4
$\begingroup$

(Update: I ended up posting two answers for this, each addressing different things. I feel the second one is more deserving to be accepted if nobody posts a better answer.)

Aside from the complex numbers, the triplex numbers, and their isomorphisms, I'd like to know which other unital algebras with dimensions greater than one follow these rules:

  1. The algebra must contain non-trivial finite cyclic groups under multiplication aside from $\{-1,1\}$. This means the algebra has more than two "roots of unity".

  2. Given any such cyclic group, the elements of the group must form some or all the vertices of a single regular convex polytope instance in the underlying vector space of the algebra. For a cyclic group $G$ that forms vertices of a polytope instance $P$, let's call $G$ a "polytopic group" of $P$. If $G$ covers all the vertices of $P$, let's call it a "full polytopic group" of $P$.

Since only $n$-simplexes, $n$-orthoplexes, and $n$-orthotopes have regular convex variants in all dimensions $n$, we can restrict ourselves to those polytope families, but let us add a third rule:

  1. If the algebra is $n$-dimensional, all regular convex simplexes, orthoplexes, and orthotopes of $n$ dimensions or less must have instances in the underlying vector space such that, for any such polytope instance $P$, there are polytopic groups of $P$ that collectively cover all its vertices.

Let us call such an algebra a "highly polytopic algebra".


For $n=2$, as implied, $\Bbb{C}$ is a highly polytopic algebra. The imaginary unit $i$ and its powers are a full polytopic group of a square (which is both a 2-orthoplex and a 2-orthotope), and the third roots of unity in the complex plane are a full polytopic group of an equilateral triangle (2-simplex).

I don't think split-complex numbers have polytopic groups that cover all three vertices of an equilateral triangle, and dual numbers don't even have that for a square, but please correct me if I'm wrong.


For $n=3$, I believe the three-dimensional triplex numbers (which, I've been told, is isomorphic to $\mathbb R\times \mathbb C$) is also highly polytopic, as I'll try to demonstrate below.

To lessen confusion, let's denote the two non-real elements of the triplex basis as $j$ and $k$. Triplex number multiplication is commutative and has $j^3=k^3=jk=1$, $j^2=k$, and $k^2=j$. We find the following polytopic groups in triplex space:

$3$-orthoplex

$1$, $j$, $k$ and their negatives form a full polytopic group of a regular octahedron, with its generator elements being $-j$ and $-k$:

$$-j\to{k}\to{-1}\to{j}\to{-k}\to{1}$$ $$-k\to{j}\to{-1}\to{k}\to{-j}\to{1}$$

$2$-orthoplex/$2$-orthotope

Side note: The non-unit circle that touches $1$, $j$, and $k$ and with radius $\sqrt{\frac{2}{3}}$ shares one important property with the unit circle on the complex plane: A full polytopic group of any regular polygon can be generated on the circle using an exponential formula. For the triplex numbers, a generator for an $n$-sided polygon is equals to $e^{\frac{2\pi}{n\sqrt{3}}(j-k)}$.

Using that formula, we find that $s=\frac{1}{3}-\frac{(\sqrt{3}-1)}{3}j+\frac{\sqrt{3}+1}{3}k$ is a generator of a square on the circle described above.

$$\left(s=\frac{1}{3}-\frac{(\sqrt{3}-1)}{3}j+\frac{\sqrt{3}+1}{3}k\right)\to{\left(s^2=-\frac{1}{3}+\frac{2}{3}j+\frac{2}{3}k\right)}\to{\left(s^3=\frac{1}{3}+\frac{\sqrt{3}+1}{3}j-\frac{(\sqrt{3}-1)}{3}k\right)}\to{\left(s^4=1\right)}$$

$3$-orthotope

The square we just described above is a face of a cube centered at the origin in triplex space. The other vertices of the cube are the negatives of $s$ and its powers. Notably, $-s=-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{(\sqrt{3}+1)}{3}k$ is equivalent to the number $c$ I mentioned in this question about what I call "cubic numbers".

This cube's vertices are covered by four polytopic groups: $\{s, s^2, s^3, 1\}$, $\{-s, s^2, -s^3, 1\}$, $\{-s^2, 1\}$, and $\{-1, 1\}$, but perhaps there's a full polytopic group of a cube there somewhere that I didn't find.

$3$-simplex

$-s$ is also a generator of a full polytopic group of one of the demicubes of the cube described above:

$$\left(-s=-\frac{1}{3}+\frac{(\sqrt{3}-1)}{3}j-\frac{(\sqrt{3}+1)}{3}k\right)\to{\left(s^2\right)}\to{\left(-s^3=-\frac{1}{3}-\frac{(\sqrt{3}+1)}{3}j+\frac{(\sqrt{3}-1)}{3}k\right)}\to{\left(1\right)}$$

A demicube is of course a regular tetrahedron

$2$-simplex

This one's the easiest: The third roots of unity $1$, $j$, and $k$ form a full polytopic group of an equilateral triangle. This could also be seen by evaluating $e^{\frac{2\pi}{3\sqrt{3}}(j-k)}$.

You could check out the above-mentioned triplex polytopes in this geogebra link. (The point labeled "A" is meant to be the multiplicative identity.)


Questions:

Is there any highly polytopic algebra in four dimensions? I know $\Bbb{H}$ is not highly polytopic because quaternion powers are always co-planar with each other and the origin, so $\Bbb{H}$ doesn't have polytopic groups of regular polyhedra and polychora aside from the 3-orthoplex and the 4-orthoplex.

I ultimately would like to know if there's an algorithm to construct highly polytopic algebras in higher dimensions.

Also, unfortunately I can't seem to find a full polytopic group of a cube in triplex space. If there is one, perhaps I could make a stronger definition of "polytopic group" that is synonymous to full polytopic groups. Is it possible that I just missed one?


Motivation: I'm mainly interested in full polytopic groups of $n$-simplexes because they relate to other ideas I've had in the past, particularly the $n$-rational numbers, but I figured that adding the other polytope families might potentially be more interesting (and perhaps more useful?) for others.

$\endgroup$
7
  • $\begingroup$ Do you require associativity, commutativity ? $\endgroup$ Sep 11, 2022 at 11:22
  • $\begingroup$ @CaptainLama I guess for now we could restrict it to commutative and associative solutions. Are there solutions outside of that? $\endgroup$ Sep 12, 2022 at 4:24
  • $\begingroup$ Did you consider tessarines (isomorphic to $\mathbb C\times\mathbb C$)? What about split-quaternions ($2\times2$ matrices), although they are not commutative? $\endgroup$
    – Anixx
    Sep 13, 2022 at 12:02
  • $\begingroup$ @Anixx I know by their Cayley tables that they have full polytopic groups of polygons just like complex numbers, but I haven't studied them enough yet to notice polyhedra or polychora (aside from trivial orthoplexes), and as four-dimensional algebras they will need to have more of those to count as per Rule 3. $\endgroup$ Sep 13, 2022 at 15:08
  • $\begingroup$ @Anixx I wrote another answer detailing a four-dimensional case. Would you mind checking it for mistakes? $\endgroup$ Sep 28, 2022 at 21:58

2 Answers 2

2
$\begingroup$

In triplex space, the circle $C_{p}$ passing through $1$, $j$, $k$ and another circle $C_{a}$ passing through $-1$, $-j$, and $-k$ both have radius $\sqrt{\frac{2}{3}}$, are parallel to each other, and form a cylindrical region $C$ where all the triplex polytopic groups we've seen exist.

If you look at the polyhedra generated in this "unit cylinder" region for long enough, you'll notice something pretty important: elements of polytopic groups appear either on $C_{p}$ (which only ever generates polygons, as it is analogous to the unit circle on the complex plane) or on $C_{a}$. I don't have proof that no other cyclic groups exist outside those two circles, but if it's true then it would explain why there don't seem to be polytopic groups for regular icosahedra and dodecahedra: it's impossible to fit all their vertices on just two parallel circles.

But we're only here for simplexes, orthoplexes, and orthotopes, anyway, so let's talk about how cubes fit in the circles of $C$. Let's look at $x = -\frac{1+\sqrt{2}}{3} + j\frac{\sqrt{6}-2+\sqrt{2}}{6} - k\frac{\sqrt{6}+2-\sqrt{2}}{6}$, which is a point on $C_{a}$ that generates a full polytopic group of a polyhedron with eight vertices:

$$\left(x = -\frac{1+\sqrt{2}}{3} + j\frac{\sqrt{6}-2+\sqrt{2}}{6} - k\frac{\sqrt{6}+2-\sqrt{2}}{6}\right)\to{\left(x^2 = s\right)}\to{\left(x^3 = \frac{\sqrt{2}-1}{3} - j\frac{2+\sqrt{2}-\sqrt{6}}{6} - k\frac{2+\sqrt{2}+\sqrt{6}}{6}\right)\to{\left(x^4 = s^2\right)}}\to{\left(x^5 = \frac{\sqrt{2}-1}{3} - j\frac{2+\sqrt{2}+\sqrt{6}}{6} - k\frac{2+\sqrt{2}-\sqrt{6}}{6}\right)}\to{\left(x^6 = s^{3}\right)}\to{\left(x^7 = -\frac{1+\sqrt{2}}{3} - j\frac{\sqrt{6}+2-\sqrt{2}}{6} + k\frac{\sqrt{6}-2+\sqrt{2}}{6}\right)}\to{\left(x^8 = 1\right)}$$

Very nice, except this polyhedron is not a cube. In fact, it's an anticube, a.k.a. a square antiprism. Incidentally, the tetrahedron and the octahedron, both generated by points on $C_{a}$, are also antiprisms that just so happen to be regular polyhedra as well.

Given the definition of the two circles, points on $C_{a}$ are negatives of points on $C_{p}$, so the even powers of any point on $C_{a}$ will always be on $C_{p}$ since $(-x)^2=x^2$ while their odd powers will always be on $C_{a}$. It's easy to see how this zigzagging between the two circles would naturally produce non-coplanar triangles that form antiprisms.

So while points on $C_{p}$ generate full polytopic groups of regular polygons, points on $C_{a}$ generate full polytopic groups of antiprisms. This is why I couldn't find a full polytopic group for a cube: Among the regular polyhedra, only the tetrahedron and the octahedron are antiprisms.

What's the significance of antiprisms? Well, I'm no mathematician, so I'll just quote from Wikipedia:

For example, as of 2001 it had been proven for only a limited number of non-trivial cases that the n-gonal antiprism is the mathematically optimal arrangement of $2n$ points in the sense of maximizing the minimum Euclidean distance between any two points on the set: in 1943 by László Fejes Tóth for 4 and 6 points (digonal and trigonal antiprisms, which are Platonic solids); in 1951 by Schütte and van der Waerden for 8 points (tetragonal antiprism, which is not a cube).[2]

So what I'm trying to get at is that I may have been aiming for the wrong goal. I feel it might be more mathematically interesting to find $n$-dimensional algebras with cyclic groups whose elements "maximize the minimum Euclidean distance between any two points" in the group for $n$ dimensions, whichever polytope their convex hull might be.

As for regular polytopes, well, at least a regular $n$-simplex, with its vertices equidistant to each other, will always be a viable solution here for $n+1$ points in $n$ dimensions.

But hey, finding hypercubes would still be awesome!

$\endgroup$
2
  • $\begingroup$ The algebra isomorphic to $n\times n$ square matrices is $n^2$ dimentional, not $n$ dimensional $\endgroup$
    – Anixx
    Sep 15, 2022 at 20:04
  • $\begingroup$ Thanks for pointing that out! I edited my answer. $\endgroup$ Sep 16, 2022 at 1:52
0
$\begingroup$

I believe I found a highly polytopic algebra in four dimensions. I'm posting this as a separate answer, and if it gains upvotes and nobody complains then I will mark this as the accepted answer. As for my previous answer, it can serve as an explanation as to why I didn't look for polytopic groups of a hypercube here.


First, let's look at some similarities between complex numbers and triplex numbers:

  • They both contain a full polytopic group of a regular $n$-orthoplex centered at the origin, where $n$ is the number of dimensions in the space. Let's call this the orthoplexic group.
  • For both algebras, the generators of the orthoplexic group are $n$th roots of $-1$, i.e., $i$ and $-i$ are second roots of $-1$ while $-j$ and $-k$ are third roots of $-1$.
  • Both algebras also have a full polytopic group of a regular $n$-simplex centered at the origin.

To generalize this for $n=4$, I conjectured that any four-dimensional unital algebra over the reals with a fourth root of $-1$ that's also a unit vector orthogonal to $1$ will not only have its own orthoplexic group but also a full polytopic group of a regular $4$-simplex centered at the origin.

So I tested my conjecture.


Let $h=\left( \begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\\end{array}\right)$. There are other matrix definitions for $h$ that produce the same results, but the important thing is that $h^{4}=-1$. This means $h^{2}$ squares to $-1$, so let's give $h^{2}$ the more familiar name $i$.

We can use $\{1,h,i,ih\}$ as the orthonormal basis of a four-dimensional unital associative commutative algebra over the reals. Let's call this algebra the "quadruplex numbers". A quadruplex number $r + ah + bi + cih$ can be seen as the following matrix:

$$\left( \begin{array}{cccc} r & -c & -b & -a \\ a & r & -c & -b \\ b & a & r & -c \\ c & b & a & r \\\end{array}\right)$$

We also see that $h^{8}=1$, and that $h$ generates a full polytopic group of a regular $4$-orthoplex centered at the origin in quadruplex space. Thus, we've found our orthoplexic group.

Now let's look for the $4$-simplex. If there exists a full polytopic group $G=\{f,f^2,f^3,f^4,1\}$ of a $4$-simplex/$5$-cell centered at the origin in quadruplex space, then $f$, $f^2$, $f^3$, and $f^4$ must all have a real component equal to $-1/4$. This is because the vertex-center-vertex angle between any two vertices of a regular $5$-cell is $\arccos(-1/4)$.

This leads to the following equation for the generator $f$ in quadruplex space (if it exists):

$$\left( \begin{array}{cccc} -\frac{1}{4} & -c & -b & -a \\ a & -\frac{1}{4} & -c & -b \\ b & a & -\frac{1}{4} & -c \\ c & b & a & -\frac{1}{4} \\\end{array}\right)^{5}=I$$

($I$ is of course the identity matrix.)

I admit that I did not try to learn how to solve this and just went straight to Wolfram Alpha, which produced these results. Looking through the real solutions for $f$, we see that there are actually two possible solutions for $G$. Let's call them $G_1$ and a $G_2$.

For $G_1$ with $f_{1}\approx{-0.25 + 0.9393474h + 0.1816356i + 0.1487780ih}$, $$f_1^{2}\approx{-0.25 - 0.5237205h + 0.7694209i + 0.2668489ih}$$ $$f_1^{3}\approx{-0.25 - 0.2668489h - 0.7694209i + 0.5237205ih}$$ $$f_1^{4}\approx{-0.25 - 0.1487780h - 0.1816356i - 0.9393474ih}$$ $$f_1^{5}=1$$

For $G_2$ with $f_2\approx{-0.25 + 0.5237205h + 0.7694209i - 0.2668489ih}$, $$f_2^2\approx{-0.25 + 0.1487780h - 0.1816356i + 0.9393474ih}$$ $$f_2^3\approx{-0.25 - 0.9393474h + 0.1816356i - 0.1487780ih}$$ $$f_2^4\approx{-0.25 + 0.2668489h - 0.7694209i - 0.5237205ih}$$ $$f_2^5=1$$

Since we're talking about cyclic groups of order 5, every element except the identity is a generator of the group, so except for $1$, all the elements of $G_1$ and $G_2$ are included in Wolfram Alpha's results.


Edit: The $4$-dimensional algebra described above is isomorphic to the tessarines, which @Anixx suggested I look at.

Given the tessarine identities $j^2=1,k^2=-1,jk=i$, we find that $j=\frac{h}{\sqrt{2}}-\frac{ih}{\sqrt{2}}$ and $k=\frac{h}{\sqrt{2}}+\frac{ih}{\sqrt{2}}$

Since the tessarines are isomorphic to $\mathbb{C}\times\mathbb{C}$, we now have the following set of "highly polytopic" algebras (though I now call them simply "polytopic algebras"):

  • $\mathbb{R}$
  • $\mathbb{C}$
  • $\mathbb{R}\times\mathbb{C}$
  • $\mathbb{C}\times\mathbb{C}$

There is a possible pattern here that I am not quite knowledgeable enough to prove, so I will refrain from saying it. Suffice to say I am trying to learn as much abstract algebra as I can so that one day I can make this answer post worthier of being marked "accepted".

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .