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According to the countable generator theorem, every ergodic invertible measure-preserving transformation has a generating partition.

What are the generating partitions of the dyadic odometer ? I don't find the answer in textbooks. Is there a "canonical" one among them ?

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    $\begingroup$ "Generating partition" and "dyadic odometer" are standard concepts in ergodic theory which do not in my view require explanation or exposition additional to that given in the question. The question asked about these objects is in my opinion reasonable and of an acceptable level for this forum. I do not think that this question is unclear and I see no reason for it to be closed. $\endgroup$ – Ian Morris Jan 30 '16 at 17:23
  • $\begingroup$ I expect Ian has made a good point, but a likely counter from some of our users is that the question ought to be given some context/motivation. If this were done, then I'd think there's probably little ground for keeping it closed (and -- sayeth the broken record here -- if it is the case that you do not know the field, then please consider whether you should be the one to cast a vote to close). $\endgroup$ – Todd Trimble Jan 30 '16 at 19:23
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    $\begingroup$ As @IanMorris says, this is completely standard. What could I do ? Not giving these standard definitions... Every ergodic measure-preserving transformation has a generating partition (Rokhlin's theorem), and it is a natural question to ask which partitions are generating. $\endgroup$ – Stéphane Laurent Jan 30 '16 at 21:48
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    $\begingroup$ @ToddTrimble I edited. Is it better ? $\endgroup$ – Stéphane Laurent Jan 30 '16 at 21:50
  • $\begingroup$ Thank you; yes, it is better. I think I like your comment (which mentions Rokhlin's theorem) even more, and would edit in your comment myself except that I don't know the field. However, I will cast a reopening vote. $\endgroup$ – Todd Trimble Jan 30 '16 at 22:12
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Quite a nice one is the two-set partition $A,A^c$, where $A$ is the set of points with an even number of terminal 0's: $$ A = \bigcup_{k \geq 0} A_k $$ where $A_k = \bigl\{(x_1, x_2, \ldots) \mid x_1=\ldots=x_{2k}=0 \text{ and } x_{2k+1}=1 \bigr\}$. Or, if you work with the odometer acting on the space $[0,1[$: $$ A = \bigcup_{k \geq 0} \left[\frac{1}{2^{2k+1}}, \frac{1}{2^{2k}} \right[ $$

This partition is generating for the following reason. Code a trajectory $x, Tx, T^2x, \ldots$ by a sequence $(v_0, v_1, \ldots)$ of $a$'s and $b$'s according to whether it's in $A$ or not.

Then you can get the first digit $x_1$ of $x$ by looking at the blocks of four consecutive terms of the $v$ sequence:

  • if the first digit of $x$ is $x_1=1$, then the block $(v_0,v_1,v_2,v_3)$ of four consecutive terms is one of $aaab$, $abaa$ or $abab$ (that is $(v_0,v_1,v_2,v_3)$ is $a*a*$, with at least one of the stars a $b$);
  • if $x_1=0$, then $(v_0,v_1,v_2,v_3)$ is $*a*a$ with at least one of the stars a $b$.

In particular, the sets of possible codes are disjoint, so that this information suffices to determine the first digit of $x$.

To get the second digit of $x$, look at blocks of eight consecutive terms:

  • if the first two digits of $x$ are 00, then the code is $*aba*aba$ with at least one of the $*$'s an $a$;
  • if the first two digits of $x$ are $01$, then the code is $ba*aba*a$ with at least one of the $*$'s an $a$;
  • if the first two digits of $x$ are 10, then the code is $aba*aba*$ with at least one of the $*$'s an $a$;
  • if the first two digits of $x$ are 11, then the code is $a*aba*ab$ with at least one of the $*$'s an $a$;

Again, these sets of possibilities are disjoint, so that one can recover the first and second digits of $x$ from 8 terms of the $v$ sequence.

And so on, looking at the block of $2^{n+1}$ consecutive terms of the $v$ sequence determines the first $n$ digits of $x$.

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  • $\begingroup$ Excuse-me, what do you mean by the number of terminal 0's ? $\endgroup$ – Stéphane Laurent Jan 30 '16 at 0:51
  • $\begingroup$ I think of the odometer as the set of left-infinite strings of 0s and 1s. The odometer transformation is adding 1, with carry to the left (just like addition of binary numbers, except these have infinitely many digits). A point in the odometer is therefore $(\ldots,x_2,x_1)$ with the $x_i$'s in $\{0,1\}$. My set is $\{x\colon \exists n\colon x_1=\ldots=x_{2n}=0; x_{2n+1}=1\}$. $\endgroup$ – Anthony Quas Jan 30 '16 at 1:21
  • $\begingroup$ Yes, this is the so-called dyadic adic machine. Ok I see now, you mean the number of 0's before the first one. I'll come back after thinking about this partition. $\endgroup$ – Stéphane Laurent Jan 30 '16 at 1:44
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    $\begingroup$ I write my odometers backwards compared to the rest of the world, so that carries go to the left (as usual in addition) rather than to the right. $\endgroup$ – Anthony Quas Jan 30 '16 at 2:12
  • $\begingroup$ I generally index my stochastic processes with negative integers, so I'm not the one who will blame you :-) Ok, I took a look at this partition. Why do you say it is nice ? Is there something special about the associated stationary process on $\{0,1\}$ ? And how do you know this partition is generating ? $\endgroup$ – Stéphane Laurent Jan 30 '16 at 2:15

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