I would like to know how compare from the algebraic point of view these two 3-dimensional hypercomplex number systems.
3-dimensional split-complex numbers
Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication.
Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$.
A number would be written in the form $a+bj+ck$. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$.
Here is a Mathematica code to experiment with:
Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} ->
x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;
Using this code one can see that
$j^2=k^2=1$
$jk=j+k-1$
$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$
$j^j=j^k=j$
$k^k=k^j=k$
$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$
$0^{j+k}=1-\frac{j}{2}-\frac{k}{2}$
The division formula would be:
$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\right)+\frac{a_1+b_1-c_1}{2 \left(a_2+b_2-c_2\right)}+\frac{a_1-b_1+c_1}{2 \left(a_2-b_2+c_2\right)}$
If we add a complex unity $i$, we will get a 6-dimensional number system.
Particularly, we will see that
$i^{j+k}=1-j-k$
and
$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$
Triplex numbers
This is a realization of triplex numbers, described in this video.
Here,
$1=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$j=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}
\right)$
$k=\left(
\begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{array}
\right)$
Mathematica code:
Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. ({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
$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b k+c j\right)-a_2 \left(b_2 \left(a k+b j+c\right)+c_2 \left(a j+b+c k\right)\right)+c_2^2 \left(a k+b j+c\right)-b_2 c_2 \left(a+b k+c j\right)+b_2^2 \left(a j+b+c k\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$
and
$N(a+bj+ck)=\sqrt[3]{a^3+b^3+c^3-3abc}$
is the analog of modulus.
If we add the complex unity, we will see that
$i^{j+k}=-\frac{j}{3}-\frac{j}{\sqrt{3}}-\frac{k}{3}+\frac{k}{\sqrt{3}}-\frac{1}{3}$
It looks that the both systems are commutative associative algebras with zero divisors but without nilpotents. Can something be said about their differences from the algebraic point of view?
Best Answer
Algebraically speaking, the second one is $\mathbb R\times\mathbb C$.
Being commutative semisimple rings, they would be considered quite tame in the big scheme of ring theory. One can say, however, they are the only two three-dimensional commutative semisimple $\mathbb R$-algebras.
Here are some differences: