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 \(\mathbb{X}_n\)?

Of course, we know that a 1-cube exists in \(\mathbb{X}_1\). That's just a line segment (same as the 1-orthoplex and the 1-simplex) and is thus covered by \(\{1,-1\}\), a cyclic group of order two.

In two dimensions, we have the happy coincidence that the 2-cube is the same as the 2-orthoplex (a square), and is covered by the 2-orthoplexic group \(\{1,i,-1,-i\}\), which is a cyclic group of order four in \(\mathbb{X}_2\).

The 3-cubic group

Things get a little interesting in \(\mathbb{X}_3\), as the cube is not a convex hull of any cyclic group there. Instead, we need to realize another coincidence: the 3-simplex is the demicube, and so we can take the union of the 3-simplexic group in \(\mathbb{X}_3\) and the additive inverses of its elements and come up with a cube. Luckily, this superset is also a group. Not a cyclic group, but still a finite abelian group. If you look closely enough, you'll notice that this cubic group is isomorphic to the direct product of cyclic groups \(C_2\times{C_4}\).

Incidentally, \(\mathbb{X}_3\) itself is also a direct product, specifically it is isomorphic to \(\mathbb{X}_1\times\mathbb{X}_2\cong\mathbb{R}\times\mathbb{C}\). We know this because the basis group of \(\mathbb{X}_3\) makes it isomorphic to the group algebra \(\mathbb{R}[C_3]\), and it is known that \(\mathbb{R}[C_n]\cong\mathbb{R}\times\mathbb{C}^{(n-1)/2}\) when n is an odd number.

So clearly it makes sense to define polytopic groups to include other finitely-generated multiplicative abelian groups in an algebra, not just cyclic groups in an algebra.

The 4-cubic group

We've already seen that the \(C_2\) in \(\mathbb{X}_1\), the \(C_4\) in \(\mathbb{X}_2\), and the \(C_2\times{C_4}\) in \(\mathbb{X_1}\times\mathbb{X_2}\) coincidentally all form n-cubes. So, where is the next n-cube? Specifically, where is the 4-cube/tesseract in \(\mathbb{X}_4\)?

Since the 4-orthoplexic group in \(\mathbb{X}_4\) is generated by a fourth root of negative one, it is isomorphic to \(\mathbb{R}[X]/(X^4+1)\cong\mathbb{X}_2\times\mathbb{X}_2\). Given the relations between cyclic groups and algebras described above, \(\mathbb{X}_2\times\mathbb{X}_2\) suggests that we look for a group isomorphic to \(C_4\times{C_4}\) in \(\mathbb{X}_4\) and see if it is a 4-cube.

In order to do this more simply and elegantly, we can use an orthonormal basis different from the 4-orthoplexic group. You see, there is another algebra isomorphic to \(\mathbb{X}_2\times\mathbb{X}_2\): The tessarines. The basis elements used in the tessarines are denoted \(\{1,i,j,k\) where \(i^2=k^2=-1,i^2=1,ij=k\). 

Given the \(\mathbb{X}_4\) basis \(\{1,\chi_4,\chi_4^2,\chi_4^3\}\) (more on this notation here), we have the following mappings from the orthoplexic basis to the Tessarine basis:

\[i=\chi_4^2\]

\[j=\frac{\chi_4-\chi_4^3}{\sqrt{2}}\]

\[k=\frac{\chi_4+\chi_4^3}{\sqrt{2}}\]

In the Tessarine basis, the following is a group isomorphic to \(C_4\times{C_4}\):

\[\{\pm 1,\pm i,\pm j,\pm k, \pm(\frac{1+i+j-k}{2}), \pm(\frac{1+i-j+k}{2}), \pm(\frac{1-i+j+k}{2}), \pm(\frac{-1+i+j+k}{2})\}\]

These sixteen group elements form the vertices of the 4-cube that we're looking for. Since \(\mathbb{X}_4\) is isomorphic to the tessarines, we know that this 4-cube exists there as well.

The n-cube conjecture

In any polytopic algebra \(\mathbb{X}_n\), there is an abelian group \(\Gamma_n\) under multiplication that's defined by the following relations:

\[\Gamma_1\cong{C_2}\text{ (cyclic)}\]
\[\Gamma_2\cong{C_4}\text{ (cyclic)}\]
\[\Gamma_n\cong\begin{cases}
C_2\times{C_4^{m}}, & \text{if $n=2m+1$ (odd)} \\
C_4^{m}, & \text{if $n=2m$ (even)}
\end{cases},\forall{m\in\mathbb{N}}\]

Here's a conjecture: In any \(\mathbb{X}_n\), the convex hull of \(\Gamma_n\) is an n-cube.