Welcome to Fractal Forums

I'm Steve, and I have been doing higher dimensional fractals starting 15 years ago.  My major interest is recursive mappings in Group Algebra spaces. This has been done to a certain extent more recently by others.

Fractals can also be done using Loops - a group construct where the associative law fails.
I.e. .
Two loops have been studied. The most interesting is the Octonion Loop in 8 dimensions where the images range from delicate to harsh.

Another unexpected group is the cyclic group C3 which has the axes, 1, i, j, and the generators . There are no complex numbers involved in the usual sense where the square of some element is unity. The images of this group however are mostly like the M-set with oddities at sharp borders.

The web site showing a few dozen of these images is given below. There was no attempt to specially color an image; all images use the same color table. The images displayed here are chosen mostly to illustrate similarities or differences from the Mandelbrot set, and not so much for aesthetics. Generally the most unusual or alien looking images are displayed at the website.

Website:
http://www.insidetheoutbox.net/recmaps/
Various groups are linked at the left side of the page.

Hello and Welcome to the forums!

Feel free to post some of these images here (in our Gallery or just in threads) as I'm sure there are folks here who would be interested, but may not pay much attention to the "Meet and Greet" section  O0

I look forward to seeing more stuff in the future!

hello and welcome to the forums,

wow, that is impressive stuff, i am sure you will make many friends here :D

Thanks Sockratease and Trifox.
Good idea.  I will be posting individual images with an explanation of why the image is interesting. 

Great collection of images. Have you ever tried to render 3D slices ?

I don't have 3-D software.  I'm curious what the C3 group looks like, because it has 3 dimensions and would show a complete rendering with 3D.

A 3d projection of A4b3n can also be interesting.

Greetings, and Welcome to this particular Forum !!!    :)

There are a few Members here that have been doing fractals for 15 years or more, and some of those appreciate the math quite a bit, so you should fit right in with the group.

I skimmed over a bit of your web pages, did not have enough time to really get into it.  It appears you wrote your own generator, is it all written in Java, or just a portion of it??
 

I'm sorry I did not reply earlier. Too busy with other things. Yes. It is all written in Java. I spent a lot of time writing an interface that could handle and display multiple images and script files. There are a lot of features for zooming, backtracking, saving, previewing, specifying dimensions and other parameters, etc.

Hello, Steve.

Here's what I get for the C3 fractal in ChaosPro:

(http://i624.photobucket.com/albums/tt323/schlega1/fractals/Steve.jpg)

Interestingly, the thickness does not seem to depend on iteration count or bailout value, and is not just cut off by a clipping plane.

I'm not familiar with ChaosPro, so I'm not sure what I am seeing. Do the colors represent a z-level? It is interesting that you get a slice that looks like the complete M-Set.  I found that slices along the coordinate axes show only truncated versions.  What intrigues me is that pieces of the M-Set even show up with group elements, 1, i, i².   Where i³ = 1.  That is quite different than the complex i² = -1 and i⁴=1.

I haven't put much effort into uderstanding fractal coloring methods yet, so I can't really help you with what the colors represent. I just used the default coloring method.

I think I have figured out why it's just a mandelbrot with thickness though. It turns out that there are several solutions to v²=v in this system (where v = a + bi +cj). I haven't explicitly found the vectors yet, but there seems to be pairs of vectors (v,w) with the property that v²=v and w² = -v.

Here's some 3d sections of the C4 version:
http://www.youtube.com/watch?v=Fdc34HKiZfw

Very interesting.  I have not looked into algebraic properties of C3.  I made a series of slices through C3 and you are right, it appears to be a M set with thickness, but it is like a tower where each slice up the tower is the M set slid over a bit and truncated on one side or another. The complete M set is never represented in a single slice, but a composite of the different slices comprise a complete M set.

My problem in understanding C3 is that there is no good definition of a negative number. I think you hit on the solution to why C3 is so related to the M set.  If you could define pairs of C3 elements (w,v)  and define a new group operation between pairs such that (v,w)⁴ = 1, then (v,w)² would define a sort of -1 as long as (v,w)⁴ is distinct from (v,w)².

As far as C4, it is probably easier to understand.  There still is no definition of "negative" but it is straight-forward to define since complex numbers are more simply embedded in C4. I had to look at your video about 20 times before I could come to grips with it. It seems that it is also a tower in more than one dimension.

Actually, it sounds like I might be doing something slightly different than you are describing. The vectors I used were not elements of C3 (or C4). The group just represents the relationship between the axes. The  coefficients I used were just ordinary numbers, so -(a +bi +cj) = -a -bi -cj. If you meant a system where the coefficients must be nonnegative, then I think you might get interesting results by arranging the C3 axes around a plane, and the C4 axes could be arranged in a tetrahedron.

Right negative coefficients are allowed. I haven't been thinking clearly about the problem. What I meant is there is no way of getting negative coefficients if they all start out positive.

This is what is now obvious in retrospect. For those coordinates where the coefficients a, b, and c are all positive, then the square has all positive coefficients:

Ie., if x = a +bi +cj then x² = 1(a² +2bc) +i( c²+2ab) +j(b²+2ac), which remain positive.

And they will always be positive when squaring over many iterations, so all orbits of x are positive..

This is unlike the complex plane where, when squared, the real part is a² -b² (potentially becoming negative even though both coefficients start out positive.)

If at least one coefficient in C3 is negative, then upon iterated squaring, negative values can eventually mix into all coefficients.  In this case orbits can perhaps wander almost anywhere in the space.

Now I think I have answered my question:
(http://www.insidetheoutbox.net/recmaps/c3/0-c3-matrix_tn.jpg) 

The top two images are that of C3 sliced through the i and j axes respectively. These shapes are sharply truncated in the upper right because the coefficients are all too positive, and cause divergence at the sharp boundary.

Now the unanswered question is why does the remainder of the images look so close to the M-set?  The orbits all wander around in 3 dimensions and still give the same results as the M-set. My conjecture is that a projection of the 3-D orbits onto the 2-D planes are the same as the M-set orbits. Otherwise it is too coincidental. To prove that is going to be an interesting exercise.

An even harder problem is why does C11 have cross-sections that look like the M-set? See
http://www.insidetheoutbox.net/recmaps/recmap-index.php?genre=c11

Schlega:
You were interested in idempotent operations (v² = v). I didn't get the connection till now, that a projection in linear algebra is an idempotent operator.  I succeeded in finding a projection to map the C3 space to a plane that contains the Mandelbrot recursions. It is a -45 degree projection (perpendicular to the i plane) of the j plane onto the real/i plane.  The M-set will lie on the real/i plane, and the j plane disappears.

Or mathematically the projection is defined as follows:
Let a point in C3 be x = 1(a) + i(b) + j(c)
Define projection P(x) = 1(a-c) + i(b) + j(0)
(The real component has the j component subtracted from it because of the -45 degree projection)
(The j component is set to zero because the cross section we are looking at is at j=0)
It is obvious that P(P(x)) = P(x) and is idempotent.

Squaring any C3 point, x, is given by
x² = 1(a² +2bc) + i(c²+2ab) + j(b²+2ac)
(using the relations i² = j and j³ = 1)

But since we are looking at a j=0 cross section, c is zero,
x² = 1(a²) + i(2ab) + j(b²)
Now apply the projection operator:
P(x²) = 1(a²-b²) + i(2ab)
This is identical to the square of the complex number and thus leads to the M-set.

However the 45 degree projection causes an elongation. This can be seen in the slices of C3 - the circular bulbs become ellipses.
(http://www.insidetheoutbox.net/recmaps/c3/0-c3-matrix.jpg)

I'm not sure if this is the full explanation just yet. x² is no longer in the c = 0 plane, so continuing to the next iteration would no longer project to the right place.

As for C11 and the other fractals on your site, if you are willing to share the code you use to generate them I'd love to look at some 3D slices. (Of course, if you would prefer to do that yourself I'll just wait to see your results)

Yes, actually I proved it for the first iteration. Other iterations would have to be proved by induction: if the n^th projection of the orbit of C3 is equal to the n^th M-set orbit, then the n+1 projection is also equal etc.  I don't know how to do it yet.  It seems like you would have to equate n^th order polynomials rather than iterations, and that sounds messy, but there might be a simpler way.

I can give you the inner loop computations for C11, but while there are only 20 floating point ops for each iteration of C3, there are 164 Flops for C11.  Rather than continue that on the "Hello, I'm new Here" board it should be on a math or theory board. I will post whatever groups you want. I presume you want only the inner loops, because my software is all in Java, and I don't know what you need.  Let me know.

Thank you. The groups I'd like to look at first are C11 and A4b3p. The inner loops are all I need, but I would also like to look at how you generate them automatically. I'm not much of a programmer, but I try to learn.

You're right about this needing its own thread, so I'll stop replying in this topic now.