The Wikipedia article n hypercomplex numbers completely lacks any mention of 3-dimensional and 6-dimensional systems, making impression they are impossible.
Intuition. Basically, if you add two complex dimensions to reals, say $i$ and $j$, you automatically get a fourth dimension $ij$ because this number cannot be expressed using only the three dimensions. The system you get then is called bicomplex numbers and 4-dimensional.
On the other hand, if you add two split-complex dimensions to reals, say $j$ and $k$, you do not get a fourth dimension automatically because we can define $jk=j+k-1$, which can be expressed in the already existing 3 dimensions. Thus, you get a 3D algebra.
It seems that each of the two added split-complex dimensions are isomorphic to the classic split-complex axis.
Construction.
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
$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$
That said, what are the algebraic properties of these 3D and 6D number systems?
Where they ever studied?
Best Answer
This is an understandable impression because of a confluence of two factors: the way ring theory developed in the early days, out of geometry and real numbers, and the need to explain what the old-timers meant to a modern day person who is comfortable with algebras over fields.
Hypercomplex numbers were, originally, were "things like the quaternions" meaning that in general you took real numbers, extra symbols like $i,j,k$ with certain algebraic relations, and then generated an algebra with them. I do not know for certain in what cases application of the algebra preceded the invention of it, but I assume that it happened both ways for many of the classical so-called hypercomplex number systems. This in part explains the limited scope of what was given as examples in the wiki article.
On the other hand, the wiki wants to express what hypercomplex numbers are concisely, and so it does so with more modern terminology that was not in the minds of the old-timers: a finite dimensional $\mathbb R$-algebra. With this more modern term there are of course a multitude of hypercomplex number systems including product rings $\mathbb R^n$ for any $n\in\mathbb N^+$ which have dimension $n$.
As for your specific examples, the six dimensional one is isomorphic to the product ring $\mathbb C^3$.
Algebraically the properties of $\mathbb R^3$ and $\mathbb C^3$ are very boring, both being commutative semisimple rings. Their ideal structure and module categories are easy to describe and understand.
Whether or not they have interesting applications or are viable or have special analytic properties are completely different questions which depend on the resourcefulness of the researcher presenting evidence.