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 $A \in GL_d(\mathbb{Z})$, (that is if it has determinant $\pm1$) then it is a unitary operator. I was wandering, 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 ?

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 ?