Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Here is a Bratelli-Vershik graph: alt text

This graph should (I do not master this topic) define a Vershik adic transformation $T$ on its path space.

  • What are the invariant measure(s) on the path space for which $T$ is ergodic ?

  • Is $T$ (isomorphic to) a known example of transformation in ergodic theory ?

EDIT 26/08/2012: I finally don't understand why $T$ is isomorphic to the dyadic odometer, as claimed in RW's answer below. Below are some observations about this graph.

  • To be sure of the definition of the graph, using Vershik terminology it is given by the Markov compactum $(M_n)$ where $M_n$ is the $n\times (n+1)$-matrix obtained by concatenating the $n \times n$-identity matrix with a column of ones: $$M_n=\begin{array}{cccccc} 1 & 0 & \cdots &\cdots & 0 & 1 \\\ 0 & \ddots & \ddots & & \vdots & \vdots \\\ \vdots & \ddots & \ddots & & \vdots & \vdots \\\ \vdots & & & & 0 & \vdots \\\ 0 & \cdots & & 0 & 1 & 1 \end{array}$$

    • In figure below this is an example of the action of $T$, the blue path becomes the red path: alt text
  • According to the link between adic and cut-and-stack the transformation $T$ should correspond to a cutting and stacking construction looking like:alt text

Is there something wrong in my understanding ? Could you elaborate a little the isomorphism between $T$ and the dyadic odometer ? Perhaps I could convince myself that the cutting and stacking construction asymptotically coincides with the odometer, but I don't see the isomorphism between the paths space of this adic graph and the paths space of the usual adic graph of the odometer.

EDIT (later): I have checked that the cut-and-stack procedure is indeed an approximation of the dyadic odometer. Hence RW is right but still I don't see the isomorphism between the paths space.

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

There are two edges going out of every vertex of this graph; label the vertical one with 0, and the other one with 1. Thus, paths in your graph can be parametrized by the space of 0,1 sequences. If one removes the vertical path corresponding to the sequence (0,0,0,...), then the tail equivalence relation and the lexicographic order on its classes for the punctured path space coincide with the corresponding objects for the punctured space of sequences. Thus, your transformation in precisely the classical 2-adic shift with all due consequences.

share|improve this answer
    
Please what is the "2-adic shift" ? Is it Bernoulli shift on $\{0,1\}^\mathbb{Z}$ ? –  Stéphane Laurent Aug 20 '12 at 5:31
    
After googling I find that the dyadic shift is $Tx=2x \text{ (mod $1$)}$. Please do you know a textbook or an online course on ergodic theory which gives the properties of $T$ ? –  Stéphane Laurent Aug 20 '12 at 9:07
    
... in particular I would like to see the corresponding cut-and-stack construction. –  Stéphane Laurent Aug 20 '12 at 11:48
    
Sorry, I shouldn't have called it a shift (although it should be clear from my description that it's by no means a Bernoulli shift - one-sided or two-sided). What I mean is the "2-adic odometer transformation", i.e., $z\mapsto z+1$ on 2-adic integers, or, in other words, the adic transformation on the Bratteli diagram, for which each level consists just of a single point, and there are two edges between any two consecutive levels. –  R W Aug 20 '12 at 11:49
    
Are you sure ? This is not the usual Bratelli-Vershik graph associated to the odometer, isn't it (showed on page 17 here math.unc.edu/Faculty/petersen/lecturespdf.pdf )? –  Stéphane Laurent Aug 20 '12 at 13:34
show 15 more comments

Let me make some trivial remarks about your question. I suppose you are looking for an automorphism of a measure space that is isomorphic to the transformation $T$ defined by the above Bratteli diagram ($T$ is the so called Bratteli-Vershik transformation that acts on the path space $X$ of this diagram). Staying in the context of measurable dynamics, you should clearly define a measure $\mu$ invariant (or quasi-invariant) with respect to $T$. In other words, $T$ must be considered on the support of $\mu$ what is not necessary the whole space $X$. Another thing is that $\mu$ could be non-unique. The concept of isomorphism in measurable dynamics is actually $\rm{mod}\ 0$ isomorphism so that you can ignore sets of measure zero.

I am doubtful that your transformation $T$ is isomorphic to an odometer. In the other question of yours when the right-most edges of the diagram are multiple you will get that an induced transformation $T_A$ is isomorphic to a non-stationary odometer. I suppose this is rather obvious so that let me omit details.

share|improve this answer
    
Thanks. The invariant measures is the first question of my post. But I think there is a natural fully supported invariant measure and this is the invariant measure in which I am interested: start from the vertex at the top of the graph and go to one of the two possible vertices at next level with probabilities $1/2$ and $1/2$. –  Stéphane Laurent Aug 26 '12 at 12:37
    
@ Stéphane! You cannot start with an arbitrary probability vector (say (1/2, 1/2) and expect to have an invariant measure (see arxiv.org/pdf/0812.1088.pdf for instance). Maybe the following paper would be also interesting for you arxiv.org/pdf/0812.1088.pdf –  SIB Aug 26 '12 at 13:18
    
Correction: arxiv.org/pdf/1204.1621.pdf –  SIB Aug 26 '12 at 13:20
    
Thanks for the reference. Sorry but doing mathematics is not currently my daily job hence I don't have time enough to read many papers. Do you mean $T$ is not invariant with the mesure I proposed ? –  Stéphane Laurent Aug 26 '12 at 14:31
    
Correction: sorry, obviously I meant $T$ preserves this measure, not T is invariant... –  Stéphane Laurent Aug 26 '12 at 15:05
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.