Metric entropy for $\beta$-transformation $x\mapsto \beta x\pmod 1$.

Question

I'm asked to calculate the metric entropy for the $\beta$-expansion $T:[0,1)\to[0,1):x\mapsto\beta x\pmod1$, where $\beta=\frac{1+\sqrt5}2$ is the golden mean, w.r.t. the invariant measure $\mu$ given by \begin{equation} \mu(A)= \int_A g(x)d\lambda(x), \end{equation} with $$g(x)=\begin{cases} \frac{5+3\sqrt{5}}{10}, \text{ } 0\leq x < \frac1\beta, \\ \frac{5+\sqrt{5}}{10} \text{ , } \frac1\beta\leq x <1. \end{cases}$$ in the same way as in this question, using Shannon-McMillan-Breiman theory. In the only answer to that question, an alternative solution is proposed, giving a metric entropy $\log\beta$, which is also stated by other sources.

When I try to use Shannon-McMillan-Breiman theory, I first note that for all $A\in\mathcal B$ (the Borel $\sigma$-algebra), $$\frac{5+\sqrt{5}}{10}\lambda(A)\leq\mu(A)\leq\frac{5+3\sqrt{5}}{10}\lambda(A),$$ hence Shannon-McMillan-Breiman gives $h_\mu(\alpha,T)=\lim_{n\to\infty}-\frac1n\log\lambda(\alpha_n(x))$, where $\alpha_n(x)$ denotes the element of the partition $\bigvee_{i=0}^{n-1}T^{-i}\alpha$ containing $x$. Now if we take $\alpha=\{[0,\frac1\beta),[\frac1\beta,1)\}$, it is not hard to see that a typical element of $\bigvee_{i=0}^{n-1}T^{-i}\alpha$ is an interval $\Delta(i_1,\ldots,i_n):=\{x\in[0,1):b_1(x)=i_1,\ldots,b_n(x)=i_n\}$, where $x=\sum_{i\geq1}b_i(x)/\beta^i$, which has Lebesgue measure $$\left(\frac1\beta\right)^{n-\sum_{i=1}^ni_j}\left(\frac1{\beta^2}\right)^{\sum_{i=1}^ni_j},$$ whence $$-\log\lambda(\Delta(i_1,\ldots,i_n))=(n+\sum_{i=1}^ni_j)\log\beta.$$ Now since the set endpoints of intervals $\Delta(i_1,\ldots,i_n)$ lie dense in $[0,1)$, it follows $\alpha$ is a generating partition w.r.t. $T$, so by Kolmogorov-Sinai $h_\mu(T)=h_\mu(\alpha,T)=\lim_{n\to\infty}-\frac1n\log\lambda(\alpha_n(x))$. From here it comes down to calculating $\lim_{n\to\infty}\frac{1}n\sum_{i=1}^ni_j$, which equals $\mu([\frac1\beta,1))\neq0$ by the Birkhoff ergodic theorem. But this means that I get a different solution (more precisely, $(\frac32-\frac{\sqrt5}{10})\log\beta$) for the metric entropy than the one given in the linked answer and in other sources.

I hope someone can point me out what I did wrong.

Answer 1

I believe your mistake lies in your estimate of the length of your typical intervals, $\Delta(i_1,\ldots,i_n)$.

Since $\alpha$ defines a Markov partition in which applying $T$ to any interval scales its length by $\beta$, so too do does $\bigvee_{i=0}^{n-1}T^{-i}\alpha$, for each $n\in \mathbb N$.

That means any (closed) $I \in \bigvee_{i=0}^{n}T^{-i}\alpha$ satisfies either

• $T^n(I) = [0,\beta^{-1}] \implies \beta^n\lambda(I) = \beta^{-1} \implies \lambda(I) = \beta^{-n-1}$, or
• $T^n(I) = [\beta^{-1},1] \implies \beta^n\lambda(I) = 1-\beta^{-1} = \beta^{-2} \implies \lambda(I) = \beta^{-n-2}$.

Therefore, up to a constant multiple, each interval is of length $\beta^{-n}$, from which you can easily obtain the accepted value.

Remark: Interestingly enough, the numbers of long and short intervals at the $n$th level are the $n$th and $(n-1)$th Fibonacci numbers, respectively. (This relates the fact that for this value of $\beta$, all $\beta$-expansion in $[0,1]$ can be chosen uniquely to have coefficients in $\{0,1\}$ and with no consecutive $1$s — compare to the linked answer in the question.)