I understand if you are discouraged by these results and rather don't want to spend your time exploring these further. That's fine.

Nevertheless, let me try to address your concerns about the particular form these formulas take.

The all-positive nature of these is correct. It's a consequence of how these tables arise. It's more of an extension of split-complex numbers than complex ones:

(a1 + aj j)(b1 + bj j) = a1 b1 + aj bj + j (a1 bj + aj b1) - this, to my knowledge, doesn't give a circle or egg, but rather, which of course is also boring, a square. Its Buddhabrot take looks somewhat interesting though. - all the orbits are along 45° angles of the axis but the densities vary in non-trivial ways.

I didn't expect super high detail interesting results from these but simply an "egg", and for all three of these no less, is surprising to me.

I'm not entirely sure if there is an obvious natural way to introduce negatives. I mean you could just randomly flip signs and define i² = -1 and/or k² = -1 or something but then the geometric meaning would be lost.

As for the lack of "cycling", I'm not quite sure what you mean by that, but the two that correspond to an edge pair that's open on one end certainly have a degeneracy to them:

The k-move turns this:

--.<-

into this:

->.--

and vice-versa, but if it's pointing outside:

<-.--

it will go around all the way and end up in the same position so then k does nothing.

The i-move, meanwhile, always works:

<-.--

turns into this:

->.--

and after that you can use the k-move as above.

Since k sometimes changes things and sometimes does not, that results in the degeneracies evident in the multiplication table / formula.

This is also why "Edges outfacing", i.e. calling either one of the following the "1-position",

<-.--

or

--.->

doesn't have k as its basis, but rather mixed moves like iki which does this:

<-.--

turns into

--.->

and vice versa.

(Basically, k rotates around the tip of a vector whereas iki turns around the shaft)

Putting numbers onto this is perhaps kind of weird but I was hoping something more interesting would happen. In principle, any group should be able to be made into this. For instance, here is a group with two elements, the integers modulo 2: (in the following,

**bold** numbers aren't the same as your normal numbers. They will rather be the labels of the individual directions in a given system.)

**0**+

**0** =

**0****0**+

**1** =

**1****1**+

**0** =

**1****1**+

**1** =

**0**aka the operation

**or**.

This is slightly confusing in that the neutral element of addition is 0 and thus the neutral dimension that is equivalent to a scalar also is the one labeled

**0**. The corresponding multiplication table is:

so then you'd get a 2-dimensional system:

(a

**0**+a

**1** **1**)(b

**0**+b

**1** **1**) = (a

**0** b

**0** + a

**1** b

**1**)

**0** + (a

**0** b

**1** + a

**1** b

**0**)

**1**which, of course, is just the same as the split-complex numbers:

**0** = 1

**1** = j

There is also the

**and** table, or multiplication over the integers modulo 2:

**0***

**0** =

**0****0***

**1** =

**0****1***

**0** =

**0****1***

**1** =

**1**which actually can't quite be written in the same form as the other table. Gotta expand it to include a head column/row:

with would, if used in the analogous way, give an incredibly degenerate number system:

(a

**0**+a

**1** **1**)(b

**0**+b

**1** **1**) = (a

**0** b

**0** + a

**0** b

**1** + a

**1** b

**0**)

**0** + (a

**1** b

**1**)

**1**which is just silly: In such a system, what ever is in the

**1**-direction, stays in the

**1**-direction.

I think this one isn't actually a group though: You cannot reach

**1** from

**0** so

**0***

**1** does not have an inverse which would be a requirement for a group.

All these groups are not originally about numbers and so there is no reason for them to ever "be negative" as it were, so they will inherently never "accidentally" mimic the complex numbers. There might be something to change that though. Adding in a negative term is bound to make things more interesting seeing as to how j²=1 is the least interesting of the three major cases (the other two being i²=-1 and ε²=0 with (a1 + aε ε)(b1 + bε ε) = a1 b1 + (a1 bε + aε b1) ε and the aε bε term just vanishing), but I wonder if there is a way to do this "reasonably mathematically" rather than by throwing it in there randomly.