Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a finite collection of diffeomorphisms $g_1,\cdots,g_n$ taking the unit interval $I$ to disjoint subintervals $I_1, I_2,\cdots,I_n$. If $G$ is the semigroup they generate, the limit set of $G$ (also called the attractor of the IFS) is a Cantor set, and under suitable hypotheses, Bowen (and under slightly weaker hypotheses, Urbanski) showed that the Hausdorff dimension of this Cantor set is the smallest zero of the pressure function $P$, defined by $$P(t) = \lim_{n \to \infty} \frac 1 n \log \sum_{w \in G_n} \|w'\|^t$$ where $G_n$ is the set of elements in $G$ of word length $n$, and $\|\cdot\|$ is the sup norm (the hypotheses for Bowen's theorem is that the $g_i$ are uniformly contracting; Urbanski proves the same theorem when the $g_i$ are allowed to have neutral fixed points; my examples have such points).

This is all well and good, but how do I actually estimate the least zero of the pressure function for an explicit example? (yes, I mean numerically) My $g_i$ are all given by the restrictions of explicit polynomial functions of low degree, but the computational bottleneck seems to be the large number of elements in $G_n$.

Or is there a better method to estimate the Hausdorff dimension in practice? Note that although I just want to estimate the dimension, I would like to be able to (computer-assisted if necessary) give rigorous bounds on the error.

share|improve this question

1 Answer 1

Write $\Lambda_n(t) = \sum_{w\in G_n} |f'(w)|^{-t}$, where $f\colon \bigcup_j I_j \to I$ is the interval map whose inverse branches are the maps $g_j$, and where $|(f^n)'(w)|$ is the supremum of $|(f^n)'(x)|$ taken over all $x$ in the basic interval corresponding to the word $w$.

Suppose that $f$ is uniformly expanding (the $g_j$ are uniformly contracting) and $f'$ is Holder continuous. Then there exists $V\in \mathbb{R}$ such that $|(f^n)'(x)/(f^n)'(y)| \leq e^V$ whenever $x$ and $y$ are in the same basic interval (of any order). In this case standard estimates yield $$ (1) \qquad \qquad e^{nP(t)} \leq \Lambda_n(t) \leq e^V e^{nP(t)} $$ for every $n$. This doesn't get rid of the fact that there are a large number of elements in $G_n$, but it at least gives you the rigorous bound $|P(t) - \frac 1n \log \Lambda_n(t)| \leq \frac Vn$ for any given $n$. Versions of (1) can be found in (Rufus Bowen, "Some systems with unique equilibrium states", Math. Systems Theory 8 (1974/75), no. 3, 193–202).

In the non-uniformly expanding case (Urbanski's), it's not quite so easy, since in general there may not be any $V$ with the bounded distortion property described above. Nevertheless, certain partial bounded distortion results should still be available, and at least for $t$ less than the Hausdorff dimension of the limit set, I believe the techniques in (Climenhaga and Thompson, "Equilibrium states beyond specification and the Bowen property", arXiv:1106.3575), particularly Section 4.3 and Proposition 5.3, will suffice to show that there is a constant $V$ such that (1) holds. I'm not sure how easy that constant will be to compute, or how it may decay as $t$ approaches the Hausdorff dimension of the limit set -- however, if you can use that to get $P(t)>0$ for some $t$ that you suspect is a very good estimate, then you have the lower bound on Hausdorff dimension, which is typically the harder one, and the upper bound may be accessible by other means.

share|improve this answer
1  
Reading through what I just wrote, I'm not sure how useful it will actually be, since to get within $\delta$ of the pressure you need $n=V/\delta$, which may be quite large. I suppose it depends on how big the bounded distortion constant $V$ ends up being, which is basically a measure of how far your IFS is from being linear. Oh well... I'll leave it up here anyway... –  Vaughn Climenhaga Oct 1 '11 at 1:36
    
Thanks for the references and the suggestions. I am in fact interested in the non-uniformly contracting (OK, expanding) case, so I'll take a look at the Climenhaga-Thompson paper. –  Danny Calegari Oct 1 '11 at 6:16
    
Just out of curiosity, what did you have in mind when you say "the upper bound may be accessible by other means"? –  Danny Calegari Oct 1 '11 at 6:20
    
If you want to show that $t_1\leq dim_HZ\leq t_2$, getting $\dim_H Z\geq t_1$ requires you to prove something about every possible $\epsilon$-cover of $Z$, which is one reason Bowen's equation and the thermodynamic formalism is a useful tool for computing Hausdorff dimension, because it lets you sidestep that need. Proving $dim_H Z\leq t_2$, on the other hand, just requires proving something for particular good $\epsilon$-covers of $Z$, which for something like the limit set of an IFS can be chosen in a very canonical way, in this case as being composed of basic intervals. –  Vaughn Climenhaga Oct 1 '11 at 16:43
    
So the "other means" I meant were to pick a $t$ that you think is an upper bound, and then compute an upper bound for the $t$-dimensional Hausdorff measure by using covers by basic intervals. In the end, though, I suppose that just comes down to computing the sum $\sum_{w\in G_n} \|w'\|^t$ and showing that it goes to $0$, which is more or less equivalent to showing that $P(t)\leq 0$. So this is probably another example of something that's pretty straightforward from the theoretical side but can be problematic numerically. –  Vaughn Climenhaga Oct 1 '11 at 16:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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