A follow up question related to entropy - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:55:54Z http://mathoverflow.net/feeds/question/71636 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy A follow up question related to entropy Mahdi Majidi-Zolbanin 2011-07-30T05:31:12Z 2012-01-28T22:33:25Z <p>For a self-map $\varphi:X\longrightarrow X$ of a space $X$, many important notions of entropy are defined through a limit of the form $$\lim_{n\rightarrow\infty}\frac{1}{n}\log a_n,$$ where in each case $a_n$ represents some appropriate quantity (see, for example, <a href="http://mathoverflow.net/questions/69218/if-you-were-to-axiomatize-the-notion-of-entropy/69231#69231" rel="nofollow">this answer</a> to one of my previous questions.) Let $h(\varphi)$ denote a typical entropy that is defined by a limit as above and <strong>after</strong> Ian's example, assume that $h(\varphi)>0$. Does anyone know if limits of the form $$\lim_{n\rightarrow\infty}\ \ \frac{a_n}{\exp(n\cdot h(\varphi))}$$ have been studied anywhere? I will appreciate any possible information about such limits. For example, is there a known case where the limit exists? If so, what is the limit called? etc.</p> <p><strong>EDIT</strong>: As pointed out later by Ian, even if we assume $h(\varphi)>0$ this limit may not exist. I was curios to know if there were instances where the limit is known to exist. Or even better, can one characterize self-maps $\varphi$ for which the limit exists? </p> http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy/71665#71665 Answer by Vaughn Climenhaga for A follow up question related to entropy Vaughn Climenhaga 2011-07-30T17:02:45Z 2011-07-30T17:02:45Z <p>The limit exists for the first two examples that come to mind, namely topological entropy on the full shift and on certain simple Markov shifts.</p> <p>If $X \subset \Sigma_d^+ = \{1,2, \dots, d\}^{\mathbb{N}}$ and $\sigma$ is the shift map, then for the topological entropy the quantity $a_n$ denotes the number of words of length $n$ that appear in some sequence $x\in X$. If $X$ is the full shift, then $a_n = d^n$, the entropy is $h = \log d$, and we quickly see that $a_n / e^{nh} = 1$ for all $n$.</p> <p>Slightly more interesting is when you have a Markov shift, say $X\subset \Sigma_2^+$ determined by the transition matrix <code>$\begin{pmatrix} 1 &amp; 1 \\ 1 &amp; 0 \end{pmatrix}$</code>.</p> <p>In this case it's not hard to show that the sequence $a_n$ is actually the Fibonacci sequence, and thus writing $\phi = \frac {1+\sqrt 5}2$ and $\psi = \frac { 1-\sqrt 5}2$, we have $$a_n = \frac 1{\sqrt 5} (\phi^{n+2} - \psi^{n+2}).$$ Since $|\psi|&lt;1$, this shows that the limit of $a_n / e^{nh}$ exists.</p> <p>My guess is that a similar argument works for other Markov shifts and shows that the limit exists in those cases, based on obtaining a recurrence relation for $a_n$ and then an exact formula using standard tools for solving such sequences.</p> <p>All that said, it's not immediately clear what the significance of the limit is, and I don't know of any name for it. For other interesting shifts, such as sofic shifts or shifts with specification, I'd be surprised if the limit always exists. </p> <p>What <strong>is</strong> certainly quite important is to have conditions under which the ratio $a_n / e^{nh}$ is bounded away from $0$ and $\infty$. Such estimates are a significant part of arguments on the uniqueness of a measure of maximal entropy (and more generally uniqueness of equilibrium states), in particular the proof that such a measure satisfies a Gibbs property. For example, see Bowen's 1975 paper "<a href="http://www.springerlink.com/content/uu2n4xh52463rn14/" rel="nofollow">Some systems with unique equilibrium states</a>". (Dan Thompson and I also struggled with this not too long ago in Section 5.1 of <a href="http://arxiv.org/abs/1011.2780" rel="nofollow">this paper</a>.)</p> <p>It turns out that in the general setting, one of those bounds is immediate -- the sequence $a_n$ is submultiplicative, and so it's not hard to show that $a_n \geq e^{nh}$ for all $n$, whatever other properties the shift space has. Getting an upper bound on $a_n / e^{nh}$ is harder and requires some sort of specification property.</p> http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy/71669#71669 Answer by Ian Morris for A follow up question related to entropy Ian Morris 2011-07-30T17:52:55Z 2011-07-30T18:05:27Z <p>For the topological entropy of a subshift on finitely many symbols, I think that this limit will typically be infinite. Here is an example where this is the case.</p> <p>Let $\Sigma_2= \{0,1\}^{\mathbb{N}}$ with the infinite product topology and let $T \colon \Sigma_2 \to \Sigma_2$ denote the shift transformation given by <code>$T[(x_i)_{i=1}^\infty]:=(x_{i+1})_{i=1}^\infty$</code>. The map $T$ is a continuous surjective transformation of the compact metrisable space $\Sigma_2$. Let us define a <em>cylinder set</em> of depth $n$ to be a set $Z \subseteq \Sigma_2$ having the form $$Z=\{(x_i) \in \Sigma_2 \colon x_j=z_j \text{ for all }j\text{ such that }1 \leq j \leq n\}$$ for some finite sequence of symbols $z_1,\ldots,z_n \in \{0,1\}$. If $K$ is a nonempty compact subset of $\Sigma_2$ such that $TK \subseteq K$, then the topological entropy of $T$ restricted to $K$ admits the following description: if for each $n \geq 1$ we let $a_n$ be the number of distinct cylinder sets of depth $n$ which have nonempty intersection with $K$, then <code>$h_{top}(K) = \lim_{n \to \infty} \frac{1}{n} \log a_n$</code>. This holds because the cylinder sets form the smallest-growing family of open covers in the sense of the usual definition of topological entropy on a compact space.</p> <p>Now, let $K \subset \Sigma_2$ be a compact $T$-invariant set of Sturmian words with some specified irrational slope (for the definition and fundamental properties of Sturmian words see e.g. Wikipedia). Such sets exist and satisfy $a_n=n+1$ for all $n \geq 1$. In particular the restriction of the shift map $T$ to $K$ has topological entropy zero and $a_ne^{-nh}=n+1 \to \infty$.</p> <p>More generally, a little further thought shows that $a_ne^{-nh} \to \infty$ for every nonempty compact minimal invariant subset of $\Sigma_2$ which has zero topological entropy and is not equal to a periodic orbit.</p> http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy/86916#86916 Answer by Joe Silverman for A follow up question related to entropy Joe Silverman 2012-01-28T20:45:31Z 2012-01-28T20:45:31Z <p>Let $\phi:\mathbb{P}^N\to\mathbb{P}^N$ be a rational map. The <em>algebraic entropy</em> of $\phi$ is the quantity <code>$$h_{alg}(\phi) = \limsup_{n\to\infty} \frac{1}{n}\log \deg(\phi^n).$$</code></p> <p>Suppose now that $\phi$ is defined over $\overline{\mathbb{Q}}$. Since you're using $h$ for entropy, I will let $w:\mathbb{P}^N(\overline{\mathbb{Q}})\to[0,\infty)$ denote the (absolute logarithmic) Weil height. The <em>arithmetic entropy</em> of $(\phi,P)$ is the quantity <code>$$h_{arith}(\phi,P)=\limsup_{n\to\infty} \frac{1}{n}\log w(\phi^n(P)).$$</code> In the arithmetic setting, the quantity you're asking about is more-or-less what's called the canonical height. In particular, if we assume that $\phi$ is a morphism, then $h_{alg}(\phi)=\log(d)$, and <code>$$\lim_{n\to\infty} \frac{w(\phi^n(P))}{\exp(nh_{alg}(\phi)} = \lim_{n\to\infty} \frac{w(\phi^n(P))}{d^n}$$</code> exists and is called the <em>$\phi$-canonical height of $P$</em>. (It's usually denoted $\hat{h}_\phi(P)$.) For the case of morphisms, this is all well known, see for example [2]. Algebraic entropy for rational maps is a subject of current research, see for example [1], and arithmetic entropy is defined (and studied for monomial maps) in [3].</p> <ol> <li>Degree-growth of monomial maps. <em>Ergodic Theory Dynam. Systems</em>, 27(5):1375--1397, 2007.</li> <li>Canonical heights on varieties with morphisms. <em>Compositio Math.</em>, 89(2):163-205, 1993.</li> <li><a href="http://arxiv.org/abs/1111.5664" rel="nofollow">http://arxiv.org/abs/1111.5664</a></li> </ol> http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy/86931#86931 Answer by Barbara Schapira for A follow up question related to entropy Barbara Schapira 2012-01-28T22:33:25Z 2012-01-28T22:33:25Z <p>In the case of the geodesic flow acting on the unit tangent bundle of a compact negatively curved manifold, if $a_n$ is the number of closed geodesics of length at most $n$, and $h$ the topological entropy of the geodesic flow, Margulis proved that $a_n$ is equivalent to $\frac{e^{hn}}{hn}$. </p>