Fractals in the C3, C11 and A4B3p Groups Algebras

[Steve]

This is a continuation of a discussion in the "Meet & Greet" board (http://www.fractalforums.com/meet-and-greet/hello-t3307/15/) that Schlega and I are pursuing. This board is more appropriate for that topic.

@ Schlega:
I still have a strong gut feeling that there is a relation between C3 and the M-set, but it turns out I was wrong in thinking it was a simple projection . From the pictures of C3 I think it is at least an affine transformation that maps C3 to the complex plane.

In the following code point[] is the coordinate point being iterated. For a 2-D slice, 8 of the coordinates are held constant, and the other two span the image. If you are able to do 3-D ray tracing then 7 would remain constant.

Here is the Java inner loop for Group Algebra C11:
for (kk = 0; kk < loops; kk++) {
   e0=E[0]; e1=E[1]; e2=E[2]; e3=E[3]; e4=E[4]; e5=E[5]; e6=E[6]; e7=E[7]; e8=E[8];
                 e9=E[9]; e10=E[10];

   E[0] = point[0] + e0*e0 + 2.0f*( + e1*e10 + e2*e9 + e3*e8 + e4*e7 + e5*e6 );
   E[1] = point[1] + e6*e6 + 2.0f*( + e0*e1 + e2*e10 + e3*e9 + e4*e8 + e5*e7 );
   E[2] = point[2] + e1*e1 + 2.0f*( + e0*e2 + e3*e10 + e4*e9 + e5*e8 + e6*e7 );
   E[3] = point[3] + e7*e7 + 2.0f*( + e0*e3 + e1*e2 + e4*e10 + e5*e9 + e6*e8 );
   E[4] = point[4] + e2*e2 + 2.0f*( + e0*e4 + e1*e3 + e5*e10 + e6*e9 + e7*e8 );
   E[5] = point[5] + e8*e8 + 2.0f*( + e0*e5 + e1*e4 + e2*e3 + e6*e10 + e7*e9 );
   E[6] = point[6] + e3*e3 + 2.0f*( + e0*e6 + e1*e5 + e2*e4 + e7*e10 + e8*e9 );
   E[7] = point[7] + e9*e9 + 2.0f*( + e0*e7 + e1*e6 + e2*e5 + e3*e4 + e8*e10 );
   E[8] = point[8] + e4*e4 + 2.0f*( + e0*e8 + e1*e7 + e2*e6 + e3*e5 + e9*e10 );
   E[9] = point[9] + e10*e10 + 2.0f*( + e0*e9 + e1*e8 + e2*e7 + e3*e6 + e4*e5 );
   E[10] = point[10] + e5*e5 + 2.0f*( + e0*e10 + e1*e9 + e2*e8 + e3*e7 + e4*e6 );

   magn = E[0]*E[0]+E[1]*E[1]+E[2]*E[2]+E[3]*E[3]+E[4]*E[4]+E[5]*E[5]
         +E[6]*E[6]+E[7]*E[7]+E[8]*E[8] +E[9]*E[9]+E[10]*E[10];
   fpCounter++;
   if ( magn > escape)  break;
}


And this is the inner loop for the group A4B3p

for (kk = 0; kk < loops; kk++) {
   e0=E[0]; e1=E[1]; e2=E[2]; e3=E[3]; e4=E[4]; e5=E[5];

   E[0] = point[0] + e0*e0 - e1*e1 + 2.0f*( + e2*e4 - e3*e5 );
   E[1] = point[1] + 2.0f*( + e0*e1 + e2*e5 + e3*e4 );
   E[2] = point[2] + e4*e4 - e5*e5 + 2.0f*( + e0*e2 - e1*e3 );
   E[3] = point[3] + 2.0f*( + e0*e3 + e1*e2 + e4*e5 );
   E[4] = point[4] + e2*e2 - e3*e3 + 2.0f*( + e0*e4 - e1*e5 );
   E[5] = point[5] + 2.0f*( + e0*e5 + e1*e4 + e2*e3 );

   magn = E[0]*E[0]+E[1]*E[1]+E[2]*E[2]+E[3]*E[3]+E[4]*E[4]+E[5]*E[5];
   fpCounter++;
   if ( magn > escape)  break;
}

The Java program that generates the above code from a simple script file is 1000 lines long. If you want it I can get it to you.

In the script file, the generator rule for C11 is simply C^11=1
The rule for A4B3p is A^4=1, A^2=-1, B^3=1,  AB=BA

[Schlega]

I still have a strong gut feeling that there is a relation between C3 and the M-set, but it turns out I was wrong in thinking it was a simple projection . From the pictures of C3 I think it is at least an affine transformation that maps C3 to the complex plane.

My feeling is that there is something more general going on. It looks like there is a connection between the M-set and all of the groups on your site.

The Java program that generates the above code from a simple script file is 1000 lines long. If you want it I can get it to you.

Well, I almost certainly won't understand how it works, but I'll PM you my email address anyway.

Thank you for posting the loop code. It would have been a nightmare to do by hand. Here's the (0,1,2) cross section of C11:
at 25 iterations


at 100 iterations:


and a composite image of a few different iteration depths:


I'll have more up tomorrow, but for now I'll just post the formula file so you can explore it too.

Code:

cp_63303(QUATERNION) {
parameter real bailout, c4;
parameter quaternion c5678;
parameter complex c910;
int ndx;
real e0,e1,e2,e3,e4,e5,e6,e7,e8,e9,e10,magn,p0,p1,p2,p3,p4,p5,p6,p7,p8,p9,p10;
real E[11];


	void init(void)
	{
		p0 = real(pixel);
		p1 = imag(pixel);
		p2 = part_j(pixel);
		p3 = part_k(pixel);
		p4 = c4;
		p5 = real(c5678);
		p6 = imag(c5678);
		p7 = part_j(c5678);
		p8 = part_k(c5678);
		p9 = real(c910);
		p10 = imag(c910);
		
		for(ndx = 0; ndx <= 10; ndx = ndx+1){ E[ndx] = 0.0;}
		
	}
	void loop(void)
	{
		e0=E[0]; e1=E[1]; e2=E[2]; e3=E[3]; e4=E[4]; e5=E[5]; e6=E[6]; e7=E[7]; e8=E[8];
                 e9=E[9]; e10=E[10];

		E[0] =  p0 + e0*e0 + ( e1*e10 + e2*e9 + e3*e8 + e4*e7 + e5*e6 )*2;
		E[1] =  p1 + e6*e6 + ( e0*e1 + e2*e10 + e3*e9 + e4*e8 + e5*e7 )*2;
		E[2] =  p2 + e1*e1 + ( e0*e2 + e3*e10 + e4*e9 + e5*e8 + e6*e7 )*2;
		E[3] =  p3 + e7*e7 + ( e0*e3 + e1*e2 + e4*e10 + e5*e9 + e6*e8 )*2;
		E[4] =  p4 + e2*e2 + ( e0*e4 + e1*e3 + e5*e10 + e6*e9 + e7*e8 )*2;
		E[5] =  p5 + e8*e8 + ( e0*e5 + e1*e4 + e2*e3 + e6*e10 + e7*e9 )*2;
		E[6] =  p6 + e3*e3 + ( e0*e6 + e1*e5 + e2*e4 + e7*e10 + e8*e9 )*2;
		E[7] =  p7 + e9*e9 + ( e0*e7 + e1*e6 + e2*e5 + e3*e4 + e8*e10 )*2;
		E[8] =  p8 + e4*e4 + ( e0*e8 + e1*e7 + e2*e6 + e3*e5 + e9*e10 )*2;
		E[9] =  p9 + e10*e10 + ( e0*e9 + e1*e8 + e2*e7 + e3*e6 + e4*e5 )*2;
		E[10] = p10 + e5*e5 + ( e0*e10 + e1*e9 + e2*e8 + e3*e7 + e4*e6 )*2;

		for(ndx = 0; ndx <= 10; ndx = ndx+1){ E[ndx] = E[ndx];}
		
		magn = E[0]*E[0]+E[1]*E[1]+E[2]*E[2]+E[3]*E[3]+E[4]*E[4]+E[5]*E[5]+E[6]*E[6]+E[7]*E[7]+E[8]*E[8] +E[9]*E[9]+E[10]*E[10];
	}
	bool bailout(void)
	{
		return(magn <= bailout);
	}

	void description(void)
	{
		this.title = "C11";

		bailout.caption = "Bailout Value";
		bailout.default = 4.0;
		bailout.min = 1.0;
		bailout.hint = "Defines the bailout radius: As soon as a pixel falls outside a circle with this radius, the iteration stops.";
		
		c4.default = 0.0;
		c5678.default = (0.0,0.0,0.0,0.0);
		c910.default = (0.0,0.0);
	}
}

[matsoljare]

I posted about a related subject a while ago here:

http://www.fractalforums.com/general-discussion-b77/e8-and-other-lie-groups

[Steve]

My feeling is that there is something more general going on. It looks like there is a connection between the M-set and all of the groups on your site.

I agree. That's why I am focussing on the C3 group - because it is simplest and can be done on one sheet of paper - maybe. A solution may indicate a general approach. As you would guess, groups do have embedded M-sets if the order of the group has divisor 4.  However C11 does not have any slices of M-set where the cross section goes through the origin. Groups of prime numbers may not always contain M-sets, although C3 partly does. I will have to look at C5 and C7 sometime.

Looking at your pictures of C11, the group in 3-D is wackier than ever.

[Schlega]

However C11 does not have any slices of M-set where the cross section goes through the origin. Groups of prime numbers may not always contain M-sets, although C3 partly does. I will have to look at C5 and C7 sometime.

Are you sure that it doesn't have one hidden at some odd angle? There's quite a few "ripped" portions that resemble parts of the M-set. There might be a plane where the whole thing comes through.

I haven't had much time to explore yet, but here's a few more slices:

(0,1,3):


(1,2,3):


This one is not aligned with the coordinate axes:

[Steve]

There might be a plane where the whole thing comes through.

You have more optimism than I do. Of course it seems that if there is a full M-set slice it would most likely be a plane through the origin because the M-set is a connected set.  Off-origin planes tend to become disconnected.

It should be a simple search to look at tilts in C3. However, my software is lacking in the user interface to look at tilted planes because that much flexibility in the system would make it harder to control. However, I can do simple custom code as work-arounds for simple groups.

I am trying to picture how the 2-D slices cut through your 3-D shapes. What you have looks like origami.

It would also be interesting if you could render dimensions 0, 1, 10, because there is bilateral symmetry about the vertical axis in 0 and 10, and bilateral symmetry about a 45 degree axis in 1 and 10.  The same can be said for 0, 2, 9 and 0, 3, 8, etc. Which should give shapes identical to 0, 1, 10.

This matrix shows orthogonal slices through each pair of the 11 dimensions.

[Schlega]

Here's the (0,1,10) slice:
10 iter:


25:


100:

[bib]

These last pictures remind me the first frames of this quaternion video :

<a href="http://www.youtube.com/v/RMQNbUrka7Y&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/RMQNbUrka7Y&rel=1&fs=1&hd=1</a>