How to estimate the pressure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:44:50Z http://mathoverflow.net/feeds/question/76890 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76890/how-to-estimate-the-pressure How to estimate the pressure? Danny Calegari 2011-09-30T22:05:22Z 2011-10-01T01:34:47Z <p>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).</p> <p>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 $G_n$. </p> <p>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. </p> http://mathoverflow.net/questions/76890/how-to-estimate-the-pressure/76904#76904 Answer by Vaughn Climenhaga for How to estimate the pressure? Vaughn Climenhaga 2011-10-01T01:34:47Z 2011-10-01T01:34:47Z <p>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 $w$.</p> <p>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", <em>Math. Systems Theory</em> <strong>8</strong> (1974/75), no. 3, 193–202).</p> <p>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", <a href="http://arxiv.org/abs/1106.3575" rel="nofollow">arXiv:1106.3575</a>), 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.</p>