Suppose I have some nicely defined "fractal" subset of (to make life simpler) Euclidean space $\mathbb{E}^n,$ of some arbitrary Hausdorff dimension $s,$ such that the corresponding Hausdorff measure $H_s$ has positive mass. The first question is whether it makes sense to talk of a "uniform sample" (or Poisson point process) with respect to $H_s$ (since it is only an outer measure), and second question is: if it makes sense, have people figured out how to generate uniform $H_s$ variates (in simple cases, like the von Koch snowflake, or the more interesting cases, like limit sets of Kleinian groups)?
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$\begingroup$ What do you mean "only an outer measure"? Isn't it a measure on the borel sets? Also, does "variate" here just mean an algorithm to produce uniform random variables? If so, it is easy in the case of the snowflake. $\endgroup$– Ori Gurel-GurevichCommented Sep 17, 2011 at 19:12
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$\begingroup$ What Ori said. Use a Borel set. All the usual "nicely fractal" sets in Euclidean space are Borel sets. Indeed, they are compact or at least sigma-compact. $\endgroup$– Gerald EdgarCommented Sep 17, 2011 at 19:16
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$\begingroup$ I am a babe in the woods of measure theory... As for variates, yes, an algorithm. In particular, what is the "easy method" for the snowflake? And how do you go about proving that some algorithm gives uniform points (for snowflake or dynamically defined sets)? $\endgroup$– Igor RivinCommented Sep 17, 2011 at 19:20
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$\begingroup$ Other people have answered, but still: To pick a uniform point on one side (out of three) of the snowflake, you just have to pick 1 of the 4 parts of the curve, then 1 of 4 parts of that part, etc. It is really no different than picking a uniform point on $[0,1]$, by iteratively choosing halves of the interval. $\endgroup$– Ori Gurel-GurevichCommented Sep 19, 2011 at 18:44
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$\begingroup$ @Ori: I sort of guessed that, but why is that true? What is the general version of the construction? Presumably, it's some iterated function system thing, but if I knew all there was to know about such, I would not be asking the question... $\endgroup$– Igor RivinCommented Sep 19, 2011 at 19:25
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3 Answers
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There is a standard map from the interval $[0,1]$ onto the snowflake curve, that maps Lebesgue measure into (constant multiple of) Hausdorff measure. So take a sample in $[0,1]$ and map it to the snowflake.
This works for all of the standard "self-similar" IFS constructions of fractals, when the IFS satisfies the open set condition. There is a standard "parameter space" or "code space" with a well-understood measure on it that maps onto the fractal, and the image measure is (constant multiple of) Hausdorff measure, and also (constant multiple of) packing measure.
answered Sep 17, 2011 at 20:17
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$\begingroup$ @Gerald: thanks, but saying that the construction is "standard" does not tell me what it is. Would you happen to have any references you would particularly recommend? $\endgroup$ Commented Sep 17, 2011 at 20:20
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$\begingroup$ Ken Falconer's books are a great place to start. $\endgroup$ Commented Sep 19, 2011 at 14:21
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$\begingroup$ @BSteinhurst: thanks, I have started looking at one of them already... $\endgroup$ Commented Sep 19, 2011 at 15:21
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$\begingroup$ I could also plug my own book, Measure, Topology, and Fractal Geometry $\endgroup$ Commented Sep 19, 2011 at 18:31
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$\begingroup$ @Gerald: I don't think there is anything morally wrong about suggesting your own book (especially if it is the best book on the subject :)) $\endgroup$ Commented Sep 19, 2011 at 19:26
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Given $A$ a loc. compact space and $\mu$ a loc. finite measure on $A$, there exists a unique Poisson point process $\eta$ with control measure $\mu$. (see for instance th.3.2.1 in "stochastic geometry", by Schneider & Weil, if there are no atoms, but it holds in full generality) .
If $A$ is your fractal and $\mu$ Hausdorff measure, then $\eta$ is what you are looking for, isn't it?
For the construction, everything depends on how you define your fractal set. The difficulty is exactly the same as drawing a uniform random point in a subset that has finite Hausdorff measure.
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$\begingroup$ Firstly, thanks for the reference! As for "nicely defined", the canonical example is a limit set of a kleinian group, but again, the snowflake is a nice simple example... $\endgroup$ Commented Sep 19, 2011 at 15:19
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If one interprets "nicely defined" correctly, the answer is "yes, duh!".
One possible interpretation is to be the attractor of a Horseshoe map. Then the dynamics on this attractor can be conjugated to symbolic dynamics and it is clear what "drawing a random point" means, and how to generate them numerically. Of course, one also needs to check that the invariant measure has maximal Hausdorff dimension.
In order to illustrate this, let me give the simplest example. Consider the map $f(x) = x^2 - c$ for $c > 4$. It is well known, that there exists an invariant set $\Sigma = f^{-1}(\Sigma)$ with Hausdorff dimension $\in (0,1)$. $\Sigma$ can be described as $$ \Sigma = \{x:\quad \exists C>0 \forall n\geq 1, |f^n(x)|\leq C\} $$ so the set of points with bounded orbit. If you define for $x\in \Sigma$ the sequence $$ x_n = \begin{cases} 1 & f^n(x) > 0, \\\ - 1,& f^n(x) < 0 \end{cases} $$ Then standard results imply that the Bernouilli measure on $\{-1,1\}^{\mathbb{N}}$ is conjugate to the measure of maximal Hausdorff dimension, which is ergodic with respect to $(\Sigma, f)$. Hence, picking a random point just amounts to choosing a random string of $\pm 1$.
