- #1

Anixx

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- TL;DR Summary
- What are the properties of the 3D split-complex numbers? Were they ever considered?

For some reason the 3-dimensional hypercomplex numbers were not touched in the most of overviews of hypercomplex numbers.

But I think this is not deseved.

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 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:

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}##

But I think this is not deseved.

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 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:

Code:

```
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}##