User andré caldas - MathOverflow

System with invariant measure, but no ergodic measure.

2011-10-03T11:30:01Z

Question

Examples of continuous transformations $T: X \to X$ such that the family of invariant probability measures $M(T)$ is NOT empty but there is no ergodic measure ($E(T) = \emptyset$).
Notice that the measures considered are defined over the Borel sets of $X$.
Example of a dynamical system where the following inequality is strict: $$\sup_{m \in E(T)} h_m(T) < \sup_{\mu \in M(T)} h_\mu(T)$$.

Background

Consider $T(x) = x + 1$ over the set of integers $\mathbb{Z}$. In this case, $E(T) = M(T) = \emptyset$. The first question asks for a $\emptyset = E(T) \subsetneq M(T)$ example.

In the locally-compact metrizable case, the set of positive invariant measures $\mu$ with $0 \leq \mu(X) \leq 1$ is compact (weak* topology) with extremals with total measures equal to $0$ or $1$. That is, according to Krein-Milman Theorem, if $M(T) \neq \emptyset$, then $E(T) \neq \emptyset$. So, an answer to Question 1 is not supposed to be locally-compact metrizable.

[Edit: The question only makes sense if the $\sigma$-algebra is fixed. So the post was edited, making $X$ a topological space, $T$ continuous and the $\sigma$-algebra is the family of Borel sets.]

Supremum amongst Kolmogorov-Sinai entropies: ergodic or just invariant measures.

2011-10-01T03:38:31Z

Cases where

$sup_{\mu \in E(T)} h_\mu(T) \neq \sup_{\mu \in M(T)} h_\mu(T)$ .

Background

For a topological space $X$, let $T: X \to X$ be a continuous application. Then, call the set of $T$-invariant probability measures $M(T)$ , and call the set of $T$-ergodic (probability) measures $E(T)$ . It is evident that $E(T) \subset M(T)$ . But it may happen that $M(T) = \emptyset$ . For example, take $X = \mathbb{R}$ and $T(x) = x+1$ .

Since an ergodic measure is invariant, it is immediate that

$ \begin{equation*} \sup_{\mu \in E(T)} h_\mu(T) \leq \sup_{\mu \in M(T)} h_\mu(T). \end{equation*} $

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 $\mu \in M(T)$ , there exists a measure $\tau$, over the set $E(T)$ , such that

$ \begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m). \end{equation*} $

When $X$ is compact (locally compact, in fact), the above equation is a consequence of Choquet Representation Theorem and the Krein-Milman Theorem. (See, for example, Theorem 8.4 from Walters, P. An Introduction to Ergodic Theory)

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 Topological Pressure and the Variational Principle for Noncompact Sets, at the end of page 310: (I will rename the spaces and applications in order to conform to this post's notation.)

We may assume that measure $\mu$ is ergodic. In fact, consider the partition $\eta$ of $X$ into ergodic components $X_s,\, s \in S$ , of measure $\mu$. Denote by $\mu_s$ the measures on $X_s$ (then $T * \mu_s = \mu_s$ ), and by $\nu$ the measure on the quotient space $X / \eta$. Then $h_\mu(T) = \int_{Y/\eta} h_m(T) d\nu(m)$ .

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

$ \begin{equation*} h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m), \end{equation*} $

which in turns implies the equality

$ \begin{equation*} \sup_{\mu \in M(T)} h_\mu(T) = \sup_{\mu \in E(T)} h_\mu(T). \end{equation*} $

In Pitskel' and Pesin's paper, $T$ is not even supposed to be the restriction to $X$ of a continuous transformation $\widetilde{T}: \widetilde{X} \to \widetilde{X}$ .

Questions

How do I prove that when $X$ is a Borel subset of a compact metrizable space $\widetilde{X}$ and $T$ is a continuous application $T: X \to X$ , then for any $\mu \in M(T)$ , there exists a measure $\tau$ over $E(T)$ such that $h_\mu(T) = \int_{E(T)} h_m(T) d\tau(m)$ ?
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 $\widetilde{T}: \widetilde{X} \to \widetilde{X}$ ?
Do you know nice examples of transformations of measurable spaces where $E(T) = \emptyset$ while $M(T) \neq \emptyset$ ?

PS: This is my first post to MathOverflow. This is really exciting! :-)

Answer by André Caldas for locally connected versus locally compact

2011-10-13T04:01:52Z

I, myself, am revolted with such a definition, too.

It seems, according to Wikipedia, that there is no consensus on the definition. And probably, few people care about it, because in Hausdorff spaces, all the definitions are equivalent. In addition to the definition you present, one could also say that a space is locally compact when every point has a closed compact neighbohood. In general, even if a neighborhood is compact, it does not mean its closure will be compact as well.

In the Hausdorff case, the closure of subsets of compact sets are compact, since every compact set is closed in this case. Also, in the Hausdorff case it is true that if $K$ is a compact neighborhood of $x$, then $x$ has a neighborhood filter base made out of compact sets. This is because $K$ is a normal space.

Compact group extension of a zero entropy system.

2011-10-05T03:18:02Z

Suppose $T: X \to X$ is a continuous map and $\mu$ a $T$-ergodic probability measure over the Borel sets of $X$. Now, suppose $K \subset \mathrm{Hom}(X)$ is a compact group of measure-preserving homeomorphisms of $X$ commuting with $T$. Consider the factor $Y = K \backslash X = \{Kx | x \in X\}$ . Then, defining $\hat{T}(Kx) = T(Kx) = KT(x)$, we have that the canonical projection factors $T$ into $\hat{T}$.

Call $\hat{\mu}$ the $\hat{T}$-invariant probability measure induced on the Borel sets of $Y$. How do I prove that $$h_{\hat{\mu}}(\hat{T}) = 0 \Rightarrow h_\mu(T) = 0?$$

I am trying to make sense of the demonstration of Lemma 7 at Asymptotic Pairs in Positive-Entropy Systems.

Notation: $h_\mu(T)$ is the Kolmogorov-Sinai entropy.

Notice: It is evident that $h_\mu(T) = 0 \Rightarrow h_{\hat{\mu}}(\hat{T}) = 0$.

Notice: The factor $K \backslash X$ is intended to be a group extension.

Edit: Added the fact that elements of $K$ must be measure-preserving.

Edit: Changed $\mu$ from $T$-invariant to $T$-ergodic.

Furstenberg-Zimmer Theorem: non-invertible systems.

2011-10-03T03:26:48Z

Questions

Is there a version of Furstenber-Zimmer Theorem for non-invertible measure preserving systems?
Where can I find it?
What is the precise statement?

Background

In many works that reference the Furstenberg-Zimmer Theorem, the theorem itself is not stated. Authors usually cite the works of Furstenberg (The structure of distal flows and/or Ergodic behavior of diagonal measures and a theorem of Szemerédi on arithmetic progressions) and Zimmer (Extensions of ergodic group actions and/or Extensions of ergodic actions and generalized discrete spectrum). The point is that in many places, the theorem is being used for non-invertible systems. This happens, for instance in On Li-Yorke Pairs, where the systems are assumed to be surjective, but not necessarily invertible. In this paper, for the proof of Theorem 2.1, the authors use Furstenber-Zimmer Theorem.

As far as I understood, Zimmer's work deals with group actions. That is, invertible systems. And for Furstenberg's Ergodic behaviour of diagonal measures [...], he deals with regular measure preserving systems.

Unfortunately, Furstenberg and Zimmer (obviously) did not call their result the Furstenberg-Zimmer Theorem. In fact, it seems to me that Furstenberg didn't even call it a theorem. :-P

I could find a precise statement of the theorem for the invertible case at a Terry Tao's post. But I could not find any precise statement for the non-invertible case.

Answer by André Caldas for Poincare Recurrence and Dense Sets

2011-10-04T18:37:52Z

Since you are considering $\mu$ to be the Lebesgue measure, you have that given a set $F$ with $\mu(F) = 1$, then $F$ must be dense. This is because any set with non-empty interior has positive measure, and $F^\complement$ has measure $0$. So, $F^\complement$ has empty interior and consequently, $F$ is dense.

The above easily gives a partial answer to your first question:

If $\mu(E) = 1$, then $\mu(M) = \mu(E) = 1$. As explained, $M$ is dense in $X$.

In a similar fashion, for the second question:

If $E$ is such that for any open set $A$, $$A \cap E \neq \emptyset \Rightarrow \mu(A \cap E) \neq 0,$$ then $M$ is dense in $E$.
In fact, $\mu( M^\complement \cap E ) = 0$, so, the assumption on $E$ implies that $M^\complement \cap E$ has empty interior in $E$. That is, $M$ is dense in $E$.

Now, using this fact, we can improve the answer to the first question:

If $E$ is dense in $X$, $\mu(E) > 0$ and $$A \cap E \neq \emptyset \Rightarrow \mu(A \cap E) \neq 0,$$ then $M$ is dense in $E$, and consequently, dense in $X$.
It is obvious that if $E$ is not dense in $X$, then $M$ is not dense either.

Answer by André Caldas for Examples of common false beliefs in mathematics.

2011-10-03T12:32:12Z

In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$ , then

$G_1 / H_1 \cong G_2 / H_2$ .

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$ . The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quocient.

Comment by André Caldas

2012-01-03T13:41:46Z

How do you prove that $\mu(X) = 1$ implies $m(X) = 1$ for $\tau$-a.e. $m$? Remember that having an "ergodic decomposition" means that for every continuous $\phi: \widetilde{X} \to \mathbb{R}$, $\int \phi d\mu = \int_{E(T)} \int \phi dm d\tau$. Also, notice that it might happen that $X \neq T^{-1}X$.

Comment by André Caldas

2011-10-13T18:57:30Z

This is a very neat way to justify my intuitive but not so elaborate and well justified belief that any neighborhood should have the property. Note taken! :-) I think of a neighborhood as a "sufficiently large set". Large enough to be considered a neighborhood. Often, one needs to choose a "small" set, but at the same time, big enough to make the small step big enough to achieve the goal... "path connect two points", for instance.

Comment by André Caldas

2011-10-13T15:06:58Z

I liked your question, Leandro! I had never thought of $\mathbb{R}$ as a infinite dimensional space with compact unit ball! You opened my mind a bit! :-) +1

Comment by André Caldas

2011-10-13T14:52:18Z

Which one is your "standard 'locally <blank>' definition"? :-)

Comment by André Caldas

2011-10-11T19:54:09Z

The extension is unique, are you just asking if the extension is or not an open surjection?

Comment by André Caldas

2011-10-07T14:16:38Z

@Asaf: Thank you very much for the help. I will study your answer. :-)

Comment by André Caldas

2011-10-05T17:05:02Z

By the way, a similar argument to "Lemma 7" is used in Theorem 2.3 of "On Li-Yorke Pairs": imath.kiev.ua/~skolyada/LY.pdf "Since an isometric extension of a zero-entropy system has still entropy zero [...]"

Comment by André Caldas

2011-10-05T16:46:37Z

@Asaf: Oops... (again!) Sorry for not paying the attention you deserve! Theorem 8.2 reads: If $(X, \mathcal{B}, \mu, T)$ is an ergodic isometric extension of $(Y, \mathcal{D}, \nu, T)$ then $(X, \mathcal{B}, \mu, T)$ is a factor of an ergodic group extension of $(Y, \mathcal{D}, \nu, T)$. In Furstenberg's paper, the definition for isometric extensions is quite complicated, but there is a comment (p. 236) saying: Extensions for which $\mathcal{E}(X|Y,T) = L^2(X)$ will be called isometric extensions.

Comment by André Caldas

2011-10-05T14:39:22Z

@Asaf: I will have to study a bit more to fully understand your comments... :-) I guess I am forgetting the fact that $\mu$ is supposed to be ergodic. I will edit (again!) the post. By the way, I am trying to make sense of the demonstration of Lemma 7 at math.u-psud.fr/~ruette/articles/asympent.pdf

Comment by André Caldas

2011-10-05T12:01:50Z

@Asaf: Oops... you are right. It is a very good example. I will fix the post.

Comment by André Caldas

2011-10-05T11:46:54Z

I do not want to prove the entropy decreases. It is quite the opposite. I want to prove that if the entropy in the quotient is $0$ then it will be $0$ in the original system.

Comment by André Caldas

2011-10-05T11:44:25Z

Yes, $G$ means $K$... :-) I will correct it.

Comment by André Caldas

2011-10-04T19:09:21Z

Thank you very much, Asaf! This is my new canonical reference for ergodic theory. I am ordering this book. I had a pick, and what I really needed was Theorem 7.21. It is stated for not necessarily invertible "Borel Probability Spaces" (Definition 5.13).

Comment by André Caldas

2011-10-04T17:43:22Z

@Asaf: Tao assumes it in Lecture 1: terrytao.wordpress.com/2008/01/08/… For the reference you pointed (Manfred), I think this is exactly what I needed. I didn't know about this book. If I may, I'd like to suggest you to post it as an answer. :-) Manfred's book is what I should be reading!! Thank you very very much!

Comment by André Caldas

2011-10-04T17:24:06Z

@Asaf: Subspaces of locally-compact spaces might not be locally-compact. The rationals (or irrationals) are not locally-compact. As for the Krein-Milman theorem, what one needs is the topology to be generated by a separating family of linear functionals, and the set in question to be compact in this topology.