# Comparing fractals

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:

fractal 2:

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!

• 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. – user_of_math Oct 14 '14 at 4:11
• 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. – zudumathics Nov 26 '16 at 7:32

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.
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.