Abstract Algebra Adventures Part 6: The n-cubic groups

When it comes to Group Theory, this series has so far focused on cyclic groups. We've talked much about the orthoplexic groups and the simplexic groups, both of which are cyclic.

But we cannot talk of two infinite polytope families (the orthoplexes and the simplexes) without talking about the third infinite polytope family: the n-cubes. It's just that, well, in the context of polytopic algebras, n-cubes are a bit more special, and it's taken me much longer to figure out how they fit in our puzzle.

So, how do we go about constructing groups whose convex hulls are n-cubes? Do such groups even exist in \(\mathbb{X}_n\)?

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: \(\bf{\Bbb{X}_3}\)

Abstract Algebra Adventures Part 5: Homogeneous Forms of \(\Bbb{X}_{n}\)

Previously I discussed a family of algebras with \(n\) dimensions with finite cyclic subgroups that also span \(n\) dimensions. These are what I call polytopic algebras (denoted as \(\Bbb{X}_{n}\)) because the convex hull of those cyclic subgroups are \(n\)-polytopes.

"Polytopic group" is what I call any finite cyclic group that exists in a unital algebra. A polytopic group has \(n\) dimensions if its elements span an \(n\)-dimensional subspace. An \(n\)-dimensional algebra is only polytopic if it has \(n\)-dimensional polytopic groups.

I also showed a general way of finding polytopic groups in \(\Bbb{X}_{n}\), and I claimed that we could use \(n\)-dimensional polytopic groups as the basis of \((n+1)\)-rational numbers. In this post I'll try to prove that claim.

Abstract Algebra Adventures Part 4: Polytopic Algebras

In the complex plane, a regular convex \(n\)-sided polygon is the convex hull of a finite cyclic group of order \(n\) under multiplication. This cyclic group is of course the \(n\)th roots of unity in \(\Bbb{C}\).

Given this relationship between convex \(2\)-polytopes and roots of unity in that \(2\)-dimensional algebra, let's try to construct a family of algebras such that if a member algebra is \(n\)-dimensional, it has cyclic subgroups under multiplication whose convex hull is an \(n\)-polytope.

Aside from the aforementioned case with polygons, mathematicians don't usually associate cyclic groups with polytopes. Rather, it is the symmetry groups that commonly comes to mind whenever polytopes are mentioned in the context of abstract algebra. But I'd like to show that there is a very natural notion of dimensionality within finite cyclic groups under any unital algebra's multiplication operation, and that it's not always just the two dimensions of the roots of unity in the complex plane.

Polytopic Groups and Algebras

Let's define a "polytopic group" to be a finite cyclic group whose group operation is the multiplication operation of a certain algebra. The span of the group's elements within the algebra's vector space determines the polytopic group's dimensionality, and its convex hull is a polytope with the same dimensionality.

Let's also define an "\(\bf{n}\)-polytopic algebra" as an \(n\)-dimensional real algebra with \(n\)-dimensional polytopic subgroups. Meaning, the dimensionality of the algebra and its highest-dimensional polytopic subgroup is the same.

We can thus say that \(\Bbb{C}\) is a \(2\)-polytopic algebra and \(\Bbb{R}\) is a \(1\)-polytopic algebra. Unfortunately, \(\Bbb{H}\) (the quaternions) and Cayley-Dickson constructions derived from it are NOT polytopic algebras, but they do have \(2\)-polytopic subalgebras. Also, all unital algebras contain the \(1\)-polytopic subalgebra \(\Bbb{R}\).

The following is my attempt at constructing other polytopic algebras. Note that I will be focusing on algebras over the field of real numbers, but you are free to try this exercise for other fields.