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.

Multirationals and Polytopic Groups

Any \((n+1)\)-rational number can be written in the following equivalent forms:

\[r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n}=r^1*d_1^s*d_2^{s^2}*\dots*d_{n}^{s^{n}}\]

where \({\sum_{k=0}^{n}s^k}=0\) so that the scaling rule for multirationals applies (i.e. \(x^1*x^s*x^{s^2}*\dots*x^{s^{n}}=1\) for any positive non-zero real number \(x\)). This means the elements of the set \(\{1,s,s^2,\dots,s^{n}\}\) are equally distributed and centered around the origin.

I already showed in Part 3 of this series that for \(n>2\), this set cannot contain pairs of additive inverses \(s^k\) and \(-s^k\) or pairs of conjugates \(s^j\) and \((\overline{s^j})\). Thus, the \(n\)th roots of unity in \(\Bbb{C}\) will NOT suffice here.

There is, however, one family of cyclic groups of order \(n\) that contains no pairs of additive inverses for \(n>1\): the polytopic groups of regular simplexes. This is because for any regular simplex with more than one dimension, there is no such thing as an opposite pair of vertices. As long as the simplex is centered at the origin, there is no way for any two vertices to be the additive inverse of each other unless the simplex is just a line segment (\(1\)-simplex).

Incidentally, we already know that multirationals use simplexes for \(n=1\) and \(n=2\): The \(2\)-rationals are based on the polytopic group of a \(1\)-simplex, while the \(3\)-rationals are based on that of a regular \(2\)-simplex.

In other words, the set \(\{1,s,s^2,\dots,s^{n}\}\) used for \((n+1)\)-rational numbers is a polytopic group of a regular \(n\)-simplex centered at the origin. Let us call this an \(\bf{n}\)-simplexic group.

Simplexic Groups in \(\Bbb{X}_{n}\)

Let us look for \(n\)-simplexic groups in \(\Bbb{X}_{n}\). We know that such a group has order \(n+1\), so raising an element of the simplexic group to the \((n+1)\)st power just gives \(1\). In matrix form:

\[\left(
\begin{array}{ccccc} c_1 & -c_n & \dots & -c_3 & -c_2 \\ c_2 & c_1 & -c_n & \dots & -c_3\\ \vdots & c_2 & c_1 & \ddots & \vdots \\ c_{n-1} & & \ddots & \ddots & -c_n \\ c_n & c_{n-1} & \dots & c_2 & c_1\\\end{array}\right)^{n+1}=\left(
\begin{array}{ccccc} 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 0 & \dots & 0\\ \vdots & 0 & 1 & \ddots & \vdots \\ 0 & & \ddots & \ddots & 0 \\ 0 & 0 & \dots & 0 & 1\\\end{array}\right)\]

Where \((c_1,c_2,c_3,\dots,c_{n})\) are the Cartesian coordinates of an element of the simplexic group.

Since that equation could have more than one solution, we need to use methods of narrowing these coordinates down to the solution representing an \(n\)-simplex. For example, we know that the elements of the simplexic group must be centered at the origin. If a regular \(n\)-simplex is centered at the origin with one vertex at \(1\), then the other vertices must have the same real component \(\bf{-\frac{1}{n}}\), because the vertex-center-vertex angle of any regular \(n\)-simplex is \(\arccos(-\frac{1}{n})\). So we have \(c_1=-\frac{1}{n}\).

Once we obtain the solutions for \((c_1,c_2,c_3,\dots,c_{n})\), we can pick one generator of \(n+1\) affinely independent vectors centered at the origin. We can use these \(n+1\) group elements to create an \(n\)-rational number:

\[r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n}=r^1*d_1^s*d_2^{s^2}*\dots*d_{n}^{s^{n}}\]

Example: Looking for the \(\bf{3}\)-simplexic group in \(\boldsymbol{\Bbb{X}_3}\)

For \(n=3\), we know that \(c_1=-\frac{1}{3}\), so we have this equation:

\[\left(
\begin{array}{ccc} -\frac{1}{3} & c_3 & c_2 \\ c_2 & -\frac{1}{3} & c_3\\ c_3 & c_2 & -\frac{1}{3}\\\end{array}\right)^4=\left(
\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1\\\end{array}\right)\]

If you entered that in Wolfram Alpha, you'll get this:

Again, for each of those three solutions, \(c_1=-\frac{1}{3}\). These three, plus the trivial solution \(1\), form the \(3\)-simplexic group in \(\Bbb{X}_{3}\). One of the generators is the following:

\begin{align}
s&=-\frac{1}{3}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3}\right)j + \left(-\frac{1}{3}-\frac{1}{\sqrt{3}}\right)k \\
&=-\frac{1}{3}+\frac{\sqrt{3}-1}{3}j-\frac{\sqrt{3}+1}{3}k
\end{align}

where \(\{1,j,k\}\) is the basis group of \(\Bbb{X}_{3}\).

With a \(3\)-simplexic group in hand, we can create a \(4\)-rational number for any quadruplet ratio \((r:d_1:d_2:d_3)\) using this equation:

\[r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}d_3=r^1*d_1^s*d_2^{s^2}*d_{3}^{s^{3}}\]

Other Examples

For \(\Bbb{X}_{1}\) or \(\Bbb{R}\), the \(1\)-orthoplexic group \(\langle{-1}\rangle\) is also a \(1\)-simplexic group that can be used to create \(2\)-rational numbers (fractions of real numbers). \(\langle{-1}\rangle\) of course is just \(\{-1,1\}\).

For \(\Bbb{X}_{2}\) or \(\Bbb{C}\), the third roots of unity \(\langle{-\frac{1}{2} + i\frac{\sqrt{3}}{2}}\rangle\) form a \(2\)-simplexic group that can be used to create \(3\)-rational numbers.

For \(\Bbb{X}_{3}\) with basis \(\{1,j,k\}\), we get the aforementioned \(\left\langle{-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{\sqrt{3}+1}{3}k}\right\rangle\).

For \(\Bbb{X}_{4}\) with basis \(\{1,h,i,ih\}\), there are (surprisingly enough) two cyclic groups of order five that are \(4\)-simplexic groups, namely

\[\left\langle{-\frac{1}{4} + \frac{q}{4}h + \left(\frac{1}{4} - \frac{11q^2}{8} + \frac{9q^4}{20} - \frac{q^6}{40}\right)i + \left(\frac{23q}{8} - 4q^3 + \frac{39q^5}{40} - \frac{q^7}{20}\right)ih}\right\rangle\]

where \(q=\sqrt{5 + \sqrt{5} \pm \sqrt{5(5 + 2\sqrt{5})}}\). These can be used to create two variants of \(5\)-rational numbers.

I don't understand four-dimensional coordinate geometry enough to be super confident about this, though. It's unfortunate that Wolfram Alpha doesn't seem powerful enough to calculate a \(5\)-simplexic group generator so I couldn't check if \(\Bbb{X}_{5}\) has multiple simplexic groups as well, so my journey of simplexic group discovery is paused for now.

To Be Continued...

Now we have a framework where basis elements are part of an orthoplexic group and multirationals are created from simplexic groups. I actually have an idea about how to incorporate the third family of regular polytopes (the hypercubes) into this framework. This idea is related to exponentials and I plan to include it in the next part. But I think for now I'll try to fill in some gaps in my knowledge first and see if I could improve my existing posts.