Roughly speaking, a theorem by Vershik says that every ergodic invertible measure-preserving transformation is isomorphic to some "adic" transformation on the spaces of paths of a Bratelli-Vershik graph. What is the adic transformation corresponding to a Bernoulli shift ?
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2 Answers
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The short answer is that nobody knows. The reason is that Vershik's proof uses Rokhlin's towers and is thus virtually non-constructive.
As far as I know, the only known examples of explicit adic realizations are substitutional dynamical systems and the irrational rotations of the circle. Even for a simple ergodic rotation of the 2-torus this is an open question, let alone Bernoulli shifts.
answered Feb 16, 2012 at 13:18
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$\begingroup$ I believe I know a way to construct an adic realization of a rotation of the 2-torus. I can send you my notes about that if you are interested. $\endgroup$ May 20, 2016 at 16:54
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$\begingroup$ This has been done, I believe. A paper in Monatshefte a few years ago, don't remember the author. $\endgroup$ May 20, 2016 at 17:44
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$\begingroup$ More generally, I believe I know a way to construct an adic realization of $T \times S$ once one has an adic realization of $S$. But I'm not sure it works. $\endgroup$ May 20, 2016 at 18:30
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$\begingroup$ Well, good luck. I look forward to reading about this one day. $\endgroup$ May 20, 2016 at 23:35
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$\begingroup$ I try to explain this construction on an example here. Another example here. I discovered this construction while I was working on something else (currently I do not intend to publish this work). $\endgroup$ May 22, 2016 at 15:01
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I think you can realize the Bernoulli shift on $k$ symbols as an adic transformation on the following Bratelli-Vershik diagram: put $k$ nodes on the first level. Suppose levels 1 to $n$ have been defined, and call $L_n$ the set of nodes in the $n$-th level. Then nodes on the $(n+1)$-th level are pairs $(i,j)\in L_n\times L_n$, where $(i,j)$ is connected to $i$ and to $j$ (in this order) to the $n$-th level. Does not this work?
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$\begingroup$ Thanks to both of you. @Thierry, hence the graph is "homogeneous", in the sense that the number of paths to some node at level $n$ to a node at level $1$ is the same for all nodes at level $n$ ? $\endgroup$ Feb 16, 2012 at 15:18
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$\begingroup$ In fact I am not very easy with the adic representation. Is it easy to check that Thierry's proposal is fine ? $\endgroup$ Feb 16, 2012 at 15:55
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$\begingroup$ Indeed, the graph I proposed is homogeneous: Each node of level n has exactly k^n paths to the root of the diagram. But then I must admit that this poses some problem, since this introduces eigenvalues for the associated adic transformation. So this adic transformation is rather an extension of the Bernoulli shift (probably a direct product with an odometer). $\endgroup$ Feb 16, 2012 at 20:03