Whipped cream look of quaternians an artifact of projection?

[kjknohw]

One of the many problems with 3d fractals is the "whipped cream", example below shows a quaternion:


Whipped cream is bad because that means there are regions of unbounded distortion. If you zoom in on one of these regions, you lose detail in the direction of the stretch.

In the mandelbulb the mapping function has "polar pinch" singularities due to spherical coordinates. However, I believe quaternions are not distorted since the rules of quaternion arithmetic are "natural" like complex numbers and singularity-free.

Instead, the distortion comes from the projection of 4d into 3d. Imagine you looking at a cube. You will see a projection of the 3d cube into 2d. All the faces are perfect squares but faces that are seen near edge-on look distorted into thin rectangles or parallelograms. I think a similar thing is happening with the quaternions, in which the 4d to 3d projection is causing the distortion.

What are your thoughts?

[Tglad]

Unfortunately quaternion mandelbrots/julias are stretched. The q*q operation is non-conformal, meaning it introduces stretch... and since the operation happens an infinite number of times... whipped cream.
There are only 5 operations that don't introduce stretch in 3d (or higher dimensions)- translation, scale, rotations, reflections and sphere inversion. Or any combination of those, which is always reducible to just one of each operation. Liouville's theorum.

[David Makin]

Unfortunately quaternion mandelbrots/julias are stretched. The q*q operation is non-conformal, meaning it introduces stretch... and since the operation happens an infinite number of times... whipped cream.
There are only 5 operations that don't introduce stretch in 3d (or higher dimensions)- translation, scale, rotations, reflections and sphere inversion. Or any combination of those, which is always reducible to just one of each operation. Liouville's theorum.

Mad thoughts - prompted by lack of formal math beyond "A" level:

In Euclidian space is the restricted conformality of Rn (n>2) directly related to the lack of true algebraic fields in Rn (n>2) ?

I'm guessing yes

Also I would suggest that a true secondary "imaginary" vector/constant/function (j) is required to solve this issue at least for R3 or R4, probably such that j acts in two dimensions instead of one.....Hmmmm

[M Benesi]

  I tend to think that the "stretch" is caused by variable "interference" with one another (for the Mandelbulb, at the very least).

  You've got 3 components- 1 of which is influenced by the other 2.  In certain areas, the chaotic effects of the one variable cancel out the chaotic effects of the other 2- so you get a boring stretchy section. 

  I don't know if it's the same for quaternions.  I tend to see them as... lacking magnitude management- more on that if I look at the math later.  Just had an idea to pursue. 

[Tglad]

I should add that it is Liouville's proved theorem. So, unless an operator uses the 5 mentioned transforms, it will have stretch.
It cannot be unproved by using a new squaring operator on 3d or 4d vectors, nor by a different management of magnitudes, or by a inventing new 'hypercomplex' numbers; it is just a fundamental fact of 3d (or greater) geometry.

[M Benesi]

  I'm not finding any reference to a proof of  "stretch" anywhere???  In fact, all I can find is the proof of the conformal mapping constraints.

  Anyways...  The additional variables interact to cancel out chaos in certain areas... unless set up properly. 

[Tglad]

Yes, its a shame that the encyclopaedias don't explain very clearly... but conformal means that tiny spheres transformed will still look like spheres. Anti-conformal means that a tiny letter p transformed will look like a letter q, reflections are anti-conformal.
If it isn't conformal or anti-conformal then tiny spheres transformed will become ellipsoids, they will be stretched, or squashed depending on which word you want to use.

[miner49er]

Nothing of value to add but I have to say this is a fascinating discussion.

[msltoe]

I've been thinking about the definition of conformality lately and wrote a small program in Octave (Matlab clone) to compute the "stretchiness" for an arbitrary function.
Here's another way to think about it:
- To be conformal, the Jacobian of the function should be a rotation matrix times a scalar. A Jacobian is a NxN matrix (where N=the dimension). Its matrix entries are the derivatives of the x,y,z functions by x,y, and z.
- Another way to think about the Jacobian of a conformal function: every derivative = infinitesimal vector must be orthogonal to every other (dx . dy = 0, dy . dz = 0, dx . dz = 0) and every derivative's magnitude must be 1 (dx . dx = 1, dy . dy = 1, dz . dz = 1) times a scalar for all entries.
- The limits of conformal functions in N>2 must have to do with the fact that having the Jacobian = a rotation is akin to solving a series of differential equations, but there's too many equations and not enough flexibility for the function.

example, the Mandelbrot function,

fx = x*x - y*y;
fy = 2*x*y;

J = (2x -2y)
      (2y  2x)

by multiplication,

J = 2*sqrt(x*x+y*y) * (cos -sin)
                                   sin  cos)

J = scalar * rotation matrix

[s31415]

I haven't seen it mentioned yet, so I'll mention it. The true definition of a conformal transformation, valid in any dimension, is a transformation that preserves angles. If you have any two smooth curves intersecting at a point, there is a well defined notion of angle between the two curves, namely the angle between their tangents at this point. A conformal transformation will map these curves to different ones, but their angle of intersection will be preserved.

As was mentioned, in dimension greater than 2, the conformal group is finite dimensional, generated by translations, rotations, scale transformations and inversion. In two dimensions, at least if we are interested in conformal transformations defined locally, the conformal group is infinite dimensional. Seeing the plane as the complex plane, any holomorphic change of coordinate is conformal. Holomorphic meaning that the new coordinate w is a function of the old coordinate z but not of its complex conjugate \bar{z}. For instance the function iterated to produce the Mandelbrot set z -> z^2 + p is conformal, because there is only explicit dependence on z on the right hand side. From this example, we see that dimension 2 is very special, because it allows conformal transformations that are not gobally one to one. This fact is the root of many visual features of the Mandelbrot set. That's why, no matter how interesting it is by itself, the Mandelbulb cannot be a true 3d analogue of the Mandelbrot set. And why I believe there is no such thing. But from the richness of these new 3d fractals, it's well worth looking for it...

And I agree that whipped cream effects are a direct consequence of the presence of a non-conformal transformation.

[M Benesi]

The true definition of a conformal transformation, valid in any dimension, is a transformation that preserves angles. If you have any two smooth curves intersecting at a point, there is a well defined notion of angle between the two curves, namely the angle between their tangents at this point. A conformal transformation will map these curves to different ones, but their angle of intersection will be preserved.

 Highlighted an important statement.... and on to an explanation of what I thinks been going on (with the word "conformal"):

  I think here, at least if I remember correctly, Tglad and maybe others were using conformal to mean that a transformation of 3 points (a triangle) with the equation z^2 + c would result in an angle preserved triangle in a new location (perhaps within the Mandelbrot set, or outside the set specifically??).  I never bothered to check the math, so can't say that the M-set transform is conformal or not, and I haven't read a proof either- I've just seen it said numerous times.  I'm actually interested in a link.... but I suppose a simple triangle test (both inside and outside the set, as I don't recall whether it's supposed to be one or the other) would work as well:

A =(0,0)  B= (0, .1i ) C= (-.1,.1i)       right triangle
A'= (0,0)  B'= (-.01,.1i) C'=(-.1,-.08i)   not a right triangle... angles ain't preserved.  NC

outside the set:

A=(1,0)   B= (2,0)  C= (2,1i)    right...
A'=(2,0)  B'= (6,0)  C'= (5,5i)   not right...  NC  

    So I don't really know why.... or how the idea came about that the M-set transform is conformal?

    Any links to the original proof or statement about it?   There is this one...

[msltoe]

Matt,

 For the M-set, the angles are preserved in the infinitesimal limit, x = x + delta, y=y + delta, which is all that's required for conformality.
 My previous post in this thread has the simple proof. (J = Jacobian)

-mike

[Tglad]

Just to add to that, if it helps, here are some equivalent definitions for conformal:
-Angle between any two curves is preserved
-The Jacobian of the derivative of the transform only contains rotation and scale.
-Small triangles map to small triangles with the same internal angles
-Small spheres map to small spheres

If you are testing with triangles then the key word here is 'small' which is used in a very restricted sense, it means infinitely small. Or to write it in long hand- As you decrease the size of a triangle to 0, the transformed triangle's internal angles converge towards those of the original triangle.

In these conformal transforms of a clock, you can see that the large clocks can have weird shapes, but at the tiniest scales the clocks are all circular, so 'small' shapes are preserved-
http://en.wikipedia.org/wiki/Conformal_pictures

[David Makin]

Another question about Louiville's conformality theorem - I hope this makes sense:

The theorem states that the conformal mappings in Rn (n>2) are limited to those as described in this thread and here:

http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28conformal_mappings%29

My question is, is it possible that other transfoms in Euclidian space of Rn (n>2) will be conformal if we only consider a lower dimensional slice of that space, e.g. a slice in Rm (m<n).
I'm pretty sure that the answer is yes ?
If so then I still think it is possible to achieve the elusive "perfect" 3D Mandelbrot equivalent but only as a 3D slice of a Mandelbrot in Rn (n>3).
This would mean that there's no point really persuing anything in plain R3, but IMHO it's still possible that there may be an R4 (or higher) algebra such that a 3D slice gives the holy grail.

[Tglad]

If you mean performing a transformation in only say two of the dimensions then the answer is no...
For example, a transform that did Z = Z^2 on the r and i axes, and J=J on the j axis is not conformal.
If you mean literally only transforming a 2d plane that happens to be in 3d space (e.g. only when j=0), then yes the transformation is conformal, it is an exception because Liouville's theorem applies to smooth transforms, and that wouldn't be smooth or continuous.
It does not follow that 4d space can contain a 3d non-mobius conformal transform, regardless of what algebra you use, the theorem is a property of euclidean geometry, not any particular algebra.