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Running from... Diamonds?: Mandelbrot beyond ii
===============================================

This post assumes you have read the previous post, which discusses some more implicit assumptions when generating the Mandelbrot set. The core assumption is in how multiplication works in a number system, and this is the manner in which I will proceed to generalize the set to 3D.


First Steps in 3D
-----------------


The previous post ended just as I began discussing numbers with two imaginary components *i* and *j*. If we add in the real component again, then numbers are now three-dimensional and multiplication can be characterized as

$$
\begin{gather*}
\begin{matrix} w_1 w_2 = (a + bi + cj) (d + ei + fj) =&
ad + (bd + ae)\boldsymbol{i} + (af + cd)\boldsymbol{j} \\
&+\ be\boldsymbol{i^2} + bf\boldsymbol{ij} + ce\boldsymbol{ji} + cf\boldsymbol{j^2}
\end{matrix} \\ \\
= \Sigma \left(
\begin{pmatrix}
ad & ae & af \\
bd & be & bf \\
cd & ce & cf
\end{pmatrix}
\odot
\begin{pmatrix}
1 & i & j \\
i & i^2 & ij \\
j & ji & j^2
\end{pmatrix}
\right )
= \Sigma \left(
P_3 \odot
\begin{pmatrix}
1 & \begin{matrix}i & j \end{matrix} \\
\begin{matrix}i \\ j \end{matrix} & M
\end{pmatrix}
\right )
\end{gather*}
$$

...where M is the matrix block which controls how the imaginary terms i and j multiply. Previously, *M* was the single element $i^2$. It is now a 2×2 matrix, so there is much more freedom in the choice of its elements.

With more freedom comes the need to make decisions which narrow the focus. Most obviously, *M* should be symmetric so that multiplication is commutative. Next, expressions such as $ij = i$, which eliminate an imaginary term arbitrarily, imply the lack of an inverse element or that an imaginary axis is equivalent to a real one. I glossed over this issue earlier when rotating $i^2$ about the unit circle. Now that *M* is 2×2, we can only populate the off-diagonal with real components. Considering otherwise for 3D algebras is out of the scope of this post, as it is not sufficiently interesting.

The above decisions do not affect the main diagonal, except in that $i^2 \neq i$ and $j^2 \neq j$. Squares are still allowed to "cycle" between imaginary axes, and may even change in sign (e.g., $i^2 = -j$).

To generate the set, I will apply the iteration 7 times and use the Euclidean norm. In 3D, our range is the cube from that extends from -2 to 2 in all dimensions. I will visualize the set in two ways: a scatterplot of points which remain after the iterations, and as a video of slices along planes parallel to the *ij* plane. The scatterplot will be useful to identify similar shapes, while the slices will show the halos that correspond to number of iterations.


Roots of Unity: Part Deux
-------------------------

[Roots of unity](https://en.wikipedia.org/wiki/Root_of_unity) are the most well-suited tool to extending a number system. I mentioned previously that -1 is the second root of unity, and it creates negative numbers. *i* is the fourth root of unity, and it creates the complex numbers. Continuing along even numbers, the *sixth* root of unity should correspond to 3D numbers. But didn't we already discuss this when we sheared the set in the previous post?

Not necessarily. Previously, we expressed this root as a complex number. It can be described completely algebraically, which avoids using square roots. Just as the fourth roots of unity lie on the vertices of a (tilted) square, the *sixth* roots lie on a regular hexagon, with 0 at the center. One of these vertices must correspond to 1, and the point opposite it to -1. The real axis forms the line that connects them.

![]()

This leaves two pairs of vertices, each of which can correspond to an "imaginary" axis. There are four possible ways to orient these axes, by "picking" a root for the positive-going direction.


### Upper and Lower Halves

Two of these choices are equivalent: +*i* and +*j* on the upper half of the hexagon, and +*i* and +*j* on the lower half of the hexagon. Even though this basis is not orthogonal like the fourth root of unity, the components still multiply in the same way that their angle on the unit circle adds, and in either case,

$$
i^2 = j,~ j^2 = -i,~ ij = ji = -1
~\implies~
M =
\begin{pmatrix} j & -1 \\ -1 & -i \end{pmatrix}
$$

Alternatively, +i and +j may both lie on the right half of the figure, near 1. In this case,

$$
i^2 = -j,~ j^2 = -i,~ ij = ji = 1
~\implies~
M =
\begin{pmatrix} -j & 1 \\ 1 & -i \end{pmatrix}
$$

Both of these definitions of M have the commonality that $i^3 = -1$.


### Left Half

However, by picking the two roots on the left half of the figure, $i^3 = j^3 = 1$, i.e. they are both third roots of unity. These can be imagined as the rotations on a triangle with labelled vertices which preserve the orientation, but permute the labels.

:::: {layout-ncol = "2"}
::: {}
![]()
:::

::: {}
$$
j^2 = i,~ i^2 = j,~ ij = ji = 1
\\ M =
\begin{pmatrix} j & 1 \\ 1 & i \end{pmatrix}
$$
:::
::::


### Right Half

The reflection of i and j in the above case makes both of them sixth roots of unity, i.e., $i^3 = j^3 =  -1$.

$$
j^2 = i,~ i^2 = -j,~ ij = ji = -1
~\implies~
M =
\begin{pmatrix} -j & -1 \\ -1 & i \end{pmatrix}
$$


### Visualizations

Despite there being four different ways in which the sixth root defines *M*, the Mandelbrot set makes the same shape in all cases. In fact, its shape scarcely changes from its familiar appearance:

This shape is a cylinder of the the original Mandelbrot set. In fact, from another angle, it resembles the split-complex Mandelbrot set (the diamond). This implies that within each system created by the sixth root of unity, there are still numbers which behave similarly to the standard complex numbers (after all, the basis is not orthogonal) and the split-complex numbers.

$$
M_6 = \begin{pmatrix}
j & -1 \\
-1 & -i
\end{pmatrix}
\equiv \begin{pmatrix}
-j & 1 \\
1 & -i
\end{pmatrix}
\equiv \begin{pmatrix}
j & 1 \\
1 & i
\end{pmatrix}
\equiv \begin{pmatrix}
-j & -1 \\
-1 & i
\end{pmatrix}
$$

Taking slices of the first multiplication (which corresponds to the upper left diagram), the familiar set appears to appear only in parts at a time. There are some regions which appear outside of the familiar shape, but these appear to be artifacts of the "imprecision" with which these numbers approximate the complex numbers.

![]()


Mutual Complexity
-----------------

With the restraints on *M* given before, four of the eight "cyclic" systems which have been discussed. Three of the remaining are:

$$
M_+ = \begin{pmatrix}
j & -1 \\
-1 & i
\end{pmatrix},~
M_* = \begin{pmatrix}
-j & 1 \\
1 & i
\end{pmatrix},~
M_- = \begin{pmatrix}
-j & -1 \\
-1 & -i
\end{pmatrix}
$$

The remaining is equivalent to $M_*$, with the signs on j and i swapped. Each of these systems is in some way mutually complex or split-complex. Unfortunately, in all them, associativity breaks down:

$$
\begin{matrix}
M_+: &
(i^2)j = j^2 = -i \neq i = i(ij)
\\
M_*: &
(i^2)j = -j^2 = -i \neq i = i(ij)
\\
M_-: &
i(j^2) = -i^2 = j \neq -j = (ij)j
\end{matrix}
$$

Just as all the sixth roots correspond to the same figure, so too do all of these nonassociative variants.

![]()

This figure is distinct from the original set, appearing more fish-like. The slices (which correspond to the upper left diagram are) appear to bubble and burst as planes intersect the figure.

![]()


Doubly Complex
--------------

*M* need not contain complex members. The entries 1 and -1 may freely appear on the main diagonal. Four more matrices can be generated if commutativity is maintained

$$
\begin{matrix}
M_{-1} = \begin{pmatrix}
-1 & -1 \\
-1 & -1
\end{pmatrix}, &
M_{\pm 1} = \begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}, \\
M_{\mp 1} = \begin{pmatrix}
-1 & 1 \\
1 & -1
\end{pmatrix}, &
M_{+1} = \begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
\end{matrix}
$$

Assuming associativity, each multiplication states that one of *i* or *j* is redundant.

$$
\begin{matrix}
M_{-1}: &
(i^2)j = -j \neq -i = i(ij) \\
M_{\pm 1}: &
(i^2)j = j \neq -i = i(ij) \\
M_{\mp 1}: &
(i^2)j = -j \neq i = i(ij) \\
M_{+1}: &
(i^2)j = j \neq i = i(ij)
\end{matrix}
$$

Running the Mandelbrot iteration reveals pairs of these to be similar. However, I believe this is an artifact of the range/iteration limits. It may be the case that the thin sliver of points near the negative real boundary is the same in all schemes. I can think of no other explanation for the diagonal pair to appear different.

![]()

::: {}
![]()
$M_{-1}$
:::

::: {}
![]()
$M_{-1}$
:::

::: {}
![]()
$M_{+1}$
:::


Impotent Products
-----------------

Thus far, the product of two imaginary terms of magnitude 1 produces a third term of magnitude 1. However, it is also possible that the product has magnitude 0, a sort of infinitesimal. Without making *M* a matrix of all 0's, there are 6 possible variations.

The first four are produced when *i* and *j* are "mutually" infinitesimal, i.e., their product is 0, but their squares are not. This means that in *M*, the main diagonal is 1 or -1, and the off diagonal is 0:

$$
M = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},~
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} \equiv
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix},~
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
$$

Notably, all of these products satisfy either of the quadratic equations $i^2 + j^2 = \pm 2$ or $i^2\ - j^2 = \pm 2$. Therefore, it might be unsurprising that the Mandelbrot set is in the shape of conic sections.

![]()

The upper-left shape is two cones that lie base-on-base, the upper-right and lower-left are distinctly hyperbolic, and the lower-right is a solid of revolution of the original Mandelbrot set. The upper left can also be thought of as a solid of revolution, but of the Mandelbrot set produced by the *split-complex numbers*, which is why it appears so diamond-like. While it may be the case that the remaining two sets demonstrate the properties of both, it is not readily visible, at least not within the range shown. Slices of each are as follows:

:::: {layout-ncol = "2"}
::: {}
$$
M_\circ = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
:::

::: {}
![]()
:::
::::

:::: {layout-ncol = "2"}
::: {}
$$
\begin{align*} M_H = \begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix} \\
\equiv \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\end{align*}
$$
:::

::: {}
![]()
:::
::::

:::: {layout-ncol = "2"}
::: {}
$$
M_\bullet = \begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
$$
:::

::: {}
![]()
:::
::::


### Other Hyperbolic Cases

There are two additional products if the squares of *i* and *j* are nilpotent, but their product is either 1 or -1:

![]()

These turn out to be reflections of one another. The slices are similar to the hyperbolae in $M_H$ above, but corresponding to the hyperbola $xy = 1$ instead.

![]()


### Permanently Imaginary

Beyond the six already given, there are also the products which are mutually infinitesimal, but independently cyclic. This multiplication is the same as the "point-swap" kind from the previous post, where the imaginary part of the product is pointwise, then swapped between axes. Similarly to previous cases, all four figures produced are the same.

![]()

To me, this shape looks like a fancy wine jug, complete with a loop to attach a string. With respect to other shapes, it appears to resemble the [Burning Ship fractal](https://en.wikipedia.org/wiki/Burning_Ship_fractal), bisected by a plane. Slices reveal bubbles of exclusion inside, as well as a rather erratic section past $\Re(z) = 0$.

![]()


Pro-Working from Home
---------------------

We can also try relaxing the commutativity rule. If the sign changes when multiplicands are swapped, such behavior is called anticommutativity. For the 2×2 *M*, this means that the off-diagonal elements are nonzero and have opposite sign from each other. Without compromising associativity, two anticommutative elements necessitate a third:

$$
\begin{gather*}
ij = -ji = \alpha \\
iij = i(ij) = -(ij)i = -iji \\
i\alpha = -\alpha i
\end{gather*}
$$

$\alpha$ cannot contain either *i* or *j* without relaxing the inverse rule.

Since squaring a number is part of the Mandelbrot map, these multiplication schemes are identical to those with zeros on the off-diagonal, due to cancellation. When *M* is totally real, we form the familiar conics-shaped figures again. Similarly, cyclic squares correspond to the same burning ship shape. If we allowed 0 on the main diagonal, *M* would be equivalent to a zero matrix, and be unsatisfactory for our purposes.

:::: {}
::: {}
![]()
:::

::: {}
![]()
:::
::::

Since all of these schemes are nonassociative, starting with cubes, there is a choice to left-multiply or right-multiply first. Until four dimensions, anticommutativity isn't a well-founded property.


Summary and Closing
-------------------

Rather than discussing the analogues of the pointwise-like products in 3D, I will stop here. Enough data has been collected to discern the various *M*.

Despite properties like inverses and commutativity being readily apparent from the shape of *M*, associativity is rather elusive. The kind of associativity being violated is apparent in the Mandelbrot set itself. When the multiplication scheme includes 0 terms, one can deduce that $i = 0$ or $j = 0$ by assuming associativity. In terms of chaotic behavior in the Mandelbrot set, this is better-behaved than the any of the cyclic types which do not correspond to sixth roots of unity, where the deduction is instead $i = \pm j$.

The determinant of *M* is only sometimes relevant. For all of the types which are sixth roots of unity, $|M| = 0$. Contrast this with the cyclic types which are not, for which $|M| = -2$. All of the totally real M also have determinant 0, as do the cyclic, mutually infinitesimal products. Worse than that, if M has 0's on the off-diagonal, then there exists a similar anticommutative scheme with a different determinant.

To address both of these problems, and the growing vastness of higher-dimensional space, I will add more restrictions in the [next, final post]() regarding 4D Mandelbrot analogues.
