3
$\begingroup$

Is there a way to compare if two fractals are "isomorphic"? I'll give an example of what I mean. Consider the following two fractals. First we have the Sierpinski triangle, and next we have the "Sierpinski diamond".

fractal 1:

Sierpinski triangle

fractal 2:

Sierpinski diamond

Now it seems to me that while the two fractals clearly aren't identical, they ought to be the same in every way that matters, i.e. isomorphic somehow. If someone told me that they had just invented a new fractal and showed me the diamond, I would not be impressed because it is essentially the "same" fractal even though the number of triangles per iteration are different. Is this reasonable? I'm picturing this as something like how there is a bijection between $\mathbb{N}$ and $\mathbb{Q}$.

As for two fractals that would be not isomorphic, I think that the Mandelbrot set and Sierpinski's triangle ought to be "different".

One guess is that Sierpinski's triangle is a "discrete" fractal in that it corresponds to the limit of a sequence indexed with the natural numbers, whereas the Mandelbrot set is like a "continuous" process. This then would be like how $\mathbb{N}$ and $\mathbb{R}$ are not isomorphic.

Is there any mathematical support for this intuition? I'd really like to know the proper terms for this sort of thing, find some accessible references, and hear any comments that you might have. Thanks!

$\endgroup$
  • 2
    $\begingroup$ If two fractals have different Hausdorff dimensions they are clearly different fractals, but if they have the same Hausdorff dimension they are not necessarily the same. $\endgroup$ – user_of_math Oct 14 '14 at 4:11
  • 1
    $\begingroup$ I don't think your "diamond fractal" is even a fractal. You see, only the triangles at the "mirror" boundary constitute the diamond, while all the other triangles do not. In other words, the recursive rule breaks down globally. $\endgroup$ – zudumathics Nov 26 '16 at 7:32
4
$\begingroup$

By "isomorphic" I would say that we have some class of functions mapping sets to sets that preserve relevant properties of those.

In topology, for example, two sets are said to be homeomorphic if there is a continuous bijection with a continuous inverse mapping one to the other. Any two line segments in the plane are homeomorphic, for example. A circle is homeomorphic to an ellipse, or to any smooth, closed curve that doesn't self-intersect. A circle is not homeomorphic to a segment, however. This is all interesting because a homeomorphism preserves topological properties. An open subset maps to another open subset, for example. A continuous function on one set can be transferred to a continuous function on the other. It's important to think about what properties you want to preserve when talking isomorphism.

I don't know of a universally accepted notion of isomorphism in fractal geometry. However, I think that one reasonable candidate would be to say two fractals are isomorphic if they are bi-lipschitz equivalent. More specifically, if $X$ and $Y$ are subsets of Euclidean space (or even metric spaces), then a function $f:X\rightarrow Y$ is called bi-lipschitz if there are positive numbers $m$ and $M$ such that $$m|x-y| \leq |f(x)-f(y)| \leq M|x-y|.$$ The sets $X$ and $Y$ are bi-lipschitz equivalent, if such a map exists. Bi-lipschitz maps are sometimes called maps of bounded distortion. Here's an image of the Sierpinski triangle together with a bi-lipschitz image.

enter image description here

Note that a bi-lipschitz map is automatically a homeomorphism. Thus, they preserve topological properties. Bi-lipschitz is a stronger notion, however, and also preserves fractal dimension - an important consideration in this context. Thus, the distorted image of the Sierpinski triangle above has dimension $\log(3)/\log(2)$, which it presumably should.

Now, your "diamond fractal" is not bi-lipschitz" equivalent to the Sierpinski triangle. Indeed, I would say that it is unreasonable to call these equivalent under any circumstances. They are not even homeomorphic! Now I agree that it might not be very impressive to simply toss out the second set as a groovy picture and say "Hey! Look what I made!!" On the other hand, suppose you are studying deeper properties of these objects. For example, there is a well understood theory of differential equations on the Sierpinski triangle which allows to discuss how heat might conduct through such a set or how a wave might propagate through it. Much of this theory has been extended to similar types of sets - the so-called finitely ramified self-similar sets. Now your diamond fractal is not bi-lipschitz equivalent to any such set and it might be quite a task to develop such a theory. In short, if someone came up and said "I have a rigorous development of a Laplacian on the diamond fractal", I think that would be impressive and would be substantively different from the similar theory on the Sierpinski triangle.

$\endgroup$
  • 1
    $\begingroup$ Thank you very much. This is really helpful. $\endgroup$ – jld Oct 14 '14 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.