of one of its edges, and the additive inverse is the midpoint of the opposite edge. There is therefore a deep connection between this cube and the exponential map of the circle group of \(\Bbb{X}_3\).

Other coincidences can be seen when we look at Casual Graphman's other diagonal basis elements:

1. \(\frac{1+j+k}{3}\) is the center of one of the cube's faces.
2. \(\frac{2-j-k}{3}\) is the midpoint of yet another edge of the cube.

The \(n\)-cube of \(\Bbb{X}_{n}\)

Those coincidences regarding the \(\Bbb{X}_3\) cube raises the question of whether there are \(n\)-cubes in other \(\Bbb{X}_{n}\) that are somehow related to exponentiation.

Of course, the connection between complex exponentiation and the fourth roots of unity (which form a \(2\)-cube, i.e. a square, in \(\Bbb{X}_2\)) is pretty obvious. That's just Euler's formula as it is most commonly known.

But how about in \(\Bbb{X}_4\)? Is there a hypercube in this four-dimensional algebra that is related to the exponential formula? Well, we know that the vertices of a hypercube can also be seen as the vertices of two \(4\)-orthoplexes, so perhaps the orthoplexic group of \(\Bbb{X}_4\) is just half of a multiplicative hypercube group. And since \(i\) is also in the orthoplexic group of \(\Bbb{X}_4\), we know that Euler's formula might once again connect the \(n\)-cube and the circle group in this space.

(Edit: I finally found the hypercube in \(\Bbb{X}_4\), and they are connected to the tessarine basis)

The Three Forms of Polytopic Algebras

This idea about \(n\)-cubes could be the final piece of a puzzle that I've been trying to solve for months now about how \(\Bbb{X}_n\) relates to the three families of regular polytopes.

In geometry, there are three families of regular polytopes that exist in any \(n\) dimensions:

• The \(n\)-simplex
• The \(n\)-orthoplex
• The \(n\)-cube

We've seen that a group whose convex hull is an \(n\)-orthoplex can be used to determine the orthonormal basis of \(\Bbb{X}_n\) and therefore define a Cartesian form.

Next, we've seen how another group whose convex hull is an \(n\)-simplex can be used to generate \((n+1)\)-rational numbers in \(\Bbb{X}_n\) and therefore define a homogeneous form.

Finally, if my hunch is correct, another group whose convex hull is an \(n\)-cube can be used to define a polar form for \(\Bbb{X}_n\).

Again, this puzzle is yet unsolved by me. I have no idea if this is a known problem in abstract algebra. I am not a mathematician. I wish an actual mathematician could help me with this.