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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 and properties

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

$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$


So, I wonder whether anyone ever described such a system separately from just $\mathbb{R}^3$ (to which it is isomorphic similarly to how usual split-complex numbers are isomorphic to $\mathbb{R}^2$)? Maybe in the 19th century?

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    $\begingroup$ This does not really look like a historical question to me. $\endgroup$ Apr 30 at 16:47

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