I have trouble filling with details this proof. In fact at the end of the proof it says "we leave the details to the interested reader." The only parts without details is step 3.
Actually I have also some question about step 3
Question 1: What is mean exactly that $ \rho $ is a point-mass ? That there exsits some $ \{ y \} \in \mathbb{S}^{\mathbb{N}}/Y $ such that $ \rho(\{ y\}) > 0 $ ? So $ C = \{ y \} $? If yes why we can say that $ \rho(\{ y\}) <1 $ ?
Question 2: Why $A = \pi^{-1} C$ is a set that is invariant (I suppose $S$-invariant) ?
Theorem: Let $(X,\mathcal{A},\mu,T)$ be a measure preserving system, where $(X,\mathcal{A},\mu)$ is a standard probability space. Then $(X,\mathcal{A},\mu,T)$ is a Kronecker system if and only if it is isomorphic to an ergodic group rotation.
Part of the proof with details:
Step 1: We proved that every group rotation has discrete spectrum and is ergodic. Completing one direction.
Notation Given an orthonormal basis $\chi_n$ of $L^2(X)$ consisting of eigenfunctions with associated eigenvalues $\lambda_n$ we have defined a map $\phi : X \to \mathbb{S}^{\mathbb{N}} $ given by $\phi(x) = (\chi_n(x))_{n\in \mathbb{N}}$, $\lambda =(\lambda_n)_{n \in \mathbb{N}} $ and we have defined $Y=\overline{ \{ \lambda^n : n \in \mathbb{Z} \} } $. Then if $ \mathcal{B} $ is the $\sigma$-algebra of Borel sets on $Y$, $\nu$ the Haar measure on $Y$ and $S:Y \to Y $ the transformation $S(y) = \lambda y $ for all $y\in Y$ we have that $(Y,\mathcal{B},\nu,S)$ is a group rotation.
Step 2: We have proved that $(X,\mathcal{A}, \mu,T)$ is isomorphic to $( Y^{\ast}, \mathcal{B}^{\ast}, \nu^{\ast}, S) $ where $Y^{\ast}=\phi(X)$, $\mathcal{B}^{\ast}$ is the Borel stes in $\mathbb{S}^{\mathbb{N}} $ restricted to $Y^{\ast}$, and $ \nu^{\ast} = \phi \mu $.
Step 3 We have proven with a lack of detail that $( Y^{\ast}, \mathcal{B}^{\ast}, \nu^{\ast}, S) $ and $(Y, \mathcal{B}, \nu, S)$ are isomorphic.
I write what we have said in step 3
Since $(Y,\cdot)$ is a subgroup of $(\mathbb{S}^{\mathbb{N}}, \cdot) $ we consider $ \pi : \mathbb{S}^{\mathbb{N}} \to \mathbb{S}^{\mathbb{N}}/Y $ to be the natural quotient map, and let $ \rho $ be the push-forward of $ \nu^{\ast} $ under $ \pi$, i.e. $ \rho = \pi \nu^{\ast} $. If $ \rho $ is not a point-mass then exist a set $C \subseteq \mathbb{S}^{\mathbb{N}}/Y$ with $ 0 < \rho(C) < 1 $. The set $A = \pi^{-1} C$ is a subset of $Y^{\ast} $ that is invariant satisfying $0 < \nu^{\ast}(A) < 1$ which is impossible because the system is ergodic. Hence $ \rho$ must be a point-mass. Let $ u \cdot Y$ denote its support. It is now striaghtforward to show that the map $ \psi : Y \to Y^{\ast} $ given by $ \psi(y) = u \cdot y $ is an isomorphism.
Details need to fill it I need to prove that $ \psi $ is a factor map. So three condition
- That $ \psi(Y)$ has full measure
- That $ \psi \nu = \nu^{\ast} $
- That $ (S \circ \psi)(y) = (\psi \circ S)(y)$, for $\nu$-almost every $ y \in Y$.
Is very easy to prove 3 since $ (S \circ \psi)(y) = \lambda u y = (\psi \circ S)(y)$. I think that 1 holds since $ u\cdot Y $ is the support of $ \nu^{\ast} $, then $ \psi(Y) $ has full measure (but not sure actually). For point 2 I have no idea.
I need to show also that $\psi$ is invertible, to do that i think it suffice to take $ u^{-1} \cdot y$ but not sure.