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.
\(\boldsymbol{\Bbb{X}_{n}}\) and the Orthoplexic Groups
In geometry, an \(n\)-orthoplex is an \(n\)-polytope with \(2n\) vertices where each vertex has an opposite vertex, and each vertex is orthogonal to all other vertices except its opposite.
Let us define \(\{1, x_1, x_2, \dots, x_{n-1}\}\) as an orthonormal basis for an \(n\)-dimensional unital algebra where \(\{-x_{n-1},\dots,-x_2,-x_1, -1, 1, x_1, x_2, \dots, x_{n-1}\}\) is a cyclic group under multiplication.
This cyclic group contains \(2n\) elements: the \(n\) elements of the orthonormal basis including \(1\), and the \(n\) elements including \(-1\) that are additive inverses of the basis elements. Seen as vectors, each element is orthogonal to all other elements except their opposite. Thus, this is a polytopic group of an \(n\)-orthoplex. Let's call it an "\(\bf{n}\)-orthoplexic group".
Let \(\bf{\Bbb{X}_{n}}\) be the algebra of the \(n\)-orthoplexic group. We know that \(\Bbb{X}_{n}\) is an \(n\)-polytopic algebra because it is \(n\)-dimensional and it has an \(n\)-orthoplexic group that is, by definition, also \(n\)-dimensional.
We just found an infinite family of polytopic algebras. Hurray!
The next step is to discover how to multiply in \(\Bbb{X}_{n}\). This can be done by analyzing the generators of its orthoplexic group. We need to handle two cases: odd dimensions (\(\Bbb{X}_{2m+1}\)) and even dimensions (\(\Bbb{X}_{2m}\)). Let's take each case in turn.
Multiplication in \(\boldsymbol{\Bbb{X}_{2m+1}}\)
The motivating example for this is the one-dimensional algebra \(\Bbb{R}\), which we'll also call \(\Bbb{X}_{1}\) since \(\langle{-1}\rangle\) is a \(1\)-orthoplexic group in \(\Bbb{R}\).
A \((2m+1)\)-orthoplexic group \(Y\subset\Bbb{X}_{2m+1}\) is a cyclic group of order \(4m+2\) under multiplication, meaning if \(Y=\langle{-y}\rangle\) then \(-y\) is a primitive \((4m+2)\)-nd root of unity.
\[(-y)^{4m+2}=1\]
I'm marking the generator as negative for the sake of consistency, since the generator of \(\Bbb{X}_1\) is negative.
\[(-y)^{4m+2}=y^{4m+2}\text{ since the exponent is even, but}\]
\[(-y)^{2m+1}\boldsymbol{\neq}{y^{2m+1}}\text{ since the exponent is odd}\]
Both \((-y)^{2m+1}\) and \(y^{2m+1}\) must be second roots of unity, but since \((-y)^{2m+1}\neq{y^{2m+1}}\), only one of them could be equal to \(-1\), because there can only be one element of order two in a cyclic group of even order. So for these two elements of a cyclic group of order \(4m+2\), one will be equal \(-1\) and the other will be equal to the non-primitive second root of unity: \(+1\) itself.
Since \(-y\) is a primitive \((4m+2)\)-nd root of unity, \((-y)^{2m+1}\) cannot be \(+1\).
\[\therefore{(-y)^{2m+1}=-1},\quad{y^{2m+1}=+1}\]
Meaning \(-y\) is a \((2m+1)\)-st root of negative one, while \(y\) is a primitive \((2m+1)\)-st root of unity. In other words, \(y\) generates a cyclic subgroup of order \(2m+1\) under multiplication. Specifically, \(\langle{y}\rangle\) is an orthonormal basis of \(\Bbb{X}_{2m+1}\). This group is a polytopic group of a \(2m\)-simplex. Unlike the orthoplexic group generated by \(\langle{-y}\rangle\), this particular "simplexic" group generated by \(\langle{y}\rangle\) is not centered at the origin and its dimensionality is one less than that of \(\langle{-y}\rangle\).
With these facts, one can construct a Cayley table containing \(-y\) and its powers in order to define multiplication in \(\Bbb{X}_{2m+1}\).
& 1 & -y & y^2 & \dots & y^{2m} & -1 & y & -(y^2) & \dots & y^{4m} & -y^{4m+1} \\ \hline
1 & 1 & -y & y^2 & \dots & y^{-1} & -1 & y & -(y^2) & \dots & y^{-2} & -y^{-1} \\
-y & -y & y^2 & -y^3 & \dots & -1 & y & -(y^2) & y^3 & \dots & -y^{-1} & 1 \\
y^2 & y^2 & -y^3 & y^4 & \dots & y & -(y^2) & y^3 & -(y^4) & \dots & 1 & -y \\
\vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\
y^{2m} & y^{-1} & -1 & y & \dots & y^{-2} & -y^{-1} & 1 & -y & \dots & y^{-3} & -(y^{-2}) \\
-1 & -1 & y & -(y^2) & \dots & -y^{-1} & 1 & -y & y^2 & \dots & -(y^{-2}) & y^{-1} \\
y & y & -(y^2) & y^3 & \dots & 1 & -y & y^2 & -y^3 & \dots & y^{-1} & -1 \\
-(y^2) & -(y^2) & y^3 & -(y^4) & \dots & -y & y^2 & -y^3 & y^4 & \dots & -1 & y \\
\vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots & \vdots & \vdots & \dots & \vdots \\
y^{4m} & y^{-2} & -y^{-1} & 1 & \dots & y^{-3} & -(y^{-2}) & y^{-1} & -1 & \dots & y^{-4} & -y^{-3} \\
-y^{4m+1} & -y^{-1} & 1 & -y & \dots & -(y^{-2}) & y^{-1} & -1 & y & \dots & -y^{-3} & y^{-2} \\
\end{array}
Multiplication in \(\boldsymbol{\Bbb{X}_{2m}}\)
Just like in the odd case, the even case has a motivating example: \(\Bbb{X}_{2}=\Bbb{C}\).
A \(2m\)-orthoplexic group \(Z\subset\Bbb{X}_{2m}\) is a cyclic group of order \(4m\) under multiplication, meaning if \(Z=\langle{z}\rangle\) then \(z\) is a primitive \(4m\)-th root of unity.
\[z^{4m}=1\]
This means that \(z^{2m}\) is a second root of unity. Again, there can only be one element of order two in a cyclic group of even order, so \(z^{2m}=-1\), i.e. \(z\) is a \(2m\)-th root of negative one. This implies that \(z^m=\sqrt{-1}\) and that \(\Bbb{C}\subseteq\Bbb{X}_{2m}\). If \(m=1\), we get the motivating example \(\Bbb{X}_{2}=\Bbb{C}\).
We also see that \(z^{2m+1}=-a\) is also a generator of the orthoplexic group since \(2m+1\) and \(4m\) are relatively prime. So everything mentioned above regarding the generator \(z\) also applies to its additive inverse \(-z\).
Since \(z\) generates a cyclic group of order \(4m\) and is also an element of the orthonormal basis \(\{1,z,z^2,\dots,z^{2m-1}\}\) which has \(2m\) elements, this means an orthonormal basis of \(\Bbb{X}_{2m}\) does not form a group under multiplication. In fact, the multiplicative inverse of \(z\) is not in the basis:
\[z^{-1}=z^{4m-1}=-z^{2m-1}\]
With these facts, one can construct a Cayley table containing \(z\) and its powers in order to define multiplication in \(\Bbb{X}_{2m}\).
\begin{array}{r|rrrrrrrrrrrr}
& 1 & z & z^2 & \dots & z^{m-1} & \sqrt{-1} & \dots & z^{2m-1} & -1 & -z & \dots & z^{4m-1} \\ \hline
1 & 1 & z & z^2 & \dots & z^{m-1} & \sqrt{-1} & \dots & z^{2m-1} & -1 & -z & \dots & -z^{2m-1} \\
z & z & z^2 & z^3 & \dots & \sqrt{-1} & z^{m+1} & \dots & -1 & -z & -(z^2) & \dots & 1 \\
z^2 & z^2 & z^3 & z^4 & \dots & z^{m+1} & z^{m+2} & \dots & -z & -(z^2) & -z^3 & \dots & z \\
\vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots & \dots & \vdots & \vdots & \vdots & \dots & \vdots \\
z^{m-1} & z^{m-1} & \sqrt{-1} & z^{m+1} & \dots & z^{2m-2} & z^{2m-1} & \dots & -(z^{m-2}) & -(z^{m-1}) & -\sqrt{-1} & \dots & z^{m-2} \\
\sqrt{-1} & \sqrt{-1} & z^{m+1} & z^{m+2} & \dots & z^{2m-1} & -1 & \dots & -(z^{m-1}) & -\sqrt{-1} & -(z^{m+1}) & \dots & z^{m-1} \\
\vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots & \dots & \vdots & \vdots & \vdots & \dots & \vdots \\
z^{2m-1} & z^{2m-1} & -1 & -z & \dots & -(z^{m-2}) & -(z^{m-1}) & \dots & -(z^{2m-2}) & -z^{2m-1} & 1 & \dots & z^{2m-2} \\
-1 & -1 & -z & -(z^2) & \dots & -(z^{m-1}) & -\sqrt{-1} & \dots & -z^{2m-1} & 1 & z & \dots & z^{2m-1} \\
-z & -z & -(z^2) & -z^3 & \dots & -\sqrt{-1} & -(z^{m+1}) & \dots & 1 & z & z^2 & \dots & -1 \\
\vdots & \vdots & \vdots & \vdots & \dots & \vdots & \vdots & \dots & \vdots & \vdots & \vdots & \dots & \vdots \\
z^{4m-1} & -z^{2m-1} & 1 & z & \dots & z^{m-2} & z^{m-1} & \dots & z^{2m-2} & z^{2m-1} & -1 & \dots & -(z^{2m-2}) \\
\end{array}
Matrix representations of \(\boldsymbol{\Bbb{X}_{n}}\)
Let \(v\in\Bbb{X}_{n}\) be of the following form:
\[v=c_1*1 + c_2*x + c_3*x^2 + \dots + c_{n}*x^{n-1}\]
where \(\{1,x,x^2,\dots,x^{n-1}\}\) is an orthonormal basis from the \(n\)-orthoplexic group and \((c_1,c_2,\dots,c_{n})\) are Cartesian coordinates of \(v\).
If \(n\) is even, then \(x\) and \(v\) can be seen as \(n\times{n}\) Toeplitz matrices of the following form:
\[x=\left(
\begin{array}{ccccc} 0 & 0 & \dots & 0 & -1 \\ 1 & 0 & 0 & \dots & 0\\ 0 & 1 & 0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & 0 & 1 & 0\\\end{array}\right)\]
\begin{array}{ccccc} 0 & 0 & \dots & 0 & -1 \\ 1 & 0 & 0 & \dots & 0\\ 0 & 1 & 0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & 0 & 1 & 0\\\end{array}\right)\]
\begin{array}{ccccc} c_1 & -c_n & \dots & -c_3 & -c_2 \\ c_2 & c_1 & -c_n & \dots & -c_3\\ c_3 & c_2 & c_1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & -c_n \\ c_n & \dots & c_3 & c_2 & c_1\\\end{array}\right)\]
The classic example of this is of course \(\Bbb{X}_{2}=\Bbb{C}\), where \(c_1+c_2i\) has the matrix form \(\left(\begin{array}{cc} c_1 & -c_2 \\ c_2 & c_1\\\end{array}\right)\).
On the other hand, if \(n\) is odd, \(x\) and \(v\) can instead be represented by circulant matrices (so still Toeplitz but of a simpler kind):
\[x=\left(
\begin{array}{ccccc} 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0\\ 0 & 1 & 0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & 0 & 1 & 0\\\end{array}\right)\]
\begin{array}{ccccc} 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0\\ 0 & 1 & 0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & \dots & 0 & 1 & 0\\\end{array}\right)\]
\[v=\left(
\begin{array}{ccccc} c_1 & c_n & \dots & c_3 & c_2 \\ c_2 & c_1 & c_n & \dots & c_3\\ c_3 & c_2 & c_1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & c_n \\ c_n & \dots & c_3 & c_2 & c_1\\\end{array}\right)\]
\begin{array}{ccccc} c_1 & c_n & \dots & c_3 & c_2 \\ c_2 & c_1 & c_n & \dots & c_3\\ c_3 & c_2 & c_1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & c_n \\ c_n & \dots & c_3 & c_2 & c_1\\\end{array}\right)\]
Determinant and non-uniqueness
For any \(\Bbb{X}_{n}\), the determinant of an orthonormal basis element is always \(1\).
Take note that the matrices above are not unique. For example, the odd case could also be represented by the following:
\begin{array}{ccccc} 0 & 0 & \dots & 0 & 1 \\ -1 & 0 & 0 & \dots & 0\\ 0 & -1 & 0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & \dots & -1 & 0\\\end{array}\right)\]
\begin{array}{ccccc} c_1 & -c_n & \dots & -c_3 & c_2 \\ -c_2 & c_1 & -c_n & \dots & -c_3\\ c_3 & -c_2 & c_1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & -c_n \\ c_n & \dots & c_3 & -c_2 & c_1\\\end{array}\right)\]
Where diagonals alternate between positives and negatives. Here the determinant of \(x\) is still \(1\).
Other polytopic groups in \(\boldsymbol{\Bbb{X}_{n}}\)
Now that we know how to construct \(\Bbb{X}_{n}\) using an \(n\)-orthoplexic group, let's look for other polytopic groups in \(\Bbb{X}_{n}\).
Given an orthonormal basis \(\{1,x,x^2,\dots,x^{n-1}\}\) in \(\Bbb{X}_{n}\), let \((c_1, c_2, c_3,\dots,c_{n})\) be the Cartesian coordinates (using the given basis) of a generator \(s\in\Bbb{X}_{n}\) of a group of order \(m\):
s&=c_1 + c_2*x + c_3*x^2 + \dots + c_{n}*x^{n-1} \\
&=\sum_{k=1}^{n}c_{k}*x^{k-1}
\end{align}
(Where \(\langle{s}\rangle=\{1,s,s^2,\dots,s^{m-1}\}\).)
So \(\langle{s}\rangle\) is a polytopic group of a polytope with \(m\) vertices. We know the properties of the basis given the parity of \(n\), but in order to find values for \(s\) for a given \(m\) we will also need to determine coordinates that fit the formula.
I'm not a mathematician and I have no idea how to get the coordinates by hand, but you can use Wolfram Alpha to solve this problem via a straightforward method: Just use the matrix form of \(s\) and see what Wolfram Alpha spits out. For example, for \((n=3,m=4)\):
\begin{array}{ccc} c_1 & c_3 & c_2 \\ c_2 & c_1 & c_3\\ c_3 & c_2 & c_1\\\end{array}\right)^4=\left(
\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1\\\end{array}\right)\]
This is just the matrix form of the expression \(s^4=1\). Put something equivalent to that in Wolfram Alpha and it will provide eight different real solutions. Those solutions all generate groups whose order is a factor of \(4\) and can therefore be said to be "fourth roots of unity", but if you graph them you'll see that their dimensionality differs. For example, the convex hull of \(\langle{\frac{1}{3}+\frac{1+\sqrt{3}}{3}x+\frac{1-\sqrt{3}}{3}x^2}\rangle\) is a square (2-polytope), while that of \(\langle{-\frac{1}{3}+\frac{\sqrt{3}-1}{3}x-\frac{\sqrt{3}+1}{3}x^2}\rangle\) is a tetrahedron (3-polytope).
Wolfram Alpha finds it harder to solve this problem for higher values of \(n\). In fact, even the Pro version does not seem powerful enough to solve for \((n=5,m=6)\) (for example), which is a shame. Perhaps someone with a better understanding of these kinds of matrices can help me out. 😁
Significance and Use
For \(n>2\), it is a sad fact that \(\Bbb{X}_{n}\) is not a composition algebra. This is because a non-degenerate quadratic form (which is the "norm" of a composition algebra) requires all basis elements to square to a real number, which of course does not apply for all polytopic algebras above \(\Bbb{C}\). This might make higher dimension polytopic algebras less interesting to some mathematicians.
There's a video about what is essentially \(\Bbb{X}_{3}\), where the author pointed out that the determinant of the matrix representation can very well be considered a "norm" for these kinds of algebras, and that a sort of "conjugate" can be obtained from that. He also derived a closed form exponential function for \(\Bbb{X}_{3}\) that some commenters praised for its significance to differential equations.
All of that is interesting. However, there is one particular reason why I even developed the idea of polytopic algebras in the first place.
I hinted in Part 3 that we can create multirational numbers in \(\Bbb{X}_{n}\). To me, this is the main motivation for everything I've written above. The section on finding other polytopic groups in \(\Bbb{X}_{n}\) already provides a hint, but in the next part of this series I will describe how exactly we can use polytopic groups to create multirational numbers.
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