Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:08:59Z http://mathoverflow.net/feeds/question/76908 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76908/supremum-amongst-kolmogorov-sinai-entropies-ergodic-or-just-invariant-measures Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. André Caldas 2011-10-01T03:38:31Z 2012-04-29T01:22:00Z <p>Cases where</p> <p><code>$sup_{\mu \in E(T)} h_\mu(T) \neq \sup_{\mu \in M(T)} h_\mu(T)$</code>.</p> <h2>Background</h2> <p>For a topological space $X$, let <code>$T: X \to X$</code> be a continuous application. Then, call the set of $T$-invariant probability measures <code>$M(T)$</code>, and call the set of $T$-ergodic (probability) measures <code>$E(T)$</code>. It is evident that <code>$E(T) \subset M(T)$</code>. But it may happen that <code>$M(T) = \emptyset$</code>. For example, take <code>$X = \mathbb{R}$</code> and <code>$T(x) = x+1$</code>.</p> <p>Since an ergodic measure is invariant, it is immediate that</p> <p><code>$\begin{equation*} \sup_{\mu \in E(T)} h_\mu(T) \leq \sup_{\mu \in M(T)} h_\mu(T). \end{equation*}$</code></p> <p>The question is whether equality holds or not. When $X$ is compact, it is well known that equality holds. In this case, it is a consequence of Jacobs' Theorem, which states that for any <code>$\mu \in M(T)$</code>, there exists a measure $\tau$, over the set <code>$E(T)$</code>, such that</p> <p><code>$\begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m). \end{equation*}$</code></p> <p>When $X$ is compact (locally compact, in fact), the above equation is a consequence of <a href="http://en.wikipedia.org/wiki/Choquet_theory" rel="nofollow" title="Theorem about generalized convex combinations">Choquet Representation Theorem</a> and the <a href="http://en.wikipedia.org/wiki/Krein%E2%80%93Milman_theorem" rel="nofollow" title="Theorem on existence of extremal points">Krein-Milman Theorem</a>. (See, for example, Theorem 8.4 from Walters, P. <em>An Introduction to Ergodic Theory</em>)</p> <p>Now, when $X$ is not necessarily compact, but it is a Borel subset of a compact metrizable set $\widetilde{X}$, Pesin and Pitskel' argue in their <em>Topological Pressure and the Variational Principle for Noncompact Sets</em>, at the end of page 310: (I will rename the spaces and applications in order to conform to this post's notation.)</p> <blockquote> <p>We may assume that measure $\mu$ is ergodic. In fact, consider the partition $\eta$ of $X$ into ergodic components <code>$X_s,\, s \in S$</code>, of measure $\mu$. Denote by <code>$\mu_s$</code> the measures on <code>$X_s$</code> (then <code>$T * \mu_s = \mu_s$</code>), and by $\nu$ the measure on the quotient space $X / \eta$. Then <code>$h_\mu(T) = \int_{Y/\eta} h_m(T) d\nu(m)$</code>.</p> </blockquote> <p>As far as I understand, $Y/\eta$ is just the same as $E(T)$, since each ergodic component is associated with an ergodic measure. And for the same reason, $\nu$ is just our $\tau$. So, what is being stated is the validity of</p> <p><code>$\begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m), \end{equation*}$</code></p> <p>which in turns implies the equality</p> <p><code>$\begin{equation*} \sup_{\mu \in M(T)} h_\mu(T) = \sup_{\mu \in E(T)} h_\mu(T). \end{equation*}$</code></p> <p>In Pitskel' and Pesin's paper, $T$ is not even supposed to be the restriction to $X$ of a continuous transformation <code>$\widetilde{T}: \widetilde{X} \to \widetilde{X}$</code>.</p> <h2>Questions</h2> <ol> <li><p>How do I prove that when $X$ is a Borel subset of a compact metrizable space <code>$\widetilde{X}$</code> and $T$ is a continuous application <code>$T: X \to X$</code>, then for any <code>$\mu \in M(T)$</code>, there exists a measure $\tau$ over $E(T)$ such that <code>$h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m)$</code>?</p></li> <li><p>In case the answer to question "1" is negative, is there a prove for the specific case where $T$ is the restriction of a continuous application <code>$\widetilde{T}: \widetilde{X} \to \widetilde{X}$</code>?</p></li> <li><p>Do you know <em>nice</em> examples of transformations of measurable spaces where <code>$E(T) = \emptyset$</code> while <code>$M(T) \neq \emptyset$</code>?</p></li> </ol> <hr> <p>PS: This is my first post to MathOverflow. This is really exciting! :-)</p> http://mathoverflow.net/questions/76908/supremum-amongst-kolmogorov-sinai-entropies-ergodic-or-just-invariant-measures/77006#77006 Answer by Daniel Mansfield for Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. Daniel Mansfield 2011-10-03T03:07:08Z 2011-10-03T03:07:08Z <p>Welcome to mathoverflow!</p> <p>I believe I can answer question 3. My reference here is Walters, P. <em>An introduction to Ergodic Theory</em>. Chapter 6.2.</p> <p>When $T: X \mapsto X$ is a continuous transformation of a compact metrisable space $X$, then there will always be a measure for which $T$ is measure preserving. Hence $M(T)$ is never empty. (Corollary 6.9.1)</p> <p>The space $M(T)$ is compact, convex and nonempty. Hence it has an extreme point by <a href="http://mathoverflow.net/questions/15654/extreme-point-compact-convex-set" rel="nofollow">this argument</a>. </p> <p>The extreme points are the ergodic measures (Walters theorem 6.10(iii))</p> <p>Hence $E(T) \neq \emptyset$.</p> http://mathoverflow.net/questions/76908/supremum-amongst-kolmogorov-sinai-entropies-ergodic-or-just-invariant-measures/84748#84748 Answer by Pengfei for Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. Pengfei 2012-01-02T14:49:52Z 2012-01-04T03:04:58Z <p>I think the answer is positive for the special case (2). </p> <p>In this case every $(X,T)$-invariant measure $\mu$ is also $(\widetilde{X},T)$-invariant and hence admits an ergodic decompostion, say $\tau=\tau_\mu$. Since $\mu(X)=1$, $m(X)=1$ for $\tau$-a.e. $m\in E(\widetilde{X},T)$. In other words, the ergodic decomposition of $\mu$ is 'localized' on $(X,T)$. Note that $h_m(X,T)=h_m(\widetilde{X},T)$ for each $m$ with $m(X)=1$. Therefore </p> <p>$h_\mu(X,T)=h_\mu(\widetilde{X},T)=\int_{E(T)}h_m(\widetilde{X},T)d\tau(m)=\int_{E(T)}h_m(X,T)d\tau(m)$.</p> <hr> <p>After André Caldas: I take a snapshot (<a href="http://www.freeimagehosting.net/e2770" rel="nofollow">link</a>) of Theorem 6.4 in Chapter II of Mane's book:</p> <p><img src="http://www.freeimagehosting.net/newuploads/e2770.jpg" alt="alt text"></p> http://mathoverflow.net/questions/76908/supremum-amongst-kolmogorov-sinai-entropies-ergodic-or-just-invariant-measures/87545#87545 Answer by R W for Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures. R W 2012-02-04T20:00:12Z 2012-02-04T20:00:12Z <p>You make a pretty common mistake here - your question has nothing to do with continuity, compactness etc. and belongs entirely to the measure category. What matters here is that you have a measure preserving transformation of a probability Lebesgue space (sometimes these spaces are also called standard probability spaces - there is a nice wikipedia article about them) and a measurable partition of this space pointwise invariant with respect to the transformation (particular, and, essentially, the only interesting case: the partition into ergodic components of the transformation). Then the entropy of the transformation is the integral of the entropies of its restrictions to the elements of the partition (the formula which appears in Pesin and Pitskel'). In fact, it should not be attributed to them - a much better reference (actually, for the whole entropy theory as well) is Rokhlin's lecture notes on entropy theory in Russian Math Surveys 1967 (the formula we discuss is in section 8.11 there). It takes care of your questions (1) and (2). As for (3), any invariant probability measure has a unique (mod 0) decomposition into ergodic invariant measures (once again, this is an entirely measure category property), so that there are <b>no</b> (nice or not so nice) examples of transformations $T$ with $E(T)=\emptyset$ while $M(T)\neq\emptyset$.</p>