Let $J$ be an arc in $\mathbb{S}^{1}\subset\mathbb{C}$ (no matter open or closed) and $\alpha\in(0,2\pi)$ be an angle such that $\alpha/\pi$ is irrational. Consider in $\mathbb{S}^{1}$ the sequence $z_{n}=e^{in\alpha}$. Then this sequence is dense in $\mathbb{S}^{1}$ by Kronecker's Theorem or by ergodicity. Let's associate with the arc $J$ its "indicator sequence" $s(J)={s_{n}\}$ of zeroes and ones defined as follows:

$s_{n}=1$ if $z_{n}\in J$ and $s_{n}=0$ if $z_{n}\notin J$.

So, we get something like 0 0 1 1 1 0 0 0 1 1 0 0 1. . . Suppose that we are given such a sequence $s(J)$ for some $J$ and some $\alpha$. By the Ergodic Theorem one gets the measure of arc $J$ as the limit

$\mathtt{meas}(J)=2\pi\underset{n\rightarrow\infty}{\lim}\frac{\sigma_{n}}{n}$ where $\sigma_{n}$ is the number of 1's in ${s_{1},s_{2},...,s_{n}}$.

OK, but is it possible to detect the "frequency" $\alpha$ only by the 0-1 data contained in the sequence $s_{n}$? More precisely, my question is:

Let $\{s_{n}\}$ be a sequence of 0's and 1's and we know that it is an "indicator sequence" for some arc $J\subset\mathbb{S}^{1}$ and some angle $\alpha$. Is it then possible to get $\alpha$ by some formula similar to the above one for the measure of $J$? This would be something like a "rotation number" of sequence $\{s_{n}\}$.

Similar question may be posed for the torus $\mathbb{T}^{n\text{ }}$and an open set $J\subset\mathbb{T}^{n\text{ }}$. Then we should detect not only the frequencies $\alpha_{1},\alpha_{2},...$ but also the "dimension" $n$ of the sequence. Here $\alpha_{1},...,\alpha_{n},\pi$ have to be independent over $\mathbb{Z}$.

[I know that the "indicator sequence" is a standard construction in symbolic dynamics, but I am not very involved in the topic, so references are welcome.]

P.S. Curly brackets {} are not displayed in math mode. How to fix the problem?

infinitetype. Subshifts of finite type are those subshifts which can be obtained from the full shift by "forbidding" finitely many words, and have special properties not enjoyed by Sturmian shifts. (These days it is usually assumed that the forbidden words are all of length two, which can be achieved by re-coding the alphabet.) $\endgroup$ – Ian Morris Sep 11 '11 at 21:00