Is there any where I can find info about Triplex Numbers?

Question

The triplex numbers as a quotient ring of the polynomial algebra (and a field/clifford algebra) are defined as:

$$ \{ a+b j_1+c j_2 | a, b, c \in \mathbb{R}, j_1^2 = j_2, j_2^2 = j_1, j_1 j_2 = j_2 j_1 =1 \} $$ (Or in simplicial coordinates:

This would correspond to the space of all points within a tetrahedron with vertices at the origin, $j_1$, $j_2$, and the point corresponding to $1$) but can also be defined as $R[X]/\langle X^3 - 1 \rangle$ (or using the cartesian product of rings, $\mathbb{R} \times \mathbb{C}$)

In the context of a simplex, a point P can be represented by simplicial (or barycentric) coordinates relative to the vertices of the simplex. For a tetrahedron, each point P within the tetrahedron can be represented by a set of four non-negative weights $(w_0, w_1, w_2, w_3)$, where each weight corresponds to a vertex of the tetrahedron, and the weights sum to 1. The position of the point is determined by these weights, which indicate the relative influence or proximity of each vertex on the point P.

We can then represent the triplex number $a + bj_1 + cj_2$ as a point P in the tetrahedron with simplicial coordinates $(w_0, w_1, w_2, w_3)$, where $w_0 = 1 - a - b - c$, $w_1 = a$, $w_2 = b$, and $w_3 = c$.

Now, when we talk about operations on these triplex numbers, we are actually talking about operations on these points within the tetrahedron. For instance, when we add two triplex numbers, we are essentially finding the resultant point in the tetrahedron.

Addition is defined as

$$ (c+d j_1+f j_2) + (g+h j_1+l j_2) + (m+n j_1+o j_2) = (c+d+f)+(g+h+l) j_1+(m+n+o) j_2 $$

(Or in simplicial coordinates:

We can consider addition as the process of taking the vector sum within the tetrahedron. If $P$, $Q$, and $R$ are points in the tetrahedron, the sum $P+Q+R$ is the point that corresponds to the vector sum of $\overrightarrow{V_0P}$, $\overrightarrow{V_0Q}$, and $\overrightarrow{V_0R}$.)

Subtraction is defined as

$$ z-x = z+(-x) $$

The identity elements are

$$ 1 + j_1 + j_2, 2 - j_1 - j_2 $$ and have zero divisors

with the familiar $1$ and $0$ also serving as identity elements (Or in simplicial coordinates:

These identity elements could correspond to the vertices of the tetrahedron.).

The additive inverse of any number y is:

$$ -y $$

The multiplicative inverse of any number D is:

The multiplicative inverse of a point $P$ could correspond to the inversion of $P$ through the circumcenter of the tetrahedron (or as defined in Somos's answer).

$j_2$ and $j_1$ are defined as

$$ j_2 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, j_1 = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} $$

The norm function N(x) is defined as:

$$ N(x) = a^2 + b^2 + c^2 $$

(Or in simplicial coordinates:

The norm of a point $P$ could be defined as the squared distance from the point to the origin $V_0$, $N(P) = ||\overrightarrow{V_0P}||^2$.)

The modified norm function M(x) is defined as:

$$ M(x) = N(x)^\frac{2}{3} $$

x' and x* are defined as:

$$ x' = \frac{N(x)}{x}, x* = \frac{M(x)}{x} $$

The exponential function is defined as

$$ \exp(a+b i_1+c j_2) = A(a) + B(b) j_1+C(c) j_2 $$

where

$$ A(x) = \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}, B(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)!}, C(x) = \sum_{n=0}^{\infty} \frac{x^{3n+2}}{(3n+2)!} $$

The logarithm function is defined as

$$ \ln(a+b j_1+c j_2) = \exp^{-1}(a+b j_1+c j_2) = t_1+t_2 j_1+t_3 j_2 $$

Multiplication is defined (using triplets) as: \[ [a_0, a_1, a_2] * [b_0, b_1, b_2] = [a_0b_0+a_1b_2+a_2b_1, a_0b_1+a_1b_0+a_2b_2, a_0b_2+a_1b_1+a_2b_0]. \] \[ j_3 = [0, 1, 0] \] \[ [a_0, a_1, a_2] = a_0 + a_1j_3 + a_2j_3^2 \]

(Or in simplicial coordinates: Let's denote the simplicial coordinates of two points P and Q in the tetrahedron as $[a_0, a_1, a_2]$ and $[b_0, b_1, b_2]$, respectively.

To find the product $P*Q$, you would perform a specific operation which I denote g(x) on the simplicial coordinates of $P$ and $Q$ to yield the simplicial coordinates of the result.)

We can calculate:

$$ \tan^{-1}(a+b j_1+c j_2) = t_3, \sin t_2 \, \text{rad} = \frac{c}{exp(t_2)}, t_1 = \ln(a), g = \ln(\sqrt{b^2+c^2}), h = \text{atan2}(b, c), \cos h \, \text{rad} = \frac{b}{exp(g)}, j_1^{t_4} = \exp( x \ln(j_1)) , N(x') = N(x)^{2} $$

These are the definitions and operations as specified in the triplex numbers system. , I want to know where I can find documentation about this as I've only found documentation about it in video form (of course simplicial coordinates are kind of documentation but they're also kind of not), the accepted answer is by Anixx, also includes the division formula in his answer and for a matrix representation of triplex numbers, dual numbers or the diagonal basis go to Somos's answer which is at time of editing most liked, also for info about simplicial coordinates go to one for Somos's comments. Note: I've finally found more documentation (kind of)! On this 3blue1brown video: https://www.youtube.com/watch?v=kYB8IZa5AuE&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=3&t=0s .

Answer 1

The question asks

I want to know where I can find documentation about this

where "this" refers to "triplex numbers". There is a YouTube video about the numbers which is quite good as far as it goes. The main point is about the usefulness of the "diagonal basis". The Wikipedia article on split-complex number is a good place to start reading about the concepts involved. Note the section at the end "Synonyms" with over a dozen different names used for the split-complex numbers. A similar situation exists for "triplex numbers". The multiplicity of names causes confusion. For example, there a programming language called "triplex" from an 1982 article "Triplex: A system for interval arithmetic" which is totally unrelated.

The linked MSE question by Torsten Schoeneberg is good for much background information about the history of $n$-dimensional hypercomplex number systems.

As the video explains, the "triplex numbers" can be understood by using a matrix representation. Accordingly, consider the three matrix units defined by (the video uses $1,j,k$ instead)

$$ i_1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right), \quad i_2 =\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right), \quad i_3=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right). \tag1 $$

The first unit $\,i_1\,$ is the $3\times 3$ identity matrix. They form a cyclic multiplicative group. That is,

$$ i_1i_1= i_2i_3 =i_3i_2 =i_1,\;\; i_1i_2 = i_2i_1 = i_3i_3= i_2,\;\; i_1i_3=i_3i_1=i_2i_2=i_3. \tag2 $$

A triplex number is a linear combination of the three units. That is,

$$ [a_1,a_2,a_3] := a_1i_1+a_2i_2+a_3i_3 = \left( \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_3 & a_1 & a_2 \\ a_2 & a_3 & a_1 \\ \end{array} \right). \tag3 $$

Define the triplex number $\,z := [1,1,1] = i_1+i_2+i_3.\,$ The determinant of $\,z\,$ is zero and therefore $\,z\,$ has no multiplicative inverse. However, it is not the zero matrix. In general, if a triplex number has a multiplicative inverse, then

$$ [a_1,a_2,a_3]^{-1} =[a_1^2-a_2a_3, a_3^2-a_1a_2, a_2^2-a_1a_3]/ (a_1^3+a_2^3+a_3^3-3a_1a_2a_3) \tag4 $$

as explained in the video. This implies that zero divisors exist in the system. In order to deal with this, the video suggests using another basis for the triplex system. The "diagonal basis" is defined by

$$ e_1 := (i_1+i_2+i_3)/3,\quad e_2 := (2i_1-i_2-i_3)/3,\quad e_3 := (i_2-i_3)/\sqrt{3}. \tag5 $$

In this new basis the multiplication facts are

$$ e_1e_1=e_1,\;\;e_1e_2=e_1e_3=0,\;\; e_2e_2=-e_3e_3=e_2,\;\; e_2e_3=e_3. \tag6 $$

The implication of these facts is that the triplex numbers split into two parts. The first part with basis $\,e_1\,$ is just like the real numbers. The second part with basis $\,e_2,e_3\,$ is just like the complex numbers. In other words, the triplex numbers are isomorphic to the direct product of the ring of real numbers and the ring of complex numbers. Another way to state this fact is the triplex numbers are isomorphic to the quotient of the polynomial ring $\,\mathbb{R}[x]\,$ by the polynomial $\,x^3-1\,$ with the mapping $\,[a_1,a_2,a_3] \mapsto a_1+a_2x+a_3x^2.$

This example of quotient of the polynomial ring $\,\mathbb{R}[x]\,$ produces triplex numbers. The quotient by the polynomial $\,x^2+1\,$ is complex numbers. Use $\,x^2-1\,$ to get split-complex numbers. Other examples include the dual numbers with polynomial $\,x^2.\,$ You might find it interesting to try using the polynomial $\,x^4-1=(x^2+1)(x^2-1).\,$

Answer 2

Triplex numbers (as described in this video to which OP refers) have the following properties.

Here,

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$

• Triplex numbers are isomorphic to $\mathbb{R}\times\mathbb{C}$ via isomorphism:

$$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) \to (1, e^{2 i \pi/3}), \left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) \to (1, e^{-2 i \pi/3});$$

$$(x,a+bi)\to\frac{1}{3} \left(x+2 a\right)+j \left(\frac{b}{\sqrt{3}}-\frac{a}{3}+\frac{x}{3}\right)+k \left(-\frac{b}{\sqrt{3}}-\frac{a}{3}+\frac{x}{3}\right)$$

• Triplex numbers have zero divisors and idempotents. They do not have nilpotents. They do not include imaginary unit, although can be extended with it to form a 6-dimensional algebra.

Code for the Mathematica system, implementing triplex numbers:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Post = (# /. ({j -> {1, E^(2 I \[Pi]/3)}, 
          k -> {1, E^(-2 I \[Pi]/3)}}) /. {x_, y_} -> 
        FullSimplify[(x/3 + Im[y]/Sqrt[3] - Re[y]/3) j + (x/3 - 
             Im[y]/Sqrt[3] - Re[y]/3) k + 
          1/3 (x + y + Conjugate[y])] // FullSimplify // Expand) &;

Particularly, we will see that

$j^2=k$, $k^2=j$, $jk=1$

$j^k=-\frac{1}{3} 2 e^{\frac{\pi }{\sqrt{3}}} j+\frac{j}{3}+\frac{1}{3} e^{\frac{\pi }{\sqrt{3}}} k+\frac{k}{3}+\frac{e^{\frac{\pi }{\sqrt{3}}}}{3}+\frac{1}{3}$

$0^{j + k + 1}=-\frac{j}{3}-\frac{k}{3}+\frac{2}{3}$

$\log j = \frac{2 \pi j}{3 \sqrt{3}}-\frac{2 \pi k}{3 \sqrt{3}}$

$\log(j+k)=\frac{\pi j}{\sqrt{3}}+\frac{1}{3} j \log (2)-\frac{\pi k}{\sqrt{3}}+\frac{1}{3} k \log (2)+\frac{\log (2)}{3}$

The division formula is

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b j_2+c j_1\right)-a_2 \left(b_2 \left(a j_2+b j_1+c\right)+c_2 \left(a j_1+b+c j_2\right)\right)+c_2^2 \left(a j_2+b j_1+c\right)-b_2 c_2 \left(a+b j_2+c j_1\right)+b_2^2 \left(a j_1+b+c j_2\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$

If we add complex unity, we will see that

$i^{j+k}=-\frac{j}{3}-\frac{j}{\sqrt{3}}-\frac{k}{3}+\frac{k}{\sqrt{3}}-\frac{1}{3}$