Timeline for What is the adic realization of a Bernoulli shift ?
Current License: CC BY-SA 3.0
4 events
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| Feb 16, 2012 at 20:03 | comment | added | Thierry de la Rue | 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). | |
| Feb 16, 2012 at 15:55 | comment | added | Stéphane Laurent | In fact I am not very easy with the adic representation. Is it easy to check that Thierry's proposal is fine ? | |
| Feb 16, 2012 at 15:18 | comment | added | Stéphane Laurent | 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 $n$ ? | |
| Feb 16, 2012 at 14:51 | history | answered | Thierry de la Rue | CC BY-SA 3.0 |