Let $T$ be a measure preserving bijection of a probability space $(X,\nu)$. Consider the Koopman representation of $\mathbb{Z}$ on $L^2(X,\nu)$ given by $[z.f](x) = f(T^{-z}(x))$. The question is: can I tell from the representation whether $(X,\nu,T)$ has zero entropy?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I would think you can tell, since for any set $A$ you can realise in $\mathsf{L}^2$ by its characteristic function $1_A$. But you might be asking which property of the Koopman representation imply (or are implied by) zero entropy? An example in this vein: ergodicity (if my memory serves me well) is equivalent to the coefficients of the representation to have zero mean. [As a side note: I think you need to restrict $\mathsf{L}^2$ to functions whose integral w.r.t. $\mu$ is 0. Otherwise you get lots of fixed vectors for the representation: the constant functions] $\endgroup$– ARGCommented Nov 3, 2021 at 15:25
-
$\begingroup$ Indeed, if you consider just the zero-mean subspace then ergodicity is equivalent to having no invariant vectors. And yes: I am asking which properties of the representation imply or are implied by zero entropy. $\endgroup$– VladimirCommented Nov 3, 2021 at 16:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
There is an example (due to Newton - Parry and also attributed by Rokhlin to Girsanov) of a zero entropy measure preserving transformation with a countable Lebesgue spectrum (and mixing of all orders). Therefore, the associated Koopman operator is unitarily equivalent to that of any Bernoulli shift.
