Wold decomposition of toral endomorphisms

Question

Suppose that $A\in M_d(\mathbb{Z})$ is a $d \times d$ matrix with non zero determinant and suppose that $\mathbb{T}^d$ is the $d$-dimensional torus. Then one can define an operator on $L^2(\mathbb{T}^d,dx)$ as follows \begin{equation} T_Af (x) : = f(Ax), \forall x\in \mathbb{T}^d. \end{equation} This is of course an isometric operator, while if $A \in \operatorname{GL}_d(\mathbb{Z})$ (that is if it has determinant $\pm1$), then it is a unitary operator. I was wondering, what is the Wold decomposition of this operator in the general case. That is, what is the part of $T_A$ in the Wold decomposition which is a unilateral shift, for $A$ non invertible ?

Answer 1

The Hilbert space $H := L^2(\mathbb{T}^d,dx)$ can be canonically identified with the subspace of all (classes of) $\mathbb{Z}^d$-periodic locally square-integrable functions on $\mathbb{R}^d$. Then, for every integer $n \ge 0$, $T_A^n(H)$ is identified with the subspace of all $A^{-n}\mathbb{Z}^d$-periodic functions.

Observe that $(A^{-n}\mathbb{Z}^d)_{n \ge 0}$ is an increasing sequence of subgroups of $\mathbb{Z}^d$. Thus $$G := \bigcup_{n \ge 0}A^{-n}\mathbb{Z}^d$$ is a subgroup of $\mathbb{R}^d$. Moreover, the intersection of $T_A^n(H)$ over all $n \ge 0$ is the subspace of all $G$-periodic, or equivalently all $\overline{G}$-periodic functions.

If $G$ is dense in $\mathbb{R}^d$, then the intersection of $T_A^n(H)$ over all $n \ge 0$ is the null space, so $T_A$ itself is a unilateral shift. Otherwise, $T_A$ has a non-trivial unitary component in the Wold decomposition.