## Is there an intrinsic definition of fractal (i.e. not embedded in euclidean space)?

Long ago, manifolds were embedded subsets of euclidean space defined by polynomials. Later, using the gluing of open sets, people realized they could define manifolds intrinsically. And in certain cases, this lead to new manifolds which could not be realized as subsets of euclidean spaces. Eg non-orientable surfaces cannot be embedded in $R^3$, non-algebraic manifolds etc.

Nowadays, fractals are found as iterations of certain polynomial maps on euclidean space. (or some other ways, as wikipedia says, but still embdedded). Are we missing out on some fractals? Is there an intrinsic definition of fractals? (This would also have to come with an intrinsic definition of dimension. Haudsorff dimension is still an embedded dimensions, using covering of balls in euclidean space.)

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 All smooth manifolds can be realized as subsets of Euclidean space of a suitable dimension. – Qfwfq May 18 2010 at 14:14 I believe you need the addition there of "locally". Any open ball can be mapped to an open set in Euclidean space – Michael Hoffman May 18 2010 at 16:46

Topological dimension (say, covering dimension) $\dim_\mathrm{T}$ and Hausdorff dimension $\dim_\mathrm{H}$ both make sense for metric spaces. Benoit Mandelbrot defined $A$ to be a fractal iff $\dim_\mathrm{T} A < \dim_\mathrm{H} A$. The packing dimension $\dim_\mathrm{P}$ also makes sense in metric space. James Taylor defined $A$ to be a fractal iff $\dim_\mathrm{H} A = \dim_\mathrm{P} A$. Also making sense for metric space is the definition of Michael Barnsley ... A fractal is an element of the hyperspace $\mathbb{H}(K)$ of a compact metric space $K$. Perhaps you have your own, different, definition?

Definitions for all these are in my book Integral, Probability, and Fractal Measures

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thanks edgar. and all these definitions of dimension are invariant under isometry/hemeomorphism? – Colin Tan May 19 2010 at 3:01
for instance any simply connected proper domain of the plane (which could be a fractal) is biholomorphic to the open disc (which is not a fractal). but yet biholomorphism implies homeomorphism. – Colin Tan May 19 2010 at 3:06
@Colin: I prefer to say not that the domain is fractal, but the boundary is. And that biholomorphism generally is not Lipschitz on the boundary. – Gerald Edgar May 19 2010 at 14:40

To the best of my knowledge there is no universally agreed upon precise definition of the word "fractal", so it's not clear to me exactly what would or would not constitute an example of a fractal that is not embedded in Euclidean space.

However, the various quantities referred to as "fractal dimension" -- Hausdorff dimension, box dimension, etc. -- do not actually require an ambient Euclidean space for their definition. All you need is a metric on the set X under consideration -- this is enough to define "balls of radius r", and once you can do that the definition of Hausdorff dimension, box dimension, etc. goes through exactly as in the Euclidean case.

In fact, there's a very general framework for all these dimensional quantities (for me the standard reference is "Dimension Theory in Dynamical Systems" by Yakov Pesin), which can be formulated in a setting completely independent of Euclidean space.

As a possible example of a "non-Euclidean fractal", I would consider the symbolic space $\Sigma_2^+ = \{0,1\}^\mathbb{N}$ with a metric such as $d(x,y) = 2^{-t(x,y)}$, where $t(x,y)$ is the first coordinate in which x and y differ. This is homeomorphic to the Cantor set but not embedded in Euclidean space.

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