0
$\begingroup$

Where can I find the details of constructing singular continuous ergodic measures for the map $z \to z^2$ on the unit circle? I know that it was done by Furstenberg, but I could not find it explicitly in his book.

$\endgroup$

1 Answer 1

2
$\begingroup$

$z \mapsto z^2$ is conjugated to $x \mapsto 2x$ mod 1 on ${\bf R}/{\bf Z}$. This map is in turn semiconjugated to the shift map $\sigma(\{a_n\}_{n\in {\bf N}} = \{a_{n+1}\}_{n\in {\bf N}}$ on $\{0,1\}^{\bf N}$ through base 2 decomposition. $$ \{a_n\} \mapsto \sum_n {a_n \over 2^{n+1}} $$ So you can push any ergodic probability measure invariant by the shift to get a continuous measure which is singular from Lebesgue as soon as it is not Lebesgue. So for example, consider for some $p \in (0,1)$ not equal to $1/2$, $$ \mu_p = (p\delta_{0} + (1-p)\delta_{1})^{\otimes {\bf N}} $$ and push this measure to the circle with the map $$ \matrix{\varphi : & \{0,1\}^{\bf N} & \longrightarrow & S^1 \cr &\{a_n\} & \mapsto & exp\Bigl(2\pi i \sum_n {a_n \over 2^{n+1}}\Bigr) \cr} $$ There are many ergodic probability measures invariant by the shift which have full support, far more than just the $\mu_p$. They are built using thermodynamic formalism. A reference is the book of Bowen, LNM 470, Equilibrium states for Axiom A diffeomorphisms. Any book about hyperbolic or symbolic dynamics should fit. Brin Stuck, Introduction to dynamical systems, Katok-Hasselblatt Modern theory of dynamical systems etc.

$\endgroup$
1
  • 1
    $\begingroup$ To add to @coudy's answer, there is exactly one absolutely continuous ergodic invariant measure, namely Lebesgue. All other ergodic measures are mutually singular with respect to Lebesgue. Some of these are the equilibrium states described in the answer. But there are also many interesting measures that are very different. For instance, the Sturmian measures have zero entropy. These are studied in the paper "Le poisson n'a pas d'arêtes" by T. Bousch. $\endgroup$ Commented Nov 11 at 5:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .