Is there an invertible measure-preserving transformation (preferably a nice one) admitting every irrational rotation as a factor ? I guess the spectrum is the relevant tool to address this question but I do not master spectral theory. Of course, if it is true, I am interested in a "minimal" such transformation. It is understood I consider only the mpt on a Lebesgue space.
$\begingroup$
$\endgroup$
5
-
$\begingroup$ The naively intuitive thing to do would be to "enumerate" the irrational rotations as $T_\alpha \colon X_\alpha \to X_\alpha$ where each $X_\alpha$ is the circle, $T_\alpha$ is rotation by angle $\alpha \in [0,1)$, and $\alpha$ runs over the uncountable set $[0,1)\setminus \mathbb{Q}$. Now let $X$ be the (uncountably infinite) product over the $X_\alpha$'s and let $T$ be the product transformation of the $T_\alpha$'s. It's clear that each irrational rotation is a topological factor. The measure theory on $X$ looks like a bit of a headache, but perhaps there are no genuine obstacles...? $\endgroup$– Ian MorrisCommented Jan 20, 2016 at 15:37
-
1$\begingroup$ @IanMorris I had the same idea, but an uncountable product of a Lebesgue space is not a Lebesgue space. $\endgroup$– Stéphane LaurentCommented Jan 20, 2016 at 15:51
-
1$\begingroup$ I think any finite measure $L^2$ space is separable, and so there can be only countably many eigenfunctions. $\endgroup$– Anthony QuasCommented Jan 20, 2016 at 15:56
-
$\begingroup$ This can't work for the reason pointed out by Anthony: $L^2(X,\mu)$ is separable if $X$ is a Lebesgue space (I'm not sure about general $X$ with $\mu(X)=1$), so there can't be uncountably many eigenvalues. $\endgroup$– Christian RemlingCommented Jan 20, 2016 at 17:29
-
$\begingroup$ Ok, thank you. @ChristianRemling: $L^2((\Omega, {\cal A}, \mathbb{P}); \mathbb{R})$ is separable if and only if the $\sigma$-field ${\cal A}$ is $\mathbb{P}$-essentially separable. $\endgroup$– Stéphane LaurentCommented Jan 20, 2016 at 22:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
8
There is an ergodic system $\mathbf Z = (Z,m,R)$, such that $\mathbf Z$ admits every irrational rotation as a factor.
As mentioned in the comments, such a system $\mathbf Z$ cannot have $L^2(Z,m)$ separable, since the eigenspaces are mutually orthogonal, and there are uncountably many eigenspaces. However, $\mathbf Z$ has some nice properties: it is a rotation on a compact separable group $b\mathbb Z$, known as the Bohr compactification of the integers. There are several ways to construct $b\mathbb Z$, but the important properties of $b\mathbb Z$ are
- (B1) $b\mathbb Z$ is a compact abelian group (hence it has finite Haar measure $m$, which may be normalized to have $m(b\mathbb Z) = 1$).
- (B2) $\mathbb Z$ embeds densely in $b\mathbb Z$: there is a homomorphism $e:\mathbb Z \to b\mathbb Z$ having dense image in $b\mathbb Z$. Fixing such a homomorphism, we may consider $\mathbb Z$ to be a subgroup of $b\mathbb Z$.
- (B3) If $G$ is a compact abelian group and $\rho: \mathbb Z \to G$ is a homomoprhism, then $\rho$ extends to a continuous homomorphism $\rho: b\mathbb Z \to G$.
(See Rudin's Fourier Analysis on Groups for some exposition of the Bohr compactification.)
We may consider the measure preserving system $(Z,m,R)$, where $Z = b\mathbb Z$, $m=$ normalized Haar measure on $Z$, and $R z = z+1$. Then $(Z,m,R)$ is an ergodic group rotation, as the orbit of a point $z$ is $\{z+n : n\in \mathbb Z\}$, which is dense in $Z$, by (B2).
Claim. Every ergodic group rotation is a factor of $(Z,m,R)$.
Proof. Let $(G,m_G,R_\alpha)$ be an ergodic group rotation, so that $\{n\alpha : n \in \mathbb Z\}$ is dense in the compact abelian group $G$. The map $n\mapsto n\alpha$ is a homomorphism from $\mathbb Z$ to $G$ so Property (B3) provides a continuous homomorphism $\rho: b\mathbb Z\to G$ such that $\rho(n) = n\alpha$ for $n\in \mathbb Z.$ The continuity of $\rho$ then implies $\rho(b\mathbb Z) = G$, and it is easy to check that $\rho$ is actually a factor map. $\square$
It is natural to ask if $\mathbf Z$ is minimal among ergodic systems admitting every irrational rotation as a factor. The natural attempt to prove minimality would rely on the following two purported theorems, both of which are standard in the settings where $L^2(X,\mu)$ is separable, but I am not aware of their status in the non-separable setting.
(purported) Theorem 1. If $\mathbf X = (X,\mathcal B,\mu, T)$ is a measure preserving system and $\mathcal A$ is a $T$-invariant sub-sigma algebra of $\mathcal B$, then there is a factor $(Y,\mathcal D,\nu,S)$ of $\mathbf X$ with factor map $\phi: X\to Y$ such that
(i) $\phi^{-1}(D) \in \mathcal A$ (up to $\mu$-measure 0) for all $D\in \mathcal D$
(ii) For every $B\in \mathcal A$, there is $D\in \mathcal D$ such that $\mu(\phi^{-1}(D) \triangle B) = 0.$
(purported) Theorem 2. If $\mathbf Y = (Y,\nu,S)$ is an ergodic measure preserving system such that $L^2(Y,\nu)$ is spanned by the eigenfunctions of $\mathbf Y$, then $\mathbf Y$ is isomorphic to an ergodic group rotation.
Purported Theorem 2 is a generalization of the Halmos-von Neumann theorem to the non-separable setting.
answered Feb 7, 2016 at 20:57
-
$\begingroup$ Thank you, very interesting. On Lebesgue spaces, it is known (unless I misunderstood) that $n \mapsto f(T^nx)$ is uniformly almost periodic for almost every $x$ when $T$ has discrete spectrum, and $f \in L^2$. Do you know whether it is true for this transformation ? $\endgroup$ Commented Feb 7, 2016 at 23:42
-
$\begingroup$ In the measure preserving setting, there are discrete spectrum transformations $T$ and bounded functions $f$ such that for all $x$, the map $n\mapsto f(T^n x)$ is not almost periodic. For example, take an irrational circle rotation, and let $f$ be the characteristic function of a compact totally disconnected set having positive positive measure. Then for almost all $x$, the set $\{n : f(T^n x) > 1/2\}$ is nonempty and is not syndetic, which is enough to show that $n\mapsto f(T^n x)$ is not uniformly almost periodic. $\endgroup$ Commented Feb 8, 2016 at 5:03
-
$\begingroup$ However, for group rotations $(G,m_G, R_\alpha)$ and continuous functions $f$, it is the case that $n \mapsto f(R_\alpha^n x)$ is uniformly almost periodic for every $x\in G$. $\endgroup$ Commented Feb 8, 2016 at 5:05
-
$\begingroup$ In fact, this is something I saw in an old paper, but this fact is not explicitely written, so I had to guess the statement. I don't know where to find this statement with a proof. I should open another post instead of talking about that here. $\endgroup$ Commented Feb 8, 2016 at 9:00
-
1$\begingroup$ Dear John, this is what I meant for the almost periodicity (it is not the uniform one). $\endgroup$ Commented Feb 10, 2016 at 21:01