Spectrum and tower decomposition

Question

I'm trying to read "Partitions of Lebesgue space in trajectories defined by ergodic automorphisms" by Belinskaya (1968). In the beginning of the proof of theorem 2.7, the author considers an ergodic automorphism $R$ of a Lebesgue space whose spectrum contains all $2^i$-th roots of unity ($i=1,2,\ldots$), and then he/she claims that for each $i$ there exists a system of sets ${\{D_i^k\}}_{k=1}^{2^i}$ such that $R D_i^k=D_i^{k+1}$ for $k=1, \ldots, 2^i-1$ and $R D_i^{2^i}=D_i^1$. I'm rather a newbie in ergodic theory and I don't know where this claim comes from. I would appreciate any explanation.

Answer 1

Let $R$ be an ergodic automorphism of a Lebesgue space. Let $\omega$ be a root of unity in the spectrum of $R$, and $n$ be the smallest positive integer such that $\omega^n = 1$.

Let $f$ be a non-zero eigenfunction corresponding to the eigenvalue $\omega$. Then:

$f^n \circ R = (f \circ R)^n = (\omega f)^n = f^n,$

and $f^n$ is an eigenfunction for the eigenvalue $1$. Without loss of generality, we can assume that $f^n$ is bounded (you just have to take $f/|f|$ instead whenever $f \neq 0$). Then, since the transformation is ergodic, $f^n$ is constant. Up to multiplication by a constant, we can assume that $f^n \equiv 1$.

Thus $f$ takes its values in the group generated by $\omega$. For all $0 \leq k < n$, let $A_k$ be the set $\{f = \omega^k\}$. This is a partition of the whole space. Moreover, if $x \in A_k$, then $f \circ R (x) = \omega f(x) = \omega \cdot \omega^k = \omega^{k+1}$, so $R (x) \in A_{k+1}$ (with the addition taken modulo $n$), and conversely. Hence, $R A_k = A_{k+1}$.