I have been reading some papers recently, in particular, Blanchard's paper *$\beta$-expansions and symbolic dynamics* which state that a $\beta$-shift $S\_{\beta}$ is a synchronised shift if and only if the orbit of the greedy $\beta$-expansion of $1$ is not dense in $S\_{\beta}$.


The reference given for this result is a preprint due to A. Bertrand-Mathis, *Questions diverses relatives aux systemes codes: applications au $\beta$-shift*. Do you know any published reference where the proof can be found? Or, do you know a place where the preprint can be downloaded? 


I believe that the proof of this fact is similar to the proof for the fact that $S$-gap shifts are synchronised. If $\beta \in (1,2)$, must work in the same way. However, I am not sure if a similar argument will holds for $\beta > 2$.


Thanks in advance. 



Synchronised $\beta$-shifts
I do not know a reference to a published proof, most sources refer to Blanchard's paper. The proof is relatively easy once you remember the structure of a canonical graph representation of $\beta$-shifts. Assume that $\beta>1$ is such that the greedy $\beta$-expansion $(\omega\_i)\_{i\ge 1}$ of $\mathbf 1$ is not eventually periodic (as otherwise the $\beta$-shift is sofic, hence synchronized). The canonical representation of this $\beta$-shift is a countable graph with vertices $v\_0,v\_1,v\_2,\ldots$ and **right edges** $v\_{i-1}\to v\_i$ labeled with $\omega\_i$ for $i=1,2,\ldots$ and for each vertex $v\_{i-1}$ which is an initial vertex of some right edge labeled with nonzero symbol (that is, for each $i\ge 1$ such that $\omega\_i>0$) we add **backward edges**, that is, we add $\omega\_i$ different edges (each of them starts at $v\_{i-1}$ and ends at $v\_0$) labeled by $0,1,\ldots,\omega\_i-1$ (that is, there is $\omega\_i$ such edges). Then a finite word $w=w\_1\ldots w\_n$ is in the language of the $\beta$-shift $S\_\beta$ if and only if there is a path in our graph starting at $v\_0$ and labeled with $w$. Now if the greedy $\beta$-expansion $(\omega\_i)\_{i\ge 1}$ of $\mathbf 1$ is not dense in $S\_\beta$ then there exists a word $u=u\_1\ldots u\_k$ in the language of $S\_\beta$ which does not appear as a label for a path consisting of right edges only. Take a shortest such word. It means that $u\_1\ldots u\_{k-1}$ is a label for a path consisting of right edges only. Consider a path starting at $v\_0$ following right edges until the first appearance of $u\_1\ldots u\_{k-1}$ and then returning to $v\_0$ so that the word $u$ is a suffix of the label of that path. Let $\bar u=\bar u\_1\ldots \bar u\_\ell$ be the label of that path. There is exactly one backward edge on that path. Note that it follows that $\bar u\_1\ldots \bar u\_{\ell-1}$ is a ``lexicographically maximal'' label and such a label can only appear over a path consisting of right edges only (this is a defining property of $\beta$-shifts). I claim that $\bar u$ is a synchronizing word. This is because each path labeled by $\bar u$ and appearing somewhere in the graph must end at $v\_0$ as otherwise $\bar u$ would appear in the greedy $\beta$-expansion of $\mathbf 1$. Let $w\_1\bar u$, $\bar uw\_2$ be two words in the language of $S\_\beta$. Then $w\_1\bar uw\_2$ is a label for a path starting at $v\_0$ then labeled by $w\_1 \bar u$, and since this path ends at $v\_0$, then it may be followed by a path labeled by $w\_2$. On the other hand it is easy to see that if the greed $\beta$-expansion is dense then each word has to appear in it infinitely often, so each word is a label appearing over infinitely many different paths in the graph representation. No such word can be a synchronizing word.


